Split (1)

train · 398 rows

id stringlengths 10 10	source stringlengths 74 74	problem stringlengths 57 1.29k	solutions listlengths 0 9
IMO-2009-5	https://artofproblemsolving.com/wiki/index.php/2009_IMO_Problems/Problem_5	Determine all functions $f$ from the set of positive integers to the set of positive integers such that, for all positive integers $a$ and $b$, there exists a non-degenerate triangle with sides of lengths \[ a,f(b) \] and \[ f(b+f(a)-1). \] (A triangle is non-degenerate if its vertices are not collinear.)	[ "Answer: The only such function is \$f(x)=x\$.\n\nIt is easy to see that this function satisfy the condition. We are going to proof that this is the only such function.\n\nWe start with\n\nLemma. If 1, \$a\$, \$b\$ are sides of a non-degenerate triangle then \$a=b\$.\n\nProof. In this case \$a<b+1\$, therefore \$a \\le b\$. By the same reason, \$b \\le a\$. Therefore \$a=b\$.\n\nLet \$u=f(1)\$. Now consider the given condition for \$a=1\$. By Lemma we get that\n\n\\[\nf(b)=f(b+u-1)\n\\]\n\nfor any\n\n\\[\nb\n\\]\n\n.\n\nFirst, suppose \$u>1\$. Then \$f\$ is a periodic function. But then \$f\$ is bounded by a constant \$M\$, i.e. \$f(x)<M\$. Then take, \$a>2M\$. We get that \$a,f(b)\$ and \$f(b+f(a)-1)\$ are sides of the triangle, but the first number is greater than \$2M\$ and other two are less than \$M\$, which is imposible. We get the contradiction, so \$u\$ could not be greater than 1.\n\nSo \$u=f(1)=1\$.\n\nProperty 1. For any \$x\$\n\n\\[\nf(f(x)) = x\n\\]\n\nProof. Consider the given condition for \$a=x\$, \$b=1\$ and use Lemma.\n\nProperty 2. For any \$x\$ and \$y\$\n\n\\[\nf(x+1)+f(y) > f(x+y)\n\\]\n\nProof. Consider the given condition for \$a=f(x+1)\$, \$b=y\$ and use triangle inequality and Property 1.\n\nLet \$g(x)=f(x+1)-1\$.\n\nProperty 3. For any \$x\$ and \$y\$\n\n\\[\ng(x)+g(y) \\ge g(x+y)\n\\]\n\nProof. Follows from Property 2.\n\nProperty 4. For any \$x\$, \$y\$, \$m\$, \$n\$\n\n\\[\ng(nx)+g(my) \\ge g(nx+my)\n\\]\n\nProof. Follows from Property 3.\n\nProperty 5. For any \$n\$, there is \$k>n\$, s.t. \$g(k) \\ge k\$.\n\nProof. Because of the Property 1.\n\nProperty 6. \$g(x)=x\$.\n\nProof. Suppose, for some \$x\$, \$g(x)=y\\ne x\$. Without lost of generality we can assume that \$x<y\$. Then there is a number \$n\$, such than any \$k>n\$ could be represented as \$k=ax+by\$, where \$a>b\$. Then \$g(k)=g(ax+by)\\ge ag(x)+bg(y)=ay+bx>ax+by=k\$. That contradicts to the Property 5.\n\nTherefore by definion of \$g\$, \$f(x)=x\$." ]
IMO-2009-6	https://artofproblemsolving.com/wiki/index.php/2009_IMO_Problems/Problem_6	Let $a_1,a_2,\ldots,a_n$ be distinct positive integers and let $M$ be a set of $n-1$ positive integers not containing $s=a_1+a_2+\ldots+a_n$. A grasshopper is to jump along the real axis, starting at the point $0$ and making $n$ jumps to the right with lengths $a_1,a_2,\ldots,a_n$ in some order. Prove that the order can be chosen in such a way that the grasshopper never lands on any point in $M$.	[ "We will use strong induction on \$n\$. When \$n = 1\$, there are no elements in \$M\$, so the one jump can be made without landing on a point in \$M\$. When \$n = 2\$, we consider two cases. If \$a_1\$ is not in \$M\$, then the order \$a_1, a_2\$ will work. If \$a_1\$ is in \$M\$, then \$a_2\$ is not in \$M\$ and the order \$a_2, a_1\$ will work. We will assume that the order can be chosen in such a way for all integers \$n < k\$ for \$k \\ge 3\$.\n\nWhen \$n = k\$, we can assume WLOG that \$a_1<a_2< \\ldots <a_k\$. We can also assume that the elements of \$M\$ are distinct, since if two elements are identical, we can treat them as one and use induction on \$n = k-1\$. We now consider four cases:\n\nCase 1: \$a_k\$ is in \$M\$ but \$a_1, a_2, \\ldots, a_{k-1}\$ are not.\n\nThere are at most \$k-2\$ elements in \$M\$ greater than \$a_k\$. Then, from our assumption, there exists an order of jumps starting from \$a_k\$ of lengths \$a_1, a_2, \\ldots, a_{k-1}\$ where the first jump is \$a_i\$ such that the grasshopper will not land in any point in \$M\$. We can then switch the jumps \$a_k\$ and \$a_i\$. Since \$a_i\$ is not in \$M\$ and no subsequent jumps result in the grasshopper landing in a point in \$M\$, this sequence is valid.\n\nCase 2: \$a_k\$ is not in \$M\$ but at least one of \$a_1, a_2, \\ldots, a_{k-1}\$ is.\n\nThere are at most \$k-2\$ elements in \$M\$ greater than \$a_k\$. Then, from our assumption, there exists an order of jumps starting from \$a_k\$ of lengths \$a_1, a_2, \\ldots, a_{k-1}\$ such that the grasshopper will not land in any point in \$M\$. Adding \$a_k\$ at the beginning of this sequence then results in a valid sequence.\n\nCase 3: \$a_k\$ is in \$M\$ and at least one of \$a_1, a_2, \\ldots, a_{k-1}\$ is.\n\nConsider the following pairs of distinct integers \$(a_1, a_1+a_k), (a_2, a_2+a_k), \\ldots, (a_{k-1}, a_{k-1}+a_k)\$. Since at most \$k-2\$ of these points are in \$M\$, by the pigeonhole principle, there exists an integer \$i\$ such that \$a_i, a_i + a_k\$ are both not in \$M\$. Since there are at most \$k-3\$ elements greater than \$a_k\$, at most \$k-3\$ elements are greater than \$a_i + a_k\$. Then, from our assumption, there exists an order of jumps starting from \$a_i + a_k\$ of lengths \$a_1, a_2, \\ldots, a_{i-1}, a_{i+1}, \\ldots, a_{k-1}\$ such that the grasshopper will not land in any point in \$M\$. Adding \$a_i, a_k\$ at the beginning of this sequence then results in a valid sequence.\n\nCase 4: None of \$a_1, a_2, \\ldots, a_k\$ are in \$M\$.\n\nAssume that there exists no valid sequence - that any order of jumps will result in the grasshopper landing on at least one element in \$M\$. We can select any element \$m_i\$ in \$M\$ and, from our assumption, construct a sequence beginning at some \$a_j\$ containing \$n-1\$ jumps \$a_1, a_2, \\ldots, a_{j-1}, a_{j+1}, \\ldots, a_k\$ that will not land on any of the other \$n-2\$ elements in \$M\$. Since no valid sequence exists, during this sequence of jumps, the grasshopper must land on \$m_i\$.\n\nWLOG, we let \$m_1 < m_2 < \\ldots < m_{k-1}\$. We let \$i = 1\$ and \$j = k\$ and use the corresponding sequence that only lands on \$m_1\$ and no other element in \$M\$. Let the jump immediately after landing on \$m_1\$ be \$a_x\$, where \$x < n\$. We swap jumps \$a_x\$ and \$a_n\$. Now, since \$a_n > a_x\$, all jumps before \$a_n\$ will land on a integer less than \$m_1\$. Since \$m_1\$ is the smallest element in \$M\$, none of the jumps before \$a_n\$ will land on a point in \$M\$. Since no jumps after \$a_n\$ will land on an element in \$M\$ by the construction, the grasshopper must then land on an element \$m_y > m_1\$ in \$M\$ after making the jump \$a_n\$. However, this would imply that the original construction would land on another element \$m_y\$ different from \$m_1\$, but this is a contradiction. So, a valid sequence must exist.\n\nSince we have an induction on \$n\$, the statement must be true for all \$n\$, and we are done." ]
IMO-2010-1	https://artofproblemsolving.com/wiki/index.php/2010_IMO_Problems/Problem_1	Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all $x,y\in\mathbb{R}$ the following equality holds \[ f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor \] where $\left\lfloor a\right\rfloor$ is greatest integer not greater than $a.$	[ "Put \$x=y=0\$. Then \$f(0)=0\$ or \$\\lfloor f(0) \\rfloor=1\$.\n\n\$\\bullet\$ If \$\\lfloor f(0) \\rfloor=1\$, putting \$y=0\$ we get \$f(x)=f(0)\$, that is f is constant. Substituing in the original equation we find \$f(x)=0, \\ \\forall x \\in \\mathbb{R}\$ or \$f(x)=a, \\ \\forall x \\in \\mathbb{R}\$, where \$a \\in [1,2)\$.\n\n\$\\bullet\$ If \$f(0)=0\$, putting \$x=y=1\$ we get \$f(1)=0\$ or \$\\lfloor f(1) \\rfloor=1\$.\n\nFor \$f(1)=0\$, we set \$x=1\$ to find \$f(y)=0 \\ \\forall y\$, which is a solution.\n\nFor \$\\lfloor f(1) \\rfloor=1\$, setting \$y=1\$ yields \$f(\\lfloor x \\rfloor)=f(x), \\ ()\$.\n\nPutting \$x=2, y=\\frac{1}{2}\$ to the original we get \$f(1)=f(2)\\lfloor f(\\frac{1}{2}) \\rfloor\$. However, from \$()\$ we have \$f(\\frac{1}{2})=f(0)=0\$, so \$f(1)=0\$ which contradicts the fact \$\\lfloor f(1) \\rfloor=1\$.\n\nSo, \$f(x)=0, \\ \\forall x\$ or \$f(x)=a, \\ \\forall x, \\ a \\in [1,2)\$. ( By socrates[1])", "Substituting \$y=0\$ we have \$f(0) = f(x) [f(0)]\$. If \$[f0)] \\ne 0\$ then \$f(x) = \\frac{f(0)}{[f(0)]}\$. Then \$f(x)\$ is constant. Let \$f(x)=c\$. Then substituting that in (1) we have \$c=c[c] \\Rightarrow c(1-[c])=0 \\Rightarrow c=0\$, or \$[c]=1\$. Therefore \$f(x)=c\$ where \$c=0\$ or \$c \\in [1,2)\$\n\nIf \$[f(0)] = 0\$ then \$f(0)=0\$. Now substituting \$x=1\$ we have \$f(y)=f(1)[f(y)]\$. If \$f(1) \\ne 0\$ then \$[f(y)] = \\frac{f(y)}{f(1)}\$ and substituting this in (1) we have \$f([x]y)=\\frac{f(x)f(y)}{f(1)}\$. Then \$f([x]y)=f(x[y])\$. Substituting \$x=1/2, y=2\$ we get \$f(0)=f(1)\$. Then \$f(1)=0\$, which is a contradiction Therefore \$f(1)=0\$. and then \$f(y)=0\$ for all \$R\$\n\nThen the only solutions are \$f(x)=0\$ or \$f(x)=c\$ where \$c \\in [1,2)\$.( By m.candales [2])", "Let \$y=0\$, then \$f(0)=f(x)\\left\\lfloor f(0)\\right\\rfloor\$.\n\nCase 1: \$\\left\\lfloor f(0)\\right\\rfloor\\neq 0\$\n\nThen \$f(x)=\\frac{f(0)}{\\left\\lfloor f(0)\\right\\rfloor}\$ is a constant. Let \$f(x)=k\$, then \$k=k\\left\\lfloor k \\right\\rfloor \\Leftrightarrow k=0 \\vee 1\\leq k<2\$. It is easy to check that this are solutions.\n\nCase 2: \$\\left\\lfloor f(0)\\right\\rfloor= 0\$\n\nIn this case we conclude that \$\\left\\lfloor f(0)\\right\\rfloor= 0\\Rightarrow f(0)=0\$\n\nLemma:If \$y\$ is such that \$0\\leq f(y)<1\$, \$f(y)=0\$\n\nProof of the Lemma: If \$x=1\$ we have that \$f(\\left\\lfloor x\\right\\rfloor y)=f(y)=f(x)\\left\\lfloor f(y)\\right\\rfloor =0\$, as desired.\n\nLet \$0\\leq x<1\$, so that we have: \$0=f(0)=f(\\left\\lfloor x\\right\\rfloor y)=f(x)\\left\\lfloor f(y)\\right\\rfloor\\Rightarrow\$ \$\\Rightarrow f(x)=0 \\vee 0\\leq f(y)<1 \\Rightarrow f(x)=0 \\vee f(y)=0\$, using the lemma.\n\nIf \$f\$ is not constant and equal to \$0\$, letting \$y\$ be such that \$f(y)\\neq 0\$ implies that \$f(x)=0, \\forall 0\\leq x<1\$.\n\nNow it's enough to notice that any real number \$x\$ is equal to \$ky\$, where \$k\\in \\mathbb{Z}\$ and \$0\\leq y< 1\$, so that \$f(x)=f(ky)=f(\\left\\lfloor k\\right\\rfloor y)=f(k)\\left\\lfloor f(y)\\right\\rfloor=0\$. Since \$x\$ was arbitrary, we have that \$f\$ is constant and equal to \$0\$.\n\nWe conclude that the solutions are \$f(x)=k\$, where \$k=0 \\vee 1\\leq k<2\$.( By Jorge Miranda [3] )", "Clearly \$f(\\left\\lfloor x\\right\\rfloor y) = f(\\left\\lfloor \\lfloor x \\rfloor \\right\\rfloor y) = f(\\lfloor x \\rfloor)\\left\\lfloor f(y)\\right\\rfloor\$, so \$(f(x) - f(\\lfloor x \\rfloor))\\left\\lfloor f(y)\\right\\rfloor = 0\$ for all \$x,y\\in\\mathbb{R}\$.\n\nIf \$\\left\\lfloor f(y)\\right\\rfloor = 0\$ for all \$y \\in \\mathbb{R}\$, then by taking \$x=1\$ we get \$f(y)=f(1)\\left\\lfloor f(y)\\right\\rfloor = 0\$, so \$f\$ is identically null (which checks).\n\nIf, contrariwise, \$\\left\\lfloor f(y_0)\\right\\rfloor \\neq 0\$ for some \$y_0 \\in \\mathbb{R}\$, it follows \$f(x) = f(\\lfloor x \\rfloor)\$ for all \$x \\in \\mathbb{R}\$.\n\nNow it immediately follows \$f(x) = f(\\lfloor x \\rfloor \\cdot 1) = f(x)\\lfloor f(1) \\rfloor\$, hence \$f(x)(1 - \\lfloor f(1) \\rfloor) = 0\$.\n\nFor \$x=y_0\$ this implies \$\\lfloor f(1) \\rfloor = 1\$. Assume \$\\lfloor f(0) \\rfloor=0\$; then \$1 \\leq f(1) = f\\left ( 2\\cdot \\dfrac {1} {2} \\right ) = f(2)\\left \\lfloor f \\left ( \\dfrac {1} {2} \\right ) \\right \\rfloor = f(2)\\left \\lfloor f \\left ( \\left \\lfloor \\dfrac {1} {2} \\right \\rfloor \\right ) \\right \\rfloor = f(2)\\left \\lfloor f(0) \\right \\rfloor = 0\$, absurd.\n\nTherefore \$\\lfloor f(0) \\rfloor \\neq 0\$, and now \$y=0\$ in the given functional equation yields \$f(0) = f(x)\\lfloor f(0) \\rfloor\$ for all \$x \\in \\mathbb{R}\$, therefore \$f(x) = c \\neq 0\$ constant, with \$\\lfloor c \\rfloor = \\lfloor f(1) \\rfloor = 1\$, i.e. \$c \\in [1,2)\$ (which obviously checks).( By mavropnevma [4])", "Let \$x=1\$ to get \$f(y) = f(1) \\lfloor f(y) \\rfloor\$. Substituting \$x\$ for \$y\$, this implies\n\n\\[\nf(\\lfloor x \\rfloor y) = f(1) \\lfloor f(x) \\rfloor \\lfloor f(y) \\rfloor = f(x \\lfloor y \\rfloor).\n\\]\n\nSetting \$x = 1/2, y = 2\$ here gives \$f(0) = f(1)\$. If \$f(0) \\neq 0\$ then \$f(0) = f(x) \\lfloor f(0) \\rfloor\$ gives \$f(x) = c\$. Hence \$c = c \\lfloor c \\rfloor\$ which implies \$c \\in [1,2)\$ since \$c \\neq 0\$. If \$f(0) = f(1) = 0\$ then \$f(y) = f(\\lfloor 1 \\rfloor y) = f(1) \\lfloor f(y) \\rfloor = 0\$. So either \$f(x) = c \\in [1,2)\$ or \$f(x) = 0\$.\n\n~not_detriti" ]
IMO-2010-2	https://artofproblemsolving.com/wiki/index.php/2010_IMO_Problems/Problem_2	Given a triangle $ABC$, with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$. Let $E$ be a point on arc $BDC$, and $F$ a point on the segment $BC$, such that $\angle BAF=\angle CAE< \frac12\angle BAC$. If $G$ is the midpoint of $IF$, prove that the intersection of lines $EI$ and $DG$ lies on $\Gamma$.	[ "Note that it suffices to prove alternatively that if \$EI\$ meets the circle again at \$J\$ and \$JD\$ meets \$IF\$ at \$G\$, then \$G\$ is the midpoint of \$IF\$. Let \$JD\$ meet \$BC\$ at \$K\$.\n\nObservation 1. D is the midpoint of arc \$BDC\$ because it lies on angle bisector \$AI\$.\n\nObservation 2. \$AI\$ bisects \$\\angle{FAE}\$ as well.\n\nKey Lemma. Triangles \$DKI\$ and \$DIJ\$ are similar. Proof. Because triangles \$DKB\$ and \$DBJ\$ are similar by AA Similarity (for \$\\angle{KBD}\$ and \$\\angle{BJD}\$ both intercept equally sized arcs), we have \$BD^2 = BK \\cdot BJ\$. But we know that triangle \$DBI\$ is isosceles (hint: prove \$\\angle{BID} = \\angle{IBD}\$), and so \$BI^2 = BK \\cdot BJ\$. Hence, by SAS Similarity, triangles \$DKI\$ and \$DIJ\$ are similar, as desired.\n\nObservation 3. As a result, we have \$\\angle{KID} = \\angle{IJD} = \\angle{DAE} = \\angle{FAD}\$.\n\nObservation 4. \$IK // AF\$.\n\nObservation 5. If \$AF\$ and \$JD\$ intersect at \$L\$, then \$AJLI\$ is cyclic.\n\nObservation 6. Because \$\\angle{ALI} = \\angle{AJE} = \\angle{AJC} + \\angle{CJE} = \\angle{B} + \\angle{AEC} = \\angle{B} + \\angle{BAF} = \\angle{AFC}\$, we have \$LI // FK\$.\n\nObservation 7. \$LIKF\$ is a parallelogram, so its diagonals bisect each other, so \$G\$ is the midpoint of \$FI\$, as desired.", "Let \$A'\$ be A-excenter \$\\triangle ABC \\implies\$\n\n\\[\nDI = DB = DC = DA', AB \\cdot AC = AI \\cdot AA'.\n\\]\n\n\\[\n\\angle BAF = \\angle EAC, \\angle ABF = \\angle ABC = \\angle AEC \\implies \\triangle ABF \\sim \\triangle AEC \\implies\n\\]\n\n\\[\n\\frac {AB}{AF} = \\frac {AE}{AC} \\implies AF \\cdot AE = AB \\cdot AC = AI \\cdot AA' \\implies \\frac {AF}{AA'} = \\frac {AI}{AE}.\n\\]\n\n\\[\n\\angle FAA' = \\angle IAE \\implies \\triangle FAA' \\sim \\triangle IAE \\implies \\angle FA'A = \\angle IEA.\n\\]\n\n\\[\nIG = GF, ID = DA' \\implies GD \|\| FA' \\implies \\angle GDA = \\angle FA'A = \\angle IEA \\implies\n\\]\n\nthe intersection of lines \$EI\$ and \$DG\$ lies on \$\\Gamma\$.\n\nAfter pavel kozlov vladimir.shelomovskii@gmail.com, vvsss" ]
IMO-2010-3	https://artofproblemsolving.com/wiki/index.php/2010_IMO_Problems/Problem_3	Find all functions $g:\mathbb{N}\rightarrow\mathbb{N}$ such that $\left(g(m)+n\right)\left(g(n)+m\right)$ is a perfect square for all $m,n\in\mathbb{N}.$	[ "Suppose such function \$g\$ exist then:\n\nLemma 1) \$g(m) \\ne g(m+1)\$\n\nbut \$\\left(g(m)+m\\right)^2<\\left(g(m+1)+m\\right)\\left(g(m)+m+1\\right)<\\left(g(m)+m+1\\right)^2\$.\n\nLemma 2) \$\|g(m)-g(m+1)\| = 1\$ (we have show that it can't be 0)\n\nThen there must exist a prime number \$p\$ such that \$g(m)\$ and \$g(m+1)\$ are in the same residue class modulo \$p\$.\n\nIf \$\|g(m)-g(m+1)\| = p^aq\$ where \$q\$ is not divisible by \$p\$.\n\nHence, that number is not a perfect square.\n\nIf \$g(m)-g(m+1) = 1\$, then \$g(x) = -x + k\$, \$k\\in\\mathbb{N}\$.\n\nIf \$g(m)-g(m+1) = -1\$, then \$g(x) = x + k\$, \$k\\in\\mathbb{N}\$." ]
IMO-2010-4	https://artofproblemsolving.com/wiki/index.php/2010_IMO_Problems/Problem_4	Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$). The lines $AP$, $BP$ and $CP$ meet again its circumcircle $\Gamma$ at $K$, $L$, respectively $M$. The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$. Show that from $SC = SP$ follows $MK = ML$.	[ "Without loss of generality, suppose that \$AS > BS\$. By Power of a Point, \$SP^2 = SC^2 = SB \\cdot SA\$, so \$\\overline{SP}\$ is tangent to the circumcircle of \$\\triangle ABP\$. Thus, \$\\angle KPS = 180 - \\angle SPA = \\widehat{AP}/2 = \\angle ABP\$. It follows that after some angle-chasing,\n\n\\[\n\\begin{aligned} \\widehat{ML} &= \\widehat{MA} + 2\\angle AKL \\\\ &= \\left(2\\angle CPK - \\widehat{KC}\\right) + 2\\angle ABL \\\\ &= 2\\left(\\angle CPK + \\angle KPS\\right) - \\widehat{KC} \\\\ &= 2\\angle PCS - \\widehat{KC} \\\\ &= \\widehat{MK}, \\end{aligned}\n\\]\n\nso \$ML = MK\$ as desired.\n\n\\[\n[asy] import graph; size(10.71cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(9); defaultpen(dps); pen ds=black; real xmin=-2.22,xmax=4.48,ymin=-1.99,ymax=3; pen ccqqqq=rgb(0.8,0,0), qqzzqq=rgb(0,0.6,0), evevff=rgb(0.9,0.9,1); filldraw(arc((0.08,1.23),0.37,-17.48,12.32)--(0.08,1.23)--cycle,evevff,blue); filldraw(arc((0,0),0.37,0,29.8)--(0,0)--cycle,evevff,blue); filldraw(arc((1.42,0.81),0.37,-179.98,-150.2)--(1.42,0.81)--cycle,evevff,blue); draw((2,2.55)--(4,0),linewidth(1.6)+ccqqqq); draw((2,2.55)--(0,0),linewidth(1.6)+ccqqqq); draw(circle((2,0.49),2.06),linewidth(1.2)); draw((0.76,-1.16)--(3.43,1.97)); draw((3.43,1.97)--(0.08,1.23)); draw((0.08,1.23)--(0.76,-1.16)); draw((3.43,1.97)--(xmin,0.22xmin+1.22)); draw((0.08,1.23)--(4,0)); draw((1.42,0.81)--(-1.85,0.81),linewidth(1.6)+qqzzqq); draw((0,0)--(4,0)); draw(circle((1.42,3.15),2.34),linetype(\"4 4\")); draw(arc((0.08,1.23),0.37,-17.48,12.32),blue); draw(arc((0.08,1.23),0.31,-17.48,12.32),blue); draw(arc((0,0),0.37,0,29.8),blue); draw(arc((0,0),0.31,0,29.8),blue); draw(arc((1.42,0.81),0.37,-179.98,-150.2),blue); draw(arc((1.42,0.81),0.31,-179.98,-150.2),blue); draw((2,2.55)--(0.76,-1.16)); draw((0,0)--(3.43,1.97)); draw((-1.85,0.81)--(xmax,-0.75xmax-0.58)); draw((-1.85,0.81)--(0.76,-1.16),linewidth(1.6)+qqzzqq); dot((0,0),ds); label(\"$K$\",(-0.30,0.06),NElsf); dot((4,0),ds); label(\"$L$\",(4.2,0.09),NElsf); dot((2,2.55),ds); label(\"$M$\",(2.05,2.62),NElsf); dot((1.42,0.81),ds); label(\"$P$\",(1.27,0.96),NElsf); dot((3.43,1.97),ds); label(\"$A$\",(3.30,2.17),NElsf); dot((0.76,-1.16),ds); label(\"$C$\",(0.91,-1.19),NElsf); dot((0.08,1.23),ds); label(\"$B$\",(-0.11,1.49),NElsf); dot((-1.85,0.81),ds); label(\"$S$\",(-1.8,0.89),NElsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]\n\\]", "Let the tangent at \$M\$ to \$\\Gamma\$ intersect \$SC\$ at \$X\$. We now have that since \$\\triangle{XMC}\$ and \$\\triangle{SPC}\$ are both isosceles, \$\\angle{SPC}=\\angle{SCP}=\\angle{XMC}\$. This yields that \$MX \\\| PS\$.\n\nNow consider the power of point \$S\$ with respect to \$\\Gamma\$.\n\n\\[\nSC^2 = SP^2 =SA \\cdot SB \\quad \\Rightarrow \\quad \\frac{SP}{SA}=\\frac{SB}{SP}\n\\]\n\nHence by AA similarity, we have that \$\\triangle{SPA} \\sim \\triangle{SBP}\$. Combining this with the arc angle theorem yields that \$\\angle{SAP}=\\angle{SPB}=\\angle{PLK}\$. Hence \$PS \\\| LK\$.\n\nThis implies that the tangent at \$M\$ is parallel to \$LK\$ and therefore that \$M\$ is the midpoint of arc \$LK\$. Hence \$MK=ML\$.\n\n\\[\n[asy] import graph; size(10.71cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(9); defaultpen(dps); pen ds=black; real xmin=-5.2,xmax=4.48,ymin=-1.99,ymax=3; pen ccqqqq=rgb(0,1,0), qqzzqq=rgb(0,0,1), evevff=rgb(0.9,0.9,1); draw((2,2.55)--(4,0),linewidth(1.3)+ccqqqq); draw((2,2.55)--(0,0),linewidth(1.3)+ccqqqq); draw(circle((2,0.49),2.06),linewidth(1.3)); draw((0.76,-1.16)--(3.43,1.97)); draw((3.43,1.97)--(0.08,1.23)); draw((0.08,1.23)--(0.76,-1.16)); draw((0.08,1.23)--(4,0)); draw((1.42,0.81)--(-1.85,0.81),linewidth(1.3)+qqzzqq); draw((0,0)--(4,0),linewidth(1.3)+qqzzqq); draw((2,2.55)--(0.76,-1.16)); draw((0,0)--(3.43,1.97)); draw((-1.85,0.81)--(xmax,-0.75xmax-0.58)); draw((2,2.55)--(-4.16,2.55),linetype(\"0 4\")); draw((0.76,-1.16)--(-1.85,0.81)); draw((-1.85,0.81)--(-4.16,2.55),linetype(\"0 4\")); draw((3.43,1.97)--(-1.85,0.81)); dot((0,0),ds); label(\"$K$\",(-0.30,0.06),NElsf); dot((4,0),ds); label(\"$L$\",(4.2,0.09),NElsf); dot((2,2.55),ds); label(\"$M$\",(2.05,2.62),NElsf); dot((1.42,0.81),ds); label(\"$P$\",(1.27,0.96),NElsf); dot((3.43,1.97),ds); label(\"$A$\",(3.30,2.17),NElsf); dot((0.76,-1.16),ds); label(\"$C$\",(0.91,-1.19),NElsf); dot((0.08,1.23),ds); label(\"$B$\",(-0.11,1.49),NElsf); dot((-1.85,0.81),ds); label(\"$S$\",(-1.8,0.89),NElsf); dot((-4.16,2.55),ds); label(\"$X$\",(-4.16, 2.65),NElsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]\n\\]", "Since \$SP = SC\$, we can construct circle \$\\omega\$ tangent to \$SP\$ and \$SC\$ at \$P\$ and \$C\$ respectively. Then, we see that the circumcircle of \$\\triangle ABP\$ must be tangent to \$SP\$ at \$P\$ by the converse of the radical axis theorem, since \$\\omega\$ is internally tangent to \$\\Gamma\$ at \$C\$, and \$(ABP)\$ meets \$\\Gamma\$ at \$A\$ and \$B\$. Thus, \$\\angle SPB = \\angle BAK = \\angle BLK\$, so lines \$SP\$ and \$KL\$ are parallel.\n\nIt then follows that \$\\angle SCP = \\angle SPC = \\angle KDC = \\angle KLC + \\angle LCD\$. Since \$SC\$ is tangent to \$\\Gamma\$, we see that \$\\angle KLC = \\angle SCK\$, and thus \$\\angle KCD = \\angle LCD = \\angle SCP - \\angle KLC\$. Thus since \$CM\$ bisects \$\\angle KCL\$ in cyclic quad \$CKML\$, we have \$MK = ML\$. \$\\blacksquare\$" ]
IMO-2010-5	https://artofproblemsolving.com/wiki/index.php/2010_IMO_Problems/Problem_5	Each of the six boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$, $B_6$ initially contains one coin. The following operations are allowed Type 1) Choose a non-empty box $B_j$, $1\leq j \leq 5$, remove one coin from $B_j$ and add two coins to $B_{j+1}$; Type 2) Choose a non-empty box $B_k$, $1\leq k \leq 4$, remove one coin from $B_k$ and swap the contents (maybe empty) of the boxes $B_{k+1}$ and $B_{k+2}$. Determine if there exists a finite sequence of operations of the allowed types, such that the five boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$ become empty, while box $B_6$ contains exactly $2010^{2010^{2010}}$ coins.	[ "Let the notation \$[a_1,a_2,a_3,a_4,a_5,a_6]\$ be the configuration in which the \$x\$-th box has \$a_x\$ coin,\n\nLet \$T=2010^{2010^{2010}}\$.\n\nOur starting configuration is \$[1,1,1,1,1,1]\$\n\nCompound move 1: \$[a,0]\\rightarrow[0,2a]\$, this is just repeated type 1 move on all \$a\$ coins.\n\nCompound move 2: \$[a,0,0]\\rightarrow[0,2^a,0]\$, apply type 1 move on 1 of the coin to get \$[a-1,2,0]\$, then apply compound move 1 to the 2 coins to get \$[a-1,0,2^2]\$, apply type 2 move and get \$[a-2,2^2,0]\$, and repeat compound move 1 and type 2 move until \$[0,2^a,0]\$ is achieve.\n\nCompound move 3: \$[a,b,0,0]\\rightarrow[a-1,2^b,0,0]\$, apply compound move 2 to obtain \$[a,0,2^b,0]\$ and use type 2 move to get \$[a-1,2^b,0,0]\$\n\nCompound move 4: \$[a,0,0,0]\\rightarrow[0,2^{2^{.^{.^{.^2}}}},0,0]\$ with \$a\$ \$2\$'s. Apply compound move 3 \$a\$ times.\n\nCompound move 5: \$[a,0,0]\\rightarrow[b,0,0]\$ with \$a>b\$, use type 2 move \$(a-b)\$ times.\n\nLet's follow this move:\n\nUsing Compound move 1, 4 and 5, We can obtain:\n\n\\[\n[0,1,19,0,0,0]\\rightarrow[0,1,0,X,0,0]\\rightarrow[0,0,X,0,0,0]\\rightarrow[0,0,0,Y,0,0]\\rightarrow[0,0,0,T/4,0,0]\\rightarrow[0,0,0,0,T/2,0]\\rightarrow[0,0,0,0,0,T]\n\\]\n\n, where \$X, Y = 2^{2^{.^{.^{.^2}}}}\$ where X has \$19\$ 2's and Y has \$X\$ 2's, and \$Y\$ is clearly bigger then \$T/4\$" ]
IMO-2010-6	https://artofproblemsolving.com/wiki/index.php/2010_IMO_Problems/Problem_6	Let $a_1, a_2, a_3$ be a sequence of positive real numbers, and $s$ be a positive integer, such that \[ a_n = \max \{ a_k + a_{n-k} \mid 1 \leq k \leq n-1 \} \ \textrm{ for all } \ n > s. \] Prove there exist positive integers $\ell \leq s$ and $N$, such that \[ a_n = a_{\ell} + a_{n - \ell} \ \textrm{ for all } \ n \geq N. \]	[ "So for solving This Problem, we need to take a assumption that,\n\n- \$n = t_{1}+...+t_{s}\$\n- \$t_{i} \\text{is divisibe by i for every i}\$\n\n - \$\\text{there exist indices i} \\not\\ge \\text{j with i+j} \\not\\le \\text{s,}t_{i}\\not\\le\\text{i and }t_{j} \\ge \\text{j}\$\n - \$\\text{there exists an index i}\\not\\le\\text{s/2 with} t_{i} \\ge \\text{2i}\$" ]
IMO-2011-1	https://artofproblemsolving.com/wiki/index.php/2011_IMO_Problems/Problem_1	Given any set $A = \{a_1, a_2, a_3, a_4\}$ of four distinct positive integers, we denote the sum $a_1 +a_2 +a_3 +a_4$ by $s_A$. Let $n_A$ denote the number of pairs $(i, j)$ with $1 \leq i < j \leq 4$ for which $a_i +a_j$ divides $s_A$. Find all sets $A$ of four distinct positive integers which achieve the largest possible value of $n_A$.	[ "Firstly, if we order \$a_1 \\ge a_2 \\ge a_3 \\ge a_4\$, we see \$2(a_3 + a_4) \\ge (a_1+a_2)+(a_3+a_4) = s_A \\geq 0\$, so \$(a_3, a_4)\$ isn't a couple that satisfies the conditions of the problem. Also, \$2(a_4 + a_2) = (a_4 + a_4) + (a_2 + a_2) \\ge (a_4+a_3)+(a_2+a_1) = s_A \\ge 0\$, so again \$(a_2, a_4)\$ isn't a good couple. We have in total 6 couples. So \$n_A \\leq 6-2=4\$.\n\nWe now find all sets \$A\$ with \$n_A = 4\$. If \$(a,b)\$ and \$(c,d)\$ are both good couples, and \$A=\\{a, b, c, d\\}\$, we have \$a+b=c+d=s_A/2\$. So WLOG \$A=\\{a,b,a+x,b-x\\}\$ with \$x > 0\$ and \$a < b, b-x, a+x\$. It's easy to see \$a=a_1\$ and since \$(a_2, a_4),(a_3,a_4)\$ are bad, all couples containing \$a\$ must be good. Obviously \$(a,b)\$ and \$(a+x,b-x)\$ are good (\$s_A=2(a+b)\$). So we have \$2a+x \| 2a+2b\$ and \$a+b-x\|2a+2b \\Rightarrow a+b-x\|2x\$.\n\nUsing the second equation, we see that if \$y=a+b\$, \$y-x\|2x \\Rightarrow yk_1-xk_1=2x \\Rightarrow yk_1 = x(2+k_1) \\Rightarrow y=x((2+k_1)/k_1)\$, for some \$k_1\$ a positive integer.\n\nSo now we use the first equation to get \$2ak_2 + xk_2 = 2y = 2x(2+k_1)/k_1 \\Rightarrow 2ak_2 = x(\\frac{4+2k_1}{k_1}-k_2) \\Rightarrow 2a=x(\\frac{4+2k_1}{k_1k_2} - 1)\$, for a natural \$k_2\$.\n\nFinally, we obtain \$k_1 \| 4+2k-1 \\Rightarrow k_1 \| 4 \\Rightarrow k_1=\$ 1, 2 or 4. We divide in cases:\n\nCASE I: \$k_1=1\$. So \$y=3x\$ and \$2a=x((\\frac{6}{k_2}) -1)\$. But \$a < b-x \\Rightarrow 2a < y-x=2x \\Rightarrow (6/k_2) - 1 < 2 \\Rightarrow k_2 > 2 \\Rightarrow k_2 =\$3, 4,5 or 6. \$k_2=6\$ implies \$a=0\$, impossible. \$a=x\$ when \$k_2=3\$. We easily see \$b=3x=3x\$ and \$A=\\{x, 3x, 3x-x, 2x\\}\$, impossible since \$3x-x=2x\$. When \$k_2=4\$, \$a=x/4=y/12\$, and we get \$\\{11a, a, 5a, 7a\\}\$.Uf \$k_2=5\$, \$a=x/5=y/15\$ and we get \$\\{a, 14a, 6a, 9a\\}\$.\n\nCASE II and III:\$k_1=\$2, 4. Left to the reader.\n\nANSWER: \$\\{11a, a, 5a, 7a\\}\$,\$\\{a, 11a, 19a, 29a\\}\$, for any positive integer \$a\$.\n\n(Note: The above solution looks generally correct, but the actual answer should be \$\\{11a, a, 5a, 7a\\}\$,\$\\{a, 11a, 19a, 29a\\}\$. You can check that \$\\{a, 14a, 6a, 9a\\}\$ doesn't actually work. -Someone who didn't write up the above solution but solved the problem in a similar way)", "Similarly to the first solution, let us order the terms in set \$A\$ by \$0 < a_1 < a_2 < a_3 < a_4\$ since it is given that these four integers must be distinct and positive. Obviously, \$s_a > 0\$ from this. We know that pair \$(a_3, a_4)\$ cannot work because \$2(a_3 + a_4) > a_1 + a_2 + a_3 + a_4 = s_A.\$ Additionally, pair \$(a_2, a_4)\$ cannot work for the same reason: \$2(a_2 + a_4) = a_4 + a_4 + a_2 + a_2 > s_A.\$\n\nHence, since there were only \${4 \\choose 2} = 6\$ pairs \$(i, j)\$ in total, and two pairs are guaranteed not to work, we are to find sets \$A\$ satisfying \$n_A = 6 - 2 = 4.\$ Due to our making \$a_4\$ the largest term of set \$A\$, the only way to have a pair \$(a_i, a_4)\$ that exists such that \$a_i + a_4 \| s_A\$ is to make \$i\$ equal to 1 and find an \$a_1\$ such that \$\\frac{s_A}{a_1 + a_4} = 2.\$ When this happens, we know that \$\\frac{s_A}{a_2 + a_3} = 2\$, as well. It clearly follows that we need pairs \$(a_1, a_2)\$ and \$(a_1, a_3)\$ that exist such that both \$a_1 + a_2\$ and \$a_1 + a_3\$ divide \$s_A.\$\n\nTo satisfy conditions \$\\frac{s_A}{a_1 + a_4} = 2\$ and \$\\frac{s_A}{a_2 + a_3} = 2\$, choose a set \$A\$ with four distinct positive integer elements \$a - x, a, b, b + x\$ with \$0 < a - x < a < b < b + x.\$ Obviously, \$x < a < b\$ for our set to consist solely of positive integers. The problem becomes to find \$x, a,\$ and \$b\$ such that \$2a - x \| 2(a + b)\$ and \$(a + b) - x \| 2a + 2b.\$\n\nWe realize that \$x\$ can be any positive even integer. This allows \$2a - x\$ and \$2(a + b)\$ to have a common factor of 2, simplifying our problem a bit. Hence, \$a - \\frac{x}{2}\$ must divide \$(a + b).\$ Now, we can look at the other condition, \$(a + b) - x \| 2a + 2b\$, to solve this problem. To find \$a, b, x\$ satisfying \$(a + b) - x \| 2a + 2b\$, it suffices to define a positive integer \$n\$ that makes \$n(a + b - x) = 2(a + b) \\Rightarrow 0 = (n - 2)a + (n - 2)b - nx \\Rightarrow (n - 2)(a + b) = nx.\$ Note that \$x\$ must be less than \$\\frac{a + b}{2}.\$ If \$x \\ge \\frac{a + b}{2}\$, \$(n - 2)(a + b) = nx\$ would have solutions \$x, a,\$ and \$b\$ that will make \$a - x < 0\$, since \$a < \\frac{a + b}{2}\$ due to \$a < b.\$ This is a contradiction with the condition that each element in set \$A\$ has to be positive. Despite this, We move on to some cases.\n\n- \$0 < n \\le 2\$ cannot work because this makes at least one of the solutions \$a, b, x\$ to \$(n - 2)(a + b) = nx\$ zero or negative, another contradiction.\n- For \$n = 3\$, we have that \$a + b = 3x.\$ We need to find possible values for \$a\$ such that \$x < a < \\frac{a + b}{2}\$ that satisfy both \$a + b = 3x\$ and \$a - \\frac{x}{2} \| (a + b) \\Rightarrow a - \\frac{x}{2} \| 3x.\$ Note that because \$(a + b) = 3\$, the sum of our set \$A\$ is \$2(a + b) = 6x.\$ Hence, it suffices to call our terms \$a -x, a, 3x - a,\$ and \$4x - a.\$ The only values for \$a\$ and \$x\$ that work are those such that \$5 \| a\$ and \$4 \| x\$ at the same time, and \$11 \| a\$ and \$10 \| x\$ at the same time. Hence, we get that \$a - x = \\frac{a}{5}\$ or \$\\frac{a}{11}.\$ Moreover, it follows that \$x = \\frac{4a}{5}\$ or \$\\frac{10a}{11.}\$ On a side note, because we are constrained to \$x < a < \\frac{a + b}{2}\$, \$\\frac{a}{5} = \\frac{x}{4}\$ or \$\\frac{a}{11} = \\frac{x}{10}\$. Note that this completely satisfies \$a - \\frac{x}{2} \| 3x\$ because the \$a\$ and \$x\$ that we chose exists such that \$4(a - \\frac{x}{2}) = 3x\$ for \$5 \| a\$, \$4 \| x\$, and \$\\frac{a}{5} = \\frac{x}{4}\$. At the same time, \$5(a - \\frac{x}{2}) = 3x\$ for \$11 \| a\$, \$10 \| x\$, and \$\\frac{a}{11} = \\frac{x}{10}\$. In turn, we have the two sets \$A = \\{\\frac{a}{5}, a, \\frac{7a}{5}, \\frac{11a}{5}\\}\$ and \$\\{\\frac{a}{11}, a, \\frac{19a}{11}, \\frac{29a}{11}\\}.\$ Since \$a\$ must be a multiple of 5 or 11, we can scale these two sets by multiplying by 5 and 11, respectively, to get sets \$A = \\{p, 5p, 7p, 11p\\}\$ and \$\\{p, 11p, 19p, 29p\\}\$ with \$p\$ being any \$\\frac{a}{5}\$ or \$\\frac{a}{11}\$ that results in positive integers for the first and second set, respectively. Remember that \$a\$ cannot be zero because our set must contain positive integers.\n- \$n \\ge 4\$ cannot work because we have established that \$x\$ cannot be greater than or equal to \$\\frac{a + b}{2}.\$ Solving the inequality \$\\frac{n - 2}{n}(a + b) \\ge \\frac{a + b}{2}\$, \$a + b\$ cancels, resulting in \$\\frac{n - 2}{n} \\ge \\frac{1}{2} \\Rightarrow 2n - 4\\ge n \\Rightarrow n \\ge 4.\$ This shows that any \$n\$ greater than or equal to 4 will not work for \$x = \\frac{n - 2}{n}(a + b)\$ .\n\nThus, the only possible sets \$A\$ we have for satisfying the maximum \$n_A\$, which is 4, are \$A = \\{p, 5p, 7p, 11p\\}\$ and \$\\{p, 11p, 19p, 29p\\}\$, with \$p\$ being any positive integer since it is obvious that any \$a\$ that is a positive multiple of 5 or 11 makes \$\\frac{a}{5}\$ and \$\\frac{a}{11}\$ positive integers, respectively." ]
IMO-2011-2	https://artofproblemsolving.com/wiki/index.php/2011_IMO_Problems/Problem_2	Let $\mathcal{S}$ be a finite set of at least two points in the plane. Assume that no three points of $\mathcal S$ are collinear. A windmill is a process that starts with a line $\ell$ going through a single point $P \in \mathcal S$. The line rotates clockwise about the pivot $P$ until the first time that the line meets some other point belonging to $\mathcal S$. This point, $Q$, takes over as the new pivot, and the line now rotates clockwise about $Q$, until it next meets a point of $\mathcal S$. This process continues indefinitely. Show that we can choose a point $P$ in $\mathcal S$ and a line $\ell$ going through $P$ such that the resulting windmill uses each point of $\mathcal S$ as a pivot infinitely many times.	[ "Choose a coordinate system so that all points in \$\\mathcal S\$ have distinct x-coordinates. Number the points \$P_i=(x_i,y_i)\$ of \$\\mathcal S\$ by increasing x-coordinates: \$x_1<x_2<\\ldots<x_N\$.\n\nIn order to divide the set \$\\mathcal S\$ into two halves, define \$n\$ so that \$N=2n+1+d\$ where \$d=0\$ for an odd number of points and \$d=1\$ for an even number of points.\n\nStart the \"windmill\" process with the line \$\\ell\$ going vertically through the point \$P_{n+1}\$. Attach a down-up direction to this line so that we can color all points as follows: Points to the left of \$\\ell\$ (with lower x-coordinates) are blue, the pivot point on \$\\ell\$ is white and point to the right of \$\\ell\$ (with higher x-coordinates) are red. We have now \$n\$ blue points, one white point and \$n+d\$ red points.\n\nAfter processing the \"windmill\" by 180 degrees, the line \$\\ell\$ goes vertically up-down. Now, points with lower x-coordinates are to the right of \$\\ell\$ and colored red; points with higher x-coordinates are to the left of \$\\ell\$ and colored blue.\n\nNote that at each pivot exchange, the old pivot point enters the same side of \$\\ell\$ where the new pivot point came from. This means that throughout the \"windmill\" process, the number of blue points and the number of red points stay constant, respectively: We still have \$n\$ blue points, one white point and \$n+d\$ red points. This means that the current pivot point is \$P_{n+d}\$.\n\nNote that all blue and all red points changed their color from the start of the \"windmill\" process. This implies that every point was a pivot at some stage of the rotation.\n\nFor every 180 degrees of \"windmill\" rotation, the same argument applies: all colored points must change their color and hence be a pivot at some stage. Infinitely many rotations imply infinitely many color changes. This completes the proof.\n\n## Interactive Graph\n\nSee here (By u/basuboss on reddit)" ]
IMO-2011-3	https://artofproblemsolving.com/wiki/index.php/2011_IMO_Problems/Problem_3	Let $f: \mathbb R \to \mathbb R$ be a real-valued function defined on the set of real numbers that satisfies \[ f(x + y) \le yf(x) + f(f(x)) \] for all real numbers $x$ and $y$. Prove that $f(x) = 0$ for all $x \le 0$.	[ "Let \$P(x,y)\$ be the given assertion. Comparing \$P(x,f(y)-x)\$ and \$P(y,f(x)-y)\$ yields,\n\n\\[\nxf(x)+yf(y)\\leq 2f(x)f(y).\n\\]\n\n\$y\\mapsto 2f(x)\\implies xf(x)\\leq 0. \\qquad ()\$\n\n\\[\n\\textbf{Claim: }f(k)\\leq 0~~\\forall k.\n\\]\n\n\$Proof.\$ Suppose \$\\exists k:f(k)>0,\$ then\n\n\\[\nf(k+y)\\leq yf(k)+f(f(k)).\n\\]\n\nNow \$y\\to -\\infty\$ implies that \$\\lim_{x\\to -\\infty} f(x)=-\\infty.\$ \$P(x,z-x)\\implies f(z)\\leq (z-x)f(x)+f(f(x)).\$\n\nThen \$x\\to -\\infty,\$ yields a contradiction. \$\\blacksquare\$\n\nFrom \$()\$ we get \$f(x)=0,\\forall x<0.\$ \$P(0,f(0))\\implies f(0)\\geq 0,\$ thus we get \$f(0)=0,\$ as desired. \$\\square\$" ]
IMO-2011-4	https://artofproblemsolving.com/wiki/index.php/2011_IMO_Problems/Problem_4	Let $n > 0$ be an integer. We are given a balance and $n$ weights of weight $2^0,2^1, \cdots ,2^{n-1}$. We are to place each of the $n$ weights on the balance, one after another, in such a way that the right pan is never heavier than the left pan. At each step we choose one of the weights that has not yet been placed on the balance, and place it on either the left pan or the right pan, until all of the weights have been placed. Determine the number of ways in which this can be done.	[ "Call our answer \$W(n)\$. We proceed to prove \$W(n)=(2n-1)!!\$.\n\nIt is evident \$W(1)=1\$.\n\nNow, the key observation is that smaller weights can never add up to the weight of a larger weight, ie which side is heavier is determined completely by the heaviest weight currently placed. It follows, therefore, that the number of ways to place \$n\$ weights on the balance according to the rule is the same no matter which \$n\$ distinct powers of two are the weights, as each weight completely overpowers any smaller weight and is completely overpowered by any larger weight. That is, there is the 1st heaviest weight, the 2nd heaviest, the 3rd, ..., the n-th heaviest, and each weight is negligible compared to any heavier weight. Thus, any valid placement of \$n\$ weights of weight \$2^0,2^1, \\cdots ,2^{n-1}\$ can changed by replacing \$2^i\$ with the \$(n-i)\$-th heaviest weight in the set \${2^{a_k}}\$, where \$a_k \\in \\mathbb{Z}\$, and vice versa, forming a \$1:1\$ relation. With this in mind, we use recursion upon the last weight placement. There are \$2n-1\$ choices; namely, you can put any weight on either side except for the heaviest weight on the right. For the first \$n-1\$ weight placements, the answer reduces to \$W(n-1)\$. We can reduce \$W(n-1)\$ in the same way.\n\n\\[\nW(n)=(2n-1)W(n-1)=(2n-1)(2n-3)W(n-2)=...=(2n-1)!!W(1)=(2n-1)!!\n\\]\n\n\\[\n\\text{QED}\n\\]", "We can compute the answer \$W(n)\$ by conditioning on the position of the heaviest weight in the order of placement :\n\n\\[\nW(n)=\\sum_{k=1}^{n}W(k-1)2^{n-k}\\frac{(n-1)!}{(k-1)!}\n\\]\n\nThe heaviest weight can only go to the left pan. \$W(k-1)\$ is the number of ways to place the first \$(k-1)\$ weights which got placed before the heaviest weight at position \$k\$. We used the fact that the number of valid ways does not depend on the actual weights because each weight is heavier than the sum of all the weights lighter than it. There are \$\\frac{(n-1)!}{(k-1)!}\$ ways to select \$(n-k)\$ weights ( the order matters ) after the heaviest one out of \$(n-1)\$ other weights . This is because there are \$(n-1)\$ ways to select the 1st weight after the heaviest, \$(n-2)\$ to select the next one etc. Each of these \$(n-k)\$ weights can go to the left or the right pan so there are \$2^{n-k}\$ ways to create all left-right combinations. Rearange as recursion:\n\n\\[\nW(n)=2(n-1)\\sum_{k=1}^{n-1}W(k-1)2^{n-1-k}\\frac{(n-2)!}{(k-1)!}+W(n-1) = (2n-1)W(n-1)\n\\]\n\nThat is:\n\n\\[\nW(n)=(2n-1)(2n-3)...1 = (2n-1)!!\n\\]\n\nas \$W(1)=1\$ since if you have just one weight it can only go to the left pan.\n\n--alexander_skabelin 9:24, 13 July 2023 (EST)" ]
IMO-2011-5	https://artofproblemsolving.com/wiki/index.php/2011_IMO_Problems/Problem_5	Let $f$ be a function from the set of integers to the set of positive integers. Suppose that, for any two integers $m$ and $n$, the difference $f(m) - f(n)$ is divisible by $f(m - n)$. Prove that, for all integers $m$ and $n$ with $f(m) \leq f(n)$, the number $f(n)$ is divisible by $f(m)$.	[ "Solution 1 Let \$f\$ be a function from the set of integers to the set of positive integers. Suppose that, for any two integers \$m\$ and \$n\$, the difference \$f(m) - f(n)\$ is divisible by \$f(m - n)\$. Prove that, for all integers \$m\$ and \$n\$ with \$f(m) \\leq f(n)\$, the number \$f(n)\$ is divisible by \$f(m)\$.\n\nSolution 2\n\nWe first note the following facts:\n\n1. \$f(m)\|f(0)\$ for all \$m \\in \\mathbb{Z}\$: Since \$f(m) \|( f(m) - f(0))\$.\n2. \$f(m) = f(-m)\$ for all \$m \\in \\mathbb{Z}\$: Since \$f(m) \| (f(0) - f(-m))\$, we get \$f(m) \| f(-m)\$ from above. This holds for all \$m\$, so \$f(m) = f(-m)\$ for all \$m\$.\n3. \$f(m)\|f(km)\$ for all \$k \\in \\mathbb{Z}\$. Because of the the above observations, we need to show this only for \$k > 0\$. When \$k = 1\$, this is clearly true. We now use induction, along with the observation that \$f(m)\|(f((k+1)m) - f(km))\$, so that \$f(m)\|f(km) \\implies f(m)\|f((k+1)m)\$.\n4. If \$f(a) \| f(ax + b)\$, then \$f(a) \| f(b)\$. We have from the hypotheses that \$f(ax)\|(f(ax+b)-f(b))\$ which implies that \$f(a)\|(f(ax + b) - f(b))\$ and therefore \$f(a) \| f(b)\$ (here we used the last observation).\n\nFrom the first three observations, we get the following lemma:\n\nLemma 1: Suppose \$m \\neq n\$, and \$f(m) \\leq f(n)\$. If \$\|n-m\|\$ divides \$n\$, then \$f(m) \| f(n)\$.\n\nProof: Let \$d = \|n-m\|\$. Using the second observation above, we get that\n\n\\[\nf(n)\|(f(d) - f(m)).\\;\\;\\;\\;\\;\\;(1)\n\\]\n\nNow, since \$d\|n\$, we get that \$f(d) \| f(n)\$ (from the third observation above), and hence \$f(d) \\leq f(n)\$. Since \$f(m) \\leq f(n)\$ as well, and the range of \$f\$ is positive integers, equation (1) can hold only if \$f(m) = f(d)\$. But \$f(d) \| f(n)\$, so \$f(m) \| f(n)\$, as required.\n\nWe can now complete the proof. Notice that because of the first and second observations above, we can assume without loss of generality that \$m, n > 0\$. So, let \$m,n\$ be positive integers, and let \$g = {\\rm gcd (n, m)}\$. We now show that if \$f(m) \\leq f(n)\$ then \$f(m)\|f(g)\$, and hence \$f(m) \| f(n)\$.\n\nBy the Euclidean algorithm, there exist positive integers \$x\$ and \$y\$ such that \$mx = ny + g\$. Notice that \$g = mx - ny\$ divides both \$mx\$ and \$ny\$. We now have two cases:\n\nCase 1: \$f(mx) \\leq f(ny)\$. In this case, by Lemma 1, \$f(mx) \| f(ny)\$, and hence by the third and fourth observations above, \$f(m) \| f(mx - g)\$ which implies that \$f(m) \| f(g)\$. This immediately implies \$f(m) \| f(n)\$ by the third observation above, since \$g \| n\$.\n\nCase 2: \$f(mx) > f(ny)\$. In this case, by Lemma 1, \$f(ny) \| f(mx)\$, and hence by the third and fourth observations above \$f(n) \| f(g)\$. However, \$g\$ divides both \$m\$ and \$n\$, so by the third observation above, we get that \$f(g)\|f(n)\$ and \$f(g)\|f(m)\$. Thus, using the fact that \$f(m) \\leq f(n)\$, we get \$f(g) = f(m) = g(n)\$ and hence \$f(m) \| f(n)\$.\n\n--Mahamaya 20:05, 21 May 2012 (EDT)" ]
IMO-2011-6	https://artofproblemsolving.com/wiki/index.php/2011_IMO_Problems/Problem_6	Let $ABC$ be an acute triangle with circumcircle $\Gamma$. Let $\ell$ be a tangent line to $\Gamma$, and let $\ell_a, \ell_b$ and $\ell_c$ be the lines obtained by reflecting $\ell$ in the lines $BC$, $CA$ and $AB$, respectively. Show that the circumcircle of the triangle determined by the lines $\ell_a, \ell_b$ and $\ell_c$ is tangent to the circle $\Gamma$.	[ "Without loss of generality, let \$\\Gamma\$ be the unit circle and let \$\\ell\$ be the line \$y=1\$.\n\nDenote the coordinates of \$A,B,C\$ by \$(x_a, y_a)\$ and similarly for B and C.\n\nWe get \$x_a^2+y_a^2=1\$, ...\n\nThe equation for line \$AB\$ is \$\\frac{x-x_a}{y-y_a}=\\frac{x-x_b}{y-y_b}\$\n\nThrough a little bash, (0,1) reflects to \$\\left(\\frac{(-x_a+x_b+x_by_a)(y_b-_a)}{(x_a-x_b)^2+(y_a-y_b)^2}, \\frac{(x_by_a-x_ay_b)(x_a-x_b)+(y_a-y_b)^2}{(x_a-x_b)^2+(y_a-y_b)^2}\\right)\$. \$(x_a-x_b)^2+(y_a-y_b)^2\$ simplifies to \$-(2x_ax_b+2y_ay_b)\$. The terms for the other 2 are symmetric. The intersection point must reflect to itself, and the equation is \$\\left(\\frac{x_by_a-x_ay_b-x_a+x_b}{y_b-y_a}, 1\\right)\$.\n\nIt is trivial to find the intersections of a,b and their perpendicular bisectors, so this is left to the reader as an exercise.\n\nRegardless, the circumcenter and an intersection of the circles are collinear with (0,0), so it is a tangency.\n\n-Trex4days" ]
IMO-2012-1	https://artofproblemsolving.com/wiki/index.php/2012_IMO_Problems/Problem_1	Given triangle $ABC$ the point $J$ is the centre of the excircle opposite the vertex $A.$ This excircle is tangent to the side $BC$ at $M$, and to the lines $AB$ and $AC$ at $K$ and $L$, respectively. The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G.$ Let $S$ be the point of intersection of the lines $AF$ and $BC$, and let $T$ be the point of intersection of the lines $AG$ and $BC.$ Prove that $M$ is the midpoint of $ST$.	[ "First, \$BK = BM\$ because \$BK\$ and \$BM\$ are both tangents from \$B\$ to the excircle \$J\$. Then \$BJ \\bot KM\$. Call the \$X\$ the intersection between \$BJ\$ and \$KM\$. Similarly, let the intersection between the perpendicular line segments \$CJ\$ and \$LM\$ be \$Y\$. We have \$\\angle XBM = \\angle XBK = \\angle FBA\$ and \$\\angle XMB = \\angle XKB\$. We then have, \$\\angle XBM + \\angle XBK + \\angle XMB + \\angle XKB = \\angle MBK + \\angle XMB + \\angle XKB\$ \$= \\angle MBK + \\angle KMB + \\angle MKB = 180^{\\circ}\$. So \$\\angle XBM = 90^{\\circ} - \\angle XMB\$. We also have \$180^{\\circ} = \\angle FBA + \\angle ABC + \\angle XBM = 2\\angle XBM + \\angle ABC = 180^{\\circ} - 2\\angle XMB\$ \$+ \\angle ABC\$. Then \$\\angle ABC = 2\\angle XMB\$. Notice that \$\\angle XFM = 90^{\\circ} - \\angle XMB - \\angle BMF = 90^{\\circ} - \\angle XMB - \\angle YMC\$. Then, \$\\angle ACB = 2\\angle YMC\$. \$\\angle BAC = 180^{\\circ} - \\angle ABC - \\angle ACB = 180^{\\circ} - 2(\\angle XMB + \\angle YMC)\$ \$= 2(90^{\\circ} - (\\angle XMB + \\angle YMC) = 2\\angle XFM\$. Similarly, \$\\angle BAC = 2\\angle YGM\$. Draw the line segments \$FK\$ and \$GL\$. \$\\triangle FXK\$ and \$\\triangle FXM\$ are congruent and \$\\triangle GYL\$ and \$\\triangle GYM\$ are congruent. Quadrilateral \$AFJL\$ is cyclic because \$\\angle JAL = \\frac{\\angle BAC}{2} = \\angle XFM = \\angle JFL\$. Quadrilateral \$AFKJ\$ is also cyclic because \$\\angle JAK = \\frac{\\angle BAC}{2} = \\angle XFM = \\angle XFK = \\angle JFK\$. The circumcircle of \$\\triangle AFJ\$ also contains the points \$K\$ and \$J\$ because there is a circle around the quadrilaterals \$AFJL\$ and \$AFKJ\$. Therefore, pentagon \$AFKJL\$ is also cyclic. Finally, quadrilateral \$AGLJ\$ is cyclic because \$\\angle JAL = \\frac{\\angle BAC}{2} = \\angle YGM = \\angle YGL = \\angle JGL\$. Again, \$\\triangle AJL\$ is common in both the cyclic pentagon \$AFKJL\$ and cyclic quadrilateral \$AGLJ\$, so the circumcircle of \$\\triangle AJL\$ also contains the points \$F\$, \$K\$, and \$G\$. Therefore, hexagon \$AFKJLG\$ is cyclic. Since \$\\angle AKJ\$ and \$\\angle ALJ\$ are both right angles, \$AJ\$ is the diameter of the circle around cyclic hexagon \$AFKJLG\$. Therefore, \$\\angle AFJ\$ and \$\\angle AGJ\$ are both right angles. \$\\triangle BFS\$ and \$\\triangle BFA\$ are congruent by ASA congruency, and so are \$\\triangle CGT\$ and \$\\triangle CGA\$. We have \$SB = AB\$, \$TC = AC\$, \$BM = BK\$, and \$CM = CL\$. Since \$AK\$ and \$AL\$ are tangents from \$A\$ to the circle \$J\$, \$AK = AL\$. Then, we have \$AK = AL\$, which becomes \$AB + BK = AC + CL\$, which is \$SB + BM = TC + CM\$, or \$SM = TM\$. This means that \$M\$ is the midpoint of \$ST\$.\n\nQED\n\n--Aopsqwerty 21:19, 19 July 2012 (EDT)", "For simplicity, let \$A, B, C\$ written alone denote the angles of triangle \$ABC\$, and \$a\$, \$b\$, \$c\$ denote its sides.\n\nLet \$R\$ be the radius of the A-excircle. Because \$CM = CL\$, we have \$CML\$ isosceles and so \$\\angle{CML} = \\dfrac{\\angle{C}}{2}\$ by the Exterior Angle Theorem. Then because \$\\angle{FBS} = 90^\\circ - \\dfrac{B}{2}\$, we have \$\\angle{BFM} = \\dfrac{\\angle{A}}{2}\$, again by the Exterior Angle Theorem.\n\nNotice that \$\\angle{BJM} = \\dfrac{\\angle{B}}{2}\$ and \$\\angle{CJM} = \\dfrac{\\angle{C}}{2}\$, and so\n\n\\[\na = R \\tan \\frac{B}{2} + R \\tan \\frac{C}{2} = R \\frac{\\sin \\frac{B+C}{2}}{\\cos \\frac{B}{2} \\cos \\frac{C}{2}}\n\\]\n\nafter converting tangents to sine and cosine. Thus,\n\n\\[\nR = a \\cos \\frac{B}{2} \\cos \\frac{C}{2} \\sec \\frac{A}{2}.\n\\]\n\nIt follows that \$BM = a \\sin \\dfrac{B}{2} \\cos \\dfrac{C}{2} \\sec \\frac{A}{2}\$. By the Law of Sines on triangle \$BFM\$ and \$ABC\$ and the double-angle formula for sine, we have\n\n\\[\nBF = BM \\cdot \\frac{\\sin \\frac{C}{2}}{\\sin \\frac{A}{2}} = a \\sin \\frac{B}{2} \\cdot \\frac{\\sin C}{\\sin A} = c \\sin \\frac{B}{2}.\n\\]\n\nTherefore, triangle \$BFA\$ is congruent to a right triangle with hypotenuse length \$c\$ and one angle of measure \$90^\\circ - \\dfrac{B}{2}\$ by SAS Congruence, and so \$\\angle{BFA} = 90^\\circ\$. It then follows that triangles \$BFS\$ and \$BFA\$ are congruent by \$ASA\$, and so \$AF = FS\$. Thus, \$J\$ lies on the perpendicular bisector of \$AS\$. Similarly, \$J\$ lies on the perpendicular bisector of \$AT\$, and so \$J\$ is the circumcenter of \$ATS\$. In particular, \$J\$ lies on the perpendicular bisector of \$ST\$, and so, because \$JM\$ is perpendicular to \$ST\$, \$M\$ must be the midpoint of \$ST\$, as desired.\n\n--Suli 17:53, 8 February 2015 (EST)", "Same as Solution 2, except noticing that (letting \$s = \\dfrac{a + b + c}{2}\$ be the semi-perimeter):\n\n\\[\nFB = BM \\cdot \\frac{\\sin \\frac{C}{2}}{\\sin \\frac{A}{2}} = (s - c) \\cdot \\frac{\\sqrt{\\frac{(s-a)(s-b)}{ab}}}{\\sqrt{\\frac{(s-b)(s-c)}{bc}}} = c \\sqrt{\\frac{(s-a)(s-c)}{ac}} = c \\sin \\frac{B}{2}.\n\\]\n\n--Suli 18:21, 8 February 2015 (EST)", "As before in Solution 2, we find that \$\\angle{JFL} = \\dfrac{\\angle{A}}{2}.\$ But it is clear that \$AJ\$ bisects \$\\angle{KAL}\$, so \$\\angle{JAL} = \\dfrac{\\angle{A}}{2} = \\angle{JFL}\$ and hence \$AFJL\$ is cyclic. In particular, \$\\angle{AFJ} = \\angle{ALJ} = 90^\\circ\$, and continue as in Solution 2." ]
IMO-2012-2	https://artofproblemsolving.com/wiki/index.php/2012_IMO_Problems/Problem_2	Let ${{a}_{2}}, {{a}_{3}}, \cdots, {{a}_{n}}$ be positive real numbers that satisfy ${{a}_{2}}\cdot {{a}_{3}}\cdots {{a}_{n}}=1$ . Prove that \[ \left(a_2+1\right)^2\cdot \left(a_3+1\right)^3\cdots \left(a_n+1\right)^n\gneq n^n \]	[ "The inequality between arithmetic and geometric mean implies\n\n\\[\n{{\\left( {{a}_{k}}+1 \\right)}^{k}}={{\\left( {{a}_{k}}+\\frac{1}{k-1}+\\frac{1}{k-1}+\\cdots +\\frac{1}{k-1} \\right)}^{k}}\\ge {{k}^{k}}\\cdot {{a}_{k}}\\cdot \\frac{1}{{{\\left( k-1 \\right)}^{k-1}}}=\\frac{{{k}^{k}}}{{{\\left( k-1 \\right)}^{k-1}}}\\cdot {{a}_{k}}\n\\]\n\nThe inequality is strict unless \$a_k=\\frac1{k-1}\$. Multiplying analogous inequalities for \$k=2,\\text{ 3, }\\cdots \\text{, }n\$ yields\n\n\\[\n\\left(a_2+1\\right)^2\\cdot \\left(a_3+1\\right)^3\\cdots \\left(a_n+n\\right)^n\\gneq \\frac{2^2}{1^1}\\cdot\\frac{3^3}{2^2}\\cdot \\frac{4^4}{3^3}\\cdots \\frac{n^n}{(n-1)^{n-1}}\\cdot a_2\\cdot a_3\\cdots a_n=n^n\n\\]" ]
IMO-2012-3	https://artofproblemsolving.com/wiki/index.php/2012_IMO_Problems/Problem_3	The liar’s guessing game is a game played between two players A and B. The rules of the game depend on two positive integers $k$ and $n$ which are known to both players. At the start of the game the player A chooses integers $x$ and $N$ with $1 \le x \le N$. Player A keeps $x$ secret, and truthfully tells $N$ to the player B. The player B now tries to obtain information about $x$ by asking player A questions as follows: each question consists of B specifying an arbitrary set $S$ of positive integers (possibly one specified in some previous question), and asking A whether $x$ belongs to $S$. Player B may ask as many questions as he wishes. After each question, player A must immediately answer it with yes or no, but is allowed to lie as many times as she wants; the only restriction is that, among any $k+1$ consecutive answers, at least one answer must be truthful. After B has asked as many questions as he wants, he must specify a set $X$ of at most $n$ positive integers. If $x \in X$, then B wins; otherwise, he loses. Prove that: (a) If $n \ge 2^k$ then B has a winning strategy. (b) There exists a positive integer $k_0$ such that for every $k \ge k_0$ there exists an integer $n \ge 1.99^k$ for which B cannot guarantee a victory.	[ "It suffices to show that there is a winning strategy for \$n = 2^k-1\$, as a winning strategy for any \$n \\ge 2^k\$ easily gives a winning strategy for \$n = 2^k-1\$.\n\nPlayer B first divides all integers \$1\$ through \$N\$ into \$2^k\$ nonempty sets \$T_s\$, where \$s\$ is some binary string of length \$k\$. On the \$i^{th}\$ question for \$1 \\le i \\le k\$, player B asks whether the set containing \$x\$ is labeled with a string that has a \$1\$ in the \$i^{th}\$ position. After asking these questions, there is one set that player A must indicate does not contain \$x\$ at \$k\$ times in a row; WLOG we may assume this set is \$T_{00 \\dots 0}\$.\n\nOn the next question, B asks if \$x \\in T_{00 \\dots 0}\$. If A says 'no', then B can rule out \$T_{00 \\dots 0}\$. If A says 'yes', then B asks if \$x \\in T_{100 \\dots 0}\$. If A says 'no', then B can rule out \$T_{100 \\dots 0}\$. If A says 'yes', then B asks if \$x \\in T_{0100 \\dots 0}\$. If A says 'no', then B can rule out \$T_{0100 \\dots 0}\$, while if A says 'yes', B can rule out \$T_{1100 \\dots 0}\$.\n\nIn any case, B can ascertain that \$x \\not \\in T_s\$ for some string \$s\$, thus reducing the number of possibilities for \$x\$. Player B can repeat this procedure as long as at least \$2^k\$ elements remain, eventually reducing the number of possible values of \$x\$ to at most \$2^k - 1\$.", "Take \$k\$ sufficiently large so that there is some positive integer \$n\$ with \$n \\ge 1.99^k\$ but \$n+1 \\le .0011.999^{k+1}\$.\n\nPlayer A sets \$N=n+1\$. Let \$f(i,m)\$ denote the number of times in a row that A has just indicated that \$i \\neq x\$. In other words, if A indicates on the \$m^{th}\$ step that \$x\$ was in a set containing \$i\$, then \$f(i,m)=0\$, while if A indicates otherwise, we would have \$f(i,m)=f(i,m-1)+1\$.\n\nAt the \$m^{th}\$ step, player A makes a decision so as to minimize the sum:\n\n\\[\ng(m)=\\sum_{i=1}^N 1.999^{f(i,m)}\n\\]\n\nI claim that, by using this strategy, A will preserve the invariant that \$g(m) < 1.999^{k+1}\$. We can prove this by induction on \$m\$. Note that initially, \$g(0)=N<1.999^{k+1}\$.\n\nSuppose that \$g(m-1) < 1.999^{k+1}\$. Note that if B asks if \$x \\in S\$ for some \$S \\subset \\{1,2,\\dots,N\\}\$, then the two possibilities for \$g(m)\$ are:\n\n\\[\ng_\\text{in}(m)=N-\|S\|+1.999\\sum_{i \\in S} 1.999^{f(i,m-1)}\n\\]\n\nand\n\n\\[\ng_\\text{out}(m)=\|S\|+1.999\\sum_{i \\not \\in S} 1.999^{f(i,m-1)}\n\\]\n\nNote that \$g_\\text{in}(m)+g_\\text{out}(m)=N + 1.999g(m-1)\$. Therefore, we have that:\n\n\\[\n\\min(g_{\\text{in}}(m), g_{\\text{out}}(m)) \\le \\frac{g_{\\text{in}}(m) + g_{\\text{out}}(m)}{2} = \\frac{N + 1.999g(m-1)}{2} < \\frac{.0011.999^{k+1} +1.999^{k+2}}{2} = 1.999^{k+1}\n\\]\n\nThis completes the induction. In particular, since \$g(m)<1.999^{k+1}\$ at all times, A will never indicate that \$x\\neq i\$ more that \$k\$ times in a row. So player B will never be able to rule out any element of \$\\{1,2,\\dots,N\\}\$ and submit a set of \$n\$ integers that are guaranteed to contain \$x\$." ]
IMO-2012-4	https://artofproblemsolving.com/wiki/index.php/2012_IMO_Problems/Problem_4	Find all functions $f: \mathbb{Z} \to \mathbb{Z}$ such that, for all integers $a, b,$ and $c$ that satisfy $a + b+ c = 0$, the following equality holds: \[ f(a)^2 + f(b)^2 + f(c)^2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a). \] (Here $\mathbb{Z}$ denotes the set of integers.)	[ "Consider \$a = b = c = 0.\$ Then \$f(0)^2 + f(0)^2 + f(0)^2 = 2f(0)f(0) + 2f(0)f(0) + 2f(0)f(0) \\Rightarrow 3f(0)^2 = 6f(0)^2 \\Rightarrow\$\n\n\\[\nf(0) = 0.\n\\]\n\nNow we look at \$b = -a, c = 0.\$ \$f(a)^2 + f(-a)^2 + f(0)^2 = 2f(a)f(-a) + 2f(-a)f(0) + 2f(0)f(a) \\Rightarrow\$ \$f(a)^2 + f(-a)^2 = 2f(a)f(-a) \\Rightarrow\$ \$f(a)^2 - 2f(a)f(-a) + f(-a)^2 = 0 \\Rightarrow\$ \$(f(a) - f(-a))^2=0 \\Rightarrow\$\n\n\\[\nf(a)=f(-a).\n\\]\n\nWe can write \$f(c)^2 - 2f(c)(f(a)+f(b)) + (f(a)-f(b))^2 = 0 \\Rightarrow\$\n\n\\[\nf(c) = f(-c) = f(a+b) =\\frac{2(f(a)+f(b)) \\pm \\sqrt{4(f(a)+f(b))^2 - 4(f(a)-f(b))^2}}{2}\n\\]\n\n\\[\n\\Rightarrow f(a+b) = f(a) + f(b) \\pm 2\\sqrt{f(a)f(b)}\n\\]\n\nIf \$f(b) = 0\$, then\n\n\\[\nf(a+b) = f(a) = f((a)mod(b))\n\\]\n\nCase 1: \$f(1) = 0 \\Rightarrow f(x)= 0\$ \$\\forall\$ \$x.\$ \$\\Box\$\n\nCase 2: \$f(1) \\not= 0\$, we will have \$f(2) = f(1) + f(1) \\pm 2\\sqrt{f(1)f(1)} \\Rightarrow f(2) = 0\$ or \$f(2) = 4f(1)\$\n\nCase 2.1: \$f(1) \\not= 0, f(2) = 0 \\Rightarrow f(x) = f(x\$ \$mod\$ \$2) \\Rightarrow f(x) = f(1)\$ if \$x\$ is odd, \$f(x) = 0\$ if \$x\$ is even. \$\\Box\$\n\nCase 2.2: \$f(1) \\not= 0, f(2) = 4f(1) \\Rightarrow f(3) = f(2) + f(1) \\pm 2\\sqrt{f(2)f(1)}\$\n\n\$\\Rightarrow f(3) = 5f(1) \\pm 4f(1) \\Rightarrow f(3) = f(1)\$ or \$9f(1)\$\n\nCase 2.2.1: \$f(1) \\not= 0, f(2) = 4f(1), f(3) = f(1).\$\n\n\$\\Rightarrow f(4) = f(1) + f(3) \\pm 2\\sqrt{f(1)f(3)}\$ and \$f(4) = f(2) + f(2) \\pm 2\\sqrt{f(2)f(2)}\$\n\n\$\\Rightarrow f(4) = f(1)\$ or \$0\$ and \$f(4) = 16f(1)\$ or \$0\$\n\n\$\\Rightarrow f(4) = 0 \\Rightarrow f(x) = f(x\$ \$mod\$ \$4).\\Box\$\n\nCase 2.2.2: \$f(1) \\not= 0, f(2) = 4f(1), f(3) = 9f(1).\$\n\n\$\\Rightarrow f(4) = f(1 + 3) = f(1) + f(3) \\pm 2\\sqrt{f(1)f(3)} = 16f(1)\$ or \$4f(1)\$\n\nand \$f(4) = f(2) + f(2) \\pm 2\\sqrt{f(2)f(2)} = 16f(1)\$ or \$0. \\Rightarrow f(4) = 16f(1).\$\n\nIf \$x \\le 4\$ then \$f(x) = f(1)x^2.\$\n\nWe will prove by induction \$f(x) = f(1)x^2\$ \$\\forall\$ \$x.\$\n\nIf \$x \\le m\$ then \$f(x) = f(1)x^2.\$ \$\\forall\$ \$x\$ is true for some \$m\$.\n\nand if the statement is true for \$m=k\$\n\n\$\\Rightarrow f(k+1) = f(k) + f(1) \\pm 2\\sqrt{f(k)f(1)} = f(1)(k+1)^2\$ or \$f(1)(k-1)^2\$\n\nand \$f(k+1) = f(k-1) + f(2) \\pm 2\\sqrt{f(k-1)f(2)} = f(1)(k+1)^2\$ or \$f(1)(k-3)^2\$\n\n\\[\n\\Rightarrow f(k+1) = f(1)(k+1)^2.\n\\]\n\n\$\\Rightarrow\$ the statement is true for \$m=k+1\$ as well.\n\nAs the statement is true for \$m = 4\$, by mathematical induction we can conclude\n\n\$f(x) = f(1)x^2\$ \$\\forall\$ \$x.\\Box\$\n\nSo, Case 2.1, Case 2.2.1 and Case 2.2.2 are the three independent possible solutions.\n\n--Dineshram" ]
IMO-2012-5	https://artofproblemsolving.com/wiki/index.php/2012_IMO_Problems/Problem_5	Let $ABC$ be a triangle with $\angle BCA=90^{\circ}$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK=BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL=AC$. Let $M=\overline{AL}\cap \overline{BK}$. Prove that $MK=ML$.	[ "Let \$\\Gamma\$, \$\\Gamma'\$, \$\\Gamma''\$ be the circumcircle of triangle \$ABC\$, the circle with its center as \$A\$ and radius as \$AC\$, and the circle with its center as \$B\$ and radius as \$BC\$, Respectively. Since the center of \$\\Gamma\$ lies on \$BC\$, the three circles above are coaxial to line \$CD\$.\n\nLet Line \$AX\$ and Line \$BX\$ collide with \$\\Gamma\$ on \$P\$ (\$\\neq A\$) and \$Q\$ (\$\\neq B\$), Respectively. Also let \$R = AQ \\cap BP\$.\n\nThen, since \$\\angle AYB = \\angle AZB = 90^{\\circ}\$, by ceva's theorem, \$R\$ lies on \$CD\$.\n\nSince triangles \$ABC\$ and \$ACD\$ are similar, \$AL^2 = AC^2 = AD \\cdot AB\$, Thus \$\\angle ALD = \\angle ABL\$. In the same way, \$\\angle BKD = \\angle BAK\$.\n\nTherefore, \$\\angle ARD = \\angle ABQ = \\angle ALD\$ making \$(A, R, L, D)\$ concyclic. In the same way, \$(B, R, K, D)\$ is concyclic.\n\nSo \$\\angle ADR = \\angle ALR = 90^{\\circ}\$, and in the same way \$\\angle BKR = 90^{\\circ}\$. Therefore, the lines \$RK\$ and \$RL\$ are tangent to \$\\Gamma'\$ and \$\\Gamma''\$, respectively.\n\nSince \$R\$ is on \$CD\$, and \$CD\$ is the concentric line of \$\\Gamma'\$ and \$\\Gamma''\$, \$RK^2 = RL^2\$, Thus \$RK = RL\$. Since \$RM\$ is in the middle and \$\\angle ADR = \\angle BKR = 90^{\\circ}\$, we can say triangles \$RKM\$ and \$RLM\$ are congruent. Therefore, \$KM = LM\$.\n\nEdit: I believe that this solution, which was posted on IMO 2012-4's page, was meant to be posted here.\n\n~ Latex edit by Kscv" ]
IMO-2012-6	https://artofproblemsolving.com/wiki/index.php/2012_IMO_Problems/Problem_6	Find all positive integers $n$ for which there exist non-negative integers $a_1, a_2, \ldots, a_n$ such that \[ \frac{1}{2^{a_1}} + \frac{1}{2^{a_2}} + \cdots + \frac{1}{2^{a_n}} = \frac{1}{3^{a_1}} + \frac{2}{3^{a_2}} + \cdots + \frac{n}{3^{a_n}} = 1. \]	[]
IMO-2013-1	https://artofproblemsolving.com/wiki/index.php/2013_IMO_Problems/Problem_1	Prove that for any pair of positive integers $k$ and $n$, there exist $k$ positive integers $m_1,m_2,...,m_k$ (not necessarily different) such that $1+\frac{2^k-1}{n}=(1+\frac{1}{m_1})(1+\frac{1}{m_2})...(1+\frac{1}{m_k})$.	[ "We prove the claim by induction on \$k\$.\n\nBase case: If \$k = 1\$ then \$1 +\\frac{2^1-1}{n} = 1 + \\frac{1}{n}\$, so the claim is true for all positive integers \$n\$.\n\nInductive hypothesis: Suppose that for some \$m \\in \\mathbb{Z}^{+}\$ the claim is true for \$k = m\$, for all \$n \\in \\mathbb{Z}^{+}\$.\n\nInductive step: Let \$n\$ be arbitrary and fixed. Case on the parity of \$n\$:\n\n[Case 1: \$n\$ is even] \$1 + \\frac{2^{m+1} - 1}{n} = \\left( 1 + \\frac{2^{m} - 1}{\\frac{n}{2}} \\right) \\cdot \\left( 1 + \\frac{1}{n + 2^{m+1} - 2} \\right)\$\n\n[Case 2: \$n\$ is odd] \$1 + \\frac{2^{m+1} - 1}{n} = \\left( 1 + \\frac{2^{m}-1}{\\frac{n+1}{2}} \\right) \\cdot \\left( 1 + \\frac{1}{n} \\right)\$\n\nIn either case, \$1 + \\frac{2^{m+1} - 1}{n} = \\left( 1 + \\frac{2^m - 1}{c} \\right) \\cdot \\left( 1 + \\frac{1}{a_{m+1}} \\right)\$ for some \$c, a_{m+1} \\in \\mathbb{Z}^+\$.\n\nBy the induction hypothesis we can choose \$a_1, ..., a_m\$ such that \$\\left( 1 + \\frac{2^m - 1}{c} \\right) = \\prod_{i=1}^{m} (1 + \\frac{1}{a_i})\$.\n\nTherefore, since \$n\$ was arbitrary, the claim is true for \$k = m+1\$, for all \$n\$. Our induction is complete and the claim is true for all positive integers \$n\$, \$k\$.", "We will prove the claim by constructing telescoping product out of positive integers:\n\n\\[\n\\frac{a_2}{a_1}\\cdot\\frac{a_3}{a_2}\\cdot\\frac{a_4}{a_3}\\cdot \\cdot \\frac{a_{k+1}}{a_k} = \\frac{a_{k+1}}{a_1} = \\frac{\\left(n+2^{k}-1\\right)}{n}\n\\]\n\nwhere \$a_1=n\$ and where each fraction \$\\frac{a_{i+1}}{a_i}=\\frac{a_{i}+\\Delta_i}{a_i}\$ can also be written as \$\\frac{m_i+1}{m_i}\$ for some positive integer \$m_i\$. All \$\\Delta_i\$ will be different with \$\\Delta_i=2^j\$, \$0\\le j \\le (k-1)\$. \$\\sum_{i=1}^{k}{\\Delta_i}=\\sum_{i=1}^{k}{2^{i-1}}=2^{k}-1\$.\n\nAscend: make telescoping product of fractions with a sequence of \$\\Delta_i\$ that increases in magnitude up to the maximum \$2^{k-1}\$. If the maximum \$\\Delta=2^{k-1}\$ is reached go to the descend phase. Pull out factors of \$2\$ (up to the maximum \$2^{k-1}\$). \$n=2^j(2z+1)\$, \$z+1=2^q(2r+1)\$ etc where \$j\\ge 0\$, \$q\\ge 0\$ ... Construct telescoping as\n\n\\[\n\\frac{2^j(2z+2)}{2^j(2z+1)}\\cdot \\frac{2^{j+q+1}(2r+2)}{2^{j+q+1}(2r+1)} ...\n\\]\n\n. The corresponding differences \$\\Delta\$ are \$2^j, 2^{j+q+1}\$ ... Every \$\\Delta_i\$ is bigger then the previous \$\\Delta_{i-1}\$ by at least factor \$2\$. Therefore, the biggest needed \$\\Delta=2^{k-1}\$ can be created telescoping at most \$k\$ fractions. After we constructed the fraction \$\\frac{2^{k-1}(h+1)}{2^{k-1}h}\$ with the biggest needed \$\\Delta=2^{k-1}\$ we can construct any remaining needed \$\\Delta_i\$ as described below.\n\nDescend: If we need \$\\Delta=2^{d}\$ where \$d<k-1\$ we can use as the next telescoping fraction \$\\frac{2^{d}(2^{k-1-d}(h+1)+1)}{2^{d}(2^{k-1-d}(h+1))}\$. We can construct all the remaining nedded \$\\Delta_i\$ in the decreasing order of their magnitude by repeating the same step.\n\n--alexander_skabelin 9:24, 11 July 2023 (EST)", "Write\n\n\\[\nS_j = n\\Big(1+\\frac{1}{m_1}\\Big) \\cdots \\Big(1+\\frac{1}{m_j}\\Big)\n\\]\n\nand \$S_0 = n\$. Note if \$l\|r\$ then \$r\\Big(1+\\frac{1}{r/l} \\Big) = r+l\$. Therefore if \$S_j, j<k\$, is an integer, if \$l\|S_j\$ then \$m_{j+1}\$ can be chosen such that \$S_{j+1} = S_j + l\$. Thus one way to solve the problem is by finding a sequence \$S_0,S_1,...,S_k = n+2^k-1\$ such that \$S_{j+1} = S_j + l_j\$ where \$l_j \| S_j\$ for all \$j < m\$.\n\nWe claim by induction that one can choose \$S_0,...,S_k=n+2^k-1\$ such that \$\\{l_0,...,l_{k-1}\\} = \\{1,2,...,2^{k-1}\\}\$.\n\nBase case \$k=1\$: We may choose \$l_0 = 1 \| S_0\$ so that \$S_0 = n, S_1 = n+1 = n + 2^1 - 1\$.\n\nInduction case: Suppose \$S_0,...,S_{k-1} = n + 2^{k-1} - 1\$ and \$\\{l_0,...,l_{k-2}\\} = \\{1,2,...,2^{k-2}\\}\$. Say \$l_j = 2^{k-2}\$. Then either \$S_j\$ or \$S_{j+1}=S_j+2^{k-2}\$ will be divisible by \$2^{k-1}\$. Therefore our new sequence \$S_0',...,S_k' = n+2^k-1\$ can be the same sequence as\n\n\\[\nS_0,...,S_j,S_j+2^{k-1},S_{j+1}+2^{k-1},...,S_{k-1} + 2^{k-1}\n\\]\n\nor\n\n\\[\nS_0,...,S_{j+1},S_{j+1}+2^{k-1},S_{j+2}+2^{k-1},...,S_{k-1} + 2^{k-1}\n\\]\n\naccordingly to which term is divisible by \$2^{k-1}\$. Of course if \$l_p = 2^q\$ where \$q<{k-1}\$ then \$l_p \| S_p + 2^{k-1}\$ so that the sequences will be valid.\n\n~not_detriti" ]
IMO-2013-2	https://artofproblemsolving.com/wiki/index.php/2013_IMO_Problems/Problem_2	A configuration of $4027$ points in the plane is called Colombian if it consists of $2013$ red points and $2014$ blue points, and no three of the points of the configuration are collinear. By drawing some lines, the plane is divided into several regions. An arrangement of lines is good for a Colombian configuration if the following two conditions are satisfied: - no line passes through any point of the configuration; - no region contains points of both colours. Find the least value of $k$ such that for any Colombian configuration of $4027$ points, there is a good arrangement of $k$ lines.	[ "We can start off by imagining the points in their worst configuration. With some trials, we find \$2013\$ lines to be the answer to the worst cases. We can assume the answer is \$2013\$. We will now prove it.\n\nWe will first prove that the sufficient number of lines required for a good arrangement for a configuration consisting of \$u\$ red points and \$v\$ blue points, where \$u\$ is even and \$v\$ is odd and \$u - v = 1\$, is \$v\$.\n\nNotice that the condition \"no three points are co-linear\" implies the following: No blue point will get in the way of the line between two red points and vice versa. What this means, is that for any two points \$A\$ and \$B\$ of the same color, we can draw two lines parallel to, and on different sides of the line \$AB\$, to form a region with only the points \$A\$ and \$B\$ in it.\n\nNow consider a configuration consisting of u red points and v blue ones (\$u\$ is even, \$v\$ is odd, \$u>v\$). Let the set of points \$S = \\{A_1, A_2, ... A_k\\}\$ be the out-most points of the configuration, such that you could form a convex k-gon, \$A_1 A_2 A_3 ... A_k\$, that has all of the other points within it.\n\nIf the set S has at least one blue point, there can be a line that separates the plane into two regions: one only consisting of only a blue point, and one consisting of the rest. For the rest of the blue points, we can draw parallel lines as mentioned before to split them from the red points. We end up with \$v\$ lines.\n\nIf the set \$S\$ has no blue points, there can be a line that divides the plane into two regions: one consisting of two red points, and one consisting of the rest. For the rest of the red points, we can draw parallel lines as mentioned before to split them from the blue points. We end up with \$u-1 = v\$ lines.\n\nNow we will show that there are configurations that can not be partitioned with less than \$v\$ lines.\n\nConsider the arrangement of these points on a circle so that between every two blue points there are at least one red point (on the circle).\n\nThere are no less than \$2v\$ arcs of this circle, that has one end blue and other red (and no other colored points inside the arc) - one such arc on each side of each blue point. For a line partitioning to be good, each of these arcs have to be crossed by at least one line, but one line can not cross more than \$2\$ arcs on a circle - therefore, this configuration can not be partitioned with less than \$v\$ lines!\n\nOur proof is done, and we have our final answer: \$2013\$.", "I claim that the answer is \$2013\$.\n\nI will first make some definitions:\n\nCall a tube around two points to be the set of two lines parallel to the line connecting the two points that are arbitrarily close to the two points.\n\nDefine a convex hull of a set of points to be the convex polygon that has all of the points in the set either inside or on the boundary of the polygon.\n\nWe will first prove that we can get at most \$2013\$ lines, then we will prove that we can get at least \$2013\$, proving that we must have \$2013\$ lines.\n\nConsider the convex hull of the Colombian configuration. If there is a red point on it, then cut it off from the rest of the set with a line. This leaves 2012 red points left. We can pair these 2012 red points into 1006 pairs. Then, put tubes around each pair, which leaves us with 1006 tubes, or 2012 lines. Adding the line that separated the first red point, we get 2013 lines.\n\nNow, consider a circle. Evenly space 4027 points on the circle, and label them 1 to 4027. Now, color all the points labeled with an even number red and the rest blue. Note that each line connecting a red point and a blue point must intersect one of the lines forming the good arrangement. Also note that each line forming the good arrangement can only form at intersect 2 of the lines connecting the points which form the convex hull. Therefore, since we have 4026 such lines connecting the points, we get at least \$2013\$ lines, and we are done.\n\n-Yiyj1" ]
IMO-2013-3	https://artofproblemsolving.com/wiki/index.php/2013_IMO_Problems/Problem_3	Let the excircle of triangle $ABC$ opposite the vertex $A$ be tangent to the side $BC$ at the point $A_1$. Define the points $B_1$ on $CA$ and $C_1$ on $AB$ analogously, using the excircles opposite $B$ and $C$, respectively. Suppose that the circumcentre of triangle $A_1B_1C_1$ lies on the circumcircle of triangle $ABC$. Prove that triangle $ABC$ is right-angled. Proposed by Alexander A. Polyansky, Russia	[ "\\[\n[asy] unitsize(2.5cm); void b() { pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10); pen dd=linetype(\"4 8\"); /* Define the excenter / pair excenter(pair A=(0,0), pair B=(0,0), pair C=(0,0)) { return extension(A,bisectorpoint(C,A,B),B,rotate(90,B)bisectorpoint(A,B,C)); } /* Draw points / pair A=D(\"A\",dir(115),N), B=D(\"B\",(-1,0),W), C=D(\"C\",(1,0),E), I=D(incenter(A,B,C)), Ia=D(\"I_a\",excenter(A,B,C),S), Ib=D(\"I_b\",excenter(B,C,A),E), Ic=D(\"I_c\",excenter(C,A,B),W), A1=D(\"A_1\",foot(Ia,B,C),SSW), B1=D(\"B_1\",foot(Ib,C,A),E), C1=D(\"C_1\",foot(Ic,A,B),N), Ma=D(\"M_a\",midpoint(Ib--Ic),N), Mb=D(\"M_b\",midpoint(Ic--Ia),SW), Mc=D(\"M_c\",midpoint(Ia--Ib),SE), V=D(\"V\",circumcenter(Ia,Ib,Ic),SE); / Draw paths */ D(unitcircle,heavyblue); D(circumcircle(B,C,V),linetype(\"2 2\")+rgb(0.6,0,1)); D(circumcircle(B,C1,V),linetype(\"2 2\")+rgb(0.6,0,1)); D(A--B--C--cycle); D(Ia--Ib--Ic--cycle,gray(0.3)+linewidth(1)); D(A1--B1--C1--cycle); D(I--Ia,dd+red); D(I--Ib,dd+red); D(I--Ic,dd+red); D(V--Ia,dd+heavygreen); D(V--Ib,dd+heavygreen); D(V--Ic,dd+heavygreen); } b(); pathflag=false; b(); [/asy]\n\\]\n\nLet the excenters opposite \$A,B,C\$ be \$I_a,I_b,I_c\$. Let the midpoint of \$\\overline{I_bI_c}\$ be \$M_a\$, which lies on \$(ABC)\$, the nine-point circle of \$\\triangle I_aI_bI_c\$; analogously define \$M_b,M_c\$.\n\n\$M_aB=M_aC\$ and \$BC_1=s-a=B_1C\$, so \$\\triangle M_aBC_1\\cong\\triangle M_aCB_1\$ (SAS), thus \$M_a\$ is equidistant from \$B_1,C_1\$, with analogous results for \$M_b,M_c\$. It follows that the circumcentre of \$\\triangle A_1B_1C_1\$ is one of \$M_a,M_b,M_c\$; WLOG, suppose it is \$M_a\$.\n\nBy isogonal conjugacy, \$I_aA_1,I_bB_1I_cC_1\$ concur at the Bevan point \$V\$ of \$\\triangle ABC\$. \$M_aM_b\$ is the common perpendicular bisector of \$\\overline{C_1A_1}\$ and \$\\overline{I_cC}\$, so \$C_1A_1\\parallel I_cC\$. \$(A_1C_1M_b)\$ is the circle on diameter \$\\overline{VB}\$, so by Reim's theorem, \$V \\in (I_bI_cBC)\$.\n\nHence \$\\angle I_cI_aI_b=\\tfrac{1}{2}\\angle I_cVI_b=45^{\\circ}\\implies\\angle CAB=180^{\\circ}-2\\angle I_cI_aI_b=90^{\\circ}\$, as required." ]
IMO-2013-4	https://artofproblemsolving.com/wiki/index.php/2013_IMO_Problems/Problem_4	Let $ABC$ be an acute triangle with orthocenter $H$, and let $W$ be a point on the side $BC$, lying strictly between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$, respectively. Denote by $\omega_1$ is the circumcircle of $BWN$, and let $X$ be the point on $\omega_1$ such that $WX$ is a diameter of $\omega_1$. Analogously, denote by $\omega_2$ the circumcircle of triangle $CWM$, and let $Y$ be the point such that $WY$ is a diameter of $\omega_2$. Prove that $X, Y$ and $H$ are collinear.	[ "\\[\n[asy] //Original diagram by suli, August 2014. Feel free to make edits, but please leave this comment in place. import olympiad; import math; unitsize(10); pair A = (15,25), B = (0,0), C = (20,0), W = (12,0); pair H = orthocenter(A, B, C); pair N = extension(C,H, A,B); pair M = extension(B,H, A,C); pair L = extension(A,H, B,C); path p1 = circumcircle(N, B, W); path p2 = circumcircle(C, M, W); pair X = intersectionpoints(B--(0,10), p1)[1]; pair Y = intersectionpoints(C--(20,10), p2)[1]; pair T = intersectionpoints(p1,p2)[0]; //Time to start drawing dot(A); dot(B); dot(C); dot(W); dot(H); dot(M); dot(L); dot(X); dot(N); dot(Y); dot(T); dot(circumcenter(N,B,W)); dot(circumcenter(C,M,W)); draw(p1); draw(p2); draw(A--B--C--Y--W--X--B); draw(C--A); draw(A--L); draw(B--M); draw(C--N); label(\"A\", A, E); label(\"B\", B, S); label(\"C\", C, E); label(\"L\", L, S); label(\"M\", M, S); label(\"N\", N, S); label(\"H\", H, E); label(\"T\", T, E); label(\"W\", W, S); label(\"X\", X, E); label(\"Y\", Y, E); [/asy]\n\\]\n\nLet \$T\$ be the intersection of \$\\omega_1\$ and \$\\omega_2\$ other than \$W\$.\n\nLemma 1: \$T\$ is on \$XY\$.\n\nProof: We have \$\\angle{XTW} = \\angle{YTW} = 90^\\circ\$ because they intercept semicircles. Hence, \$\\angle{XTY} = \\angle{XTW} + \\angle{YTW} = 180^\\circ\$, so \$XTY\$ is a straight line.\n\nLemma 2: \$T\$ is on \$AW\$.\n\nProof: Let the circumcircles of \$NBW\$ and \$MWC\$ be \$\\omega_1\$ and \$\\omega_2\$, respectively, and, as \$BNMC\$ is cyclic (from congruent \$\\angle{BNC} = \\angle{BMC} = 90^\\circ\$), let its circumcircle be \$\\omega_3\$. Then each pair of circles' radical axises, \$BN, TW,\$ and \$MC\$, must concur at the intersection of \$BN\$ and \$MC\$, which is \$A\$.\n\nLemma 3: \$YT\$ is perpendicular to \$AW\$.\n\nProof: This is immediate from \$\\angle{YTW} = 90^\\circ\$.\n\nLet \$AH\$ meet \$BC\$ at \$L\$, which is also the foot of the altitude to that side. Hence, \$\\angle{ALB} = 90^\\circ.\$\n\nLemma 4: Quadrilateral \$THLW\$ is cyclic.\n\nProof: We know that \$NHLB\$ is cyclic because \$\\angle{BNH}\$ and \$\\angle{BLH}\$, opposite and right angles, sum to \$180^\\circ\$. Furthermore, we are given that \$NTWB\$ is cyclic. Hence, by Power of a Point,\n\n\\[\nAT * AW = AN * AB = AH * AL.\n\\]\n\nThe converse of Power of a Point then proves \$THLW\$ cyclic.\n\nHence, \$\\angle{WTH} = 180^\\circ - \\angle{WLH} = 90^\\circ\$, and so \$HT\$ is perpendicular to \$AW\$ as well. Combining this with Lemma 3's statement, we deduce that \$T, H, Y\$ are collinear. But, as \$X\$ is on \$YT\$ (from Lemma 1), \$X, Y, H\$ are collinear. This completes the proof.\n\n\\[\n\\blacksquare\n\\]\n\n--Suli 13:51, 25 August 2014 (EDT)", "Probably a simpler solution than above.\n\nAs above, let \$T = \\omega_1 \\cap \\omega_2 \\neq W.\$ By Miquel \$MTHN\$ is cyclic. Then since \$\\angle WCY = \\angle WBX = 90^{\\circ}\$ we know, because \$W,B,X,T \\in \\omega_1\$ and \$W,C,Y,T \\in \\omega_2,\$ that \$\\angle WTY = \\angle WTX = 90^{\\circ},\$ thus \$X,T,Y\$ are collinear. There are a few ways to finish.\n\n(a)\n\n\\[\nBX \\perp BC \\perp AH \\iff \\angle NBX = \\angle NAH\n\\]\n\n\\[\n\\iff \\angle NTX = \\angle NTH \\iff H \\in TX\n\\]\n\nso \$H,X,Y\$ are collinear, as desired \$\\square\$\n\n(b) Since \$BNMC\$ is cyclic we know \$AN \\cdot AB = AM \\cdot AC\$ which means \$p(A,\\omega_1) = p(A, \\omega_2)\$ so \$A\$ is on the radical axis, \$TW,\$ hence\n\n\\[\n\\angle ATX = \\angle XBW = 90^{\\circ} = \\angle AMH = \\angle ATH\n\\]\n\nso \$H\$ lies on this line as well and we may conclude \$\\square\$\n\n--mathguy623 03:10, 12 August 2016 (EDT)", "\\[\n[asy] unitsize(0.8cm); draw((0,0)--(14,0)--cycle); draw((0,0)--(5,12)--cycle); draw((5,12)--(14,0)--cycle); draw((2.071,4.97)--(14,0)--cycle); draw((8.96,6.72)--(0,0)--cycle); draw((2.071,4.97)--(9,0)--cycle); draw((8.96,6.72)--(9,0)--cycle); draw((0,0)--(0,2.083)--cycle); draw((14,0)--(14,6.75)--cycle); draw((0,2.083)--(14,6.75)--cycle); draw((0,2.083)--(2.071,4.97)--cycle); draw((0,2.083)--(9,0)--cycle); draw((8.96,6.72)--(14,6.75)--cycle); draw((14,6.75)--(9,0)--cycle); draw(circle((4.5,1.0415),4.61895)); draw(circle((11.5,3.375),4.20007)); label((0,0),\"$C$\",SW); label((14,0),\"$B$\",SE); label((5,12),\"$A$\",N); label((8.96,7),\"$N$\",N); label((2.071,4.97),\"$M$\",NW); label((8.25,0),\"$W$\",S); label((5,4),\"$H$\",N); label((0,2.083),\"$Y$\",NW); label((14,6.75),\"$X$\",NE); [/asy]\n\\]\n\nFor any point \$P,\$ let \$p\$ denote the complex number for point \$P.\$ First off, let \$C\$ be the origin. Now, since \$WY\$ is the diameter of the circumcircle of triangle \$CMW\$, we must have \$\\angle{YMW}=\\angle{YCH}=90^{\\circ}.\$ Since angles inscribed in the same arc are congruent, \$\\angle{MYW}=\\angle{MCW}=\\angle{C}.\$ This means that \$\\frac{\|w-m\|}{\|y-m\|}=\\tan{\\angle{C}}.\$ Combining this with the fact that \$\\angle{YMW}\$ is right, we find that\n\n\\[\ni(y-m)\\tan{\\angle{C}}=(w-m).\n\\]\n\nSolving, we find that\n\n\\[\ny=\\frac{w+(i\\tan{\\angle{C}}-1)m}{i\\tan{\\angle{C}}}.\n\\]\n\nWe wish to simplify \$(i\\tan{\\angle{C}}-1)m\$ first. Note that\n\n\\[\nm=\\frac{\|CM\|}{\|CA\|}\\cdot(a)=\\frac{(\|BC\|)\\cos{\\angle{C}}}{\|CA\|}\\cdot(\|CA\|)(\\cos{\\angle{C}}+i\\sin{\\angle{C}})=(\|BC\|)\\cos{\\angle{C}}(\\cos{\\angle{C}}+i\\sin{\\angle{C}}).\n\\]\n\nThis means that\n\n\\[\n\\begin{align} (i\\tan{\\angle{C}}-1)m&=(i\\tan{\\angle{C}}-1)((\|BC\|)\\cos{\\angle{C}}(\\cos{\\angle{C}}+i\\sin{\\angle{C}}))\\\\ &=\\left(\\frac{i\\sin{\\angle{C}}}{\\cos{\\angle{C}}}-1\\right)((\|BC\|)\\cos{\\angle{C}}(\\cos{\\angle{C}}+i\\sin{\\angle{C}}))\\\\ &=(i\\sin{\\angle{C}}-\\cos{\\angle{C}})(\|BC\|)(\\cos{\\angle{C}}+i\\sin{\\angle{C}})\\\\ &=-\|BC\| \\end{align}\n\\]\n\nThis means that\n\n\\[\ny=\\frac{w-\|BC\|}{i\\tan{\\angle{C}}}=\\frac{\|BC\|-w}{\\tan{\\angle{C}}}i.\n\\]\n\nNow, since \$WX\$ is a diameter of the circumcircle of triangle \$NBW,\$ we must have \$\\angle{WNX}=\\angle{WBX}=90^{\\circ}.\$ Since angles inscribed in the same arc are congruent, \$\\angle{NXW}=\\angle{NBW}=\\angle{B}.\$ This means that \$\\frac{\|x-n\|}{\|w-n\|}=\\cot{\\angle{B}}.\$ Combining this with the fact that \$\\angle{WNX}\$ is right, we find that\n\n\\[\n(x-n)=i\\cot{\\angle{B}}(w-n).\n\\]\n\nSolving, we find that\n\n\\[\nx=wi\\cot{\\angle{B}}+n(1-i\\cot{\\angle{B}}).\n\\]\n\nWe wish to simplify \$n(1-i\\cot{\\angle{B}})\$ first. Note that\n\n\\[\nn=e^{(90-\\angle{B})i}\|CN\|=(\\cos{(90-\\angle{B})}+i\\sin{(90-\\angle{B})})(\|BC\|)\\sin{\\angle{B}}=(\|BC\|)\\sin{\\angle{B}}(\\sin{\\angle{B}}+i\\cos{\\angle{B}})).\n\\]\n\nThis means that\n\n\\[\n\\begin{align} n(1-i\\cot{\\angle{B}})&=(\|BC\|)\\sin{\\angle{B}}(\\sin{\\angle{B}}+i\\cos{\\angle{B}}))(1-i\\cot{\\angle{B}})\\\\ &=(\|BC\|)\\sin{\\angle{B}}(\\sin{\\angle{B}}+i\\cos{\\angle{B}}))\\left(1-\\frac{\\cos{\\angle{B}}}{\\sin{\\angle{B}}}i\\right)\\\\ &=(\|BC\|)(\\sin{\\angle{B}}+i\\cos{\\angle{B}})(\\sin{\\angle{B}}-i\\cos{\\angle{B}})\\\\ &=(\|BC\|)(1) \\end{align}\n\\]\n\nThis means that\n\n\\[\nx=\|BC\|+wi\\cot{\\angle{B}}=\|BC\|+\\frac{w}{\\tan{\\angle{B}}}i.\n\\]\n\nThis means that the line through complex numbers \$x\$ and \$y\$ satisfy the equation\n\n\\[\n\\Im{(z)}=\\left(\\frac{\\frac{w}{\\tan{\\angle{B}}}-\\frac{\|BC\|-w}{\\tan{\\angle{C}}}}{\|BC\|}\\right)\\Re{(z)}+\\frac{\|BC\|-w}{\\tan{\\angle{C}}}.\n\\]\n\nIf there is a fixed point to the line, then the real and imaginary values of the point must not contin \$w\$. If the fixed point is \$c,\$ then we have\n\n\\[\n\\left(\\frac{\\frac{1}{\\tan{\\angle{B}}}+\\frac{1}{\\tan{\\angle{C}}}}{\|BC\|}\\right)\\Re{(c)}-\\frac{1}{\\tan{\\angle{C}}}=0,\n\\]\n\nafter comparing the coefficient of \$w.\$ This means that\n\n\\[\n\\begin{align} \\Re{(c)}&=\\frac{\\frac{1}{\\tan{\\angle{C}}}}{\\frac{\\frac{1}{\\tan{\\angle{B}}}+\\frac{1}{\\tan{\\angle{C}}}}{\|BC\|}}=\\frac{\\frac{\\cos{\\angle{C}}}{\\sin{\\angle{C}}}}{\\frac{\\frac{\\cos{\\angle{B}}}{\\sin{\\angle{B}}}+\\frac{\\cos{\\angle{C}}}{\\sin{\\angle{C}}}}{\|BC\|}}\\\\ &=\\frac{\\frac{\\cos{\\angle{C}}}{\\sin{\\angle{C}}}}{\\frac{\\frac{\\cos{\\angle{B}}\\sin{\\angle{C}}+\\sin{\\angle{B}}\\cos{\\angle{C}}}{\\sin{\\angle{B}}\\sin{\\angle{C}}}}{\|BC\|}}=\\frac{\\frac{\\cos{\\angle{C}}}{\\sin{\\angle{C}}}}{\\frac{\\sin{(\\angle{B}+\\angle{C})}}{\|BC\|\\sin{\\angle{B}}\\sin{\\angle{C}}}}\\\\ &=\\frac{\\frac{\\cos{\\angle{C}}}{\\sin{\\angle{C}}}}{\\frac{\\sin{(180-\\angle{A})}}{\|BC\|\\sin{\\angle{B}\\sin{\\angle{C}}}}}=\\frac{\|BC\|\\cos{\\angle{C}}\\sin{\\angle{B}}\\sin{\\angle{C}}}{\\sin{\\angle{A}\\sin{\\angle{C}}}}\\\\ &=\\frac{\|BC\|}{\\sin{\\angle{A}}}\\cdot(\\cos{\\angle{C}}\\sin{\\angle{B}}). \\end{align}\n\\]\n\nFrom the Law of Sines, we find that\n\n\\[\n\\frac{\|BC\|}{\\sin{\\angle{A}}}=\\frac{\|AC\|}{\\sin{\\angle{B}}}.\n\\]\n\nSubstituting this, we find that\n\n\\[\n\\begin{align} \\Re{(c)}&=\\frac{\|BC\|}{\\sin{\\angle{A}}}\\cdot(\\cos{\\angle{C}}\\sin{\\angle{B}})\\\\ &=\\frac{\|AC\|}{\\sin{\\angle{B}}}\\cdot(\\cos{\\angle{C}}\\sin{\\angle{B}})\\\\ &=\|AC\|\\cos{\\angle{C}}. \\end{align}\n\\]\n\nThis means that\n\n\\[\n\\begin{align} \\Im{(c)}&=\\left(\\frac{\\frac{w}{\\tan{\\angle{B}}}-\\frac{\|BC\|-w}{\\tan{\\angle{C}}}}{\|BC\|}\\right)\|AC\|\\cos{\\angle{C}}+\\frac{\|BC\|-w}{\\tan{\\angle{C}}}. \\end{align}\n\\]\n\nSince the \$w\$'s cancel out, we can just discard everything with \$w\$ in it. Thus,\n\n\\[\n\\begin{align} \\Im{(c)}&=\\left(\\frac{-\\frac{\|BC\|}{\\tan{\\angle{C}}}}{\|BC\|}\\right)\|AC\|\\cos{\\angle{C}}+\\frac{\|BC\|}{\\tan{\\angle{C}}}\\\\ &=\\left(-\\frac{1}{\\tan{\\angle{C}}}\\right)\|AC\|\\cos{\\angle{C}}+\\frac{\|BC\|}{\\tan{\\angle{C}}}\\\\ &=\\frac{1}{\\tan{\\angle{C}}}\\left(\|BC\|-\|AC\|\\cos{\\angle{C}}\\right). \\end{align}\n\\]\n\nSince\n\n\\[\n\|BC\|=\|AC\|\\cos{\\angle{C}}+\|AB\|\\cos{\\angle{B}},\n\\]\n\nwe have\n\n\\[\n\|BC\|-\|AC\|\\cos{\\angle{C}}=\|AB\|\\cos{\\angle{B}}.\n\\]\n\nThus,\n\n\\[\n\\Im{(c)}=\\frac{1}{\\tan{\\angle{C}}}\\cdot\|AB\|\\cos{\\angle{B}}.\n\\]\n\nIn conclusion,\n\n\\[\nc=\|AC\|\\cos{\\angle{C}}+\\left(\\frac{1}{\\tan{\\angle{C}}}\\cdot\|AB\|\\cos{\\angle{B}}\\right)i,\n\\]\n\nwhich is the fixed point \$XY\$ always passes through. However, by inspection, \$c=H=\\text{the orthocenter}.\$ Therefore, we conclude that \$X,Y,H\$ are collinear for all acute triangles \$ABC.\$\n\n~pinkpig" ]
IMO-2013-5	https://artofproblemsolving.com/wiki/index.php/2013_IMO_Problems/Problem_5	Let $\mathbb Q_{>0}$ be the set of all positive rational numbers. Let $f:\mathbb Q_{>0}\to\mathbb R$ be a function satisfying the following three conditions: (i) for all $x,y\in\mathbb Q_{>0}$, we have $f(x)f(y)\geq f(xy)$; (ii) for all $x,y\in\mathbb Q_{>0}$, we have $f(x+y)\geq f(x)+f(y)$; (iii) there exists a rational number $a>1$ such that $f(a)=a$. Prove that $f(x)=x$ for all $x\in\mathbb Q_{>0}$.	[]
IMO-2013-6	https://artofproblemsolving.com/wiki/index.php/2013_IMO_Problems/Problem_6	Let $n \ge 3$ be an integer, and consider a circle with $n + 1$ equally spaced points marked on it. Consider all labellings of these points with the numbers $0, 1, ... , n$ such that each label is used exactly once; two such labellings are considered to be the same if one can be obtained from the other by a rotation of the circle. A labelling is called beautiful if, for any four labels $a < b < c < d$ with $a + d = b + c$, the chord joining the points labelled $a$ and $d$ does not intersect the chord joining the points labelled $b$ and $c$. Let $M$ be the number of beautiful labelings, and let N be the number of ordered pairs $(x, y)$ of positive integers such that $x + y \le n$ and $\gcd(x, y) = 1$. Prove that \[ M = N + 1. \]	[]
IMO-2014-1	https://artofproblemsolving.com/wiki/index.php/2014_IMO_Problems/Problem_1	Let $a_0<a_1<a_2<\cdots \quad$ be an infinite sequence of positive integers, Prove that there exists a unique integer $n\ge1$ such that \[ a_n<\frac{a_0+a_1+\cdots + a_n}{n}\le a_{n+1}. \]	[ "Define \$f(n) = a_0 + a_1 + \\dots + a_n - n a_{n+1}\$. (In particular, \$f(0) = a_0.\$) Notice that because \$a_{n+2} \\ge a_{n+1}\$, we have\n\n\\[\na_0 + a_1 + \\dots + a_n - n a_{n+1} > a_0 + a_1 + \\dots + a_n + a_{n+1} - (n+1) a_{n+2}.\n\\]\n\nThus, \$f(n) > f(n+1)\$; i.e., \$f\$ is monotonic decreasing. Therefore, because \$f(0) > 0\$, there exists a unique \$N\$ such that \$f(N-1) > 0 \\ge f(N)\$. In other words,\n\n\\[\na_0 + a_1 + \\dots + a_{N-1} - (N-1) a_N > 0\n\\]\n\n\\[\na_0 + a_1 + \\dots + a_N - n a_{N+1} \\le 0.\n\\]\n\nThis rearranges to give\n\n\\[\na_N < \\frac{a_0 + a_1 + \\dots + a_N}{N} \\le a_{N+1}.\n\\]\n\nDefine \$g(n) = a_0 + a_1 + \\dots + a_n - n a_n\$. Then because \$a_{n+1} > a_n\$, we have\n\n\\[\na_0 + a_1 + \\dots + a_n - n a_n > a_0 + a_1 + \\dots + a_n + a_{n+1} - (n+1) a_{n+1}.\n\\]\n\nTherefore, \$g\$ is also monotonic decreasing. Note that \$g(N+1) = a_0 + a_1 + \\dots + a_{N+1} - (N+1) a_{N+1} \\le 0\$ from our inequality, and so \$g(k) \\le 0\$ for all \$k > N\$. Thus, the given inequality, which requires that \$g(n) > 0\$, cannot be satisfied for \$n > N\$, and so \$N\$ is the unique solution to this inequality.\n\n--Suli 22:38, 7 February 2015 (EST)", "It is more convenient to work with differences \$d_i=a_i-a_{i-1}\$, \$i\\ge 1\$. \$d_i\\ge 1\$. Instead of using \$a_i=a_0+d_1+\\ldots+d_i\$ the inequalities can be rewritten in terms of \$d_i\$ as\n\n\\[\n0 < V_n \\le nd_{n+1}\n\\]\n\nwhere \$V_n=a_0-d_2-2d_3-3d_4-\\ldots - (n-1)d_n\$. \$V_n\$ is strictly monotonically decreasing. \$V_1=a_0 >0\$. That is the left inequality is satisfied for \$n=1\$. Lets take a look at the time step \$(n+1)\$ which is right after \$n\$: \$n \\rightarrow n+1\$\n\n\\[\n0< V_n-nd_{n+1}\\le (n+1)d_{n+2}\n\\]\n\nThe condition \$V_n \\le nd_{n+1}\$for breaking the left inequality at some step \$n+1\$ is exactly the condition for satisfying the right inequality at step \$n\$. Once left inequality is broken at step \$(n+1)\$ it will remain broken for future steps as \$V_n\$ is strictly decreasing. The right inequality will be satisfied for some \$n\$ as \$V_n\$ is strictly decreasing integer sequence and the right hand side \$nd_{n+1}\$ of the right inequality is bounded by \$1\$ from below. In summary, the left inequality is satisfied initially and as soon as the right inequality is satisfied, which will happen for some \$n\$, the left inequality will break at the very next step and will remain broken for all future steps. That is \$n\$ when both inequalities are satisfied exists and unique.\n\n--alexander_skabelin 9:24, 7 July 2023 (EST)" ]
IMO-2014-2	https://artofproblemsolving.com/wiki/index.php/2014_IMO_Problems/Problem_2	Let $n\ge2$ be an integer. Consider an $n\times n$ chessboard consisting of $n^2$ unit squares. A configuration of $n$ rooks on this board is $\textit{peaceful}$ if every row and every column contains exactly one rook. Find the greatest positive integer $k$ such that, for each peaceful configuration of $n$ rooks, there is a $k\times k$ square which does not contain a rook on any of its $k^2$ squares.	[ "We claim the answer is \$k = \\lceil \\sqrt{n}\\rceil - 1\$, where \$\\lceil n\\rceil\$ is the ceiling function of \$n\$; i.e., the least integer greater than or equal to \$n\$. Notice that \$\\lceil n\\rceil < n + 1\$.\n\nFirst, we shall show that each \$n \\times n\$ chessboard with a peaceful configuration of \$n\$ rooks contains a valid \$k \\times k\$ square. Consider firstly the rook \$R\$ on the top row of the board. Because \$k < n\$, there exists a set \$C\$ of \$k\$ consecutive columns, one of which contains rook \$R\$. Consider the \$n - k + 1\$ \$k \\times k\$ squares in \$C\$. Of them, only one contains the rook \$R\$ on one of its squares. Furthermore, each of the other \$k - 1\$ rooks in \$C\$ can only make up to \$k\$ of the \$k \\times k\$ squares have it on one of its squares. Therefore, because\n\n\\[\nn - k + 1 = n - \\lceil \\sqrt{n}\\rceil + 2 > n - \\sqrt{n} + 1 = \\sqrt{n} (\\sqrt{n} - 1) + 1 > (\\lceil \\sqrt{n}\\rceil - 1) (\\lceil \\sqrt{n}\\rceil - 2) + 1 = k(k - 1) + 1,\n\\]\n\nby the Pigeonhole Principle there exists a \$k \\times k\$ square in \$C\$ not covered by any rook in \$C\$. Clearly, that square cannot be covered by any rook outside of \$C\$, and so it is a valid choice.\n\nIt remains to show that there exists a chessboard without a \$(k+1) \\times (k+1)\$ square. Indeed, such a board exists. First, place a rook at the upper-left corner of the board. Next, place a rook 1 space to the right and \$(k+1)\$ spaces down of the first rook. Then, place a rook 1 space to the right and \$(k+1)\$ spaces down of the second rook, and so on, until the placement of a new rook will be under the lower boundary of the board. In that case, place a rook in the same unoccupied column and in the first unoccupied row, and continue placement of subsequent rooks. This arrangement of rooks is clearly peaceful, and because \$(k+1)^2 > n\$ it has the property that any \$(k+1) \\times (k+1)\$ square in the board will be occupied by at least one rook, completing the proof." ]
IMO-2014-3	https://artofproblemsolving.com/wiki/index.php/2014_IMO_Problems/Problem_3	Convex quadrilateral $ABCD$ has $\angle{ABC}=\angle{CDA}=90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ lies inside $\triangle{SCT}$ and \[ \angle{CHS}-\angle{CSB}=90^{\circ},\quad \angle{THC}-\angle{DTC} = 90^{\circ}. \] Prove that line $BD$ is tangent to the circumcircle of $\triangle{TSH}.$	[ "\\[\n[asy] import cse5; import graph; import olympiad; dotfactor = 3; unitsize(1.5inch); path circle = Circle(origin, 1); // draw(circle); pair A = (0,1), C=(0,-1); pair Oo = (0,-0.05); pair Bb = rotate(-8,Oo)(2,-0.05), Dd =rotate(-8, Oo)(-2,-0.05); pair B = IP(Dd--Bb, circle, 1); pair D = IP(Dd--Bb, circle, 0); pair H = foot(A, Dd, Bb); pair H1 = bisectorpoint(C,H); pair H2 = foot(H1,C,H); pair K = extension(A,B,H1,H2); pair L = extension(A,D,H1,H2); path circle3 = Circle(K,length(K-C)); path circle4 = Circle(L,length(L-C)); // draw(circle3, dashed); // draw(circle4, dashed); pair T = IP(D--A,circle4, 0); pair S = IP(A--B, circle3, 0); pair O = circumcenter(H,S,T); path circle2 = Circle(O, length(O-H)); pair Q = IP(A--D,circle2, 0); pair R = IP(B--A, circle2, 0); pair G = IP(A--H, circle2, 0); pair Q = rotate(180,D)C; pair P = rotate(180,B)C; path arc1 = arc(L,length(L-H), 110,0); path arc2 = arc(K, length(K-H), 225, 45); pair M = extension(C,D,L,K); dot(\"$C$\", C, SE); dot(\"$D$\", D, W); dot(\"$A$\", A, dir(40));dot(\"$B$\", B, E);dot(\"$H$\", H, SW); dot(\"$T$\", T, W); dot(\"$S$\", S, E); dot(\"$K$\",K,E); dot(\"$L$\",L,W); // dot(\"$R$\",R,NE); dot(\"$Q$\",Q,NW); dot(\"$G$\", G, SE); dot(\"$Q$\",Q,NW); dot(\"$P$\",P,NE); dot(\"$M$\",M,S); draw(arc1,dashed); draw(arc2, dashed); draw(circle2); draw(A--D--B--cycle); // draw(A--C, dashed); draw(A--H); draw(C--S--H--cycle); draw(C--T--H); draw(D--C--B); draw(D--L--K--B, dashed); draw(T--G, dashed); draw(K--H, dashed); draw(D--Q--P--B,dashed); draw(Q--H--P,dashed); draw(C--L--H, dashed); draw(C--K,dashed); draw(T--S,dashed); // draw(anglemark(H,S,K)); // draw(anglemark(K,H,S)); label(\"$x$\",C+(0,0.1),N); label(\"$y$\",C+(-0.1,0.3),N); label(\"$w$\",H+(-0.03,0.1), N); label(\"$z$\", H+(0.06,0.12), N ); label(\"$u$\",T+(-0.1,-0.2), S); label(\"$v$\", S+(0,-0.2), S); label(\"$t$\",T+(0.1,-0.3), S); label(\"$s$\", S+(-0.08,-0.2), S); [/asy]\n\\]\n\nDenote \$\\angle{HSB}=v\$, \$\\angle{HTD}=u\$, \$\\angle{HSC}=s\$, \$\\angle{HTC}=t\$, \$\\angle{HCS}=x\$, \$\\angle{HCT}=y\$, \$\\angle{AHS}=z\$, \$\\angle{AHT}=w\$. Since \$\\angle{CHS}-\\angle{CSB} =90\$ and \$\\angle{CHT}-\\angle{CTD}=90\$, we have \$\\angle{CSA}=x+90\$, \$\\angle{CTA}=y+90\$.\n\nSince \$\\angle{CHS} = \\angle{CSB}+90\$, the tangent of the circumcircle of \$\\triangle{CSH}\$ at point \$S\$ is perpendicular to \$SB\$; therefore, the circumcenter of \$\\triangle{CSH}\$ (point \$K\$) is on \$AB\$. Similarly, the circumcenter of \$\\triangle{CTH}\$ (point \$L\$) is on \$AD\$. In addition, \$KL\$ is the perpendicular bisector of \$CH\$.\n\nExtend \$CB\$ to meet circumcircle of \$\\triangle{CSH}\$ at \$P\$, and extend \$CD\$ to meet circumcircle of \$\\triangle{CTH}\$ at \$Q\$. Then, since \$\\angle{ADC}=\\angle{ABC}=90\$, \$AD\$ and \$AB\$ are the perpendicular bisector of \$CQ\$ and \$CP\$, respectively; hence \$A\$ is the circumcenter of \$\\triangle{PCQ}\$. Since \$B\$ and \$D\$ are midpoints on \$CP\$ and \$CQ\$, \$PQ \\parallel BD\$; also, \$AH \\perp BD\$, so \$AH \\perp PQ\$. Since \$A\$ is the circumcenter, \$AH\$ is also the perpendicular bisector of \$PQ\$. Hence,\n\n\\[\nHP=HQ\n\\]\n\nWe have\n\n\\[\nu=\\angle{CHT}-90=(90-w+\\angle{DHC}) - 90\n\\]\n\n\\[\nv=\\angle{CHS}-90 = (90-z+\\angle{BHC}) - 90\n\\]\n\nHence, \$u+v = -w-z+\\angle{DHC}+\\angle{BHC} = 180 -(w+z)\$, or\n\n\\[\nw+z=180-u-v\n\\]\n\nSince quadrilaterals \$QTHC\$ and \$PSHC\$ are cyclic, we have \$\\angle{THQ}=\\angle{TCQ}=90-u\$, \$\\angle{SHP}=\\angle{SCP}=90-v\$; so,\n\n\\[\n\\angle{PHQ}=\\angle{THQ}+\\angle{SHP}+w+z=90-u+90-v+w+z = 2(w+z)\n\\]\n\n\\[\nw+z=\\frac{1}{2}\\angle{PHQ}=\\angle{AHQ}=w+\\angle{THQ}=w+\\angle{TCQ}\n\\]\n\nHence,\n\n\\[\n\\angle{TCQ}=z \\qquad \\qquad\n\\]\n\nSimilarly,\n\n\\[\n\\angle{SCP}=w\n\\]\n\nNow we apply law of Sines repeatedly on pairs of triangles. For \$\\triangle{QCH}\$ and \$\\triangle{PCH}\$, \$\\angle{HQC}=\\angle{HTC}=t\$, \$\\angle{HCQ}=y+z\$, \$\\angle{HPC}=s\$, \$\\angle{HCP}=x+w\$; hence,\n\n\\[\n\\frac{\\sin{t}}{\\sin{(y+z)}}=\\frac{HC}{HQ}=\\frac{HC}{HP}=\\frac{\\sin{s}}{\\sin{(x+w)}} \\qquad \\qquad (1)\n\\]\n\nFor \$\\triangle{LHK}\$, \$\\angle{HLK}=\\frac{1}{2}\\angle{HLC}=t\$, \$\\angle{HKL}=\\frac{1}{2}\\angle{HKC}=s\$; hence,\n\n\\[\n\\frac{\\sin{s}}{\\sin{t}}=\\frac{LH}{KH}=\\frac{LT}{KS} \\qquad \\qquad (2)\n\\]\n\nFor \$\\triangle{LAK}\$, \$\\angle{ALK}=90-\\angle{DML}=90-\\angle{CMK}=\\angle{MCH}=y+z\$, and similarly, \$\\angle{AKL}=w+x\$; hence,\n\n\\[\n\\frac{\\sin{(w+x)}}{\\sin{(y+z)}}=\\frac{AL}{AK} \\qquad \\qquad (3)\n\\]\n\nCombining \$(1), (2), (3)\$, we have\n\n\\[\n\\frac{AL}{AK}=\\frac{LT}{KS}=\\frac{AL-AT}{AK-AS}=\\frac{AT}{AS}\n\\]\n\nTherefore, \$TS \\parallel KL\$, and \$\\angle{ATS} = \\angle{ALK}=y+z\$. Let the circumcircle of \$\\triangle{THS}\$ meets \$AH\$ at \$G\$. We have,\n\n\\[\n\\angle{ATG}=\\angle{ATS}-\\angle{GTS}=(y+z)-\\angle{AHS}= y\n\\]\n\nAnd,\n\n\\[\n\\angle{GTH}=\\angle{ATH}-\\angle{ATG}=(90+y) - y = 90\n\\]\n\nThis proves \$HG\$ is the diameter of the circle and the center of the circle is on AH. \$\\square\$\n\nSolution by \$Mathdummy\$." ]
IMO-2014-4	https://artofproblemsolving.com/wiki/index.php/2014_IMO_Problems/Problem_4	Points $P$ and $Q$ lie on side $BC$ of acute-angled $\triangle{ABC}$ so that $\angle{PAB}=\angle{BCA}$ and $\angle{CAQ}=\angle{ABC}$. Points $M$ and $N$ lie on lines $AP$ and $AQ$, respectively, such that $P$ is the midpoint of $AM$, and $Q$ is the midpoint of $AN$. Prove that lines $BM$ and $CN$ intersect on the circumcircle of $\triangle{ABC}$.	[ "\\[\n[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra / import graph; size(10.60000000000002cm); real labelscalefactor = 0.5; / changes label-to-point distance / pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); / default pen style / pen dotstyle = black; / point style / real xmin = -4.740000000000007, xmax = 16.46000000000002, ymin = -7.520000000000004, ymax = 4.140000000000004; / image dimensions / pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); pen qqwuqq = rgb(0.000000000000000,0.3921568627450985,0.000000000000000); draw((1.800000000000002,3.640000000000004)--(-0.2200000000000002,-1.200000000000001)--(7.660000000000009,-1.140000000000001)--cycle, zzttqq); draw(arc((7.660000000000009,-1.140000000000001),0.6000000000000009,140.7958863822920,180.4362538499006)--(7.660000000000009,-1.140000000000001)--cycle, red); draw(arc((-0.2200000000000002,-1.200000000000001),0.6000000000000009,0.4362538499004549,67.34652063545073)--(-0.2200000000000002,-1.200000000000001)--cycle, qqwuqq); draw(arc((1.800000000000002,3.640000000000004),0.6000000000000009,-106.1141136177082,-39.20411361770813)--(1.800000000000002,3.640000000000004)--cycle, qqwuqq); draw(arc((1.800000000000002,3.640000000000004),0.6000000000000009,-112.6534793645495,-73.01347936454944)--(1.800000000000002,3.640000000000004)--cycle, red); / draw figures / draw((1.800000000000002,3.640000000000004)--(-0.2200000000000002,-1.200000000000001), zzttqq); draw((-0.2200000000000002,-1.200000000000001)--(7.660000000000009,-1.140000000000001), zzttqq); draw((7.660000000000009,-1.140000000000001)--(1.800000000000002,3.640000000000004), zzttqq); draw(arc((7.660000000000009,-1.140000000000001),0.6000000000000009,140.7958863822920,180.4362538499006), red); draw(arc((7.660000000000009,-1.140000000000001),0.5000000000000008,140.7958863822920,180.4362538499006), red); draw(arc((-0.2200000000000002,-1.200000000000001),0.6000000000000009,0.4362538499004549,67.34652063545073), qqwuqq); draw(arc((-0.2200000000000002,-1.200000000000001),0.5000000000000008,0.4362538499004549,67.34652063545073), qqwuqq); draw(arc((-0.2200000000000002,-1.200000000000001),0.4000000000000006,0.4362538499004549,67.34652063545073), qqwuqq); draw(arc((1.800000000000002,3.640000000000004),0.6000000000000009,-106.1141136177082,-39.20411361770813), qqwuqq); draw(arc((1.800000000000002,3.640000000000004),0.5000000000000008,-106.1141136177082,-39.20411361770813), qqwuqq); draw(arc((1.800000000000002,3.640000000000004),0.4000000000000006,-106.1141136177082,-39.20411361770813), qqwuqq); draw(arc((1.800000000000002,3.640000000000004),0.6000000000000009,-112.6534793645495,-73.01347936454944), red); draw(arc((1.800000000000002,3.640000000000004),0.5000000000000008,-112.6534793645495,-73.01347936454944), red); draw((-1.022670636276736,-6.130338243877306)--(7.660000000000009,-1.140000000000001)); draw((4.740746205921980,-5.986847110107199)--(-0.2200000000000002,-1.200000000000001)); draw((1.800000000000002,3.640000000000004)--(4.740746205921980,-5.986847110107199)); draw((1.800000000000002,3.640000000000004)--(-1.022670636276736,-6.130338243877306)); draw(circle((3.711084749329270,0.0008695880898521494), 4.110415438128883)); / dots and labels / dot((1.800000000000002,3.640000000000004),dotstyle); label(\"$A$\", (1.880000000000002,3.760000000000004), NE labelscalefactor); dot((-0.2200000000000002,-1.200000000000001),dotstyle); label(\"$B$\", (-0.1400000000000008,-1.080000000000000), NE * labelscalefactor); dot((7.660000000000009,-1.140000000000001),dotstyle); label(\"$C$\", (7.740000000000012,-1.020000000000000), NE * labelscalefactor); dot((0.3886646818616330,-1.245169121938651),dotstyle); label(\"$Q$\", (0.4600000000000001,-1.120000000000000), NE * labelscalefactor); dot((3.270373102960991,-1.173423555053598),dotstyle); label(\"$P$\", (3.360000000000004,-1.060000000000000), NE * labelscalefactor); dot((4.740746205921980,-5.986847110107199),dotstyle); label(\"$M$\", (4.820000000000006,-5.860000000000003), NE * labelscalefactor); dot((-1.022670636276736,-6.130338243877306),dotstyle); label(\"$N$\", (-0.9400000000000020,-6.020000000000004), NE * labelscalefactor); dot((2.709057008802497,-3.985539257126989),dotstyle); label(\"$D$\", (2.780000000000003,-3.860000000000002), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture / [/asy]\n\\]\n\nWe are trying to prove that the intersection of \$BM\$ and \$CN\$, call it point \$D\$, is on the circumcircle of triangle \$ABC\$. In other words, we are trying to prove \$\\angle {BDC} + \\angle {BAC} = 180\$. Let the intersection of \$BM\$ and \$AN\$ be point \$E\$, and the intersection of \$AM\$ and \$CN\$ be point \$F\$. Let us assume \$\\angle {BDC} + \\angle {BAC} = 180\$. Note: This is circular reasoning. If \$\\angle {BDC} + \\angle {BAC} = 180\$, then \$\\angle {BAC}\$ should be equal to \$\\angle {BDN}\$ and \$\\angle {CDM}\$. We can quickly prove that the triangles \$ABC\$, \$APB\$, and \$AQC\$ are similar, so \$\\angle {BAC} = \\angle {AQC} = \\angle {APB}\$. We also see that \$\\angle {AQC} = \\angle {BQN} = \\angle {APB} = \\angle {CPF}\$. Also because angles \$BEQ\$ and \$NED, MFD\$ and \$CFP\$ are equal, the triangles \$BEQ\$ and \$NED\$, \$MDF\$ and \$FCP\$ must be two pairs of similar triangles. Therefore we must prove angles \$CBM\$ and \$ANC, AMB\$ and \$BCN\$ are equal. We have angles \$BQA = APC = NQC = BPM\$. We also have \$AQ = QN\$, \$AP = PM\$. Because the triangles \$ABP\$ and \$ACQ\$ are similar, we have \$\\dfrac {EC}{EN} = \\dfrac {BF}{FM}\$, so triangles \$BFM\$ and \$NEC\$ are similar. So the angles \$CBM\$ and \$ANC, BCN\$ and \$AMB\$ are equal and we are done.", "Let \$L\$ be the midpoint of \$BC\$. Easy angle chasing gives \$\\angle{AQP} = \\angle{APQ} = \\angle{BAC}\$. Because \$P\$ is the midpoint of \$AM\$, the cotangent rule applied on triangle \$MBA\$ gives us\n\n\\[\n\\cot \\angle{MBC} - \\cot \\angle{ABC} = 2\\cot \\angle{BAC}.\n\\]\n\nHence, by the cotangent rule on \$ABC\$, we have\n\n\\[\n\\cot \\angle{BAL} = 2\\cot \\angle{BAC} + \\cot \\angle{ABC} = \\cot \\angle{MBC}.\n\\]\n\nBecause the period of cotangent is \$180^\\circ\$, but angles are less than \$180^\\circ\$, we have \$\\angle{BAL} = \\angle{MBC}.\$\n\nSimilarly, we have \$\\angle{LAC} = \\angle{NCB}.\$ Hence, if \$BM\$ and \$CN\$ intersect at \$Z\$, then \$\\angle{BZC} = 180^\\circ - \\angle{BAC}\$ by the Angle Sum in a Triangle Theorem. Hence, \$BACZ\$ is cyclic, which is equivalent to the desired result.\n\n--Suli 23:27, 7 February 2015 (EST)", "Let \$L\$ be the midpoint of \$BC\$. By AA Similarity, triangles \$BAP\$ and \$BCA\$ are similar, so \$\\dfrac{BA}{AP} = \\dfrac{BC}{CA}\$ and \$\\angle{BPA} = \\angle{BAC}\$. Similarly, \$\\angle{CQA} = \\angle{BAC}\$, and so triangle \$AQP\$ is isosceles. Thus, \$AQ = AP\$, and so \$\\dfrac{BA}{AQ} = \\dfrac{BC}{CA}\$. Dividing both sides by 2, we have \$\\dfrac{BA}{AN} = \\dfrac{BL}{AC}\$, or\n\n\\[\n\\frac{BA}{BL} = \\frac{AN}{AC}.\n\\]\n\nBut we also have \$\\angle{ABL} = \\angle{CAQ}\$, so triangles \$ABL\$ and \$NAC\$ are similar by \$SAS\$ similarity. In particular, \$\\angle{ANC} = \\angle{BAL}\$. Similarly, \$\\angle{BMA} = \\angle{CAL}\$, so \$\\angle{ANC} + \\angle{BMA} = \\angle{BAC}\$. In addition, angle sum in triangle \$AQP\$ gives \$\\angle{QAP} = 180^\\circ - 2\\angle{A}\$. Therefore, if we let lines \$BM\$ and \$CN\$ intersect at \$T\$, by Angle Sum in quadrilateral \$AMTN\$ concave \$\\angle{NTM} = 180^\\circ + \\angle{A}\$, and so convex \$\\angle{BTC} = 180^\\circ - \\angle{A}\$, which is enough to prove that \$BACT\$ is cyclic. This completes the proof.\n\n\\[\n[asy] size(250); defaultpen(fontsize(8pt)); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair Pp = rotate(50, A)B; pair P = extension(A,Pp,B,C); pair Qp = rotate(-70, A)C; pair Q = extension(A,Qp,B,C); pair M = rotate(180,P)A; pair N = rotate(180,Q)*A; path c1 = circumcircle(A,B,C); pair T = IP(B--M,C--N); pair L = midpoint(B--C); draw(A--B--C--cycle^^B--Q--A--P^^Q--N--C^^P--M--B^^A--L); draw(c1); dot(\"$A$\", A, dir(100)); dot(\"$B$\", B, dir(-110)); dot(\"$C$\", C, dir(-40)); dot(\"$P$\", P, dir(50)); dot(\"$Q$\", Q, dir(-170)); dot(\"$M$\", M, dir(-50)); dot(\"$N$\", N, dir(-140)); dot(\"$T$\", T,dir(-90)); dot(\"$L$\", L, dir(-120)); [/asy]\n\\]\n\n--Suli 10:38, 8 February 2015 (EST)", "Let \$D_1\$ be the second intersection of \$NC\$ with the circumcircle of \$\\triangle ABC,\$ and \$D_2\$ the second intersection of \$MB\$ with the circumcircle of \$\\triangle ABC.\$ By inscribed angles, the tangent at \$C\$ is parallel to \$AN.\$ Let \$P_{\\infty}\$ denote the point at infinity along line \$AN.\$ Note that\n\n\\[\n(A,D_1;B,C)\\stackrel{C}{=}(A,N;Q,P_{\\infty})=-1.\n\\]\n\nSo, \$ABD_1C\$ is harmonic. Similarly, we can find \$ABD_2C\$ is harmonic. Therefore, \$D_1=D_2,\$ which means that \$BM\$ and \$CN\$ intersect on the circumcircle. \$\\blacksquare\$", "We use barycentric coordinates. Due to the equal angles, \$AC\$ is tangent to the circumcircle of \$ABQ\$ and \$AB\$ is tangent to the circumcircle of the \$APC.\$ Therefore, we can use power of a point to solve for side ratios. We have\n\n\\[\nA=(1,0,0), B=(0,1,0), C=(0,0,1)\n\\]\n\n\\[\nP=(0:a^2-c^2:c^2),Q=(0:b^2:a^2-b^2)\n\\]\n\n\\[\nM=(-a^2:2a^2-2c^2:2c^2),N=(-a^2:2b^2:2a^2-2b^2)\n\\]\n\nTherefore, \$D=(-a^2:2b^2:2c^2),\$ as \$BM\$ and \$CN\$ are cevians. Note that \$(x,y,z)\$ lies on the circumcircle iff \$a^2yz+b^2xz+c^2xy=0.\$ Substituting the values in, we have\n\n\\[\n-4a^2b^2c^2+2a^2b^2c^2+2a^2b^2c^2=0,\n\\]\n\nso we are done. \$\\blacksquare\$", "Note that the givens immediately imply that \$\\triangle{ABC} \\sim \\triangle{QAC} \\sim \\triangle{PBA}\$, hence \$\\angle{AQP}=\\angle{APQ}=\\angle{A}\$. Let \$D\$ be the midpoint of BC, \$E\$ be the midpoint of \$AC\$, and \$F\$ the midpoint of \$AB\$. By the similar triangles, we have \$\\angle{BAD}=\\angle{AQE}=\\angle{AMC}\$. We also have \$\\angle{BAD}=\\angle{BPF}=\\angle{MNB}\$, so we find \$\\angle{AMC}=\\angle{MNB}\$. We note that \$\\angle{AMC}+\\angle{CMN}=\\angle{AMN}=\\angle{AQP}=\\angle{A}\$, so \$\\angle{CMN}+\\angle{MNB}=A\$, which gives that \$\\angle{BKC}=180-\\angle{A}\$ and we are done.\n\nAs an addition, \$AK\$ is the A-symmedian in \$\\triangle{ABC}\$.", "\\[\nup\n\\]\n\nLet \$w\$ be the circumcircle of \$\\triangle{ABC}\$. Let \$E\$ and \$F\$ be the intersection of \$w\$ with \$AQ\$ and \$AP\$, respectively. By basic angle chasing, we have \$\\angle{ABC} = \\angle{CBE}\$ and \$\\angle{ACB} = \\angle{BCF}\$. So if the intersection of \$BE\$ and \$CF\$ is \$D\$, \$BC\$ bisects \$AD\$. And we know that \$BC\$ bisects \$AN\$ and \$AM\$, that means \$N\$, \$D\$ and \$M\$ are collinear. Now, we define the point \$K\$ which is the intersection of \$BM\$ and \$w\$. And let us say \$X\$ to the intersection of \$CK\$ and \$AE\$. By Pascal Theorem at \$FCKBEA\$:\n\n\$D\$, \$X\$ and \$M\$ are collinear. That means \$X\$ is on \$DM\$, \$CK\$ and \$AE\$. Therefore \$X = N\$ and this ends the proof because the intersection of \$CN\$ and \$BM\$ is the point \$K\$ which is on \$w\$. \$\\blacksquare\$\n\n~Ege Saribas" ]
IMO-2014-5	https://artofproblemsolving.com/wiki/index.php/2014_IMO_Problems/Problem_5	For each positive integer $n$, the Bank of Cape Town issues coins of denomination $\tfrac{1}{n}$. Given a finite collection of such coins (of not necessarily different denominations) with total value at most $99+\tfrac{1}{2}$, prove that it is possible to split this collection into $100$ or fewer groups, such that each group has total value at most $1$.	[ "The bound is not tight. We'll prove the result for at most \$k - \\frac{k}{2k+1}\$ with \$k\$ groups.\n\nFirst, perform the following optimizations. - If any coin of size \$\\frac{1}{2m}\$ appears twice, then replace it with a single coin of size \$\\frac{1}{m}\$. - If any coin of size \$\\frac{1}{2m+1}\$ appears \$2m+1\$ times, group it into a single group and induct downwards. Apply this operation repeatedly until it cannot be done anymore.\n\nNow construct boxes \$B_0\$, \$B_1\$, ...., \$B_{k-1}\$. In box \$B_0\$ put any coins of size \$\\tfrac 12\$ (clearly there is at most one). In the other boxes \$B_m\$, put coins of size \$\\frac{1}{2m+1}\$ and \$\\frac{1}{2m+2}\$ (at most \$2m\$ of the former and at most one of the latter). Note that the total weight in the box is less than \$1\$. Finally, place the remaining ``light coins of size at most \$\\frac{1}{2k+1}\$ in a pile.\n\nThen just toss coins from the pile into the boxes arbitrarily, other than the proviso that no box should have its weight exceed \$1\$. We claim this uses up all coins in the pile. Assume not, and that some coin remains in the pile when all the boxes are saturated. Then all the boxes must have at least \$1 -\\frac{1}{2k+1}\$, meaning the total amount in the boxes is strictly greater than\n\n\\[\nk \\left( 1 - \\frac{1}{2k+1} \\right)\n\\]\n\nwhich is a contradiction. (The inequality is strict since the pile has a coin leftover.)" ]
IMO-2014-6	https://artofproblemsolving.com/wiki/index.php/2014_IMO_Problems/Problem_6	A set of lines in the plane is in $\textit{general position}$ if no two are parallel and no three pass through the same point. A set of lines in general position cuts the plane into regions, some of which have finite area; we call these its $\textit{finite regions}$. Prove that for all sufficiently large $n$, in any set of $n$ lines in general position it is possible to colour at least $\sqrt{n}$ of the lines blue in such a way that none of its finite regions has a completely blue boundary.	[ "Call the \$\\binom{n}{2}\$ intersections, well, points. Then each line will have \$n-1\$ points. We call 2 points (on a line) neighbors if there are no other points on the line segment joining those 2. Then each finite region has to be a convex polygon whose any pair of neighboring vertices are, well, neighbors. Now start by coloring any line blue, and all points are uncolored (neither red nor blue).\n\nSuppose there is an uncolored line which does not pass through any red points. Color that line blue. Now consider each of the intersections that line makes with the other blue lines. For each intersection point \$X\$ of the 2 blue lines \$l_1,l_2\$, color it blue (it was originally uncolored). Now consider the neighbors of \$X\$ on \$l_1,l_2\$, call them \$A_1,A_2\$ on \$l_1\$ and \$B_1,B_2\$ on \$l_2\$ (if there is only 1 neighbor, let the other point be an uncolored dummy). If neither \$A_1\$ nor \$A_2\$ is blue, we colour them both red. Else if neither \$B_1\$ nor \$B_2\$ is blue, we colour them both red. Otherwise, we colour the uncolored ones (out of the 4 points) red. Anyway, we would have colored at most 2 points red. Once we have done this for all the intersection points, proceed back to the start of this paragraph until no such line exists.\n\nNow to show it works. Suppose there is a finite region polygon \$P_1...P_m\$ with all blue boundaries (with \$P_{m+1}=P_1\$). Then all the points must be blue. Consider the first point colored blue, WLOG let it be \$P_2\$. Suppose the previous line to be colored blue is \$P_1P_2\$. Then \$P_2P_3\$ has to be colored blue before that, and \$P_3P_4\$ has to be uncolored (otherwise \$P_3\$ will be colored blue first). So \$P_3\$ is uncolored. Then if \$P_1\$ is blue, \$P_3\$ will be colored red by our coloring algorithm. \$P_1\$ has to be uncolored and by our coloring algorithm at least one of \$P_1,P_3\$ will be colored red. Note that no blue line will pass through a red point. Thus there are no finite region with blue boundaries.\n\nNow suppose \$k\$ lines are blue. By our algorithm, each blue point intersection will introduce at most 2 red points. Since we can't color any more lines blue, those red points will cover the remaining lines. Thus \$2\\binom{k}{2}\\geq n-k\\implies k\\geq \\sqrt{n}\$ exact.\n\nNote: this works for all \${n}\\geq{2}\$, provided there is no mistake." ]
IMO-2015-1	https://artofproblemsolving.com/wiki/index.php/2015_IMO_Problems/Problem_1	We say that a finite set $\mathcal{S}$ in the plane is balanced if, for any two different points $A$, $B$ in $\mathcal{S}$, there is a point $C$ in $\mathcal{S}$ such that $AC=BC$. We say that $\mathcal{S}$ is centre-free if for any three points $A$, $B$, $C$ in $\mathcal{S}$, there is no point $P$ in $\mathcal{S}$ such that $PA=PB=PC$. 1. Show that for all integers $n\geq 3$, there exists a balanced set consisting of $n$ points. 2. Determine all integers $n\geq 3$ for which there exists a balanced centre-free set consisting of $n$ points.	[ "Part (a): We explicitly construct the sets \$\\mathcal{S}\$. For odd \$n\$, \$\\mathcal{S}\$ can be taken to be the vertices of regular polygons \$P_n\$ with \$n\$ sides: given any two vertices \$A\$ and \$B\$, one of the two open half-spaces into which \$AB\$ divides \$P_n\$ contains an odd number of \$k\$ of vertices of \$P_n\$. The \$((k+1)/2)^{th}\$ vertex encountered while moving from \$A\$ to \$B\$ along the circumcircle of \$P_n\$ is therefore equidistant from \$A\$ and \$B\$.\n\nIf \$n \\geq 4\$ is even, choose \$m\\geq 0\$ to be the largest integer such that\n\n\\[\nx:=(n-2)(\\pi/3)/2^m \\geq 2\\pi/3.\n\\]\n\nHence \$x < 4\\pi/3 < 2\\pi\$. Consider a circle \$K\$ with centre \$O\$, and let \$A_1, \\ldots, A_{n-1}\$ be distinct points placed counterclockwise (say) on \$K\$ such that \$\\angle A_iOA_{i+1}=\\pi/3/2^m\$ (for \$i=1,\\ldots,n-2\$). Hence for any line \$OA_i\$, there is a line \$OA_j\$ such that \$\\angle A_iOA_j=\\pi/3\$ (using the facts that \$2\\pi > x=\\angle A_1OA_{n-1} \\geq 2\\pi/3\$, and \$n-1\$ odd). Thus \$O\$, \$A_i\$ and \$A_j\$ form an equilateral triangle. In other words, for arbitrary \$A_i\$, there exists \$A_j\$ equidistant to \$O\$ and \$A_i\$. Also given any \$i,j\$ such that \$1 \\leq i, j \\leq n-1\$, \$O\$ is equidistant to \$A_i\$ and \$A_j\$. Hence the \$n\$ points \$O, A_1, \\ldots, A_{n-1}\$ form a balanced set.\n\nPart (b): Note that if \$n\$ is odd, the set \$\\mathcal{S}\$ of vertices of a regular polygon \$P_n\$ of \$n\$ sides forms a balanced set (as above) and a centre-free set (trivially, since the centre of the circumscribing circle of \$P_n\$ does not belong to \$\\mathcal{S}\$).\n\nFor \$n\$ even, we prove that a balanced, centre free set consisting of \$n\$ points does not exist. Assume that \$\\mathcal{S}=\\{A_i: 1\\leq i \\leq n\\}\$ is centre-free. Pick an arbitrary \$A_i \\in \\mathcal{S}\$, and let \$n_i\$ be the number of distinct non-ordered pairs of points \$(A_j,A_k)\$ (\$j\\neq k\$) to which \$A_i\$ is equidistant. Any two such pairs are disjoint (for, if there were two such pairs \$(A_r,A_s)\$ and \$(A_r, A_t)\$ with \$r, s, t\$ distinct, then \$A_i\$ would be equidistant to \$A_r\$, \$A_s\$, and \$A_t\$, violating the centre-free property). Hence \$n_i \\leq (n-2)/2\$ (we use the fact that \$n\$ is even here), which means \$\\sum_i n_i \\leq n(n-2)/2 = n(n-1)/2 -n/2\$. Hence there are at least \$n/2\$ non-ordered pairs \$(A_j, A_k)\$ such that no point in \$\\mathcal{S}\$ is equidistant to \$A_j\$ and \$A_k\$." ]
IMO-2015-2	https://artofproblemsolving.com/wiki/index.php/2015_IMO_Problems/Problem_2	Determine all triples of positive integers $(a,b,c)$ such that each of the numbers \[ ab-c,\; bc-a,\; ca-b \] is a power of 2. (A power of 2 is an integer of the form $2^n$ where $n$ is a non-negative integer ).	[ "The solutions for \$(a,b,c)\$ are \$(2,2,2)\$, \$(2,2,3)\$, \$(2,6,11)\$, \$(3,5,7)\$, and permutations of these triples.\n\nThroughout the proof, we assume \$a \\leq b \\leq c\$, so that \$ab-c = 2^m\$, \$ca-b = 2^n\$, \$bc-a=2^p\$, with \$m \\leq n \\leq p\$. Note that \$a>1\$ since otherwise \$b-c=2^m\$, which is impossible. Hence \$2^n = ac-b \\geq (a-1)c \\geq 2\$, i.e., \$n\$ and \$p\$ are positive.\n\nObserve that if \$a=b\\geq 3\$, we get \$a(c-1)=2^n\$, so \$a\$ and \$c-1\$ are (even and) powers of \$2\$. Hence \$c\$ is odd and \$a^2-c=2^m=1\$. Hence \$c+1=a^2\$ is also a power of \$2\$, which implies \$c=3\$. But \$a=b=c=3\$ is not a solution; hence \$a=b\\geq 3\$ is infeasible. We consider the remaining cases as follows.\n\nCase 1: \$a=2\$. We have\n\n\\[\n2b-c=2^m,\\; 2c-b=2^n,\\; bc-2=2^p.\n\\]\n\nFrom the second equation, \$b\$ is even. From the third equation, if \$p=1\$, then \$b=c=2\$; if \$p>1\$, then \$c\$ is odd, which implies that \$m=0\$. Hence \$3b=2^n+2\$ (so \$n\\geq 2\$), \$3c=2^{n+1}+1\$, and \$(2^{n-1}+1)(2^{n+1}+1)=9(2^{p-1}+1)\$. Hence \$1\\equiv 9 \\mod 2^{n-1} \\implies n \\leq 4\$. Hence \$n\$ is 2 or 4, and \$(b,c)\$ equals \$(2,3)\$ or \$(6,11)\$. Thus the solutions for \$(a,b,c)\$ are \$(2,2,2)\$, \$(2,2,3)\$ or \$(2,6,11)\$.\n\nCase 2: \$3\\leq a<b\\leq c\$. Since \$(a-1)c \\leq 2^n\$, we have \$c\\leq 2^{n-1}\$. Hence\n\n\\[\nb+a < 2c\\leq 2^{n+1}/(a-1) \\leq 2^n,\n\\]\n\n\\[\nb-a < c \\leq 2^{n-1}.\n\\]\n\nHence \$b-a\$ is not divisible by \$2^{n-1}\$, and \$b+a\$ is not divisible by \$2^{n-1}\$ for \$a\\geq 5\$. Adding and subtracting \$ac-b=2^n\$ and \$bc-a=2^p\$, we get\n\n\\[\n(c-1)(b+a) = 2^p+2^n,\n\\]\n\n\\[\n(c+1)(b-a) = 2^p-2^n.\n\\]\n\nFrom the latter equation, \$c+1\$ is divisible by \$4\$. Hence \$c-1\$ is not divisible by \$4\$, which implies that \$b+a < 2^n\$ is a multiple of \$2^{n-1}\$. Hence \$a \\leq 4\$ and \$b+a=2^{n-1}\$.\n\nConsider \$a=3\$, which implies \$3b-c=2^m\$, \$3c-b=2^n\$, \$b=2^{n-1}-3\$. Hence \$2^{n-1}-3=3.2^{m-3}+2^{n-3}\$, or \$2^{n-3}=2^{m-3}+1\$. Hence \$m=3\$, \$n=4\$, \$b=5\$ and \$c=7\$.\n\nFinally, consider \$a=4\$, \$4c-b=2^n\$, \$b=2^{n-1}-4\$. Hence \$c=3.2^{n-3}-1\$. But \$b \\leq c\$ implies \$2^{n-3} \\leq 3\$ and \$a<b\$ implies \$2^{n-3}>2\$. Hence there are no solutions with \$a=4\$.\n\nWe obtain \$(a,b,c)=(3,5,7)\$ as the only solution with \$3 \\leq a < b \\leq c\$." ]
IMO-2015-3	https://artofproblemsolving.com/wiki/index.php/2015_IMO_Problems/Problem_3	Let $ABC$ be an acute triangle with $AB>AC$. Let $\Gamma$ be its circumcircle, $H$ its orthocenter, and $F$ the foot of the altitude from $A$. Let $M$ be the midpoint of $BC$. Let $Q$ be the point on $\Gamma$ such that $\angle HKQ=90^\circ$. Assume that the points $A$, $B$, $C$, $K$, and $Q$ are all different, and lie on $\Gamma$ in this order. Prove that the circumcircles of triangles $KQH$ and $FKM$ are tangent to each other.	[ "\\[\nup\n\\]\n\nWe know that there is a negative inversion which is at \$H\$ and swaps the nine-point circle and \$\\Gamma\$. And this maps:\n\n\$A \\longleftrightarrow F\$. Also, let \$M \\longleftrightarrow Q`\$. Of course \$\\triangle HFM \\sim \\triangle HQ'A\$ so \$\\angle HQ'A = 90\$. Hence, \$Q' = Q\$. So:\n\n\$M \\longleftrightarrow Q\$. Let \$HA\$ and \$HQ\$ intersect with nine-point circle \$T\$ and \$Q\$, respectively. Let's define the point \$L\$ such that \$TNML\$ is rectangle. We have found \$M \\longleftrightarrow Q\$ and if we do the same thing, we find:\n\n\$L \\longleftrightarrow K\$. Now, we can say:\n\n\$(KQH) \\longleftrightarrow ML\$ and \$(FKM) \\longleftrightarrow (ALQ)\$. İf we manage to show \$ML\$ and \$(ALQ)\$ are tangent, the proof ends.\n\nWe can easily say \$TN \|\| AQ\$ and \$AQ = 2.TN\$ because \$T\$ and \$N\$ are the midpoints of \$HA\$ and \$HQ\$, respectively.\n\nBecause of the rectangle \$TNML\$, \$TN \|\| ML\$ and \$TN = ML\$.\n\nHence, \$ML \|\| AQ\$ and \$AQ = 2.ML\$ so \$L\$ is on the perpendecular bisector of \$AQ\$ and that follows \$\\triangle ALQ\$ is isoceles. And we know that \$ML \|\| AQ\$, so \$ML\$ is tangent to \$(ALQ)\$. We are done. \$\\blacksquare\$\n\n~ EgeSaribas" ]
IMO-2015-4	https://artofproblemsolving.com/wiki/index.php/2015_IMO_Problems/Problem_4	Triangle $ABC$ has circumcircle $\Omega$ and circumcenter $O$. A circle $\Gamma$ with center $A$ intersects the segment $BC$ at points $D$ and $E$, such that $B$, $D$, $E$, and $C$ are all different and lie on line $BC$ in this order. Let $F$ and $G$ be the points of intersection of $\Gamma$ and $\Omega$, such that $A$, $F$, $B$, $C$, and $G$ lie on $\Omega$ in this order. Let $K$ be the second point of intersection of the circumcircle of triangle $BDF$ and the segment $AB$. Let $L$ be the second point of intersection of the circumcircle of triangle $CGE$ and the segment $CA$. Suppose that the lines $FK$ and $GL$ are different and intersect at the point $X$. Prove that $X$ lies on the line $AO$. Proposed by Silouanos Brazitikos and Evangelos Psychas, Greece	[ "Lemma (On three chords). If two lines pass through different endpoints of two circles' common chord, then the other two chords cut by these lines on the circles are parallel. Proof The second and the third chords are anti-parallel to the first (common) chord with respect to the given lines, so they are parallel to each other. \$\\Box\$ To solve this problem, it is sufficient to apply the lemma 5 times. Indeed, let the lines \$FD, GE, FK, GL\$ meet \$\\Omega\$ second time at \$H, I, M, N\$ respectively. One of the circles that figure in lemma is always \$\\Omega\$, while the other is one of three other circles from the problem statement. Applying the lemma to the lines \$FDH\$ and \$GEI\$, \$FKM\$ and \$BDC\$, \$FDH\$ and \$BKA\$, \$GLN\$ and \$CEB\$, \$GEI\$ and \$CLA\$, we get \$DE \\parallel IH\$, \$KD \\parallel MC\$, \$KD \\parallel AH\$, \$LE \\parallel NB\$, \$LE \\parallel AI\$, respectively. From this, \$BC \\parallel IH\$, \$MC \\parallel AH\$, \$NB \\parallel AI\$. Therefore, \$AN=IB=HC=AM\$. This means that \$N\$ and \$M\$ are symmetric wrt \$AO\$, a diameter of \$\\Omega\$ through \$A\$. So are \$F\$ and \$G\$, as \$AF=AG\$. Therefore, the lines \$FM\$ and \$GN\$ are symmetric wrt \$AO\$ and meet on it." ]
IMO-2015-5	https://artofproblemsolving.com/wiki/index.php/2015_IMO_Problems/Problem_5	Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f$:$\mathbb{R}\rightarrow\mathbb{R}$ satisfying the equation \[ f(x+f(x+y))+f(xy) = x+f(x+y)+yf(x) \] for all real numbers $x$ and $y$. Proposed by Dorlir Ahmeti, Albania	[ "\$f(x+f(x+y)) + f(xy) = x + f(x+y) + yf(x)\$ for all real numbers \$x\$ and \$y\$.\n\n(1) Put \$x=y=0\$ in the equation, We get\$f(0 + f(0)) + f(0) = 0 + f(0) + 0\$ or \$f(f(0)) = 0\$ Let \$f(0) = k\$, then \$f(k) = 0\$\n\n(2) Put \$x=0, y=k\$ in the equation, We get \$f(0 + f(k)) + f(0) = 0 + f(k) + kf(0)\$ But \$f(k) = 0\$ and \$f(0) = k\$ so, \$f(0) + f(0) = f(0)^2\$ or \$f(0)[f(0) - 2] = 0\$ Hence \$f(0) = 0, 2\$\n\nCase \$1\$ : \$f(0) = 0\$\n\nPut \$x=0, y=x\$ in the equation, We get \$f(0 + f(x)) + f(0) = 0 + f(x) + xf(0)\$ or, \$f(f(x)) = f(x)\$ Say \$f(x) = z\$, we get \$f(z) = z\$\n\nSo, \$f(x) = x\$ is a solution -- fallacy\n\nCase \$2\$ : \$f(0) = 2\$ Again put \$x=0, y=x\$ in the equation, We get \$f(0 + f(x)) + f(0) = 0 + f(x) + xf(0)\$ or, \$f(f(x)) + 2 = f(x) + 2x\$\n\nWe observe that \$f(x)\$ must be a polynomial of power \$1\$ as any other power (for that matter, any other function) will make the \$LHS\$ and \$RHS\$ of different powers and will not have any non-trivial solutions. -- fallacy\n\nAlso, if we put \$x=0\$ in the above equation we get \$f(2) = 0\$\n\n\$f(x) = 2-x\$ satisfies both the above.\n\nHence, the solutions are \$\\boxed{\\color{red}{f(x) = x}}\$ and \$\\boxed{\\color{red}{f(x) = 2-x}}\$." ]
IMO-2015-6	https://artofproblemsolving.com/wiki/index.php/2015_IMO_Problems/Problem_6	The sequence $a_1,a_2,\dots$ of integers satisfies the conditions: (i) $1\le a_j\le2015$ for all $j\ge1$, (ii) $k+a_k\neq \ell+a_\ell$ for all $1\le k<\ell$. Prove that there exist two positive integers $b$ and $N$ for which \[ \left\vert\sum_{j=m+1}^n(a_j-b)\right\vert\le1007^2 \] for all integers $m$ and $n$ such that $n>m\ge N$. Proposed by Ivan Guo and Ross Atkins, Australia.	[ "We can prove the more general statement. Theorem Let \$T\$ be a non-negative integer parameter. If given a sequence \$a_1,a_2,\\dots\$ that satisfies the conditions: (i) \$1 \\le a_j \\le T+1\$ for all \$j \\le 1;\$ (ii) \$k+a_k \\ne \\ell+a_\\ell\$ for all \$1 \\le k<\\ell,\$ then there exist two integers \$v\$ and \$N\$, \$0 \\le v \\le T\$ and \$N>0\$, for which\n\n\\[\n\\left\\vert\\sum_{j=m+1}^n(a_j-(v+1))\\right\\vert\\le (T-v)\\,v \\le \\left(\\frac T 2\\right)^2\n\\]\n\nfor all integers \$m\$ and \$n\$ such that \$n>m\\ge N\$. Proof Consider the set of points on a plane: \$\\{(x,y) \\mid x,y \\in \\mathbb N,\\,y \\le T+1\\}\$ (let's call it the base strip). The sequence is represented on it by the subset \$\\{(i,a_i) \\mid i \\in \\mathbb N\\}\$ which is painted red. Each line \$L_n\$ of kind \$x+y=n\$, where \$n \\ge 2\$ is an integer, will be called a carrier line. Note that the condition (ii) states that each carrier line is occupied by at most one red point. Every such line with exactly one red point on it will be called occupied while the rest carriers will be called free. First we prove that the set of free carriers is finite. Indeed, all the carriers \$L_n\$ with \$2 \\le n \\le T+m+1\$, where \$m \\in \\mathbb N\$, cover the part of the base strip having \$x \\le m\$, so there are at least \$m\$ red points lying on them. Thus, the number of free carriers among them is at most \$T+m-m=T\$. Since \$m\$ is arbitrary, the total number of free cariers doesn't exceed \$T\$. Take \$N\$ such that all the carriers \$L_n\$ with \$n \\ge N+2\$ are occupied. Next, take any red point and consider the points on the base strip which are lying on the point's carrier below it (i.e. having strictly greater \$x\$). Call it the point's trace and paint black. Finally, the set \$P_x=\\{\\,y \\mid (x,y) \\text { is black}\\}\$ will be called a pattern, the number elements \$\|P_x\|\$ in it will be called its volume and the sum \$w_x\$ of all elements in \$P_x\$ will be called its weight. Now let \$j>N\$. Lets track how \$P_j\$, its volume and its weight change when \$j\$ increases by 1. Obviously \$P_{j+1}\$ is \$P_j\$ plus added \$a_j\$ (red point) and shifted down by \$1\$; a point that goes to \$0\$ vanishes. This means that the volume doesn't change at all. Indeed, if \$1 \\notin P_j\$, then \$a_j=1\$ or the carrier \$L_{j+1}\$ will remain unoccupied, so one point always vanishes going to \$0\$, and exactly one point is added before the shift (it may be the same point). Let \$v\$ be this constant volume. As a pattern never contains \$T+1\$, \$v\$ ranges from \$0\$ to \$T\$. Now it is clearly that \$w_{j+1}\$ is \$w_j\$ plus \$a_j\$ (red point added) and minus \$v+1\$ (all elements are shifted by \$1\$), i.e. \$a_j-(v+1)=w_{j+1}-w_j\$. Thus, when \$n>m\\ge N\$:\n\n\\[\n\\sum_{j=m+1}^n(a_j-(v+1))=w_{n+1}-w_{m+1}\\,.\n\\]\n\nTo finish, it remains to note that the weight ranges from \$1+2+\\dots+v\$ to \$(T-v+1)+(T-v+2)+\\dots+T\$, so its swing is:\n\n\\[\n\\sum_{i=1}^v (T-v+i) - \\sum_{i=1}^v i=\\sum_{i=1}^v (T-v)=(T-v)\\,v\\;. \\quad \\blacksquare\n\\]" ]
IMO-2016-1	https://artofproblemsolving.com/wiki/index.php/2016_IMO_Problems/Problem_1	Triangle $BCF$ has a right angle at $B$. Let $A$ be the point on line $CF$ such that $FA=FB$ and $F$ lies between $A$ and $C$. Point $D$ is chosen so that $DA=DC$ and $AC$ is the bisector of $\angle{DAB}$. Point $E$ is chosen so that $EA=ED$ and $AD$ is the bisector of $\angle{EAC}$. Let $M$ be the midpoint of $CF$. Let $X$ be the point such that $AMXE$ is a parallelogram. Prove that $BD,FX$ and $ME$ are concurrent. \[ 2016IMOQ1.jpg \]	[ "\\[\n2016IMOQ1Solution.jpg\n\\]\n\nThe Problem shows that \$\\angle DAC = \\angle DCA = \\angle CAD\$, it follows that \$AB \\parallel CD\$. Extend \$DC\$ to intersect \$AB\$ at \$G\$, we get \$\\angle GFA = \\angle GFB = \\angle CFD\$. Making triangles \$\\triangle CDF\$ and \$\\triangle AGF\$ similar. Also, \$\\angle FDC = \\angle FGA = 90^\\circ\$ and \$\\angle FBC = 90^\\circ\$, which points \$D\$, \$C\$, \$B\$, and \$F\$ are concyclic.\n\nAnd \$\\angle BFC = \\angle FBA + \\angle FAB = \\angle FAE = \\angle AFE\$. Triangle \$\\triangle AFE\$ is congruent to \$\\triangle FBM\$, and \$AE = EF = FM = MB\$. Let \$MX = EA = MF\$, then points \$B\$, \$C\$, \$D\$, \$F\$, and \$X\$ are concyclic.\n\nFinally \$AD = DB\$ and \$\\angle DAF = \\angle DBF = \\angle FXD\$. \$\\angle MFX = \\angle FXD = \\angle FXM\$ and \$FE \\parallel MD\$ with \$EF = FM = MD = DE\$, making \$EFMD\$ a rhombus. And \$\\angle FBD = \\angle MBD = \\angle MXF = \\angle DXF\$ and triangle \$\\triangle BEM\$ is congruent to \$\\triangle XEM\$, while \$\\triangle MFX\$ is congruent to \$\\triangle MBD\$ which is congruent to \$\\triangle FEM\$, so \$EM = FX = BD\$.\n\n~Athmyx", "Let \$\\angle FBA = \\angle FAB = \\angle FAD = \\angle FCD = \\angle DAE = \\angle ADE = \\alpha\$. And WLOG, \$MF = 1\$. Hence, \$CF = 2\$,\n\n\$\\implies\$ \$BF = CF.cos(2\\alpha) = 2.cos(2\\alpha) = FA\$,\n\n\$\\implies\$ \$DA = \\frac{AC}{2cos(\\alpha)} = \\frac{CF+FA}{2cos(\\alpha)} = \\frac{2+2cos(2\\alpha)}{2cos(\\alpha)} = \\frac{1+cos(2\\alpha)}{cos(\\alpha)}\$ and\n\n\$\\implies\$ \$DE = AE = \\frac{DA}{2cos(\\alpha)} = \\frac{1+cos(2\\alpha)}{2.(cos(\\alpha))^2} = 1\$.\n\nSo \$MX = DE = 1\$ which means \$B\$, \$C\$, \$X\$ and \$F\$ are concyclic. We know that \$DE \|\| MC\$ and \$DE = 1 = MC\$, so we conclude \$MCDE\$ is parallelogram. So \$\\angle AME = \\alpha\$. That means \$MDEA\$ is isosceles trapezoid. Hence, \$MD = EA = 1\$. By basic angle chasing,\n\n\$\\angle MBF = \\angle MFB = 2\\alpha\$ and \$\\angle MXD = \\angle MDX = 2\\alpha\$ and we have seen that \$MB = MF = MD = MX\$, so \$BFDX\$ is isosceles trapezoid. And we know that \$ME\$ bisects \$\\angle FMD\$, so \$ME\$ is the symmetrical axis of \$BFDX\$.\n\n\$B\$ and \$X\$, \$D\$ and \$E\$ are symmetrical respect to \$ME\$. Hence, the symmetry of \$BD\$ with respect to \$ME\$ is \$FX\$. And we are done \$\\blacksquare\$.\n\n~EgeSaribas" ]
IMO-2016-2	https://artofproblemsolving.com/wiki/index.php/2016_IMO_Problems/Problem_2	Find all integers $n$ for which each cell of $n \times n$ table can be filled with one of the letters $I,M$ and $O$ in such a way that: Note. The rows and columns of an $n \times n$ table are each labelled $1$ to $n$ in a natural order. Thus each cell corresponds to a pair of positive integer $(i,j)$ with $1 \le i,j \le n$. For $n>1$, the table has $4n-2$ diagonals of two types. A diagonal of first type consists all cells $(i,j)$ for which $i+j$ is a constant, and the diagonal of this second type consists all cells $(i,j)$ for which $i-j$ is constant.	[ "Here is a solution using counting in two ways.\n\nIt's obvious that \$3 \\mid n\$. We consider all the squares indexed \$(3k+2,3l+2)\$ and call it [i]good[/i] square. Let \$a\$ be number of [i]good[/i] squares that are filled with \$I\$. We can see that every good square lies on both type of diagonals. So if we call \$D_1,D_2\$ be the set of all squares in first type, second type of diagonal that are filled with \$I\$, respectively. We will have \$\|D_1 \\cap D_2\|=a\$ and \$\|D_1\|=\|D_2\|= \\tfrac 19 n^2\$. Hence,\n\n\\[\n\|D_1 \\cup D_2\|=\|D_1\|+\|D_2\|-\|D_1 \\cap D_2\|= \\tfrac 29 n^2-a.\n\\]\n\nHence, number of squares filled with \$I\$ that either lie on first type of diagonal or second type but not in both is \$\\tfrac 29 n^2-2a\$.\n\nNow, these squares counted above doesn't fill the columns indexed \$3k+2\$ and rows indexed \$3k+2\$. So we need to fill these squares with \$I\$. Let \$C\$ be the set of squares in columns indexed \$3k+2\$ that are filled with \$I\$, similar to set \$R\$ of rows indexed \$3k+2\$. We have\n\n\\[\n\|C \\cup R\|=\|C\|+\|R\|-\|C \\cap R\|= \\tfrac 29 n^2-a.\n\\]\n\nFrom these, we find that the total number of squares filled with \$I\$ is \$\\tfrac 49 n^2-3a\$. Note that this is also equal to \$\\tfrac 13 n^2\$ so this means \$3a= \\tfrac 19 n^2\$ implies \$9 \\mid n\$, which is the final answer.\n\nThis argument gives us a way to construct a \$9 \\times 9\$ table, which is first we fill all the \"good\" squares in a way such that one third of them are \$I\$, one third are \$O\$ and one third are \$M\$. After that, we fill all the diagonals and then fill the remain squares." ]
IMO-2016-3	https://artofproblemsolving.com/wiki/index.php/2016_IMO_Problems/Problem_3	Let $P = A_1A_2 \cdots A_k$ be a convex polygon in the plane. The vertices $A_1,A_2,\dots, A_k$ have integral coordinates and lie on a circle. Let $S$ be the area of $P$. An odd positive integer $n$ is given such that the squares of the side lengths of $P$ are integers divisible by $n$. Prove that $2S$ is an integer divisible by $n$.	[ "Note that \$2S\$ is always an integer for any lattice polygon, so it remains to show that it is divisible by \$n\$. It clearly suffices to prove the problem for when \$n=p^m\$ is a prime power. We proceed using induction on \$k\$, with the base case of \$k=3\$ settled by Heron's formula: If \$a,b,c\$ are the side lengths of the triangle, then the square of the area is \$S^2=\\frac{2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4}{16}\$. As \$n\\mid a^2,b^2,c^2\$, we have that \$n^2\\mid S^2\\implies n\\mid S\$, as desired.\n\nFor the inductive step, we claim that there exists a diagonal whose length squared is also divisible by \$n\$. Then, we may split \$P\$ into two polygons with less vertices and areas divisible by \$n\$ by assumption. Let \$3\\le i\\le k-1\$ be such that \$v_p(A_1A_i^2)=q\$ is minimized. By Ptolemy's Theorem on cyclic quadrilateral \$A_1A_{i-1}A_iA_{i+1}\$, we have that \$\\sqrt{A_1A_{i-1}^2\\cdot A_iA_{i+1}^2}+\\sqrt{A_1A_{i+1}^2\\cdot A_iA_{i-1}^2}=\\sqrt{A_1A_i^2\\cdot A_{i-1}A_{i+1}^2}\$, or \$\\sqrt{\\frac{A_1A_{i-1}^2\\cdot A_iA_{i+1}^2}{np^q}}+\\sqrt{\\frac{A_1A_{i+1}^2\\cdot A_iA_{i-1}^2}{np^q}}=\\sqrt{\\frac{A_1A_i^2\\cdot A_{i-1}A_{i+1}^2}{np^q}}\$. As we have \$p^q\\mid AA_{i-1}^2,AA_{i+1}^2\$ and \$n\\mid A_iA_{i+1}^2,A_iA_{i-1}^2\$, the terms under the square roots on the LHS are integers, so the LHS is an algebraic integer. This implies that the term under the square root on the RHS is also an integer, so \$np^q\\mid A_1A_i^2\\cdot A_{i-1}A_{i+1}^2\\implies n\\mid A_{i-1}A_{i+1}^2\$, as desired." ]
IMO-2016-4	https://artofproblemsolving.com/wiki/index.php/2016_IMO_Problems/Problem_4	A set of positive integers is called fragrant if it contains at least two elements and each of its elements has a prime factor in common with at least one of the other elements. Let $P(n)=n^2+n+1$. What is the least possible positive integer value of $b$ such that there exists a non-negative integer $a$ for which the set $\{P(a+1),P(a+2),\ldots,P(a+b)\}$ is fragrant?	[ "Consider \$P(x)\$ and \$P(x+y)\$. We note that in order to \$p \\mid P(x)\$ and \$p \\mid P(x+y)-P(x)\$ we must have \$p \\mid x^2+x+1\$ and \$p \\mid y(2x+y+1)\$. It is obvious that \$p \\equiv 1 \\pmod{6}\$ since \$p \\mid n^2+n+1 \\mid (2n+1)^2+3\$ or \$\\left( \\tfrac{-3}{p} \\right)=1\$.\n\nIf \$y=1\$ then \$p \\mid 2x+2\$ implies \$p \\mid x+1\$. WLOG, we can let \$x=p-1\$ and see that \$p \\nmid x^2+x+1\$. So there doesn't exists \$x\$ so that \$\\gcd \\left( P(x),P(x+1) \\right)>1\$.\n\nIf \$y=2\$, we find \$p \\mid 2x+3\$ and we let \$x=\\tfrac{p-3}{2}\$. Hence, from \$p \\mid x^2+x+1\$ we get \$p=7\$. So there is one prime \$p=7\$ such that \$\\gcd \\left( P(x),P(x+2) \\right)>1\$.\n\nIf \$y=3\$, it is obvious that \$p=3\$ satisfy and is the only answer.\n\nIf \$y=4\$, we do the similar thing and get \$p \\mid 2x+5\$ and \$p \\mid x^2+x+1\$ so \$p=19\$. ___________________________ Now, back to the problem, since there doesn't exists any prime \$p\$ for \$y=1\$ so \$b \\ge 3\$. The only prime for \$y=2\$ is \$p=7\$ so if we choose \$7 \\mid P(x+1), 7 \\mid P(x+3)\$ then there will be a number \$7 \\nmid P(x+2)\$ remains. This means \$b \\ge 5\$ since we need a prime \$p \\mid P(x+2)\$ and \$p \\mid P(x+y)\$ but \$y \\ge 5\$ since \$p \\ne 7\$.\n\nIf \$b=5\$, we consider two cases, where there are two numbers that are divisible by \$19\$ (which means \$19 \\mid P(x+1), 19 \\mid P(x+5)\$), the middle-three numbers \$P(x+2),P(x+3),P(x+4)\$ we can't find a way make each two of them have common prime factor. If no two are divisible by \$19\$ then they can only be divisible by \$7,3\$, but it can't cover all \$5\$ \"consecutive\" numbers.\n\nIf \$b=6\$ then we can pick \$19 \\mid P(x+2),P(x+6), 3 \\mid P(x+1), P(x+4), 7 \\mid P(x+3), 7 \\mid P(x+5)\$.\n\nSo the final answer is \$b=6\$." ]
IMO-2016-5	https://artofproblemsolving.com/wiki/index.php/2016_IMO_Problems/Problem_5	The equation \[ (x-1)(x-2)\cdots(x-2016)=(x-1)(x-2)\cdots (x-2016) \] is written on the board, with $2016$ linear factors on each side. What is the least possible value of $k$ for which it is possible to erase exactly $k$ of these $4032$ linear factors so that at least one factor remains on each side and the resulting equation has no real solutions?	[]
IMO-2016-6	https://artofproblemsolving.com/wiki/index.php/2016_IMO_Problems/Problem_6	There are $n\ge 2$ line segments in the plane such that every two segments cross and no three segments meet at a point. Geoff has to choose an endpoint of each segment and place a frog on it facing the other endpoint. Then he will clap his hands $n-1$ times. Every time he claps,each frog will immediately jump forward to the next intersection point on its segment. Frogs never change the direction of their jumps. Geoff wishes to place the frogs in such a way that no two of them will ever occupy the same intersection point at the same time. (a) Prove that Geoff can always fulfill his wish if $n$ is odd. (b) Prove that Geoff can never fulfill his wish if $n$ is even.	[ "Beautiful problem. Despite being placed a bit higher in both the Shortlist and the actual exam, kudos to the proposer for involving combi, geo and NT in one single problem. Anyway, here's my solution: Color each of the \$2n\$ endpoints [color=#00f]blue[/color] and all the \$\\binom{n}{2}\$ intersection points [color=#f00]red[/color]. Join the blue points with a smooth curve \$\\cal{C}\$ such that none of the line segments intersect \$\\cal{C}\$ more than twice (i.e. all the red points lie inside \$\\cal{C}\$). Call the [i]inverse[/i] of a blue point \$\\cal{B}\$, the point which lies on both \$\\cal{C}\$ and the line segment through \$\\cal{B}\$, and denote it by \$\\cal{B'}\$. Also, we define the [i]neighbor[/i] of a blue point \$\\cal{B}\$ as the point which lies on the right of \$\\cal{B}\$ (if we stand facing the inverse of \$\\cal{B}\$) and nearest to it. Finally, if Geoff (from now on we) can place a frog (in such a way that it satisfies his wish) at some blue point, then mark a circle around it; otherwise mark a cross at that point (similar to the game tic tac toe). We first make an important observation, which will provide the main crux of the problem (in all cases, we try to consider the optimal situation)-\n\n[b]Observation[/b] If a blue point is circled, then we cannot circle its neighbor.\n\n[u]PROOF[/u] Consider a blue point \$\\cal{B}\$ and its inverse \$\\cal{B'}\$. Let \$\\cal{A}\$ be the neighbor of \$\\cal{B}\$. Then it's easy to see that \$\\cal{A'}\$ is the neighbor of \$\\cal{B'}\$. Let \$P=\\cal{BB'} \\cap \\cal{AA'}\$, and assume to the contrary that both \$\\cal{B}\$ and \$\\cal{A}\$ can be circled. Suppose there are \$x\$ red points on the segment \$\\cal{B} P\$, and \$y\$ red points on the segment \$\\cal{A} P\$ (excluding \$P\$). Then, we must have \$x \\neq y\$. However, each line segment passing through the \$x\$ red points on \$\\cal{B} P\$ must also pass through the \$y\$ red points on \$\\cal{A} P\$ (since otherwise the condition that \$\\cal{B_n}\$ is the neighbor of \$\\cal{B}\$ gets violated). This gives \$x=y\$, a contradiction. Thus, our observation is true.\n\n[b]Part (b)[/b] Circle any arbitrary blue point \$\\cal{B}\$, and cross its inverse \$\\cal{B'}\$. Then, by our Observation, there must be a cross at its neighbor, and a circle at the inverse of its neighbor. Repeat this process as long as possible. Note that there are exactly \$n-1\$ blue points on both sides of \$\\cal{BB'}\$ (since otherwise all the red points do not lie inside \$\\cal{C}\$). However, for the alternating sequence of circles and crosses staring from \$\\cal{B}\$ and ending at \$\\cal{B'}\$ (where we move in anticlockwise direction) to go on [i]peacefully[/i], there must be an even number of blue points on the arc \$\\widehat{\\cal{BB'}}\$, which gives that \$n-1\$ is even, a contradiction. Thus, Geoff can never fulfill his wish if \$n\$ is even.\n\n[b]Part (a)[/b] As in Part (b), we get a sequence of alternating circles and crosses, and we have to show that this configuration will make Geoff happy. Suppose to the contrary that, for some odd \$n\$, two frogs meet at the same time (despite our exhaustive strategies :D). Consider the line segments on which these frogs were kept, and call them \$\\cal{AA'}\$ and \$\\cal{BB'}\$, where \$A\$ and \$B\$ were circled. Let \$P=\\cal{AA'} \\cap \\cal{BB'}\$. Suppose there are \$k\$ blue points on the arc \$\\widehat{\\cal{AB}}\$ not containing \$A'\$ and \$B'\$. As both \$A\$ and \$B\$ are circled, so \$k\$ must be odd. By our assumption, there exist an equal number of red points on the segments \$\\cal{A} P\$ and \$\\cal{B} P\$ (excluding \$P\$). Now, suppose there exists a line segment passing through a red point on segment \$\\cal{A} P\$ but not passing through any red point on segment \$\\cal{B} P\$. As there are an equal number of red points on these two segments, so there must also exist a line segment passing through one of the red points on segment \$\\cal{B} P\$, but not passing through any red point on segment \$\\cal{A} P\$. Both of these line segments will then contribute to the number of blue points present on arc \$\\widehat{\\cal{AB}}\$ (i.e. while counting \$k\$). Thus all the blue points on \$\\widehat{\\cal{AB}}\$ come in pairs, which gives that \$k\$ is even, a contradiction. Otherwise, no line segment intersects \$\\widehat{\\cal{AB}}\$, which gives \$k=0\$, again not possible. Thus, our assumption must be wrong, and so Geoff can always fulfill his wish if \$n\$ is odd." ]
IMO-2017-1	https://artofproblemsolving.com/wiki/index.php/2017_IMO_Problems/Problem_1	For each integer $a_0 > 1$, define the sequence $a_0, a_1, a_2, \ldots$ for $n \geq 0$ as \[ a_{n+1} = \begin{cases} \sqrt{a_n} & \text{if } \sqrt{a_n} \text{ is an integer,} \\ a_n + 3 & \text{otherwise.} \end{cases} \] Determine all values of $a_0$ such that there exists a number $A$ such that $a_n = A$ for infinitely many values of $n$.	[ "First we observe the following:\n\nWhen we start with \$a_0=3\$, we get \$a_1=6\$, \$a_2=9\$, \$a_3=3\$ and the pattern \$3,6,9\$ repeats.\n\nWhen we start with \$a_0=6\$, we get \$a_1=9\$, \$a_2=3\$, \$a_3=6\$ and the pattern \$3,6,9\$ repeats.\n\nWhen we start with \$a_0=9\$, we get \$a_1=3\$, \$a_2=6\$, \$a_3=9\$ and the pattern \$3,6,9\$ repeats.\n\nWhen we start with \$a_0=12\$, we get \$a_1=15\$, \$a_2=15\$,..., \$a_8=36\$, \$a_9=6\$, \$a_{10}=9\$, \$a_{11}=3\$ and the pattern \$3,6,9\$ repeats.\n\nWhen this pattern \$3,6,9\$ repeats, this means that there exists a number \$A\$ such that \$a_n = A\$ for infinitely many values of \$n\$ and that number \$A\$ is either \$3,6,\$ or \$9\$.\n\nWhen we start with any number \$a_0\\not\\equiv 0\\; mod\\; 3\$, we don't see a repeating pattern.\n\nTherefore the claim is that \$a_0=3k\$ where \$k\$ is a positive integer and we need to prove this claim.\n\nWhen we start with \$a_0=3k\$, the next term if it is not a square is \$3k+3\$, then \$3k+6\$ and so on until we get \$3k+3p\$ where \$p\$ is an integer and \$(k+p)=3q^2\$ where \$q\$ is an integer. Then the next term will be \$\\sqrt{9q^2}=3q\$ and the pattern repeats again when \$q=k\$ or when \$q=3\$ or \$6\$.\n\nIn order for these patterns to repeat, any square in the sequence need to be a multiple of 3.\n\nTo try the other two cases where \$a_0\\not\\equiv 0\\; mod\\; 3\$, we can try \$a_0=3k\\pm 1\$ then the next terms will be in the form \$3k+3p\\pm 1 = 3(k+p) \\pm 1\$.\n\nWhen \$3(k+p) \\pm 1\$ is a square, it will not be a multiple of \$3\$ because \$3(k+p) \\pm 1\$ is not a multiple of \$3\$ and \$3(k+p) \\pm 1 \\ne 9q^2\$ because \$3(k+p) \\pm 1 \\equiv \\pm 1\\; mod\\; 3\$ and \$q^2\$ would have to be \$\\frac{(k+p)}{3} \\pm \\frac{1}{9}\$ which is not an integer even if \$k+p\$ is a multiple of \$3\$.\n\nTherefore the pattern doesn't repeat for any of the other cases where \$a_0=3k\\pm 1\$ and only repeats when \$a_0\\equiv 0\\; mod\\; 3\$\n\nSo, the answer to this problem is \$a_0=3k\\;\\forall k \\in \\mathbb{Z}^{+}\$ and \$A=3,6,\$ or \$9\$.\n\n~Tomas Diaz. orders@tomasdiaz.com" ]
IMO-2017-2	https://artofproblemsolving.com/wiki/index.php/2017_IMO_Problems/Problem_2	Let $\mathbb{R}$ be the set of real numbers , determine all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for any real numbers $x$ and $y$ \[ f(f(x)f(y)) + f(x+y)=f(xy) \]	[ "Looking at the equation one can deduce that the functions that will work will be linear. That is, a polynomial of at most a degree of 1.\n\nThus, \$f\$ is in the form \$f(x)=mx+b\$\n\nTherefore,\n\n\\[\nf((mx+b)(my+b))+m(x+y)+b=mxy+b\n\\]\n\n\\[\nf(m^2xy+mb(x+y)+b^2)+m(x+y)+b=mxy+b\n\\]\n\n\\[\nm(m^2xy+mb(x+y)+b^2)+b+m(x+y)+b=mxy+b\n\\]\n\n\\[\nm^3xy+m^2b(x+y)+mb^2+m(x+y)+b=mxy\n\\]\n\n\\[\nm(m^2-1)xy+m(bm+1)(x+y)+b(bm+1)=0\n\\]\n\nTherefore,\n\n\$m(m^2-1)=0\$ [Equation 1]\n\n\$m(bm+1)=0\$ [Equation 2]\n\n\$b(bm+1)\$ [Equation 3]\n\nFrom [Equation 1] we have, \$m=0,\\pm 1\$\n\nFrom [Equation 2] we have, \$m=0, bm=-1\\pm 1\$\n\nFrom [Equation 3] we have, \$b=0, bm=-1\\pm 1\$\n\nWhen \$m=0\$, \$b=0\$, then \$f(x)=0\$\n\nWhen \$bm=-1\$, \$b=\\frac{-1}{m}\$, then since \$m=\\pm 1\$, then \$b=\\mp 1\$\n\nWhen \$bm=-1\$, \$b=\\frac{-1}{m}\$, then since \$m=\\pm 1\$, then \$b=\\mp 1\$ then \$f(x)=\\pm x \\mp1\$ which gives these two functions:\n\n\$f(x)=x-1\$ and \$f(x)=1-x\$, which with \$f(x)=0\$ provide all the three functions for this problem.\n\n~Tomas Diaz. orders@tomasdiaz.com" ]
IMO-2017-3	https://artofproblemsolving.com/wiki/index.php/2017_IMO_Problems/Problem_3	A hunter and an invisible rabbit play a game in the Euclidean plane. The rabbit's starting point, $A_0$, and the hunter's starting point, $B_0$, are the same. After $n-1$ rounds of the game, the rabbit is at point $A_{n-1}$ and the hunter is at point $B_{n-1}$. In the nth round of the game, three things occur in order. (i) The rabbit moves invisibly to a point $A_n$ such that the distance between $A_{n-1}$ and $A_n$ is exactly 1. (ii) A tracking device reports a point $P_n$ to the hunter. The only guarantee provided by the tracking device is that the distance between $P_n$ and $A_n$ is at most 1. (iii) The hunter moves visibly to a point $B_n$ such that the distance between $B_{n-1}$ and $B_n$ is exactly 1. Is it always possible, no matter how the rabbit moves, and no matter what points are reported by the tracking device, for the hunter to choose her moves so that after $10^9$ rounds she can ensure that the distance between her and the rabbit is at most 100?	[ "Answer: No. There is no such strategy for the hunter. The rabbit will always “Win”\n\nProof: Suppose on the contrary that the answer is Yes. Therefore, there exists a strategy for the hunter to always “win” no matter how the rabbit moved or how the radar pinged. We will show that with bad luck from radar pings, the hunter cannot guarantee that the distance stays below \$100\$ after \$10^9\$ moves.\n\nLet \$d_n\$ be the distance be the distance between the hunter and the rabbit after \$n\$ rounds. If \$d_n \\geq 100, n < 10^9,\$ the rabbit has won as all it needs to do is to move straight away from the hunter and the distance between the two will be kept at or above \$100\$ thereon. Now we tackle the other case, \$d_n \\le 100\$. We will show that whatever strategy the hunter follows, after \$200\$ rounds, the rabbit can increase \$d_n^2\$ by at least \$\\frac{1}{2}\$ with lucky radar pings. This way, \$d_n^2\$ will reach \$10^4\$ in less than \$2\\cdot 200 \\cdot 10^4 = 4 \\cdot 10^6 < 10^9\$ rounds, in which the rabbit wins. Suppose the hunter is at \$H_n\$ and the rabbit is at \$R_n\$. Suppose the rabbit \$\\textit{reveals}\$ its location (this allow us to ignore all information from previous radar pings). Let \$r\$ be the line \$H_n R_n,\$ and let \$Y_1\$ and \$Y_2\$ be points which are \$1\$ unit away from \$r\$ and \$200\$ units away from \$R_n\$, as in the figure below.\n\nThe rabbit’s plan is to simply choose one of the points \$Y_1\$ or \$Y_2\$ and hop \$200\$ rounds straight towards it. Since all hops stay within \$1\$ distance unit from \$r\$, it is possible that all radar pings stay on \$r\$. In particular, in this case, the hunter has no way of determining whether the rabbit is heading for \$Y_1\$ or \$Y_2\$.\n\nLooking at these pings, what can the hunter do? His best strategy is to go \$200\$ rounds straight to the right, ending up at point \$H’\$ in the figure because the hunter will always be to the left of \$H’\$ after the \$200\$ rounds, and if the hunter is above \$r\$, then he will be further away from \$Y_2\$, and if he is below \$r\$, then he will be further away from \$Y_1\$. In short, he can never be sure that the distance from him and the rabbit will be less than \$y = H’Y_1 = H’Y_2\$ after these \$200\$ rounds. To estimate \$y^2\$, we take \$Z\$ as the midpoint of segment \$Y_1Y_2\$, we take \$R’\$ as a point \$200\$ units to the right of \$R_n\$, and define \$\\varepsilon = ZR’\$ (Note that \$H’R’ = d_n\$). Then\n\n\\[\ny^2 = 1 + (H’Z)^2 = 1+(d_n-\\varepsilon)^2\n\\]\n\nWhere\n\n\\[\n\\varepsilon = 200 - R_nZ = 200 - \\sqrt{200^2-1} = \\frac{1}{200+\\sqrt{200^2-1}} > \\frac{1}{400}\n\\]\n\n. In particular, \$\\varepsilon^2 + 1 = 400\\varepsilon\$, so\n\n\\[\ny^2 = d_n^2 - 2\\varepsilon d_n + \\varepsilon^2+1 = d_n^2 + \\varepsilon(400-2d_n)\n\\]\n\nSince \$\\varepsilon > \\frac{1}{400}\$ and we assumed \$d_n < 100\$, this shows that \$y^2 > d_n^2 + \\frac{1}{2}\$. So ,as we claimed, with this list of radar pings, no matter what the hunter does, the rabbit might achieve\n\n\\[\nd_{n+200}^2 > d_n^2 + \\frac{1}{2}\n\\]\n\nThe rabbit wins. ~Archieguan" ]
IMO-2017-4	https://artofproblemsolving.com/wiki/index.php/2017_IMO_Problems/Problem_4	Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$.	[ "We construct inversion which maps \$KT\$ into the circle \$\\omega_1\$ and \$\\Gamma\$ into \$\\Gamma.\$ Than we prove that \$\\omega_1\$ is tangent to \$\\Gamma.\$\n\nQuadrangle \$RJSK\$ is cyclic \$\\implies \\angle RSJ = \\angle RKJ.\$\n\nQuadrangle \$AJST\$ is cyclic \$\\implies \\angle RSJ = \\angle TAJ \\implies AT\|\|RK.\$\n\nWe construct circle \$\\omega\$ centered at \$R\$ which maps \$\\Gamma\$ into \$\\Gamma.\$\n\nLet \$C = \\omega \\cap RT \\implies RC^2 = RS \\cdot RT.\$ Inversion with respect to \$\\omega\$ swap \$T\$ and \$S \\implies \\Gamma\$ maps into \$\\Gamma (\\Gamma = \\Gamma').\$\n\nLet \$O\$ be the center of \$\\Gamma.\$\n\nInversion with respect to \$\\omega\$ maps \$K\$ into \$K'\$. \$K\$ belong \$KT \\implies\$ circle \$K'SR = \\omega_1\$ is the image of \$KT\$. Let \$Q\$ be the center of \$\\omega_1.\$\n\n\$K'T\$ is the image of \$\\Omega\$ at this inversion, \$l = AR\$ is tangent line to \$\\Omega\$ at \$R,\$ so \$K'T\|\|AR.\$\n\n\$K'\$ is image K at this inversion \$\\implies K \\in RK' \\implies RK'\|\|AT \\implies ARK'T\$ is parallelogram.\n\n\$S\$ is the midpoint of \$RT \\implies S\$ is the center of symmetry of \$ATK'R \\implies\$ \$\\triangle RSK'\$ is symmetrical to \$\\triangle TSA\$ with respect to \$S \\implies\$ \$\\omega_1\$ is symmetrical to \$\\Gamma\$ with respect to \$S \\implies\$ \$O\$ is symmetrycal \$Q\$ with respect to \$S.\$\n\n\$S\$ lies on \$\\Gamma\$ and on \$\\omega_1 \\implies \\Gamma\$ is tangent to \$\\omega_1 \\implies\$ line \$KT\$ is tangent to \$\\Gamma.\$\n\nvladimir.shelomovskii@gmail.com, vvsss", "We use the tangent-chord theorem: the angle formed between a chord and a tangent line to a circle is equal to the inscribed angle on the other side of the chord.\n\nQuadrangle \$RJSK\$ is cyclic \$\\implies \\angle RSJ = \\angle RKJ.\$\n\nQuadrangle \$AJST\$ is cyclic \$\\implies \\angle RSJ = \\angle TAJ\$\n\n\\[\n\\implies AT\|\|RK.\n\\]\n\n(One can use Reim’s theorem – it is shorter way.)\n\nLet \$B\$ be symmetric to \$A\$ with respect to \$S \\implies\$ \$ATBR\$ is parallelogram.\n\n\\[\n\\angle KST = \\angle SRK + \\angle SKR = \\angle KRA\n\\]\n\n\$\\angle RBT = \\angle RAT \\implies \\angle KST + \\angle KBT = 180^\\circ\$ \$\\implies SKBT\$ is cyclic.\n\n\\[\n\\angle SBK = \\angle STK = \\angle SAT \\implies\n\\]\n\nInscribed angle of \$\\Gamma (\\angle SAT)\$ is equal to angle between \$KT\$ and chord \$ST \\implies\$\n\n\$KT\$ is tangent to \$\\Gamma\$ by the inverse of tangent-chord theorem.\n\nvladimir.shelomovskii@gmail.com, vvsss" ]
IMO-2017-5	https://artofproblemsolving.com/wiki/index.php/2017_IMO_Problems/Problem_5	An integer $N \ge 2$ is given. A collection of $N(N + 1)$ soccer players, no two of whom are of the same height, stand in a row. Sir Alex wants to remove $N(N - 1)$ players from this row leaving a new row of $2N$ players in which the following $N$ conditions hold: ($1$) no one stands between the two tallest players, ($2$) no one stands between the third and fourth tallest players, \[ \;\;\vdots \] ($N$) no one stands between the two shortest players. Show that this is always possible.	[ "The answer is \$\\boxed{2016}\$. Clearly, if we erase less than \$2016\$ terms, then some term will appear on both sides by the pigeonhole principle, thereby causing a real root. Now, we show how we can erase \$2016\$ terms and have no real roots.\n\nLet \$f(x)=(x-1)(x-4)\$ and \$g(x)=(x-2)(x-3)\$. It is not hard to see that we can erase \$2016\$ terms to get the equation\n\n\\[\nf(x)f(x-4)\\cdots f(x-2012) = g(x)g(x-4)\\cdots g(x-2012).\n\\]\n\nWe now show this has no real solutions. Clearly, none of \$1,2,\\ldots,2016\$ are solutions, since plugging them in causes one side to be \$0\$ and one side to not be \$0\$. Therefore, letting \$h(x)=g(x)/f(x)\$, our solution must satisfy\n\n\\[\nS:=h(x)h(x-4)\\cdots h(x-2012)=1.\n\\]\n\nIt is not hard to see that \$h(x)>1\$ for \$x<1\$ and \$x>4\$, and the maximum value of \$h\$ in \$(1,4)\$ is \$1/9\$. Now, if our root \$x\$ is an interval of the form \$(4k,4k+1)\$, then all the \$h(x-4j)\$ values are bigger than \$1\$, which can't be. Thus, we have that \$x\\in(4k+1,4k+4)\$ for some \$k\$. Also, its not too hard to show that \$h(x-4j)\\le h(4\|k-j\|+1)\$ (we do casework on whether \$k>j\$ or \$k<j\$). But\n\n\\[\n\\alpha(\|k-j\|):=h(4\|k-j\|+1) = 1 + \\frac{1}{2\|k-j\|(4\|k-j\|-3)}.\n\\]\n\nWe see that \$\\alpha\$ is a decreasing function, so the product of all the \$h(x-4j)\$ terms besides \$h(x-4k)\$ is at most\n\n\\[\n\\alpha(1)^2\\alpha(2)^2\\cdots\\alpha(1007)^2\\alpha(1008),\n\\]\n\nso the entire product \$S\$ is at most\n\n\\[\n\\frac{1}{9}\\alpha(1)^2\\alpha(2)^2\\cdots\\alpha(1007)^2\\alpha(1008).\n\\]\n\nIt suffices to show that this is less than \$1\$. Note that \$\\log(1+y)<y\$ for all positive \$y\$, so we have that\n\n\\[\n\\log(\\alpha(x)) < \\frac{1}{2x(4x-3)}=:\\beta(x).\n\\]\n\nNote that \\begin{align} \\log S &\\le -\\log 9 + 2\\sum_{x=1}^{1008}\\log(\\alpha(x)) \\\\ &< -\\log 9 + \\sum_{x=1}^{2018}\\frac{1}{x(4x-3)} \\\\ &< -\\log 9 + 1+\\sum_{x=2}^{\\infty}\\frac{1}{x(4x-4)} \\\\ &= -\\log 9 + 1 + \\frac{1}{4} \\\\ &= -\\log 9 + 5/4, \\end{align} so \$S<\\frac{1}{9}e^{5/4}<1\$. Therefore, \$S\\not=1\$, so there are no roots, as desired" ]
IMO-2017-6	https://artofproblemsolving.com/wiki/index.php/2017_IMO_Problems/Problem_6	An ordered pair $(x, y)$ of integers is a primitive point if the greatest common divisor of $x$ and $y$ is $1$. Given a finite set $S$ of primitive points, prove that there exist a positive integer $n$ and integers $a_0, a_1, \ldots , a_n$ such that, for each $(x, y)$ in $S$, we have: \[ a_0x^n + a_1x^{n-1} y + a_2x^{n-2}y^2 + \cdots + a_{n-1}xy^{n-1} + a_ny^n = 1. \]	[ "The proof goes by induction in the number \$n=\|S\|\$ of points of the set. The base case is trivial by Bézout's Theorem. Write \$S =\\{(a_1,b_1)., \\ldots, (a_n,b_n), (a_{n_1},b_{n+1})\\}\$ and let \$g(x,y)\$ be a homogeneous polynomial of degree \$\\ell\$ such that \$g(a_i,b_i)=1\$ for \$1 \\leq i \\leq n\$ (by the induction hypothesis). We will construct a homogeneous polynomial \$f(x,y)\$ of the form\n\n\\[\nf(x,y)=g(x,y)^k + \\prod_{i=1}^n (b_ix - a_iy) \\cdot h(x,y),\n\\]\n\nwhere \$k\$ and \$h(x,y)\$ will be chosen so that the conditions are fulfilled. Note that \$f(a_i,b_i) = 1\$ for \$1 \\leq i \\leq n\$, so we need to ensure that \$f\$ is a homogeneous polynomial and that \$f(a_{n+1},b_{n+1}) = 1\$. We need that\n\n\\[\n1 = f(a_{n+1},b_{n+1}) = g(a_{n+1},b_{n+1})^k + \\prod_{i=1}^n (b_ia_{n+1} - a_ib_{n+1})\\cdot h(a_{n+1},b_{n+1}).\n\\]\n\nIf \$\\gcd\\left(\\prod_{i=1}^n (b_ia_{n+1} - a_ib_{n+1}),g(a_{n+1},b_{n+1})\\right) = 1\$, we can use Euler-Fermat's Theorem to guarantee that \$h(a_{n+1},b_{n+1})\$ is an integer. If not, there would be a prime \$p\$ such that \$p\$ divides \$g(a_{n+1},b_{n+1})\$ and wlog \$b_1a_{n+1}-a_1b_{n+1}\$. Working modulo \$p\$, we have that\n\n\\[\n0 \\equiv b_1^{\\ell}g(a_{n+1},b_{n+1}) \\equiv g(b_1a_{n+1},b_1b_{n+1})\\equiv g(a_1b_{n+1},b_1b_{n+1}) \\equiv b_{n+1}^{\\ell}g(a_1,b_1) \\equiv b_{n+1}^{\\ell}\\mod p.\n\\]\n\nAnalogously, we have that \$a_{n+1}^{\\ell} \\equiv 0\$, which is a contradiction, because \$(a_{n+1},b_{n+1})\$ is a primitive point. In order to finish, it's enough to prove that for any integer \$d\\ge 0\$ and any integer \$M\$, there is a homogeneous polynomial \$T(x,y)\$ of degree \$d\$ such that \$T(a_{n+1},b_{n+1}) = M\$. For this, just take \$T(x,y) = M(\\alpha x + \\beta y)^d\$, where \$\\alpha\$ and \$\\beta\$ are integers such that \$\\alpha a_{n+1} + \\beta b_{n+1} = 1\$." ]
IMO-2018-1	https://artofproblemsolving.com/wiki/index.php/2018_IMO_Problems/Problem_1	Let $\Gamma$ be the circumcircle of acute triangle $ABC$. Points $D$ and $E$ are on segments $AB$ and $AC$ respectively such that $AD = AE$. The perpendicular bisectors of $BD$ and $CE$ intersect minor arcs $AB$ and $AC$ of $\Gamma$ at points $F$ and $G$ respectively. Prove that lines $DE$ and $FG$ are either parallel or they are the same line.	[ "http://wiki-images.artofproblemsolving.com/5/5d/FB_IMG_1531446409131.jpg\n\nThe diagram is certainly not to scale, but the argument is sound (I believe) and involves re-ordering the construction as specified in the original problem so that an identical state of affairs results, yet in so doing differently it is made clear that the line segments in question are parallel.\n\nConstruct a right-angled triangle ABC'. Select an arbitrary point H along the segment BC', and from point H select an arbitrary point F such that the segment HF is perpendicular to the segment AB. Mark the distance from the intersection of HF and AB to B at B\" (i.e., HF is a perpendicular bisector). It follows that the triangle B\"FB is isosceles. Construct an isosceles triangle FHG. Mark the distance of AB\" along AC' at C\". (From here, a circle can be constructed according to the sets of points A, B, F, and A, B, G. Points F and G may be repositioned to allow for these circles to coincide; also, point H may be repositioned so that point C falls on the coinciding circle, understood that HG is the other perpendicular bisector.)\n\nAssign the angle BAC the value α. Hence, the angle FHG has the value 180°-α, and the angle HFG (also, HGF) has the value α/2. Assign the angle BFH the value β. Hence, the angle B'FB\" has the value β-α/2. Consequently, the angle FB'B\" has the value 180°-(β-α/2)-(90°-β) = 90°+α/2, and so too its vertical angle BB'G. As the triangle B\"AC\" is isosceles, and its subtended angle has the value α, the angles BB\"C\" and CC\"B\" both have the value 90°+α/2. It follows therefore that segments B\"C\" and FG are parallel.\n\n(N.B. Points D and E, as given in the wording of the original problem, have been renamed B\" and C\" here.)", "The essence of the proof is using a rhombus formed by the perpendicular bisectors of the segments \$BD, CE, AB\$ and \$AC\$ and the parallelism of its diagonal and the base of the triangle formed by the perpendiculars from one vertex of the rhombus.\n\nThe perpendicular bisectors of the segments \$BD, CE, AB\$ and \$AC\$ intersect at points \$O\$ (the circumcenter \$ABC), H, H',\$ and \$Q.\$ Let \$AD = AE = 2x, AC = 2b.\$ The distance from the line \$HO\$ to point \$C\$ is \$b,\$ from \$QH'\$ to \$C\$ is \$b - x.\$ Therefore, the distance between lines \$HO\$ and \$QH'\$ is \$x.\$ Similarly, the distance between the lines \$HQ\$ and \$OH'\$ is \$x.\$\n\nThe quadrilateral \$OHQH'\$ is formed by the intersection of two pairs of equidistant lines \$\\implies\$ \$OHQH'\$ is a parallelogram with equal heights \$\\implies\$ \$OHQH'\$ is a rhombus.\n\nLet \$I\$ and \$I'\$ be the feet of the heights of the rhombus from the vertex \$O.\$ In isosceles triangles \$\\triangle ADE\$ and \$\\triangle OII'\$ the sides are parallel, hence the bases of these triangles are parallel, \$DE\|\|II'.\$ The feet of the heights \$I\$ and \$I'\$ divide the sides of the rhombus in the same ratio, which means that the diagonal \$HH'\$ of the rhombus is parallel to the segments \$II' \|\| DE.\$\n\nThe angles between the diagonal of the rhombus and its sides are the same, so \$\\angle HQO = \\angle H'QO.\$ The lines \$QF\$ and \$QG\$ are symmetrical with respect to the diameter \$OQ,\$ so \$QF = QG, QF' = QG'.\$ The homothety of triangles \$QHH', QFG,\$ and \$QF'G'\$ centered at \$Q\$ implies \$FG \|\| F'G' \|\| HH'.\$ Therefore, these three lines are parallel to \$DE.\$\n\nNote that triangle \$ABC\$ may be obtuse.\n\nvladimir.shelomovskii@gmail.com, vvsss" ]
IMO-2018-2	https://artofproblemsolving.com/wiki/index.php/2018_IMO_Problems/Problem_2	Find all numbers $n \ge 3$ for which there exists real numbers $a_1, a_2, ..., a_{n+2}$ satisfying $a_{n+1} = a_1, a_{n+2} = a_2$ and \[ a_{i}a_{i+1} + 1 = a_{i+2} \] for $i = 1, 2, ..., n.$	[ "We find at least one series of real numbers for \$n = 3,\$ for each \$n = 3k\$ and we prove that if \$n = 3k \\pm 1,\$ then the series does not exist.\n\nCase 1\n\nLet \$n = 3.\$ We get system of equations\n\n\\[\n\\begin{cases} a_1 a_2 + 1 = a_3 \\\\a_2 a_3 + 1 = a_1 \\\\a_3 a_1 + 1 = a_2 \\end{cases}\n\\]\n\nWe subtract the first equation from the second and get:\n\n\\[\na_2 (a_3 – a_1) = (a_1 – a_3).\n\\]\n\nSo \$a_2 = – 1 \\implies a_1 = 2, a_3 = – 1.\$\n\nCase 1a\n\nLet \$n = 3k, k={1,2,...}.\$ Real numbers \$a_1 =a_4 =...=2, a_2 = a_3 = a_5=...=-1\$ satisfying \$a_{n+1} = a_1, a_{n+2} = a_2\$ and \$a_{i}a_{i+1} + 1 = a_{i+2}\$.\n\nCase 2\n\nLet \$n = 4.\$ We get system of equations\n\n\\[\n\\begin{cases} a_1 a_2 + 1 = a_3 \\\\a_2 a_3 + 1 = a_4 \\\\a_3 a_4 + 1 = a_1\\\\a_4 a_1 + 1 = a_2 \\end{cases}\n\\]\n\nWe multiply each equation by the number on the right-hand side and get:\n\n\\[\n\\begin{cases} a_1 a_2 a_3 + a_3 = a_3^2 \\\\a_2 a_3 a_4 + a_4 = a_4^2 \\\\a_3 a_4 a_1 + a_1 = a_1^2 \\\\a_4 a_1 a_2 + a_2 = a_2^2 \\end{cases}\n\\]\n\nWe multiply each equation by a number that precedes a pair of product numbers in a given sequence \$a_1, a_2, a_3, a_4, a_1, a_2.\$ So we multiply the equation with product \$a_1 a_2\$ by \$a_4\$, we multiply the equation with product \$a_4 a_1\$ by \$a_3\$ etc. We get:\n\n\\[\n\\begin{cases} a_1 a_2 a_4 + a_4 = a_3 a_4 \\\\a_2 a_3 a_1 + a_1 = a_4 a_1 \\\\a_3 a_4 a_2 + a_2 = a_1 a_2 \\\\a_4 a_1 a_3 + a_3 = a_2 a_3 \\end{cases}\n\\]\n\nWe add all the equations of the first system, and all the equations of the second system. The sum of the left parts are the same! It includes the sum of all the numbers \$a_i\$ and the sum of the triples of consecutive numbers \$a_i a_{i + 1} a_{i + 2}.\$ Hence, the sums of the right parts are equal, that is,\n\n\\[\na_1^2 + a_2^2 + ... + a_4^2 – a_1 a_2 – a_2 a_3 – a_3 a_4 – a_1 a_4 = 0.\n\\]\n\nIt is known that this expression is doubled\n\n\\[\n(a_1 – a_2)^2 + (a_2 – a_3)^2 + ... + (a_4 – a_1)^2 = 0 \\implies a_1 = a_2 = a_3 = a_4 .\n\\]\n\nSubstituting into any of the initial equations, we obtain the equation \$a_1^2 + 1 = a_1,\$ which does not have real roots. Hence, there are no such real numbers.\n\nCase 2a\n\nLet \$n = 5.\$ We get system of equations\n\n\\[\n\\begin{cases} a_1 a_2 + 1 = a_3 \\\\a_2 a_3 + 1 = a_4 \\\\a_3 a_4 + 1 = a_5\\\\a_4 a_5 + 1 = a_1 \\\\a_5 a_1 + 1 = a_2\\end{cases}\n\\]\n\nWe repeat all steps of Case 2 and get: there are no such real numbers.\n\nCase 2b\n\nLet \$n = 3k \\pm 1.\$ We repeat all steps of cases \$2\$ and \$2a\$ and get: there are no such real numbers.\n\nvladimir.shelomovskii@gmail.com, vvsss" ]
IMO-2018-3	https://artofproblemsolving.com/wiki/index.php/2018_IMO_Problems/Problem_3	An anti-Pascal triangle is an equilateral triangular array of numbers such that, except for the numbers in the bottom row, each number is the absolute value of the difference of the two numbers immediately below it. For example, the following is an anti-Pascal triangle with four rows which contains every integer from $1$ to $10$ \[ 4 \] \[ 2\quad 6 \] \[ 5\quad 7 \quad 1 \] \[ 8\quad 3 \quad 10 \quad 9 \] Does there exist an anti-Pascal triangle with $2018$ rows which contains every integer from $1$ to $1 + 2 + 3 + \dots + 2018$?	[ "Trivially it is required that every positive integer from \$1\$ to \$1+2+3+\\cdots+2018\$ appears exactly once.\n\nLet \$M_n\$ denote the maximum number in the \$n\$th row and let \$m_n\$ denote the minimum number in the \$n\$th row.\n\nNow assume \$n\\leq 2017\$ and consider the numbers directly below \$M_n\$. Call these \$a\$ and \$b\$ where w.l.o.g. \$a>b\$. Then \$a-b=M_n\$. Since \$a\\leq M_{n+1}\$ and \$b\\geq m_{n+1}\$, we obtain that \$M_{n+1}\\geq M_n+m_{n+1}\$.\n\nThus, for \$1\\leq i<j\\leq 2018\$,\n\n\\[\nM_j\\geq M_i+\\sum_{k=i+1}^j m_i\n\\]\n\nIn particular, since \$M_1=m_1\$,\n\n\\[\nM_{2018}\\geq \\sum_{k=1}^{2018} m_k\n\\]\n\nThus \$M_{2018}\$ is a sum of 2018 distinct positive integers. However, \$M_{2018}\\leq 1+2+3+\\cdots+2018\$. Thus \$M_{2018}=1+2+3+\\cdots+2018\$ and \$\\{m_1,m_2,m_3,\\dots,m_{2018}\\}\$ is a permutation of \$\\{1,2,3,\\dots,2018\\}\$. Also, this implies that the other inequalities are also equalities and, for any \$1\\leq j\\leq 2018\$,\n\n\\[\nM_j=\\sum_{k=1}^j m_k\n\\]\n\nNow let any positive integer \$n\\leq 2018\$ be \"small\" and let any positive integer \$1+2+3+\\cdots+2017\\leq n\\leq 1+2+3+\\cdots+2018\$ be \"large\". Since \$\\{m_1,m_2,m_3,\\dots,m_{2018}\\}\$ is a permutation of \$\\{1,2,3,\\dots,2018\\}\$, there is exactly one small number in each row.\n\nIf \$n\\leq 1954\$, we have\n\n\\[\nM_n=\\sum_{k=1}^n m_k \\leq 2018+2017+2016+\\cdots+65\n\\]\n\n\\[\n=(1+2+3+\\cdots+2018)-(1+2+3+\\cdots+64)\n\\]\n\n\\[\n=(1+2+3+\\cdots+2018)-2080\n\\]\n\n\\[\n<1+2+3+\\cdots+2017\n\\]\n\nso the \$n\$th row cannot contain any large numbers.\n\nIf \$1955\\leq n\\leq 2017\$, let \$l\$ be a large number in the \$n\$th row. Let the numbers directly below \$l\$ be \$a\$ and \$b\$ where w.l.o.g. \$a>b\$. We have \$b=a-l\$ and \$a\\leq 1+2+3+\\cdots+2018\$ so, because \$l\$ is large, \$b\\leq 2018\$ so \$b\$ is small. Thus \$b=m_{n+1}\$ so \$l\$ is directly above \$m_{n+1}\$. Thus there are at most 2 large numbers in the \$n\$th row.\n\nThus there are at most 126 large numbers outside the bottom row. Since there are 2019 large numbers, there are at least 1893 large numbers in the bottom row so at most 125 non-large numbers in the bottom row. Now there are 2017 pairs of adjacent large numbers in the bottom row. We remove the pair directly beneath \$m_{2017}\$ and at most 250 other pairs containing a non-large number. Thus we can find a pair of adjacent large numbers in the bottom row, not directly beneath \$m_{2017}\$. However, their difference is small and in the 2017th row but not \$m_{2017}\$, which is a contradiction. Thus there is no such anti-Pascal triangle." ]
IMO-2018-4	https://artofproblemsolving.com/wiki/index.php/2018_IMO_Problems/Problem_4	A site is any point $(x, y)$ in the plane such that $x$ and $y$ are both positive integers less than or equal to 20. Initially, each of the 400 sites is unoccupied. Amy and Ben take turns placing stones with Amy going first. On her turn, Amy places a new red stone on an unoccupied site such that the distance between any two sites occupied by red stones is not equal to $\sqrt{5}$. On his turn, Ben places a new blue stone on any unoccupied site. (A site occupied by a blue stone is allowed to be at any distance from any other occupied site.) They stop as soon as a player cannot place a stone. Find the greatest $K$ such that Amy can ensure that she places at least $K$ red stones, no matter how Ben places his blue stones.	[ "The maximal K is 100. Amy can reach at least 100 by playing only on sites for which x+y is even. There are 200 such sites, none are of distance \$\\sqrt{5}\$ from each other and Ben can occupy at most half of them. On the other hand Ben can prevent Amy from reaching more than 100 using the following strategy: Picture the sites as a 20 by 20 board and divide it into 25 non overlapping 4-by-4 squares. We label each site in the square as follows:\n\n\\[\n1, 2, 3, 4\n\\]\n\n\\[\n5, 6, 7, 8\n\\]\n\n\\[\n8, 7, 6, 5\n\\]\n\n\\[\n4, 3, 2, 1\n\\]\n\nWhenever Amy plays in a square Ben plays in the same square and in the site with the same label. In each square Amy can place at most 2 stones in sites labeled 1,4,6,7 (no three sites with labels from this set are free from distance \$\\sqrt{5}\$ and Amy can play one stone on each label since Ben plays the other). Likewise for the sites labeled 2,3,5,8. So in total Amy can place at most 4 stones in each of the 25 squares for a total of 100 stones." ]
IMO-2018-5	https://artofproblemsolving.com/wiki/index.php/2018_IMO_Problems/Problem_5	Let $a_1, a_2, \dots$ be an infinite sequence of positive integers. Suppose that there is an integer$N > 1$ such that, for each $n \geq N$, the number $\frac{a_1}{a_2}+\frac{a_2}{a_3}+\dots +\frac{a_{n-1}}{a_n}+\frac{a_n}{a_1}$ is an integer. Prove that there is a positive integer $M$ such that $a_m = a_{m+1}$ for all $m \geq M.$	[ "The condition implies that the difference \$S(n) = \\frac{a_{n+1}}{a_1} - \\frac{a_n}{a_1} + \\frac{a_n}{a_{n+1}}\$ is an integer for all \$n > N\$. We proceed by \$p\$-adic valuation only henceforth.\n\n[b][color=red]Claim:[/color][/b] If \$p \\nmid a_1\$, then \$\\nu_p(a_{n+1}) \\le \\nu_p(a_n)\$ for \$n \\ge N\$.\n\n[i]Proof.[/i] The first two terms of \$S(n)\$ have nonnegative \$\\nu_p\$, so we need \$\\nu_p(\\frac{a_n}{a_{n+1}}) \\ge 0\$. \$\\blacksquare\$\n\n[b][color=red]Claim:[/color][/b] If \$p \\mid a_1\$, then \$\\nu_p(a_n)\$ is eventually constant.\n\n[i]Proof.[/i] By hypothesis \$\\nu_p(a_1) > 0\$. We consider two cases. [list] []First assume \$\\nu_p(a_k) \\ge \\nu_p(a_1)\$ for some \$k > N\$. We claim that for any \$n \\ge k\$ we have:\n\n\\[\n\\nu_p(a_1) \\le \\nu_p(a_{n+1}) \\le \\nu_p(a_n).\n\\]\n\nThis is just by induction on \$n\$; from \$\\nu(\\frac{a_n}{a_1}) \\ge 0\$, we have\n\n\\[\n\\nu_p\\left( \\frac{a_{n+1}}{a_1} + \\frac{a_n}{a_{n+1}} \\right) \\ge 0\n\\]\n\nwhich implies the displayed inequality (since otherwise exactly one term of \$S(n)\$ has nonnegative \$\\nu_p\$). Thus once we reach this case, \$\\nu_p(a_n)\$ is monotic but bounded below by \$\\nu_p(a_1)\$, and so it is eventually constant.\n\n[]Now assume \$\\nu_p(a_k) < \\nu_p(a_1)\$ for every \$k > N\$. Take any \$n > N\$ then. We have \$\\nu_p\\left(\\frac{a_{n+1}}{a_1}\\right) < 0\$, and also \$\\nu_p\\left(\\frac{a_n}{a_1}\\right) < 0\$, so among the three terms of \$S(n)\$, two must have equal \$p\$-adic valuation. We consider all three possibilities: \\begin{align} \\nu_p\\left(\\frac{a_{n+1}}{a_1}\\right) = \\nu_p\\left(\\frac{a_n}{a_1}\\right) &\\implies \\boxed{\\nu_p(a_{n+1}) = \\nu_p(a_{n})} \\\\ \\nu_p\\left(\\frac{a_{n+1}}{a_1}\\right) = \\nu_p\\left(\\frac{a_n}{a_{n+1}}\\right) &\\implies \\boxed{\\nu_p(a_{n+1}) = \\frac{\\nu_p(a_n) + \\nu_p(a_1)}{2}} \\\\ \\nu_p\\left(\\frac{a_{n}}{a_1}\\right) = \\nu_p\\left(\\frac{a_n}{a_{n+1}}\\right) &\\implies \\nu_p(a_{n+1}) = \\nu_p(a_1),\\text{ but this is impossible}. \\end{align} Thus, \$\\nu_p(a_{n+1}) \\ge \\nu_p(a_n)\$ and \$\\nu_p(a_n)\$ is bounded above by \$\\nu_p(a_1)\$. So in this case we must also stabilize. \\qedhere [/list] \$\\blacksquare\$\n\nSince the latter claim is applied only to finitely many primes, after some time \$\\nu_p(a_n)\$ is fixed for all \$p \\mid a_1\$. Afterwards, the sequence satisfies \$a_{n+1} \\mid a_n\$ for each \$n\$, and thus must be eventually constant.\n\n[b][color=red]Remark:[/color][/b] This solution is almost completely \$p\$-adic, in the sense that I think a similar result if one replaces \$a_n \\in {\\mathbb Z}\$ by \$a_n \\in {\\mathbb Z}_p\$ for any particular prime \$p\$. In other words, the primes almost do not talk to each other.\n\nThere is one caveat: if \$x_n\$ is an integer sequence such that \$\\nu_p(x_n)\$ is eventually constant for each prime then \$x_n\$ may not be constant. For example, take \$x_n\$ to be the \$n\$th prime! That's why in the first claim (applied to co-finitely many of the primes), we need the stronger non-decreasing condition, rather than just eventually constant" ]
IMO-2018-6	https://artofproblemsolving.com/wiki/index.php/2018_IMO_Problems/Problem_6	A convex quadrilateral $ABCD$ satisfies $AB\cdot CD=BC \cdot DA.$ Point $X$ lies inside $ABCD$ so that $\angle XAB = \angle XCD$ and $\angle XBC = \angle XDA.$ Prove that $\angle BXA + \angle DXC = 180^{\circ}$	[ "We want to find the point \$X.\$ Let \$E\$ and \$F\$ be the intersection points of \$AB\$ and \$CD,\$ and \$BC\$ and \$DA,\$ respectively. The poinx \$X\$ is inside \$ABCD,\$ so points \$E,A,X,C\$ follow in this order.\n\n\$\\angle XAB = \\angle XCD \\implies \\angle XAE + \\angle XCE = 180^\\circ\$ \$\\implies AXCE\$ is cyclic \$\\implies X\$ lie on circle \$ACE.\$\n\nSimilarly, \$X\$ lie on circle \$BDF.\$\n\nPoint \$X\$ is the point of intersection of circles \$ACE\$ and \$\\Omega = BDF.\$\n\nSpecial case\n\nLet \$AD = CD\$ and \$AB = BC \\implies AB \\cdot CD = BC \\cdot DA.\$\n\nThe points \$B\$ and \$D\$ are symmetric with respect to the circle \$\\theta = ACEF\$ (Claim 1).\n\nThe circle \$BDF\$ is orthogonal to the circle \$\\theta\$ (Claim 2).\n\n\$\\hspace{10mm} \\angle FCX = \\angle BCX = \\frac {\\overset{\\Large\\frown} {XAF}}{2}\$ of \$\\theta.\$ \$\\hspace{10mm} \\angle CBX = \\angle XDA = \\frac {\\overset{\\Large\\frown} {XBF}}{2}\$ of \$\\Omega.\$\n\n\$\\overset{\\Large\\frown} {XAF} + \\overset{\\Large\\frown} {XBF} = 180^\\circ\$ (Claim 3) \$\\implies\$ \$\\angle XCB + \\angle XBC = 90^\\circ \\implies \\angle CXB = 90^\\circ.\$\n\nSimilarly, \$\\angle AXD = 90^\\circ \\implies\$\n\n\\[\n\\angle BXA + \\angle DXC = 360^\\circ -\\angle AXD -\\angle CXB = 180^\\circ.\n\\]\n\nCommon case\n\nDenote by \$O\$ the intersection point of \$BD\$ and the perpendicular bisector of \$AC.\$ Let \$\\omega\$ be a circle (red) with center \$O\$ and radius \$OA = R.\$\n\nWe will prove \$\\sin\\angle BXA =\\sin \\angle DXC\$ using point \$Y\$ symmetric to \$X\$ with respect to \$\\omega.\$\n\nThe points \$B\$ and \$D\$ are symmetric with respect to \$\\omega\$ (Claim 1).\n\nThe circles \$BDF\$ and \$BDE\$ are orthogonal to the circle \$\\omega\$ (Claim 2).\n\nCircles \$ACF\$ and \$ACE\$ are symmetric with respect to the circle \$\\omega\$ (Lemma).\n\nDenote by \$Y\$ the point of intersection of circles \$BDF\$ and \$ACF.\$\n\nQuadrangle \$BYDF\$ is cyclic \$\\implies \\angle CBY = \\angle ADY.\$\n\nQuadrangle \$AYCF\$ is cyclic \$\\implies \\angle YAD = \\angle YCB.\$\n\nThe triangles \$\\triangle YAD \\sim \\triangle YCB\$ by two angles, so\n\n\\[\n\\frac {BC}{AD} = \\frac {CY}{AY} = \\frac {BY} {DY} \\hspace{10mm} (1).\n\\]\n\nThe points \$X\$ and \$Y\$ are symmetric with respect to the circle \$\\omega\$, since they lie on the intersection of the circles \$ACF\$ and \$ACE\$ symmetric with respect to \$\\omega\$ and the circle \$BDF\$ orthogonal to \$\\omega.\$\n\nThe point \$B\$ is symmetric to \$D\$ with respect to \$\\omega \\implies\$\n\n\\[\n\\triangle OBC \\sim \\triangle OCD \\implies \\frac {OB}{OC} = \\frac {BC}{CD} = \\frac {OC}{OD},\n\\]\n\n\\[\n\\frac {OB}{OD} = \\frac {OB}{OC} \\cdot \\frac {OC}{OD} = \\frac{BC^2}{CD^2} = \\frac{BC}{CD} \\cdot \\frac {AB}{AD}.\n\\]\n\nThe point \$B\$ is symmetric to \$D\$ and the point \$X\$ is symmetric to \$Y\$ with respect to \$\\omega,\$ hence\n\n\\[\n\\frac {BX}{DY} = \\frac {R^2}{OD \\cdot OY} ,\\frac {DX}{BY} = \\frac{R^2}{OB \\cdot OY}.\n\\]\n\n\\[\n\\frac{BX}{DX} =\\frac{DY}{BY} \\cdot \\frac {OB}{OD} = \\frac{AD}{BC} \\cdot \\frac{BC}{CD} \\cdot \\frac{AB}{AD} = \\frac{AB}{CD}.\n\\]\n\nDenote \$\\angle XAB = \\angle XCD = \\alpha, \\angle BXA = \\varphi, \\angle DXC = \\psi.\$\n\nBy the law of sines for \$\\triangle ABX,\$ we obtain \$\\frac {AB}{\\sin \\varphi} = \\frac{BX}{\\sin \\alpha}.\$\n\nBy the law of sines for \$\\triangle CDX,\$ we obtain \$\\frac {CD}{\\sin \\psi} = \\frac {DX}{\\sin \\alpha}.\$\n\nHence we get \$\\frac{\\sin \\psi} {\\sin \\varphi}= \\frac {CD}{DX} \\cdot \\frac{BX}{AB} = 1.\$\n\nIf \$\\varphi = \\psi,\$ then \$\\triangle XAB \\sim \\triangle XCD \\implies \\frac {CD}{AB} = \\frac {BX}{DX} = \\frac{AX}{CX} = \\frac {AD}{BC}.\$ \$CD \\cdot BC = AB \\cdot AD \\implies AD = CD, AB = BC.\$ This is a special case.\n\nIn all other cases, the equality of the sines follows \$\\varphi + \\psi = 180^\\circ .\$\n\nClaim 1 Let \$A, C,\$ and \$E\$ be arbitrary points on a circle \$\\omega, l\$ be the perpendicular bisector to the segment \$AC.\$ Then the straight lines \$AE\$ and \$CE\$ intersect \$l\$ at the points \$B\$ and \$D,\$ symmetric with respect to \$\\omega.\$\n\nClaim 2 Let points \$B\$ and \$D\$ be symmetric with respect to the circle \$\\omega.\$ Then any circle \$\\Omega\$ passing through these points is orthogonal to \$\\omega.\$\n\nClaim 3 The sum of the arcs between the points of intersection of two perpendicular circles is \$180^\\circ.\$ In the figure they are a blue and red arcs \$\\overset{\\Large\\frown} {CD}, \\alpha + \\beta = 180^\\circ.\$\n\nLemma The opposite sides of the quadrilateral \$ABCD\$ intersect at points \$E\$ and \$F\$ (\$E\$ lies on \$AB\$). The circle \$\\omega\$ centered at the point \$O\$ contains the ends of the diagonal \$AC.\$ The points \$B\$ and \$D\$ are symmetric with respect to the circle \$\\omega\$ (in other words, the inversion with respect to \$\\omega\$ maps \$B\$ into \$D).\$ Then the circles \$ACE\$ and \$ACF\$ are symmetric with respect to \$\\omega.\$\n\nProof We will prove that the point \$G,\$ symmetric to the point \$E\$ with respect to \$\\omega,\$ belongs to the circle \$ACF\$ becouse \$\\angle AGC = \\angle AFC.\$\n\nA circle \$BDE\$ containing points \$B\$ and \$D\$ symmetric with respect to \$\\omega,\$ is orthogonal to \$\\omega\$ (Claim 2) and maps into itself under inversion with respect to the circle \$\\omega.\$ Hence, the point \$E\$ under this inversion passes to some point \$G,\$ of the same circle \$BDE.\$\n\nA straight line \$ABE\$ containing the point \$A\$ of the circle \$\\omega,\$ under inversion with respect to \$\\omega,\$ maps into the circle \$OADG.\$ Hence, the inscribed angles of this circle are equal \$\\angle ADB = \\angle AGE.\$ \$\\angle OCE = \\angle CGE (CE\$ maps into \$CG)\$ and \$\\angle OCE = \\angle BCD (BC\$ maps into \$DC).\$ Consequently, the angles \$\\angle AFC = \\angle ADB – \\angle FBD = \\angle AGE - \\angle CGE = \\angle AGC.\$ These angles subtend the \$\\overset{\\Large\\frown} {AC}\$ of the \$ACF\$ circle, that is, the point \$G,\$ symmetric to the point \$E\$ with respect to \$\\omega,\$ belongs to the circle \$ACF.\$\n\nvladimir.shelomovskii@gmail.com, vvsss" ]
IMO-2019-1	https://artofproblemsolving.com/wiki/index.php/2019_IMO_Problems/Problem_1	Let $\mathbb{Z}$ be the set of integers. Determine all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that, for all integers $a$ and $b$, \[ f(2a) + 2f(b) = f(f(a + b)). \]	[ "The only solutions are \$f(x)=0, 2x+c.\$ For some integer \$c.\$\n\nObviously these work. We prove these are the only linear solutions. Plug \$a=0\$ and \$b=0\$ separately to get that \$f(2x)=2f(x)-f(0).\$ Plug \$(0, a+b)\$ to see \$f(0)+f(a+b)=f(a)+f(b),\$ and subtracting \$2f(0)\$ from both sides shows \$f(a)-f(0)\$ to be additive thus linear by Cauchy since this is on integers. Thus, \$f(a)\$ is linear, and so we are done since it can be easily shown that \$0\$ and \$2x+c\$ are the only linear solutions by plugging \$mx+n\$ into the equation.", "We claim the only solutions are \$f\\equiv0\$ and \$f(x)=2x+c\$ for some integer \$c\$., which obviously work. Plugging in \$(0,n)\$ and \$(1,n-1)\$ give \$f(0)+2f(n)=f(f(n))=f(2)+2f(n-1)\$, so \$f(n)-f(n-1)=\\frac{f(2)-f(0)}2\$. Since this difference is constant and \$f\\colon\\mathbb Z\\rightarrow\\mathbb Z\$, we must have \$f\$ is linear (finite differences or induction). It is easy to see the only linear solutions are those specified above. \$\\blacksquare\$\n\n~ Ezra Guerrero", "We just plug values to find basic properties of the function. If \$a=0\$,\n\n\\[\nf(0)+2f(b)=f(f(b)).\n\\]\n\nWe know that \$f(0)\$ is constant and we can see that \$f(b)\$ is not constant (because it varies depends on the value of b), so let \$f(b)\$ be a variable. This means we are making this a function of f(b). This yields that\n\n\\[\nf((f(b)))=2(f(b))+f(0)\n\\]\n\n. This is a linear function (in terms of f(b))! However, we are not done yet. We have to show that this works. When \$b=0\$ and plugging in our function yields:\n\n\\[\n4a+2f(0)= f(f(a) \\implies 4a+2f(0)= f(2a+f(0)) \\implies 4a+2f(0)=4a+2f(0)\n\\]\n\nand the last part is indeed true so we are good. Don’t forget the 0 function this works also. Let us verify that the 0 function works. This means \$0+0=0\$, which is true and a so called “trivial” solution. Now let us prove this rigorously. Take \$(a,b)\$ when they equal \$(-1,x)\$ and \$(0,x-1)\$ (I chose it that way since they are 1 apart from each other; Really you could have chosen \$(y,x)\$ and \$(y+1,x-1)\$ and that would work). We will show that this is linear or below. Substitute in the equations for both yields\n\n\\[\nf(0)+2f(x-1)=f(f(x-1)) \\qquad \\text{and} \\qquad f(-1)+ 2f(x)=f(f(x-1))\n\\]\n\nEquating and then subtracting yields\n\n\\[\n2(f(x)-f(x-1))=f(0)-f(-1)\n\\]\n\n. From this equation how do we know this is linear? Well treat the f(x) and f(x-1) like variables and dividing by two we can see that \$(f(-1)-f(0))/2\$ is constant since -1 and 0 are fixed values. And since the difference is constant and \$x\$ and \$x-1\$ are one apart this tells you that it is a linear function as the difference is constant, i.e 5x+4 is linear since the first positive x’s yields are 4, 9, 14… and we can see the difference is the same. Therefore we proved that the functions we said are good.\n\nSide note: I am referring to the linear function \$2x+b\$ where \$b \\in \\mathbb{R}\$ and the constant function which is \$0\$.\n\n~EaZ_Shadow" ]
IMO-2019-2	https://artofproblemsolving.com/wiki/index.php/2019_IMO_Problems/Problem_2	In triangle $ABC$, point $A_1$ lies on side $BC$ and point $B_1$ lies on side $AC$. Let $P$ and $Q$ be points on segments $AA_1$ and $BB_1$, respectively, such that $PQ$ is parallel to $AB$. Let $P_1$ be a point on line $PB_1$, such that $B_1$ lies strictly between $P$ and $P_1$, and $\angle PP_1C=\angle BAC$. Similarly, let $Q_1$ be the point on line $QA_1$, such that $A_1$ lies strictly between $Q$ and $Q_1$, and $\angle CQ_1Q=\angle CBA$. Prove that points $P,Q,P_1$, and $Q_1$ are concyclic.	[ "The essence of the proof is to build a circle through the points \$P, Q,\$ and two additional points \$A_0\$ and \$B_0,\$ then we prove that the points \$P_1\$ and \$Q_1\$ lie on the same circle.\n\nWe assume that the intersection point of \$AP\$ and \$BQ\$ lies on the segment \$PA_1.\$ If it lies on segment \$AP,\$ then the proof is the same, but some angles will be replaced with additional ones up to \$180^\\circ\$.\n\nLet the circumcircle of \$\\triangle ABC\$ be \$\\Omega\$. Let \$A_0\$ and \$B_0\$ be the points of intersection of \$AP\$ and \$BQ\$ with \$\\Omega\$. Let \$\\angle BAP = \\delta.\$\n\n\\[\nPQ\|\|AB \\implies \\angle QPA_0 = \\delta.\n\\]\n\n\$\\angle BAP = \\angle BB_0A_0 = \\delta\$ since they intersept the arc \$BA_0\$ of the circle \$\\Omega\$.\n\n\$\\angle QPA_0 = \\angle QB_0A_0 \\implies QPB_0A_0\$ is cyclic (in circle \$\\omega.\$)\n\nLet \$\\angle BAC = \\alpha, \\angle AA_0B_0 = \\varphi.\$\n\n\$\\angle PP_1C = \\alpha, \\angle BB_0C = \\alpha\$ since they intersept the arc \$BC\$ of the circle \$\\Omega.\$ So \$B_0P_1CB_1\$ is cyclic.\n\n\$\\angle ACB_0 = \\angle AA_0B_0 = \\varphi\$ (they intersept the arc \$A_0B_0\$ of the circle \$\\Omega).\$\n\n\$\\angle B_1CB_0 = \\varphi.\$ \$\\angle B_1P_1B_0 = \\angle B_1CB_0 = \\varphi\$ (since they intersept the arc \$B_1B_0\$ of the circle \$B_0P_1CB_1).\$\n\nHence \$\\angle PA_0B_0 = \\angle PP_1B_0 = \\varphi,\$ the point \$P_1\$ lies on \$\\omega.\$\n\nSimilarly, point \$Q_1\$ lies on \$\\omega.\$\n\nvladimir.shelomovskii@gmail.com, vvsss" ]
IMO-2019-3	https://artofproblemsolving.com/wiki/index.php/2019_IMO_Problems/Problem_3	A social network has $2019$ users, some pairs of whom are friends. Whenever user $A$ is friends with user $B$, user $B$ is also friends with user $A$. Events of the following kind may happen repeatedly, one at a time: Three users $A$, $B$, and $C$ such that $A$ is friends with both $B$ and $C$, but $B$ and $C$ are not friends, change their friendship statuses such that $B$ and $C$ are now friends, but $A$ is no longer friends with $B$, and no longer friends with $C$. All other friendship statuses are unchanged. Initially, $1010$ users have $1009$ friends each, and $1009$ users have $1010$ friends each. Prove that there exists a sequence of such events after which each user is friends with at most one other user.	[ "Let \$G\$ be a graph with \$2019\$ vertices representing the users and edges corresponding to their friendships.\n\nWe have the following properties:\n\n1. \$G\$ is connected, since two non-adjacent vertices with no common friends will have at least \$1009+1009=2018\$ vertices neighboring them, which is more than the rest of the vertices \$2019-2=2017\$.\n\n2. \$G\$ has vertices of odd degree, and this property remains invariant after an event. Note that the parity of the degrees of the vertices is invariant after an event.\n\n3. The total number of edges strictly decreases after an event.\n\n4. \$G\$ does not contain all possible edges (it is not a clique), and this property remains invariant after an event, since events decrease the number of edges.\n\nAlgorithm: Repeat the following operations, while possible.\n\n- If there is a cycle, and the smallest cycle is not a triangle, take \$a\$ not in the cycle but adjacent to it, \$b\$ adjacent to it in the cycle, and \$c\$ a adjacent to \$b\$ in the cycle. Apply that event. Note that this preserves Property 1.\n\n- If the smallest cycle is a triangle, let \$K\$ be a maximal clique in \$G\$. By Property 4, \$K\\neq G\$. Take \$a\$ not in \$K\$ but adjacent to it, \$b\$ in \$K\$ adjacent to \$a\$, and \$c\$ in \$K\$ not adjacent to \$a\$. The \$c\$ must exist, since \$K\$ is maximal. Apply the event to \$a,b,c\$. Note that this preserves Property 1.\n\nThis algorithm stops due to Property 3. When it stops, the resulting \$G\$ must still satisfy Property 1 (it is connected). Since no more steps can be applied, then there cannot be cycles. It is a tree.\n\n5. Note that events can be applied to a tree whenever there is a vertex of degree at least 2, and that after the event, we get two trees.\n\nApply events until no more are possible. Due to Property 3, this iteration stops. Due to Property 5, it stops with a forest of trees with \$2\$ or \$1\$ vertices each." ]
IMO-2019-4	https://artofproblemsolving.com/wiki/index.php/2019_IMO_Problems/Problem_4	Find all pairs $(k,n)$ of positive integers such that \[ k!=(2^n-1)(2^n-2)(2^n-4)\dots(2^n-2^{n-1}). \]	[ "\\[\nLHS\n\\]\n\n\$k! = 1\$ (when \$k = 1\$), \$2\$ (when \$k = 2\$), \$6\$ (when \$k = 3\$)\n\n\$RHS = 1\$(when \$n = 1\$), \$6\$ (when \$n = 2\$)\n\nHence, \$(1,1)\$, \$(3,2)\$ satisfy\n\nFor \$k = 2: RHS\$ is strictly increasing, and will never satisfy \$k\$ = 2 for integer n since \$RHS = 6\$ when \$n = 2\$.\n\n\\[\nk!=(2^n-1)(2^n-2)(2^n-4)\\dots(2^n-2^{n-1}) ... (1)\n\\]\n\nIn all solutions, for any prime \$p\$ and positive integer \$N\$, we will denote by\n\n\\[\nv_p(N)\n\\]\n\nthe exponent of the largest power of \$p\$ that divides \$N\$. The right-hand side of \$(1)\$ will be denoted by\n\n\\[\nL_n;\n\\]\n\nthat is,\n\n\\[\nL_n = (2^n-1)(2^n-2)(2^n-4)\\dots(2^n-2^{n-1})\n\\]\n\n\\[\n(2^{1+2+3+\\dots+(n-1)})(2^n-1)(2^{n-1}-1)(2^{n-2}-1)\\dots(2^1-1)\n\\]\n\n=\n\n\\[\nv_2(L_n) = {\\frac{n(n-1)}{2}}\n\\]\n\nOn the other hand,\n\n\\[\nv_2(k!)\n\\]\n\nis expressed by the \$Legendre\$ \$formula\$ as\n\n\\[\nv_2(k!) < \\sum_{i=1}^{\\infty} (\\frac{k}{2^i})) = k\n\\]\n\nThus, \$k! = L_n\$ implies the inequality\n\n\\[\n\\frac{n(n-1)}{2} < k ... (2)\n\\]\n\nIn order to obtain an opposite estimate, observe that\n\n\\[\nL_n = (2^n-1)(2^n-2)(2^n-4)\\dots(2^n-2^{n-1})\n\\]\n\nWe claim that\n\n\\[\n2^{n^2} < (\\frac{n(n-1)}{2})! ... (3)\n\\]\n\nfor all \$n \\geq 6\$\n\nFor \$n \\geq 6\$ the estimate (3) is true because\n\n\\[\n2^{6^2} < (6.9)(10^{10})\n\\]\n\nand\n\n\\[\n(\\frac{n(n-1)}{2})! = 15! > (1.3)(10^{12})\n\\]\n\n~flamewavelight and ~phoenixfire\n\n## Remark\n\nContinuing as above, we realize that \$L_n < 2^{n^{2}}\$. But this is absurd for all \$n \\geq 6\$. So the claim holds. Since we have already checked \$n = 1, 2\$ it remains to check \$n = 3, 4, 5\$ and we see neither of them work so \$(1,1) and (3,2)\$ are the unique two solutions.\n\n~ilikemath247365" ]
IMO-2019-5	https://artofproblemsolving.com/wiki/index.php/2019_IMO_Problems/Problem_5	The Bank of Bath issues coins with an $H$ on one side and a $T$ on the other. Harry has $n$ of these coins arranged in a line from left to right. He repeatedly performs the following operation: If there are exactly $k > 0$ coins showing $H$, then he turns over the $k^{th}$ coin from the left; otherwise, all coins show $T$ and he stops. For example, if $n = 3$ the process starting with the configuration $THT$ would be $THT \rightarrow HHT \rightarrow HTT \rightarrow TTT$, which stops after three operations. (a) Show that, for each initial configuration, Harry stops after a finite number of operations. (b) For each initial configuration $C$, let $L(C)$ be the number of operations before Harry stops. For example, $L(THT) = 3$ and $L(TTT) = 0$. Determine the average value of $L(C)$ over all $2^n$ possible initial configurations $C$.	[ "Don't worry, this is way simpler than it's length may suggest :)\n\nClaim: The expected value of \$L(C)\$ is \$\\frac{n(n+1)}{4}\$.\n\nWe prove parts A and B simultaneously using induction.\n\nBase case: \$n=1\$\n\n\$L(T)=0\$ and \$L(H)=1\$, and these are the only possibilities, so clearly the process terminates and \$E[L(C)] = \\frac{1}{2} = \\frac{12}{4}\$.\n\nInductive step: Assume that for \$n-1\$ coins, the process always terminates, and \$E[L(C)] = \\frac{n(n-1)}{4}\$. Define \$E_{n-1} = E[L(C)]\$ here. Now, for \$n\$ coins, define \$E_{n} = E[L{C}]\$ across all \$2^n\$ configurations. Consider cases on the last coin:\n\nCase 1: Final coin is \$T\$\n\nIf the last coin is \$T\$, then we simply can forget about it, as we can flip it to heads only if we have n heads, which is impossible. This occurs with probablity \$P_{T} = 1/2\$, and will be identical to the general \$n-1\$ case; thus, this process terminates, and we will have an expected number of moves here of \$E_{n-1}\$.\n\nCase 2: Final coin is \$H\$\n\nThe probability that this occurs is \$P_{H} = 1/2\$. The only way we have for the process to terminate now is for all coins to show heads, after which we can progressively flip each coin from right to left (this will take \$n\$ moves). Thus, we now have a new goal for the first \$n-1\$ coins (given the head at the end); get all heads.\n\nWe claim that this process is symmetrical to the original process on \$n-1\$ coins, and thus it terminates and has the expected number of moves \$E_{n-1}\$.\n\nProof: Let us have \$k \\ge 1\$ heads in our group of n coins, of which \$k-1\$ are in the first n-1. We are supposed to flip the \$k\$'th coin from the left in the first n-1 coins, which is equivalent to flipping the \$(n-k)\$'th coin from the right (in the first \$n-1\$ coins, NOT all \$n\$ coins), but there are \$n-k\$ tails in the first \$n-1\$ coins! We now have a new process defined as follows:\n\nConsider a configuration of \$n-1\$ coins with \$x \\ge 1\$ tails, and flip the x'th coin from the right, and keep flipping until all coins are heads.\n\nThis is clearly symmetric to the original process, and will terminate if and only if the original process on \$n-1\$ coins does, and has the same expected number of moves as the original. Thus, the claim is proven.\n\nBack to the case: We now have a process to get to all heads that will terminate with an expected number of steps of \$E_{n-1}\$, and then an n-step algorithm to reach the finish. Thus, here also, the process terminates, and we have a total expected number of steps as \$E_{n-1}+n\$.\n\nThus, the overall process on n coins will terminate, since the process terminates in both the representative cases on n-1 coins. \$\\boxed{Q.E.D}\$ Also, the expected value of steps can be easily computed now; \$E_{n} = (P_{T})(E_{n-1})+(P_{H})(E_{n-1}+n) = E_{n-1}+n/2 = \\frac{n(n-1}{4}+\\frac{n}{2} = \\boxed{\\frac{n(n+1)}{4}}\$ as desired. We are now done.\n\n-Solution by thanosaops", "We define a sequence \$\\{a_i\\}_{i=1}^n\$: \$a_i=1\$ if the \$i\$'th coin is \$H\$ and \$a_i=0\$ if the \$i\$'th coin is \$T\$. Then, \$k\$ will be the number of \$1\$'s in \$\\{a_i\\}\$ and after each operation, \$a_k\$ will be changed to \$1-a_k\$. Let \$f(n)=\\sum L\\{a_i\\}=\$ sum of \$L\\{a_i\\}\$ over all possible \$\\{a_i\\}\$ (of length \$n\$).\n\nWe shall prove that the process terminates and find the value of \$f(n)\$ using strong induction.\n\nBase case: Clearly, for \$n=1\$, the process terminates.\n\nInduction hypothesis: For \$n=1, 2, 3, \\cdots, k\$, the process terminates.\n\nInductive step: We claim that for \$n=k+1\$, the process terminates as well.\n\nCase 1: \$a_1=1\$\n\nThis means that \$\\{a_i\\}=\\{1, a_2, a_3, \\cdots, a_n\\}\$. With the proof of the claim in Solution 1, we can similarly prove that operating on \$\\{a_i\\}\$ is equivalent to operating on \$\\{a_2, a_3, \\cdots, a_n\\}\$.\n\nBy our induction hypothesis, this process will terminate, and hence \$\\{a_i\\}\$ will be changed to \$\\{1, 0, 0, \\cdots, 0\\}\$ at the end of this process. Now, we operate one more time to change \$a_1\$ from \$1\$ to \$0\$. Hence, the process terminates.\n\nWe can also see that \$L\\{a_i\\}=L\\{a_2, a_3, \\cdots, a_n\\}+1\$. Hence, sum of \$L\\{a_i\\}\$ over all possible \$\\{a_i\\}\$ in this case\n\n\\[\n=\\sum L\\{a_i\\}=\\sum \\left[L\\{a_2, a_3, \\cdots, a_n\\}+1\\right]=f(n-1)+2^{n-1}\n\\]\n\nCase 2: \$a_1=0\$ and \$a_n=1\$\n\nThis means that \$\\{a_i\\}=\\{0, a_2, a_3, \\cdots, a_{n-1}, 1\\}\$. Operating on \$\\{a_i\\}\$ is now equivalent to operating on \$\\{a_2, a_3, \\cdots, a_{n-1}\\}\$.\n\nBy our induction hypothesis, this process will terminate, and hence \$\\{a_i\\}\$ will be changed to \$\\{0, 0, 0, \\cdots, 0, 1\\}\$ at the end of this process. Now, since \$k=1\$, \$a_1\$ will become \$1\$. Now, \$k=2\$, so \$a_2\$ will become \$1\$ as well. This process will repeat until \$k=n\$ and \$\\{a_i\\}\$ has been changed to \$\\{1, 1, 1, \\cdots, 1\\}\$. Now, since \$k=n\$, \$a_n\$ will become \$0\$. Now, \$k=n-1\$, so \$a_{n-1}\$ will become \$0\$ as well. This process will repeat again until \$k=0\$ and \$\\{a_i\\}\$ has been changed to \$\\{0, 0, 0, \\cdots, 0\\}\$. Hence, the process terminates.\n\nWe can also see that \$L\\{a_i\\}=L\\{a_2, a_3, \\cdots, a_{n-1}\\}+(2n-1)\$. Hence, sum of \$L\\{a_i\\}\$ over all possible \$\\{a_i\\}\$ in this case\n\n\\[\n=\\sum L\\{a_i\\}=\\sum \\left[L\\{a_2, a_3, \\cdots, a_{n-1}\\}+(2n-1)\\right]=f(n-2)+2^{n-2}\\cdot(2n-1)\n\\]\n\nCase 3: \$a_n=0\$\n\nThis means that \$\\{a_i\\}=\\{a_1, a_2, a_3, \\cdots, a_{n-1}, 0\\}\$. Operating on \$\\{a_i\\}\$ is now equivalent to operating on \$\\{a_1, a_2, \\cdots, a_{n-1}\\}\$.\n\nBy our induction hypothesis, this process will terminate, and hence \$\\{a_i\\}\$ will be changed to \$\\{0, 0, 0, \\cdots, 0, 0\\}\$ at the end of this process. Hence, the process terminates.\n\nWe can also see that \$L\\{a_i\\}=L\\{a_1, a_2, \\cdots, a_{n-1}\\}\$. Hence, sum of \$L\\{a_i\\}\$ over all possible \$\\{a_i\\}\$ in this case\n\n\\[\n=\\sum L\\{a_i\\}=\\sum \\left[L\\{a_1, a_2, \\cdots, a_{n-1}\\}\\right]=f(n-1)\n\\]\n\nThis is already sufficient to prove part A, but we want to find \$f(n)\$ too, and we have overcounted!\n\nCase 4: \$a_1=1\$ and \$a_n=0\$ (This is what we have overcounted)\n\nThis means that \$\\{a_i\\}=\\{1, a_2, a_3, \\cdots, a_{n-1}, 0\\}\$. Operating on \$\\{a_i\\}\$ is now equivalent to operating on \$\\{a_2, a_3, \\cdots, a_{n-1}\\}\$.\n\nBy our induction hypothesis, this process will terminate, and hence \$\\{a_i\\}\$ will be changed to \$\\{1, 0, 0, \\cdots, 0, 0\\}\$ at the end of this process. Now, we operate one more time to change \$a_1\$ from \$1\$ to \$0\$. Hence, the process terminates.\n\nWe can also see that \$L\\{a_i\\}=L\\{a_2, a_3, \\cdots, a_{n-1}\\}+1\$. Hence, sum of \$L\\{a_i\\}\$ over all possible \$\\{a_i\\}\$ in this case\n\n\\[\n=\\sum L\\{a_i\\}=\\sum \\left[L\\{a_1, a_2, \\cdots, a_{n-1}\\}+1\\right]=f(n-2)+2^{n-2}\n\\]\n\nHence, we have proven that for all configurations, the process terminates, \$\\boxed{Q.E.D.}\$ Now, all that remains is to calculate \$f(n)\$.\n\n\\[\n\\begin{align} f(n) &= f(n-1)+2^{n-1}+f(n-2)+2^{n-2}\\cdot(2n-1)+f(n-1)-\\left[f(n-2)+2^{n-2}\\right] \\\\ &= 2f(n-1)+2^{n-1}\\cdot n \\end{align*}\n\\]\n\nThis can be rearranged to give us\n\n\\[\nf(n)-n(n+1)\\cdot2^{n-2}=2\\left[f(n-1)-n(n-1)\\cdot2^{n-3}\\right]\n\\]\n\nWhen \$n=1\$, \$f(n)-n(n+1)\\cdot2^{n-2}=0\$. This is good, because now we know that \$\\forall n\\in\\mathbb{Z}^{+}, f(n)-n(n+1)\\cdot2^{n-2}=0\$ i.e. \$f(n)=n(n+1)\\cdot2^{n-2}\$.\n\nNow the average will be \$\\frac{n(n+1)\\cdot2^{n-2}}{2^n}=\\boxed{\\frac{1}{4}n(n+1)}\$\n\n~IraeVid13" ]
IMO-2019-6	https://artofproblemsolving.com/wiki/index.php/2019_IMO_Problems/Problem_6	Let $I$ be the incenter of acute triangle $ABC$ with $AB \neq AC$. The incircle $\omega$ of $ABC$ is tangent to sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. The line through $D$ perpendicular to $EF$ meets $\omega$ again at $R$. Line $AR$ meets ω again at $P$. The circumcircles of triangles $PCE$ and $PBF$ meet again at $Q$. Prove that lines $DI$ and $PQ$ meet on the line through $A$ perpendicular to $AI$.	[ "Step 1\n\nWe find an auxiliary point \$S.\$\n\nLet \$G\$ be the antipode of \$D\$ on \$\\omega, GD = 2R,\$ where \$R\$ is radius \$\\omega.\$\n\nWe define \$A' = PG \\cap AI.\$\n\n\$RD\|\|AI, PRGD\$ is cyclic \$\\implies \\angle IAP = \\angle DRP = \\angle DGP.\$\n\n\$RD\|\|AI, RD \\perp RG, RI=GI \\implies \\angle AIR = \\angle AIG \\implies\$\n\n\\[\n\\triangle AIR \\sim \\triangle GIA' \\implies \\frac {AI}{GI} = \\frac {RI}{A'I}\\implies A'I \\cdot AI = R^2.\n\\]\n\nAn inversion with respect \$\\omega\$ swap \$A\$ and \$A' \\implies A'\$ is the midpoint \$EF.\$\n\nLet \$DA'\$ meets \$\\omega\$ again at \$S.\$ We define \$T = PS \\cap DI.\$\n\nOpposite sides of any quadrilateral inscribed in the circle \$\\omega\$ meet on the polar line of the intersection of the diagonals with respect to \$\\omega \\implies DI\$ and \$PS\$ meet on the line through \$A\$ perpendicular to \$AI.\$ The problem is reduced to proving that \$Q \\in PST.\$\n\nStep 2\n\nWe find a simplified way to define the point \$Q.\$\n\nWe define \$\\angle BAC = 2 \\alpha \\implies \\angle AFE = \\angle AEF = 90^\\circ – \\alpha \\implies\$ \$\\angle BFE = \\angle CEF = 180^\\circ – (90^\\circ – \\alpha) = 90^\\circ + \\alpha = \\angle BIC\$ \$(AI, BI,\$ and \$CI\$ are bisectrices).\n\nWe use the Tangent-Chord Theorem and get\n\n\\[\n\\angle EPF = \\angle AEF = 90^\\circ – \\alpha.\n\\]\n\n\$\\angle BQC = \\angle BQP + \\angle PQC = \\angle BFP + \\angle CEP =\$ \$=\\angle BFE – \\angle EFP + \\angle CEF – \\angle FEP =\$ \$= 90^\\circ + \\alpha + 90^\\circ + \\alpha – (90^\\circ + \\alpha) =\$ \$90^\\circ + \\alpha = \\angle BIC \\implies\$\n\nPoints \$Q, B, I,\$ and \$C\$ are concyclic.\n\nStep 3\n\nWe perform inversion around \$\\omega.\$ The straight line \$PST\$ maps onto circle \$PITS.\$ We denote this circle \$\\Omega.\$ We prove that the midpoint of \$AD\$ lies on the circle \$\\Omega.\$\n\nIn the diagram, the configuration under study is transformed using inversion with respect to \$\\omega.\$ The images of the points are labeled in the same way as the points themselves. Points \$D,E,F,P,S,\$ and \$G\$ have saved their position. Vertices \$A, B,\$ and \$C\$ have moved to the midpoints of the segments \$EF, FD,\$ and \$DE,\$ respectively.\n\nLet \$M\$ be the midpoint \$AD.\$\n\nWe define \$\\angle MID = \\beta, \\angle MDI = \\gamma \\implies\$ \$\\angle IMA = \\angle MID + \\angle MDI = \\beta + \\gamma = \\varphi.\$ \$DI = IS \\implies \\angle ISD = \\gamma.\$\n\n\$MI\$ is triangle \$DAG\$ midline \$\\implies MI \|\| AG \\implies\$\n\n\\[\nMI \|\| PG \\implies \\angle MAP = \\angle AMI = \\varphi.\n\\]\n\n\\[\n\\angle DPA = 90^\\circ \\implies PM = MA \\implies\n\\]\n\n\\[\n\\angle PMA = \\angle PMS = 180^\\circ – 2 \\varphi.\n\\]\n\n\$PI = IS \\implies \\angle PIS = 180^\\circ – 2 \\varphi =\\angle DPA \\implies\$ point \$M\$ lies on \$\\Omega.\$ \$ABDC\$ is parallelogram \$\\implies M\$ is midpoint \$BC.\$\n\nStep 4\n\nWe prove that image of \$Q\$ lies on \$\\Omega.\$\n\nIn the inversion plane the image of point \$Q\$ lies on straight line \$BC\$ (It is image of circle \$BIC)\$ and on circle \$PCE.\$\n\n\\[\n\\angle PQM = \\angle PQC = \\angle PEC = \\angle PED = \\angle PSD = \\angle PSM \\implies\n\\]\n\npoint \$Q\$ lies on \$\\Omega\$.\n\nvladimir.shelomovskii@gmail.com, vvsss" ]
IMO-2020-1	https://artofproblemsolving.com/wiki/index.php/2020_IMO_Problems/Problem_1	Consider the convex quadrilateral $ABCD$. The point $P$ is in the interior of $ABCD$. The following ratio equalities hold: \[ \angle PAD : \angle PBA : \angle DPA = 1 : 2 : 3 = \angle CBP : \angle BAP : \angle BPC. \] Prove that the following three lines meet in a point: the internal bisectors of angles $\angle ADP$ and $\angle PCB$ and the perpendicular bisector of segment $\overline{AB}$.	[ "Let the perpendicular bisector of \$AP,BP\$ meet at point \$O\$, those two lines meet at \$AD,BC\$ at \$N,M\$ respectively.\n\nAs the problem states, denote that \$\\angle{PBC}=\\alpha, \\angle{BAP}=2\\alpha, \\angle {BPC}=3\\alpha\$. We can express another triple with \$\\beta\$ as well. Since the perpendicular line of \$BP\$ meets \$BC\$ at point \$M\$, \$BM=MP, \\angle {BPM}=\\alpha, \\angle {PMC}=2\\alpha\$, which means that points \$A,P,M,B\$ are concyclic since \$\\angle{PAB}=\\angle{PMC}\$\n\nSimilarly, points \$A,N,P,B\$ are concyclic as well, which means five points \$A,N,P,M,B\$ are concyclic., \$ON=OP=OM\$\n\nMoreover, since \$\\angle{CPM}=\\angle{CMP}\$, \$CP=CM\$ so the angle bisector if the angle \$MCP\$ must be the perpendicular line of \$MP\$, so as the angle bisector of \$\\angle{ADP}\$, which means those three lines must be concurrent at the circumcenter of the circle containing five points \$A,N,P,M,B\$ as desired\n\n~ bluesoul and \"Shen Kislay kai\" ~ edits by Pearl2008", "The essence of the proof is the replacement of the bisectors of angles by the perpendicular bisectors of the sides of the cyclic pentagon.\n\nLet \$O\$ be the circumcenter of \$\\triangle ABP, \\angle PAD = \\alpha, OE\$ is the perpendicular bisector of \$AP,\$ and point \$E\$ lies on \$AD.\$ Then\n\n\\[\n\\angle APE = \\alpha, \\angle PEA = \\pi - 2\\alpha, \\angle ABP = 2\\alpha \\implies\n\\]\n\n\$\\hspace{33mm} ABPE\$ is cyclic.\n\n\\[\n\\angle PED = 2\\alpha = \\angle DPE \\implies\n\\]\n\nthe bisector of the \$\\angle ADP\$ is the perpendicular bisector of the side \$EP\$ of the cyclic \$ABPE\$ that passes through the center \$O.\$\n\nA similar reasoning can be done for \$OF,\$ the perpendicular bisector of \$BP.\$\n\nvladimir.shelomovskii@gmail.com, vvsss" ]
IMO-2020-2	https://artofproblemsolving.com/wiki/index.php/2020_IMO_Problems/Problem_2	The real numbers $a$, $b$, $c$, $d$ are such that $a \geq b \geq c \geq d > 0$ and $a + b + c + d = 1$. Prove that \[ (a + 2b + 3c + 4d) a^a b^b c^c d^d < 1. \]	[ "Using Weighted AM-GM we get\n\n\\[\n\\frac{a\\cdot a +b\\cdot b +c\\cdot c +d\\cdot d}{a+b+c+d} \\ge \\sqrt[a+b+c+d]{a^a b^b c^c d^d}\n\\]\n\n\\[\n\\implies a^a b^b c^c d^d \\le a^2 +b^2 +c^2 +d^2\n\\]\n\nSo,\n\n\\[\n(a+2b+3c+4d) a^ab^bc^cd^d \\le (a+2b+3c+4d)(a^2+b^2+c^2+d^2)\n\\]\n\nNow notice that\n\nSo, we get\n\n\\[\n\\begin{split}&~~~~(a+2b+3c+4d)(a^2+b^2+c^2+d^2) \\\\ &= a^2(a+2b+3c+4d)+b^2(a+2b+3c+4d)+c^2 (a+2b+3c+4d) +d^2 (a+2b+3c+4d)\\\\ &\\le a^2(a+3b+3c+3d)+b^2(3a+b+3c+3d)+c^2 (3a+3b+c+3d) +d^2 (3a+3b+3c+d)\\\\ &<(a+b+c+d)^3 \\\\&=1\\end{split}\n\\]\n\nNow, for equality we must have \$a=b=c=d=\\frac{1}{4}\$\n\nIn that case we get\n\n\\[\n(a+2b+3c+4d) a^ab^bc^cd^d \\le (a+2b+3c+4d)(a^2+b^2+c^2+d^2) =\\frac{5}{8} <1\n\\]\n\n~Shen Kislay kai" ]
IMO-2020-3	https://artofproblemsolving.com/wiki/index.php/2020_IMO_Problems/Problem_3	There are $4n$ pebbles of weights $1, 2, 3, . . . , 4n$. Each pebble is colored in one of $n$ colors and there are four pebbles of each color. Show that we can arrange the pebbles into two piles so that the following two conditions are both satisfied: - The total weights of both piles are the same. - Each pile contains two pebbles of each color.	[]
IMO-2020-4	https://artofproblemsolving.com/wiki/index.php/2020_IMO_Problems/Problem_4	There is an integer $n > 1$. There are $n^2$ stations on a slope of a mountain, all at different altitudes. Each of two cable car companies, $A$ and $B$, operates $k$ cable cars; each cable car provides a transfer from one of the stations to a higher one (with no intermediate stops). The $k$ cable cars of $A$ have $k$ different starting points and $k$ different finishing points, and a cable car that starts higher also finishes higher. The same conditions hold for $B$. We say that two stations are linked by a company if one can start from the lower station and reach the higher one by using one or more cars of that company (no other movements between stations are allowed). Determine the smallest positive integer k for which one can guarantee that there are two stations that are linked by both companies.	[]
IMO-2020-5	https://artofproblemsolving.com/wiki/index.php/2020_IMO_Problems/Problem_5	A deck of $n > 1$ cards is given. A positive integer is written on each card. The deck has the property that the arithmetic mean of the numbers on each pair of cards is also the geometric mean of the numbers on some collection of one or more cards. For which $n$ does it follow that the numbers on the cards are all equal?	[ "Claim : For all n > 1, all numbers must be equal Contradiction: Let us assume this is not true and for a certain n, there are k distinct positive integers which can be written in ascending order as follows :\n\nzk > zk-1 > zk-2 > … > z1\n\nSince zk is the largest of the numbers, it has to be greater than 1. This implies that there will be a prime p1 that divides zk.\n\nNow we know that the arithmetic mean of zk and zk-1 is more than zk-1, thus the geometric mean which it is equivalent to must include the term zk. Since we know the arithmetic mean is a rational number, this means the root of the products of the n numbers of the geometric mean must be positive, and this will also divide p.\n\nWe can use similar argument for the sum of zk-1 and zk-2 will be greater than zk-2 and thus the geometric mean is once again divisible by p. We can repeat this all the way to z1, which would mean that by dividing each term by p, we can reach another set of n cards which fulfil the conditions. We can keep applying this until zk. However this is a contradiction.\n\nThus, for all n > 1, all the terms on the card will be equal.\n\nQED" ]
IMO-2020-6	https://artofproblemsolving.com/wiki/index.php/2020_IMO_Problems/Problem_6	Prove that there exists a positive constant $c$ such that the following statement is true: Consider an integer $n > 1$, and a set $S$ of n points in the plane such that the distance between any two different points in $S$ is at least $1$. It follows that there is a line $\ell$ separating $S$ such that the distance from any point of $S$ to $\ell$ is at least $cn^{- \frac{1}{3}}$. (A line $\ell$ separates a set of points $S$ if some segment joining two points in $S$ crosses $\ell$.) Note. Weaker results with $cn^{- \frac{1}{3}}$ replaced by $cn^{- \alpha}$ may be awarded points depending on the value of the constant $\alpha > \frac{1}{3}$.	[ "For any unit vector \$v\$, let \$a_v=\\min_{p\\in S} p \\cdot v\$ and \$b_v = \\max_{p\\in S} p\\cdot v\$. If \$b_v - a_v\\geq n^{2/3}\$ then we can find a line \$\\ell\$ perpendicular to \$v\$ such that \$\\ell\$ separates \$S\$, and any point in \$S\$ is at least \$\\Omega(n^{2/3}/n) = \\Omega(n^{-1/3})\$ away from \$\\ell\$.\n\nSuppose there is no such direction \$v\$, then \$S\$ is contained in a box with side length \$n^{2/3}\$ by considering the direction of \$(1, 0)\$ and \$(0, 1)\$, respectively. Hence, \$S\$ is contained in a disk with radius \$n^{2/3}\$. Now suppose that \$D\$ is the disk with the minimum radius, say \$r\$, which contains \$S\$. Then, \$r=O(n^{2/3})\$. Since the distance between any two points in \$S\$ is at least \$1\$, \$r=\\Omega(\\sqrt{n})\$ too.\n\nLet \$p\$ be any point in \$S\$ on the boundary of \$D\$. Let \$\\ell_1\$ be the line tangent to \$D\$ at \$p\$, and \$\\ell_2\$ the line obtained by translating \$\\ell_1\$ by distance \$1\$ towards the inside of \$D\$. Let \$H\$ be the region sandwiched by \$\\ell_1\$ and \$\\ell_2\$. It is easy to show that both the area and the perimeter of \$H\\cap D\$ is bounded by \$O(\\sqrt{r})\$ (since \$r=\\Omega(\\sqrt{n})\$). Hence, there can only be \$O(\\sqrt{r})=O(n^{1/3})\$ points in \$H\\cap S\$, by that any two points in \$S\$ are distance \$1\$ apart. Since the width of \$H\$ is \$1\$, there must exist a line \$\\ell\$ parallel to \$\\ell_1\$ such that \$\\ell\$ separates \$S\$, and any point in \$S\$ is at least \$1/O(n^{1/3}) = \\Omega(n^{-1/3})\$ away from \$\\ell\$. Q.E.D.\n\nNote. One can also show that \$\\Omega(n^{-1/3})\$ is best possible. ~Shen Kislay kai" ]
IMO-2021-1	https://artofproblemsolving.com/wiki/index.php/2021_IMO_Problems/Problem_1	Let $n \geq 100$ be an integer. Ivan writes the numbers $n, n+1, \ldots, 2 n$ each on different cards. He then shuffles these $n+1$ cards, and divides them into two piles. Prove that at least one of the piles contains two cards such that the sum of their numbers is a perfect square.	[ "If we can guarantee that there exist \$3\$ cards such that every pair of them sum to a perfect square, then we can guarantee that one of the piles contains \$2\$ cards that sum to a perfect square. Assume the perfect squares \$p^2\$, \$q^2\$, and \$r^2\$ satisfy the following system of equations:\n\n\\[\n\\usepackage{amsmath} \\begin{align} a+b &= p^2 \\\\ b+c &= q^2 \\\\ a+c &= r^2 \\end{align}\n\\]\n\nwhere \$a\$, \$b\$, and \$c\$ are numbers on three of the cards. Solving for \$a\$, \$b\$, and \$c\$ in terms of \$p\$, \$q\$, and \$r\$ tells us that \$a = \\frac{p^2 + r^2 - q^2}{2}\$, \$b=\\frac{p^2 + q^2 - r^2}{2}\$, and \$c=\\frac{q^2 + r^2 - p^2}{2}\$. We can then substitute \$p^2 = (2e-1)^2\$, \$q^2 = (2e)^2\$, and \$r^2 = (2e+1)^2\$ to cancel out the \$2\$s in the denominatior, and simplifying gives \$a = 2e^2 + 1\$, \$b = 2e(e-2)\$, and \$c = 2e(e+2)\$. Now, we have to prove that there exists three numbers in these forms between \$n\$ and \$2n\$ when \$n \\ge 100\$. Notice that \$b\$ will always be the least of the three and \$c\$ will always be the greatest of the three. So it is sufficient to prove that there exists numbers in the form \$2e(e-2)\$ and \$2e(e+2)\$ between \$n\$ and \$2n\$.\n\nFor two numbers in the form of \$2e(e-2)\$ and \$2e(e+2)\$ to be between \$n\$ and \$2n\$, the inequalities\n\n\\[\n\\usepackage{amsmath} \\begin{align} 2e(e-2) &\\ge n \\\\ 2e(e+2) &\\le 2n \\\\ \\end{align}\n\\]\n\nmust be satisfied. We can then expand and simplify to get that\n\n\\[\n\\usepackage{amsmath} \\begin{align} e^2 - 2e - \\frac{n}{2} &\\ge 0 \\\\ e^2 + 2e - n &\\le 0. \\\\ \\end{align}\n\\]\n\nThen, we can complete the square on the left sides of both inequalities and isolate \$e\$ to get that\n\n\\[\n\\usepackage{amsmath} \\begin{align} e &\\ge \\sqrt{1 + \\frac{n}{2}} + 1 \\\\ e &\\le \\sqrt{1 + n} - 1 \\\\ \\end{align}\n\\]\n\nNotice that \$e\$ must be an integer, so there must be an integer between \$\\sqrt{1 + n} - 1\$ and \$\\sqrt{1 + \\frac{n}{2}} + 1\$. If \$\\sqrt{1 + n} - 1\$ and \$\\sqrt{1 + \\frac{n}{2}} + 1\$ differ by at least \$1\$, then we can guarantee that there is an integer between them (and those integers are the possible values of \$e\$). Setting up the inequality \$\\sqrt{1 + n} - \\sqrt{1 + \\frac{n}{2}} - 2 \\ge 1\$ and solving for \$n\$ tells us that \$n \\in [107, \\infty)\$ always works. Testing the remaining \$7\$ numbers (\$100\$ to \$106\$) manually tells us that there is an integer between \$\\sqrt{1 + n} - 1\$ and \$\\sqrt{1 + \\frac{n}{2}} + 1\$ when \$n \\ge 100\$. Therefore, there exists a triplet of integers \$(a,b,c)\$ with \$a, b, c \\in \\{n, n+1, ..., 2n\\}\$ when \$n \\ge 100\$ such that every pair of the numbers sum to a perfect square. By the pigeonhole principle, we know that \$2\$ of the numbers must be on cards in the same pile, and hence, when \$n \\ge 100\$, there will always be a pile with \$2\$ numbers that sum to a perfect square. \$\\square\$\n\n~Mathdreams", "Claim: If \$n \\geq 100\$, then there exist at least three perfect squares between \$n/2 + 1\$ and \$n + 1\$ inclusive.\n\nProof: If \$100 \\leq n \\leq 125\$, then the perfect squares \$64\$, \$81\$, and \$100\$ are between \$n/2 + 1\$ and \$n + 1\$.\n\nWhat if \$n \\geq 126\$? Let \$f(t) = \\sqrt{t + 1} - \\sqrt{t/2 + 1}\$. Note that\n\n\\[\nf(126) = \\sqrt{127} - \\sqrt{64} > 11 - 8 = 3.\n\\]\n\nMoreover, \$f\$ is increasing because\n\n\\[\nf'(t) = \\frac{1}{\\sqrt{4t + 4}} - \\frac{1}{\\sqrt{8t + 16}} > 0.\n\\]\n\nSo \$f(n) \\geq f(126) > 3\$. Thus there are at least three distinct integers between \$\\sqrt{n/2 + 1}\$ and \$\\sqrt{n + 1}\$, and their squares will lie between \$n/2 + 1\$ and \$n + 1\$. This proves the claim.\n\nNow given any \$n \\geq 100\$, it follows from the claim that there exist three consecutive squares \$(k - 1)^2\$, \$k^2\$, and \$(k + 1)^2\$ such that\n\n\\[\n\\frac{n}{2} + 1 \\leq (k - 1)^2, k^2, (k + 1)^2 \\leq n + 1,\n\\]\n\nand therefore\n\n\\[\nn \\leq 2k^2 - 4k, 2k^2 + 1, 2k^2 + 4k \\leq 2n.\n\\]\n\nThe three numbers \$2k^2 - 4k\$, \$2k^2 + 1\$, and \$2k^2 + 4k\$ have the property that the sum of any two of them is a perfect square:\n\n\\begin{align} (2k^2 - 4k) + (2k^2 + 1) &= (2k - 1)^2, \\\\ (2k^2 - 4k) + (2k^2 + 4k) &= (2k)^2, \\\\ (2k^2 + 1) + (2k^2 + 4k) &= (2k + 1)^2. \\end{align}\n\nBy the pigeonhole principle, cards showing two of these numbers will end up in the same pile, and the sum of those cards’ numbers will be a perfect square." ]
IMO-2021-2	https://artofproblemsolving.com/wiki/index.php/2021_IMO_Problems/Problem_2	Show that the inequality \[ \sum_{i=1}^n \sum_{j=1}^n \sqrt{\|x_i-x_j\|} \le \sum_{i=1}^n \sum_{j=1}^n \sqrt{\|x_i+x_j\|} \] holds for all real numbers $x_1,x_2,\dots,x_n$.	[ "then,\n\n\\[\n\\sum \\sum x_i^2+x_j^2-2x_ix_j \\leq \\sum \\sum x_i^2+x_j^2+2x_ix_j \\to \\sum \\sum 4x_ix_j\\geq 0,\n\\]\n\ntherefore we have to prove that\n\n\\[\n\\sum \\sum a_ia_j\\geq 0\n\\]\n\nfor every list \$x_i\$, and we can describe this to\n\n\\[\n\\sum \\sum a_ia_j=\\sum a_i^2 + \\sum\\sum a_ia_j(i\\neq j)\n\\]\n\nwe know that\n\n\\[\n\\frac{a_i^2}{2}+\\frac{a_j^2}{2} \\geq \|a_ia_j\|\n\\]\n\ntherefore,\n\n\\[\na_i^2+a_j^2 \\geq -(a_ia_j+a_ja_i)\n\\]\n\n\\[\n\\to \\sum a_i^2 + \\sum\\sum a_ia_j \\geq 0\n\\]\n\n\\[\nQ.E.D.\n\\]\n\n-Mathhyhyhye" ]
IMO-2021-3	https://artofproblemsolving.com/wiki/index.php/2021_IMO_Problems/Problem_3	Let $D$ be an interior point of the acute triangle $ABC$ with $AB > AC$ so that $\angle DAB= \angle CAD$. The point $E$ on the segment $AC$ satisfies $\angle ADE= \angle BCD$, the point $F$ on the segment $AB$ satisfies $\angle FDA= \angle DBC$, and the point $X$ on the line $AC$ satisfies $CX=BX$. Let $O_1$ and $O_2$ be the circumcentres of the triangles $ADC$ and $EXD$ respectively. Prove that the lines $BC$, $EF$, and $O_1 O_2$ are concurrent.	[ "We prove that circles \$ACD, EXD\$ and \$\\Omega_0\$ centered at \$P\$ (the intersection point \$BC\$ and \$EF)\$ have a common chord.\n\nLet \$P\$ be the intersection point of the tangent to the circle \$\\omega_2 = BDC\$ at the point \$D\$ and the line \$BC, A'\$ is inverse to \$A\$ with respect to the circle \$\\Omega_0\$ centered at \$P\$ with radius \$PD.\$ Then the pairs of points \$F\$ and \$E, B\$ and \$C\$ are inverse with respect to \$\\Omega_0\$, so the points \$F, E,\$ and \$P\$ are collinear. Quadrilaterals containing the pairs of inverse points \$B\$ and \$C, E\$ and \$F, A\$ and \$A'\$ are inscribed, \$FE\$ is antiparallel to \$BC\$ with respect to angle \$A\$ (see \$\\boldsymbol{Claim}\$).\n\nConsider the circles \$\\omega = ACD\$ centered at \$O_1, \\omega' = A'BD,\$ \$\\omega_1 = ABC, \\Omega = EXD\$ centered at \$O_2 , \\Omega_1 = A'BX,\$ and \$\\Omega_0.\$\n\nDenote \$\\angle ACB = \\gamma\$. Then \$\\angle BXC = \\angle BXE = \\pi – 2\\gamma,\$ \$\\angle AA'B = \\gamma (AA'CB\$ is cyclic), \$\\angle AA'E = \\pi – \\angle AFE = \\pi – \\gamma (AA'EF\$ is cyclic, \$FE\$ is antiparallel), \$\\angle BA'E = \\angle AA'E – \\angle AA'B = \\pi – 2\\gamma = \\angle BXE \\implies\$\n\n\$\\hspace{13mm}E\$ is the point of the circle \$\\Omega_1.\$\n\nLet the point \$Y\$ be the radical center of the circles \$\\omega, \\omega', \\omega_1.\$ It has the same power \$\\nu\$ with respect to these circles. The common chords of the pairs of circles \$A'B, AC, DT,\$ where \$T = \\omega \\cap \\omega',\$ intersect at this point. \$Y\$ has power \$\\nu\$ with respect to \$\\Omega_1\$ since \$A'B\$ is the radical axis of \$\\omega', \\omega_1, \\Omega_1.\$ \$Y\$ has power \$\\nu\$ with respect to \$\\Omega\$ since \$XE\$ containing \$Y\$ is the radical axis of \$\\Omega\$ and \$\\Omega_1.\$ Hence \$Y\$ has power \$\\nu\$ with respect to \$\\omega, \\omega', \\Omega.\$\n\nLet \$T'\$ be the point of intersection \$\\omega \\cap \\Omega.\$ Since the circles \$\\omega\$ and \$\\omega'\$ are inverse with respect to \$\\Omega_0,\$ then \$T\$ lies on \$\\Omega_0,\$ and \$P\$ lies on the perpendicular bisector of \$DT.\$ The power of a point \$Y\$ with respect to the circles \$\\omega, \\omega',\$ and \$\\Omega\$ are the same, \$DY \\cdot YT = DY \\cdot YT' \\implies\$ the points \$T\$ and \$T'\$ coincide.\n\nThe centers of the circles \$\\omega\$ and \$\\Omega\$ (\$O_1\$ and \$O_2\$) are located on the perpendicular bisector \$DT'\$, the point \$P\$ is located on the perpendicular bisector \$DT\$ and, therefore, the points \$P, O_1,\$ and \$O_2\$ lie on a line, that is, the lines \$BC, EF,\$ and \$O_1 O_2\$ are concurrent.\n\n\\[\n\\boldsymbol{Claim}\n\\]\n\nLet \$AK\$ be bisector of the triangle \$ABC\$, point \$D\$ lies on \$AK.\$ The point \$E\$ on the segment \$AC\$ satisfies \$\\angle ADE= \\angle BCD\$. The point \$F\$ on the segment \$AB\$ satisfies \$\\angle ADF= \\angle CBD.\$ Let \$P\$ be the intersection point of the tangent to the circle \$BDC\$ at the point \$D\$ and the line \$BC.\$ Let the circle \$\\Omega_0\$ be centered at \$P\$ and has the radius \$PD.\$\n\nThen the pairs of points \$F\$ and \$E, B\$ and \$C\$ are inverse with respect to \$\\Omega_0\$ and \$EF\$ and \$BC\$ are antiparallel with respect to the sides of an angle \$A.\$\n\n\\[\n\\boldsymbol{Proof}\n\\]\n\nLet the point \$E'\$ is symmetric to \$E\$ with respect to bisector \$AK, E'L \|\| BC.\$ Symmetry of points \$E\$ and \$E'\$ implies \$\\angle AEL = \\angle AE'L.\$\n\n\\[\n\\angle DCK = \\angle E'DL, \\angle DKC = \\angle E'LD \\implies\n\\]\n\n\\[\n\\triangle DCK \\sim \\triangle E'DL \\implies \\frac {E'L}{KD}= \\frac {DL}{KC}.\n\\]\n\n\\[\n\\triangle ALE' \\sim \\triangle AKB \\implies \\frac {E'L}{BK}= \\frac {AL}{AK}\\implies\n\\]\n\n\\[\n\\frac {AL}{DL} = \\frac {AK \\cdot DK}{BK \\cdot KC}.\n\\]\n\nSimilarly, we prove that \$FL\$ and \$BC\$ are antiparallel with respect to angle \$A,\$ and the points \$L\$ in triangles \$\\triangle EDL\$ and \$\\triangle FDL\$ coincide. Hence, \$FE\$ and \$BC\$ are antiparallel and \$BCEF\$ is cyclic. Note that \$\\angle DFE = \\angle DLE – \\angle FDL = \\angle AKC – \\angle CBD\$ and \$\\angle PDE = 180^o – \\angle CDK – \\angle CDP – \\angle LDE = 180^o – (180^o – \\angle AKC – \\angle BCD) – \\angle CBD – \\angle BCD\$ \$\\angle PDE = \\angle AKC – \\angle CBD = \\angle DFE,\$ so \$PD\$ is tangent to the circle \$DEF.\$\n\n\$PD^2 = PC \\cdot PB = PE \\cdot PF,\$ that is, the points \$B\$ and \$C, E\$ and \$F\$ are inverse with respect to the circle \$\\Omega_0.\$\n\nvladimir.shelomovskii@gmail.com, vvsss" ]
IMO-2021-4	https://artofproblemsolving.com/wiki/index.php/2021_IMO_Problems/Problem_4	Let $\Gamma$ be a circle with centre $I$, and $ABCD$ a convex quadrilateral such that each of the segments $AB, BC, CD$ and $DA$ is tangent to $\Gamma$. Let $\Omega$ be the circumcircle of the triangle $AIC$. The extension of $BA$ beyond $A$ meets $\Omega$ at $X$, and the extension of $BC$ beyond $C$ meets $\Omega$ at $Z$. The extensions of $AD$ and $CD$ beyond $D$ meet $\Omega$ at $Y$ and $T$, respectively. Prove that \[ AD + DT + T X + XA = CD + DY + Y Z + ZC \]	[ "Let \$O\$ be the centre of \$\\Omega\$.\n\nFor \$AB=BC\$ the result follows simply. By Pitot's Theorem we have\n\n\\[\nAB + CD = BC + AD\n\\]\n\nso that, \$AD = CD.\$ The configuration becomes symmetric about \$OI\$ and the result follows immediately.\n\nNow assume WLOG \$AB < BC\$. Then \$T\$ lies between \$A\$ and \$X\$ in the minor arc \$AX\$ and \$Z\$ lies between \$Y\$ and \$C\$ in the minor arc \$YC\$. Consider the cyclic quadrilateral \$ACZX\$. We have \$\\angle CZX = \\angle CAB\$ and \$\\angle IAC = \\angle IZC\$. So that,\n\n\\[\n\\angle CZX - \\angle IZC = \\angle CAB - \\angle IAC\n\\]\n\n\\[\n\\angle IZX = \\angle IAB\n\\]\n\nSince \$I\$ is the incenter of quadrilateral \$ABCD\$, \$AI\$ is the angular bisector of \$\\angle DBA\$. This gives us,\n\n\\[\n\\angle IZX = \\angle IAB = \\angle IAD = \\angle IAY\n\\]\n\nHence the chords \$IX\$ and \$IY\$ are equal. So \$Y\$ is the reflection of \$X\$ about \$OI\$. Hence,\n\n\\[\nTX = YZ\n\\]\n\nand now it suffices to prove\n\n\\[\nAD + DT + XA = CD + DY + ZC\n\\]\n\nLet \$P, Q, N\$ and \$M\$ be the tangency points of \$\\Gamma\$ with \$AB, BC, CD\$ and \$DA\$ respectively. Then by tangents we have, \$AD = AM + MD = AP + ND\$. So \$AD + DT + XA = AP + ND + DT + XA = XP + NT\$. Similarly we get, \$CD + DY + ZC = ZQ + YM\$. So it suffices to prove,\n\n\\[\nXP + NT = ZQ + YM\n\\]\n\nConsider the tangent \$XJ\$ to \$\\Gamma\$ with \$J \\ne P\$. Since \$X\$ and \$Y\$ are reflections about \$OI\$ and \$\\Gamma\$ is a circle centred at \$I\$ the tangents \$XJ\$ and \$YM\$ are reflections of each other. Hence\n\n\\[\nXP = XJ = YM\n\\]\n\nBy a similar argument on the reflection of \$T\$ and \$Z\$ we get \$NT = ZQ\$ and finally,\n\n\\[\nXP + NT = ZQ + YM\n\\]\n\nas required. \$QED\$\n\n~BUMSTAKA", "Denote \$AD\$ tangents to the circle \$I\$ at \$N\$, \$CD\$ tangents to the same circle at \$M\$; \$XB\$ tangents at \$F\$ and \$ZB\$ tangents at \$J\$. We can get that \$AD=AM+MD;CD=DN+CN\$.Since \$AM=AF,XA=XF-AM;ZC=ZJ-CN\$ Same reason, we can get that \$DT=TN-DM;DY=YM-DM\$ We can find that \$AD+XA+DT=XF+TN;CD+DY+ZC=ZJ+YM\$. Connect \$IM,IT,IT,IZ,IX,IN,IJ,IF\$ separately, we can create two pairs of congruent triangles. In \$\\triangle{XIF},\\triangle{YIM}\$, since \$\\widehat{AI}=\\widehat{AI},\\angle{FXI}=\\angle{MYI}\$ After getting that \$\\angle{IFX}=\\angle{IMY};\\angle{FXI}=\\angle{MYI};IF=IM\$, we can find that \$\\triangle{IFX}\\cong \\triangle{IMY}\$. Getting that \$YM=XF\$, same reason, we can get that \$ZJ=TN\$. Now the only thing left is that we have to prove \$TX=YZ\$. Since \$\\widehat{IX}=\\widehat{IY};\\widehat{IT}=\\widehat{IZ}\$ we can subtract and get that \$\\widehat{XT}=\\widehat{YZ}\$,means \$XT=YZ\$ and we are done ~bluesoul", "We use the equality of the tangent segments and symmetry.\n\nUsing Claim 1 we get \$\\overset{\\Large\\frown} {TX}\$ symmetric to \$\\overset{\\Large\\frown} {ZY}\$ with respect \$IO.\$\n\nTherefore \$\\hspace{10mm} TX = ZY.\$\n\nLet \$P, Q, N\$ and \$M\$ be the tangency points of \$\\Gamma\$ with \$AB, BC, CD,\$ and \$DA,\$ respectively.\n\nUsing Claim 2 we get \$TM = QZ, PX = NY.\$\n\n\\[\nAD + DT + XA = AN+ND + TM – MD +XP-PA =\n\\]\n\n\\[\n= XP + TM = QZ + NY = MC + ZC + MD + DY =\n\\]\n\n\\[\n=CD + ZC + DY.\n\\]\n\nClaim 1\n\nLet \$O\$ be the center of \$\\Omega.\$ Then point \$T\$ is symmetric to \$Z\$ with respect \$IO,\$ point \$X\$ is symmetric to \$Y\$ with respect \$IO.\$\n\nProof\n\nLet \$\\angle BAD =2\\alpha, \\angle ABC =2\\beta, \\angle BCD =2\\gamma, \\angle ADC =2\\delta.\$\n\nWe find measure of some arcs:\n\n\\[\n\\overset{\\Large\\frown} {IT}= 2\\angle ICT = 2\\gamma,\n\\]\n\n\\[\n\\overset{\\Large\\frown} {IY}= 2\\angle IAY = 2\\alpha,\n\\]\n\n\\[\n\\overset{\\Large\\frown} {XY}= 2\\angle XAY = 2\\pi - 4\\alpha,\n\\]\n\n\$\\overset{\\Large\\frown} {IX}= 2\\pi - \\overset{\\Large\\frown} {IY} - \\overset{\\Large\\frown} {XY} = 2\\alpha =\\overset{\\Large\\frown} {IY}\\implies\$ symmetry \$X\$ and \$Y.\$\n\n\\[\n\\overset{\\Large\\frown} {TZ}= 2\\angle DCZ = 2\\pi – 4\\gamma,\n\\]\n\n\$\\overset{\\Large\\frown} {IZ}= 2\\pi - \\overset{\\Large\\frown} {IT} - \\overset{\\Large\\frown} {TZ}= 2\\gamma= \\overset{\\Large\\frown} {IT}\\implies\$ symmetry \$T\$ and \$Z.\$\n\nClaim 2\n\nLet circles \$\\omega\$ centered at \$I\$ and \$\\Omega\$ centered at \$O\$ be given. Let points \$A\$ and \$A'\$ lies on \$\\Omega\$ and \$A\$ be symmetric to \$A'\$ with respect \$OI.\$ Let \$AC\$ and \$A'B\$ be tangents to \$\\omega\$. Then \$AC = A'B.\$\n\nProof\n\n\\[\nAI = A'I, IB = IC, \\angle ACI = \\angle A'BI = 90^\\circ \\implies\n\\]\n\n\\[\n\\triangle AIC = \\triangle A'IB \\implies A'B = AC.\n\\]\n\nvladimir.shelomovskii@gmail.com, vvsss" ]
IMO-2021-5	https://artofproblemsolving.com/wiki/index.php/2021_IMO_Problems/Problem_5	Two squirrels, Bushy and Jumpy, have collected 2021 walnuts for the winter. Jumpy numbers the walnuts from 1 through 2021, and digs 2021 little holes in a circular pattern in the ground around their favourite tree. The next morning Jumpy notices that Bushy had placed one walnut into each hole, but had paid no attention to the numbering. Unhappy, Jumpy decides to reorder the walnuts by performing a sequence of 2021 moves. In the $k$-th move, Jumpy swaps the positions of the two walnuts adjacent to walnut $k$. Prove that there exists a value of $k$ such that, on the $k$-th move, Jumpy swaps some walnuts $a$ and $b$ such that $a < k < b$.	[ "We will start by introducing some notation.\n\n- Let the holes be denoted by \$H_1, H_2, \\dots, H_{2021}\$. The index \$i\$ in \$H_i\$ is considered modulo \$2021\$\n- Let the nuts be denoted by \$N_1, N_2, \\dots, N_{2021}\$ and define \$N_i < N_j\$ when \$i < j\$.\n- Let a nut \$n\$ in hole \$H_i\$ be a good nut if the two neighboring nuts \$a, b\$ in holes \$H_{i-1}\$ and \$H_{i+1}\$ satisfy either \$a,b < n\$ or \$n < a,b\$. A nut \$n\$ is a bad nut if it is not good. When good or bad is used for a hole, it refers to the nut in the hole.\n- The status of a nut is wether it is good or bad.\n- Two adjacent nuts \$a,b\$ in holes \$H_i\$ and \$H_{i+1}\$ respectively are an increasing pair if \$a < b\$ and a decreasing pair if \$a > b\$.\n- Just before the \$k\$'th move we denote all nuts \$N_i\$ with \$i\\geq k\$ to be upcomming nuts.\n- During move \$k\$, \$N_k\$ is the current nut.\n\nThe proof works by counting the parity of upcomming bad nuts and hinges on the fact that 2021 is odd. We start by proving that at any point in time there are an odd number of bad nuts. Let \$B\$ be the number of bad nuts and \$G\$ the number of good nuts. A bad nut is either part of two increasing or two decreasing pairs. Let the number of increasing bad nuts be \$B_1\$ and decreasing bad nuts be \$B_2\$. A good nut is part of one increasing and one decreasing pair. Let \$I\$ and \$D\$ be the number of increasing and decreasing pairs respectively. Then\n\n\\[\n2I = G + 2B_1\n\\]\n\n\\[\n2D = G + 2B_2\n\\]\n\nsince we are double-counting each pair. Therefore \$G\$ must be even, and since \$2021 = G + B\$ is odd, \$B\$ must be odd.\n\nIf we can prove the existence of a bad current nut, we are done. We show that when a current nut is good, the number of bad upcomming nuts does not change parity after the move. Since there are and odd number of upcomming bad nuts before the first move (every nut is upcomming) and 0 upcomming bad nuts after the last move (there are 0 upcomming nuts), this will show not all current nuts can be good which completes the proof.\n\nWe now show that when the current nut is good, the number of bad upcomming nots does not change parity. Consider the \$k\$'th move and assume the current \$n\$ is good and lies in hole \$H_i\$. After the move, the current nut is no longer upcomming, but since it was asummed to be good, this does not contrubute to the number of bad upcomming nuts. The only nuts whose status can possibly change is the nuts in holes \$H_{i-2}, H_{i-1}, H_{i+1}, H_{i+2}\$. Since the current nut is good, its two neighbors \$a,b\$ in \$H_{i-1}, H_{i+1}\$ respectively are either both smaller or larger than \$n\$. We tackle the two cases seperately.\n\nCase \$a,b < n\$\n\nIn this case neither \$a\$ nor \$b\$ are upcomming nuts at the time of the move and their status after the move is irrelevant for the parity of bad upcomming nuts. Consider the nut \$c\$ in hole \$H_{i-2}\$. If \$c\$ is not upcomming, it is irrelevant. Assume \$c\$ is upcomming. Then \$c > n\$ and thus \$c > a,b\$. Therefore the nut pair \$H_{i-2}, H_{i-1}\$ is decreasing both before and after the move, so \$c\$ cannot change status. One can make an analogous argument for the nut on \$H_{i+2}\$. This completes the proof the the parity of upcomming bad nuts are unchanged in this case.\n\nCase \$a,b > n\$\n\nIn this case the nuts \$a,b\$ are upcomming, so their status matters. Let \$c\$ be the nut in \$H_{i-2}\$. If the status of either of the holes \$H_{i-2}, H_{i-1}\$ change, it changes for both of them. This is because the pair \$H_{i-1}, H_i\$ is decreasing both before and after the move, so for a status change to occur it must be the case that it is the pair \$H_{i-2}, H_{i-1}\$ which changes direction (which in turn changes the status for both holes). If swapping \$a\$ and \$b\$ changes the direction of \$H_{i-2}, H_{i-1}\$, then either \$a < c < b\$ or \$b < c < a\$ is true. In both cases, \$c > n\$, so \$c\$ is an upcomming nut. Since the nuts in both \$H_{i-1}\$ and \$H_{i+1}\$ are upcomming both before and after the move, changing the status of either \$H_{i-2}\$ or \$H_{i-1}\$ yields a status change for two upcomming nuts. A completely analogous argument can be made for \$H_{i+1}, H_{i+2}\$.\n\nThis shows that if some upcomming nut changes status after the move, then an even number of upcomming nuts change status. This preserves the parity of the number of upcomming bad nuts and completes the proof." ]
IMO-2021-6	https://artofproblemsolving.com/wiki/index.php/2021_IMO_Problems/Problem_6	Let $m \ge 2$ be an integer, $A$ be a finite set of (not necessarily positive) integers, and $B_1, B_2, B_3 , \ldots, B_m$ be subsets of $A$. Assume that for each $k = 1, 2,...,m$ the sum of the elements of $B_k$ is $m^k$. Prove that $A$ contains at least $m/2$ elements.	[]
IMO-2022-1	https://artofproblemsolving.com/wiki/index.php/2022_IMO_Problems/Problem_1	The Bank of Oslo issues two types of coin: aluminium (denoted A) and bronze (denoted B). Marianne has $n$ aluminium coins and $n$ bronze coins, arranged in a row in some arbitrary initial order. A chain is any subsequence of consecutive coins of the same type. Given a fixed positive integer $k\le 2n$, Marianne repeatedly performs the following operation: she identifies the longest chain containing the $k^{th}$ coin from the left, and moves all coins in that chain to the left end of the row. For example, if $n = 4$ and $k = 4$, the process starting from the ordering AABBBABA would be AABBBABA → BBBAAABA → AAABBBBA → BBBBAAAA → BBBBAAAA → ... Find all pairs $(n, k)$ with $1 \le k \le 2n$ such that for every initial ordering, at some moment during the process, the leftmost $n$ coins will all be of the same type.	[ "We call a chain basic when it is the largest possible for the coins it consists of. Let \$A=[i,j]\$ be the basic chain with the \$i\$-th and \$j\$-th coins being the first and last, respectively.\n\nClaim:\n\n\\[\nk \\notin \\{1, 2, \\ldots, n-1\\} \\cup \\{\\lceil \\frac{3n}{2} \\rceil + 1, \\ldots, 2n\\}.\n\\]\n\nProof: For \$k < n\$, it is easy to see that the arrangement \$A\\ldots AB\\ldots BA\$ remains the same.\n\nFor \$k > \\lceil \\frac{3n}{2} \\rceil = 2n - \\lfloor \\frac{n}{2} \\rfloor\$, we obtain the arrangement \$A\\ldots AB\\ldots BA\\ldots AB\\ldots B\$, where each basic chain consists of \$\\lfloor \\frac{n}{2} \\rfloor, \\lceil \\frac{n}{2} \\rceil, \\lceil \\frac{n}{2} \\rceil, \\lfloor \\frac{n}{2} \\rfloor\$ coins, respectively. Since the number of coins in the last chain is \$\\geq \\lfloor \\frac{n}{2} \\rfloor\$, it follows that \$k\$ is greater than the number of the remaining coins, or in other words, it is always contained in the last chain.\n\nHowever, we have a loop:\n\n\\[\nA\\ldots AB\\ldots BA\\ldots AB\\ldots B\\rightarrow B\\ldots BA\\ldots AB\\ldots BA\\rightarrow \\ldots \\blacksquare\n\\]\n\nWe will prove that in any other case, the number of basic chains decreases by a constant, which proves the claim.\n\nFor \$k \\in B=[l, m]\$, where \$l > 1\$ and \$m < 2n\$, the basic chains \$B_1=[l{'}, l-1]\$ and \$B_2=[m+1, m']\$ merge into one, and we are done since it is impossible to increase.\n\nFor \$k \\in C=[1, l]\$, where \$n + 1 > l \\geq k \\geq n\$, it holds that \$l = n\$, which is what we need to prove.\n\nFor \$k \\in D=[m, 2n]\$, we will prove that the basic chains are of quantity 2 or 3 (two are obtained with one move): Indeed, if there are at least 4 basic chains, from the beginning of the pigeonhole, we have at least one chain with a number of coins < \$\\lfloor \\frac{n}{2} \\rfloor + 1 \\leq 2n - k + 1\$. Therefore, \$k\$ does not belong to this chain when it is the last, and then the number of basic chains decreases, which completes the proof. \$\\blacksquare\$\n\nIn conclusion, such pairs are \$(n, k)\$, where \$k \\in \\{n, n+1, \\ldots, \\lceil \\frac{3n}{2} \\rceil\\}\$." ]
IMO-2022-2	https://artofproblemsolving.com/wiki/index.php/2022_IMO_Problems/Problem_2	Let $\mathbb{R}^+$ denote the set of positive real numbers. Find all functions $f : \mathbb{R}^+ \to \mathbb{R}^+$ such that for each $x \in \mathbb{R}^+$, there is exactly one $y \in \mathbb{R}^+$ satisfying \[ xf (y) + yf (x) \le 2 \] .	[ "The unique solution is the function \$ f(x) = \\frac{1}{x} \$ for every \$ x \\in \\mathbb{R}^+ \$. This function clearly satisfies the required property since the expression \$ xf(y) + yf(x) = \\frac{x}{y} + \\frac{y}{x} \$ is greater than 2 for every \$ y \\neq x \$ (directly from AM-GM) and equal to 2 (with equality) for the unique value \$ y = x \$.\n\nProof: Let's consider a solution based on some ideas we encountered in the preparation classes for the Olympiad, specifically involving auxiliary sets and functions with specific properties.\n\nThe fact that for every \$ x \\in \\mathbb{R}^+ \$, there exists a unique \$ y \\in \\mathbb{R}^+ \$ that satisfies the equation \$ xf(y) + yf(x) \\leq 2 \$ can be equivalently expressed as follows: there exists a well-defined function \$ g: \\mathbb{R}^+ \\to \\mathbb{R}^+ \$ given by \$ g(x) := y \$, where \$ y \$ is the one mentioned above. The well-definedness of this function is evident due to the existence and uniqueness, and it satisfies the equation \$ P(x): \\quad xf(g(x)) + f(x)g(x) \\leq 2 \$ while applying the same property for \$ x \\mapsto g(x) \$ gives another unique \$ y := g(g(x)) \$ such that \$ g(x)f(y) + yf(g(x)) \\leq 2 \$. Therefore, we have \$ xf(y) + yf(x) > 2 \$ for all \$ y \\neq g(x) \$.\n\nSince this inequality holds for \$ y = x \$ (from \$ xf(y) + yf(x) > 2 \$), the uniqueness assumption implies that \$ g(g(x)) = x \$, making \$ g \$ an involution (hence bijective).\n\nGenerally, working with an involution naturally leads us to consider its fixed points, especially since we aim to show that \$ g(x) = x \$ identically (which holds for the solution \$ f(x) = \\frac{1}{x}\$). Let's define the set of fixed points of \$ g \$ as \$ \\mathcal{S} := \\{ x \\in \\mathbb{R}^+ \\mid g(x) = x \\} \$ and show that \$ \\mathcal{S} = \\mathbb{R}^+ \$ is the entire domain.\n\nAssume for a contradiction that some \$ x \\notin \\mathcal{S} \$ is not a fixed point, i.e., \$ x \\neq g(x) \$. Then, the inequality \$ 2xf(x) > 2 \$ (derived from \$ y \\mapsto x \$) holds, implying \$ f(x) > \\frac{1}{x} \$. Similarly, \$ x \\notin \\mathcal{S} \$ implies \$ g(x) \\notin \\mathcal{S} \$ (otherwise \$ g(x) \\in \\mathcal{S} \$ implies \$ x = g(g(x)) = g(x) \$, a contradiction), leading to \$ f(g(x)) > \\frac{1}{g(x)} \$.\n\nApplying these inequalities to \$ P(x) \$ gives \$ xf(g(x)) + f(x)g(x) < 2 \$, which is clearly a contradiction as \$ \\frac{x}{g(x)} + \\frac{g(x)}{x} \\geqslant 2 \$, e.g., from the AM-GM inequality. Therefore, we must have \$ x \\in \\mathcal{S} \$ for every \$ x \\in \\mathbb{R}^+ \$, i.e., \$ g(x) = x \$.\n\nSubstituting this relationship into the original equation, we obtain \$ P(x): \\quad xf(x) + f(x)x \\leq 2 \\implies xf(x) \\leq 1 \\implies f(x) \\leq \\frac{1}{x} \$ for every \$ x \\in \\mathbb{R}^+ \$. Applying \$ yf(y) \\leq 1 \$ to the equation \$ xf(y) + yf(x) > 2 \$ (since \$ g(x) = x \$) yields \$ f(x) > \\frac{2}{y} - \\frac{x}{y^2} \$, and taking the limit \$ y \\to x \$ from either side results in \$ f(x) \\geq \\frac{1}{x} \$.\n\nCombining the results, we have \$ f(x) \\leq \\frac{1}{x} \$ and \$ f(x) \\geq \\frac{1}{x} \$, implying \$ f(x) = \\frac{1}{x} \$ as desired. \$\\blacksquare\$\n\nNote: This solution is written more extensively and with more details than necessary for a competition, especially since I include comments at certain points to encourage understanding of the ideas and explain the solution. In practice, this idea would take up only a few lines." ]
IMO-2022-3	https://artofproblemsolving.com/wiki/index.php/2022_IMO_Problems/Problem_3	Let $k$ be a positive integer and let $S$ be a finite set of odd prime numbers. Prove that there is at most one way (up to rotation and reflection) to place the elements of $S$ around a circle such that the product of any two neighbours is of the form $x^2 + x + k$ for some positive integer $x$.	[]
IMO-2022-4	https://artofproblemsolving.com/wiki/index.php/2022_IMO_Problems/Problem_4	Let $ABCDE$ be a convex pentagon such that $BC = DE$. Assume that there is a point $T$ inside $ABCDE$ with $TB = TD$, $TC = TE$ and $\angle ABT = \angle TEA$. Let line $AB$ intersect lines $CD$ and $CT$ at points $P$ and $Q$, respectively. Assume that the points $P, B, A, Q$ occur on their line in that order. Let line $AE$ intersect lines $CD$ and $DT$ at points $R$ and $S$, respectively. Assume that the points $R, E, A, S$ occur on their line in that order. Prove that the points $P, S, Q, R$ lie on a circle.	[ "\\[\nTB = TD, TC = TE, BC = DE \\implies\n\\]\n\n\\[\n\\triangle TBC = \\triangle TDE \\implies \\angle BTC = \\angle DTE.\n\\]\n\n\\[\n\\angle BTQ = 180^\\circ - \\angle BTC = 180^\\circ - \\angle DTE = \\angle STE\n\\]\n\n\\[\n\\angle ABT = \\angle AET \\implies \\triangle TQB \\sim \\triangle TSE \\implies\n\\]\n\n\\[\n\\angle PQC = \\angle EST, \\hspace{18mm}\\frac {QT}{ST}= \\frac {TB}{TE} \\implies\n\\]\n\n\\[\nQT \\cdot TE =QT \\cdot TC = ST \\cdot TB= ST \\cdot TD \\implies\n\\]\n\n\$\\hspace{28mm}CDQS\$ is cyclic \$\\implies \\angle QCD = \\angle QSD.\$\n\n\\[\n\\angle QPR =\\angle QPC = \\angle QCD - \\angle PQC =\n\\]\n\n\\[\n\\angle QSD - \\angle EST = \\angle QSR \\implies\n\\]\n\n\$\\hspace{43mm}PRQS\$ is cyclic.\n\nvladimir.shelomovskii@gmail.com, vvsss" ]
IMO-2022-5	https://artofproblemsolving.com/wiki/index.php/2022_IMO_Problems/Problem_5	Find all triples $(a,b,p)$ of positive integers with $p$ prime and \[ a^p = b! + p \]	[ "Case 1: \$b < p\$\n\n- Since \$b!\$ is indivisible by \$p\$, then \$a\$ must also be indivisible by \$p\$.\n\n- If \$a \\le b\$, then \$a^p-b!\$ is divisible by \$a\$, so \$a\$ must be a divisor of \$p\$, but \$a=1\$ obviously has no solutions and we ruled out \$a=p\$ already. For \$a > b\$, let's show that there are no solutions using simple inequalities.\n\n- If \$b < a < p\$, then \$b! \\le (a-1)! \\le (a-1)^{a-1}\$ and \$a^p = (a-1+1)^p \\ge (a-1)^p + p (a-1)^{p-1}\$ by throwing away the remaining (non-negative) terms of binomial theorem. For any solution, \$(a-1)^p + p (a-1)^{p-1} \\le a^p = b!+p \\le (a-1)^{a-1} + p\$, which is impossible for \$a-1 > 1\$. That leaves us with \$a=2\$ and \$b=1\$, but \$2^n > n+1\$ for any integer \$n \\ge 2\$ (proof by induction), so there are no solutions.\n\n- If \$b < p < a\$, RHS is at most \$(p-1)^{p-1}+p\$ and LHS is at least \$(p+1)^p \\ge p^p+p\$ (again from binomial theorem), which gives no solutions as well.\n\nCase 2: \$p \\le b < 2p\$\n\n- Since \$b!\$ is divisible by \$p\$, then \$a\$ must also be divisible by \$p\$.\n\n- In addition, RHS is at most \$(2p-1)!+p = p \\prod_{i=1}^{p-1} (p-i)(p+i) + p < p^{2p-1} + p \\le p^{2p}\$, so \$a < p^2\$. We may write \$a=pc\$, where \$1 \\le c < p\$.\n\n- Since \$c\$ is a divisor of \$a^p\$ and \$b!\$ it must also be a divisor of \$p\$, so \$c=1\$ and \$a=p\$. We're looking for solutions of \$p^{p-1}-1 = b!/p\$.\n\n- Let's factorise \$p^{p-1}-1\$: if \$p-1 = 2^k \\cdot n\$ with \$n\$ odd and \$k > 0\$, it's\n\n\\[\n(p^n-1) \\cdot (p^n+1) \\cdot (p^{2n}+1) \\cdot \\ldots \\cdot (p^{2^{k-1} n}+1) = (p-1) \\cdot (1+p+\\ldots+p^{n-1}) \\cdot (p+1) \\cdot (1-p+\\ldots-p^{n-2}+p^{n-1}) \\cdot (p^{2n}+1) \\cdot \\ldots \\cdot (p^{2^{k-1} n}+1) \\,.\n\\]\n\n- Since \$(1+p+\\ldots+p^{n-1})\$ and \$(1-p+\\ldots-p^{n-2}+p^{n-1})\$ contain an odd number of odd terms (remember the assumption \$k > 0\$ aka \$p \\neq 2\$), they're odd. Also, \$p^2 \\equiv 1\$ modulo \$4\$, so \$(p^{2n}+1)\$ and each following term is even but indivisible by \$4\$. The highest power of \$2\$ dividing \$p^{p-1}-1\$ is therefore \$2^{k + l + k-1}\$ where \$2^l\$ is the highest power dividing \$p+1\$.\n\n- In comparison, \$(p+1)!/p\$ has factors \$(p+1)\$, \$(2^k n)\$ \$(2^{k-1} n)\$ etc (up to \$2n\$), and \$(p-1)/2-k\$ other even factors, so it's divisible at least by \$2^{l+k(k+1)/2+(p-1)/2-k}\$. Since \$l+k(k+1)/2+(p-1)/2-k > l+2k-1\$ for \$p \\ge 7\$, the only possible solutions have \$p < 7\$ or \$b = p\$.\n\n- If \$b=p\$, we reuse the inequalities \$p^{p-1} \\ge (p-1)^{p-1}+p-1\$ and \$b!/p \\le (p-1)^{p-1}\$ to show that there are no solutions for \$p > 2\$.\n\n- Finally, \$5^5-5 = 3120\$ isn't a factorial, \$3^3-3 = 24 = 4!\$ and \$2^2-2 = 2 = 2!\$.\n\nCase 3: \$2p \\le b\$\n\nJust like in case 2, \$b!\$ is divisible by \$p\$ so \$a\$ must also be divisible by \$p\$. However, \$b!\$ and \$a^p\$ are also both divisible by \$p^2\$, so remainders modulo \$p^2\$ tell us that no solutions exist.\n\nConclusion:\n\nThe only solutions are \$(a,b,p) = (2,2,2), (3,4,3)\$.", "I considered the cases:\n\n1) If \$a\$ is even, then it must:\n\n2) If \$a\$ is odd, then it must:\n\nExamining 1.a), we end up with an equation of the form \$a^p = p+1\$, which has no integer solutions.\n\nIn case 1.b), \$b!\$ can take values: 2, 6, 24, 120, 720, ..., so \$b!+2\$ takes values: 4, 8, 26, 122, 722, .... We observe that the only perfect square is 4 among the possible cases, as for \$b \\geq 5\$, the result ends in 2, which is not a perfect square. Therefore, we have the triple \$(2,2,2)\$.\n\nCase 2.a) yields \$a^2 = 3\$, which is rejected.\n\nExamining the last case 2.b), we have for \$b!\$ the values: 2, 6, 24, 120, 720, 5040,\n\nThe analysis of the last case is incomplete, which is why I wasn't initially sure about the number of triples. Therefore, with this approach (which is not strictly documented), we find the triples: \$(2,2,2), (3,4,3)\$.", "Consider \$b \\geq a\$. Then, \$b!\$ must have a factor of \$a\$. Since \$a\\mid a^{p}\$ and \$a\\mid b!\$, \$a\\mid p\$. But \$p\$ is prime so \$a\$ can only be \$1\$ or the set of prime numbers. If \$a = 1\$, then \$b! + p = 1\$ which is impossible since \$b\$ is a positive integer and so is \$p\$. Therefore, \$a\$ must be the set of prime numbers specifically \$a = p\$. This means \$a^{a} = b! + a\$. We can rearrange this to solve this Diophantine Equation: \$a^{a} - a = b!\$ with \$a\$ being a prime number. Wilson's Theorem states that if \$p\$ is a prime number, then \$(p - 1)! \\equiv -1(\\mod p)\$. This motivates us to consider different cases.\n\nCase 1: \$b = k - 1\$ where \$k\$ is a prime\n\nThis means \$a^{a} - a \\equiv -1(\\mod k) = kq - 1 \\implies a\\mid kq - 1 \\implies kq \\equiv 1(\\mod a) = ta + 1\$. Note that \$a^{a} - a\$ is always even. Thus, \$kq\$ is odd. Let \$kq = 2c + 1\$. Here, we clearly see that \$2c = ta\$. Here, we only consider \$a = 2, c = t\$. Note that \$a = 2\$ does indeed lead a solution of \$(a, b, p) = (2, 2, 2)\$ so we have found our first solution triple.\n\nNow, we go back to our original equation: \$a^{a} - a = b!, b = k - 1\$, \$k\$ is a prime. Note that \$k \\geq 5\$. Consider \$k = 5\$. Then, \$a^{a} - a = 24 \\implies a\\mid 24\$. It is immediately obvious that \$a = 3\$ is the only solution which yields a solution of \$(a, b, p) = (3, 4, 3)\$. We will prove that is the second and last solution triple.\n\nCase 1.1: \$a > k\$\n\nThen, \$a^{a} - a > k^{k} - k \\implies kq - 1 > k^{k} - k \\implies kq + k - 1 > k^{k}\$. Since \$k\$ is a prime, we have \$q + 1 - \\frac{1}{k} > k^{k - 1}\$. Note that if \$q < k\$, then this inequality will obviously not hold as all terms are strictly less than \$k\$. If \$q \\geq k\$, then \$k^{2} + k - 1 > k^{k}\$. Using the same trick and dividing by \$k^{2}\$, we get a similar contradiction. Therefore, if \$a > k\$, we have no solution.\n\nCase 1.2: \$a < k\$\n\nNote that \$a = k\$ will lead a contradiction in the modular arithmetic. Note that \$k \\geq 7\$. We have \$a^{a} - a \\leq 7q - 1\$. Notice \$a^{a} - a < k^{k} - k\$. It is immediately obvious that only \$7q - 1 \\leq k^{k} - k\$ will hold solutions(If \$7q - 1 > k^{k} - k\$, we can use a similar trick as above to prove that nothing will work). But note that \$k^{k} - k \\geq 7^{7} - 7\$ which means \$7q - 1 \\leq 7^{7} - 7\$. It is now obvious that \$q \\leq 7^{6} - \\frac{6}{7}\$ \\implies \$q \\leq 7^{5}\$. Recall that \$b \\geq a\$. Thus, \$k - 1 \\geq a\$. We now have \$k - 1 \\mid a(a^{a - 1} - 1) \\implies k - 1 \\mid a\$ or \$k - 1 \\mid a^{a - 1} - 1\$. Notice this just comes from \$a^{a} - a = (k - 1)!\$ as \$b = k - 1\$. But if \$k - 1 \\mid a\$, and because \$a \\leq k - 1\$, it must be that \$a = k - 1 = b\$. Therefore, we have \$a^{a} - a = a!\$ which already tells us that \$a = 2\$ is the only solution, but we have already found this solution. Next, we move on to the case where \$a^{a - 1} - 1 \\equiv 0(\\mod k - 1)\$. One thing we found was that \$q \\leq 7^{5}\$ which came from \$k \\geq 7\$. We will use this later. First, we consider if \$a - 1\$ is prime. If it is, we can apply Fermat's Little Theorem to get \$a^{a - 1} \\equiv a(\\mod a - 1) \\implies a^{a - 1} - 1 \\equiv 0(\\mod a - 1)\$. But this tells us that if \$k - 1 \\mid a^{a - 1} - 1\$, then \$a - 1 \\mid a^{a - 1} - 1\$ considering if \$a - 1\$ is prime. Note that if \$a = k\$, this is impossible because Case 1.2 specifically considers \$a < k\$. Therefore, we have that \$b\$ is actually bigger than at least one prime and we know one of them is \$a - 1\$. Recall \$b! = kq - 1\$. We have \$(k - 1)! = kq - 1\$. We also have \$(a - 1)! < kq - 1\$. Recall that if this was true, \$a \\mid kq - 1\$. Therefore, we can write \$kq - 1 = az\$. Thus, \$(a - 1)! < az\$. Because \$k \\geq 7\$, this tells us that \$az \\geq 7q - 1\$. Now, we recall that \$(a - 1)! < 7q - 1 \\implies (\\frac{7q - 1}{z} - 1)! < 7q - 1\$. We note that \$(\\frac{7q - 1}{z} - 1)! < 7q\$. Note that \$q \\leq 7^{5}\$. We are assuming that if there exists a solution to Case 1.2, there should exist a solution to this new inequality and thus, there should exist a solution to when \$q = 7^{5}\$. We see that \$(\\frac{7^{6} - 1}{z} - 1)! < 7^{6}\$. Note that \$z \\leq 7\$ because it divides \$k\$ and we are considering if equality holds, then \$k = 7\$. But this is obviously not true! Therefore, if \$a - 1\$ is prime and \$a < k\$, there exists no solutions. Now, we go to \$a - 1\$ is composite. Recall that \$k\$ itself is a prime. \$a - 1 < k - 1\$. (I have not finished proving this case.)\n\nCase 1.3: \$b\$ is anything but \$k - 1\$ where \$k\$ is a prime\n\nRecall that we have \$a^{a} - a = b!\$. Notice that \$b = k\$ will lead to a contradiction because \$a \\geq 3 \\implies a^{a} - a\$ is composite. Therefore, \$b = k + \\alpha\$ where \$\\alpha\$ is a constant of anything but \$-1, 0\$. Let \$r\$ be a prime that divides \$(k + \\alpha)!\$. Thus, \$r \\mid a^{a} - a = a(a^{a - 1} - 1)\$. Therefore, we know that \$r \\mid k + \\alpha - i\$ for \$0 \\leq i \\leq k + \\alpha - 1\$. We also know that \$r \\mid a\$ or \$r \\mid a^{a - 1} - 1\$. If we consider \$r \\mid a\$, because \$r\$ is a prime, \$r = a \\implies a\$ is a prime number. In other words, we have just proven that if there exists a prime \$r\$ that divides \$b!\$ where \$b = k + \\alpha\$, then there also may exist a prime \$a\$ that also divides \$b!\$. But because \$b \\geq a\$, \$a = b, b - 1, ..., 1\$. Therefore, at least one of \$b - j\$ where \$0 \\leq j \\leq b - 1\$, must be a prime number. Now, we have \$a^{a} - a \\leq b^{b} - b\$. Let us consider the highest power of prime \$a\$ in \$b!\$. We need to find \$v_a(b!)\$. By Legendre's Formula,\n\n\\[\ne_p(n!)=\\sum_{i=1}^{\\infty} \\left\\lfloor \\dfrac{n}{p^i}\\right\\rfloor =\\frac{n-S_{p}(n)}{p-1}\n\\]\n\n. We just need to find that \$\\sum_{i=1}^{\\infty} \\left\\lfloor \\dfrac{b}{a^i}\\right\\rfloor = a\$. If \$b\$ is a multiple of \$k\$, this will not hold true simply be plugging in. Even if \$b = a + \\beta\$, because the sum is finite, it would imply \$\\beta = 0\$ which goes back to the case where \$b\$ is a multiple of \$a\$. Therefore, the only condition for which this can hold is if \$a = 1\$ which contradicts the part where \$1\$ must divide \$b - i\$ because \$1\$ is not prime. We now have to prove that this case doesn't hold even if \$r \\mid a^{a - 1} - 1\$. In this case, \$a\$ is not necessarily prime. We can apply Fermat's Little Theorem if we assume \$a - 1\$ is a prime. Then, \$a^{a - 1} \\equiv a(\\mod a - 1)\$. This tells us that \$a^{a - 1} - 1 \\equiv 0(\\mod a - 1)\$ which tells us that \$a - 1\$ is another prime that divides it. Therefore, there are two primes that divide \$a^{a - 1} - 1\$ and those are \$r\$ and \$a - 1\$ if \$a - 1\$ is a prime. We conclude that either \$r = a - 1\$ or \$a^{a - 1} - 1\$ has more than one prime factor. If we consider \$r = a - 1\$, we can apply Legendre's again and see that it gives us the same result. If we have \$a^{a - 1} - 1\$ has at least two distinct prime factors and two of them are \$r\$ and \$a - 1\$. This can only occur if \$a = 2\$(The other case will lead to a - 1 being composite which is not what we assumed). But we already considered this case. Therefore, we don't have a new solution here. Now, onto when \$a - 1\$ is not prime.(I have not finished proving this case.)\n\nCase 2: \$p < b < a\$\n\nIf we go back to the original equation of \$a^{p} = b! + p\$, then we see \$p \\mid b! + p\$ and thus \$p \\mid a^{p}\$. Therefore, \$a^{p} \\equiv 0(\\mod p)\$. Fermat's Little Theorem states that \$a^{p} \\equiv a(\\mod p)\$ where \$p\$ is a prime. We have that exact same case here and thus this cannot hold.\n\nCase 3: \$b < p\$, \$b < a\$\n\nIf \$b = p\$, the same story as Case 2 would apply so we can discard that case. Let's consider a stronger inequality: \$b < p < a\$ or \$b < a < p\$ (Note: If a = p, we would get our original two triples so we can discard this case)\n\nCase 3.1: \$b < p < a\$\n\nThis shows that \$a^{a} > a! + a\$. We will show this holds for all integers \$a \\geq 3\$. This is our base case. We will prove this by induction. We want to show that if \$a^{a} > a! + a\$ holds true for all integers \$a \\geq 3\$, then it must hold for all integers such that \$(a + 1)^{a + 1} > (a + 1)! + (a + 1)\$. We have \$(a + 1)^{a + 1} = (a + 1)^{a} \\cdot (a + 1) > a^{a} \\cdot (a + 1)\$. By our inductive statement, \$a^{a} \\cdot (a + 1) > (a! + a)(a + 1) = (a + 1)! + a(a + 1)\$. Thus, we wish to prove \$(a + 1)! + a(a + 1) > (a + 1)! + (a + 1)\$. Remember, our base case was \$a \\geq 3\$ and thus for the inductive case, \$a + 1 \\geq 3\$ which means \$a \\geq 2\$ which satisfies this inequality. Therefore, we have proved this case.\n\nWe have shown that if \$b < p < a\$, then \$a \\geq 3\$. Then, it is obvious that \$b! + p \\geq 3^{p} \\implies b! \\geq 3^{p} - p\$. Since \$b < p \\implies b! < p! \\implies p! > 3^{p} - p\$. We see that this only holds true for \$p \\geq 7\$. But we have shown above that no prime above \$7\$ will satisfy the original equation. Therefore, this inequality doesn't hold.\n\nCase 3.2: \$b < a < p\$\n\nThe same story applies as Case 3.1 but in this case \$p \\geq 3\$. This will mean \$a^{3} = b! + 3, a^{5} = b! + 5\$ and so on. We note that \$p \\geq 7\$ will not work based on our proof for Case 2. Thus, we just need to check \$a^{5} = b! + 5\$ as the previous case led to a solution and unique solution of \$a = 3\$. Note that \$b^{5} < b! + 5\$. In fact, we will show this only holds when \$b = 0, 1\$. We will prove this by contradiction. Assume \$b^{5} < b! + 5\$ for \$b \\geq 2\$. We have \$b^{5} - 5 < b! \\implies b^{5} - 5 < b(b - 1)(b - 2)...(2)(1)\$. Dividing by \$b^{5}\$ gives \$1 - \\frac{5}{b^{5}} < \\frac{b - 1}{b} \\cdot \\frac{b - 2}{b} \\cdot \\frac{b - 3}{b} \\cdot \\frac{b - 4}{b} \\cdot (b - 5)(b - 6)...(2)(1)\$. Note that we can write this as \$1 - \\frac{5}{b^{5}} < (1 - \\frac{1}{b})(1 - \\frac{2}{b})(1 - \\frac{3}{b})(1 - \\frac{4}{b})(b - 5)...(2)(1)\$. If \$b \\geq 5\$, the \$RHS\$ is tending towards \$0\$ by limits and the \$LHS\$ is tending towards \$1\$ by limits as \$b\$ approaches infinity. Therefore, we can check \$b = 2, 3, 4\$ and see none of them hold. Thus our initial assumption was wrong and thus \$b^{5} < b! + 5\$ only holds for \$b = 0,1\$ and \$0,1\$ aren't solutions to the original equation. Therefore, this final case doesn't have a solution either.\n\nHence, the two solutions are \$\\boxed{(a,b,p) = (2,2,2),(3,4,3)}\$.\n\n~ilikemath247365" ]
IMO-2022-6	https://artofproblemsolving.com/wiki/index.php/2022_IMO_Problems/Problem_6	Let $n$ be a positive integer. A Nordic square is an $n \times n$ board containing all the integers from $1$ to $n^2$ so that each cell contains exactly one number. Two different cells are considered adjacent if they share an edge. Every cell that is adjacent only to cells containing larger numbers is called a valley. An uphill path is a sequence of one or more cells such that: (i) the first cell in the sequence is a valley, (ii) each subsequent cell in the sequence is adjacent to the previous cell, and (iii) the numbers written in the cells in the sequence are in increasing order. Find, as a function of $n$, the smallest possible total number of uphill paths in a Nordic square.	[ "The minimum total number of uphill paths in a Nordic square is \$2n(n-1)+1\$\n\nProof: For every pair of adjacent cells in the grid, there is at least one pathway that begins at the higher valued cell, steps to the lower valued cell, and then continues stepping to lower adjacent cells until (by finite descent) a valley is reached. When reversed, this forms an uphill pathway that terminates at that exactly that pair of adjacent cells.\n\nThere are \$2n(n-1)\$ such adjacent pairs (each with at least one unique uphill path). In addition, the cell containing the value 1 will always be a valley which, by itself, forms an additional (singleton) uphill path. Therefore the minimum total number of uphill paths is at least \$2n(n-1)+1\$.\n\nThe diagram below shows a general family of constructions of Nordic squares that achieve this lower bound, thus proving the claim:\n\nIn these constructions, numbers are progressively inserted into the white spaces in ascending order emanating outwards from the \"1\" in the corner in the directions of the red arrows, and the remaining j largest numbers are then inserted into the cells labelled as \"mountains\". In this manner, the pathway of finite descent from each pair of adjacent cells is unique, and all terminate at the \"1\" valley, therefore the above-derived minimum is achieved. \$\\blacksquare\$\n\n- Griphomaniac (talk) 13:16, 21 July 2022 (EDT)" ]
IMO-2023-1	https://artofproblemsolving.com/wiki/index.php/2023_IMO_Problems/Problem_1	Determine all composite integers $n>1$ that satisfy the following property: if $d_1,d_2,\dots,d_k$ are all the positive divisors of $n$ with $1=d_1<d_2<\dots<d_k=n$, then $d_i$ divides $d_{i+1}+d_{i+2}$ for every $1\le i \le k-2$.	[ "If \$n\$ has at least \$2\$ prime divisors, WLOG let \$p<q\$ be the smallest two of these primes. Then the ordered tuple of divisors is of the form \$(1,\\, p,\\, p^2 \\dots,\\, p^a,\\, q \\dots,\\, n)\$ for some integer \$a\\geq 1\$.\n\nTo prove this claim, note that \$p\$ is the smallest prime that divides \$n\$, so it is the smallest divisor not equal to \$1\$, meaning the first \$2\$ divisors are \$1\$ and \$p\$. Furthermore, the smallest divisor of \$n\$ that is not equal to a power of \$p\$ (i.e. not equal to \$(1,\\, p,\\, p^2\\dots)\$ is equal to \$q\$. This is because all other divisors either include a prime \$z\$ different from both \$q\$ and \$p\$, which is larger than \$q\$ (since \$q\$ and \$p\$ are the smallest two prime divisors of \$n\$), or don’t include a different prime \$z\$. In the first case, since \$z>q\$, the divisor is larger than \$q\$. In the second case, all divisors divisible by \$q^2\$ are also larger than \$q\$, and otherwise are of the form \$p^x \\cdot q^1\$ or \$p^x\$ for some nonnegative integer \$x\$. If the divisor is of the form \$p^x\$, then it is a power of \$p\$. If it is of the form \$p^x \\cdot q^1\$, the smallest of these factors is \$p^0 \\cdot q^1 = q\$. Therefore, (in the case where \$2\$ or more primes divide \$n\$) the ordered tuple of divisors is of the form \$(1,\\, p,\\, p^2 \\dots,\\, p^a,\\, q \\dots,\\, n)\$ for some integer \$a\\geq 1\$, since after each divisor \$p^x\$, the next smallest divisor is either \$p^{x+1}\$ or simply \$q\$.\n\nIf \$a\\geq 2\$, the condition fails. This is because \$p^{a-1} \\nmid p^a + q\$, since \$p^a\$ is divisible by \$p^{a-1}\$, but \$q\$ is not since it is a prime different from \$p\$. If \$a=1\$, then \$p^{a-1}=p^0=1\$, which does divide \$q\$. Therefore \$a\$ must equal \$1\$ for the condition to be satisfied in this case. However, we know that the ordered list of divisors satisfies \$d_i \\cdot d_{k+1-i}=n\$, meaning since the first \$3\$ divisors are \$(1, p, q)\$, then the last \$3\$ divisors are \$(\\frac{n}{q}, \\frac{n}{p}, n)\$, so \$(\\frac{n}{q})\$ must divide \$(\\frac{n}{p} + n)\$. But \$\\frac{n}{q}\$ is divisible by \$p\$, so \$\\frac{n}{p} + n\$ must also be divisible by \$p\$, but since \$a=1\$ \$\\frac{n}{p}\$ is and \$n\$ isn't.\n\nWhen \$n=p^x\$, it is easy to verify this works for all primes \$p\$ and all \$x\\geq 2\$, since \$p^y \\vert p^{y+1} + p^{y+2}\$, and the divisors are ordered as \${1,\\, p,\\, p^2…\\, p^x}\$.\n\n~SpencerD,K kai, Najeeb the math genius", "Similar argument as above but restated.\n\nLet \$n = p_1^{e_1} p_2^{d_2} \\dots p_k^{e_k}\$, with \$p_1 < p_2 < \\dots < p_k\$ primes.\n\nIf \$k=1\$, it's easy to verify that this property holds, since \$p^l \\mid p^{l+1} + p^{l+2}\$.\n\nSuppose \$k \\geq 2\$. All divisors of \$n\$ which contain a factor of \$p_k\$ for \$k \\geq 2\$ is at least as large as \$p_2\$. So, the only divisors which are possibly less than \$p_2\$ are of the form \$p_1^{j}\$. Also, by construction \$p_1\$ is less than \$p_2\$. Putting it all together, this says that the smallest factors are \$(1, p_1, \\dots, p_1^j, p_2)\$ for some \$j \\geq 1\$.\n\nIf \$j \\geq 2\$, then \$p_1^{j-1} \\nmid p_1^j + p_2\$, since the LHS has a factor of \$p_1\$ and the RHS does not since.\n\nIf \$j = 1\$, then the largest three factors are \$(\\frac{n}{p_2}, \\frac{n}{p_1}, n)\$. But \$\\frac{n}{p_2}\$ and \$n\$ have a factor of \$p_1^{e_1}\$, while \$\\frac{n}{p_1}\$ only has a factor of \$p_1^{e_1 - 1}\$, so \$\\frac{n}{p_2} \\nmid \\frac{n}{p_1} + n\$.\n\nHence the prime powers are the only composite numbers which satisfy the property. ~kislay kai approved" ]
IMO-2023-2	https://artofproblemsolving.com/wiki/index.php/2023_IMO_Problems/Problem_2	Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$.	[ "Denote the point diametrically opposite to a point \$S\$ through \$S' \\implies AS'\$ is the internal angle bisector of \$\\angle BAC\$.\n\nDenote the crosspoint of \$BS\$ and \$AS'\$ through \$H, \\angle ABS = \\varphi.\$\n\n\\[\nAE \\perp BC, SS' \\perp BC \\implies \\overset{\\Large\\frown} {AS} = \\overset{\\Large\\frown} {ES'} = 2\\varphi \\implies\n\\]\n\n\\[\n\\angle EAS' = \\varphi = \\angle ABS \\implies \\angle DAH = \\angle ABH \\implies\n\\]\n\n\\[\n\\triangle AHD \\sim \\triangle BAH \\implies \\frac {AH}{BH} = \\frac {DH}{AH} \\implies AH^2 = BH \\cdot DH.\n\\]\n\nTo finishing the solution we need only to prove that \$PH = AH.\$\n\nDenote \$F = SS' \\cap AC \\implies \\angle CBS = \\frac {\\overset{\\Large\\frown} {CS}}{2} = \\frac {\\overset{\\Large\\frown} {BS}}{2} = \\frac {\\overset{\\Large\\frown} {AB}}{2} + \\frac {\\overset{\\Large\\frown} {AS}}{2} =\$ \$=\\angle FCB + \\varphi \\implies \\angle FBS = \\angle ABS \\implies H\$ is incenter of \$\\triangle ABF.\$\n\nDenote \$T = S'B \\cap SA \\implies SB \\perp TS', S'A \\perp TS \\implies H\$ is the orthocenter of \$\\triangle TSS'.\$\n\nDenote \$G = PS' \\cap AE \\implies \\angle BPG = \\angle BPS' = \\angle BSS' = \\angle BDG\$ \$\\implies B, L, P, D,\$ and \$G\$ are concyclic.\n\n\\[\n\\angle EBS' = \\varphi, \\angle LBG = \\angle LDG = 90^\\circ = \\angle DBS' \\implies \\angle DBG = \\varphi = \\angle SBF \\implies\n\\]\n\npoints \$B, G,\$ and \$F\$ are collinear \$\\implies GF\$ is symmetric to \$AF\$ with respect \$TF.\$\n\nWe use the lemma and complete the proof.\n\nLemma 1\n\nLet acute triangle \$\\triangle ABC, AB > AC\$ be given.\n\nLet \$H\$ be the orthocenter of \$\\triangle ABC, BHD\$ be the height.\n\nLet \$\\Omega\$ be the circle \$BCD. BC\$ is the diameter of \$\\Omega.\$\n\nThe point \$E\$ is symmetric to \$D\$ with respect to \$AH.\$\n\nThe line \$BE\$ meets \$\\Omega\$ again at \$F \\neq B\$.\n\nProve that \$HF = HD.\$\n\nProof\n\nLet \$\\omega\$ be the circle centered at \$H\$ with radius \$HD.\$\n\nThe \$\\omega\$ meets \$\\Omega\$ again at \$F' \\neq D, HD = HF'.\$\n\nLet \$\\omega\$ meets \$BF'\$ again at \$E' \\neq F'\$.\n\nWe use Reim’s theorem for \$\\omega, \\Omega\$ and lines \$CDD\$ and \$BE'F'\$ and get \$E'D \|\| BC\$\n\n(this idea was recommended by Leonid Shatunov).\n\n\\[\nAH \\perp BC \\implies AH \\perp E'D \\implies\n\\]\n\nThe point \$E'\$ is symmetric to \$D\$ with respect to \$AH \\implies E' = E \\implies F' = F \\implies HF = HD.\$\n\nvladimir.shelomovskii@gmail.com, vvsss", "Let \$O\$ be the circumcenter of \$\\triangle ABC\$. We proceed with showing that \$PH=AH\$. Suppose that \$PD\$ intersects \$SS'\$ and \$\\Omega\$ at \$Q\$ and \$F \\ne A\$ respectively. Note that\n\n\\[\n\\angle CBE=\\angle DLE=\\angle DPB=\\angle FCB \\implies EF \\\| BC.\n\\]\n\nSince \$AE \\perp BC\$, we have \$AE \\perp EF\$ and hence \$AF\$ is a diameter of \$\\Omega\$. By similar triangles \$OQ=\\frac{1}{2}AD\$ and therefore\n\n\\[\n\\frac{SO}{OQ}=\\frac{SS'}{AD}=\\frac{SH}{HD} \\implies OH \\\| DQ.\n\\]\n\nSince \$AF\$ is a diameter of \$\\Omega\$, \$FP \\perp AP \\implies OH \\perp AP\$ and thus \$H\$ lies on the perpendicular bisector of \$AP\$. This proves the claim.", "Identify \$\\Omega\$ with the unit circle, and let the internal bisector of \$\\angle BAC\$ meet \$\\overleftrightarrow{BS}\$ at \$Q\$ and \$\\omega\$ again at \$T\$. We set up so that \\begin{align} \|a\|=\|b\|=\|s\|&=1 \\\\ c &= \\frac{s^2}b \\\\ t &= -s \\\\ e = -\\frac{bc}a &= -\\frac{s^2}a \\\\ d = \\frac{ae(b+s) - bs(a+e)}{ae-bs} &= \\frac{a^2b+abs+as^2-bs^2}{a(b+s)} \\\\ q = \\frac{at(b+s) - bs(a+t)}{at-bs} &= \\frac{2ab+as-bs}{a+b} \\end{align} Now we find the coordinate of \$P\$. We have \$\|p\|=1\$, and\n\n\\[\n\\frac{(b-p)(d-\\ell)}{(b-\\ell)(d-p)} \\in \\mathbb{R}\n\\]\n\nNow \$\\frac{d-\\ell}{b-c}, \\frac{b-\\ell}{b-e} \\in \\mathbb{R}\$. Thus we have\n\n\\[\n\\frac{(b-p)(b-c)}{(b-e)(d-p)} \\in \\mathbb{R}\n\\]\n\nand so\n\n\\[\n\\frac{a^2(b-p)(b+s)^2(b-s)}{b(ab+s^2)(a^2b+abs+as^2-bs^2-abp-aps)} = \\frac{as(b-p)(b+s)^2(b-s)}{b(ab+s^2)(ps^2+aps+abp-a^2p-as^2-abs)}\n\\]\n\n\\[\na(ps^2+aps+abp-a^2p-as^2-abs) = s(a^2b+abs+as^2-bs^2-abp-aps)\n\\]\n\n\\[\n2aps^2+a^2ps+a^2bp-a^3p-a^2s^2-2a^2bs-abs^2-as^3+bs^3+abps = 0\n\\]\n\n\\[\np = \\frac{2a^2bs+abs^2+as^3+a^2s^2-bs^3}{2as^2+a^2s+a^2b+abs-a^3} = \\frac{s(2ab+as-bs)}{a(2s+b-a)}\n\\]\n\nIt remains to show that \$Q\$ lies on the tangent to \$\\omega\$ at \$P\$. Now let \$O\$ be the center of \$\\omega\$. Define the vectors\n\n\\[\nb' = b-p = \\frac{(b-a)(ab+s^2)}{a(2s+b-a)}\n\\]\n\nand \\begin{align} d' = d-p &= \\frac{(2s+b-a)(a^2b+abs+as^2-bs^2) - s(b+s)(2ab+as-bs)}{a(b+s)(2s+b-a)} \\\\ &= \\frac{a^2bs+as^3-bs^3+a^2b^2-a^3b-a^2s^2+abs^2-ab^2s}{a(b+s)(2s+b-a)} \\\\ &= \\frac{(a-s)(b-a)(ab+s^2)}{a(b+s)(2s+b-a)} \\end{align} We observe that \$\\overline{b'} = -\\frac{b'}{bp}\$ and \$\\overline{d'} = \\frac{d'}{ap}\$. Thus if we define \$o' = o-p\$, we have\n\n\\[\no' = \\frac{b'd'(\\overline{b'}-\\overline{d'})}{\\overline{b'}d'-b'\\overline{d'}} = \\frac{ab'+bd'}{a+b} = \\frac{\\frac{(b-a)(ab+s^2)}{2s+b-a} + \\frac{b(a-s)(b-a)(ab+s^2)}{a(b+s)(2s+b-a)}}{a+b} = \\frac{(b-a)(ab+s^2)(2ab+as-bs)}{a(a+b)(b+s)(2s+b-a)}\n\\]\n\nMeanwhile, we compute\n\n\\[\nq-p = \\frac{2ab+as-bs}{a+b} - \\frac{s(2ab+as-bs)}{a(2s+b-a)} = \\frac{(a-s)(b-a)(2ab+as-bs)}{a(a+b)(2s+b-a)}\n\\]\n\nSo\n\n\\[\n\\frac{o-p}{q-p} = \\frac{(ab+s^2)}{(a-s)(b+s)}\n\\]\n\nwhich is pure imaginary. \$\\blacksquare\$ ~approved by Kislay kai" ]
IMO-2023-3	https://artofproblemsolving.com/wiki/index.php/2023_IMO_Problems/Problem_3	For each integer $k \geqslant 2$, determine all infinite sequences of positive integers $a_1, a_2, \ldots$ for which there exists a polynomial $P$ of the form $P(x)=x^k+c_{k-1} x^{k-1}+\cdots+c_1 x+c_0$, where $c_0, c_1, \ldots, c_{k-1}$ are non-negative integers, such that \[ P\left(a_n\right)=a_{n+1} a_{n+2} \cdots a_{n+k} \] for every integer $n \geqslant 1$.	[ "Let \$f(n)\$ and \$g(j)\$ be functions of positive integers \$n\$ and \$j\$ respectively.\n\nLet \$a_{n}=a_{1}+f(n)\$, then \$a_{n+1}=a_{1}+f(n+1)\$, and \$a_{n+k}=a_{1}+f(n+k)\$\n\nLet \$P=\\prod_{j=1}^{k}\\left ( a_{n+j} \\right ) = \\prod_{j=1}^{k}\\left ( a_{n}+g(j)) \\right )\$\n\nIf we want the coefficients of \$P(a_{n})\$ to be positive, then \$g(j)\\geq 0\$ for all \$j\$ which will give the following value for \$P\$:\n\n\\[\nP=a_{n}^{k}+C_{k-1}a_{n}^{k-1}+...+C_{1}a_{n}+\\prod_{j=1}^{k} g(j) = P(a_{n})\n\\]\n\nThus for every \$j\$ and \$n\$ we need the following:\n\n\\[\na_{n}+g(j)=a_{n+j}=a_{1}+f(n+j)\n\\]\n\nSolving for \$g(j)\$ we get:\n\n\\[\ng(j)=a_{1}+f(n+j)-a_{n}=a_{1}+f(n+j)-a_{1}-f(n)\n\\]\n\n\$g(j)=f(n+j)-f(n)\\geq 0\$ for all \$n\$ and \$j\$ because \$g(j)\$ needs to be greater than or equal to zero for all coefficients to be non-negative.\n\nThis means that \$f(n)\$ needs to be increasing with \$n\$, or staying constant, and also with \$f(1)=0\$ because \$a_{1}=a_{1}+f(1)\$.\n\nIn addition, since we need all coefficients to be integer, then all \$f(n)\$ and \$g(j)\$ must also be integers. We also need \$g(j)\$ to not be dependent of \$n\$, so in the expression \$f(n+j)-f(n)\$, the \$n\$ needs to cancel. This mean that the rate of change for \$f(n)\$ with respect to \$n\$ needs to be constant. This can only be achieved with \$f(n)\$ be the equation of a line with slope being either zero or positive integer.\n\nSo, we set \$f(n)\$ to be the equation of a line as \$f(n)=mn+b\$ with \$m\$ being the slope with a non-negative value and with \$b\$ the intercept at \$n=0\$. We know that \$f(1)=0\$ so \$f(1)=m+b=0\$ which means that \$b=-m\$ and our function becomes \$f(n)=mn-m=(n-1)m\$. Since \$f(n)\$ needs to be non-negative integer then \$m\\geq 0 \\mid m \\in \\mathbb{Z}\$ then \$f(n)\$ is increasing or constant, with \$f(1)=0\$\n\nThen, \$g(j)=f(n+j)-f(n)=(n+j-1)m-(n-1)m=jm\$\n\nThis gives:\n\n\\[\n\\prod_{j=1}^{k}\\left ( a_{n}+jm \\right )=P(a_{n})=a_{n}^{k}+C_{k-1}a_{n}^{k-1}+...+C_{1}a_{n}+k!m^{k}\n\\]\n\nwith \$C_{0}=k!m^{k}\$ and coefficients of polynomial \$\\geq 0\$\n\nThen, \$a_{n}=a_{1}+f(n)\$\n\nWhich provides the solution of all infinite sequences of positive integers as:\n\n\$a_{n}=a_{1}+(n-1)m\$, \$\\forall m\\geq 0 \\mid m \\in \\mathbb{Z}\$ and \$a_{1} \\geq 1 \\mid a_{1} \\in \\mathbb{Z}\$\n\n~ Tomas Diaz. orders@tomasdiaz.com" ]
IMO-2023-4	https://artofproblemsolving.com/wiki/index.php/2023_IMO_Problems/Problem_4	Let $x_1, x_2, \cdots , x_{2023}$ be pairwise different positive real numbers such that \[ a_n = \sqrt{(x_1+x_2+ \text{···} +x_n)(\frac1{x_1} + \frac1{x_2} + \text{···} +\frac1{x_n})} \] is an integer for every $n = 1,2,\cdots,2023$. Prove that $a_{2023} \ge 3034$.	[ "We solve for \$a_{n+2}\$ in terms of \$a_n\$ and \$x.\$ \$a_{n+2}^2 \\\\ = (\\sum^{n+2}_{k=1}x_k)(\\sum^{n+2}_{k=1}\\frac1{x_k}) \\\\ = (x_{n+1}+x_{n+2}+\\sum^{n}_{k=1}x_k)(\\frac{1}{x_{n+1}}+\\frac{1}{x_{n+2}}+\\sum^{n}_{k=1}\\frac1{x_k}) \\\\ = \\frac{x_{n+1}}{x_{n+1}} + \\frac{x_{n+1}}{x_{n+2}} + \\frac{x_{n+2}}{x_{n+1}} + \\frac{x_{n+2}}{x_{n+2}} + \\frac{1}{x_{n+1}}\\sum^{n}_{k=1}x_k + x_{n+1}\\sum^{n}_{k=1}\\frac1{x_k} + \\frac{1}{x_{n+2}}\\sum^{n}_{k=1}x_k + x_{n+2}\\sum^{n}_{k=1}\\frac1{x_k} + (\\sum^{n}_{k=1}x_k)(\\sum^{n}_{k=1}\\frac1{x_k}) \\\\ = \\frac{x_{n+1}}{x_{n+1}} + \\frac{x_{n+1}}{x_{n+2}} + \\frac{x_{n+2}}{x_{n+1}} + \\frac{x_{n+2}}{x_{n+2}} + \\frac{1}{x_{n+1}}\\sum^{n}_{k=1}x_k + x_{n+1}\\sum^{n}_{k=1}\\frac1{x_k} + \\frac{1}{x_{n+2}}\\sum^{n}_{k=1}x_k + x_{n+2}\\sum^{n}_{k=1}\\frac1{x_k} + a_n^2 \\\\ \\text{}\$\n\nAgain, by AM-GM, the above equation becomes \$a_{n+2}^2 \\ge 4 \\sqrt[4]{(\\frac{x_{n+1}}{x_{n+1}})(\\frac{x_{n+1}}{x_{n+2}})(\\frac{x_{n+2}}{x_{n+1}})(\\frac{x_{n+2}}{x_{n+2}})} + 4\\sqrt[4]{ (\\frac{1}{x_{n+1}}\\sum^{n}_{k=1}x_k)(x_{n+1}\\sum^{n}_{k=1}\\frac1{x_k})(\\frac{1}{x_{n+2}}\\sum^{n}_{k=1}x_k)(x_{n+2}\\sum^{n}_{k=1}\\frac1{x_k}) } + a_n^2 = a_n^2+4a_n+4 = (a_n+2)^2 \\\\ \\text{}\$\n\nHence, \$a_{n+2} \\ge a_{n} + 2,\$ but equality is achieved only when \$\\frac{x_{n+1}}{x_{n+1}},\\frac{x_{n+1}}{x_{n+2}},\\frac{x_{n+2}}{x_{n+1}},\$ and \$\\frac{x_{n+2}}{x_{n+2}}\$ are equal. They can never be equal because there are no two equal \$x_k.\$So \$a_{2023} \\ge a_1 + 3\\times \\frac{2023-1}{2} = 1 + 3033 = 3034\$" ]
IMO-2023-5	https://artofproblemsolving.com/wiki/index.php/2023_IMO_Problems/Problem_5	Let $n$ be a positive integer. A Japanese triangle consists of $1 + 2 + \dots + n$ circles arranged in an equilateral triangular shape such that for each $i = 1$, $2$, $\dots$, $n$, the $i^{th}$ row contains exactly $i$ circles, exactly one of which is coloured red. A ninja path in a Japanese triangle is a sequence of $n$ circles obtained by starting in the top row, then repeatedly going from a circle to one of the two circles immediately below it and finishing in the bottom row. Here is an example of a Japanese triangle with $n = 6$, along with a ninja path in that triangle containing two red circles. \[ [asy] size(4cm); pair X = dir(240); pair Y = dir(0); path c = scale(0.5)unitcircle; int[] t = {0,0,2,2,3,0}; for (int i=0; i<=5; ++i) { for (int j=0; j<=i; ++j) { filldraw(shift(iX+jY)c, (t[i]==j) ? lightred : white); draw(shift(iX+jY)c); } } draw((0,0)--(X+Y)--(2X+Y)--(3X+2Y)--(4X+2Y)--(5X+2Y),linewidth(1.5)); path q = (3,-3sqrt(3))--(-3,-3sqrt(3)); draw(q,Arrows(TeXHead, 1)); label("$n = 6$", q, S); label("$n = 6$", q, S); [/asy] \] In terms of $n$, find the greatest $k$ such that in each Japanese triangle there is a ninja path containing at least $k$ red circles.	[]
IMO-2023-6	https://artofproblemsolving.com/wiki/index.php/2023_IMO_Problems/Problem_6	Let $ABC$ be an equilateral triangle. Let $A_1,B_1,C_1$ be interior points of $ABC$ such that $BA_1=A_1C$, $CB_1=B_1A$, $AC_1=C_1B$, and \[ \angle BA_1C+\angle CB_1A+\angle AC_1B=480^\circ \] Let $BC_1$ and $CB_1$ meet at $A_2,$ let $CA_1$ and $AC_1$ meet at $B_2,$ and let $AB_1$ and $BA_1$ meet at $C_2.$ Prove that if triangle $A_1B_1C_1$ is scalene, then the three circumcircles of triangles $AA_1A_2, BB_1B_2$ and $CC_1C_2$ all pass through two common points. (Note: a scalene triangle is one where no two sides have equal length.)	[]
IMO-2024-1	https://artofproblemsolving.com/wiki/index.php/2024_IMO_Problems/Problem_1	Determine all real numbers $\alpha$ such that, for every positive integer $n$, the integer \[ \lfloor \alpha \rfloor + \lfloor 2\alpha \rfloor + \dots +\lfloor n\alpha \rfloor \] is a multiple of $n$. (Note that $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$. For example, $\lfloor -\pi \rfloor = -4$ and $\lfloor 2 \rfloor = \lfloor 2.9 \rfloor = 2$.)	[ "To solve the problem, we need to find all real numbers \$ \\alpha \$ such that, for every positive integer \$ n \$, the integer\n\n\\[\nS_n(\\alpha) = \\lfloor \\alpha \\rfloor + \\lfloor 2\\alpha \\rfloor + \\dots + \\lfloor n\\alpha \\rfloor\n\\]\n\nis divisible by \$ n \$, i.e., \$ S_n(\\alpha) \\equiv 0 \\mod n \$.\n\nStep 1: Break Down \$ \\alpha \$ into Integer and Fractional Parts\n\nLet \$ \\alpha = m + f \$, where \$ m = \\lfloor \\alpha \\rfloor \\in \\mathbb{Z} \$ and \$ f = \\{\\alpha\\} \\in [0,1) \$ is the fractional part of \$ \\alpha \$.\n\nStep 2: Express the Sum in Terms of \$ m \$ and \$ f \$\n\nUsing this, we have:\n\n\\[\n\\lfloor k\\alpha \\rfloor = \\lfloor k(m + f) \\rfloor = km + \\lfloor kf \\rfloor.\n\\]\n\nSo, the sum becomes:\n\n\\[\nS_n(\\alpha) = m \\sum_{k=1}^{n} k + \\sum_{k=1}^{n} \\lfloor kf \\rfloor = m \\frac{n(n+1)}{2} + \\sum_{k=1}^{n} \\lfloor kf \\rfloor.\n\\]\n\nStep 3: Modulo \$ n \$ Simplification\n\nWe are interested in \$ S_n(\\alpha) \\mod n \$:\n\n\\[\nS_n(\\alpha) \\equiv \\left( m \\frac{n(n+1)}{2} + \\sum_{k=1}^{n} \\lfloor kf \\rfloor \\right) \\mod n.\n\\]\n\nSince \$ m \\frac{n(n+1)}{2} \$ is divisible by \$ n \$, the expression simplifies to:\n\n\\[\nS_n(\\alpha) \\equiv \\sum_{k=1}^{n} \\lfloor kf \\rfloor \\mod n.\n\\]\n\nStep 4: Analyze the Fractional Part \$ f \$\n\nOur goal is to find all \$ f \\in [0,1) \$ such that:\n\n\\[\n\\sum_{k=1}^{n} \\lfloor kf \\rfloor \\equiv 0 \\mod n \\quad \\text{for all } n \\in \\mathbb{N}.\n\\]\n\nStep 5: Test \$ f = 0 \$\n\nIf \$ f = 0 \$, then \$ \\lfloor kf \\rfloor = 0 \$ for all \$ k \$, so:\n\n\\[\n\\sum_{k=1}^{n} \\lfloor kf \\rfloor = 0,\n\\]\n\nwhich satisfies the condition for all \$ n \$.\n\nStep 6: Consider \$ f \\in (0,1) \$\n\nFor \$ f \\in (0,1) \$, let's test small values of \$ n \$ and \$ f \$. It turns out that the sum \$ \\sum_{k=1}^{n} \\lfloor kf \\rfloor \$ does not satisfy the congruence \$ 0 \\mod n \$ for all \$ n \$. For example, if \$ f = \\frac{1}{2} \$:\n\n\\[\n\\sum_{k=1}^{2} \\lfloor k \\cdot \\frac{1}{2} \\rfloor = \\lfloor \\frac{1}{2} \\rfloor + \\lfloor 1 \\rfloor = 0 + 1 = 1 \\not\\equiv 0 \\mod 2.\n\\]\n\nThus, \$ f \\ne 0 \$ fails the condition.\n\nStep 7: Conclude that Only \$ f = 0 \$ Works\n\nSince \$ f \\ne 0 \$ does not satisfy the condition, the only possible value is \$ f = 0 \$, meaning \$ \\alpha \$ must be an integer.\n\nStep 8: Verify for Integer \$ \\alpha \$\n\nLet \$ \\alpha = m \\in \\mathbb{Z} \$. Then:\n\n\\[\nS_n(\\alpha) = m \\sum_{k=1}^{n} k = m \\frac{n(n+1)}{2}.\n\\]\n\nThis sum is divisible by \$ n \$ if and only if \$ m \$ is even.\n\nBecause \$ \\frac{n(n+1)}{2} \$ times any even integer \$ m \$ is divisible by \$ n \$.\n\nBut \$ \\frac{n(n+1)}{2} \$ times an odd integer \$ m \$ is not, if n is even.\n\nThe only real numbers \$ \\alpha \$ satisfying the condition are the even integers.\n\n~Athmyx", "If we assume that a is a whole number, you could say that the last term is automatically a multiple of n, and from there you can add the first term and the penultimate term, \$ \\alpha \$+\$ \\alpha \$ * (n-1) to get n\$ \\alpha \$ so all the previous terms can be paired up likewise with 2\$ \\alpha \$+\$ \\alpha \$ * (n-2), ... and they will always add up too n* \$ \\alpha \$. From there we know that when n is a even number such as 4 the sum is \$ \\alpha \$+2\$ \\alpha \$+3\$ \\alpha \$+4\$ \\alpha \$. The last term is once again automatically divisible, but in sequences such as this, all the other terms can add up to the last term except for the term in the middle, in this case it would be 2\$ \\alpha \$ , to solve this problem \$ \\alpha \$ must always be even, so that n/2* \$ \\alpha \$=n* \$ \\alpha \$ /2 and alpha is still a whole number. Following the pattern above, all the terms can be added to become nwhole number, and the term in the middle is n \$ \\alpha \$/2, so \$ \\alpha \$ always works when even.\n\nThe number cannot have a fraction or every few sequences it will add a number, but, if the number n is the number directly after the fraction adds one, like if the fraction was 1/3, and n was three, it would add one to the sequence, which would already be divisible by n, but one plus a multiple of n/2 (if \$ \\alpha \$ is odd it will add 1/2n* \$ \\alpha \$ to the sequence) is never a multiple of n unless n is one, or two, and if n was one, the fraction it would would accompany would be 1/1 which would be a whole number already. If n was 2 , then \$ \\alpha \$ would be, an odd number +1/2, if \$ \\alpha \$ was an odd number +1/2 then we can easily disprove it by setting n to 3, in which case it would be a multiple of n+2 , so n would have to be 2 for it to be divisible, but we already set n to 3, so it is impossible. Therefore there cannot be fractions added to the starting number.\n\n~LIUGRA001", "Let's determine all \$\\alpha\$ such that \$\\lfloor\\alpha\\rfloor + \\lfloor 2\\alpha\\rfloor + \\cdots + \\lfloor n\\alpha\\rfloor\$ is divisible by \$n\$ for every positive integer \$n\$.\n\nFirst, even integers work: if \$\\alpha = 2m\$ where \$m\$ is an integer, then\n\n\\[\n\\lfloor\\alpha\\rfloor + \\lfloor 2\\alpha\\rfloor + \\cdots + \\lfloor n\\alpha\\rfloor = 2m + 4m + \\cdots + 2mn = mn(n+1)\n\\]\n\nwhich is divisible by \$n\$.\n\nConversely, let \$\\alpha = k + \\varepsilon\$ where \$k \\in \\mathbb{Z}\$ and \$0 \\leq \\varepsilon < 1\$. Then\n\n\\[\n\\lfloor\\alpha\\rfloor + \\lfloor 2\\alpha\\rfloor + \\cdots + \\lfloor n\\alpha\\rfloor = \\frac{kn(n+1)}{2} + \\lfloor\\varepsilon\\rfloor + \\lfloor 2\\varepsilon\\rfloor + \\cdots + \\lfloor n\\varepsilon\\rfloor\n\\]\n\nCase 1: \$k\$ is even. Then \$\\frac{kn(n+1)}{2}\$ is divisible by \$n\$, so \$\\lfloor\\varepsilon\\rfloor + \\lfloor 2\\varepsilon\\rfloor + \\cdots + \\lfloor n\\varepsilon\\rfloor\$ must be divisible by \$n\$.\n\nBy induction, we can show \$\\lfloor n\\varepsilon\\rfloor = 0\$ for all \$n \\geq 1\$, which implies \$\\varepsilon = 0\$. Thus \$\\alpha\$ is an even integer.\n\nCase 2: \$k\$ is odd. Using similar reasoning, we can derive that \$\\lfloor n\\varepsilon\\rfloor = n-1\$ for all \$n \\geq 1\$, which implies \$1-\\frac{1}{n} \\leq \\varepsilon < 1\$ for all \$n\$. This is impossible.\n\nTherefore, only even integers satisfy the condition.\n\n~brandonyee" ]
IMO-2024-2	https://artofproblemsolving.com/wiki/index.php/2024_IMO_Problems/Problem_2	Find all positive integer pairs $(a,b),$ such that there exists positive integer $g,N,$ \[ \gcd (a^n+b,b^n+a)=g \] holds for all integer $n\ge N$.	[ "We will determine all pairs \$(a,b)\$ of positive integers such that \$\\gcd(a^n+b,b^n+a)=g\$ for all \$n \\geq N\$.\n\nFirst, \$(1,1)\$ works with \$g=2\$. Now for any solution \$(a,b)\$:\n\n\\begin{lemma} \$g = \\gcd(a,b)\$ or \$g = 2\\gcd(a,b)\$. \\end{lemma}\n\n\\begin{proof} Since \$g\$ divides both \$a^N+b\$ and \$a^{N+1}+b\$, it divides their difference \$a^N(a-1)\$. Similarly, \$g\$ divides \$b^N(b-1)\$. Thus \$g\$ divides \$a-b\$, so \$g\$ divides \$a^N+b-(a-b)=a^N+a\$. Hence \$g\$ divides \$\\gcd(a^N+a,a)=\\gcd(a,a-1)=1\$, a contradiction unless \$g\$ divides both \$a\$ and \$b\$. \\end{proof}\n\nLet \$d=\\gcd(a,b)\$ and write \$a=dx\$, \$b=dy\$ with \$\\gcd(x,y)=1\$. Then\n\n\\[\n\\gcd(a^n+b,b^n+a) = d \\cdot \\gcd(d^{n-1}x^n+y, d^{n-1}y^n+x)\n\\]\n\nUsing Euler's theorem, for \$n \\equiv -1 \\pmod{\\varphi(K)}\$ where \$K=d^2xy+1\$, we have:\n\n\\[\nd^{n-1}x^n+y \\equiv d^{-2}x^{-1}+y \\equiv d^{-2}x^{-1}(1+d^2xy) \\equiv 0 \\pmod{K}\n\\]\n\nSimilarly, \$d^{n-1}y^n+x \\equiv 0 \\pmod{K}\$. Since these are divisible by \$K\$, and \$K\$ must divide \$g \\leq 2d\$, we must have \$d=x=y=1\$, giving \$(a,b)=(1,1)\$.\n\n~brandonyee" ]
IMO-2024-3	https://artofproblemsolving.com/wiki/index.php/2024_IMO_Problems/Problem_3	Let $a_1, a_2, a_3, \dots$ be an infinite sequence of positive integers, and let $N$ be a positive integer. Suppose that, for each $n > N$, $a_n$ is equal to the number of times $a_{n-1}$ appears in the list $a_1, a_2, \dots, a_{n-1}$. Prove that at least one of the sequence $a_1, a_3, a_5, \dots$ and $a_2, a_4, a_6, \dots$ is eventually periodic. (An infinite sequence $b_1, b_2, b_3, \dots$ is eventually periodic if there exist positive integers $p$ and $M$ such that $b_{m+p} = b_m$ for all $m \ge M$.)	[]
IMO-2024-4	https://artofproblemsolving.com/wiki/index.php/2024_IMO_Problems/Problem_4	Let $ABC$ be a triangle with $AB < AC < BC$. Let the incentre and incircle of triangle $ABC$ be $I$ and $\omega$, respectively. Let $X$ be the point on line $BC$ different from $C$ such that the line through $X$ parallel to $AC$ is tangent to $\omega$. Similarly, let $Y$ be the point on line $BC$ different from $B$ such that the line through $Y$ parallel to $AB$ is tangent to $\omega$. Let $AI$ intersect the circumcircle of triangle $ABC$ again at $P \neq A$. Let $K$ and $L$ be the midpoints of $AC$ and $AB$, respectively. Prove that $\angle KIL + \angle YPX = 180^{\circ}$ .	[ "Part 1: Derive tangent values \$\\angle AIL\$ and \$\\angle AIK\$ with trig values of angles \$\\frac{A}{2}\$, \$\\frac{B}{2}\$, \$\\frac{C}{2}\$\n\nPart 2: Derive tangent values \$\\angle XPM\$ and \$\\angle YPM\$ with side lengths \$AB\$, \$BC\$, \$CA\$, where \$M\$ is the midpoint of \$BC\$\n\nPart 3: Prove that \$\\angle AIL + \\angle XPM = 90^\\circ\$ and \$\\angle AIK + \\angle YPM = 90^\\circ\$.", "Let point \$Z\$ be \$YZ\|\|AB, XZ\|\|AC, G \\in AC, IG\|\|BC, G' \\in AB, IG' \|\| BC.\$ \$F= AC \\cap YZ, F' = AB \\cap XZ.\$\n\n\$AFZF'\$ is the parallelogram with equal heights, so \$AFZF'\$ is rhomb \$\\implies\$\n\n\\[\n\\angle CAZ = \\angle AZY = \\angle BCP, AI = IZ.\n\\]\n\n\\[\nAL = LC, AI = IZ \\implies IL \|\| ZC \\implies \\angle LIG = \\angle BCZ.\n\\]\n\n\$\\angle AZY = \\angle BCP = \\angle YCP \\implies\$ points \$C,P, Y,\$ and \$Z\$ are concyclic.\n\nTherefore \$\\angle ZPY = \\angle LIG.\$\n\nSimilarly, \$\\angle ZPX = \\angle KIG'. \\blacksquare\$\n\nvladimir.shelomovskii@gmail.com, vvsss" ]
IMO-2024-5	https://artofproblemsolving.com/wiki/index.php/2024_IMO_Problems/Problem_5	Turbo the snail plays a game on a board with 2024 rows and 2023 columns. There are hidden monsters in 2022 of the cells. Initially, Turbo does not know where any of the monsters are, but he knows that there is exactly one monster in each row except the first row and the last row, and that each column contains at most one monster. Turbo makes a series of attempts to go from the first row to the last row. On each attempt, he chooses to start on any cell in the first row, then repeatedly moves to an adjacent cell sharing a common side. (He is allowed to return to a previously visited cell.) If he reaches a cell with a monster, his attempt ends and he is transported back to the first row to start a new attempt. The monsters do not move, and Turbo remembers whether or not each cell he has visited contains a monster. If he reaches any cell in the last row, his attempt ends and the game is over. Determine the minimum value of $n$ for which Turbo has a strategy that guarantees reaching the last row on the $n^{th}$ attempt or earlier, regardless of the locations of the monsters.	[]
IMO-2024-6	https://artofproblemsolving.com/wiki/index.php/2024_IMO_Problems/Problem_6	Let $\mathbb{Q}$ be the set of rational numbers. A function $f: \mathbb{Q} \to \mathbb{Q}$ is called $\emph{aquaesulian}$ if the following property holds: for every $x,y \in \mathbb{Q}$, \[ f(x+f(y)) = f(x) + y \quad \text{or} \quad f(f(x)+y) = x + f(y). \] Show that there exists an integer $c$ such that for any aquaesulian function $f$ there are at most $c$ different rational numbers of the form $f(r) + f(-r)$ for some rational number $r$, and find the smallest possible value of $c$.	[]
IMO-2025-1	https://artofproblemsolving.com/wiki/index.php/2025_IMO_Problems/Problem_1	A line in the plane is called sunny if it is not parallel to any of the $x$–axis, the $y$–axis, and the line $x+y=0$. Let $n\ge3$ be a given integer. Determine all nonnegative integers $k$ such that there exist $n$ distinct lines in the plane satisfying both of the following: - for all positive integers $a$ and $b$ with $a+b\le n+1$, the point $(a,b)$ is on at least one of the lines; and - exactly $k$ of the $n$ lines are sunny.	[ "Consider a valid construction for \$k \\ge 4\$.\n\n\\[\n\\text{Claim: One of the } n \\text{ lines must be } x=1, y=1, \\text{ or } x+y=n.\n\\]\n\nProof: Assume for the sake of contradiction not. Then, the following holds:\n\n\\[\n\\quad \\text{1. } x=1 \\text{ is not in our lines.}\n\\]\n\nOtherwise, two points with \$x=1\$ are on the same line. This implies that each point with \$x\$-coordinate \$1\$ must lie on distinct lines, hence there exists a bijection between the lines and points with \$x\$-coordinate \$1\$. It follows with similar reasoning that:\n\n\\[\n\\quad \\text{2. Lines are bijective with points with } y \\text{-coordinate.}\n\\]\n\n\\[\n\\quad \\text{3. Lines are bijective with points } x+y=n+1.\n\\]\n\nConsider the points on \$x+y=n+1\$ that are not \$(1,n)\$ or \$(n,1)\$. Then, because there exists a bijection, any such point must have a line through a point with \$x\$-coordinate \$1\$ and \$y\$-coordinate \$1\$ that are not \$(1,n)\$ or \$(n,1)\$ (otherwise \$x+y=n+1\$ exists). But this cannot be possible if the point is not \$(1,1)\$. Since \$n \\ge 3\$, by the Pigeonhole Principle there must be at least \$1\$ point that has to pass through this condition, hence we have proved the claim. \$\\square\$\n\n- We can find the constructions for \$0,1,3\$ easily.\n\nHence, remove one of the \$x=1, y=1,\$ or \$x+y=n+1\$ lines. We then get a valid covering for \$n-1\$ with the same number of sunny lines! Thus, any possible number of sunny lines for \$n\$ must be possible for \$n-1\$. For \$n=3\$, we have possibilities \$k=0, k=1, \\text{ or } k=3\$. By our induction above, we conclude that for any \$n\$, the possible \$k\$ is a subset of \$\\{0,1,3\\}\$. \$\\blacksquare\$\n\n~MC", "The answer is 0, 1, or 3 sunny lines.\n\nIn what follows, we draw the grid as equilateral instead of a right triangle; this has no effect on the problem statement but is more symmetric. We say a long line is one of the three lines at the edge of the grid, i.e. one of the (non-sunny) lines passing through \$n\$ points. The main claim is the following.\n\nClaim — If \$n \\geq 4\$, any set of \$n\$ lines must have at least one long line.\n\nProof. Consider the \$3(n-1)\$ points on the outer edge of the grid. If there was no long line, each of the \$n\$ lines passes through at most two such points. So we obtain \$2n \\geq 3(n-1),\$ which forces \$n \\leq 3\$. \$ \\square \$\n\nHence, by induction we may repeatedly delete a long line without changing the number of sunny lines until \$n = 3\$ (and vice-versa: given a construction for smaller \$n\$ we can increase \$n\$ by one and add a long line). We now classify all the ways to cover the \$1+2+3 = 6\$ points in an \$n=3\$ grid with 3 lines.\n\n- If there is a long line (say, the red one in the figure), the remaining \$1+2=3\$ points (circled in blue) are covered with two lines. One of the lines passes through 2 points and must not be sunny; the other line may or may not be sunny. Hence in this case the possible counts of sunny lines are 0 or 1.\n- If there is no long line, each of the three lines passes through at most 2 points. But there are 6 total lines, so in fact each line must pass through exactly two points. The only way to do this is depicted in the figure in the right. In this case there are 3 sunny lines.\n\nThis proves that 0, 1, 3 are the only possible answers.\n\nRemark. The concept of a sunny line is not that important to the problem. The proof above essentially classifies all the ways to cover the \$1+2+\\cdots+n\$ points with exactly \$n\$ lines. Namely, one should repeatedly take a long line and decrease \$n\$ until \$n=3\$, and then pick one of the finitely many cases for \$n=3\$. The count of sunny lines just happens to be whatever is possible for \$n=3\$, since long lines are not sunny.", "Let \$n \\geq 3\$ be a given integer. We want to determine all nonnegative integers \$k\$ such that there exist \$n\$ distinct lines in the plane that cover the set of points \$P_n = \\{(a,b) \\in \\mathbb{Z}^+ \times \\mathbb{Z}^+ : a+b \\leq n+1\\}\$, and exactly \$k\$ of these lines are sunny. A line is sunny if it is not parallel to the \$x\$-axis (Horizontal, H), the \$y\$-axis (Vertical, V), or the line \$x+y=0\$ (Diagonal, D, slope \$-1\$). Lines of these three types are called shady.\n\nWe will show that the possible values for \$k\$ are \$\\{0,1,3\\}\$.\n\nThe proof relies on reducing the problem to the specific case where \$n=k\$ and all lines must be sunny. Let \$C(k)\$ be the assertion that \$P_k\$ can be covered by \$k\$ distinct sunny lines. We define \$P_0 = \u000barnothing\$.\n\n### 1. The Reduction Principle\nLet \$\\mathcal{L}\$ be a set of \$n\$ distinct lines covering \$P_n\$. Let \$k\$ be the number of sunny lines. Let \$N_V, N_H, N_D\$ be the number of V, H, D lines in \$\\mathcal{L}\$, respectively. Then \$N_V + N_H + N_D = n-k\$.\n\nLemma 1 (Structural Lemma). The \$N_V\$ vertical lines in \$\\mathcal{L}\$ must be \$\\{x=1,\\ldots,x=N_V\\}\$. The \$N_H\$ horizontal lines must be \$\\{y=1,\\ldots,y=N_H\\}\$. The \$N_D\$ diagonal lines must be \$\\{x+y=s\\}\$ for \$s=n+2-N_D,\\ldots,n+1\$.\n\nProof. Consider the column \$C_a = P_n \\cap \\{x=a\\}\$. We have \$\|C_a\|=n+1-a\$. Suppose the line \$x=a\$ is not in \$\\mathcal{L}\$. The points in \$C_a\$ must be covered by the other lines in \$\\mathcal{L}\$. The \$N_V\$ vertical lines in \$\\mathcal{L}\$ are distinct from \$x=a\$, so they do not cover any point in \$C_a\$. The remaining \$n-N_V\$ non-vertical lines each cover at most one point in \$C_a\$. Thus, \$\|C_a\| \\leq n-N_V\$. \$n+1-a \\leq n-N_V\$, which implies \$a \\geq N_V+1\$. By contraposition, if \$1 \\leq a \\leq N_V\$, the line \$x=a\$ must be in \$\\mathcal{L}\$. Since there are exactly \$N_V\$ vertical lines in \$\\mathcal{L}\$, these must be \$\\{x=1,\\ldots,x=N_V\\}\$. The argument for horizontal lines is symmetric.\n\nFor diagonal lines, consider the anti-diagonal \$D_s = P_n \\cap \\{x+y=s\\}\$. We have \$\|D_s\|=s-1\$. If \$x+y=s\$ is not in \$\\mathcal{L}\$, the points in \$D_s\$ must be covered by the \$n-N_D\$ lines with slope \$\neq -1\$. Thus, \$s-1 \\leq n-N_D\$, so \$s \\leq n+1-N_D\$. By contraposition, if \$s \\geq n+2-N_D\$, the line \$x+y=s\$ must be in \$\\mathcal{L}\$. \$\\square\$\n\nTheorem 1 (Reduction Theorem). For \$n \\geq 3\$ and \$0 \\leq k \\leq n\$, a configuration of \$n\$ distinct lines covering \$P_n\$ with exactly \$k\$ sunny lines exists if and only if \$C(k)\$ is true.\n\nProof. (\$\\Rightarrow\$) Let \$\\mathcal{L}\$ be such a configuration. Let \$N=N_V+N_H+N_D=n-k\$. The counts of the shady lines are determined by Lemma 1. The remaining \$k\$ sunny lines must cover \$R=\\{(a,b)\\in P_n \\mid a>N_V, b>N_H, a+b \\leq n+1-N_D\\}\$. The \$k\$ sunny lines \$\\mathcal{S} \\subset \\mathcal{L}\$ must cover \$R\$. Consider the translation \$T(a,b)=(a-N_V,b-N_H)=(a',b')\$. If \$(a,b)\\in R\$, then \$a'\\geq 1, b'\\geq 1\$. Also, \$a'+b' = a+b-(N_V+N_H)\\leq (n+1-N_D)-(N_V+N_H)=n+1-(n-k)=k+1\$. \$T\$ maps \$R\$ bijectively to \$P_k\$. The translated lines \$T(\\mathcal{S})\$ cover \$P_k\$. Since translation preserves slopes, these \$k\$ lines are distinct and sunny. Thus \$C(k)\$ is true.\n\n(\$\\Leftarrow\$) Suppose \$C(k)\$ is true. Let \$\\mathcal{L}_k\$ be \$k\$ distinct sunny lines covering \$P_k\$. Let \$N=n-k\$. We construct a configuration for \$P_n\$. Let \$\\mathcal{N}\$ be the set of \$N\$ diagonal lines \$\\{x+y=s \\mid s=k+2,\\ldots,n+1\\}\$. Let \$\\mathcal{L}=\\mathcal{L}_k \\cup \\mathcal{N}\$. This set has \$n\$ lines. They are distinct since lines in \$\\mathcal{L}_k\$ have slope \$\neq -1\$ and lines in \$\\mathcal{N}\$ have slope \$-1\$. They cover \$P_n\$. If \$(a,b)\\in P_n\$, then \$2 \\leq a+b \\leq n+1\$. If \$a+b \\leq k+1\$, then \$(a,b)\\in P_k\$, covered by \$\\mathcal{L}_k\$. If \$k+2 \\leq a+b \\leq n+1\$, then \$(a,b)\$ is covered by \$\\mathcal{N}\$. The configuration has exactly \$k\$ sunny lines. \$\\square\$\n\n### 2. Analysis of the Core Problem \$C(k)\$\n\nWe determine the values of \$k \\geq 0\$ for which \$P_k\$ can be covered by \$k\$ distinct sunny lines.\n\n1. \$k=0\$. \$P_0=\u000barnothing\$. Covered by 0 lines. \$C(0)\$ is true. \n2. \$k=1\$. \$P_1=\\{(1,1)\\}\$. Covered by \$y=x\$ (slope 1, sunny). \$C(1)\$ is true. \n3. \$k=2\$. \$P_2=\\{(1,1),(1,2),(2,1)\\}\$. The lines connecting any pair of these points are \$x=1\$ (V), \$y=1\$ (H), or \$x+y=3\$ (D). All are shady. A sunny line can cover at most one point of \$P_2\$. To cover the 3 points, we need at least 3 sunny lines. Thus \$C(2)\$ is false. \n4. \$k \\geq 3\$. Let \$T_k\$ be the convex hull of \$P_k\$. \$T_k\$ is the triangle with vertices \$V_1=(1,1), V_2=(1,k), V_3=(k,1)\$. The edges of \$T_k\$ lie on the lines \$x=1\$ (V), \$y=1\$ (H), and \$x+y=k+1\$ (D). These lines are shady.\n\nLet \$B_k\$ be the set of points in \$P_k\$ lying on the boundary of \$T_k\$. Each edge contains \$k\$ points. Since the vertices are distinct (as \$k \\geq 2\$), the total number of points on the boundary is \$\|B_k\|=3k-3\$.\n\nSuppose \$P_k\$ is covered by \$k\$ sunny lines \$\\mathcal{L}_k\$. These lines must cover \$B_k\$. Let \$L \\in \\mathcal{L}_k\$. Since \$L\$ is sunny, it does not coincide with the lines containing the edges of \$T_k\$. A line that does not contain an edge of a convex polygon intersects the boundary of the polygon at most at two points. Thus, \$\|L \\cap B_k\|\\leq 2\$. The total coverage of \$B_k\$ by \$\\mathcal{L}_k\$ is at most \$2k\$. We must have \$\|B_k\|\\leq 2k\$. \$3k-3 \\leq 2k\$, which implies \$k \\leq 3\$.\n\nSince we assumed \$k \\geq 3\$, we must have \$k=3\$.\n\n5. \$k=3\$. We verify \$C(3)\$. \$P_3\$ consists of 6 points: \$(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)\$. We provide a covering with 3 sunny lines: \n\$L_1: y=x\$ (slope 1). Covers \$(1,1),(2,2)\$. \n\$L_2: 2x+y=5\$ (slope \$-2\$). Covers \$(1,3),(2,1)\$. \n\$L_3: x+2y=5\$ (slope \$-1/2\$). Covers \$(1,2),(3,1)\$. \nThese lines are sunny and cover \$P_3\$. \$C(3)\$ is true.\n\n### 3. Conclusion\nThe property \$C(k)\$ is true if and only if \$k \\in \\{0,1,3\\}\$. By the Reduction Theorem, for a given \$n \\geq 3\$, a configuration with \$k\$ sunny lines exists if and only if \$C(k)\$ is true and \$k \\leq n\$. Since \$n \\geq 3\$, the condition \$k \\leq n\$ is satisfied for all \$k \\in \\{0,1,3\\}\$.\n\nThe possible values for \$k\$ are 0, 1, and 3.\n" ]
IMO-2025-2	https://artofproblemsolving.com/wiki/index.php/2025_IMO_Problems/Problem_2	Let $\Omega$ and $\Gamma$ be circles with centers $M$ and $N$, respectively, such that the radius of $\Omega$ is less than the radius of $\Gamma$. Suppose circles $\Omega$ and $\Gamma$ intersect at two distinct points $A$ and $B$. Line $MN$ intersects $\Omega$ at $C$ and $\Gamma$ at $D$, such that points $C$, $M$, $N$, and $D$ lie on the line in that order. Let $P$ be the circumcenter of triangle $ACD$. Line $AP$ intersects $\Omega$ again at $E\neq A$. Line $AP$ intersects $\Gamma$ again at $F\neq A$. Let $H$ be the orthocenter of triangle $PMN$. Prove that the line through $H$ parallel to $AP$ is tangent to the circumcircle of triangle $BEF$. (The orthocenter of a triangle is the point of intersection of its altitudes.)	[ "Throughout the solution, we define\n\n\\[\n\\alpha := \\angle DCA = \\angle BCD \\implies \\angle PAD = \\angle CAB = 90^\\circ - \\alpha\n\\]\n\\[\n\\beta := \\angle ADC = \\angle CDB \\implies \\angle CAP = \\angle BAD = 90^\\circ - \\beta.\n\\]\n\nIgnore the points \$H, M, N\$ for now and focus on the remaining ones.\n\nClaim — We have \$\\overline{CE} \\parallel \\overline{AD}\$ and \$\\overline{DF} \\parallel \\overline{AC}\$.\n\nProof. \$\\angle AEC = \\angle ABC = \\angle CAB = 90^\\circ - \\alpha.\$ \$\\square\$\n\nHence, if we let \$A' := \\overline{CE} \\cap \\overline{DF}\$, we have a parallelogram \$ACA'D\$. Note in particular that \$\\overline{BA'} \\parallel \\overline{CD}\$.\n\n---\n\nNext, let \$T\$ denote the circumcenter of \$\\triangle A'EF\$. (This will be the tangency point later in the problem.)\n\nClaim — Point \$T\$ also lies on \$\\overline{BA'}\$ and is also the arc midpoint of \$\\widehat{EF}\$ on \$(BEF)\$.\n\nProof. We compute the angles of \$\\triangle A'EF\$:\n\n\\[\n\\angle FEA' = \\angle AEC = \\angle ABC = \\angle CAB = 90^\\circ - \\alpha\n\\]\n\\[\n\\angle A'FE = \\angle DFA = \\angle DBA = \\angle BAD = 90^\\circ - \\beta\n\\]\n\\[\n\\angle EA'F = \\alpha + \\beta.\n\\]\n\nThen, since \$T\$ is the circumcenter, it follows that:\n\n\\[\n\\angle EA'T = \\angle 90^\\circ - \\angle A'FE = \\beta = \\angle A'CD = \\angle CA'B.\n\\]\n\nThis shows that \$T\$ lies on \$\\overline{BA'}\$.\n\nAlso, we have \$\\angle ETF = 2 \\angle EA'F = 2(\\alpha+\\beta)\$ and\n\n\\[\n\\angle EBF = \\angle EBA + \\angle ABF = \\angle ECA + \\angle ADF\n= \\angle A'CA + \\angle ADA' = (\\alpha+\\beta)+(\\alpha+\\beta) = 2(\\alpha+\\beta),\n\\]\n\nwhich proves that \$T\$ also lies on \$\\overline{AB'}.\$ \$\\square\$\n\n---\n\nWe then bring \$M\$ and \$N\$ into the picture as follows:\n\nClaim — Point \$T\$ lies on both lines \$\\overline{ME}\$ and \$\\overline{NF}\$.\n\nProof. To show that \$F, T, N\$ are collinear, note that \$\\triangle FEA' \\sim \\triangle FAD\$ via a homothety at \$F\$. This homothety maps \$T\$ to \$N\$. \$\\square\$\n\n---\n\nWe now deal with point \$H\$ using two claims.\n\nClaim — We have \$\\overline{MH} \\parallel \\overline{AD}\$ and \$\\overline{NH} \\parallel \\overline{AC}\$.\n\nProof. Note that \$\\overline{MH} \\perp \\overline{PN}\$, but \$\\overline{PN}\$ is the perpendicular bisector of \$\\overline{AD}\$, so in fact \$\\overline{MH} \\parallel \\overline{AD}\$. Similarly, \$\\overline{NH} \\parallel \\overline{AC}\$. \$\\square\$\n\n---\n\nClaim — Lines \$\\overline{MH}\$ and \$\\overline{NH}\$ bisect \$\\angle NMT\$ and \$\\angle MNT\$. In fact, point \$H\$ is the incenter of \$\\triangle TMN\$, and \$\\angle NTH = \\angle HTM = 90^\\circ - (\\alpha+\\beta)\$.\n\nProof. Hence, \$\\angle HMN = \\angle A'CD = \\angle ADC = \\beta\$. But \$\\angle TMN = \\angle CME = 2\\angle CAE = -2(90^\\circ - 2\\beta) = 2\\beta\$. That proves \$\\overline{MH}\$ bisects \$\\angle NMT\$; the other one is similar.\n\nTo show that \$H\$ is an incenter (rather than an excenter) and get the last angle equality, we need to temporarily undirect our angles. Assume WLOG that \$\\triangle ACD\$ is directed counterclockwise. The problem condition that \$C\$ and \$D\$ are the farther intersections of line \$\\overline{MN}\$ mean that \$\\angle NM = \\angle CAD > 90^\\circ\$. We are also promised \$C, M, N, D\$ are collinear in that order. Hence the reflections of line \$\\overline{MN}\$ over lines \$\\overline{MH}\$ and \$\\overline{NH}\$, which meet at \$T\$, should meet at a point for which \$T\$ lies on the same side as \$H\$. In other words, \$\\triangle MTN\$ is oriented counterclockwise and contains \$H\$.\n\nWorking with undirected \$\\alpha = \\angle DCA\$ and \$\\beta = \\angle ADC\$ with \$\\alpha+\\beta < 90^\\circ\$,\n\n\\[\n\\angle NTH = \\angle HTM = \\tfrac{1}{2} \\angle NTM = \\tfrac{1}{2} (180^\\circ - 2(\\alpha+\\beta)) = 90^\\circ - (\\alpha+\\beta).\n\\]\n\nThis matches the claim and finishes the result. \$\\square\$\n\n---\n\nNow\n\\[\n\\angle NFA = 90^\\circ - \\angle ADF = 90^\\circ - (\\alpha+\\beta) = \\angle NTH\n\\]\n\nso \$\\overline{HT} \\parallel \\overline{AP}\$. And since \$TE = TF\$, we have the tangency requested too now, as desired. \$\\square\$\n\n---\n\nRemark. There are many other ways to describe the point \$T\$. For example, \$AMTN\$ is a parallelogram and \$MBTN\$ is an isosceles trapezoid. In coordination, we joked that it was impossible to write a false conjecture.", "### Complete Proof\n\n#### 1. Identification of \$P\$ as the Excenter of \$\\triangle AMN\$.\n\nLet \$R_1\$ and \$R_2\$ be the radii of \$\\Omega\$ (center \$M\$) and \$\\Gamma\$ (center \$N\$) respectively, with \$R_1 < R_2\$. \$P\$ is the circumcenter of \$\\triangle ACD\$, so \$PA = PC\$. Since \$A,C \\in \\Omega\$, \$MA = MC = R_1\$. Thus \$PM\$ is the perpendicular bisector of \$AC\$ and bisects \$\\angle AMC\$. The points \$C,M,N,D\$ are collinear in this order. This implies that the ray \$MC\$ is opposite to the ray \$MN\$. Therefore, \$\\angle AMC\$ and \$\\angle AMN\$ are supplementary. \$\\angle AMC\$ is the exterior angle of \$\\triangle AMN\$ at \$M\$. Since \$PM\$ bisects \$\\angle AMC\$, \$PM\$ is the external angle bisector of \$\\triangle AMN\$ at \$M\$.\n\nSimilarly, \$PA=PD\$ and \$NA=ND=R_2\$. \$PN\$ is the perpendicular bisector of \$AD\$ and bisects \$\\angle AND\$. Since \$M,N,D\$ are in order, the ray \$ND\$ is opposite to the ray \$NM\$. Thus, \$\\angle AND\$ is the exterior angle of \$\\triangle AMN\$ at \$N\$. \$PN\$ is the external angle bisector of \$\\triangle AMN\$ at \$N\$.\n\nTherefore, \$P\$ is the excenter of \$\\triangle AMN\$ opposite to \$A\$. Consequently, the line \$AP\$ is the internal angle bisector of \$\\angle MAN\$. Let \$\\angle MAN = 2\\phi\$. Since the circles intersect at two distinct points \$A\$ and \$B\$, \$\\triangle AMN\$ is non-degenerate, so \$0 < 2\\phi < 180^\\circ\$, i.e., \$0 < \\phi < 90^\\circ\$.\n\n#### 2. Determining \$\\angle EBF\$.\n\nBy symmetry with respect to the line \$MN\$, \$\\triangle BMN \\cong \\triangle AMN\$. Thus \$\\angle MBN = \\angle MAN = 2\\phi\$.\n\nWe use directed angles modulo \$180^\\circ\$. Let \$T_M(B)\$ and \$T_N(B)\$ be the tangents to \$\\Omega\$ and \$\\Gamma\$ at \$B\$, respectively. Since \$T_M(B) \\perp MB\$ and \$T_N(B) \\perp NB\$, we have \$\\angle (T_M(B), T_N(B)) = \\angle (MB, NB)\$.\n\nBy the Tangent-Chord Theorem: In \$\\Omega\$, \$\\angle (T_M(B), BE) = \\angle (AB, AE)\$. In \$\\Gamma\$, \$\\angle (T_N(B), BF) = \\angle (AB, AF)\$. Since \$A,E,F\$ are collinear on the line \$AP\$, the lines \$AE\$ and \$AF\$ are the same. Thus \$\\angle (AB, AE) = \\angle (AB, AF)\$.\n\nWe compute \$\\angle (BE, BF)\$: \n\\[\n\\angle (BE, BF) = \\angle (BE, T_M(B)) + \\angle (T_M(B), T_N(B)) + \\angle (T_N(B), BF)\n\\] \n\\[\n= -\\angle (AB, AE) + \\angle (MB, NB) + \\angle (AB, AF) = \\angle (MB, NB).\n\\] \nThus, the geometric angle \$\\angle EBF = \\angle MBN = 2\\phi\$.\n\nSince \$R_1 \\neq R_2\$, \$\\triangle AMN\$ is not isosceles, so \$AP\$ (the angle bisector) is distinct from the altitude from \$A\$. Since \$AB\$ is perpendicular to \$MN\$, \$AB\$ is the altitude line. Thus \$B\$ is not on \$AP\$. Also \$R_1 \\neq R_2\$ implies \$E \\neq F\$. Thus \$\\triangle BEF\$ is non-degenerate. Let \$\\Sigma\$ be its circumcircle.\n\n#### 3. Introduction of the Auxiliary Point \$V\$ and its properties.\n\nLet \$V\$ be the point such that \$AMVN\$ is a parallelogram. We use vectors originating from \$A\$. \$\\vec{AV} = \\vec{AM} + \\vec{AN}\$.\n\nWe calculate the lengths of \$AE\$ and \$AF\$. In \$\\triangle AME\$, \$MA = ME = R_1\$ and \$\\angle MAE = \\phi\$. Thus \$AE = 2R_1 \\cos \\phi\$. Similarly, \$AF = 2R_2 \\cos \\phi\$. Since \$R_1 < R_2\$ and \$\\cos \\phi > 0\$, \$AE < AF\$. \$A,E,F\$ are collinear in this order on \$AP\$. \$EF = AF - AE = 2(R_2 - R_1) \\cos \\phi\$.\n\nWe calculate the distances \$VE\$ and \$VF\$. \n\$\\vec{VE} = \\vec{AE} - \\vec{AV} = \\vec{AE} - (\\vec{AM} + \\vec{AN})\$. \n\\[\nVE^2 = AE^2 + AM^2 + AN^2 - 2\\vec{AE}\\cdot \\vec{AM} - 2\\vec{AE}\\cdot \\vec{AN} + 2\\vec{AM}\\cdot \\vec{AN}.\n\\] \n\$AM=R_1, AN=R_2\$. \$\\angle MAN = 2\\phi, \\angle MAE = \\angle NAE = \\phi\$. \n\$\\vec{AE}\\cdot \\vec{AM} = AE\\cdot R_1 \\cos \\phi = 2R_1^2 \\cos^2 \\phi\$. \n\$\\vec{AE}\\cdot \\vec{AN} = AE\\cdot R_2 \\cos \\phi = 2R_1 R_2 \\cos^2 \\phi\$. \n\$\\vec{AM}\\cdot \\vec{AN} = R_1 R_2 \\cos (2\\phi) = R_1 R_2 (2\\cos^2 \\phi - 1)\$.\n\nSo \n\\[\nVE^2 = (2R_1 \\cos \\phi)^2 + R_1^2 + R_2^2 - 4R_1^2 \\cos^2 \\phi - 4R_1 R_2 \\cos^2 \\phi + 2R_1 R_2 (2\\cos^2 \\phi - 1).\n\\] \n\\[\nVE^2 = R_1^2 + R_2^2 - 2R_1 R_2 = (R_2 - R_1)^2.\n\\] \nSo \$VE = R_2 - R_1\$. A similar calculation shows \$VF = R_2 - R_1\$. Thus \$VE = VF\$.\n\n#### 4. \$V\$ lies on the circumcircle \$\\Sigma\$.\n\nWe calculate \$\\angle EVF\$ using the Law of Cosines in the isosceles triangle \$\\triangle EVF\$. \n\\[\nEF^2 = VE^2 + VF^2 - 2VE \\cdot VF \\cos (\\angle EVF) = 2VE^2 (1 - \\cos (\\angle EVF)).\n\\] \n\$(2(R_2 - R_1) \\cos \\phi)^2 = 2(R_2 - R_1)^2 (1 - \\cos (\\angle EVF)).\$ \n\$4\\cos^2 \\phi = 2(1 - \\cos (\\angle EVF)).\$ \n\$\\cos (\\angle EVF) = 1 - 2\\cos^2 \\phi = -\\cos (2\\phi).\$ \nSince \$2\\phi \\in (0,180^\\circ)\$, \$\\angle EVF = 180^\\circ - 2\\phi\$.\n\nWe have \$\\angle EBF + \\angle EVF = 2\\phi + (180^\\circ - 2\\phi) = 180^\\circ\$. To conclude that \$BEVF\$ is cyclic, we must verify that \$B\$ and \$V\$ lie on opposite sides of the line \$AP\$. \n\nWe set up a coordinate system with \$A\$ at the origin \$(0,0)\$ and \$AP\$ along the positive \$x\$-axis. We can orient it such that \n\\[\nM = (R_1 \\cos \\phi, R_1 \\sin \\phi), \\quad N = (R_2 \\cos \\phi, -R_2 \\sin \\phi).\n\\]\nThen \n\\[\nV = M+N \\quad \\text{has y-coordinate } y_V = (R_1 - R_2)\\sin \\phi.\n\\]\nSince \$R_1 < R_2\$ and \$\\phi > 0\$, we have \$y_V < 0\$.\n\nPoint \$B\$ is the reflection of \$A\$ across the line \$MN\$. The line \$MN\$ has the equation\n\\[\ny - y_M = m(x - x_M), \\quad \\text{where slope } m = \\frac{-(R_1+R_2)\\sin \\phi}{(R_2-R_1)\\cos \\phi}.\n\\]\nThe y-intercept \$b\$ (intersection with the axis perpendicular to \$AP\$ through \$A\$) is\n\\[\ny_M - m x_M = R_1 \\sin \\phi - m R_1 \\cos \\phi = R_1 \\sin \\phi + \\frac{R_1(R_1+R_2)\\sin \\phi}{R_2-R_1} = \\frac{2R_1 R_2 \\sin \\phi}{R_2-R_1}.\n\\]\nSince \$R_i > 0\$ and \$\\sin \\phi > 0\$, \$b > 0\$. The line \$MN\$ passes “above” \$A\$ with respect to the y-axis. The reflection \$B\$ of \$A(0,0)\$ across the line \$y=mx+b\$ has y-coordinate\n\\[\ny_B = \\frac{2b}{m^2+1} > 0.\n\\]\n\nSo, since \$y_V < 0\$ and \$y_B > 0\$, \$V\$ and \$B\$ are on opposite sides of \$AP\$. Thus, \$BEVF\$ is cyclic, and \$V\$ lies on \$\\Sigma\$.\n\n#### 5. The Orthocenter \$H\$ and the Tangency Condition.\n\nLet \$I\$ be the incenter of \$\\triangle AMN\$. Since \$P\$ is the excenter opposite to \$A\$, the points \$A,I,P\$ are collinear on the line \$AP\$. The incenter bisector \$MI\$ and the external bisector \$MP\$ at \$M\$ are perpendicular. Similarly, \$NI \\perp NP\$. Thus, the quadrilateral \$IMPN\$ is cyclic. This circle is the circumcircle of \$\\triangle MPN\$. Let \$O\$ be its center. \$IP\$ is the diameter, so \$O\$ is the midpoint of \$IP\$.\n\n\$H\$ is the orthocenter of \$\\triangle PMN\$. By Sylvester’s theorem relating the circumcenter \$O\$ and the orthocenter \$H\$, we have \n\$\\vec{AH} = \\vec{AP} + \\vec{AM} + \\vec{AN} - 2\\vec{AO}.\$ By definition of \$V\$, \$\\vec{AV} = \\vec{AM} + \\vec{AN}.\$ \n\$\\vec{AH} = \\vec{AP} + \\vec{AV} - 2\\vec{AO}.\$\n\nThe vector from \$I\$ to \$H\$ is \$\\vec{VH} = \\vec{AH} - \\vec{AI} = \\vec{AP} - 2\\vec{AO}.\$ Since \$O\$ is the midpoint of \$IP\$, \$2\\vec{AO} = \\vec{AI} + \\vec{AP}\$. So \n\$\\vec{VH} = \\vec{AP} - (\\vec{AI} + \\vec{AP}) = -\\vec{AI} = \\vec{IA}.\$\n\nSince \$I\$ and \$A\$ lie on the line \$AP\$, the vector \$\\vec{IA}\$ is parallel to \$AP\$. Thus, the line segment \$\\overline{VH}\$ is parallel to \$AP\$. The line through \$H\$ parallel to \$AP\$ is the line \$\\overline{VH}\$.\n\nWe must show that the line \$\\overline{VH}\$ is tangent to \$\\Sigma\$. Since \$V \\in \\Sigma\$, it suffices to show that \$\\overline{VH}\$ is perpendicular to the radius at \$V\$. Let \$O_\\Sigma\$ be the center of \$\\Sigma\$. We need to show \$\\overline{VH} \\perp O_\\Sigma V\$. Since \$\\overline{VH} \\parallel AP\$, we need \$AP \\perp O_\\Sigma V\$. The points \$E,F\$ lie on \$AP\$. In Step 3, we proved \$VE=VF\$. Thus \$V\$ lies on the perpendicular bisector of chord \$EF\$. \$O_\\Sigma\$ also lies on this bisector. Therefore, the line \$O_\\Sigma V\$ is the perpendicular bisector of \$EF\$. Thus \$O_\\Sigma V \\perp EF\$. Since \$EF\$ lies on \$AP\$, \$O_\\Sigma V \\perp AP\$.\n\nWe conclude that \$\\overline{VH} \\perp O_\\Sigma V\$. Therefore, the line \$\\overline{VH}\$, which is the line through \$H\$ parallel to \$AP\$, is tangent to the circumcircle of triangle \$BEF\$ at \$V\$. \$\\square\$" ]
IMO-2025-3	https://artofproblemsolving.com/wiki/index.php/2025_IMO_Problems/Problem_3	Let $\mathbb{N}$ denote the set of positive integers. A function $f: \mathbb{N} \rightarrow \mathbb{N}$ is said to be bonza if \[ f(a) \] divides \[ b^{a} - f(b)^{f(a)} \] for all positive integers $a$ and $b$. Determine the smallest real constant $c$ such that $f(n) \leq cn$ for all bonza functions $f$ and all positive integers $n$.	[ "The answer is \$c = 4\$.\n\nLet \$P(a,b)\$ denote the given statement \$f(a) \\mid b^a - f(b)^{f(a)}\$.\n\n---\n\nClaim — We have \$f(n) \\mid n^n\$ for all \$n\$.\n\nProof. Take \$P(n,n)\$. \$\\square\$\n\n---\n\nClaim — Unless \$f = \\mathrm{id}\$, we have \$f(p) = 1\$ for all odd primes \$p\$.\n\nProof. Consider any prime \$q\$ with \$f(q) > 1\$. Then \$f(q)\$ is a power of \$q\$, and for each \$n\$ we get\n\\[\nP(q,n) \\implies q \\mid f(q) \\mid n^q - f(n)^{f(q)}.\n\\]\nFermat’s little theorem now gives \$n^q \\equiv n \\pmod{q}\$ and \$f(n)^{f(q)} \\equiv f(n) \\pmod{q}\$ (since \$f(q)\$ is a power of \$q\$), and therefore \$q \\mid n - f(n)\$. Hence, unless \$f\$ is the identity function, only finitely many \$q\$ could have \$f(q) > 1\$.\n\nNow let \$p\$ be any odd prime, and let \$q\$ be a large prime such that \$q \\not\\equiv 1 \\pmod{p}\$ (possible for all \$p > 2\$, say by Dirichlet). Then\n\\[\nP(p,q) \\implies f(p) \\mid q^p - 1^p.\n\\]\nThe RHS is \$q^p - 1 \\equiv q-1 \\not\\equiv 0 \\pmod{p}\$, so \$f(p) = 1\$. \$\\square\$\n\n---\n\nClaim — We have \$f(n) \\mid 2^\\infty\$ for all \$n\$.\n\nProof. If \$p \\mid f(n)\$ is odd then \$P(n,p)\$ gives \$p \\mid f(n) \\mid p^n - 1^n\$, contradiction. \$\\square\$\n\n(In particular, we now know \$f(n) = 1\$ for all odd \$n\$, though we don’t use this.)\n\n---\n\nClaim — We have \$f(n) \\leq 2^{\\nu_2(n)+2}\$ for all \$n\$.\n\nProof. Consider \$P(n,5) \\implies f(n) \\mid 5^n - 1^n\$. It’s well-known that \$\\nu_2(5^n - 1) = \\nu_2(n) + 2\$ for all \$n\$. \$\\square\$\n\n---\n\nThis immediately shows \$f(n) \\leq 4n\$ for all \$n\$, hence \$c = 4\$ in the problem statement works.\n\nFor the construction, the simplest one seems to be\n\\[\nf(n) = \\begin{cases}\n1 & n \\text{ odd} \\\\\n16 & n = 4 \\\\\n2 & n \\text{ even}, n \\neq 4\n\\end{cases}\n\\]\nwhich is easily checked to work and has \$f(4) = 16\$.\n\n---\n\nRemark. With a little more case analysis we can classify all functions \$f\$. The two trivial solutions are \$f(n) = n\$ and \$f(n) = 1\$; the others are described by writing \$f(n) = 2^{e(n)}\$ for any function \$e\$ satisfying\n- \$e(n) = 0\$ for odd \$n\$;\n- \$1 \\leq e(2) \\leq 2\$;\n- \$1 \\leq e(n) \\leq \\nu_2(n)+2\$ for even \$n > 2\$.\n\nThis basically means that there are almost no additional constraints beyond what is suggested by the latter two claims.", "We want to determine the smallest real constant \$c\$ such that \$f(n) \\leq cn\$ for all bonza functions \$f : \\mathbb{N} \\to \\mathbb{N}\$ and all \$n \\in \\mathbb{N}\$. A function \$f\$ is bonza if \$f(a) \\mid b^a - f(b)^{f(a)}\$ for all \$a,b \\in \\mathbb{N}\$. Let \$P(a,b)\$ denote this assertion.\n\nWe will show that the smallest constant is \$c=4\$.\n\n---\n\n### Part 1: Properties and Classification of Bonza Functions\n\nFirst, we establish some basic properties. \$P(a,a)\$ implies \$f(a) \\mid a^a\$. \$P(1,1)\$ implies \$f(1) = 1\$. Let \$S\$ be the set of primes \$p\$ such that \$f(p) > 1\$. Since \$f(p) \\mid p^p\$, if \$p \\in S\$, then \$f(p) = p^k\$ for some \$k \\geq 1\$.\n\nLemma 1. If \$p \\in S\$, then \$f(b) \\equiv b \\pmod{p}\$ for all \$b \\in \\mathbb{N}\$. \nProof. \$P(p,b) \\implies f(p) \\mid b^p - f(b)^{f(p)}\$. Since \$p \\in S\$, \$p \\mid f(p)\$. Thus \$b^p \\equiv f(b)^{f(p)} \\pmod{p}\$. By Fermat’s Little Theorem (FLT), \$b^p \\equiv b \\pmod{p}\$. Since \$f(p)\$ is a power of \$p\$, applying FLT repeatedly yields \$f(b)^{f(p)} \\equiv f(b) \\pmod{p}\$. Thus, \$b \\equiv f(b) \\pmod{p}\$. \$\\square\$\n\nLemma 2 (Classification). The set \$S\$ is either the set of all primes \$\\mathbb{P}\$, the empty set \$\\varnothing\$, or the singleton set \$\\{2\\}\$. \nProof. Case 1: \$S\$ is infinite. By Lemma 1, for any \$b \\in \\mathbb{N}\$, \$f(b) - b\$ is divisible by every prime in \$S\$. Since \$S\$ is infinite, \$f(b) - b = 0\$, so \$f(b) = b\$ for all \$b\$. Then \$f(p) = p > 1\$ for all primes \$p\$, so \$S = \\mathbb{P}\$.\n\nCase 2: \$S\$ is finite. Let \$M = \\prod_{p \\in S} p\$. (If \$S = \\varnothing\$, \$M=1\$). Let \$q\$ be a prime not in \$S\$. Then \$f(q) = 1\$. If \$S\$ is non-empty, for any \$p \\in S\$, Lemma 1 gives \$1 \\equiv f(q) \\equiv q \\pmod{p}\$. Thus \$q \\equiv 1 \\pmod{M}\$.\n\nSuppose \$S\$ is finite and non-empty. Then \$M \\geq 2\$. Suppose \$M > 2\$. Consider \$A = M-1\$. Since \$M > 2\$, \$1 < A < M\$. We have \$\\gcd(A,M) = 1\$. Let \$q_0\$ be any prime factor of \$A\$. Then \$q_0 \\nmid M\$, so \$q_0 \\notin S\$. Thus \$q_0 \\equiv 1 \\pmod{M}\$. This implies \$M \\mid q_0 - 1\$, so \$M \\leq q_0 - 1\$. Since \$q_0 \\mid A\$, \$q_0 \\leq A = M-1\$. Combining these gives \$M \\leq q_0 - 1 \\leq (M-1) - 1 = M-2\$. Contradiction. Therefore, if \$S\$ is finite and non-empty, we must have \$M=2\$. This means \$S=\\{2\\}\$. If \$S\$ is empty, \$M=1\$. \$\\square\$\n\n---\n\n### Part 2: Establishing the Upper Bound \$c \\leq 4\$\n\nWe analyze the three cases from Lemma 2.\n\nCase 1: \$S=\\mathbb{P}\$. We found \$f(n) = n\$. Then \$f(n)/n = 1\$.\n\nCase 2: \$S=\\varnothing\$. \$f(p) = 1\$ for all primes \$p\$. Let \$n \\in \\mathbb{N}\$. If \$f(n) > 1\$, let \$q\$ be a prime factor of \$f(n)\$. Since \$f(n) \\mid n^n\$, \$q \\mid n\$. \$P(n,q) \\implies f(n) \\mid q^n - f(q)^{f(n)} = q^n - 1\$. Since \$q \\notin S\$, \$f(q) = 1\$. So \$f(n) \\mid q^n - 1\$. Since \$q \\mid f(n)\$, \$q \\mid q^n - 1\$. As \$q \\mid n\$, \$q \\mid q^n\$. Thus \$q \\mid 1\$. Contradiction. So \$f(n)=1\$ for all \$n\$. \$f(n)/n \\leq 1\$.\n\nCase 3: \$S=\\{2\\}\$. \$f(2) > 1\$, and \$f(p)=1\$ for all odd primes \$p\$. First, we show \$f(n)\$ is a power of 2 for all \$n\$. Let \$q\$ be an odd prime factor of \$f(n)\$. Then \$q \\mid n\$. \$f(q)=1\$. \$P(n,q) \\implies f(n) \\mid q^n - f(q)^{f(n)} = q^n - 1\$. Since \$q \\mid f(n)\$, \$q \\mid q^n - 1\$. This is impossible as \$q \\mid n\$ implies \$q \\mid q^n\$. Thus \$f(n)\$ is a power of 2.\n\nIf \$n\$ is odd, \$f(n) \\mid n^n\$ (odd). So \$f(n)=1\$.\n\nIf \$n\$ is even. Let \$n = 2^k m\$, where \$k=\\nu_2(n) \\geq 1\$ and \$m\$ is odd. Let \$f(n)=2^e\$. Let \$b\$ be any odd integer. \$f(b)=1\$. \$P(n,b) \\implies f(n) \\mid b^n - f(b)^{f(n)} = b^n - 1\$. So \$2^e \\mid b^n - 1\$. Thus \$e \\leq \\min_{b \\text{ odd}} \\nu_2(b^n - 1)\$.\n\nWe need the following lemma to analyze the 2-adic valuation.\n\nLemma 3. Let \$X\$ be an odd integer and \$K \\geq 1\$ an integer. Then \$\\nu_2(X^{2^K} - 1) = \\nu_2(X^2 - 1) + K - 1\$. \nProof. We use induction on \$K\$. Base case \$K=1\$: \$\\nu_2(X^2-1)=\\nu_2(X^2-1)+1-1\$. True. Inductive step: Assume it holds for \$K \\geq 1\$. We check \$K+1\$. \n\$\\nu_2(X^{2^{K+1}}-1) = \\nu_2((X^{2^K}-1)(X^{2^K}+1))\$. Since \$X\$ is odd, \$X^2 \\equiv 1 \\pmod{8}\$. Since \$K \\geq 1\$, \$X^{2^K} = (X^2)^{2^{K-1}} \\equiv 1^{2^{K-1}} = 1 \\pmod{8}\$. Thus \$X^{2^K}+1 \\equiv 2 \\pmod{8}\$, so \$\\nu_2(X^{2^K}+1)=1\$. \$\\nu_2(X^{2^{K+1}}-1) = \\nu_2(X^{2^K}-1)+1 = \\nu_2(X^2-1)+K-1+1 = \\nu_2(X^2-1)+K\$. \$\\square\$\n\nNow we analyze \$\\nu_2(b^n-1)=\\nu_2(b^{2^k m}-1)\$. Let \$X=b^m\$. Since \$b,m\$ are odd, \$X\$ is odd. By Lemma 3 (with \$K=k\$), \$\\nu_2(b^n-1) = \\nu_2(X^{2^k}-1) = \\nu_2(X^2-1)+k-1\$. We want to minimize this over odd \$b\$. \$X^2-1=b^{2m}-1\$. Since \$b^m\$ is odd, \$(b^m)^2 \\equiv 1 \\pmod{8}\$, so \$\\nu_2(b^{2m}-1) \\geq 3\$. The minimum is achieved when \$b=3\$. We calculate \$\\nu_2(3^{2m}-1)=\\nu_2(9^m-1)\$. \$9^m-1=(9-1)(9^{m-1}+\\dots+1)\$. The second factor is a sum of \$m\$ odd terms. Since \$m\$ is odd, the sum is odd. \$\\nu_2(9^m-1)=\\nu_2(8)=3\$. Thus, \$\\min_{b \\text{ odd}} \\nu_2(b^n-1)=3+(k-1)=k+2\$. So \$e \\leq k+2\$.\n\nThe ratio is\n\\[\n\\frac{f(n)}{n} = \\frac{2^e}{2^k m} \\leq \\frac{2^{k+2}}{2^k m} = \\frac{4}{m}.\n\\]\nSince \$m \\geq 1\$, \$f(n)/n \\leq 4\$.\n\nIn all cases, \$f(n) \\leq 4n\$ for all bonza functions \$f\$. Thus \$c \\leq 4\$.\n\n---\n\n### Part 3: Construction and Lower Bound \$c \\geq 4\$\n\nWe construct a bonza function \$g(n)\$ that achieves the bound 4. Define \$g(n)\$ as follows:\n\\[\ng(n) = \\begin{cases}\n1 & \\text{if } n \\text{ is odd} \\\\\n16 & \\text{if } n=4 \\\\\n2 & \\text{if } n \\text{ is even and } n \\neq 4\n\\end{cases}\n\\]\n\nFor \$n=4\$, \$g(4)=16\$, so \$g(4)/4=4\$. If \$g\$ is bonza, then \$c \\geq 4\$.\n\nWe verify that \$g\$ is bonza. We check \$g(a) \\mid b^a - g(b)^{g(a)}\$.\n\nCase 1: \$a\$ is odd. \$g(a)=1\$. The condition holds trivially.\n\nCase 2: \$a=4\$. \$g(4)=16\$. We need \$16 \\mid b^4 - g(b)^{16}\$. If \$b\$ is odd, \$g(b)=1\$. We need \$16 \\mid b^4-1\$. For any odd \$b\$, \$b^2 \\equiv 1 \\pmod{16}\$, so \$b^4 \\equiv 1 \\pmod{16}\$. If \$b\$ is even, \$\\nu_2(b) \\geq 1\$. \$\\nu_2(b^4) \\geq 4\$. So \$16 \\mid b^4\$. \$g(b)\$ is either 2 or 16. \$g(b)\$ is even. \$\\nu_2(g(b)^{16}) \\geq 16\$. So \$16 \\mid g(b)^{16}\$. The condition holds.\n\nCase 3: \$a\$ is even and \$a \\neq 4\$. \$g(a)=2\$. \$a \\geq 2\$. We need \$2 \\mid b^a - g(b)^2\$. If \$b\$ is odd, \$b^a\$ is odd. \$g(b)=1\$. \$b^a - g(b)^2\$ is odd-1=even. If \$b\$ is even, \$b^a\$ is even (since \$a \\geq 2\$). \$g(b) \\in \\{2,16\\}\$. \$g(b)\$ is even, so \$g(b)^2\$ is even. \$b^a - g(b)^2\$ is even. The condition holds.\n\nThus, the function \$g(n)\$ is bonza.\n\n---\n\n### Conclusion\n\nWe have proved that \$f(n) \\leq 4n\$ for all bonza functions \$f\$, so \$c \\leq 4\$. We have constructed a bonza function \$g(n)\$ such that \$g(4)/4=4\$, so \$c \\geq 4\$. Therefore, the smallest real constant \$c\$ is 4." ]
IMO-2025-4	https://artofproblemsolving.com/wiki/index.php/2025_IMO_Problems/Problem_4	A proper divisor of a positive integer $N$ is a positive divisor of $N$ other than $N$ itself. The infinite sequence $a_1,a_2,\dots$ consists of positive integers, each of which has at least three proper divisors. For each $n\ge1$, the integer $a_{n+1}$ is the sum of the three largest proper divisors of $a_n$. Determine all possible values of $a_1$.	[ "The answer is \$a_1 = 12^e \\cdot 6 \\cdot \\ell\$ for any \$e, \\ell \\geq 0\$ with \$\\gcd(\\ell,10)=1\$.\n\nLet \$S\$ denote the set of positive integers with at least three divisors. For \$x \\in S\$, let \$\\psi(x)\$ denote the sum of the three largest ones, so that \$\\psi(a_n)=a_{n+1}\$.\n\n---\n\n¶ Proof that all such \$a_1\$ work. Let \$x = 12^e \\cdot 6 \\cdot \\ell \\in S\$ with \$\\gcd(\\ell,10)=1\$. As \$\\tfrac{1}{2}+\\tfrac{1}{3}+\\tfrac{1}{4}=\\tfrac{13}{12}\$, we get\n\\[\n\\psi(x) = \\begin{cases}\nx & e=0 \\\\\n\\frac{13}{12}x & e>0\n\\end{cases}\n\\]\nso by induction on the value of \$e\$ we see that \$\\psi(x) \\in S\$ (the base \$e=0\$ coming from \$\\psi\$ fixing \$x\$).\n\n---\n\n¶ Proof that all \$a_1\$ are of this form. In what follows \$x\$ is always an element of \$S\$, not necessarily an element of the sequence.\n\nClaim — Let \$x \\in S\$. If \$2 \\mid \\psi(x)\$ then \$2 \\mid x\$.\n\nProof. If \$x\$ is odd then every divisor of \$x\$ is odd, so \$\\psi(x)\$ is the sum of three odd numbers. \$\\square\$\n\n---\n\nClaim — Let \$x \\in S\$. If \$6 \\mid \\psi(x)\$ then \$6 \\mid x\$.\n\nProof. We consider only \$x\$ even because of the previous claim. We prove the contrapositive that \$3 \\nmid x \\implies 6 \\nmid \\psi(x)\$ (for even \$x\$).\n\n- If \$4 \\mid x\$, then letting \$d\$ be the third largest proper divisor of \$x\$,\n\\[\n\\psi(x) = \\frac{x}{2}+\\frac{x}{4}+d = \\frac{3}{4}x+d \\equiv d \\not\\equiv 0 \\pmod{3}.\n\\]\n\n- Otherwise, let \$p \\mid x\$ be the smallest prime dividing \$x\$, with \$p>3\$. If the third-smallest nontrivial divisor of \$x\$ is \$2p\$, then\n\\[\n\\psi(x) = \\frac{x}{2}+\\frac{x}{p}+\\frac{x}{2p}=\\frac{3}{2p}x+\\frac{x}{2}\\equiv \\frac{x}{2}\\not\\equiv 0 \\pmod{3}.\n\\]\n\nIf the third-smallest nontrivial divisor of \$x\$ is instead an odd prime \$q\$, then\n\\[\n\\psi(x)=\\frac{x}{2}+\\frac{x}{p}+\\frac{x}{q}\\equiv 1+0+0\\equiv 1 \\pmod{2}.\n\\] \$\\square\$\n\n---\n\nTo tie these two claims into the problem, we assert: \n\nClaim — Every \$a_i\$ must be divisible by 6.\n\nProof. The idea is to combine the previous two claims (which have no dependence on the sequence) with a size argument.\n\n- For odd \$x \\in S\$ note that \$\\psi(x) < \\bigl(\\tfrac{1}{3}+\\tfrac{1}{5}+\\tfrac{1}{7}\\bigr)x < x\$ and \$\\psi(x)\$ is still odd. So if any \$a_i\$ is odd the sequence is strictly decreasing and that’s impossible. Hence, we may assume \$a_1,a_2,\\dots\$ are all even.\n\n- If \$x \\in S\$ is even but \$3 \\nmid x\$ then \$\\psi(x) < \\bigl(\\tfrac{1}{2}+\\tfrac{1}{4}+\\tfrac{1}{5}\\bigr)x < x\$ and \$\\psi(x)\$ is still not a multiple of 3. So if any \$a_i\$ is not divisible by 3 the sequence is again strictly decreasing. \$\\square\$\n\nOn the other hand, if \$x\$ is a multiple of 6, we have the following formula for \$\\psi(x)\$:\n\\[\n\\psi(x)=\n\\begin{cases}\n\\frac{13}{12}x & 4 \\mid x \\\\\n\\frac{31}{30}x & 4 \\nmid x \\text{ but } 5 \\mid x \\\\\nx & 4 \\nmid x \\text{ and } 5 \\nmid x\n\\end{cases}\n\\]\n\nLooking back on our sequence of \$a_i\$ (which are all multiples of 6), the center case cannot happen with our \$a_i\$, because \$\\tfrac{31}{30}x\$ is odd when \$x \\equiv 2 \\pmod{4}\$. Hence in actuality\n\\[\na_{n+1}=\\tfrac{13}{12}a_n \\quad \\text{or} \\quad a_{n+1}=a_n\n\\]\nfor every \$n\$.\n\nLet \$T\$ be the smallest index such that \$a_T=a_{T+1}=a_{T+2}=\\cdots\$ (it must exist because we cannot multiply by \$\\tfrac{13}{12}\$ forever). Then we can exactly describe the sequence by\n\\[\na_n = a_1 \\cdot \\left(\\tfrac{13}{12}\\right)^{\\min(n,T)-1}.\n\\]\n\nHence \$a_1=\\left(\\tfrac{12}{13}\\right)^{T-1} a_T\$, and since \$a_T\$ is a multiple of 6 not divisible by 4 or 5, it follows \$a_1\$ has the required form.", "We want to determine all possible values of \$a_1\$ such that the sequence defined by \$a_{n+1} = S(a_n)\$ is an infinite sequence of positive integers, where \$S(N)\$ is the sum of the three largest proper divisors of \$N\$, and each \$a_n\$ has at least three proper divisors (i.e., the total number of divisors \$\\tau(a_n) \\geq 4\$).\n\nLet \$R(N) = S(N)/N\$. If \$\\tau(N) \\geq 5\$, and \$1 = d_1 < d_2 < d_3 < d_4\$ are the smallest divisors of \$N\$, then \$R(N) = \\frac{1}{d_2} + \\frac{1}{d_3} + \\frac{1}{d_4}\$.\n\n---\n\nStep 1: Proving that \$a_n\$ is even for all \$n\$.\n\nSuppose \$N\$ is odd and \$\\tau(N) \\geq 4\$. The divisors of \$N\$ are odd. If \$\\tau(N) \\geq 5\$, \$d_2 \\geq 3, d_3 \\geq 5, d_4 \\geq 7\$. \$R(N) \\leq \\frac{1}{3} + \\frac{1}{5} + \\frac{1}{7} = \\frac{71}{105} < 1\$. If \$\\tau(N) = 4\$, \$N = p^3\$ or \$N = pq\$ for odd primes \$p < q\$. The proper divisors are \$1,p,p^2\$ or \$1,p,q\$. \$S(p^3) = 1+p+p^2\$. For \$p \\geq 3\$, \$p^3 - (1+p+p^2) = p^2(p-1)-p-1 \\geq 9(2)-3-1=14 > 0\$. \$S(pq)=1+p+q\$. For \$p \\geq 3, q \\geq 5\$, \$pq-(1+p+q)=(p-1)(q-1)-2 \\geq 2 \\cdot 4-2=6>0\$. In all cases, \$S(N)<N\$. Furthermore, the three largest proper divisors are odd, so their sum \$S(N)\$ is odd.\n\nIf \$a_n\$ were odd for some \$n\$. Since \$\\tau(a_n) \\geq 4\$ by the problem statement, \$a_{n+1}=S(a_n)<a_n\$ and \$a_{n+1}\$ is odd. By induction, \$(a_k)_{k \\geq n}\$ would be a strictly decreasing infinite sequence of positive integers. This contradicts the Well-Ordering Principle. Thus, \$a_n\$ is even for all \$n\$.\n\n---\n\nStep 2: Proving that \$a_n\$ is divisible by 3 for all \$n\$.\n\nSuppose \$N\$ is even, \$\\tau(N) \\geq 4\$, and \$3 \\nmid N\$. \$d_2=2\$. Since \$3 \\nmid N\$, \$d_3 \\geq 4\$. If \$\\tau(N) \\geq 5, d_4 \\geq 5\$. \$R(N) \\leq \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{5} = \\frac{19}{20} < 1\$. If \$\\tau(N) = 4\$. \$N=8\$ or \$N=2p\$ (prime \$p \\geq 5\$). \$S(8)=7<8\$. \$S(2p)=p+3<2p\$. In all cases, \$S(N)<N\$.\n\nWe prove a lemma: Lemma: Let \$N\$ be even, \$\\tau(N) \\geq 4\$, and \$3 \\nmid N\$. If \$3 \\mid S(N)\$, then \$S(N)\$ is odd. Proof: If \$\\tau(N)=4, S(8)=7, S(2p)=p+3\$. Since \$3 \\nmid p, 3 \\nmid p+3\$. So \$3 \\nmid S(N)\$. The implication holds vacuously. If \$\\tau(N) \\geq 5. R(N)=\\frac{1}{2}+\\frac{1}{d_3}+\\frac{1}{d_4}\$. Since \$3 \\nmid N\$, \$3 \\nmid d_i\$. If \$3 \\mid S(N)\$, since \$3 \\nmid N\$, we must have \$v_3(R(N))>0\$. \$R(N)=\\frac{d_3d_4+2d_3+2d_4}{2d_3d_4}\$. The denominator is not divisible by 3. The numerator \$X=d_3d_4- d_3-d_4\$ must be divisible by 3. \$X\\equiv d_3d_4-d_3-d_4=(d_3-1)(d_4-1)-1 \\pmod{3}\$. \$X\\equiv 0 \\implies (d_3-1)(d_4-1)\\equiv 1 \\pmod{3}\$. This requires \$d_3\\equiv 2\$ and \$d_4\\equiv 2 \\pmod{3}\$. If \$4 \\mid N\$. Since \$3 \\nmid N\$, the divisors start \$1,2,4\$. So \$d_3=4\$. But \$4\\equiv 1 \\pmod{3}\$, contradiction, \$d_3\\equiv 2 \\pmod{3}\$. Thus \$v_2(N)=1\$. Thus \$N=2M\$ with \$M\$ odd, \$d_3=M\$. Let \$p\$ be the smallest prime factor of \$M (p \\geq 5)\$. \$d_3=p\$. \$p\\equiv 2 \\pmod{3}\$. Since \$d_3\\equiv 2 \\pmod{3}, p\\equiv 2 \\pmod{3}\$. So the next divisor is the next smallest divisor of \$M\$, call it \$m_3\$. Since \$d_4 \\equiv 2 \\pmod{3}, d_4 \\neq p\$. So \$d_4=m_3\$. But \$m_3\$ is an odd prime divisor of \$M\$, call it \$q\$. \$q \\equiv 2 \\pmod{3}\$. So \$N=2 \\cdot p \\cdot q \\cdot R\$. So \$d_4=q\$. We evaluate oddness of \$S(N)\$. Odd divisors of \$M, M/p\$ and \$M/d_4\$ are odd integers. So \$S(N)=\\text{Odd}+\\text{Even}+\\text{Even}=\\text{Odd}\$. The lemma is proved.\n\nNow, suppose \$3 \\nmid a_n\$ for some \$n\$. We know \$a_n\$ is even and \$a_{n+1}=S(a_n)<a_n\$. If \$3 \\mid a_{n+1}\$, by the Lemma applied to \$a_n, a_{n+1}\$ must be odd. This contradicts Step 1. Thus \$3 \\nmid a_{n+1}\$. By induction, \$(a_k)_{k \\geq n}\$ is a strictly decreasing infinite sequence of positive integers. Contradiction. Therefore, \$3 \\mid a_n\$ for all \$n\$.\n\nCombining Step 1 and Step 2, \$6 \\mid a_n\$ for all \$n\$. Note that \$6 \\mid N\$ implies \$\\tau(N) \\geq 4\$.\n\n---\n\nStep 3: Analyzing the dynamics when \$6 \\mid N\$.\n\nIf \$6 \\mid N\$, the smallest divisors are \$1,2,3\$. The fourth smallest divisor \$d_4\$ must be 4, 5, or 6. \$R(N)=\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{d_4}=\\frac{5}{6}+\\frac{1}{d_4}\$. (This holds even if \$\\tau(N)=4\$, i.e., \$N=6\$, where \$S(6)=6, R(6)=1\$, and \$d_4\$ is formally \$N=6\$).\n\nWe identify three regimes: Regime A (Growth): \$d_4=4\$. Occurs if \$12 \\mid N. R(N)=13/12\$. Regime B (Boost): \$d_4=5\$. Occurs if \$30 \\mid N\$ and \$4 \\nmid N\$ (\$v_2(N)=1\$). \$R(N)=31/30\$. Regime C (Fixed Point): \$d_4=6\$. Occurs if \$6 \\mid N, 4 \\nmid N, 5 \\nmid N. R(N)=1\$.\n\n---\n\nStep 4: Evolution of the sequence and constraints on \$a_1\$.\n\nIf \$a_n \\in B. v_2(a_n)=1. a_{n+1}=(31/30)a_n. v_2(a_{n+1})=v_2(a_n)+v_2(31/30)=1-1=0\$. \$a_{n+1}\$ is odd. This contradicts Step 1. Thus, the sequence must remain in A ∪ C.\n\nIf \$a_n \\in A. a_{n+1}=(13/12)a_n. v_2(a_{n+1})=v_2(a_n)-2. v_3(a_{n+1})=v_3(a_n)-1\$. Since \$6 \\mid a_k\$ for all \$k, v_2(a_k) \\geq 1\$ and \$v_3(a_k) \\geq 1\$. As the valuations decrease in Regime A, the sequence cannot stay in A indefinitely. It must eventually reach Regime C and stabilize there (\$a_{n+1}=a_n\$).\n\nIn Regimes A (\$R=13/12\$) and C (\$R=1\$), \$v_5(R(N))=0\$. Thus \$v_5(a_n)\$ is constant. Let \$L\$ be the stable value in C. By definition of C, \$5 \\nmid L\$. So \$v_5(L)=0\$. Therefore, \$v_5(a_1)=0\$.\n\n---\n\nStep 5: Characterization of \$a_1\$.\n\nLet \$K \\geq 0\$ be the number of steps the sequence spends in Regime A before reaching Regime C. \$a_1, \\dots, a_K \\in A\$ (if \$K \\geq 1\$) and \$a_{K+1} \\in C\$. Since \$5 \\nmid a_1, 5 \\nmid a_n\$ for all \$n\$.\n\nLet \$A=v_2(a_1)\$ and \$B=v_3(a_1)\$. \$a_{K+1}=(13/12)^Ka_1. v_2(a_{K+1})=A-2K. v_3(a_{K+1})=B-K\$. Since \$a_{K+1} \\in C\$, we must have \$v_2(a_{K+1})=1\$ (as \$6 \\mid a_{K+1}\$ and \$4 \\nmid a_{K+1}\$) and \$v_3(a_{K+1}) \\geq 1\$. \$A-2K=1 \\implies A=2K+1. B-K \\geq 1 \\implies B \\geq K+1\$.\n\nWe verify that these conditions are sufficient. We must ensure \$a_i \\in A\$ for \$1 \\leq i \\leq K\$. This means \$12 \\mid a_i\$. For \$1 \\leq i \\leq K: v_2(a_i)=A-2(i-1)=2K+1-2i+2=2(K-i)+3\$. Since \$i \\leq K, v_2(a_i) \\geq 3\$. \$v_3(a_i)=B-(i-1) \\geq (K+1)-(i-1)=K-i+2\$. Since \$i \\leq K, v_3(a_i) \\geq 2\$. Thus \$2^3 \\cdot 3^2=72\$ divides \$a_i\$. This implies \$12 \\mid a_i, so a_i \\in A\$. This also ensures that \$a_{i+1}=(13/12)a_i\$ is an integer. The sequence is valid.\n\nWe express the possible values of \$a_1\$. \$a_1=2^{2K+1}3^BM\$, where \$K \\geq 0, B \\geq K+1\$, and \$M\$ is a positive integer such that \$\\gcd(M,30)=1\$ (since \$v_5(a_1)=0\$). We rewrite this as: \$a_1=(2^{2K+1}3^{K+1}M)\\cdot(3^{B-(K+1)}M)\$. \$2^{2K+1}3^{K+1}=(2 \\cdot 4^K)(3 \\cdot 3^K)=6 \\cdot (12^K)\$. Let \$J=3^{B-K-1}M\$. \$J\$ is a positive integer. Since \$\\gcd(M,30)=1\$, \$J\$ is not divisible by 2 or 5. That is, \$\\gcd(J,10)=1\$. Conversely, any positive integer \$J\$ such that \$\\gcd(J,10)=1\$ can be represented in this form for a given \$K (B=K+1+\\nu_3(J))\$. The condition holds.\n\nThe set of all possible values of \$a_1\$ consists of integers of the form \$6J \\cdot 12^K\$, where \$K \\geq 0\$ is an integer and \$J\$ is a positive integer such that \$\\gcd(J,10)=1\$." ]
IMO-2025-5	https://artofproblemsolving.com/wiki/index.php/2025_IMO_Problems/Problem_5	Alice and Bazza are playing the inekoalaty game, a two‑player game whose rules depend on a positive real number $\lambda$ which is known to both players. On the $n$th turn of the game (starting with $n=1$) the following happens: - If $n$ is odd, Alice chooses a nonnegative real number $x_n$ such that $x_1 + x_2 + \cdots + x_n \le \lambda n$. - If $n$ is even, Bazza chooses a nonnegative real number $x_n$ such that $x_1^2 + x_2^2 + \cdots + x_n^2 \le n$ If a player cannot choose a suitable $x_n$, the game ends and the other player wins. If the game goes on forever, neither player wins. All chosen numbers are known to both players. Determine all values of $\lambda$ for which Alice has a winning strategy and all those for which Bazza has a winning strategy.	[ "The answer is that Alice has a winning strategy for \$\\lambda > 1/\\sqrt{2}\$, and Bazza has a winning strategy for \$\\lambda < 1/\\sqrt{2}\$. (Neither player can guarantee winning for \$\\lambda = 1/\\sqrt{2}\$.)\n\nWe divide the proof into two parts.\n\n¶ Alice’s strategy when \$\\lambda \\geq 1/\\sqrt{2}\$. Consider the strategy where Alice always plays \$x_{2i+1} = 0\$ for \$i = 0, \\ldots, k-1\$.\n\nIn this situation, when \$n = 2k+1\$ we have\n\\[\n\\sum_{i=1}^{2k} x_i = 0 + x_2 + 0 + x_4 + \\cdots + 0 + x_{2k}\n\\leq k \\cdot \\sqrt{\\frac{x_2^2 + \\cdots + x_{2k}^2}{k}}\n= \\sqrt{2}\\,k < \\lambda \\cdot (2k+1).\n\\]\n\nand so the choices for \$x_{2k+1}\$ are\n\\[\nx_{2k+1} \\in [0, \\, \\lambda \\cdot (2k+1) - \\sqrt{2k}]\n\\]\n\nwhich is nonempty. Hence Alice can’t ever lose with this strategy.\n\nBut suppose further \$\\lambda > 1/\\sqrt{2}\$; we show Alice can win. Choose \$k\$ large enough that\n\\[\n\\sqrt{2}\\,k < \\lambda \\cdot (2k+1) - \\sqrt{2k} + 2.\n\\]\n\nThen on the \$(2k+1)\$-st turn, Alice can (after playing 0 on all earlier turns) play a number greater than \$\\sqrt{2k}+2\$ and cause Bazza to lose.\n\n\n¶ Bazza strategy when \$\\lambda \\leq 1/\\sqrt{2}\$. Consider the strategy where Bazza always plays\n\\[\nx_{2i+2} = \\sqrt{2 - x_{2i+1}^2} \\quad \\text{for all } i = 0, \\ldots, k-1\n\\]\n(i.e. the largest possible value Bazza can play).\n\nTo analyze Bazza’s choices on each of his turns, we first need to estimate \$x_{2k+1}\$. We do this by writing\n\\[\n\\begin{aligned}\n\\lambda \\cdot (2k+1) &\\geq x_1 + x_2 + \\cdots + x_{2k+1} \\\\\n&= (x_1 + \\sqrt{2 - x_1^2}) + (x_3 + \\sqrt{2 - x_3^2}) + \\cdots + (x_{2k-1} + \\sqrt{2 - x_{2k-1}^2}) + x_{2k+1} \\\\\n&\\geq \\underbrace{\\sqrt{2} + \\cdots + \\sqrt{2}}_{k \\text{ times}} + x_{2k+1} = \\sqrt{2}\\,k + x_{2k+1},\n\\end{aligned}\n\\]\n\nwhere we have used the fact that \$t + \\sqrt{2 - t^2} \\geq 2\$ for all \$t \\geq 0\$. This means that\n\\[\nx_{2k+1} \\leq \\lambda \\cdot (2k+1) - \\sqrt{2}\\,k < \\sqrt{2}.\n\\]\n\nAnd\n\\[\nx_1^2 + x_2^2 + \\cdots + x_{2k}^2 + x_{2k+1}^2 = (2 + \\cdots + 2) + x_{2k+1}^2,\n\\]\nso Bazza can indeed choose\n\\[\nx_{2k+2} = \\sqrt{2 - x_{2k+1}^2}\n\\]\nand always has a move.\n\nBut suppose further \$\\lambda < 1/\\sqrt{2}\$. Then the above calculation also shows that Alice couldn’t have made a valid choice for large enough \$k\$, since \$\\lambda \\cdot (2k+1) - \\sqrt{2}\\,k < 0\$ for large \$k\$.\n\n\nRemark. In the strategies above, we saw that Alice prefers to always play 0 and Bazza prefers to always play as large as possible. One could consider what happens in the opposite case:\n\n- If Alice tries to always play the largest number possible, her strategy still wins for \$\\lambda > 1\$.\n- If Bazza tries to always play 0, Alice can win no matter the value for \$\\lambda > 0\$.", "\nWe determine the values of \$\\lambda\$ for which Alice has a winning strategy and those for which Bazza has a winning strategy. Let \$S_n = \\sum_{i=1}^n x_i\$ and \$Q_n = \\sum_{i=1}^n x_i^2\$. Alice (A) plays at odd \$n\$, ensuring \$x_n \\ge 0\$ and \$S_n \\le \\lambda n\$. Bazza (B) plays at even \$n\$, ensuring \$x_n \\ge 0\$ and \$Q_n \\le n\$. The critical value for \$\\lambda\$ is \$\\tfrac{1}{\\sqrt{2}}\$.\n\n---\n\nCase 1: \$0 < \\lambda < \\tfrac{1}{\\sqrt{2}}\$. Bazza has a winning strategy.\n\nLet \$\\delta = \\sqrt{2} - 2\\lambda\$. Since \$\\lambda < \\tfrac{1}{\\sqrt{2}}\$, we have \$\\delta > 0\$.\n\nBazza’s strategy (B-MaxQ) is to ensure \$Q_{2k} = 2k\$ at every turn \$n = 2k\$. This requires choosing \$x_{2k} = \\sqrt{2 - Q_{2k-1}}\$. This is feasible if \$Q_{2k-1} \\le 2k\$.\n\nLet \$C_k\$ be the budget available to Alice at the start of turn \$2k-1\$: \$C_k = \\lambda(2k-1) - S_{2k-2}\$ (with \$S_0 = 0\$). Alice must choose \$x_{2k-1} \\in [0,C_k]\$. If \$C_k < 0\$, Alice loses immediately.\n\nWe analyze the evolution of \$C_k\$, assuming the game continues and Bazza follows B-MaxQ. \n\\[\nC_{k+1} = \\lambda(2k+1) - S_{2k} = C_k + 2\\lambda - (x_{2k-1} + x_{2k}).\n\\]\n\nIf Bazza successfully follows B-MaxQ up to turn \$2k\$, then \$Q_{2k} = 2k\$ and \$Q_{2k-2} = 2k-2\$. Thus, \$x_{2k-1}^2 + x_{2k}^2 = Q_{2k} - Q_{2k-2} = 2\$. Since \$x_i \\ge 0\$, \$(x_{2k-1}+x_{2k})^2 = 2 + 2x_{2k-1}x_{2k} \\ge 2\$, so \$x_{2k-1}+x_{2k} \\ge \\sqrt{2}\$.\n\nTherefore, \n\\[\nC_{k+1} \\le C_k + 2\\lambda - \\sqrt{2} = C_k - \\delta.\n\\]\n\nWe must verify that B-MaxQ is always feasible as long as the game continues (i.e., \$C_k \\ge 0\$). We proceed by induction. \$C_1 = \\lambda\$. Since \$\\delta > 0\$, if \$C_k > 0\$, the sequence \$C_k\$ is strictly decreasing. Thus \$C_k \\le C_1 = \\lambda\$. Since \$\\lambda < 1/\\sqrt{2}\$, Alice must choose \$x_{2k-1} \\le C_k < 1/\\sqrt{2}\$.\n\nIf Bazza maintained \$Q_{2k-2} = 2k-2\$, then \$Q_{2k-1} = Q_{2k-2} + x_{2k-1}^2 = 2k-2+x_{2k-1}^2 < 2k-2+1/2 = 2k-3/2\$. Since \$Q_{2k-1} < 2k\$, Bazza can choose \$x_{2k}\$ to achieve \$Q_{2k} = 2k\$. B-MaxQ is always feasible.\n\nSince \$C_{k+1} \\le C_k - \\delta\$, the budget decreases by at least \$\\delta\$ in each round pair. \$C_k \\le C_1 - (k-1)\\delta = \\lambda - (k-1)\\delta\$. Since \$\\lambda\$ is fixed and \$\\delta > 0\$, there exists an integer \$K\$ such that \$(K-1)\\delta > \\lambda\$. For this \$K\$, \$C_K < 0\$. At turn \$2K-1\$, Alice needs to choose \$x_{2K-1} \\ge 0\$ such that \$x_{2K-1} \\le C_K\$. Since \$C_K < 0\$, no such choice exists. Bazza wins.\n\n---\n\nCase 2: \$\\lambda > \\tfrac{1}{\\sqrt{2}}\$. Alice has a winning strategy.\n\nConsider the function \$h(K) = \\tfrac{K\\sqrt{2}}{2K-1}\$ for \$K \\ge 1\$. \$h(K)\$ is strictly decreasing and \$\\lim_{K \\to \\infty} h(K) = 1/\\sqrt{2}\$. Since \$\\lambda > 1/\\sqrt{2}\$, there exists an integer \$K \\ge 1\$ such that \$\\lambda > h(K)\$. This implies \$L = \\lambda(2K-1) > K\\sqrt{2}\$.\n\nAlice’s strategy (A-Spike-K): Play \$x_{2i-1} = 0\$ for \$i = 1,\\dots,K-1\$. At turn \$2K-1\$, play the maximum possible value.\n\nFirst, we verify the feasibility. For \$i < K\$, Alice plays \$x_{2i-1} = 0\$. She needs \$S_{2i-1} = S_{2i-2} \\le \\lambda(2i-1)\$. Bazza is constrained by \$Q_{2i-2} \\le 2(i-1)\$. By the QM-AM inequality (or Cauchy-Schwarz), \$S_{2i-2} \\le \\sqrt{(i-1)Q_{2i-2}} \\le \\sqrt{(i-1)2(i-1)} = (i-1)\\sqrt{2}\$. We check the constraint: \$(i-1)\\sqrt{2} \\le \\lambda(2i-1)\$, or \$\\lambda \\ge \\tfrac{(i-1)\\sqrt{2}}{2i-1}\$. The RHS is an increasing sequence converging to \$1/\\sqrt{2}\$. Since \$\\lambda > 1/\\sqrt{2}\$, the strategy is feasible.\n\nNow we analyze the outcome. Let \$N = K-1\$. Bazza has made \$N\$ moves \$y_i = x_{2i}\$ (\$i=1,\\dots,N\$). At turn \$2K-1\$, Alice plays \$x_{2K-1} = L - S_{2N}\$. Since \$S_{2N} \\le N\\sqrt{2}\$ and \$L > K\\sqrt{2} = (N+1)\\sqrt{2}\$, \$x_{2K-1} = L - S_{2N} \\ge \\sqrt{2} > 0\$.\n\nAlice wins if Bazza cannot move at turn \$2K\$, i.e., \$Q_{2K-1} > 2K\$. \$Q_{2K-1} = Q_{2N} + (L - S_{2N})^2\$.\n\nBazza aims to minimize this quantity subject to his constraints: \$y_i \\ge 0\$ and \$\\sum_{j=1}^i y_j^2 \\le 2i\$. These constraints imply \$Q_{2N} \\le 2N\$, and consequently \$S_{2N} \\le N\\sqrt{2}\$.\n\nLet \$F(y) = Q_{2N}(y) + (L - S_{2N}(y))^2\$. Consider the strategy \$y^* = (\\sqrt{2},\\dots,\\sqrt{2})\$. This is feasible for Bazza as \$\\sum_{j=1}^i (\\sqrt{2})^2 = 2i\$. Let \$S^* = N\\sqrt{2}\$ and \$Q^* = 2N\$.\n\nLet \$y\$ be any feasible strategy for Bazza. Let \$\\Delta S = S^* - S_{2N}(y) \\ge 0\$. We compare \$F(y)\$ with \$F(y^)\$. We use the identity \$\\sum(y_i - \\sqrt{2})^2 = Q_{2N}(y) - 2\\sqrt{2}S_{2N}(y) + 2N\$. \n\$Q_{2N}(y) - Q^ = Q_{2N}(y) - 2N = \\sum(y_i-\\sqrt{2})^2 + 2\\sqrt{2}S_{2N}(y) - 4N\$. \n\$2\\sqrt{2}S_{2N}(y) = 2\\sqrt{2}(S^* - \\Delta S) = 4N - 2\\sqrt{2}\\Delta S\$. \n\$Q_{2N}(y) - Q^* = \\sum(y_i - \\sqrt{2})^2 - 2\\sqrt{2}\\Delta S\$.\n\n\\[\nF(y) - F(y^) = Q_{2N}(y) - Q^ + (L - S_{2N}(y))^2 - (L - S^)^2.\n\\] \n\\[\n(L - S_{2N}(y))^2 = (L - (S^ - \\Delta S))^2 = (L - S^* + \\Delta S)^2.\n\\]\n\n\\[\nF(y) - F(y^) = (\\sum(y_i - \\sqrt{2})^2 - 2\\sqrt{2}\\Delta S) + 2(L - S^)\\Delta S + (\\Delta S)^2.\n\\] \n\$F(y) - F(y^) = \\sum(y_i - \\sqrt{2})^2 + 2(L - S^ - \\sqrt{2})\\Delta S + (\\Delta S)^2\$.\n\nBy the choice of \$K\$, \$L > K\\sqrt{2} = (N+1)\\sqrt{2} = S^+\\sqrt{2}\$. Let \$\\epsilon = L - S^ - \\sqrt{2} > 0\$. Then \$F(y) - F(y^) = \\sum(y_i - \\sqrt{2})^2 + 2\\epsilon \\Delta S + (\\Delta S)^2\$. Since all terms are non-negative, \$F(y) \\ge F(y^)\$. The minimum value of \$Q_{2K-1}\$ is \$F(y^)\$.\n\n\\[\nQ_{2K-1} \\ge F(y^) = 2N + (L - N\\sqrt{2})^2 = 2N+(\\sqrt{2}+\\epsilon)^2 > 2.\n\\] \nSince \$\\epsilon > 0\$, \$(\\sqrt{2}+\\epsilon)^2 > 2\$. \$Q_{2K-1} > 2N+2 = 2K\$. Bazza cannot move at turn \$2K\$. Alice wins.\n\n---\n\nCase 3: \$\\lambda = \\tfrac{1}{\\sqrt{2}}\$. Neither player has a winning strategy.\n\n1. Alice’s drawing strategy (A-Zero): \$x_{2k-1} = 0\$ for all \$k\$. We verify the game continues forever. Alice’s feasibility at turn \$2k-1\$: We need \$S_{2k-2} \\le \\lambda(2k-1)\$. Bazza maximizes \$S_{2k-2}\$ subject to \$Q_{2k-2} \\le 2k-2\$, achieving at most \$(k-1)\\sqrt{2}\$. We check: \$(k-1)\\sqrt{2} \\le \\tfrac{1}{\\sqrt{2}}(2k-1) \\iff k-1 \\le 2k-1\$. True. Bazza’s survival at turn \$2k\$: We need \$Q_{2k-1} = Q_{2k-2} \\le 2k-2 < 2k\$. Bazza survives. Alice’s survival at turn \$2k+1\$: We need \$S_{2k} \\le \\lambda(2k+1)\$. Bazza maximizes \$S_{2k}\$ subject to \$Q_{2k} \\le 2k\$, achieving at most \$k\\sqrt{2}\$. We check: \$k\\sqrt{2} \\le \\tfrac{1}{\\sqrt{2}}(2k+1) \\iff 2k \\le 2k+1\$. True. The game continues forever. Bazza cannot win.\n\n2. Bazza’s drawing strategy (B-MaxQ): \$Q_{2k} = 2k\$. We verify the game continues forever. Bazza’s feasibility (survival). As shown in Case 1, if Bazza follows B-MaxQ, \$S_{2k-2} \\ge (k-1)\\sqrt{2}\$. Alice’s budget \$C_k = \\lambda(2k-1) - S_{2k-2}\$. \$C_k \\le \\tfrac{1}{\\sqrt{2}}(2k-1)-(k-1)\\sqrt{2} = \\tfrac{2k-1-2(k-1)}{\\sqrt{2}} = \\tfrac{1}{\\sqrt{2}}\$. Alice must choose \$x_{2k-1} \\le 1/\\sqrt{2}\$. Then \$Q_{2k-1} = 2k-2+x_{2k-1}^2 \\le 2k-2+1/2 < 2k\$. B-MaxQ is feasible. Bazza survives.\n\nAlice’s survival. We must show \$C_k > 0\$ for all \$k\$. \$C_1 = 1/\\sqrt{2} > 0\$. \$C_{k+1} = C_k+2\\lambda-(x_{2k-1}+x_{2k}) = C_k+\\sqrt{2}-(x_{2k-1}+x_{2k})\$. Bazza ensures \$x_{2k-1}^2+x_{2k}^2=2\$. Let \$g(t)=t+\\sqrt{2-t^2}\$. \$C_{k+1} = C_k+\\sqrt{2}-g(x_{2k-1})\$. Alice chooses \$x_{2k-1}\\in[0,C_k]\$. To ensure Alice survives, we check the minimum possible budget for the next turn. Since \$C_k \\le 1/\\sqrt{2}<1\$ and \$g(t)\$ is increasing on \$[0,1]\$ (as \$g'(t)=1+t/\\sqrt{2-t^2}>0\$ for \$t<1\$), \$g(x_{2k-1})\$ is maximized when \$x_{2k-1}=C_k\$. \$C_{k+1} \\ge C_k+\\sqrt{2}-g(C_k)=\\sqrt{2}-\\sqrt{2-C_k^2}\$. If \$C_k>0\$, then \$\\sqrt{2}-C_k^2<\\sqrt{2}\$, so \$C_{k+1}>0\$. By induction, \$C_k>0\$ for all \$k\$. Alice survives. The game continues forever. Alice cannot win.\n\n---\n\nConclusion: Alice has a winning strategy if and only if \$\\lambda > \\tfrac{1}{\\sqrt{2}}\$. Bazza has a winning strategy if and only if \$0 < \\lambda < \\tfrac{1}{\\sqrt{2}}\$. If \$\\lambda = \\tfrac{1}{\\sqrt{2}}\$, neither player has a winning strategy." ]
IMO-2025-6	https://artofproblemsolving.com/wiki/index.php/2025_IMO_Problems/Problem_6	Consider a 2025 x 2025 grid of unit squares. Matlida wishes to place on the grid some rectangular tiles, possibly of different sizes, such that each side of every tile lies on a grid line and every unit square is covered by at most one tile. Determine the minimum number of tiles Matlida needs to place so that each row and each column of the grid has exactly one unit square that is not covered by any tile.	[ "The answer is \$2112 = 2025 + 2 \\cdot 45 - 3\$. In general, the answer turns out to be \$\\lceil n + 2\\sqrt{n} - 3 \\rceil\$, but when \$n\$ is not a perfect square the solution is more complicated.\n\nRemark. The 2017 Romanian Masters in Math asked the same problem where the tiles are replaced by sticks, i.e. \$1 \\times k\$ tiles. The answer to that problem is completely different and as far as I know there is no connection at all to the present IMO problem.\n\n---\n\n### ¶ Construction\nWe show a general construction when \$n = k^2\$ illustrated below for \$k = 5\$; it generalizes readily. There are a total of \$(k-1)^2\$ tiles which are \$k \\times k\$ squares and another \$4(k-1)\$ tiles on the boundary, giving a total of\n\n\\[\n(k-1)^2 + 4(k-1) = k^2 + 2k - 3\n\\]\n\ntiles as promised.\n\n---\n\n### ¶ Bound\nThere are several approaches (all hard), but the shortest proof seems to be the following one that exploits the Erdős–Szekeres theorem; it is Solution 6 in the shortlist. The theorem being quoted is:\n\nTheorem (Erdős–Szekeres). \nLet \$n \\geq 1\$ be an integer. Given a permutation of \$(1, \\dots, n)\$, if a longest increasing subsequence (LIS) has length \$a\$ and a longest decreasing subsequence (LDS) has length \$b\$, then \$ab \\geq n\$.\n\nThis is a stronger version of the theorem compared to another version which instead just asserts that \$\\max(a,b) \\geq \\sqrt{n}\$.\n\nTo apply this to the present problem, take the \$n\$ uncovered squares which we henceforth call “black” as a permutation. Then consider both an LIS of length \$a\$ and an LDS of length \$b\$. We do the following artistic illustration:\n\n- Draw the LIS as a broken line, then connect it to the southwest and northwest corner of the board. \n- Similarly, draw the LDS as a broken line, then connect it to the northwest and southeast corner of the board. \n- These two steps partition the board into four quadrants, which we call north, east, south, west. \n- For each black cell in the north quadrant, write an N in the cell above it (for the cell in the first row, this will be off the board). Do the same for E (east), S (south), W (west). \n- Some black cells are in multiple quadrants (i.e. part of the LIS/LDS). Write all letters in that case.\n\n---\n\nWe observe that:\n\nClaim — In this algorithm, the total number of letters written is exactly\n\n\\[\nC := \n\\begin{cases}\nn + a + b + 1 & \\text{if the LIS and LDS intersect} \\\\\nn + a + b & \\text{otherwise.}\n\\end{cases}\n\\]\n\nProof. This is obvious. Each black square contributes at least one letter. Each black square on exactly one of the LIS and LDS contributes one extra letter. And a black square on both contributes 4 letters instead of \$1+1+1\$. \$\\square\$\n\nNote by AM–GM we have \$a+b \\geq 2\\sqrt{ab} \\geq 2\\sqrt{n}\$, so we have a bound of\n\n\\[\nC \\geq n + 2\\sqrt{n} + \\varepsilon \\quad \\text{where} \\quad \\varepsilon := \n\\begin{cases}\n1 & \\text{if the LIS and LDS intersect} \\\\\n0 & \\text{otherwise.}\n\\end{cases}\n\\]\n\n---\n\nTo relate \$C\$ to the number of tiles, the critical claim is the following, which is by construction:\n\nClaim — None of Matilda’s tiles can have more than one letter written in any cell.\n\nProof. This follows from the construction.\n\nWe split into two cases based on \$\\varepsilon\$.\n\n- When \$\\varepsilon = 1\$, at most four letters go off the grid (one for each direction), the number of tiles is at least \$C-4 \\geq n + 2\\sqrt{n} - 3\$. \n- Suppose \$\\varepsilon = 0\$. Then \$C-4 \\geq n + 2\\sqrt{n} - 4\$. However, we make the additional observation here that the tile where the LIS and LDS meet has no letters on it either; hence there are at least \$1+(C-4) \\geq n + 2\\sqrt{n} - 3\$ tiles.\n\n---\n\nRemark. USJL mentions that when \$n = ab\$, it is in fact always possible to guarantee \$\\varepsilon = 1\$. Moreover, when \$a \\neq b\$, the AM–GM inequality is strict. This gives a way to avoid the additional observation needed for \$\\varepsilon = 0\$ above." ]

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