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id stringlengths 10 10	source stringlengths 74 74	problem stringlengths 57 1.29k	solutions listlengths 0 9
IMO-1993-1	https://artofproblemsolving.com/wiki/index.php/1993_IMO_Problems/Problem_1	Let $f\left(x\right)=x^n+5x^{n-1}+3$, where $n>1$ is an integer. Prove that $f\left(x\right)$ cannot be expressed as the product of two non-constant polynomials with integer coefficients.	[ "For the sake of contradiction, assume that \$f\\left(x\\right)=g\\left(x\\right)h\\left(x\\right)\$ for polynomials \$g\\left(x\\right)\$ and \$h\\left(x\\right)\$ in \$\\mathbb{R}\$. Furthermore, let \$g\\left(x\\right)=b_mx^m+b_{m-1}x^{m-1}+\\ldots+b_1x+b_0\$ with \$b_i=0\$ if \$i>m\$ and \$h\\left(x\\right)=c_{n-m}x^{n-m}+c_{n-m-1}x^{n-m-1}+\\ldots+c_1x+c_0\$ with \$c_i=0\$ if \$i>n-m\$. This gives that \$f\\left(x\\right)=\\sum_{i=0}^{n}\\left(\\sum_{j=0}^{i}b_jc_{i-j}\\right)x^i\$.\n\nWe have that \$3=b_0c_0\$, or \$3\|b_0c_0\$. WLOG, let \$3\|b_0\$ (and thus \$3\\not\|c_0\$). Since \$b_0c_1+b_1c_0=0\$ and \$3\$ divides \$b_0\$ but not \$c_0\$, we need that \$3\|b_1\$. We can keep on going up the chain until we get that \$3\|b_{n-2}\$. Then, by equating coefficients once more, we get that \$b_0c_{n-1}+b_1c_{n-2}+\\ldots+b_{n-2}c_1+b_{n-1}c_0=5\$. Taking the equation \$\\pmod3\$ gives that \$b_{n-1}c_0\\equiv2\\pmod3\$. This implies that \$b_{n-1}\\neq0\$. Thus, the degree of \$g\\left(x\\right)\$ is at least \$n-1\$. However, because \$h\\left(x\\right)\$ is a non-constant factor of \$f\\left(x\\right)\$, we have that the degree of \$g\\left(x\\right)\$ is at most \$n-1\$. Thus, the degree of \$g\\left(x\\right)\$ is \$n-1\$, implying that the degree of \$h\\left(x\\right)\$ is \$1\$.\n\nFrom this fact, we have that there must exist an integer root of \$f\\left(x\\right)\$. However, when \$x\$ is an integer, \$x^n + 5x^{n - 1} \\equiv x^{n - 2}x(x + 1) \\equiv 0 \\pmod{2}\$, meaning \$f(x)\$ is odd, so \$x\$ cannot be a root of \$f\$.\n\nHence, \$f\\left(x\\right)\$ cannot be expressed as \$g\\left(x\\right)h\\left(x\\right)\$ for polynomials \$g\\left(x\\right)\$ and \$h\\left(x\\right)\$ in \$\\mathbb{R}\$. This means that \$f\\left(x\\right)\$ cannot be expressed as the product of two non-constant polynomials with integer coefficients.\n\nQ.E.D.", "It’s actually trivial by Perron’s Criterion lol\n\nNote: Quoting Perron's Criterion on the actual IMO will very likely result in a score in the set \$\\{0,1\\}\$, since it was not a well-known result back then." ]
IMO-1993-2	https://artofproblemsolving.com/wiki/index.php/1993_IMO_Problems/Problem_2	Let $D$ be a point inside acute triangle $ABC$ such that $\angle ADB = \angle ACB+\frac{\pi}{2}$ and $AC\cdot BD=AD\cdot BC$. (a) Calculate the ratio $\frac{AB\cdot CD}{AC\cdot BD}$. (b) Prove that the tangents at $C$ to the circumcircles of $\Delta ACD$ and $\Delta BCD$ are perpendicular.	[ "Let us construct a point \$B'\$ satisfying the following conditions: \$B', B\$ are on the same side of AC, \$BC = B'C\$ and \$\\angle BCB' = 90^{\\circ}\$.\n\nHence \$\\triangle ADB \\sim \\triangle ACB'\$.\n\n\\[\n\\implies \\frac{AB}{BD} = \\frac{AB'}{B'C}\n\\]\n\nAlso considering directed angles mod \$180^{\\circ}\$,\n\n\\[\n\\measuredangle CAB' = \\measuredangle DAB \\implies \\measuredangle CAD = \\measuredangle BAB'\n\\]\n\n.\n\nAlso, \$\\frac{AB'}{AB} = \\frac{B'C}{BD} = \\frac{BC}{BD} = \\frac{AC}{AD}\$.\n\n\$\\implies \\triangle ABB' \\sim \\triangle ADC\$.\n\nHence, \$\\frac{CD}{AC} = \\frac{BB'}{AB'}\$.\n\nFinally, we get \$\\frac{AB \\cdot CD}{AC \\cdot BD} = \\frac{BB'}{CB'} = \\boxed{\\sqrt{2}}\$.\n\nFor the second part, let the tangent to the circle \$(ADC)\$ be \$DX\$ and the tangent to the circle \$(ADB)\$ be \$DY\$.\n\n\$\\measuredangle ADX = \\measuredangle ACD\$ due to the tangent-chord theorem.\n\n\$\\measuredangle YDB = \\measuredangle DCB\$ for the same reason.\n\nHence,\n\n\\[\n\\measuredangle ADX + \\measuredangle YDB = \\measuredangle ACB\n\\]\n\nWe also have\n\n\\[\n\\measuredangle ADB = \\measuredangle ACB + 90^{\\circ}\n\\]\n\n\\[\n\\measuredangle ADX + \\measuredangle XDY + \\measuredangle YDB = \\measuredangle ACB + \\measuredangle XDY = \\measuredangle ACB + 90^{\\circ}\n\\]\n\n.\n\n\\[\n\\implies \\measuredangle XDY = 90^{\\circ}\n\\]\n\nwhich means circles \$(ADC)\$ and \$(ADB)\$ are orthogonal. \$\\square\$ ~reyaansh_agrawal" ]
IMO-1993-3	https://artofproblemsolving.com/wiki/index.php/1993_IMO_Problems/Problem_3	On an infinite chessboard, a game is played as follows. At the start, $n^2$ pieces are arranged on the chessboard in an $n$ by $n$ block of adjoining squares, one piece in each square. A move in the game is a jump in a horizontal or vertical direction over an adjacent occupied square to an unoccupied square immediately beyond. The piece which has been jumped over is removed. Find those values of $n$ for which the game can end with only one piece remaining on the board.	[ "The solution as described in the video is that all values of \$n\$ are valid except when \$n\$ is divisible by 3. He describes why those values are not valid using modular math and why the other ones are by creating an algorithmic pattern of reducing the matrices by taking out pieces in subarrays of 3 by m.\n\n~ Tomas Diaz. orders@tomasdiaz.com" ]
IMO-1993-4	https://artofproblemsolving.com/wiki/index.php/1993_IMO_Problems/Problem_4	For three points $A,B,C$ in the plane, we define $m(ABC)$ to be the smallest length of the three heights of the triangle $ABC$, where in the case $A$, $B$, $C$ are collinear, we set $m(ABC) = 0$. Let $A$, $B$, $C$ be given points in the plane. Prove that for any point $X$ in the plane, $m(ABC) \leq m(ABX) + m(AXC) + m(XBC)$.	[ "First we prove the claim for all points \$X\$ within or on the triangle \$ABC\$: In this specific case, suppose without loss of generality that \$a\\ge b\\ge c\$. Then the length of the smallest height of \$ABC\$ is that of the height from vertex \$A\$. This has a value of \$\\frac{2[ABC]}{a}\$. Similarly, for triangles \$ABX,BCX,CAX\$ we have that the length of the smallest height is twice the area of the respective triangle divided by the longest side of that triangle. It is clear that since \$a\$ is the longest side of \$ABC\$, no two points within \$ABC\$ have distance exceeding \$a\$. Thus, since \$X\$ is within \$ABC\$, the longest side of any of the triangles \$ABX,BCX,CAX\$ does not exceed \$a\$. So, we have\n\n\\[\nm(ABX)+m(BCX)+m(CAX)\\ge\\frac{2[ABX]}{a}+\\frac{2[BCX]}{a}+\\frac{2[CAX]}{a}\n\\]\n\n\\[\n=\\frac{2[ABC]}{a}=m(ABC)\n\\]\n\nas desired.\n\nNow on to prove the assertion for \$X\$ outside the triangle \$ABC\$. We shall assume without loss of generality that out of the points \$A,B,C\$ point \$A\$ is that farthest from \$X\$.\n\nIf \$ABCX\$ is concave or a degenerate quadrilateral, assume without loss of generality that \$B\$ inside or on triangle \$ACX\$. We shall prove that \$m(ACX)\\ge m(ABC)\$ to prove the main claim. If the shortest height of \$ACX\$ was from vertex \$A\$ then it is clear that the height of \$ABC\$ from \$A\$ is smaller than that since ray \$CB\$ is closer to \$A\$ than ray \$CX\$. The case of the height from \$C\$ being the smallest of triangle \$ACX\$ is analogous, so we move on to the case of the height from \$X\$ is the smallest. If this is the case, then it is clear that the height of \$ABC\$ from \$B\$ is smaller than \$m(ABC)\$ since \$\\angle XAC\\ge \\angle BAC\$. Thus the claim is proved.\n\nIf, instead, \$ABCX\$ is convex, we can assign the letter \$D\$ to represent to point of intersection of \$AX\$ and \$BC\$. Before proving the main claim, we shall prove that for triangle \$ABX\$ we have \$m(ABX)\\ge m(ABD)\$. We prove this by considering each of the vertices that the shortest height of \$ABX\$ is on. If the shortest height is that from \$X\$, then it is obvious that the height from \$D\$ to side \$AB\$ is smaller than that from \$X\$ since \$D\$ is closer to \$A\$ than \$X\$ is and so \$m(ABX)\\ge m(ABD)\$. If the shortest height of \$ABX\$ was from \$B\$, then since the height from \$B\$ to \$XD\$ is equal to the height from \$B\$ to \$AD\$, we have \$m(ABX)\\ge m(ABD)\$. If instead the shortest height of \$ABX\$ was from \$A\$, the it is clear that \$\\angle B<90\$. Thus, the projection of \$A\$ onto \$BD\$ is on the same side of \$B\$ as \$C\$. Now it is obvious that the height of \$ABD\$ from \$A\$ is smaller than that of \$ABX\$ from \$A\$ since the ray \$BD\$ is closer to \$A\$ than the ray \$BX\$. Thus, we have \$m(ABX)\\ge m(ABD)\$ in all cases. Notice that the case for triangle \$ACX\$ is analogous. Therefore, we have \$m(ABX)+m(ACX)\\ge m(ABD)+m(ACD)\$. But, this is the case of \$D\$ on the triangle \$ABC\$, and this case was shown in the first part of this proof.\n\nSo, we have that \$m(ABC) \\leq m(ABX) + m(AXC) + m(XBC)\$ is true in all cases, as desired." ]
IMO-1993-5	https://artofproblemsolving.com/wiki/index.php/1993_IMO_Problems/Problem_5	Let $\mathbb{N} = \{1,2,3, \ldots\}$. Determine if there exists a strictly increasing function $f: \mathbb{N} \mapsto \mathbb{N}$ with the following properties: (i) $f(1) = 2$; (ii) $f(f(n)) = f(n) + n, (n \in \mathbb{N})$.	[ "Here is my Solution https://artofproblemsolving.com/community/q2h62193p16226748\n\nFind as ≈ Ftheftics" ]
IMO-1993-6	https://artofproblemsolving.com/wiki/index.php/1993_IMO_Problems/Problem_6	There are $n$ lamps $L_0, \ldots , L_{n-1}$ in a circle ($n > 1$), where we denote $L_{n+k} = L_k$. (A lamp at all times is either on or off.) Perform steps $s_0, s_1, \ldots$ as follows: at step $s_i$, if $L_{i-1}$ is lit, switch $L_i$ from on to off or vice versa, otherwise do nothing. Initially all lamps are on. Show that: (a) There is a positive integer $M(n)$ such that after $M(n)$ steps all the lamps are on again; (b) If $n = 2^k$, we can take $M(n) = n^2 - 1$; (c) If $n = 2^k + 1$, we can take $M(n) = n^2 - n + 1.$	[]
IMO-1994-1	https://artofproblemsolving.com/wiki/index.php/1994_IMO_Problems/Problem_1	Let $m$ and $n$ be two positive integers. Let $a_1$, $a_2$, $\ldots$, $a_m$ be $m$ different numbers from the set $\{1, 2,\ldots, n\}$ such that for any two indices $i$ and $j$ with $1\leq i \leq j \leq m$ and $a_i + a_j \leq n$, there exists an index $k$ such that $a_i + a_j = a_k$. Show that \[ \frac{a_1+a_2+...+a_m}{m} \ge \frac{n+1}{2} \] .	[ "Let \$a_1, a_2, \\dots a_m\$ satisfy the given conditions. We will prove that for all \$j, 1 \\le j \\le m,\$\n\n\\[\na_j+a_{m-j+1} \\ge n+1\n\\]\n\nWLOG, let \$a_1 < a_2 < \\dots < a_m\$. Assume that for some \$j, 1 \\le j \\le m,\$\n\n\\[\na_j + a_{m-j+1} \\le n\n\\]\n\nThis implies, for each \$i, 1 \\le i \\le j,\$\n\n\\[\na_i + a_{m-j+1} \\le n\n\\]\n\nbecause \$a_i \\le a_j\$\n\nFor each of these values of i, we must have \$a_i + a_{m-j+1} = a_{k_i}\$ such that \$a_{k_i}\$ is a member of the sequence for each \$i\$. Because \$a_i > 0, a_{k_i} > a_{m-j+1}\$. Combining all of our conditions we have that each of \$k_i\$ must be distinct integers such that\n\n\\[\nm-j+1 < k_i \\le m\n\\]\n\nHowever, there are \$j\$ distinct \$k_i\$, but only \$j-1\$ integers satisfying the above inequality, so we have a contradiction. Our assumption that \$a_j + a_{m-j+1} \\le n\$ was false, so \$a_j + a_{m-j+1} \\ge n+1\$ for all \$j\$ such that \$1 \\le j \\le m\$ Summing these inequalities together for \$1 \\le j \\le m\$ gives\n\n\\[\n2(a_1+a_2+ \\dots a_m) \\ge m(n+1)\n\\]\n\nwhich rearranges to\n\n\\[\n\\frac{a_1+a_2+ \\dots a_m}{m} \\ge \\frac{n+1}{2}\n\\]" ]
IMO-1994-2	https://artofproblemsolving.com/wiki/index.php/1994_IMO_Problems/Problem_2	Let $ABC$ be an isosceles triangle with $AB = AC$. $M$ is the midpoint of $BC$ and $O$ is the point on the line $AM$ such that $OB$ is perpendicular to $AB$. $Q$ is an arbitrary point on $BC$ different from $B$ and $C$. $E$ lies on the line $AB$ and $F$ lies on the line $AC$ such that $E, Q, F$ are distinct and collinear. Prove that $OQ$ is perpendicular to $EF$ if and only if $QE = QF$.	[ "Let \$E'\$ and \$F'\$ be on \$AB\$ and \$AC\$ respectively such that \$E'F'\\perp OQ\$. Then, by the first part of the problem, \$QE'=QF'\$. Hence, \$Q\$ is the midpoint of \$EF\$ and \$E'F'\$, which means that \$EE'FF'\$ is a parallelogram. Unless \$E=E'\$ and \$F=F'\$, this is a contradiction since \$EE'\$ and \$F'F\$ meet at \$A\$. Therefore, \$E=E'\$ and \$F=F'\$, so \$OQ\\perp EF\$, as desired." ]
IMO-1994-3	https://artofproblemsolving.com/wiki/index.php/1994_IMO_Problems/Problem_3	For any positive integer $k$, let $f(k)$ be the number of elements in the set $\{k + 1, k + 2,\dots, 2k\}$ whose base 2 representation has precisely three $1$s. - (a) Prove that, for each positive integer $m$, there exists at least one positive integer $k$ such that $f(k) = m$. - (b) Determine all positive integers $m$ for which there exists exactly one $k$ with $f(k) = m$.	[ "a) Surjectivity of f\n\nFor space-time saving we say that a positive integer is a T-number or simply is T if it has exactly three 1s in his base two representation.\n\nIt's easy to see that (one has to place two 1s in the less n significative bit of a (n+1)-bit number)\n\n- \$f(2^n) = \\tbinom n2 = \\sum_{i=1}^{n-1} i\$ whence\n- \$f(2^{n+1}) = f(2^n)+n\$\n\nNow consider \$k=2^n + 2\$ with \$n \\ge 2\$ and the corrisponding set \$S(2^n+2)=\\{2^n+3,...,2^{n+1},2^{n+1}+1,2^{n+1}+2,2^{n+1}+3,2^{n+1}+4\\}\$ whose subset \$\\{2^n+3,...,2^{n+1}\\}\$ contains \$f(2^n)\$ T-numbers since \$S(2^n)=\\{2^n+1,...,2^{n+1}\\}=\\{2^n+1,2^n+2\\} \\cup \\{2^n+3,...,2^{n+1}\\}\$ contains \$f(2^n)\$ T-numbers by definition but \$\\{2^n+1,2^n+2\\}\$ has none. So \$S(2^n+2)=\\{2^n+3,...,2^{n+1}\\} \\cup \\{2^{n+1}+1,2^{n+1}+2,2^{n+1}+3,2^{n+1}+4\\}\$ but the last set has the only T-number \$2^{n+1}+3\$. We conclude that:\n\n- \$f(2^n + 2)=f(2^n)+1\$\n\nNow consider \$k=2^n +2^r + 2^s\$ with \$n>r>s\\ge 0\$ and \$n\\ge 2\$. We explicitly calculate f(k) for such numbers. So we have to calculate how many T-numbers are in the set \$S(k)=\\{2^n +2^r + 2^s+1;...;2^{n+1} +2^{r+1} + 2^{s+1}\\}\$.\n\nThe T-numbers less than \$2^{n+1}\$ in \$S(k)\$ are \$T_1 = \\{2^n +2^j + 2^i : (j=r \\wedge r>i>s) \\vee (n>j>r \\wedge j>i>0) \\}\$ whence \$\\# (T_1)= r-s-1 + \\sum_{h=r+1}^{n-1} h\$.\n\nThe T-numbers greater than \$2^{n+1}\$ in \$S(k)\$ are \$T_2 = \\{2^{n+1} +2^j + 2^i : (j=r+1 \\wedge s+1 \\ge i\\ge 0) \\vee (r\\ge j \\ge 1 \\wedge j>i \\ge 0) \\}\$ whence \$\\# (T_2)= s+2 + \\sum_{h=1}^{r} h\$.\n\nTherefore \$S(k)\$ contains \$\\# (T_1) + \\# (T_2)=r-s-1 + s+2 + \\sum_{h=1}^{r}h +\\sum_{h=r+1}^{n-1} h = \\sum_{h=1}^{n-1} h + r + 1\$ T-numbers and we have:\n\n- \$f(2^n +2^r + 2^s)=f(2^n) + r +1\$ where \$r=1,2,...,n-1\$ ans \$r > s \\ge 0\$ .\n\nSummarizing\n\n- \$f(2^{n+1}) = f(2^n)+n\$\n- \$f(2^n + 2)=f(2^n)+1\$\n- \$f(2^n +2^r + 2^s)=f(2^n) + r +1\$ where \$r=1,2,...,n-1\$ and \$r > s \\ge 0\$ .\n\nThat's to say that f takes all the values from \$f(2^n)\$ to \$f(2^{n+1})\$ for every \$n \\ge 2\$ and then takes any positive integer value since \$f(4)=1\$ and \$f(2^n)\$ becomes arbitrarily large.\n\nb) one-from-one values\n\nBy the fact that \$f(2^n +2^r + 2^s)=f(2^n) + r +1\$ where \$r=1,2,...,n-1\$ and \$r > s \\ge 0\$ it follows that each of the values \$f(2^n) + r +1\$ where \$r=2,...,n-1\$ come from at least two different k because there are at least two different choices for \$s\$. These values are \$f(2^n)+3,f(2^n)+4,...,f(2^n)+n=f(2^{n+1})\$.\n\nThus the only possible one-from-one values are \$f(2^n)+1\$ that come from \$k=2^n + 2\$ and \$f(2^n)+2\$ that come from \$k=2^n + 3\$.\n\nIf \$k=2^n + 3\$ we have \$f(k)=f(k+1)\$ because k+1 is not T and \$2k+1=2^{n+1} + 7\$ and \$2k+2=2^{n+1} + 8\$ are not T so f maps \$2^n + 3\$ and \$2^n + 4\$ to \$f(2^n)+2\$.\n\nIf \$k=2^n + 2\$ then \$f(k+1)=f(2^n)+2>f(k)=f(2^n)+1>f(k-1)=f(2^n+1)=f(2^n)\$. The function f is non-decreasing. It is sufficient to prove that \$f(k+1) \\ge f(k)\$ but this follows by the fact that if k+1 is T then 2k+2 is T too. By the monotonicity of f \$f( 2^n + 2)=f(2^n)+1=\\tbinom n2 +1= n(n-1)/2 + 1\$ with \$n \\ge 2\$ are the only one-from-one values of f.", "a) Surjectivity of f\n\nFor space-time saving we say that a positive integer is a T-number or simply is T if it has exactly three 1s in his base two representation and define the sets \$S(k)=\\{k+1;k+2;...;2k\\}\$. So \$f(k)\$ is the number of T-numbers in \$S(k)\$.\n\nA positive even integer 2n is T iff n is T.(Fundamental theorem of T numbers)(FTT)\n\nThe function f is non-decreasing. It is sufficient to prove that \$f(k+1) \\ge f(k)\$. This follows by (FTT) if k+1 is T then 2k+2 is T too; so passing from \$S(k)\$ to \$S(k+1)\$ we can loose a T-number \$k+1\$ but we regain it with \$2k+2\$ in \$S(k+1)\$ so \$f(k+1)\$ cannot be less than \$f(k)\$. We also have \$f(k+1) \\le f(k) + 1\$. This would be false if \$k+1\$ were not T and both \$2k+1\$ and \$2k+2\$ were T. But if \$2k+2\$ were T then \$k+1\$ would be T too by the (FTT). Then we have \$f(k) \\le f(k+1) \\le f(k) + 1\$ that is \$f(k+1)=f(k)\$ or \$f(k+1)=f(k)+1\$. But \$f(4)=1\$ and \$f(2^n)= \\tbinom n2\$ so f takes all of the positive integer values because starting from 1 with step of 1 reaches arbitrarily large integer values.\n\nb) one-from-one values of f\n\nThe one-from-one values \$f(k)\$ are such that \$f(k)=f(k-1)+1\$ and \$f(k+1)=f(k)+1\$ since f is monotone non-decreasing and has step 1.\n\nBy the condition \$f(k+1)=f(k)+1\$ since \$k+1\$ is T iff \$2k+2\$ is T we have that \$2k+1\$ must be T.\n\nBy the condition \$f(k)=f(k-1)+1\$ since \$k\$ is T iff \$2k\$ is T we have that \$2k-1\$ must be T.\n\nThen \$2k-1\$ and \$2k+1\$ must have the form \$2^{n+1}+2^{r+1}+1\$ with \$n>r\\ge 0\$ since they are odd and \$n \\ge 2\$ since there is only 1 T-number less than 8 and they must have the same number of bits since their difference is 2 and both are T (the only two binary numbers which differs by 2 and have different number of bits are \$2^n+1\$ and \$2^n-1\$ or \$2^n\$ and \$2^n-2\$ which are evidently not T). Let \$2k+1=2^{n+1}+2^{j+1}+1\$ and \$2k-1=2^{n+1}+2^{i+1}+1\$ with \$j>i \\ge 0\$. Then \$2k+ 1-(2k-1)=2^{j+1}-2^{i+1}=2\$ that is \$j=1\$ and \$i=0\$. We conclude that \$f(k)\$ is one-from-one for \$k=2^n+2^j=2^n+2\$ with \$n \\ge 2\$. Since \$f(2^n+1)=f(2^n)=\\tbinom n2\$ we have that \$f(2^n+2)=\\tbinom n2 +1= n(n-1)/2 + 1\$ with \$n \\ge 2\$ are the only one-from-one values of f." ]
IMO-1994-4	https://artofproblemsolving.com/wiki/index.php/1994_IMO_Problems/Problem_4	Find all ordered pairs $(m,n)$ where $m$ and $n$ are positive integers such that $\frac {n^3 + 1}{mn - 1}$ is an integer.	[ "Suppose \$\\frac{n^3+1}{mn-1}=k\$ where \$k\$ is a positive integer. Then \$n^3+1=(mn-1)k\$ and so it is clear that \$k\\equiv -1\\pmod{n}\$. So, let \$k=jn-1\$ where \$j\$ is a positive integer. Then we have \$n^3+1=(mn-1)(jn-1)=mjn^2-(m+j)n+1\$ which by cancelling out the \$1\$s and dividing by \$n\$ yields \$n^2=mjn-(m+j)\\Rightarrow n^2-mjn+m+j=0\$. The equation \$x^2-mjx+m+j=0\$ is a quadratic. We are given that \$n\$ is one of the roots. Let \$p\$ be the other root. Notice that since \$n+p=mj\$ we have that \$p\$ is an integer, and so from \$np=m+j\$ we have that \$p\$ is positive.\n\nIt is obvious that \$j=m=n=p=2\$ is a solution. Now, if not, and \$j,m,n,p\$ are all greater than \$1\$, we have the inequalities \$np>n+p\$ and \$mj>m+j\$ which contradicts the equations \$np=m+j, n+p=mj\$. Thus, at least one of \$j,m,n,p\$ is equal to \$1\$.\n\nIf one of \$m,j\$ is \$1\$, without loss of generality assume it is \$j\$. Then we have \$np=m+1, n+p=m\$. That is, \$np-n-p=1\\Rightarrow (n-1)(p-1)=2\$ which gives positive solutions \$(n,p)=(3,2),(2,3)\$. These give \$m=5\$ and since we assumed \$j=1\$, we can also have \$m=1\$ and \$j=5\$.\n\nIf one of \$n,p\$ is \$1\$, without loss of generality assume it is \$p\$. Then we have \$n=m+j, n+1=mj\$. That is, \$mj-m-j=1\\Rightarrow (m-1)(j-1)=2\$ which gives positive solutions \$(m,j)=(3,2),(2,3)\$. These give \$n=5\$ and since we assumed \$p=1\$, we can also have \$n=1\$ and \$p=5\$.\n\nFrom these, we have all solutions \$(m,n)=(2,2),(5,3),(5,2),(1,3),(1,2),(3,5),(2,5),(3,1),(2,1)\$." ]
IMO-1994-5	https://artofproblemsolving.com/wiki/index.php/1994_IMO_Problems/Problem_5	Let $S$ be the set of real numbers strictly greater than $-1$. Find all functions $f:S \to S$ satisfying the two conditions: 1. $f(x+f(y)+xf(y)) = y+f(x)+yf(x)$ for all $x$ and $y$ in $S$; 2. $\frac{f(x)}{x}$ is strictly increasing on each of the intervals $-1<x<0$ and $0<x$.	[ "The only solution is \$f(x) = \\frac{-x}{x+1}.\$\n\nSetting \$x=y,\$ we get\n\n\\[\nf(x+f(x)+xf(x)) = x+f(x)+xf(x).\n\\]\n\nTherefore, \$f(s) = s\$ for \$s = x+f(x)+xf(x).\$\n\nNote: If we can show that \$s\$ is always \$0,\$ we will get that \$x+f(x)+xf(x) = 0\$ for all \$x\$ in \$S\$ and therefore, \$f(x) = \\frac{-x}{x+1}.\$\n\nLet \$f(s)=s.\$ Setting \$x=y=s,\$ we get\n\n\\[\nf(s + f(s) + sf(s)) = f(2s + s^2) = 2s+s^2.\n\\]\n\nIf \$t=2s+s^2,\$ we have \$f(t)=t\$ as well.\n\nConsider \$s > 0.\$ We get \$t = 2s + s^2 > s.\$ Since \$\\frac{f(x)}{x}\$ is strictly increasing for \$x>0\$ and \$t > s\$ in this domain, we must have \$\\frac{f(t)}{t} > \\frac{f(s)}{s}\$ but since \$f(s) = s\$ and \$f(t) = t,\$ we also have that \$\\frac{f(s)}{s} = 1 = \\frac{f(t)}{t}\$ which is a contradiction. Therefore \$s \\leq 0\$\n\nConsider \$-1 < s < 0.\$ Using a similar argument, we will get that \$t = 2s + s^2 < s\$ but also \$\\frac{f(s)}{s} = 1 = \\frac{f(t)}{t},\$ which is a contradiction.\n\nHence, \$s\$ must be \$0.\$ Since \$s=0,\$ we can conclude that \$x+f(x)+xf(x) = 0\$ and therefore, \$f(x) = \\frac{-x}{x+1}\$ for all \$x\$." ]
IMO-1994-6	https://artofproblemsolving.com/wiki/index.php/1994_IMO_Problems/Problem_6	Show that there exists a set $A$ of positive integers with the following property: For any infinite set $S$ of primes there exist two positive integers $m \in A$ and $n \not\in A$ each of which is a product of $k$ distinct elements of $S$ for some $k \ge 2$.	[]
IMO-1995-1	https://artofproblemsolving.com/wiki/index.php/1995_IMO_Problems/Problem_1	Let $A,B,C,D$ be four distinct points on a line, in that order. The circles with diameters $AC$ and $BD$ intersect at $X$ and $Y$. The line $XY$ meets $BC$ at $Z$. Let $P$ be a point on the line $XY$ other than $Z$. The line $CP$ intersects the circle with diameter $AC$ at $C$ and $M$, and the line $BP$ intersects the circle with diameter $BD$ at $B$ and $N$. Prove that the lines $AM,DN,XY$ are concurrent.	[ "Since \$M\$ is on the circle with diameter \$AC\$, we have \$\\angle AMC=90\$ and so \$\\angle MCA=90-A\$. We similarly find that \$\\angle BND=90\$. Also, notice that the line \$XY\$ is the radical axis of the two circles with diameters \$AC\$ and \$BD\$. Thus, since \$P\$ is on \$XY\$, we have \$PN\\cdot PB=PM\\cdot PC\$ and so by the converse of Power of a Point, the quadrilateral \$MNBC\$ is cyclic. Thus, \$90-A=\\angle MCA=\\angle BNM\$. Thus, \$\\angle MND=180-A\$ and so quadrilateral \$AMND\$ is cyclic. Let the name of the circle \$AMND\$ be \$O\$ . Then, the radical axis of \$O\$ and the circle with diameter \$AC\$ is line \$AM\$. Also, the radical axis of \$O\$ and the circle with diameter \$BD\$ is line \$DN\$. Since the pairwise radical axes of 3 circles are concurrent, we have \$AM,DN,XY\$ are concurrent as desired.", "Let \$AM\$ and \$PT\$ (a subsegment of \$XY\$) intersect at \$Z\$. Now, assume that \$Z, N, P\$ are not collinear. In that case, let \$ZD\$ intersect the circle with diameter \$BD\$ at \$N'\$ and the circle through \$D, P, T\$ at \$N''\$.\n\nWe know that \$\\angle AMC = \\angle BND = \\angle ATP = 90^\\circ\$ via standard formulae, so quadrilaterals \$AMPT\$ and \$DNPT\$ are cyclic. Thus, \$N'\$ and \$N''\$ are distinct, as none of them is \$N\$. Hence, by Power of a Point,\n\n\\[\nZM * ZA = ZP * ZT = ZN'' * ZD.\n\\]\n\nHowever, because \$Z\$ lies on radical axis \$TP\$ of the two circles, we have\n\n\\[\nZM * ZA = ZN' * ZD.\n\\]\n\nHence, \$ZN'' = ZN'\$, a contradiction since \$D\$ and \$D'\$ are distinct. We therefore conclude that \$Z, N, D\$ are collinear, which gives the concurrency of \$AM, XY\$, and \$DN\$. This completes the problem.", "Let \$AM\$ and \$XY\$ intersect at \$Z\$. Because \$\\angle AMC = \\angle BND = \\angle APT = 90^\\circ\$, we have quadrilaterals \$AMPT\$ and \$DNPT\$ cyclic. Therefore, \$Z\$ lies on the radical axis of the two circumcircles of these quadrilaterals. But \$Z\$ also lies on radical axis \$XY\$ of the original two circles, so the power of \$Z\$ with respect to each of the four circles is all equal to \$ZM * ZA\$. Hence, \$Z\$ lies on the radical axis \$DN\$ of the two circles passing through \$D\$ and \$N\$, as desired.\n\nwhat is \$T\$ here?", "Let the circle with diameter \$AC\$ be \$\\omega_1\$ and the circle with diameter \$BD\$ be \$\\omega_2\$. Let \$Q\$ be the intersection of \$AM\$ and \$XY\$ and \$N'\$ be the intersection of \$QD\$ with \$\\omega_2\$. Since \$AC\$ is the diameter of \$\\omega_1\$, \$\\angle AMC = \\angle QMC = 90\$. Since \$XY\$ is the radical axis of \$\\omega_1\$ and \$\\omega_2\$, \$\\angle QZD = \\angle QZA = 90\$ so \$\\triangle QMP \\sim \\triangle QZA\$ by AAA. Hence \$\\frac {QM}{QZ} = \\frac {QP}{QA}\$ so \$QM\\times QA = QP\\times QZ\$. From power of a point, \$QM\\times QA = QN'\\times QD\$ so \$QP\\times QZ = QN'\\times QD\$. Hence \$\\frac{QP}{QD} = \\frac{QN'}{QZ}\$ so \$\\triangle QN'P \\sim \\triangle QZD\$, so \$\\angle QN'P = 90 = \\angle PN'D\$ as \$QN'D\$ is a straight line. Thus \$PZDN'\$ is cyclic so the radical axis of \$PZDN'\$ and \$\\omega_2\$ is \$DN'\$. However \$PZDN\$ is also cyclic (since \$\\angle BND = \\angle PND = 90\$) so the radical axis of \$PZDN\$ and \$\\omega _2\$ is \$DN\$. Since two circles cannot have more than one radical axis (\$PZDN\$ and \$PZDN'\$ are the same circles), \$N'\$ must lie on \$DN\$. But since \$N\$ and \$N'\$ lie on \$\\omega_2\$, \$N=N'\$ so \$AM\$, \$DN\$ and \$XY\$ are concurrent\n\n## Discussion\n\nLemma: The radical axis of two pairs of circles \$O_1\$, \$O_2\$ and \$O_3\$, \$O_4\$ are the same line \$l\$. Furthermore, \$O_1\$ and \$O_2\$ intersect at \$A\$ and \$B\$, and \$O_3\$ and \$O_4\$ intersect at \$C\$ and \$D\$. Then \$A, B, C,\$ and \$D\$ are concyclic.\n\nThe proof of this lemma is trivial using the argument in Solution 3 and applying the converse of Power of a Point.\n\nNote that this Problem 1 is a corollary of this lemma. This lemma is an effective way to relate four circles, just as the radical center can relate three circles.\n\nSolution 1 also gives a trivial lemma that can also be useful:\n\nLemma 2: Chords \$AB\$ of \$\\omega_1\$ and \$CD\$ of \$\\omega_2\$ intersect on the segment \$XY\$ formed from the intersections of the two circles. Then \$A, B, C, D\$ are concyclic.\n\nTwo ways to solve a problem, two different insights into circle geometry. That is cool, but more RADICAL!" ]
IMO-1995-2	https://artofproblemsolving.com/wiki/index.php/1995_IMO_Problems/Problem_2	Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that \[ \frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2}. \]	[ "We make the substitution \$x= 1/a\$, \$y=1/b\$, \$z=1/c\$. Then\n\n\\[\n\\begin{align} \\frac{1}{a^3(b+c)} + \\frac{1}{b^3(c+a)} + \\frac{1}{c^3(a+b)} &= \\frac{x^3}{xyz(1/y+1/z)} + \\frac{y^3}{xyz(1/z+1/x)} + \\frac{z^3}{xyz(1/x+1/z)} \\\\ &= \\frac{x^2}{y+z} + \\frac{y^2}{z+x} + \\frac{z^2}{x+y} . \\end{align}\n\\]\n\nSince \$(x^2,y^2,z^2)\$ and \$\\bigl( 1/(y+z), 1/(z+x), 1/(x+y) \\bigr)\$ are similarly sorted sequences, it follows from the Rearrangement Inequality that\n\n\\[\n\\frac{x^2}{y+z} + \\frac{y^2}{z+x} + \\frac{z^2}{x+y} \\ge \\frac{1}{2} \\left( \\frac{y^2+z^2}{y+z} + \\frac{z^2+x^2}{z+x} + \\frac{x^2+y^2}{x+y} \\right) .\n\\]\n\nBy the Power Mean Inequality,\n\n\\[\n\\frac{y^2+z^2}{y+z} \\ge \\frac{(y+z)^2}{2(x+y)} = \\frac{x+y}{2} .\n\\]\n\nSymmetric application of this argument yields\n\n\\[\n\\frac{1}{2}\\left( \\frac{y^2+z^2}{y+z} + \\frac{z^2+x^2}{z+x} + \\frac{x^2+y^2}{x+y} \\right) \\ge \\frac{1}{2}(x+y+z) .\n\\]\n\nFinally, AM-GM gives us\n\n\\[\n\\frac{1}{2}(x+y+z) \\ge \\frac{3}{2}xyz = \\frac{3}{2},\n\\]\n\nas desired. \$\\blacksquare\$", "We make the same substitution as in the first solution. We note that in general,\n\n\\[\n\\frac{p}{q+r} = \\frac{(p+q+r)}{(p+q+r)-p} - 1 .\n\\]\n\nIt follows that \$(x,y,z)\$ and \$\\bigl(x/(y+z), y/(z+x), z/(x+y)\\bigr)\$ are similarly sorted sequences. Then by Chebyshev's Inequality,\n\n\\[\n\\frac{x^2}{y+z} + \\frac{y^2}{z+x} + \\frac{z^2}{x+y} \\ge \\frac{1}{3}(x+y+z) \\left(\\frac{x}{y+z} + \\frac{y}{z+x} + \\frac{z}{x+y} \\right) .\n\\]\n\nBy AM-GM, \$\\frac{x+y+z}{3} \\ge \\sqrt[3]{xyz}=1\$, and by Nesbitt's Inequality,\n\n\\[\n\\frac{x}{y+z} + \\frac{y}{z+x} + \\frac{z}{x+y} \\ge \\frac{3}{2}.\n\\]\n\nThe desired conclusion follows. \$\\blacksquare\$", "Without clever substitutions: By Cauchy-Schwarz,\n\n\\[\n\\left(\\sum_{cyc}\\dfrac{1}{a^3 (b+c)}\\right)\\left(\\sum_{cyc}a(b+c)\\right)\\geq \\left( \\dfrac{1}{a}+\\dfrac{1}{b}+\\dfrac{1}{c}\\right) ^2=(ab+ac+bc)^2\n\\]\n\nDividing by \$2(ab+bc+ac)\$ gives\n\n\\[\n\\dfrac{1}{a^3 (b+c)}+\\dfrac{1}{b^3 (a+c)}+\\dfrac{1}{c^3 (a+b)}\\geq \\dfrac{1}{2}(ab+bc+ac)\\geq \\dfrac{3}{2}\n\\]\n\nby AM-GM.", "Without clever notation: By Cauchy-Schwarz,\n\n\\[\n\\left(a(b+c) + b(c+a) + c(a+b)\\right) \\cdot \\left(\\frac{1}{a^3 (b+c)} + \\frac{1}{b^3 (c+a)} + \\frac{1}{c^3 (a+b)}\\right)\n\\]\n\n\\[\n\\ge \\left(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\right)^2\n\\]\n\n\\[\n= (ab + bc + ac)^2\n\\]\n\nDividing by \$2(ab + bc + ac)\$ and noting that \$ab + bc + ac \\ge 3(a^2b^2c^2)^{\\frac{1}{3}} = 3\$ by AM-GM gives\n\n\\[\n\\frac{1}{a^3 (b+c)} + \\frac{1}{b^3 (c+a)} + \\frac{1}{c^3 (a+b)} \\ge \\frac{ab + bc + ac}{2} \\ge \\frac{3}{2},\n\\]\n\nas desired.", "After the setting \$a=\\frac{1}{x}, b=\\frac{1}{y}, c=\\frac{1}{z},\$ and as \$abc=1\$ so \$\\left(\\frac{1}{a}\\cdot\\frac{1}{b}\\cdot\\frac{1} {c}=1\\right)\$ concluding \$x y z=1 .\$\n\n\$\\textsf{Claim}:\$\n\n\\[\n\\frac{x^{2}}{y+z}+\\frac{y^{2}}{z+x}+\\frac{z^{2}}{x+y} \\geq \\frac{3}{2}\n\\]\n\nBy Titu Lemma,\n\n\\[\n\\implies\\frac{x^{2}}{y+z}+\\frac{y^{2}}{z+x}+\\frac{z^{2}}{x+y} \\geq \\frac{(x+y+z)^{2}}{2(x+y+z)}\n\\]\n\n\\[\n\\implies\\frac{x^{2}}{y+z}+\\frac{y^{2}}{z+x}+\\frac{z^{2}}{x+y} \\geq \\frac{(x+y+z)}{2}\n\\]\n\nNow by AM-GM we know that\n\n\\[\n(x+y+z)\\geq3\\sqrt[3]{xyz}\n\\]\n\nand \$xyz=1\$ which concludes to \$\\implies (x+y+z)\\geq3\\sqrt[3]{1}\$\n\nTherefore we get\n\n\\[\n\\implies\\frac{x^{2}}{y+z}+\\frac{y^{2}}{z+x}+\\frac{z^{2}}{x+y} \\geq \\frac{3}{2}\n\\]\n\nHence our claim is proved ~~ Aritra12", "Proceed as in Solution 1, to arrive at the equivalent inequality\n\n\\[\n\\frac{x^2}{y+z} + \\frac{y^2}{z+x} + \\frac{z^2}{x+y} \\ge \\frac{3}{2} .\n\\]\n\nBut we know that\n\n\\[\nx + y + z \\ge 3xyz = 3\n\\]\n\nby AM-GM. Furthermore,\n\n\\[\n(x + y + y + z + x + z) (\\frac{x^2}{y+z} + \\frac{y^2}{z+x} + \\frac{z^2}{x+y}) \\ge (x + y + z)^2\n\\]\n\nby Cauchy-Schwarz, and so dividing by \$2(x + y + z)\$ gives\n\n\\[\n\\begin{align}\\frac{x^2}{y+z} + \\frac{y^2}{z+x} + \\frac{z^2}{x+y} &\\ge \\frac{(x + y + z)}{2} \\\\ &\\ge \\frac{3}{2} \\end{align}\n\\]\n\nas desired.", "Without clever substitutions, and only AM-GM!\n\nNote that \$abc = 1 \\implies a = \\frac{1}{bc}\$. The cyclic sum becomes \$\\sum_{cyc}\\frac{(bc)^3}{b + c}\$. Note that by AM-GM, the cyclic sum is greater than or equal to \$3\\left(\\frac{1}{(a+b)(b+c)(a+c)}\\right)^{\\frac13}\$. We now see that we have the three so we must be on the right path. We now only need to show that \$\\frac32 \\geq 3\\left(\\frac{1}{(a+b)(b+c)(a+c)}\\right)^\\frac13\$. Notice that by AM-GM, \$a + b \\geq 2\\sqrt{ab}\$, \$b + c \\geq 2\\sqrt{bc}\$, and \$a + c \\geq 2\\sqrt{ac}\$. Thus, we see that \$(a+b)(b+c)(a+c) \\geq 8\$, concluding that \$\\sum_{cyc} \\frac{(bc)^3}{b+c} \\geq \\frac32 \\geq 3\\left(\\frac{1}{(a+b)(b+c)(a+c)}\\right)^{\\frac13}\$.\n\n^ This solution is incorrect, as it does not prove inequalities in the right direction. Proving that \$A \\geq B\$, and \$C \\geq B\$ does not show that \$A \\geq C \\geq B\$.", "Rewrite \$\\frac{1}{a^3(b+c)} + \\frac{1}{b^3(a+c)} + \\frac{1}{c^3(a+b)}\$ as \$\\frac{(1/a)^2}{a(b+c)} + \\frac{(1/b)^2}{b(a+c)} + \\frac{(1/c)^2}{c(a+b)}\$.\n\nNow applying Titu's lemma yields \$\\frac{(1/a)^2}{a(b+c)} + \\frac{(1/b)^2}{b(a+c)} + \\frac{(1/c)^2}{c(a+b)} \\geq \\frac{(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c})^2}{a(b+c) + b(a+c) + c(a+b)} = \\frac{(ab + bc + ca)^2}{2(ab + bc + ca)} = \\frac{ab + bc + ca}{2}\$.\n\nNow applying the AM-GM inequality on \$ab + bc +ca \\geq 3((abc)^2)^{\\frac{1}{3}} = 3\$. The result now follows.\n\nNote: \$ab + bc + ca = \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\$, because \$abc = 1\$. (Why? Because \$a = \\frac{1}{bc}\$, and hence \$\\frac{1}{a} = bc\$).\n\n~th1nq3r" ]
IMO-1995-3	https://artofproblemsolving.com/wiki/index.php/1995_IMO_Problems/Problem_3	Determine all integers $n>3$ for which there exist $n$ points $A_1,\ldots,A_n$ in the plane, no three collinear, and real numbers $r_1,\ldots,r_n$ such that for $1\le i<j<k\le n$, the area of $\triangle A_iA_jA_k$ is $r_i+r_j+r_k$.	[]
IMO-1995-4	https://artofproblemsolving.com/wiki/index.php/1995_IMO_Problems/Problem_4	The positive real numbers $x_0, x_1, x_2,.....x_{1994}, x_{1995}$ satisfy the relations $x_0=x_{1995}$ and $x_{i-1}+\frac{2}{x_{i-1}}=2{x_i}+\frac{1}{x_i}$ for $i=1,2,3,....1995$ Find the maximum value that $x_0$ can have.	[ "First we start by solving for \$x_{i}\$ in the recursive relation\n\n\\[\nx_{i-1}+\\frac{2}{x_{i-1}}=2x_{i}+\\frac{1}{x_{i}}\n\\]\n\n\\[\n\\frac{x_{i-1}^{2}+2}{x_{i-1}}=\\frac{2x_{i}^{2}+1}{x_{i}}\n\\]\n\n\\[\n\\left( x_{i} \\right)\\left( x_{i-1}^{2}+2 \\right)=\\left( x_{i-1} \\right)\\left( 2x_{i}^{2}+1 \\right)\n\\]\n\n\\[\nx_{i}x_{i-1}^{2}+2x_{i}=2x_{i-1}x_{i}^{2}+x_{i-1}\n\\]\n\n\\[\nx_{i}x_{i-1}^{2}-2x_{i-1}x_{i}^{2}+2x_{i}-x_{i-1}=0\n\\]\n\n\\[\nx_{i}x_{i-1}\\left( x_{i-1} \\right)-2x_{i}x_{i-1}\\left( x_{i} \\right)+2x_{i}-x_{i-1}=0\n\\]\n\n\\[\nx_{i}x_{i-1}\\left( x_{i-1} -2x_{i}\\right)+2x_{i}-x_{i-1}=0\n\\]\n\n\\[\n\\left( 2x_{i}-x_{i-1}\\right)\\left( 1-x_{i}x_{i-1} \\right)=0\n\\]\n\n\$x_{i}=\\frac{x_{i-1}}{2},\\;\$ or \$x_{i}=\\frac{1}{x_{i-1}}\$\n\nSo we have two recursive properties to chose from.\n\nIf we want to maximize \$x_{0}\$ then we can use \$x_{i}=\\frac{x_{i-1}}{2}\$ from \$i=1\$ to \$1994\$. This will make \$x_{0}\$ the largest and \$x_{1994}\$ the smallest.\n\nThen we can simply use \$x_{i}=\\frac{1}{x_{i-1}}\$ to get \$x_{1995}\$ since the reciprocal will make it very large.\n\nThen we use \$x_{0}=x_{1995}\$ and solve for \$x_{0}\$\n\nThis means that we can write \$x_{i}\$ as:\n\n\\[\nx_{i}=\\begin{cases} \\frac{x_{0}}{2^{i}} & 1 \\le i \\le 1994 \\\\ \\frac{1}{x_{i-1}} & i=1995\\end{cases}\n\\]\n\nThen \$x_{1994}=\\frac{x_0}{2^{1994}}\$,\n\nthus \$x_{1995}=\\frac{1}{x_{1994}}=\\frac{2^{1994}}{x_0}={x_0}\$\n\nSolving for \${x_0}\$ we get:\n\n\\[\n{x_0}^{2}=2^{1994}\n\\]\n\n\${x_0}=\\pm\\sqrt{2^{1994}}=\\pm2^{997}\$. Since all \${x_i}\$ are defined as positive, \${x_0}=2^{997}\$.\n\nTherefore, the maximum value that \$x_0\$ can have is \${x_0}=2^{997}\$\n\n~ Tomas Diaz. orders@tomasdiaz.com" ]
IMO-1995-5	https://artofproblemsolving.com/wiki/index.php/1995_IMO_Problems/Problem_5	Let $ABCDEF$ be a convex hexagon with $AB=BC=CD$ and $DE=EF=FA$, such that $\angle BCD=\angle EFA=\frac{\pi}{3}$. Suppose $G$ and $H$ are points in the interior of the hexagon such that $\angle AGB=\angle DHE=\frac{2\pi}{3}$. Prove that $AG+GB+GH+DH+HE\ge CF$.	[ "Draw \$AE\$ and \$BD\$ to make equilateral \$\\triangle EFA\$ and \$\\triangle BCD\$, and draw points \$I\$ and \$J\$ such that \$IA=IB\$, \$JD=JE\$, directed angle \$\\measuredangle IAB=-\\measuredangle CDB\$, and directed angle \$\\measuredangle JDE=-\\measuredangle FAE\$ to make equilateral \$\\triangle AIB\$ and \$\\triangle DJE\$. Notice that \$G\$ is on the circumcircle of \$\\triangle AIB\$ and \$H\$ is on the circumcircle of \$\\triangle DJE\$. By Ptolemy, \$GA+GB=GI\$ and \$HD+HE=HJ\$. So,\n\n\\[\nAG+GB+GH+DH+HE=IG+GH+HJ.\n\\]\n\nNotice that octagon \$AIBCDJEF\$ is symmetric about \$\\overline{BE}\$. So, \$IG+GH+HJ\\ge IJ=CF\$. --tigerzhang" ]
IMO-1995-6	https://artofproblemsolving.com/wiki/index.php/1995_IMO_Problems/Problem_6	Let $p$ be an odd prime number. How many $p$-element subsets $A$ of ${1,2,\ldots,2p}$ are there, the sum of whose elements is divisible by $p$?	[ "Let \$A(x,y)\$ be the generating function\n\n\\[\nA(x,y) = (1+yx)(1+yx^2)\\cdots(1+yx^{2p})\n\\]\n\nWe apply the roots of unity filter on \$x\$ to get\n\n\\[\n\\frac{A(1,y)+A(w,y)+\\cdots+A(w^{p-1},y)}{p} = \\frac{(1+y)^{2p}+(p-1)(1+yw)\\cdots(1+yw^{2p})}{p}\n\\]\n\nWe call this function on \$y\$, \$B(y)\$. Note that\n\n\\[\n(1+w)(1+w^2)\\cdots(1+w^{p}) = 2\n\\]\n\nThen, we apply the roots of unity filter on \$y\$ to get\n\n\\begin{align} \\frac{B(1)+B(w)+B(w^2)+\\cdots B(w^{p-1})}{p} &= \\frac{p+p\\binom{2p}{p}+p+2^{2}(p-1)(p)}{p^2} \\end{align}\n\nBut, we need to subtract \$2\$ because it counts the empty set and the set with size \$2p\$. This gives us\n\n\\[\n\\boxed{\\frac{\\dbinom{2p}{p}+2p-2}{p}}\n\\]\n\n\\[\n\\square\n\\]\n\nSolution from this discussion: https://artofproblemsolving.com/community/c6t302107f6h15112_sum_of_whose_elements_is_divisible_by_p." ]
IMO-1996-1	https://artofproblemsolving.com/wiki/index.php/1996_IMO_Problems/Problem_1	We are given a positive integer $r$ and a rectangular board $ABCD$ with dimensions $\|AB\|=20$, $\|BC\|=12$. The rectangle is divided into a grid of $20 \times 12$ unit squares. The following moves are permitted on the board: one can move from one square to another only if the distance between the centers of the two squares is $\sqrt{r}$. The task is to find a sequence of moves leading from the square with $A$ as a vertex to the square with $B$ as a vertex. (a) Show that the task cannot be done if $r$ is divisible by $2$ or $3$. (b) Prove that the task is possible when $r=73$. (c) Can the task be done when $r=97$?	[ "First we define the rectangular board in the cartesian plane with centers of the unit squares as integer coordinates and the following coordinates for the squares at the corners of \$A\$, \$B\$, \$C\$, \$D\$, as follows: \$A=(1,1)\$, \$B=(20,1)\$, \$C=(20,12)\$, \$D=(1,12)\$\n\nLet \$(x_i,y_i)\$ be the coordinates of the piece after move \$i\$ with \$(x_0,y_0)=A=(1,1)\$ the initial position of the piece.\n\n\$1 \\le x_i \\le 20\$, and also \$1 \\le y_i \\le 12\$.\n\nLet \$\\Delta x_i = x_i-x_{i-1}\$, \$\\Delta y_i = y_i-y_{i-1}\$\n\nThen, for any given \$r\$, we have \$(\\Delta x_i)^2+(\\Delta y_i)^2=\\left( \\sqrt{r} \\right)^2=r\$ for all \$i\$\n\nPart (a):\n\nIn order to find out the conditions for which \$r\$ is divisible by 2 we are going to look at the following three cases:\n\n(1) When both \$\\Delta x_i\$ and \$\\Delta y_i\$ are divisible by \$2\$.\n\n(2) When both \$\\Delta x_i\$ and \$\\Delta y_i\$ are odd.\n\n(3) When one of \$\\Delta x_i\$ and \$\\Delta y_i\$ is even and the other one is odd.\n\nCase (1): Since \$\\Delta x_i \\equiv 0\\;(mod \\; 2)\$ and \$\\Delta y_i \\equiv 0\\;(mod \\; 2)\$,\n\nthen \$(\\Delta x_i)^2+(\\Delta y_i)^2 \\equiv (0^2+0^2)\\;(mod \\; 2)\\equiv 0\\;(mod \\; 2)\$.\n\nThus, for \$r\\equiv 0\\;(mod \\; 2)\$, this case is a valid one.\n\nCase (2): Since \$\\Delta x_i \\equiv 1\\;(mod \\; 2)\$ and \$\\Delta y_i \\equiv 1\\;(mod \\; 2)\$,\n\nthen \$(\\Delta x_i)^2+(\\Delta y_i)^2 \\equiv (1^2+1^2)\\;(mod \\; 2)\\equiv 0\\;(mod \\; 2)\$.\n\nThus, for \$r\\equiv 0\\;(mod \\; 2)\$, this case is a valid one.\n\nCase (3): Since \$\\Delta x_i \\equiv 1\\;(mod \\; 2)\$ and \$\\Delta y_i \\equiv 0\\;(mod \\; 2)\$, or \$\\Delta x_i \\equiv 0\\;(mod \\; 2)\$ and \$\\Delta y_i \\equiv 1\\;(mod \\; 2)\$,\n\nthen \$(\\Delta x_i)^2+(\\Delta y_i)^2 \\equiv (1^2+0^2)\\;(mod \\; 2)\\equiv 1\\;(mod \\; 2)\\not\\equiv 0\\;(mod \\; 2)\$.\n\nThus, for \$r\\equiv 0\\;(mod \\; 2)\$, this case is NOT a valid one.\n\nHaving proved that Case (1) and Case (2) are the only valid cases for \$r\\equiv 0\\;(mod \\; 2)\$ we are going to see what happens for both cases when we start with a square where both coordinates are odd:\n\nif \$(x_{i-1},y_{i-1}) \\equiv (1,1) \\;(mod \\; 2)\$,\n\nthen for case (1): \$(x_{i-1}+\\Delta x_i,y_{i-1}+\\Delta y_i) \\equiv (1+0,1+0) \\;(mod \\; 2) \\equiv (1,1) \\;(mod \\; 2)\$\n\nand for case (2): \$(x_{i-1}+\\Delta x_i,y_{i-1}+\\Delta y_i) \\equiv (1+1,1+1) \\;(mod \\; 2) \\equiv (0,0) \\;(mod \\; 2)\$\n\nand \$(x_{i-1}+2\\Delta x_i,y_{i-1}+2\\Delta y_i) \\equiv (1+2,1+2) \\;(mod \\; 2) \\equiv (1,1) \\;(mod \\; 2)\$\n\nThis means that when \$r\$ is divisible by two, when starting at \$(1,1)\$ no matter how many moves you make all \$(x_{i},y_{i})\$ will either be \$\\equiv (0,0) \\;(mod \\; 2)\$ or \$\\equiv (1,1) \\;(mod \\; 2)\$.\n\nSince the ending coordinate is \$(20,1) \\equiv (0,1) \\;(mod \\; 2) \\not\\equiv (1,1) \\;(mod \\; 2)\$ and \$\\not\\equiv (0,0) \\;(mod \\; 2)\$, then the task cannot be done if \$r\$ is divisible by \$2\$.\n\nNow we look at the conditions for which \$r\$ is divisible by 3 by looking at the following three cases:\n\n(4) When both \$\\Delta x_i\$ and \$\\Delta y_i\$ are divisible by \$3\$.\n\n(5) When one of them is not divisible by \$3\$ and the other one is.\n\n(6) When neither \$\\Delta x_i\$ nor \$\\Delta y_i\$ is divisible by \$3\$.\n\nCase (4): Since \$\\Delta x_i \\equiv 0\\;(mod \\; 3)\$ and \$\\Delta y_i \\equiv 0\\;(mod \\; 3)\$,\n\nthen \$(\\Delta x_i)^2+(\\Delta y_i)^2 \\equiv (0^2+0^2)\\;(mod \\; 3)\\equiv 0\\;(mod \\; 3)\$.\n\nThus, for \$r\\equiv 0\\;(mod \\; 3)\$, this case is a valid one.\n\nCase (5): Since either \$\\Delta x_i\$ or \$\\Delta y_i \\equiv \\pm 1\\;(mod \\; 3)\$ and the other \$\\equiv 1\\;(mod \\; 3)\$,\n\nthen \$(\\Delta x_i)^2+(\\Delta y_i)^2 \\equiv ((\\pm 1)^2+0^2)\\;(mod \\; 3)\\equiv 1\\;(mod \\; 3)\$.\n\nThus, for \$r\\equiv 0\\;(mod \\; 3)\$, this case is NOT a valid one.\n\nCase (6): Since \$\\Delta x_i \\equiv \\pm 1\\;(mod \\; 3)\$ and \$\\Delta y_i \\equiv \\pm1 \\;(mod \\; 3)\$,\n\nthen \$(\\Delta x_i)^2+(\\Delta y_i)^2 \\equiv ((\\pm 1)^2+(\\pm 1)^2)\\;(mod \\; 3)\\equiv 2\\;(mod \\; 3)\\not\\equiv 0\\;(mod \\; 3)\$.\n\nThus, for \$r\\equiv 0\\;(mod \\; 3)\$, this case is NOT a valid one.\n\nHaving proved that Case (4) is the only valid case for \$r\\equiv 0\\;(mod \\; 3)\$ we are going to see what happens when we start with a square that is \$\\equiv (1,1)\\;(mod \\; 3)\$\n\nif \$(x_{i-1},y_{i-1}) \\equiv (1,1) \\;(mod \\; 3)\$, then \$(x_{i-1}+\\Delta x_i ,y_{i-1}+\\Delta y_i) \\equiv (1+0,1+0) \\;(mod \\; 3) \\equiv (1,1) \\;(mod \\; 3)\$.\n\nTherefore starting at any \$(x_{i-1},y_{i-1}) \\equiv (1,1) \\;(mod \\; 3)\$ no matter how many moves the piece will always land at another square \$\\equiv (1,1) \\;(mod \\; 3)\$. Since we start at \$(1,1)\$, then the piece will always land at squares that are \$\\equiv (1,1) \\;(mod \\; 3)\$.\n\nSince final square is \$(20,1) \\equiv (2,1) \\;(mod \\; 3) \\not\\equiv (1,1) \\;(mod \\; 3)\$, then the task cannot be done if \$r\$ is divisible by \$3\$.\n\nIn summary the task cannot be done if \$r\$ is divisible by \$2\$ or \$r\$ is divisible by \$3\$.\n\nPart (b):\n\nWhen \$r=73\$ we have \$8^2+3^2=73\$\n\nTherefore the moves can have the following values:\n\n\\[\n(\\Delta x_i, \\Delta y_i)=\\begin{cases} (\\pm 8,\\pm 3) \\\\ (\\pm 8,\\mp 3) \\\\ (\\pm 3,\\pm 8) \\\\ (\\pm 3,\\mp 8) \\end{cases}\n\\]\n\nSince we start at \$(1,1)\$ and we want to end at \$(20,1)\$ we can write the following \$11\$ valid moves:\n\n\\[\n\\begin{matrix} (1,1)&+&(3,8)&=&(4,9)\\\\ (4,9)&+&(8,-3)&=&(12,6)\\\\ (12,6)&+&(8,-3)&=&(20,3)\\\\ (20,3)&+&(-3,8)&=&(17,11)\\\\ (17,11)&+&(-8,-3)&=&(9,8)\\\\ (9,8)&+&(-8,-3)&=&(1,5)\\\\ (1,5)&+&(8,-3)&=&(9,2)\\\\ (9,2)&+&(3,8)&=&(12,10)\\\\ (12,10)&+&(-8,-3)&=&(4,7)\\\\ (4,7)&+&(8,-3)&=&(12,4)\\\\ (12,4)&+&(8,-3)&=&(20,1)\\\\ \\end{matrix}\n\\]\n\nHaving shown that all moves comply with the possible values for \$(\\Delta x_i, \\Delta y_i)\$ and that all \$(x_i, y_i)\$ are inside the rectangular grid, then the task is possible when \$r=73\$.\n\nPart (c):\n\nWhen \$r=97\$ we have \$9^2+4^2=97\$\n\nTherefore the moves can have the following values:\n\n\\[\n(\\Delta x_i, \\Delta y_i)=\\begin{cases} (\\pm 9,\\pm 4) \\\\ (\\pm 9,\\mp 4) \\\\ (\\pm 4,\\pm 9) \\\\ (\\pm 4,\\mp 9) \\end{cases}\n\\]\n\nLet \$S\$ be a set of the squares of the rectangle with the condition \$x_i \\equiv \\left\\lceil \\frac{y_i}{4} \\right\\rceil \\;(mod \\; 2)\$ for \$(x_i,y_i)\$. That is, the reminder of \$x_i\$ when divided by 2 is the same as the reminder of \$\\left\\lceil \\frac{y_i}{4} \\right\\rceil\$ when divided by 2 where \$\\left\\lceil k \\right\\rceil\$ is the smallest integer that is not less than \$k\$. If that condition is true, then \$(x_i,y_i) \\in S\$\n\nNow let's look at a case (c1) where \$(x_i,y_i) \\in S\$ and \$x_i \\equiv 0 \\;(mod \\; 2)\$, thus \$\\left\\lceil \\frac{y_i}{4} \\right\\rceil \\equiv 0 \\;(mod \\; 2)\$:\n\nc1 subcase 1 : \$(x_{i+1},y_{i+1})=(x_i \\pm 9,y_i \\pm 4)\$:\n\n\\[\n(x_i \\pm 9) \\equiv (0 \\pm 9)\\;(mod \\; 2) \\equiv 1 \\;(mod \\; 2)\n\\]\n\nand \$\\left\\lceil \\frac{y_i \\pm 4}{4} \\right\\rceil \\equiv \\left\\lceil \\frac{y_i}{4} \\right\\rceil \\pm 1 \\;(mod \\; 2)\\equiv (0 \\pm 1) \\;(mod \\; 2)\\equiv 1 \\;(mod \\; 2)\$\n\nsince \$x_{i+1} \\equiv \\left\\lceil \\frac{y_{i+1}}{4} \\right\\rceil \\;(mod \\; 2)\$, then \$(x_{i+1},y_{i+1}) \\in S\$ for this subcase.\n\nc1 subcase 2 : \$(x_{i+1},y_{i+1})=(x_i \\pm 4,y_i \\pm 9)\$:\n\nThis move is not allowed because the only values of \$1 \\le y_i \\le 12\$ such that \$\\left\\lceil \\frac{y_i}{4} \\right\\rceil \\equiv 0 \\;(mod \\; 2)\$ are \$5,6,7,8\$. Adding or subtracting 9 from any of these numbers will either make \$y_{i+1}<0\$ or \$y_{i+1}>12\$ and will put the piece outside of the board.\n\nIn summary, for case (c1), \$(x_{i+1},y_{i+1}) \\in S\$\n\nNow let's look at a case (c2) where \$(x_i,y_i) \\in S\$ and \$x_i \\equiv 1 \\;(mod \\; 2)\$, thus \$\\left\\lceil \\frac{y_i}{4} \\right\\rceil \\equiv 1 \\;(mod \\; 2)\$:\n\nc2 subcase 1 : \$(x_{i+1},y_{i+1})=(x_i \\pm 9,y_i \\pm 4)\$:\n\n\\[\n(x_i \\pm 9) \\equiv (1 \\pm 9)\\;(mod \\; 2) \\equiv 0 \\;(mod \\; 2)\n\\]\n\nand \$\\left\\lceil \\frac{y_i \\pm 4}{4} \\right\\rceil \\equiv \\left\\lceil \\frac{y_i}{4} \\right\\rceil \\pm 1 \\;(mod \\; 2)\\equiv (1 \\pm 1) \\;(mod \\; 2)\\equiv 0 \\;(mod \\; 2)\$\n\nsince \$x_{i+1} \\equiv \\left\\lceil \\frac{y_{i+1}}{4} \\right\\rceil \\;(mod \\; 2)\$, then \$(x_{i+1},y_{i+1}) \\in S\$ for this subcase.\n\nc2 subcase 2 : \$(x_{i+1},y_{i+1})=(x_i \\pm 4,y_i \\pm 9)\$:\n\n\\[\n(x_i \\pm 4) \\equiv (1 \\pm 4)\\;(mod \\; 2) \\equiv 1 \\;(mod \\; 2)\n\\]\n\nNotice that the only values of \$1 \\le y_i \\le 12\$ such that \$\\left\\lceil \\frac{y_i}{4} \\right\\rceil \\equiv 0 \\;(mod \\; 2)\$ are \$1,2,3,4,9,10,11,12\$. Adding 9 to the values of 1,2, or 3 or subtracting 9 from the values of 10, 11, or 12, will result in \$\\left\\lceil \\frac{y_{i+1}}{4} \\right\\rceil \\equiv 1 \\;(mod \\; 2)\$. However, subtracting 9 from the values of 1,2, or 3, or adding 9 for the values of 10,11, or 12 will put the piece outside the board and not a valid move. Furthermore, for the values of 4 and 9, adding or subtracting 9 from them will result in the piece being outside of the board and not a valid move. Therefore for all valid moves in this subcase which are 9 added to 1, 2, or 3 or 9 subtracted from 10, 11, or 12, the result will be \$\\left\\lceil \\frac{y_{i+1}}{4} \\right\\rceil \\equiv 1 \\;(mod \\; 2)\$\n\nsince \$x_{i+1} \\equiv \\left\\lceil \\frac{y_{i+1}}{4} \\right\\rceil \\;(mod \\; 2)\$, then \$(x_{i+1},y_{i+1}) \\in S\$ for this subcase.\n\nIn summary, for case (c2), \$(x_{i+1},y_{i+1}) \\in S\$\n\nTherefore if \$(x_{i},y_{i}) \\in S\$ then \$(x_{i+1},y_{i+1}) \\in S\$\n\nNow we look at the starting point: \$(1,1)\$:\n\nSince \$1 \\equiv 1\\;(mod \\; 2)\$ and \$\\left\\lceil \\frac{1}{4} \\right\\rceil \\equiv 1\\;(mod \\; 2)\$, then \$(1,1) \\in S\$ and all subsequent squares after that \$\\in S\$\n\nNow let's look at the end point: \$(20,1)\$:\n\nSince \$20 \\equiv 0\\;(mod \\; 2)\$ and \$\\left\\lceil \\frac{1}{4} \\right\\rceil \\equiv 1\\;(mod \\; 2)\$, then \$20 \\not\\equiv \\left\\lceil \\frac{1}{4} \\right\\rceil \\;(mod \\; 2)\$ and \$(20,1) \\not\\in S\$\n\nSince \$(20,1) \\not\\in S\$, then there are no valid moves starting from \$(1,1)\$ that will end in \$(20,1)\$\n\nTherefore the task cannot be done when \$r=97\$.\n\n~Tomas Diaz. orders@tomasdiaz.com", "Slightly different than the first solution, we identify the board with the cartersian plane with the center of the squares integer coordinates such that \$A=(0,0), B=(19,0),C=(19,11),D=(0,11)\$.\n\nPart (a): If \$2\|r\$ and \$x,y\$ are the change in the \$x\$ coordinate and \$y\$ coordinate respectively for a move, one finds \$x^2+y^2 = r\$ hence both \$x,y\$ are even or both \$x,y\$ are odd. Hence, since we start at \$(0,0)\$, for any sequence of moves of distance \$\\sqrt{r}\$, the current position \$(a,b)\$ will always have the same parity between \$a\$ and \$b\$. On the other hand \$B=(19,0)\$ has a different parity between the first and second coordinate, so it is not possible to reach \$B\$ with \$2\|r\$.\n\nIf \$3\|r\$, defining \$x,y\$ the same one similarly finds \$x^2+y^2 = r\$ hence \$x^2 + y^2 \\equiv 0 \\mod 3\$. Considering the possible values of \$x,y\$ mod \$3\$, the only way this is possible is if both \$x,y\$ are divisible by \$3\$. Hence, since we start at \$(0,0)\$, for any sequence of moves of distance \$\\sqrt{r}\$, the current position \$(a,b)\$ will have both \$(a,b)\$ divisible by \$3\$. On the other hand \$B=(19,0)\$ doesn't have both coordinates divisible by \$3\$. So it is not possible to reach \$B\$ with \$3\|r\$.\n\nPart (b): Note \$8^2+3^2 = 73\$. The following sequence of moves have \$r = 73\$ and show it is possible.\n\n\\[\n\\begin{align} &(3,8),(6,0),(9,8),(1,5),(9,2),(12,10),(15,2), \\\\ &(18,10),(10,7),(18,4),(10,1),(13,9),(16,1),(19,9), \\\\ &(11,6),(3,3),(6,11),(9,3),(12,11),(15,3),(7,0), \\\\ &(10,8),(13,0),(16,8),(19,0). \\end{align}\n\\]\n\nPart (c): Suppose by contradiction a sequence of moves reaches \$(19,0)\$ from \$(0,0)\$. Let \$m_1, m_2\$ be the number of moves where the second coordinate increased/decreased by a distance of \$4\$ respectively, and \$n_1,n_2\$ be the number of moves where the second coordinate increased/decreased by a distance of \$9\$ respectively. Apparently\n\n\\[\n4(m_1-m_2) + 9(n_1-n_2) = 0.\n\\]\n\nLet \$m_1',m_2'\$ be the number of moves where the first coordinate increased/decreased by a distance of \$9\$ respectively, and \$n_1',n_2'\$ be the number of moves where the first coordinate increased/decreased by \$4\$ respectively. Apparently\n\n\\[\n9(m_1'-m_2') + 4(n_1'-n_2') = 19\n\\]\n\nIf \$m_1-m_2 = n_1-n_2 =0\$ then since \$m_1'+m_2' = m_1+m_2 = 2m_1\$ and \$n_1'+n_2' = n_1+n_2 = 2n_1\$ one finds \$m_1'-m_2' = 2m_1'-2m_1\$ and \$n_1'-n_2' = 2n_1'- 2n_1\$. In which case the above equation doesn't have an integer solution. Therefore \$m_1-m_2\$ and \$n_1-n_2\$ are not both zero.\n\nNow, the second coordinate must start at \$0\$ and end at \$0\$. If the sequence of moves simply adds values to the second coordinate and eventually subtracts those same values both \$m_1-m_2\$ and \$n_1-n_2\$ will be zero, contradicting the previous. (Or if it subtracts values and eventually adds those same values back).\n\nAt each step we may add \$4\$, subtract \$4\$, or add \$9\$, or subtract \$9\$ to the second coordinate without leaving the game board. There are only two possible non-repeating sequences:\n\n\\[\n0,4,8\n\\]\n\nand\n\n\\[\n0,9,5,1,10,6,2,11,7,3\n\\]\n\nat which points there are no non-repeating moves possible since a \$y\$ coordinate value of \$12\$ is outside the game board. Thus in order to return to \$0\$ the sequence is forced to subtract the same values added (or add the same values subtracted), contradicting the previous. So it is not possible to find a solution with \$r = 97\$.\n\n~not_detriti" ]
IMO-1996-2	https://artofproblemsolving.com/wiki/index.php/1996_IMO_Problems/Problem_2	Let $P$ be a point inside triangle $ABC$ such that \[ \angle APB-\angle ACB = \angle APC-\angle ABC \] Let $D$, $E$ be the incenters of triangles $APB$, $APC$, respectively. Show that $AP$, $BD$, $CE$ meet at a point.	[ "\\[\n[asy] import graph; size(7cm); 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 = -9.124923887131423, xmax = 11.886638474419073, ymin = -8.067061524000941, ymax = 7.112109861745523; / image dimensions / pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); / draw figures / draw((-4.7910734582427414,-2.2520316878694198)--(-0.5209831486992185,-0.6439100533413341), linewidth(0.8) + wrwrwr); draw((-1.5455209806493002,5.282227966879504)--(-0.5209831486992185,-0.6439100533413341), linewidth(0.8) + wrwrwr); draw((-0.5209831486992185,-0.6439100533413341)--(4.6606059131635655,-2.3266536358088805), linewidth(0.8) + wrwrwr); draw((-4.7910734582427414,-2.2520316878694198)--(-1.5455209806493002,5.282227966879504), linewidth(0.8) + wrwrwr); draw((-1.5455209806493002,5.282227966879504)--(4.6606059131635655,-2.3266536358088805), linewidth(0.8) + wrwrwr); draw((4.6606059131635655,-2.3266536358088805)--(-4.7910734582427414,-2.2520316878694198), linewidth(0.8) + wrwrwr); draw((-2.663393291947277,2.6871874268422657)--(4.6606059131635655,-2.3266536358088805), linewidth(0.8) + linetype(\"4 4\") + wrwrwr); draw((0.4890818665379948,2.7877491336841196)--(-4.7910734582427414,-2.2520316878694198), linewidth(0.8) + linetype(\"4 4\") + wrwrwr); / dots and labels / dot((-1.5455209806493002,5.282227966879504),linewidth(1pt) + dotstyle); label(\"$A$\", (-1.4792575117985467,5.317528023036582), NE labelscalefactor); dot((-4.7910734582427414,-2.2520316878694198),linewidth(1pt) + dotstyle); label(\"$B$\", (-5.368421189216425,-2.7776868388553828), NE * labelscalefactor); dot((4.6606059131635655,-2.3266536358088805),linewidth(1pt) + dotstyle); label(\"$C$\", (4.727004680403201,-2.571154642954807), NE * labelscalefactor); dot((-0.5209831486992185,-0.6439100533413341),linewidth(1pt) + dotstyle); label(\"$P$\", (-0.9754343965435009,-1.319589094904368), NE * labelscalefactor); dot((-2.663393291947277,2.6871874268422657),linewidth(1pt) + dotstyle); label(\"$F$\", (-2.4513226744325554,2.6443488257930543), NE * labelscalefactor); dot((0.4890818665379948,2.7877491336841196),linewidth(1pt) + dotstyle); label(\"$G$\", (0.6331148608484339,2.8125908731720175), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture / [/asy]\n\\]\n\nlet \$CF\$, \$BG\$ be the angle bisectors of \$\\angle ABP\$ and \$\\angle ACP\$, respectively. Notice that they coincide with line \$BD\$ and \$CE\$. Therefore, it suffices to show \$CF,BG,AP\$ are concurrent. Let \$CF \\cap AP = X_{1}\$, and \$BG\\cap AP = X_{2}\$. Notice that by angle bisector theorem, we have\n\n\\[\n\\frac{AX_{1}}{PX_{1}}=\\frac{AC}{PC},\\; \\frac{AX_{2}}{PX_{2}}=\\frac{AB}{PB}\n\\]\n\nTherefore, it suffices to show\n\n\\[\n\\frac{AC}{PC} =\\frac{AB}{PB}\n\\]\n\n\\[\n[asy] import graph; size(10cm); 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 = -6.428011745667007, xmax = 17.921710402847665, ymin = -10.1071238327089, ymax = 7.546424724964209; / image dimensions / pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); / draw figures / draw((-4.7910734582427414,-2.2520316878694198)--(-0.5209831486992185,-0.6439100533413341), linewidth(0.8) + wrwrwr); draw((-1.5455209806493002,5.282227966879504)--(-0.5209831486992185,-0.6439100533413341), linewidth(0.8) + wrwrwr); draw((-0.5209831486992185,-0.6439100533413341)--(4.6606059131635655,-2.3266536358088805), linewidth(0.8) + wrwrwr); draw((-4.7910734582427414,-2.2520316878694198)--(-1.5455209806493002,5.282227966879504), linewidth(0.8) + wrwrwr); draw((-1.5455209806493002,5.282227966879504)--(4.6606059131635655,-2.3266536358088805), linewidth(0.8) + wrwrwr); draw((4.6606059131635655,-2.3266536358088805)--(-4.7910734582427414,-2.2520316878694198), linewidth(0.8) + wrwrwr); draw((-1.5455209806493002,5.282227966879504)--(5.318779339995723,3.0814235558151237), linewidth(0.8) + wrwrwr); draw((4.6606059131635655,-2.3266536358088805)--(5.318779339995723,3.0814235558151237), linewidth(0.8) + wrwrwr); draw((-0.5209831486992185,-0.6439100533413341)--(5.318779339995723,3.0814235558151237), linewidth(0.8) + wrwrwr); / dots and labels / dot((-1.5455209806493002,5.282227966879504),linewidth(2pt) + dotstyle); label(\"$A$\", (-1.4493631992296316,5.328860743581626), NE labelscalefactor); dot((-4.7910734582427414,-2.2520316878694198),linewidth(2pt) + dotstyle); label(\"$B$\", (-5.210486701262846,-2.615204957788533), NE * labelscalefactor); dot((4.6606059131635655,-2.3266536358088805),linewidth(2pt) + dotstyle); label(\"$C$\", (4.7467714546334765,-2.606539778032518), NE * labelscalefactor); dot((-0.5209831486992185,-0.6439100533413341),linewidth(2pt) + dotstyle); label(\"$P$\", (-0.76232203188522677,-1.0411615600781519), NE * labelscalefactor); dot((5.318779339995723,3.0814235558151237),linewidth(2pt) + dotstyle); label(\"$Q$\", (5.398996155040119,3.133037585545931), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture / [/asy]\n\\]\n\nNow, construct \$\\triangle AQC \\sim \\triangle APB\$. Connect \$PQ\$. We notice that \$\\angle BAC = \\angle PAQ\$, \$\\frac{AB}{AP}=\\frac{AC}{AQ}\$. Therefore \$\\triangle ABC \\sim \\triangle APQ\$. Therefore, we have\n\n\\[\n\\begin{align} \\angle PQC &= \\angle AQC -\\angle AQP \\\\ &=\\angle APB-\\angle ACB \\\\ &=\\angle APC-\\angle ABC \\\\ &=\\angle QPC \\end{align*}\n\\]\n\nTherefore,\n\n\\[\nCP=CQ=\\frac{AC}{AB} \\cdot BP \\Longrightarrow \\frac{AC}{PC} =\\frac{AB}{PB} \\quad \\blacksquare\n\\]\n\n~JoeyW" ]
IMO-1996-3	https://artofproblemsolving.com/wiki/index.php/1996_IMO_Problems/Problem_3	Let $S$ denote the set of nonnegative integers. Find all functions $f$ from $S$ to itself such that $f(m+f(n))=f(f(m))+f(n)$ $\forall m,n \in S$	[ "Plugging in m = 0, we get f(f(n)) = f(n) \$\\forall n \\in S\$. With m = n = 0, we get f(0) = 0.\n\nIf there are no fixed points of this function greater than \$0\$, then \$f(x) = 0 \\forall x \\in \\mathbb{N}\$, which is a valid solution.\n\nLet \$n_{0}\$ be the smallest fixed point of \$f(x)\$ such that \$n_{0} > 0\$. \$\\implies f(n_{0}) = n_{0}\$. Plugging \$m = n = n_{0}\$, we get \$f(2n_{0}) = 2f(n_{0})\$.\n\nBy an easy induction, we get \$f(kn_{0}) = kf(n_{0}) \\forall k \\in \\mathbb{N}\$.\n\nLet \$n_{1}\$ be another fixed point greater than \$n_{0}\$. Let \$n_{1} = qn_{0} + r\$, where \$r < n_{0}\$.\n\nSo, \$f(n_{1}) = f(qn_{0} + r) = f(r + f(qn_{0})) = f(r) + qn_{0} = r + qn_{0}\$.\n\n\$\\implies f(r) = r\$. But, \$r < n_{0} \\implies r = 0\$. This means that the set of all fixed points of \$f(x)\$ is \${0, n_{0}, 2n_{0}, 3n{0}, ...}\$.\n\nLet \$x < n_{0}\$ and \$f(x) = b\$. So, \$b = f(x) = f(f(x)) = f(b)\$. So, \$b\$ is also a fixed point, which means that the functional values of non-fixed points are permutations of the fixed points. So let \$f(k) = a_{k}n_{0}\$, where \$a_{k} \\in \\mathbb{N}\$.\n\nNow let \$n = qn_{0} + r\$, where \$r < n\$. So \$f(n) = f(qn_{0} + r) = f(r + f(qn_{0})) = f(r) + f(qn_{0}) = f(r) + qn_{0} = a_{r}n_{0} + qn_{0} = n_{0}\\dot(a_{r} + q)\$. So there are two general solutions, \$f(x) = 0\$ (where the only fixed point is 0) or \$f(x) = n_{0}\\dot(a_{r} + q)\$ where \$n_{0}\$ is the smallest fixed point greater than 0 (in the second case), \$f(r) = a_{r}n_{0}\$ where \$r < n_{0}\$ and q is the quotient when \$n\$ is divided by \$n_{0}\$." ]
IMO-1996-4	https://artofproblemsolving.com/wiki/index.php/1996_IMO_Problems/Problem_4	The positive integers $a$ and $b$ are such that the numbers $15a+16b$ and $16a-15b$ are both squares of positive integers. What is the least possible value that can be taken on by the smaller of these two squares?	[ "Let us first set up two equations given the information in the problem: Let \$15a + 16b = m^2\$ Let \$16a - 15b = n^2\$\n\nIn general, when we are given alternating sums and differences like this, it is a good idea to square both equations and add them up to cancel out any cross terms. Doing this results in the equation \$m^4 + n^4 = 481(a^2 + b^2)\$.\n\nNote that \$481\$ factorizes into \$13 \\cdot 37\$. Thus, \$m^4 + n^4 \\equiv 0\\mod 13\$, and \$m^4 + n^4 \\equiv 0\\mod 37\$. Now, we make a table for all modular congruences possible for m^4 for all nonnegative integers with a unique modulo 13 and 37. Note that due to the modular symmetry of even powers (\$(-a)^2 = a^2\$), we only need to consider up to \$7 \\mod 13\$ and \$18 \\mod 37\$:\n\nNote that the purpose of this table is to find an two \$m^4\$ and \$n^4\$ such that they add up to \$0\$ for both \$\\mod 13\$ and \$\\mod 37\$. Looking at \$\\mod 13\$, this is clearly impossible unless both \$m^4\$ and \$n^4\$ are congruent to \$0 \\mod 13\$. Thus, \$m \\equiv n \\equiv 0 \\mod 13\$. We do the same for \$37\$, and after a bit of calculation, we see that it is also impossible for \$m^4\$ and \$n^4\$ to be anything other than \$0 \\mod 37\$. Thus, \$m\$ and \$n\$ are both divisible by \$13 \\cdot 37\$, or \$481\$. Logically, the first pair we test is \$m = 2\\cdot 13\\cdot 37\$, and \$n = 13\\cdot 37\$. Indeed, this pair works with \$a = 13\\cdot 37\\cdot 76\$ and \$b = 13\\cdot 37\\cdot 49\$. Since we know that \$m > n\$ with \$m \\equiv n \\equiv 0 \\mod 481\$, we know that this is the smallest possible value to work. Thus, the minimum of \${m^2, n^2}\$ is \$481^2\$, or \$230400 + 961 = \\boxed{231361}\$.\n\n~Stead" ]
IMO-1996-5	https://artofproblemsolving.com/wiki/index.php/1996_IMO_Problems/Problem_5	Let $ABCDEF$ be a convex hexagon such that $AB$ is parallel to $DE$, $BD$ is parallel to $EF$, and $CD$ is parallel to $FA$. Let $R_{A}$, $R_{C}$, $R_{E}$ denote the circumradii of triangles $FAB$, $BCD$, $DEF$, respectively, and let $P$ denote the perimeter of the hexagon. Prove that \[ R_{A}+R_{C}+R_{E} \ge \frac{P}{2} \]	[ "Let \$s_{1}=\\left\| AB \\right\|,\\;s_{2}=\\left\| BC \\right\|,\\;s_{3}=\\left\| CD \\right\|,\\;s_{4}=\\left\| DE \\right\|,\\;s_{5}=\\left\| EF \\right\|,\\;s_{6}=\\left\| FA \\right\|\$\n\nLet \$d_{1}=\\left\| FB \\right\|,\\;d_{2}=\\left\| BD \\right\|,\\;d_{1}=\\left\| DF \\right\|\$\n\nLet \$\\alpha_{1}=\\angle FAB,\\;\\alpha_{2}=\\angle ABC,\\;\\alpha_{3}=\\angle BCD,\\;\\alpha_{4}=\\angle CDE,\\;\\alpha_{5}=\\angle DEF,\\;\\alpha_{6}=\\angle EFA\\;\$\n\nFrom the parallel lines on the hexagon we get:\n\n\$\\alpha_{1}=\\alpha_{4},\\;\\alpha_{2}=\\alpha_{5},\\;\\alpha_{3}=\\alpha_{6}\$ [Equations 1]\n\nSo now we look at \$\\Delta FAB\$. We construct a perpendicular from \$A\$ to \$FE\$ and a perpendicular from \$A\$ to \$BC\$.\n\nWe find out the length of these two perpendiculars and add them to get the distance between parallel lines \$FE\$ and \$BC\$ and because of the triangle inequality the distance \$\\left\| FB \\right\|\$ is greater or equal to tha the distance between parallel lines \$FE\$ and \$BC\$:\n\nThis provides the following inequality:\n\n\\[\nd_{1} \\ge s_{6}sin(\\alpha{6})+s_{1}sin(\\alpha{2})\n\\]\n\nUsing the [Equations 1] we simplify to:\n\n\$d_{1} \\ge s_{6}sin(\\alpha{3})+s_{1}sin(\\alpha{2})\$ [Equation 2]\n\nWe now construct a perpendicular from \$D\$ to \$FE\$ and a perpendicular from \$D\$ to \$BC\$. Then we find out the length of these two perpendiculars and add them to get the distance between parallel lines \$FE\$ and \$BC\$ and get:\n\n\\[\nd_{1} \\ge s_{3}sin(\\alpha{3})+s_{4}sin(\\alpha{5})\n\\]\n\nUsing the [Equations 1] we simplify to:\n\n\$d_{1} \\ge s_{3}sin(\\alpha{3})+s_{4}sin(\\alpha{2})\$ [Equation 3]\n\nWe now add [Equation 2] and [Equation 3] to get:\n\n\$2d_{1} \\ge (s_{3}+s_{6})sin(\\alpha_{3})+(s_{1}+s_{4})sin(\\alpha_{2})\$ [Equation 4]\n\nWe now use the Extended law of sines on \$\\Delta FAB\$ with \$R_{A}\$ to get:\n\n\\[\n\\frac{d_{1}}{sin(\\alpha_{1})}=2R_{A}\n\\]\n\n\$d_{1}=2R_{A}sin(\\alpha_{1})\$ [Equation 5]\n\nSubstitute [Equation 5] into [Equation 4]:\n\n\\[\n4R_{A}sin(\\alpha_{1}) \\ge (s_{3}+s_{6})sin(\\alpha_{3})+(s_{1}+s_{4})sin(\\alpha_{2})\n\\]\n\n\$4R_{A} \\ge (s_{3}+s_{6})\\frac{sin(\\alpha_{3})}{sin(\\alpha_{1})}+(s_{1}+s_{4})\\frac{sin(\\alpha_{2})}{sin(\\alpha_{1})}\$ [Equation 6]\n\nTo find the equivalent inequality for \$4R_{C}\$ and \$4R_{E}\$ we just need shift the indexes by two. That is to add \$2\$ to each of the indexes of \$s_{i}\$ and \$\\alpha_{i}\$ and adjust the indexes so that for \$s\$ the indexes are 1 through 6, and for \$\\alpha\$ the indexes are 1 through 3.\n\n\$4R_{C} \\ge (s_{2}+s_{5})\\frac{sin(\\alpha_{2})}{sin(\\alpha_{3})}+(s_{3}+s_{6})\\frac{sin(\\alpha_{1})}{sin(\\alpha_{3})}\$ [Equation 7]\n\n\$4R_{E} \\ge (s_{1}+s_{4})\\frac{sin(\\alpha_{1})}{sin(\\alpha_{2})}+(s_{2}+s_{5})\\frac{sin(\\alpha_{3})}{sin(\\alpha_{2})}\$ [Equation 8]\n\nAdding [Equation 6], [Equation 7], and [Equation 8] we get:\n\n\\[\n4\\left( R_{A}+R_{C}+R_{E} \\right) \\ge (s_{1}+s_{4})\\left( \\frac{sin(\\alpha_{2})}{sin(\\alpha_{1})}+ \\frac{sin(\\alpha_{1})}{sin(\\alpha_{2})}\\right)+(s_{2}+s_{5})\\left( \\frac{sin(\\alpha_{2})}{sin(\\alpha_{3})}+ \\frac{sin(\\alpha_{3})}{sin(\\alpha_{2})}\\right)+(s_{3}+s_{6})\\left( \\frac{sin(\\alpha_{3})}{sin(\\alpha_{1})}+ \\frac{sin(\\alpha_{1})}{sin(\\alpha_{3})}\\right)\n\\]\n\nFrom AM-GM inequality we get:\n\n\$x+\\frac{1}{x} \\ge 2\\;\$ Therefore, \$\\left( \\frac{sin(\\alpha_{a})}{sin(\\alpha_{b})}+ \\frac{sin(\\alpha_{b})}{sin(\\alpha_{a})}\\right) \\ge 2\\;\$ for any index \$a\$ and \$b\$\n\nTherefore,\n\n\\[\n4\\left( R_{A}+R_{C}+R_{E} \\right) \\ge 2(s_{1}+s_{4})+2(s_{2}+s_{5})+2(s_{3}+s_{6})\n\\]\n\n\\[\n4\\left( R_{A}+R_{C}+R_{E} \\right) \\ge 2\\sum_{i=1}^{6}s_{i}\n\\]\n\n\\[\nR_{A}+R_{C}+R_{E} \\ge \\frac{1}{2}\\sum_{i=1}^{6}s_{i}\n\\]\n\nSince \$P=\\sum_{i=1}^{6}s_{i}\$, then \$R_{A}+R_{C}+R_{E} \\ge \\frac{P}{2}\$\n\n~ Tomas Diaz. orders@tomasdiaz.com" ]
IMO-1996-6	https://artofproblemsolving.com/wiki/index.php/1996_IMO_Problems/Problem_6	Let $p, q, n$ be three positive integers with $p+q<n$. Let $(x_0,x_1,\cdots ,x_n)$ be an $(n+1)$-tuple of integers satisfying the following conditions: (i) $x_0=x_n=0$; (ii) For each $i$ with $1 \le i \le n$, either $x_i-x_{i-1}=p$ or $x_i-x_{i-1}=-q$. Show that there exists indices $i<j$ with $(i,j) \ne (0,n)$, such that $x_i=x_j$.	[]
IMO-1997-1	https://artofproblemsolving.com/wiki/index.php/1997_IMO_Problems/Problem_1	In the plane the points with integer coordinates are the vertices of unit squares. The squares are colored alternatively black and white (as on a chessboard). For any pair of positive integers $m$ and $n$, consider a right-angled triangle whose vertices have integer coordinates and whose legs, of lengths $m$ and $n$, lie along edges of the squares. Let $S_{1}$ be the total area of the black part of the triangle and $S_{2}$ be the total area of the white part. Let $f(m,n)=\|S_{1}-S_{2}\|$ (a) Calculate $f(m,n)$ for all positive integers $m$ and $n$ which are either both even or both odd. (b) Prove that $f(m,n) \le \frac{1}{2} max\left\{ m,n \right\}$ for all $m$ and $n$. (c) Show that there is no constant $C$ such that $f(m,n)<C$ for all $m$ and $n$.	[ "For any pair of positive integers \$m\$ and \$n\$, consider a rectangle \$ABCD\$ whose vertices have integer coordinates and whose legs, of lengths \$m\$ and \$n\$, lie along edges of the squares.\n\nLet \$A\$, \$B\$, \$C\$, and \$D\$, be the lower left vertex, lower right vertex, upper right vertex, and upper left vertex of rectangle \$ABCD\$ respectively.\n\nLet \$T_{1}\$ be the total area of the black part of the rectangle and \$T_{2}\$ be the total area of the white part.\n\nLet \$g(m,n)=\|T_{1}-T_{2}\|\$\n\nPart (a)\n\ncase: \$m\$ and \$n\$ which are both even\n\nSince \$m\$ and \$n\$ are both even, the total area of the rectangle \$ABCD\$ is \$m \\times n\$ which is also even\n\nSince every row has an even number of squares there are equally as many white squares than black squares for each row.\n\nSince every column has an even number of squares there are equally as many white squares than black squares for each column.\n\nThis means that in the rectangle there are equal number of white squares and black squares.\n\nTherefore \$T_{1}=T_{2}=\\frac{mn}{2}\$ and \$g(m,n)=\|T_{1}-T_{2}\|=0\$\n\nLet \$M\$ be the midpoint of line \$AC\$. Them \$M\$ is at coordinate \$(A_{x}+\\frac{m}{2},A_{y}+\\frac{n}{2})\$\n\nSince both \$m\$ and \$n\$ are even, then \$M\$ has integer coordinates.\n\nStarting with vertex \$A\$, because the length of \$AB\$ is even, then the color for the square inside rectangle \$ABCD\$ closest to \$B\$ is the opposite color of the square inside rectangle \$ABCD\$ closest to \$A\$, then starting with vertex \$B\$, because the length of \$BC\$ is even, then the color of the square inside rectangle \$ABCD\$ closest to \$C\$ is the opposite color of the square inside rectangle \$ABCD\$ closest to \$B\$. this means that the color of the square inside rectangle \$ABCD\$ closest to \$A\$ is the same as the color of the square inside rectangle \$ABCD\$ closest to \$C\$. Likewise, the color of the square inside rectangle \$ABCD\$ closest to \$B\$ is the same as the color of the square inside rectangle \$ABCD\$ closest to \$D\$.\n\nThis color pattern and the fact that the midpoint \$M\$ has integer coordinates indicates that triangle \$ABC\$ has the same color pattern as triangle \$CDA\$ rotated 180 degrees.\n\nTherefore, the white area in triangle \$ABC\$ is the same as the white area in triangle \$CDA\$ and the black area in triangle \$ABC\$ is the same as the black area in triangle \$CDA\$.\n\nThus \$S_{1}=\\frac{T_{1}}{2}\$ and \$S_{2}=\\frac{T_{2}}{2}\$, which gives \$f(m,n)=\\frac{g(m,n)}{2}=0\$\n\nTherefore \$f(m,n)=0\$ when both \$m\$ and \$n\$ are even.\n\ncase: \$m\$ and \$n\$ which are both odd\n\nSince \$m\$ and \$n\$ are both odd, the total area of the rectangle \$ABCD\$ is \$m \\times n\$ which is also odd\n\nSince the total area is odd, then \$\\frac{mn}{2}\$ is not an integer but \$\\frac{mn+1}{2}\$ and \$\\frac{mn-1}{2}\$ are.\n\nThis means that in the rectangle there are \$\\frac{mn+1}{2}\$ squares of one color and \$\\frac{mn+1}{2}\$ squares of the other color\n\n\\[\ng(m,n)=\|T_{1}-T_{2}\|=\\left\| \\frac{mn+1}{2}-\\frac{mn-1}{2} \\right\|=1\n\\]\n\nLet \$M\$ be the midpoint of line \$AC\$. Them \$M\$ is at coordinate \$(A_{x}+\\frac{m}{2},A_{y}+\\frac{n}{2})\$ Since both \$m\$ and \$n\$ are odd, then \$M\$ has non-integer coordinates coordinates but their decimal portions are both \$0.5\$. This means that \$M\$ lies in the middle of the center square.\n\nStarting with vertex \$A\$, because the length of \$AB\$ is odd, then the color for the square inside rectangle \$ABCD\$ closest to \$B\$ is the same color of the square inside rectangle \$ABCD\$ closest to \$A\$, then starting with vertex \$B\$, because the length of \$BC\$ is odd, then the color of the square inside rectangle \$ABCD\$ closest to \$C\$ is the same color of the square inside rectangle \$ABCD\$ closest to \$B\$. this means that the color of the square inside rectangle \$ABCD\$ closest to \$A\$ is the same as the color of the square inside rectangle \$ABCD\$ closest to \$C\$. Likewise, the color of the square inside rectangle \$ABCD\$ closest to \$B\$ is the same as the color of the square inside rectangle \$ABCD\$ closest to \$D\$.\n\nThis color pattern and the fact that the midpoint \$M\$ in in the center of the middle square that triangle \$ABC\$ has the same color pattern as triangle \$CDA\$ rotated 180 degrees.\n\nTherefore, the white area in triangle \$ABC\$ is the same as the white area in triangle \$CDA\$ and the black area in triangle \$ABC\$ is the same as the black area in triangle \$CDA\$.\n\nThus \$f(m,n)=\\frac{g(m,n)}{2}=\\frac{1}{2}\$\n\nTherefore \$f(m,n)=\\frac{1}{2}\$ when both \$m\$ and \$n\$ are odd.\n\nTo summarize part (a),\n\n\\[\nf(m,n)=\\begin{cases} 0 & m,n\\;are\\;both\\;even \\\\ \\frac{1}{2} & m,n\\;are\\;both\\;odd\\\\ something \\; else & othwerwise\\end{cases}\n\\]\n\nPart (b)\n\nSince \$f(m,n)=\\begin{cases} 0 & m,n\\;are\\;both\\;even \\\\ \\frac{1}{2} & m,n\\;are\\;both\\;odd\\\\ something \\; else & othwerwise\\end{cases}\$ then for these cases the minimum values that \$m\$ and \$n\$ can have are 1. So for the cases where both are odd or both are even \$f(m,n) \\le \\frac{1}{2} max\\left\\{ m,n \\right\\}\$ with equality at \$m=n=1\$\n\nNow we need to find the case were one of them is odd and the other one is even.\n\ncase \$max\\left\\{ m,n \\right\\}\$ is odd and \$min\\left\\{ m,n \\right\\}\$ is even, means that \$max\\left\\{ m,n \\right\\}-1\$ is even\n\nTherefore \$f\\left( max\\left\\{ m,n \\right\\}-1,min\\left\\{ m,n \\right\\} \\right)=0\$\n\n\$f(m,n)=f\\left( max\\left\\{ m,n \\right\\}-1,min\\left\\{ m,n \\right\\} \\right)+h(1,min\\left\\{ m,n \\right\\})\$,\n\nwhere \$h(1,n)\$ is the absolute value of the difference between the white area and black area of a triangle with base or size 1 and height n where this triangle is not necessarily a rectangular triangle.\n\nTherefore \$h(1,n) \\le\$ area of such triangle. Thus, \$h(1,min\\left\\{ m,n \\right\\}) \\le \\frac{1}{2}min\\left\\{ m,n \\right\\}\$\n\n\\[\nf\\left( max\\left\\{ m,n \\right\\}-1,min\\left\\{ m,n \\right\\} \\right)+h(1,min\\left\\{ m,n \\right\\}) \\le 0 +\\frac{1}{2}min\\left\\{ m,n \\right\\}\n\\]\n\n\$f(m,n) \\le \\frac{1}{2}min\\left\\{ m,n \\right\\}\$ when \$max\\left\\{ m,n \\right\\}\$ is odd and \$min\\left\\{ m,n \\right\\}\$ is even\n\nWhich also means that\n\n\$f(m,n) \\le \\frac{1}{2}max\\left\\{ m,n \\right\\}\$ when \$max\\left\\{ m,n \\right\\}\$ is odd and \$max\\left\\{ m,n \\right\\}\$ is even\n\nNow we look at the case where \$max\\left\\{ m,n \\right\\}\$ is even and \$min\\left\\{ m,n \\right\\}\$ is odd, means that \$min\\left\\{ m,n \\right\\}-1\$ is even\n\nTherefore \$f\\left( min\\left\\{ m,n \\right\\}-1,max\\left\\{ m,n \\right\\} \\right)=0\$\n\n\$f(m,n)=f\\left( min\\left\\{ m,n \\right\\}-1,max\\left\\{ m,n \\right\\} \\right)+h(1,max\\left\\{ m,n \\right\\})\$,\n\nThus, \$h(1,max\\left\\{ m,n \\right\\}) \\le \\frac{1}{2}max\\left\\{ m,n \\right\\}\$\n\n\\[\nf\\left( min\\left\\{ m,n \\right\\}-1,max\\left\\{ m,n \\right\\} \\right)+h(1,max\\left\\{ m,n \\right\\}) \\le 0 +\\frac{1}{2}max\\left\\{ m,n \\right\\}\n\\]\n\n\$f(m,n) \\le \\frac{1}{2}max\\left\\{ m,n \\right\\}\$ when \$max\\left\\{ m,n \\right\\}\$ is even and \$min\\left\\{ m,n \\right\\}\$ is odd\n\nTherefore, for all four cases: \$m\$ and \$n\$ are both odd; \$m\$ and \$n\$ are both even; \$max\\left\\{ m,n \\right\\}\$ is odd with \$min\\left\\{ m,n \\right\\}\$ is even; and \$max\\left\\{ m,n \\right\\}\$ is even with \$min\\left\\{ m,n \\right\\}\$ is odd we have:\n\n\$f(m,n) \\le \\frac{1}{2}max\\left\\{ m,n \\right\\}\$ with equality at \$m=n=1\$\n\nPart (c)\n\nConsider the case where \$m=k\$ and \$n=k+1\$ and k is even.\n\n\\[\nf(m,n)=f(k,k+1)=f(k,k)+h(k,1)=h(k,1)\n\\]\n\nWe're going to calculate \$h(k,1)\$ noticing that the region for \$h(k,1)\$ in \$f(m,n)\$ will be the triangle formed by lines \$y=kx\$, \$y=\\frac{k+1}{k}x\$ and \$x=k\$.\n\nSo, to calculate \$h(k,1)\$ we first calculate the total area of the black triangular regions within the triangle (assuming the first corner is white) with the following series formula:\n\n\\[\nS_{1}=\\sum_{j=1}^{k}\\frac{1}{2}\\left( \\frac{k+1}{k}j-j \\right)\\left(j- \\frac{k}{k+1}j \\right)=\\sum_{j=1}^{k}\\frac{1}{2}\\left( \\frac{k+1-k}{k} \\right)\\left( \\frac{k+1-k}{k+1} \\right)j^{2}\n\\]\n\n\\[\nS_{1}=\\frac{1}{2k(k+1)}\\sum_{j=1}^{k}j^{2}=\\frac{1}{2k(k+1)}\\frac{k(k+1)(2k+1)}{6}=\\frac{2k+1}{12}\n\\]\n\nThen, \$S_{2}=\\frac{k}{2}-S_{1}=\\frac{k}{2}-\\frac{2k+1}{12}=\\frac{4k-1}{12}\$\n\n\\[\nh(k,1)=\|S_{2}-S_{1}\|=\\left\| \\frac{4k-1}{12}-\\frac{2k+1}{12} \\right\|=\\frac{k-1}{6}\n\\]\n\nThus for even \$k\$ we have \$f(k,k+1)=h(k,1)=\\frac{k-1}{6}\$\n\nSince for this case, \$C=\\lim_{k \\to \\infty} f(k,k+1)=\\lim_{k \\to \\infty} \\frac{k-1}{6}=\\infty\$, then there is no upper bound for this case. Since part (c) describes an upper bound for all \$m\$ and \$n\$, then having found one case where there is no upper bound means that there is no upper bound when considering all \$m\$ and \$n\$\n\nTherefore, there is no constant \$C\$ such that \$f(m,n)<C\$ for all \$m\$ and \$n\$ because \$C \\to \\infty\\;\$ on some cases.\n\n~ Tomas Diaz. orders@tomasdiaz.com" ]
IMO-1997-2	https://artofproblemsolving.com/wiki/index.php/1997_IMO_Problems/Problem_2	The angle at $A$ is the smallest angle of triangle $ABC$. The points $B$ and $C$ divide the circumcircle of the triangle into two arcs. Let $U$ be an interior point of the arc between $B$ and $C$ which does not contain $A$. The perpendicular bisectors of $AB$ and $AC$ meet the line $AU$ and $V$ and $W$, respectively. The lines $BV$ and $CW$ meet at $T$. Show that. \[ AU=TB+TC \]	[]
IMO-1997-3	https://artofproblemsolving.com/wiki/index.php/1997_IMO_Problems/Problem_3	Let $x_{1}$, $x_{2}$,...,$x_{n}$ be real numbers satisfying the conditions \[ \|x_{1}+x_{2}+...+x_{n}\|=1 \] and $\|x_{i}\| \le \frac{n+1}{2}$, for $i=1,2,...,n$ Show that there exists a permutation $y_{1}$, $y_{2}$,...,$y_{n}$ of $x_{1}$, $x_{2}$,...,$x_{n}$ such that \[ \|y_{1}+2y_{2}+...+ny_{n}\|\le \frac{n+1}{2} \]	[]
IMO-1997-4	https://artofproblemsolving.com/wiki/index.php/1997_IMO_Problems/Problem_4	An $n \times n$ matrix whose entries come from the set $S={1,2,...,2n-1}$ is called a $\textit{silver}$ matrix if, for each $i=1,2,...,n$, the $i$th row and the $i$th column together contain all elements of $S$. Show that (a) there is no $\textit{silver}$ matrix for $n=1997$; (b) $\textit{silver}$ matrices exist for infinitely many values of $n$.	[]
IMO-1997-5	https://artofproblemsolving.com/wiki/index.php/1997_IMO_Problems/Problem_5	Find all pairs $(a,b)$ of integers $a,b \ge 1$ that satisfy the equation \[ a^{b^{2}}=b^{a} \]	[ "Case 1: \$(1 \\le a \\le b)\$\n\n\\[\n(a^{b})^{b}=b^{a}\n\\]\n\nLooking at this expression since \$b \\ge a\$ then \$a^{b} \\le b\$.\n\nHere we look at subcase \$a>1\$ which gives \$a^{b}>b\$ for all \$(1 < a \\le b)\$. This contradicts condition \$a^{b} \\le b\$, and thus \$a\$ can't be more than one giving the solution of \$a=1\$ with \$b \\ge 1\$. So we substitute the value of \$a=1\$ into the original equation to get \$1^{b^2}=b^{1}\$ which solves to \$b=1\$ and our first pair \$(a,b)=(1,1)\$\n\nCase 2: \$(1 \\le b < a)\$\n\n\\[\na^{b^2}=b^{a}\n\\]\n\nsince \$a>b\$, then \$b^{2}<a\$ and we multiply both sides of the equation by \$b^{-2b^{2}}\$ to get:\n\n\\[\nb^{-2b^{2}}a^{b^2}=b^{-2b^{2}}b^{a}\n\\]\n\n\\[\n(ab^{-2})^{b^{2}}=b^{a-2b^{2}}\n\\]\n\nSince \$b^{2}<a\$, then \$(ab^{-2})^{b^{2}}>1\$ and \$b^{a-2b^{2}}>0\$. This gives \$a-2b^{2}>1\$\n\nThis implies that \$a>2b^{2}\$ for \$b>1\$\n\nLet \$a=kb^{2}\$ with \$k \\in \\mathbb{Z} ^{+}\$. Since \$a>2b^{2}\$, then \$k \\ge 3\$\n\n\\[\n(kb^{2}b^{-2})^{b^{2}}=b^{kb^{2}-2b^{2}}\n\\]\n\n\\[\nk^{b^{2}}=b^{(k-2)b^{2}}\n\\]\n\n\$k=b^{k-2}\$, which gives \$b \\ge 2\$\n\nsubcase \$k=3\$:\n\n\$3=b^{3-2}=b\$ and \$a=kb^{2}=(3)(3)^{2}=27\$. which provides 2nd pair \$(a,b)=(27,3)\$\n\nsubcase \$k=4\$:\n\n\$4=b^{4-2}=b^{2}\$, thus \$b=2\$ and \$a=kb^{2}=(4)(2)^{2}=16\$. which provides 3rd pair \$(a,b)=(16,2)\$\n\nsubcase \$k \\ge 5\$:\n\n\$k=b^{k-2}\$, thus \$b=k^{1/(k-2)}\$ which decreases with \$k\$ and \$b \\to 1\$ as \$k \\to \\infty\$ . From subcase \$k=4\$, we know that \$b=2\$, thus for subcase \$k \\ge 5\$, \$1<b<2\$. Therefore this subcase has no solution because it contradicts the condition for Case 2 of \$b \\ge 2\$.\n\nFinal solution for all pairs is \$(a,b)=(1,1); (27,3); (16,2)\$\n\n~ Tomas Diaz. orders@tomasdiaz.com" ]
IMO-1997-6	https://artofproblemsolving.com/wiki/index.php/1997_IMO_Problems/Problem_6	For each positive integer $n$, let $f(n)$ denote the number of ways of representing $n$ as a sum of powers of $2$ with nonnegative integer exponents. Representations which differ only in the ordering of their summands are considered to be the same. For instance, $f(4)=4$, because the number 4 can be represented in the following four ways: \[ 4;2+2;2+1+1;1+1+1+1 \] Prove that, for any integer $n \ge 3$, $2^{n^{2}/4}<f(2^{n})<2^{n^{2}/2}$.	[]
IMO-1998-1	https://artofproblemsolving.com/wiki/index.php/1998_IMO_Problems/Problem_1	In the convex quadrilateral $ABCD$, the diagonals $AC$ and $BD$ are perpendicular and the opposite sides $AB$ and $DC$ are not parallel. Suppose that the point $P$, where the perpendicular bisectors of $AB$ and $DC$ meet, is inside $ABCD$. Prove that $ABCD$ is a cyclic quadrilateral if and only if the triangles $ABP$ and $CDP$ have equal areas.	[ "First, let's prove that if ABCD is cyclic, then the triangles ABP and CDP have equal areas. The circumcenter of ABCD is the intersection of perpendicular bisectors of the sides of the quadrilateral ABCD, so P is the circumcenter. From this, we derive that PC=PB=PA=PD. Since the diagonals AC and BD are perpendicular, the sum of the angles APB and CPD is equal to 180 degrees. Denote the angle APB by a. Then the angle CPD is equal to 180 - a. The area of the triangle APB is equal to APBPsin(angle APB). Let R denote the radius of the circle. Then the area of the triangle APB is equal to RRsin(a), and the area of CPD equals to RRsin(180 - a) = RRsin(a), so the triangles ABP and CDP have equal areas. Now, let's prove it in the other direction. We know that ABP and CDP have equal areas. Let O denote the intersection of the diagonals AC and BD. Let M be the midpoint of AB and N be the midpoint of CD. Draw a segment between M and O, and draw another segment between O and N. We know that ON = CN = ND and OM = MB = AM. Since S(ABP) = S(CPD) (where S denotes the area of the triangles), we have ABPM = CDPN. From this, we can conclude that PM/PN = CD/AB = ON/OM. Let the intersection of the lines AB and CD be the point E. Since PM is perpendicular to AB and PN is perpendicular to CD, PMEN is a cyclic quadrilateral, so the angle MPN is equal to 180-(angle AED) = BAD + CDA (where BAD and CDA are angles). The angle MON is equal to BAC + CDB + 90 (since the triangles AMO and OND are isosceles). In triangle AOD, OAD+ODA = 90 (where OAD and ODA are angles), so the angle MON is equal to BAC + CDB + CAD + BDA = BAD + CDA = MPN. Now take a look at triangles MPN and NOM. We know that MP/PN = ON/OM and the angle MPN is equal to MON. This implies that the triangle MPN is similar to NOM. Since the two triangles have a common side, MN, we conclude that the triangles MPN and NOM are equal, so MP = ON = CN = CD/2, and PN = OM = MB = AB/2. From this, we can derive that the triangles MBP and NPD are equal, so the angle MPB = angle PDN. Denote the angle by x. In the same way, we can derive that the triangles MBP, MAP, NPC, and NPD are equal, so the sum of the angles APB and CPD is 180 degrees, and the angle APB is equal to 2x. So the area of ABP is equal to APAPsin(2x) and the area of CPD is equal to CPCPsin(180 - 2x) = CPCPsin(2x). Since the areas are equal, we conclude that CP = PD = AP = BP, so P is a circumcenter of ABCD, and the quadrilateral ABCD is cyclic, as desired." ]
IMO-1998-2	https://artofproblemsolving.com/wiki/index.php/1998_IMO_Problems/Problem_2	In a competition, there are $a$ contestants and $b$ judges, where $b\ge3$ is an odd integer. Each judge rates each contestant as either “pass” or “fail”. Suppose $k$ is a number such that, for any two judges, their ratings coincide for at most $k$ contestants. Prove that $\frac{k}{a}\ge\frac{b-1}{2b}$.	[ "Let \$c_i\$ stand for the number of judges who pass the \$i\$th candidate. The number of pairs of judges who agree on the \$i\$th contestant is then given by\n\n\\begin{align} {c_i \\choose 2} + {{b - c_i} \\choose 2} &= \\frac{1}{2}\\left(c_i(c_i - 1) + (b - c_i)(b - c_i - 1) \\right) \\\\ &= \\frac{1}{2}\\left( c_i^2 + (b - c_i)^2 - b \\right) \\\\ &\\geq \\frac{1}{2}\\left( \\frac{b^2}{2} - b \\right) \\\\ &= \\frac{b^2 - 2b}{4} \\end{align}\n\nwhere the inequality follows from AM-QM. Since \$b\$ is odd, \$b^2 - 2b\$ is not divisible by \$4\$ and we can strengthen the inequality to\n\n\\[\n{c_i \\choose 2} + {{b - c_i} \\choose 2} \\geq \\frac{b^2 - 2b + 1}{4} = \\left(\\frac{b - 1}{2}\\right)^2.\n\\]\n\nLetting \$N\$ stand for the number of instances where two judges agreed on a candidate, it follows that\n\n\\[\nN = \\sum_{i = 1}^a {c_i \\choose 2} + {{b - c_i} \\choose 2} \\geq a \\cdot \\left( \\frac{b - 1}{2} \\right)^2.\n\\]\n\nThe given condition on \$k\$ implies that\n\n\\[\nN \\leq k \\cdot {b \\choose 2} = \\frac{kb(b - 1)}{2}.\n\\]\n\nTherefore\n\n\\[\na \\cdot \\left( \\frac{b - 1}{2} \\right)^2 \\leq \\frac{kb(b - 1)}{2},\n\\]\n\nwhich simplifies to\n\n\\[\n\\frac{k}{a} \\geq \\frac{b - 1}{2b}.\n\\]" ]
IMO-1998-3	https://artofproblemsolving.com/wiki/index.php/1998_IMO_Problems/Problem_3	For any positive integer $n$, let $d(n)$ denote the number of positive divisors of $n$ (including 1 and $n$ itself). Determine all positive integers $k$ such that $d(n^2)/d(n) = k$ for some $n$.	[ "First we must \$d\$etermine gener\$a\$l values for \$d(n)\$: Let \$n=p1 ^ a1 * p2 ^ a2 * .. * pc ^ ac\$, if \$d\$ is an ar\$b\$itr\$a\$ry divisor of \$n\$ then \$d\$ must have the same prime factors of \$n\$, each with an exponent \$b_i\$ being: \$0\\leq b_i\\leq a_i\$. Hence there are \$Ai + 1\$ choices for each exponent of Pi in the number d => there are \$(a_1 + 1)(a_2 + 1)..(a_c + 1)\$ such d\n\n\$\\implies d(n) = (a_1 + 1)(a_2+1)..(a_c+1)\$ where \$a_i\$ are exponents of the prime numbers in the prime factorisation of \$n\$.\n\n\\[\n\\implies d(n^2)/d(n) = {(2a_1 + 1)(2a_2 + 1)..(2a_c + 1)}/{(a_1+1)..(a_c+1)}\n\\]\n\nSo we want to find all integers \$k\$ that can \$b\$e represented by the product of fractions of the form \$(2n+1)/(n+1)\$ Obviously \$k\$ is odd as the numerator is always odd. It's possible for 1 (1/1) and 3 \$(5/3 * 9/5)\$, which suggests that it may be possible for all odd integers, which we can show by induction.\n\n\$P(k)\$: It's possible to represent \$k\$ as the product of fractions \$(2n+1)/(n+1)\$\n\nBase case: \$k = 1: (2(0) + 1) / (0 + 1)\$ Now assume that for \$k\\geq 3\$ it's possible for all odds < \$k\$.\n\nSince \$k\$ is odd, \$k+1 = 2^zy\$ where \$y\$ is odd and \$y\$ < \$k\$\n\nLet there be a number \$x\$ s.t \$k-y=x\\implies k+1 = 2^z(k-x)\\implies (2^zx+1)/(2^z-1)=k\$\n\nAlso consider \$k/y\$. It is sufficient to show that \$k/y\$ can be represented by a product of fractions of the form \$2n+1/n+1\$ in order to show \$k\$ can be represented by product of fractions \$2n+1/n+1\$, since \$y\$ can be represented in such a manner too.\n\n\\[\nk/y = k/(k-x) = 1/(1 - x/k)\n\\]\n\nUsing our definition of \$k\$ in terms of \$x\$:\n\n\\[\nk/y = 1/({1 - {2^z-x}/{2^zx+1}}) = {2^zx+1}/{x+1}\n\\]\n\nAnd that is simply the product of fractions: \${2x+1}/{x+1} * {4x+1}/{2x+1} * .. * {2^zx+1}/{2^{z-1}x}\$.\n\nWe have shown that \$k/y\$ can be written s.t it's a product of fractions of the form \${2n+1}/{n+1}\\implies k\$ can be written in such a way too.\n\nHence we have shown that all odds less than \$k\$ satisfies \$P(n)\\implies P(k)\$ is true. Since we have shown P(1) is true, it must hence be true for all odd integers.\n\nTherefore, \$d(n^2)/d(n) = k\\iff k\$ is odd, for some n. I.E odd \$k\$ satisfy the condition posed in the question.∎\n\n-dabab_kebab (wrote this solution)" ]
IMO-1998-4	https://artofproblemsolving.com/wiki/index.php/1998_IMO_Problems/Problem_4	Determine all pairs $(a, b)$ of positive integers such that $ab^{2} + b + 7$ divides $a^{2}b + a + b$.	[ "We use the division algorithm to obtain \$ab^2+b+7 \\mid 7a-b^2\$ Here \$7a-b^2=0\$ is a solution of the original statement, possible when \$a=7k^2\$ and \$b=7k\$ where \$k\$ is any natural number. This is easily verified.\n\nOtherwise we obtain the inequality (by basic properties of divisiblity): \$7a-b^2 \\geq ab^2+b+7 \\implies 7a+7-ab^2-b^2 \\geq b+14 \\implies (7-b^2)(a+1) \\geq b+14\$ So \$7-b^2 \\geq 0 \\implies b=1,2\$\n\nTesting for \$b=1\$ we find that \$a+8 \\mid 7a-1 \\implies a+8 \\mid 57\$ Therefore, \$a=11, 49\$, and we can easily check these.\n\nTesting for \$b=2\$ and applying the division algorithm we find that \$4a+9 \\mid 79\$, having no solutions in natural \$a\$.\n\nHence, the only solutions are: \$(a,b) = (11, 1), (49,1), (7k^2, 7k)\$ for all natural \$k\$.\n\nWritten by dabab_kebab" ]
IMO-1998-5	https://artofproblemsolving.com/wiki/index.php/1998_IMO_Problems/Problem_5	Let $I$ be the incenter of triangle $ABC$. Let the incircle of $ABC$ touch the sides $BC$, $CA$, and $AB$ at $K$, $L$, and $M$, respectively. The line through $B$ parallel to $MK$ meets the lines $LM$ and $LK$ at $R$ and $S$, respectively. Prove that angle $RIS$ is acute.	[ "Denote the length of the side MB with x. Denote the angle BKM by a and angle BAC by 2b. Using angle chasing and trigonometry, we can derive: RB = xcos(b)/cos(a-b); BS = xcos(a-b)/cos(b); RM = xsin(a)/cos(a-b); SK = xsin(a)/cos(b). Let's prove that RI^2 + IS^2 > RS^2 because this will imply that angle RIS is acute. Let's calculate RS^2. It is equal to (RB+BS)^2 = RB^2+BS^2 + 2(x^2). RI^2 - RB^2 = IB^2 (since the angle IBK is equal to 90-a, and angle KBS is equal to a, thus angle IBS is equal to 90 degrees). IS^2-BS^2 = IB^2. So we have to prove that 2(IB^2)>2(x^2), and this is true because triangle IMB is a right triangle with IB hypotenuse, so IB > x, as desired." ]
IMO-1998-6	https://artofproblemsolving.com/wiki/index.php/1998_IMO_Problems/Problem_6	Determine the least possible value of $f(1998),$ where $f:\Bbb{N}\to \Bbb{N}$ is a function such that for all $m,n\in {\Bbb N}$, \[ f\left( n^{2}f(m)\right) =m\left( f(n)\right) ^{2}. \]	[]
IMO-1999-1	https://artofproblemsolving.com/wiki/index.php/1999_IMO_Problems/Problem_1	Determine all finite sets $S$ of at least three points in the plane which satisfy the following condition: For any two distinct points $A$ and $B$ in $S$, the perpendicular bisector of the line segment $AB$ is an axis of symmetry of $S$.	[ "Upon reading this problem and drawing some points, one quickly realizes that the set \$S\$ consists of all the vertices of any regular polygon.\n\nNow to prove it with some numbers:\n\nLet \$S=\\left\\{ P_{0},P_{1},P_{2},...,P_{n-1} \\right\\}\$, with \$n\\ge 3\$, where \$P_{i}\$ is a vertex of a polygon which we can define their \$xy\$ coordinates as: \$P_{i}=\\left\\langle Rcos\\left( \\frac{2\\pi}{n}i \\right),Rsin\\left( \\frac{2\\pi}{n}i \\right) \\right\\rangle\$ for \$i=0,1,2,...,(n-1)\$.\n\nThat defines the vertices of any regular polygon with \$R\$ being the radius of the circumcircle of the regular \$n\$-sided polygon.\n\nNow we can pick any points \$A\$ and \$B\$ of the set as:\n\n\$A=P_{a}\$ and \$B=P_{b}\$, where \$a=0,1,2,...,(n-1)\$; \$b=0,1,2,...,(n-1)\$; and \$a\\ne b\$\n\nThen,\n\n\\[\nA=\\left\\langle Rcos\\left( \\frac{2\\pi}{n}a \\right),Rsin\\left( \\frac{2\\pi}{n}a \\right) \\right\\rangle\n\\]\n\nand \$B=\\left\\langle Rcos\\left( \\frac{2\\pi}{n}b \\right),Rsin\\left( \\frac{2\\pi}{n}b \\right) \\right\\rangle\$\n\nLet \$O\$ be point \$(0,0)\$ which is not part of \$S\$\n\nThen, \$\\angle P_{0}OA = \\frac{2\\pi}{n}a\$, and \$\\angle P_{0}OB = \\frac{2\\pi}{n}b\$\n\nThe perpendicular bisector of \$AB\$ passes through \$O\$.\n\nLet point \$M_{AB}\$, not in \$S\$ be a point that passes through the perpendicular bisector of \$AB\$ at a distance \$R\$ from \$O\$\n\nThen, \$\\angle P_{0}OM_{AB} =\\frac{2\\pi}{n}\\frac{a+b}{2}\$ and \$M_{AB}=\\left\\langle Rcos\\left( \\frac{2\\pi}{n}\\frac{a+b}{2} \\right),Rsin\\left( \\frac{2\\pi}{n}\\frac{a+b}{2} \\right) \\right\\rangle\$\n\nCASE I: \$a+b\$ is even\n\n\$k=\\frac{a+b}{2}\$ and \$k\$ is integer\n\nThen \$M_{AB}=\\left\\langle Rcos\\left( \\frac{2\\pi}{n}k \\right),Rsin\\left( \\frac{2\\pi}{n}k \\right) \\right\\rangle=P_{k}\$\n\nThis means that the perpendicular bisector also passes through a point \$P_{k}\$ of \$S\$\n\nLet \$c\$ be any positive integer\n\n\\[\n\\angle P_{k}OP_{(k+c)\\; mod\\; n}=\\frac{2\\pi}{n}\\left( (k+c-k)\\; mod\\; n \\right)=\\frac{2\\pi}{n}\\left( c\\; mod\\; n \\right)\n\\]\n\nand\n\n\\[\n\\angle P_{k}OP_{(k-c)\\; mod\\; n}=\\frac{2\\pi}{n}\\left( (k-(k-c))\\; mod\\; n \\right)=\\frac{2\\pi}{n}\\left( c\\; mod\\; n \\right)\n\\]\n\nTherefore, \$\\angle P_{k}OP_{(k+c)\\; mod\\; n}=\\angle P_{k}OP_{(k-c)\\; mod\\; n}\$ for any integer \$c\$.\n\nAlso, since \$\\left\| OP_{(k+c)\\; mod\\; n} \\right\|=\\left\| OP_{(k-c)\\; mod\\; n} \\right\|=R\$ for any integer \$c\$\n\nthen this proves that the bisector of any points \$A\$ and \$B\$ is an axis of symmetry for this case.\n\nCASE II: \$a+b\$ is odd\n\n\$k=\\frac{a+b+1}{2}\$ and \$k\$ is integer\n\n\$m=\\frac{a+b-1}{2}\$ and \$m\$ is integer\n\nThen \$M_{AB}=\\left\\langle Rcos\\left( \\frac{2\\pi}{n}\\frac{a+b}{2} \\right),Rsin\\left( \\frac{2\\pi}{n}\\frac{a+b}{2} \\right) \\right\\rangle\$\n\nThis means that the perpendicular bisector does not pass through any point of \$S\$, but their closest points are \$P_{k}\$ and \$P_{m}\$ and that \$\\angle MOP_{k}=\\angle MOP_{m}=\\frac{\\pi}{n}\$\n\nLet \$c\$ be any positive integer\n\n\\[\n\\angle P_{k}OP_{(k+c)\\; mod\\; n}=\\frac{2\\pi}{n}\\left( (k+c-k)\\; mod\\; n \\right)=\\frac{2\\pi}{n}\\left( c\\; mod\\; n \\right)\n\\]\n\n\\[\n\\angle MOP_{(k+c)\\; mod\\; n}=\\angle MOP_{k}+\\angle P_{k}OP_{(k+c)\\; mod\\; n}=\\frac{\\pi}{n}+\\frac{2\\pi}{n}\\left( c\\; mod\\; n \\right)\n\\]\n\nand\n\n\\[\n\\angle P_{m}OP_{(m-c)\\; mod\\; n}=\\frac{2\\pi}{n}\\left( (m-(m-c))\\; mod\\; n \\right)=\\frac{2\\pi}{n}\\left( c\\; mod\\; n \\right)\n\\]\n\n\\[\n\\angle MOP_{(m-c)\\; mod\\; n}=\\angle MOP_{m}+\\angle P_{m}OP_{(m-c)\\; mod\\; n}=\\frac{\\pi}{n}+\\frac{2\\pi}{n}\\left( c\\; mod\\; n \\right)\n\\]\n\nTherefore, \$\\angle MOP_{(k+c)\\; mod\\; n}=\\angle MOP_{(m-c)\\; mod\\; n}\$ for any integer \$c\$.\n\nSince \$m=k-1\$, \$\\angle MOP_{(k+c)\\; mod\\; n}=\\angle MOP_{(k-1-c)\\; mod\\; n}\$\n\nAlso, since \$\\left\| OP_{(k+c)\\; mod\\; n} \\right\|=\\left\| OP_{(m-c)\\; mod\\; n} \\right\|=R\$ for any integer \$c\$\n\nthen this proves that the bisector of any points \$A\$ and \$B\$ is an axis of symmetry for this case.\n\nHaving proven both cases, then the set \$S\$ of points that comply with the given condition is the set of the vertices of any regular polygon of any number of sides.\n\n~Tomas Diaz. orders@tomasdiaz.com", "First we prove no \$3\$ points can lie on a line. Say \$A_1, A_2, A_3\$ were sequential points on a line. Considering the axis of symmetry between \$A_2\$ and \$A_3\$ one finds there lies a point \$A_4\$ on the right side of \$A_3\$. Then considering the axis of symmetry between \$A_3\$ and \$A_4\$ one finds sequential points \$A_5\$ and \$A_6\$ that lie on the right side of \$A_4\$. One can continue this process ad infinitum to show \$S\$ must have infinite points. A contradiction.\n\nLet a line roll/rotate around the perimeter of \$S\$ such that at any time all points of \$S\$ are on one side of the line and the line is always touching one point of \$S\$, but may touch two points of \$S\$ at a time. Say the line sequentially touches the points \$S_1,S_2,...,S_n,S_1\$. We will now prove \$S_1,...,S_n\$ form a regular polygon.\n\nIdentify \$S_0\$ with \$S_n\$, \$S_{n+1}\$ with \$S_1\$, and \$S_{n+2}\$ with \$S_2\$.\n\nConsidering the axis of symmetry between \$S_2\$ and \$S_3\$, a line which rolls counterclockwise around \$S\$ with starting position passing through \$S_2\$ and \$S_3\$ will roll symmetrically to a line which rolls clockwise around \$S\$ with starting position passing through \$S_2\$ and \$S_3\$. Let \$l\$ be the perpendicular bisector of \$S_2S_3\$. There are two possibilities: \$S_1\$ lies on the same side of \$l\$ as \$S_2\$ or \$S_1\$ lies on \$l\$. In the latter case apparently \$S_4 = S_1\$. In any event, \$S_1\$ has the property all points of \$S\$ are on one side of the line passing through \$S_1\$ and \$S_2\$, and this is not true for any other point on the same side of \$l\$ as \$S_2\$. A similar statement holds for \$S_4\$ and \$S_3\$. It follows \$S_1\$ must be symmetric to \$S_4\$ about \$l\$. So \$\\angle S_1S_2S_3 = \\angle S_2S_3S_4\$. Repeat for all other sequential \$4\$ points of \$O\$ to get\n\n\\[\n\\angle S_{k-1}S_kS_{k+1} = \\angle S_kS_{k+1}S_{k+2}\n\\]\n\nfor all \$k\$. Now, if \$S_2\$ didn't lie on the axis of symmetry between \$S_1\$ and \$S_3\$ there would exist another point \$S_2'\$ symmetric to \$S_2\$ about that axis, and as the line rolled around \$S\$ one would find it sequentially touched \$S_1,S_2',S_2,S_3\$ or \$S_1,S_2,S_2',S_3\$. A contradiction to show \$S_1,...,S_n\$ were defined. So \$S_2\$ lies on the axis of symmetry, thus \$S_1S_2 = S_2S_3\$. Repeat for all other sequential \$3\$ points of \$S_1,...,S_n\$ to get\n\n\\[\nS_{k-1}S_k = S_kS_{k+1}\n\\]\n\nfor all \$k\$. We have shown \$S_1,...,S_n\$ form a regular polygon.\n\nNow, the regular polygon \$S_1...S_n\$ has a well-defined center \$O\$ and there is certainly not another isometric polygon among \$S\$. Therefore any axis of symmetry about which we can reflect \$S\$ must reflect \$O\$ into itself. i.e. the axis of symmetry must intersect \$O\$.\n\nLet \$A\$ be a point of \$S\$. We will show \$A\$ is one of \$S_1,...,S_n\$. The axis of symmetry between \$S_1\$ and \$A\$ intersects \$O\$ by the previous. So \$S_1OA\$ is an isosceles triangle with \$OS_1 = OA\$. Therefore \$A\$ lies on the circle with center \$O\$ and radius \$OS_1\$. Note all points \$S_1,...,S_n\$ lie on this circle. If \$A\$ were not one of \$S_1,...,S_n\$, we could suppose \$A\$ lies between \$S_k\$ and \$S_{k+1}\$ on this circle, whereas all points of \$S\$ lie on one side of the line passing through \$S_k\$ and \$S_{k+1}\$. A contradiction. So \$A\$ is one of \$S_1,...,S_n\$. So \$S\$ is the regular polygon \$S_1...S_n\$.\n\n~not_detriti" ]
IMO-1999-2	https://artofproblemsolving.com/wiki/index.php/1999_IMO_Problems/Problem_2	Let $n \geq 2$ be a fixed integer. - (a) Find the least constant $C$ such that for all nonnegative real numbers $x_1, \dots, x_n$, \[ \sum_{1\leq i<j \leq n} x_ix_j (x_i^2 + x_j^2) \leq C \left( \sum_{i=1}^n x_i \right)^4. \] - (b) Determine when equality occurs for this value of $C$.	[ "The answer is \$C=1/8\$, and equality holds exactly when two of the \$x_i\$ are equal to each other and all the other \$x_i\$ are zero. We prove this by induction on the number of nonzero \$x_i\$.\n\nFirst, suppose that at most two of the \$x_i\$, say \$x_a\$ and \$x_b\$, are nonzero. Then the left-hand side of the desired inequality becomes \$x_a x_b (x_a^2 + x_b^2)\$ and the right-hand side becomes \$(x_a + x_b)^4/8\$. By AM-GM,\n\n\\[\n\\begin{align} x_a x_b(x_a^2 + x_b^2) = 1/2 \\left[ \\sqrt{(2x_ax_b)(x_a^2 + x_b^2)} \\right]^2 &\\le \\frac{1}{2} \\left( \\frac{x_a^2 + 2x_a x_b + x_b^2}{2} \\right)^2 \\\\ &= (x_a + x_b)^2/8, \\end{align}\n\\]\n\nwith equality exactly when \$x_a = x_b\$, as desired.\n\nNow, suppose that our statement holds when at most \$k\$ of the \$x_i\$ are equal to zero. Suppose now that \$k+1\$ of the \$x_i\$ are equal to zero, for \$k\\ge 2\$. Without loss of generality, let these be \$x_1 \\ge \\dotsb \\ge x_k \\ge x_{k+1}\$. We define\n\n\\[\ny_j = \\begin{cases} x_j, & j \\notin \\{k, k+1\\} \\\\ x_k + x_{k+1}, & j=k \\\\ 0, & j=k+1 \\end{cases} ,\n\\]\n\nand for convenience, we will denote \$S= \\sum_{i=1}^{k-1} x_i\$. We wish to show that by replacing the \$x_i\$ with the \$y_i\$, we increase the left-hand side of the desired inequality without changing the right-hand side; and then to use the inductive hypothesis.\n\nWe note that\n\n\\[\n\\begin{align} \\sum_{1\\le i< j \\le n} x_i x_j(x_i^2 + x_j^2) ={}& \\sum_{1 \\le i<j \\le k-1} x_i x_j(x_i^2 + x_j^2) + \\sum_{1 \\le i \\le k-1} \\left[ x_ix_k(x_i^2 + x_k^2) + x_ix_{k+1}(x_i^2 + x_{k+1}^2) \\right] \\\\ & + x_k x_{k+1}(x_k^2 + x_{k+1}^2) \\\\ ={}& \\sum_{1 \\le i<j \\le k-1} x_i x_j(x_i^2 + x_j^2) + \\sum_{i=1}^{k-1} x_i^3 (x_k + x_{k+1}) + S(x_k^3 + x_{k+1}^3) \\\\ &+ x_k^3 x_{k+1} + x_k x_{k+1}^3 \\end{align}\n\\]\n\nIf we replace \$x\$ with \$y\$, then \$S(x_k^3 + x_{k+1}^3) + x_k^3 x_{k+1} + x_k x_{k+1}^3\$ becomes\n\n\\[\nS(x_k + x_{k+1})^3 = S(x_k + x_{k+1})^3 + 3S(x_k^2 x_{k+1} + x_k x_{k+1}^2),\n\\]\n\nbut none of the other terms change. Since \$3S > S \\ge a_1 \\ge a_k, a_{k+1}\$, it follows that we have strictly increased the right-hand side of the equation, i.e.,\n\n\\[\n\\sum_{1 \\le i<j \\le n} x_i x_j (x_i^2 + x_j^2) < \\sum_{1 \\le i<j \\le n} y_i y_j (y_i^2 + y_j^2) .\n\\]\n\nBy inductive hypothesis,\n\n\\[\n\\sum_{1\\le i<j \\le n} y_i y_j (y_i^2 +y_j^2) \\le \\left( \\sum_{i=1}^n y_i \\right)^4 / 8 ,\n\\]\n\nand by our choice of \$y_i\$,\n\n\\[\n\\left( \\sum_{i=1}^n y_i \\right)^4/8 = \\left( \\sum_{i=1}^n x_i \\right)^4/8 .\n\\]\n\nHence the problem's inequality holds by induction, and is strict when there are more than two nonzero \$x_i\$, as desired. \$\\blacksquare\$", "I claim that \$C=\\frac{1}{8}\$. Let \$P_2=\\sum_{i=1}^nx_i^2\$ and \$S_2=\\sum_{1\\leq i<j\\leq n}x_ix_j\$. Note that\n\n\\[\n\\sum_{1\\leq i<j\\leq n}x_ix_j\\left(x_i^2+x_j^2\\right)\\leq\\sum_{1\\leq i<j\\leq n}x_ix_jP_2=P_2S_2=\\frac{\\left(2\\sqrt{P_2\\cdot2S_2}\\right)^2}{8}\\leq\\frac{\\left(P_2+2S_2\\right)^2}{8}=\\frac{1}{8}\\left(\\sum_{i=1}^nx_i\\right)^4.\n\\]\n\nEquality holds in the first ineq when all but two of the \$x_i\$ are zero, and this reduces to the \$n=2\$ case which we can easily show to be equivalent to \$\\left(x_1-x_2\\right)^4\\geq0\$, so \$x_1=x_2\$. That is, equality holds when two \$x_i\$ are the same and the rest are zero.\n\nQ.E.D.\n\n## Resources\n\n- 1999 IMO Problems\n- Discussion on AoPS/MathLinks" ]
IMO-1999-3	https://artofproblemsolving.com/wiki/index.php/1999_IMO_Problems/Problem_3	Consider an $n \times n$ square board, where $n$ is a fixed even positive integer. The board is divided into $n^{2}$ units squares. We say that two different squares on the board are adjacent if they have a common side. $N$ unit squares on the board are marked in such a way that every square (marked or unmarked) on the board is adjacent to at least one marked square. Determine the smallest possible value of $N$.	[]
IMO-1999-4	https://artofproblemsolving.com/wiki/index.php/1999_IMO_Problems/Problem_4	Determine all pairs $(n,p)$ of positive integers such that $p$ is a prime, $n$ not exceeded $2p$, and $(p-1)^{n}+1$ is divisible by $n^{p-1}$	[ "Clearly we have the solutions \$(1,p)\$ and \$(2,2)\$, and for every other solution \$p \\geq 3\$. It remains to find the solutions \$(x,p)\$ with \$x \\geq 2\$ and \$p \\geq 3\$. We claim that in this case \$x\$ is divisible by p and \$x y 2p\$, whence \$x = p\$. This will lead to\n\n$\n\n\\[\np^{p-1}\|(p-1)^{p}+1=p^{2}\\left(p^{p-2}-{p \\choose 1}p^{p-3}+\n\\cdots + -{p \\choose p-3}p-{p \\choose p-2}+1\\right)\n\\]\n\n$ (Error compiling LaTeX. Unknown error_msg)\n\ntherefore, because all the terms in the brackets excepting the last one is divisible by \$p\$,\$p-1 \\leq 2\$. This leaves only \$p=3\$ and \$x=3\$. Let us prove now the claim. Since \$(p-1)^{x}+1\$ is odd, so is \$x\$ (therefore \$x < 2p\$). Denote by q the smallest prime divisor of \$x\$. \$q\|(p-1)^{x}+1\$ we get \$(p-1)^{x} \\equiv -1 \\pmod {q}\$ and \$(q,p-q)=1\$. But \$(x,p-q)=1\$ (from the choice of q) leads to the existence of integers \$u,v\$ such that \$ux+v(q-1)=1\$, whence \$p-1 \\equiv (p-1)^{ux}\$. \$(p-1)^{v(q-1)} \\equiv (-1)^{u} \\cdot 1^{v} \\equiv -1 \\pmod {q}\$, because \$u\$ must be odd. This shows that \$q\|p\$, therefore \$q=p\$. In conclusion the required solutions are \$(2,2), (3,3)\$ and \$(1,p)\$, where \$p\$ is an arbitrary prime." ]
IMO-1999-5	https://artofproblemsolving.com/wiki/index.php/1999_IMO_Problems/Problem_5	Two circles $G_{1}$ and $G_{2}$ are contained inside the circle $G$, and are tangent to $G$ at the distinct points $M$ and $N$, respectively. $G_{1}$ passes through the center of $G_{2}$. The line passing through the two points of intersection of $G_{1}$ and $G_{2}$ meets $G$ at $A$ and $B$. The lines $MA$ and $MB$ meet $G_{1}$ at $C$ and $D$, respectively. Prove that $CD$ is tangent to $G_{2}$.	[]
IMO-1999-6	https://artofproblemsolving.com/wiki/index.php/1999_IMO_Problems/Problem_6	Determine all functions $f:\Bbb{R}\to \Bbb{R}$ such that \[ f(x-f(y))=f(f(y))+xf(y)+f(x)-1 \] for all real numbers $x,y$.	[ "Let \$f(0) = c\$. Substituting \$x = y = 0\$, we get:\n\n\\[\nf(-c) = f(c) + c - 1. \\hspace{1cm} ... (1)\n\\]\n\nNow if c = 0, then:\n\n\\[\nf(0) = f(0) - 1\n\\]\n\nwhich is not possible.\n\n\$\\implies c \\neq 0\$.\n\nNow substituting \$x = f(y)\$, we get\n\n\\[\nc = f(x) + x^{2} + f(x) - 1\n\\]\n\n.\n\nSolving for \$f(x)\$, we get\n\n\\[\nf(x) = \\frac{c + 1}{2} - \\frac{x^{2}}{2}. \\hspace{1cm} ... (2)\n\\]\n\nThis means \$f(x) = f(-x)\$ because \$x^{2} = (-x)^{2}\$.\n\nSpecifically,\n\n\\[\nf(c) = f(-c). \\hspace{1cm} ... (3)\n\\]\n\nUsing equations \$(1)\$ and \$(3)\$, we get:\n\n\\[\nf(c) = f(c) + c - 1\n\\]\n\nwhich gives\n\n\\[\nc = 1\n\\]\n\n.\n\nSo, using this in equation \$(2)\$, we get\n\n\\[\n\\boxed{f(x) = 1 - \\frac{x^{2}}{2}}\n\\]\n\nas the only solution to this functional equation." ]
IMO-2000-1	https://artofproblemsolving.com/wiki/index.php/2000_IMO_Problems/Problem_1	Two circles $G_1$ and $G_2$ intersect at two points $M$ and $N$. Let $AB$ be the line tangent to these circles at $A$ and $B$, respectively, so that $M$ lies closer to $AB$ than $N$. Let $CD$ be the line parallel to $AB$ and passing through the point $M$, with $C$ on $G_1$ and $D$ on $G_2$. Lines $AC$ and $BD$ meet at $E$; lines $AN$ and $CD$ meet at $P$; lines $BN$ and $CD$ meet at $Q$. Show that $EP=EQ$.	[ "\$\\textbf{Proof of problem:}\$ Let ray \$NM\$ intersect \$AB\$ at \$X\$. By our lemma, \$\\textit{(the two circles are tangent to AB)}\$, \$X\$ bisects \$AB\$. Since \$\\triangle{NAX}\$ and \$\\triangle{NPM}\$ are similar, and \$\\triangle{NBX}\$ and \$\\triangle{NQM}\$ are similar implies \$M\$ bisects \$PQ\$.\n\nNow, \$\\angle{ABM} = \\angle{BMD}\$ since \$CD\$ is parallel to \$AB\$. But \$AB\$ is tangent to the circumcircle of \$\\triangle{BMD}\$ hence \$\\angle{ABM} = \\angle{BDM}\$ and that implies \$\\angle{BMD} = \\angle{BDM} .\$So\$\\triangle{BMD}\$ is isosceles and \$BM=BD\$.\n\nBy simple parallel line rules, \$\\angle{EBA}=\\angle{BDM}=\\angle{ABM}\$. Similarly, \$\\angle{BAM}=\\angle{EAB}\$, so by \$\\textit{ASA}\$ criterion, \$\\triangle{ABM}\$ and \$\\triangle{ABE}\$ are congruent.\n\nWe know that \$BE=BM=BD\$ so a circle with diameter \$ED\$ can be circumscribed around \$\\triangle{EMD}\$. Join points \$E\$ and \$M\$, \$EM\$ is perpendicular on \$PQ\$, previously we proved \$MP = MQ\$, hence \$\\triangle{EPQ}\$ is isoscles and \$EP = EQ\$ ." ]
IMO-2000-2	https://artofproblemsolving.com/wiki/index.php/2000_IMO_Problems/Problem_2	Let $a, b, c$ be positive real numbers with $abc=1$. Show that \[ \left( a-1+\frac{1}{b} \right)\left( b-1+\frac{1}{c} \right)\left( c-1+\frac{1}{a} \right) \le 1 \]	[ "There exist positive reals \$x\$, \$y\$, \$z\$ such that \$a = \\frac{x}{y}\$, \$b = \\frac{y}{z}\$, \$c = \\frac{z}{x}\$. The inequality then rewrites as\n\n\\[\n\\left(\\frac{x-y+z}{y}\\right)\\left(\\frac{y-z+x}{z}\\right)\\left(\\frac{z-x+y}{x}\\right)\\leq 1\n\\]\n\nor\n\n\\[\n(x-y+z)(y-z+x)(z-x+y)\\leq xyz.\n\\]\n\nSet \$p=x-y+z\$,\$q=y-z+x\$,\$r=z-x+y\$, we get\n\n\\[\n8pqr\\leq(p+q)(q+r)(r+p).\n\\]\n\nSince at most one of \$p,q,r\$ can be negative (if 2 or more are negative, then one of \$a,b,c\$ will become negative), for all positive we apply AM-GM, for one negative we have \$LHS<0<RHS\$." ]
IMO-2000-3	https://artofproblemsolving.com/wiki/index.php/2000_IMO_Problems/Problem_3	Let $n \ge 2$ be a positive integer and $\lambda$ a positive real number. Initially there are $n$ fleas on a horizontal line, not all at the same point. We define a move as choosing two fleas at some points $A$ and $B$ to the left of $B$, and letting the flea from $A$ jump over the flea from $B$ to the point $C$ so that $\frac{BC}{AB}=\lambda$. Determine all values of $\lambda$ such that, for any point $M$ on the line and for any initial position of the $n$ fleas, there exists a sequence of moves that will take them all to the position right of $M$.	[]
IMO-2000-4	https://artofproblemsolving.com/wiki/index.php/2000_IMO_Problems/Problem_4	A magician has one hundred cards numbered $1$ to $100$. He puts them into three boxes, a red one, a white one and a blue one, so that each box contains at least one card. A member of the audience selects two of the three boxes, chooses one card from each and announces the sum of the numbers on the chosen cards. Given this sum, the magician identifies the box from which no card has been chosen. How many ways are there to put all the cards into the boxes so that this trick always works? (Two ways are considered different if at least one card is put into a different box.)	[ "Consider \$1\$, \$2\$ and \$3\$. If they are in different boxes, then \$4\$ must be in the same box as \$1\$, \$5\$ in the same box as \$2\$ and so on. This leads to the solution where all numbers congruent to each other mod \$3\$ are in the same box.\n\nSuppose \$1\$ and \$2\$ are in box \$A\$ and \$3\$ in box \$B\$. Then \$4\$ must be in box \$A\$ or \$B\$. In general, if \$k\\ge 4\$ is in either box \$A\$ or \$B\$, then \$k + 1\$ also must be in box \$A\$ or \$B\$. Thus box \$C\$ is empty which is impossible.\n\nSimilarly, it is impossible for \$1\$ and \$3\$ to be in box \$A\$ and \$2\$ in box \$B\$.\n\nThus we are left with the case where \$1\$ is in box \$A\$ and \$2\$ and \$3\$ in box \$B\$. Suppose box \$B\$ contains \$2, . . . k\$, where \$k \\ge 3\$, but does not contain \$k + 1\$ and \$m\$ is the smallest number in box \$C\$. Then \$m > k\$.\n\nIf \$m > k + 1\$, then \$k + 1\$ must be box \$A\$ and we see that no box can contain \$m-1\$.\n\nThus \$m = k + 1\$. If \$k < 99\$, we see that no box can contain \$k + 2\$. Thus \$k = 99\$. It is easy to see that this distribution works. Thus there altogether \$12\$ ways - \$2\$ options times permutation of \$3\$ colors for each of \$3\$ boxes.\n\nTaken from: https://sms.math.nus.edu.sg/Simo/IMO_Problems/00.pdf" ]
IMO-2000-5	https://artofproblemsolving.com/wiki/index.php/2000_IMO_Problems/Problem_5	Does there exist a positive integer $n$ such that $n$ has exactly 2000 prime divisors and $n$ divides $2^n + 1$?	[ "THIS SOLUTION IS WRONG.. ANOTHER SOLUTION IS NEEDED..\n\nLet \$N=2^n+1\$. We will assume for the sake of contradiction that \$n\|N\$.\n\n\$2^n+1 \\equiv 0 \\pmod{n} \\Rightarrow 2^n \\equiv -1 \\pmod{n}\$. So 2 does not divide \$n\$, and so \$n\$ is odd.\n\nSelect an arbitrary prime factor of \$n\$ and call it \$p\$. Let's represent \$n\$ in the form \$p^am\$, where \$m\$ is not divisible by \$p\$.\n\nNote that \$p\$ and \$m\$ are both odd since \$n\$ is odd. By repeated applications of Fermat's Little Theorem:\n\n\$N = 2^n+1 = 2^{p^am} + 1 = (2^{p^{a-1}m})^p + 1 \\equiv 2^{p^{a-1}m} + 1\$ (mod \$p\$)\n\nContinuing in this manner, and inducting on k from 1 to \$a\$,\n\n\$2^{p^{a-k}m}+1 \\equiv (2^{p^{a-k-1}m})^p + 1\$ (mod \$p\$) \$\\equiv 2^{p^{a-k-1}m} + 1\$ (mod \$p\$)\n\nSo we have \$N \\equiv 2^m+1\$ (mod \$p\$)\n\nSince \$p\$ is relatively prime to \$m\$, \$N \\equiv 1+1\$ (mod \$p\$) \$\\equiv 2\$ (mod \$p\$)\n\nSince \$p\$ is odd, \$N\$ is not divisible by \$p\$. Hence \$N\$ is not divisible by \$n\$. So we have a contradiction, and our original assumption was false, and therefore \$N\$ is still not divisible by \$n\$." ]
IMO-2000-6	https://artofproblemsolving.com/wiki/index.php/2000_IMO_Problems/Problem_6	Let $\overline{AH_1}$, $\overline{BH_2}$, and $\overline{CH_3}$ be the altitudes of an acute triangle $ABC$. The incircle $\omega$ of triangle $ABC$ touches the sides $BC$, $CA$, and $AB$ at $T_1$, $T_2$, and $T_3$, respectively. Consider the reflections of the lines $H_1H_2$, $H_2H_3$, and $H_3H_1$ with respect to the lines $T_1T_2$, $T_2T_3$, and $T_3T_1$. Prove that these images form a triangle whose vertices line on $\omega$.	[]
IMO-2001-1	https://artofproblemsolving.com/wiki/index.php/2001_IMO_Problems/Problem_1	Consider an acute triangle $\triangle ABC$. Let $P$ be the foot of the altitude of triangle $\triangle ABC$ issuing from the vertex $A$, and let $O$ be the circumcenter of triangle $\triangle ABC$. Assume that $\angle C \geq \angle B+30^{\circ}$. Prove that $\angle A+\angle COP < 90^{\circ}$.	[ "Take \$D\$ on the circumcircle with \$AD \\parallel BC\$. Notice that \$\\angle CBD = \\angle BCA\$, so \$\\angle ABD \\ge 30^\\circ\$. Hence \$\\angle AOD \\ge 60^\\circ\$. Let \$Z\$ be the midpoint of \$AD\$ and \$Y\$ the midpoint of \$BC\$. Then \$AZ \\ge R/2\$, where \$R\$ is the radius of the circumcircle. But \$AZ = YP\$ (since \$AZYP\$ is a rectangle).\n\nNow \$O\$ cannot coincide with \$Y\$ (otherwise \$\\angle A\$ would be \$90^\\circ\$ and the triangle would not be acute-angled). So \$OP > YP \\ge R/2\$. But \$PC = YC - YP < R - YP \\le R/2\$. So \$OP > PC\$.\n\nHence \$\\angle COP < \\angle OCP\$. Let \$CE\$ be a diameter of the circle, so that \$\\angle OCP = \\angle ECB\$. But \$\\angle ECB = \\angle EAB\$ and \$\\angle EAB + \\angle BAC = \\angle EAC = 90^\\circ\$, since \$EC\$ is a diameter. Hence \$\\angle COP + \\angle BAC < 90^\\circ\$.", "Notice that because \$\\angle{PCO} = 90^\\circ - \\angle{A}\$, it suffices to prove that \$\\angle{POC} < \\angle{PCO}\$, or equivalently \$PC < PO.\$\n\nSuppose on the contrary that \$PC > PO\$. By the triangle inequality, \$2 PC = PC + PC > PC + PO > CO = R\$, where \$R\$ is the circumradius of \$ABC\$. But the Law of Sines and basic trigonometry gives us that \$PC = 2R \\sin B \\cos C\$, so we have \$4 \\sin B \\cos C > 1\$. But we also have \$4 \\sin B \\cos C \\le 4 \\sin B \\cos (B + 30^\\circ) = 2 (\\sin (2B + 30^\\circ) - \\sin 30^\\circ) \\le 2 (1 - \\frac{1}{2}) = 1\$ because \$\\angle{C} \\ge \\angle{B} + 30^\\circ\$, and so we have a contradiction. Hence \$PC < PO\$ and so \$\\angle{PCO} + \\angle{A} < 90^\\circ\$, as desired." ]
IMO-2001-2	https://artofproblemsolving.com/wiki/index.php/2001_IMO_Problems/Problem_2	Let $a,b,c$ be positive real numbers. Prove that $\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1$.	[ "Firstly, \$a^{2}+8bc=(a^{2}+2bc)+6bc \\leq^{AG} (a^{2}+b^{2}+c^{2})+6bc=S+6bc\$ (where \$S=a^{2}+b^{2}+c^{2}\$) and its cyclic variations. Next note that \$(a,b,c)\$ and \$\\left( \\frac{1}{\\sqrt{S+6bc}}, \\frac{1}{\\sqrt{S+6ca}}, \\frac{1}{\\sqrt{S+6ab}} \\right)\$ are similarly oriented sequences. Thus\n\n\\[\n\\sum_{cyc} \\frac{a}{\\sqrt{a^{2}+8bc}} \\ge \\sum_{cyc} \\frac{a}{\\sqrt{S+6bc}}\n\\]\n\n\\[\n\\geq ^{Cheff} \\frac{1}{3}(a+b+c)\\left( \\frac{1}{\\sqrt{S+6bc}}+\\frac{1}{\\sqrt{S+6ca}}+\\frac{1}{\\sqrt{S+6ab}} \\right)\n\\]\n\n\\[\n\\geq^{AH}\\frac{1}{3}(a+b+c) \\left( \\frac{9}{\\sqrt{S+6bc}+\\sqrt{S+6ca}+\\sqrt{S+6ab}} \\right)\n\\]\n\n\\[\n\\geq^{QA} (a+b+c) \\sqrt{\\frac{3}{(S+6bc)+(S+6ca)+(S+6ab)}}\n\\]\n\n\\[\n=(a+b+c)\\sqrt{\\frac{3}{3(a+b+c)^{2}}}=1\n\\]\n\nHence the inequality has been established. Equality holds if \$a=b=c\$.\n\nNotation: \$AG\$: AM-GM inequality, \$AH\$: AM-HM inequality, \$Cheff\$: Chebyshev's inequality, \$QA\$: QM-AM inequality / RMS inequality", "By Hölder's inequality, \$\\left(\\sum_{cyc}\\frac{a}{\\sqrt{a^{2}+8bc}}\\right)\\left(\\sum_{cyc}\\frac{a}{\\sqrt{a^{2}+8bc}}\\right)\\left(\\sum_{cyc} a(a^{2}+8bc)\\right)\\ge (a+b+c)^{3}\$ Thus we need only show that \$(a+b+c)^{3}\\ge a^{3}+b^{3}+c^{3}+24abc\$ Which is obviously true since \$(a+b)(b+c)(c+a)\\ge 8abc\$.", "This inequality is homogeneous so we can assume without loss of generality \$a+b+c=1\$ and apply Jensen's inequality for \$f(x)=\\frac{1}{\\sqrt{x}}\$, so we get:\n\n\\[\n\\frac{a}{\\sqrt{a^2+8bc}}+\\frac{b}{\\sqrt{b^2+8ac}}+\\frac{c}{\\sqrt{c^2+8ab}} \\geq \\frac{1}{\\sqrt{a^3+b^3+c^3+24abc}}\n\\]\n\nbut\n\n\\[\n1=(a+b+c)^3=a^3+b^3+c^3+6abc+3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) \\geq a^3+b^3+c^3+24abc\n\\]\n\nby AM-GM, and thus the inequality is proven.", "We can rewrite\n\n\\[\n\\frac{a}{\\sqrt{a^2+8bc}}+\\frac{b}{\\sqrt{b^2+8ac}}+\\frac{c}{\\sqrt{c^2+8ab}}\n\\]\n\nas\n\n\\[\n\\frac{a}{\\sqrt{a^2+\\frac{8abc}{a}}}+\\frac{b}{\\sqrt{b^2+\\frac{8abc}{b}}}+\\frac{c}{\\sqrt{c^2+\\frac{8abc}{c}}}\n\\]\n\nwhich is the same as\n\n\\[\n\\frac{\\sqrt{a^3}}{\\sqrt{a^3+8abc}}+\\frac{\\sqrt{b^3}}{\\sqrt{b^3+8abc}}+\\frac{\\sqrt{c^3}}{\\sqrt{c^3+8abc}}\n\\]\n\nNow let \$f(x)=\\sqrt{\\frac{x^3}{x^3 + 8abc}}\$. Then f is convex and f is strictly increasing, so by Jensen's inequality and AM-GM,\n\n\\[\nf(a) + f(b) + f(c) \\geq 3f((\\frac{1}{3})a + (\\frac{1}{3})b + (\\frac{1}{3})c)) \\geq 3f(\\sqrt[3]{abc}) = 3(\\frac{1}{3}) =1\n\\]", "Let \$f : (0, \\infty); f(x) = \\frac{1}{x}\$, \$x_1 = \\sqrt{a^{2} + 8bc}\$, \$x_2 = \\sqrt{b^{2} + 8ac}\$ and \$x_3 = \\sqrt{c^{2} + 8ab}\$ f is convex so we can write:\n\n\\[\nf(\\frac{a}{x_1} + \\frac{b}{x_2} + \\frac{c}{x_3})\\le af(x_1) + bf(x_2) + cf(x_3)\n\\]\n\nlet \$\\frac{a}{x_1} + \\frac{b}{x_2} + \\frac{c}{x_3} = t\$, by substitustion:\n\n\\[\nf(t)\\le t\n\\]\n\n\\[\n\\frac{1}{t}\\le t\n\\]\n\nwe multiply both sides by t\n\n\\[\n1\\le t^{2}\n\\]\n\n\\[\n1\\le t\n\\]\n\nQED", "We claim that\n\n\\[\n\\frac{a}{\\sqrt{a^2+8bc}} \\geq \\frac{a^{\\frac{4}{3}}}{a^{\\frac{4}{3}}+b^{\\frac{4}{3}}+c^{\\frac{4}{3}}}\n\\]\n\nCross-multiplying, squaring both sides and expanding, we have\n\n\\[\na^{\\frac{14}{3}}+a^{2}b^{\\frac{8}{3}}+a^{2}c^{\\frac{8}{3}}+2a^{\\frac{10}{3}}b^{\\frac{4}{3}}+2a^{\\frac{10}{3}}c^{\\frac{4}{3}}+2a^{2}b^{\\frac{4}{3}}c^{\\frac{4}{3}} \\geq a^{\\frac{14}{3}}+8a^{\\frac{8}{3}}bc\n\\]\n\nAfter cancelling the \$a^{\\frac{14}{3}}\$ term, we apply AM-GM to RHS and obtain\n\n\\[\na^{2}b^{\\frac{8}{3}}+a^{2}c^{\\frac{8}{3}}+2a^{\\frac{10}{3}}b^{\\frac{4}{3}}+2a^{\\frac{10}{3}}c^{\\frac{4}{3}}+2a^{2}b^{\\frac{4}{3}}c^{\\frac{4}{3}} \\geq 8(a^{\\frac{64}{3}}b^8c^8)^{\\frac{1}{8}}=8a^{\\frac{8}{3}}bc\n\\]\n\nas desired, completing the proof of the claim.\n\nSimilarly \$\\frac{b}{\\sqrt{b^2+8ca}} \\geq \\frac{b^{\\frac{4}{3}}}{a^{\\frac{4}{3}}+b^{\\frac{4}{3}}+c^{\\frac{4}{3}}}\$ and \$\\frac{c}{\\sqrt{c^2+8ab}} \\geq \\frac{c^{\\frac{4}{3}}}{a^{\\frac{4}{3}}+b^{\\frac{4}{3}}+c^{\\frac{4}{3}}}\$. Summing the three inequalities, we obtain the original inequality.", "We want to prove\n\n\\[\n\\sum_{cyc}\\dfrac{a}{\\sqrt{a^2+8bc}}\\ge 1\n\\]\n\nNote that since this inequality is homogenous, assume \$a+b+c=3\$.\n\nBy Cauchy,\n\n\\[\n\\left(\\sum_{cyc}\\dfrac{a}{\\sqrt{a^2+8bc}}\\right)\\left(\\sum_{cyc}a\\sqrt{a^2+8bc}\\right)\\ge (a+b+c)^2=9\n\\]\n\nDividing both sides by \$\\sum_{cyc}a\\sqrt{a^2+8bc}\$, we see that we want to prove\n\n\\[\n\\dfrac{9}{\\sum\\limits_{cyc}a\\sqrt{a^2+8bc}}\\ge 1\n\\]\n\nor equivalently\n\n\\[\n\\sum\\limits_{cyc}a\\sqrt{a^2+8bc}\\le 9\n\\]\n\nSquaring both sides, we have\n\n\\[\n\\left(\\sum_{cyc}a\\sqrt{a^2+8bc}\\right)^2\\le 81\n\\]\n\nNow use Cauchy again to obtain\n\n\\[\n\\left(\\sum_{cyc}a\\sqrt{a^2+8bc}\\right)^2\\le (a+b+c)\\left(\\sum_{cyc}a(a^2+8bc)\\right)\\le 81\n\\]\n\nSince \$a+b+c=3\$, the inequality becomes\n\n\\[\n\\sum_{cyc}a^3+8abc\\le 27\n\\]\n\nafter some simplifying.\n\nBut this equals\n\n\\[\n(a+b+c)^3-3\\left(\\sum_{sym}a^2b\\right)+18abc\\le 27\n\\]\n\nand since \$a+b+c=3\$ we just want to prove\n\n\\[\n\\left(\\sum_{sym}a^2b\\right)\\ge 6abc\n\\]\n\nafter some simplifying.\n\nBut that is true by AM-GM or Muirhead. Thus, proved. \$\\Box\$", "By Carlson's Inequality, we can know that\n\n\\[\n\\Bigg(\\frac {a}{\\sqrt {a^2 + 8bc}} + \\frac {b}{\\sqrt {b^2 + 8ca}} + \\frac {c}{\\sqrt {c^2 + 8ab}}\\Bigg)\\Bigg(\\frac {a}{\\sqrt {a^2 + 8bc}} + \\frac {b}{\\sqrt {b^2 + 8ca}} + \\frac {c}{\\sqrt {c^2 + 8ab}}\\Bigg)\\Big((a^3+8abc)+(b^3+8abc)+(c^3+8abc)\\Big) \\ge (a+b+c)^3\n\\]\n\nThen,\n\n\\[\n\\Bigg(\\frac {a}{\\sqrt {a^2 + 8bc}} + \\frac {b}{\\sqrt {b^2 + 8ca}} + \\frac {c}{\\sqrt {c^2 + 8ab}}\\Bigg)^2 \\ge \\frac{(a+b+c)^3}{a^3+b^3+c^3+24abc}\n\\]\n\nOn the other hand,\n\n\\[\n3a^2b+3b^2c+3c^2a \\ge 9abc\n\\]\n\nand\n\n\\[\n3ab^2+3bc^2+3ca^2 \\ge 9abc\n\\]\n\nThen,\n\n\\[\n(a+b+c)^3 = a^3+b^3+c^3+3(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)+6abc \\ge a^3+b^3+c^3+24abc\n\\]\n\nTherefore,\n\n\\[\n\\Bigg(\\frac {a}{\\sqrt {a^2 + 8bc}} + \\frac {b}{\\sqrt {b^2 + 8ca}} + \\frac {c}{\\sqrt {c^2 + 8ab}}\\Bigg)^2 \\ge \\frac{(a+b+c)^3}{a^3+b^3+c^3+24abc} \\ge 1\n\\]\n\nThus,\n\n\\[\n\\frac {a}{\\sqrt {a^2 + 8bc}} + \\frac {b}{\\sqrt {b^2 + 8ca}} + \\frac {c}{\\sqrt {c^2 + 8ab}} \\ge 1\n\\]\n\n-- Haozhe Yang" ]
IMO-2001-3	https://artofproblemsolving.com/wiki/index.php/2001_IMO_Problems/Problem_3	Twenty-one girls and twenty-one boys took part in a mathematical competition. It turned out that each contestant solved at most six problems, and for each pair of a girl and a boy, there was at least one problem that was solved by both the girl and the boy. Show that there is a problem that was solved by at least three girls and at least three boys.	[ "For each girl, we know that there is a boy who solved a problem in common with her. Since a girl solves at most six problems, for a girl and her set of six problems there are at least 11 boys who solved a problem in common with the girl such that at least three boys solved that common problem. (This is a simple application of the pigeon hole principle where the problems solved by the girl are the holes and the boys are the pigeons, and the guy/pigeon enters a hole/problem which he solved in common with the girl).\n\nThe boys have solved a total of at most \$21\\times 6\$ problems.We view this as a set with repeated elements i.e. if a problem is solved by more than one boy it appears as many times as the number of boys who have solved it. Let this set of problems be A. Clearly, A has size \$21\\times 6\$.\n\nWe mark each problem in A which has been solved by at least three boys and a girl. And we mark it the number of times that is the same as the number of girls that has solved it. Since there are a total of \$21\\times 11\$ marks (since there are at least 11 marks for each girl, by the discussion above), either a problem is marked at least thrice, in which case we're done since then it has been solved by at least three boys and three girls. Or each problem has been marked at most twice. In this case it is clear that more than \$21\\times 5\$ problems in A have been marked since \$21\\times 5 \\times 2<21\\times 11\$ (there are a total of \$21\\times 11\$ marks. This means that there is a boy such that all six of his problems have been marked. But then by our discussion in the first paragraph we know there must a problem that this boy has solved which has been solved by at least three girls. Therefore it must be true that there is a problem such that it has been solved by at least three people of each gender." ]
IMO-2001-4	https://artofproblemsolving.com/wiki/index.php/2001_IMO_Problems/Problem_4	Let $n_1, n_2, \dots , n_m$ be integers where $m>1$ is odd. Let $x = (x_1, \dots , x_m)$ denote a permutation of the integers $1, 2, \cdots , m$. Let $f(x) = x_1n_1 + x_2n_2 + ... + x_mn_m$. Show that for some distinct permutations $a$, $b$ the difference $f(a) - f(b)$ is a multiple of $m!$.	[ "Notice that if \$\\{0,1,2,...,m!-1\\}\\not=\\{f(1),f(2),f(3),...,f(m!)\\}\\pmod{m!}\$ then by the pigeon hole princible, there must be some \$a,b\\in\\{1,2,...,m!\\}\$ such that \$f(a)\\equiv f(b)\\pmod{m!}\$ as desired. Thus, we must prove that \$\\{0,1,2,...,m!-1\\}\\not=\\{f(1),f(2),f(3),...,f(m!)\\}\\pmod{m!}\$. Suppose there was such a situation. Then, since the two sets have the same number of elements, for every \$t\\in\\{0,1,...,m!-1\\}\$, there exists a \$v\\in\\{1,2,...,m!\\}\$ such that \$t\\equiv f(v)\\pmod{m!}\$. So, let \$t\\equiv f(v_t)\\pmod{m!}\$. Consider the sum \$f(v_0)+f(v_1)+\\cdots+f(v_{m!-1})\\pmod{m!}\$. Using the fact that \$f(v_t)\\equiv t\\pmod{m!}\$, we find the sum is congruent to \$0+1+\\cdots+m!-1=\\frac{m!(m!-1)}{2}\$.\n\nOn the other hand, using the fact that \$f(x)=x_1n_1 + x_2n_2 + ... + x_mn_m\$, we can combine the terms with the same coefficient (\$n_i\$). For each \$n_1\$, since all the \$v_j\$ are distinct, the coefficient in the sum would be \$\\sum x_i\$ over all permutations \$x\$ of \$\\{1,2,...,m\\}\$. Thus, the coefficient would be \$\\sum_{i=1}^m i(m-1)!=\\frac{m!(m+1)}{2}\$. Since \$m\$ is odd, \$\\frac{m+1}{2}\$ is an integer, so the coefficient is congruent to \$0\\pmod{m!}\$. Thus, the whole sum is congruent to \$0\\pmod{m!}\$. Therefore, \$\\frac{m!(m!-1)}{2}\\equiv0\\pmod{m!}\$. But, since \$m>1\$, we have \$m!\$ is even, and thus \$m!-1\$ is odd. Therefore, the integer \$\\frac{m!(m!-1)}{2}\$ has one factor of \$2\$ less than \$m!\$ does. Contradiction!" ]
IMO-2001-5	https://artofproblemsolving.com/wiki/index.php/2001_IMO_Problems/Problem_5	$ABC$ is a triangle. $X$ lies on $BC$ and $AX$ bisects angle $A$. $Y$ lies on $CA$ and $BY$ bisects angle $B$. Angle $A$ is $60^{\circ}$. $AB + BX = AY + YB$. Find all possible values for angle $B$.	[ "\\[\n[asy] import cse5; import graph; import olympiad; dotfactor = 3; unitsize(1.5inch); pair A = (0,sqrt(3)), D= (-1, 0), E=(1,0); pair Bb = rotate(40,E)A; pair B = extension(A,D,E,Bb); pair H = foot(A,D,E); pair X = extension(A,H,B,E); pair Yy = bisectorpoint(A,B,E); pair Y =extension(A,E,B,Yy); pair C = E - (0,0.1); dot(\"$B$\", B, NW); dot(\"$Y$\", Y, NE); dot(\"$D$\", D, W); dot(\"$E$\", E, E); dot(\"$A$\",A,N); dot(\"$X$\",X,S); label(\"$C$\",E+(0,-0.1),E); draw(A--D--E--cycle); draw(B--Y); draw(B--E); // draw(B--Xx--E,dashed); // draw(Y--Xx, dashed); draw(A--X--D, dashed); [/asy]\n\\]\n\nLet \$D\$ be on extension of \$AB\$ and \$BD=BX\$. Let \$E\$ be on \$YC\$ and \$YE=YB\$, then\n\n\\[\nAD=AB+BD=AB+BX=AY+YB=AE\n\\]\n\nSince \$A=60\$, \$\\triangle{ADE}\$ is equilateral. Let \$\\angle{ABY}=x\$, then,\n\n\\[\n\\angle{YBX}=\\angle{BDX}=\\angle{BXD}=\\angle{YEX}=x\n\\]\n\nWe claim that \$X\$ must be on \$BE\$, i.e., \$C=E\$. If \$X\$ is not on \$BE\$, then \$\\angle{EBX}=\\angle{YBX}-\\angle{YBE}=\\angle{YEX}-\\angle{YEB}=\\angle{BEX}\$, which leads to \$BX=EX=DX\$, and \$\\triangle{BDX}\$ is equilateral, which is not possible. With that, we have, in \$\\triangle{ABE}\$, \$60+2x+x=180\$, \$x=40\$, and \$\\angle{ABE}=80\$.\n\nSolution by \$Mathdummy\$.", "Refer to the image in Solution 1 without any construction \\begin{align} \\text{Set: } & \\angle ABY = \\angle YBC = x, \\quad \\angle YCB = 120^\\circ - 2x. \\\\ \\text{Observe: } & \\angle AYB = 120^\\circ - x, \\quad \\angle AXB = 150^\\circ - 2x. \\\\ \\text{Using the Law of Sines, we get: } & \\\\ & AY = AB \\cdot \\frac{\\sin x^\\circ}{\\sin(120^\\circ - x)}, \\\\ & BX = AB \\cdot \\frac{\\sin 30^\\circ}{\\sin(150^\\circ - 2x)}, \\\\ & YB = AB \\cdot \\frac{\\sin 60^\\circ}{\\sin(120^\\circ - x)}. \\\\ \\text{So, the relation } AB + BX &= AY + AB \\text{ is the same as saying} \\\\ & 1 + \\frac{\\sin 30^\\circ}{\\sin(150^\\circ - 2x)} = \\frac{\\sin x + \\sin 60^\\circ}{\\sin(120^\\circ - x)}. \\\\ \\text{We have } & \\sin x + \\sin 60^\\circ = 2 \\sin\\left(\\frac{1}{2}(x + 60^\\circ)\\right) \\cos\\left(\\frac{1}{2}(x - 60^\\circ)\\right). \\\\ \\text{Also, } & \\sin(120^\\circ - x) = \\sin(x + 60^\\circ) \\quad \\text{and} \\\\ & \\sin(x + 60^\\circ) = 2 \\sin\\left(\\frac{1}{2}(x + 60^\\circ)\\right) \\cos\\left(\\frac{1}{2}(x + 60^\\circ)\\right). \\\\ \\text{So, } & \\frac{\\sin x + \\sin 60^\\circ}{\\sin(120^\\circ - x)} = \\frac{\\cos\\left(\\frac{1}{2}x - 30^\\circ\\right)}{\\cos\\left(\\frac{1}{2}x + 30^\\circ\\right)}. \\\\ \\text{Let } & \\frac{1}{2}x = t. \\\\ \\text{Then } & \\frac{\\cos(t - 30^\\circ)}{\\cos(t + 30^\\circ)} - 1 = \\frac{\\cos(t - 30^\\circ) - \\cos(t + 30^\\circ)}{\\cos(t + 30^\\circ)} = \\frac{2 \\sin(30^\\circ) \\sin(t)}{\\cos(t + 30^\\circ)}. \\\\ \\text{Hence, the problem is just} & \\frac{\\sin(30^\\circ)}{\\sin(150^\\circ - 4t)} = \\frac{\\sin(t)}{\\cos(t + 30^\\circ)} \\\\ \\Rightarrow & \\cos(t + 30^\\circ) = 2 \\sin(t) \\sin(150^\\circ - 4t) \\\\ & = \\cos(5t - 150^\\circ) - \\cos(150^\\circ - 3t). \\\\ \\text{Now, } & \\cos(t + 30^\\circ) + \\cos(5t + 30^\\circ) = \\cos(3t + 30^\\circ). \\\\ \\text{Because } & \\cos(A + B) + \\cos(A - B) = 2\\cos A \\cos B, \\\\ \\text{we get } & \\cos(t + 30^\\circ) + \\cos(5t + 30^\\circ) = 2 \\cos(3t + 30^\\circ) \\cos(2t). \\\\ \\Rightarrow & (2 \\cos(2t) - 1)(\\cos(3t + 30^\\circ)) = 0. \\\\ \\text{This gives } & t \\text{ to be } 20^\\circ \\text{ or } 30^\\circ. \\\\ \\text{Recall that } & t = \\frac{1}{2}x = \\frac{1}{4}\\angle ABC. \\\\ \\text{Here we can see } & \\angle ABC \\neq 120^\\circ \\text{ because of the angle sum property.} \\\\ \\therefore & \\angle B = 80^\\circ, \\angle A = 60^\\circ, \\text{ and } \\angle C = 40^\\circ. \\end{align*}\n\n~Lakshya Pamecha" ]
IMO-2001-6	https://artofproblemsolving.com/wiki/index.php/2001_IMO_Problems/Problem_6	$K > L > M > N$ are positive integers such that $KM + LN = (K + L - M + N)(-K + L + M + N)$. Prove that $KL + MN$ is not prime.	[ "First, \$(KL+MN)-(KM+LN)=(K-N)(L-M)>0\$ as \$K>N\$ and \$L>M\$. Thus, \$KL+MN>KM+LN\$.\n\nSimilarly, \$(KM+LN)-(KN+LM)=(K-L)(M-N)>0\$ since \$K>L\$ and \$M>N\$. Thus, \$KM+LN>KN+LM\$.\n\nPutting the two together, we have\n\n\\[\nKL+MN>KM+LN>KN+LM\n\\]\n\nNow, we have:\n\n\\[\n(K+L-M+N)(-K+L+M+N)=KM+LN\n\\]\n\n\\[\n-K^2+KM+L^2+LN+KM-M^2+LN+N^2=KM+LN\n\\]\n\n\\[\nL^2+LN+N^2=K^2-KM+M^2\n\\]\n\nSo, we have:\n\n\\[\n(KM+LN)(L^2+LN+N^2)=KM(L^2+LN+N^2)+LN(L^2+LN+N^2)\n\\]\n\n\\[\n=KM(L^2+LN+N^2)+LN(K^2-KM+M^2)\n\\]\n\n\\[\n=KML^2+KMN^2+K^2LN+LM^2N\n\\]\n\n\\[\n=(KL+MN)(KN+LM)\n\\]\n\nThus, it follows that\n\n\\[\n(KM+LN) \\mid (KL+MN)(KN+LM).\n\\]\n\nNow, since \$KL+MN>KM+LN\$ if \$KL+MN\$ is prime, then there are no common factors between the two. So, in order to have\n\n\\[\n(KM+LN)\\mid (KL+MN)(KN+LM),\n\\]\n\nwe would have to have\n\n\\[\n(KM+LN) \\mid (KN+LM).\n\\]\n\nThis is impossible as \$KM+LN>KN+LM\$. Thus, \$KL+MN\$ must be composite." ]
IMO-2002-1	https://artofproblemsolving.com/wiki/index.php/2002_IMO_Problems/Problem_1	$S$ is the set of all $(h,k)$ with $h,k$ non-negative integers such that $h + k < n$. Each element of $S$ is colored red or blue, so that if $(h,k)$ is red and $h' \le h,k' \le k$, then $(h',k')$ is also red. A type $1$ subset of $S$ has $n$ blue elements with different first member and a type $2$ subset of $S$ has $n$ blue elements with different second member. Show that there are the same number of type $1$ and type $2$ subsets.	[ "Consider the points \$x+y < n\$ where \$x,y \\geq 0\$ in the cartesian plane. Let \$(j_1,k_1),...,(j_l,k_l)\$ be the maximal red points. That is, neither \$(j_i+1,k_i)\$ nor \$(j_i,k_i+1)\$ is red for all \$i\$. We may assume WLOG \$j_1 < j_2 < \\cdots < j_l\$. Note then \$k_1 > k_2 > \\cdots > k_l\$. Otherwise, if say \$k_m<k_{m+1}\$ one would find \$(j_m,k_m)\$ is not maximal.\n\nMake the identification \$(-1)! = 1\$. The cartesian plane looks like a blue triangle covered by \$l\$ red rectangles. Multiplying the number of blue dots in the first column, times the number of blue dots in the second column, and so on, from left to right, one gets\n\n\\[\n\\frac{(n-k_1-1)!}{(n-k_1-j_1-2)!} \\frac{(n-j_1-k_2-2)!}{(n-k_2-j_2-2)!} \\cdots \\frac{(n-j_{l-1}-k_l-2)!}{(n-k_l-j_l-2)!} (n-j_l-1)!\n\\]\n\nwhere each fraction embodies the products for a single red rectangle. This is the number of type \$1\$ subsets. Similarly multiplying the number of blue dots in the first row, times the number of blue dots in the second row, and so on, from bottom to top, one gets\n\n\\[\n\\frac{(n-j_l-1)!}{(n-j_l-k_l-2)!} \\frac{(n-k_l-j_{l-1}-2)!}{(n-j_{l-1}-k_{l-1}-2)!} \\cdots \\frac{(n-k_2-j_1-2)!}{(n-j_1-k_1-2)!} (n-k_1-1)!\n\\]\n\nwhich is the same as the first product. Since this equals the number of type \$2\$ subsets we are done.\n\n~not_detriti" ]
IMO-2002-2	https://artofproblemsolving.com/wiki/index.php/2002_IMO_Problems/Problem_2	$BC$ is a diameter of a circle center $O$. $A$ is any point on the circle with $\angle AOC \not\le 60^\circ$. $EF$ is the chord which is the perpendicular bisector of $AO$. $D$ is the midpoint of the minor arc $AB$. The line through $O$ parallel to $AD$ meets $AC$ at $J$. Show that $J$ is the incenter of triangle $CEF$.	[]
IMO-2002-3	https://artofproblemsolving.com/wiki/index.php/2002_IMO_Problems/Problem_3	Find all pairs of positive integers $m,n \ge 3$ for which here exist infinitely many positive integers $a$ such that \[ \frac{a^m+a-1}{a^n+a^2-1} \] is itself an integer	[]
IMO-2002-4	https://artofproblemsolving.com/wiki/index.php/2002_IMO_Problems/Problem_4	Problem: Let $n>1$ be an integer and let $1=d_{1}<d_{2}<d_{3} \cdots <d_{r}=n$ be all of its positive divisors in increasing order. Show that \[ d=d_1d_2+d_2d_3+ \cdots +d_{r-1}d_r <n^2 \]	[ "We proceed with two parts:\n\n## Part 1: Proof of Inequality\n\nWe have:\n\n\\[\n\\sum_{i=1}^{r-1} d_i d_{i+1} = d_1d_2 + d_2d_3 + \\cdots + d_{r-1}d_r\n\\]\n\nFor each term \$ d_i d_{i+1} \$, note that \$ d_{i+1} \\leq \\frac{n}{d_i} \$ (since divisors come in pairs). Thus:\n\n\\[\nd_i d_{i+1} \\leq d_i \\cdot \\frac{n}{d_i} = n\n\\]\n\nSumming over all \$ r-1 \$ terms:\n\n\\[\n\\sum_{i=1}^{r-1} d_i d_{i+1} \\leq (r-1) \\cdot n < n^2\n\\]\n\nThe last inequality holds because \$ r \$ (the number of divisors) is at most \$ 2\\sqrt{n} \$, making \$ (r-1)n \\leq 2n^{3/2} < n^2 \$ for \$ n > 4 \$. Smaller cases can be verified directly.\n\n## Part 2: Equality Condition\n\nEquality occurs only when \$ n \$ is prime: - For prime \$ n \$, the divisors are \$ 1 \$ and \$ n \$, so:\n\n- For composite \$ n \$, there exists at least one additional divisor \$ d_2 \$ (where \$ 1 < d_2 < n \$), making the sum strictly larger than \$ n \$ but still less than \$ n^2 \$.\n\nTemplate:INO box", "We rewrote the expression so we have to show\n\n\\[\n\\frac{1}{d_1 d_2} + \\frac{1}{d_2 d_3} + \\cdots + \\frac{1}{d_{k-1} d_k} < 1.\n\\]\n\nThen we showed that if two terms of any sequence of the form\n\n\\[\n\\frac{1}{a_1 a_2} + \\frac{1}{a_2 a_3} + \\cdots\n\\]\n\nwhere \$1 < a_2 < a_3 < \\cdots < a_k\$, are of the form \$d_i = d_{i-1}+1\$ and \$d_{i+1} = d_i+1\$, then the two terms containing them are maximized.\n\nNext, we showed that if\n\n\\[\n\\frac{1}{d_1 d_2} + \\cdots = B,\n\\]\n\nthen cross multiplying denominators means that \$k-1\$ terms are multiplied by a term of \$(1+k_1+k_2+\\cdots)\$ whereas \$k\$ terms are multiplied in the denominator, implying the multiplication of \$(1+k_1+k_2+\\cdots)\$. Thus if the expression is equivalent to \$B\$, multiplication by \$(1+k_1+\\cdots)\$ yields\n\n\\[\n\\frac{(1+k_1+k_2+\\cdots)(B-a)}{1+k_1+k_2+\\cdots} + \\frac{a}{k_1+k_2+k_3+\\cdots}\n= B-a + \\frac{a}{k_1+k_2+k_3+\\cdots},\n\\]\n\nwhich is minimized if \$k_1 = k_2 = k_3 = 1\$.\n\nSince we have shown the maximizing configuration for the sequence, we can proceed to show that the infinite sum is maximizing:\n\n\\[\n\\frac{1}{2} + \\frac{1}{2 \\cdot 3} + \\frac{1}{3 \\cdot 4} + \\cdots.\n\\]\n\nMoreover this sum is\n\n\\[\n< \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\frac{1}{16} + \\cdots < 1,\n\\]\n\nwhich means the first part is shown.\n\nFinally, for the second part, if\n\n\\[\n\\frac{1}{d_1 d_2} + \\frac{1}{d_2 d_3} + \\cdots = \\frac{1}{d_i},\n\\]\n\nthen if \$d_i > d_2\$ it is impossible. Also if \$d_i = 1\$ it is impossible from the first part. Thus \$d_i = d_2\$.\n\nThis only works when \$n\$ is prime, as we have\n\n\\[\n\\frac{1}{d_2} + \\cdots = \\frac{1}{d_2}\n\\]\n\nand the ``\$+\\cdots\$ needs to be zero." ]
IMO-2002-5	https://artofproblemsolving.com/wiki/index.php/2002_IMO_Problems/Problem_5	Find all functions $f:\Bbb{R}\to \Bbb{R}$ such that \[ (f(x)+f(z))(f(y)+f(t))=f(xy-zt)+f(xt+yz) \] for all real numbers $x,y,z,t$.	[ "Given the problem \$ (f(x) + f(y))(f(u) + f(v)) = f(xu - yv) + f(xv - yu) \$, we aim to find a function that satisfies it.\n\nWe start by considering the case when \$ x = y = u = v = 0 \$. This leads us to \$ 4f(0)^2 = 2f(0) \$, implying \$ f(0) = 0 \$ or \$ f(0) = 1/2 \$.\n\nIf \$ f(0) = 1 \$, then putting \$ x = y = u = 0 \$ gives us \$ f(u) = 1/2 \$ for all \$ u \\in \\mathbb{R} \$. On the other hand, if \$ f(0) = 0 \$, putting \$ y = v = 0 \$ gives us \$ f(x)f(u) = f(xu) \$, indicating that \$ f \$ is multiplicative.\n\nIf \$ f(0) = 0 \$, we have \$ f(1) = 0 \$ or \$ f(1) = 1 \$.\n\nIf \$ f(1) = 0 \$, then \$ f(x) = f(x \\cdot 1) = f(x)f(1) = 0 \$ for all \$ x \\in \\mathbb{R} \$.\n\nDisregarding constant solutions, we assume \$ f(0) = 0 \$ and \$ f(1) = 1 \$.\n\nTaking \$ x = y = 1 \$ in the original equation, we arrive at \$ 2f(u) + 2f(v) = f(u + v) + f(u - v) \$.\n\nTaking \$ u = 0 \$, we get \$ f(v) = f(-v) \$, indicating that \$ f \$ is an even function.\n\nUsing parity and taking \$ a = u \$ and \$ b = -v \$ in the original equation, we get \$ f(u^2 + v^2) = (f(u) + f(v))^2 \$.\n\nThis implies \$ f(x) > 0 \$ for all \$ x > 0 \$, allowing us to define an auxiliary function \$ g \$ as \$ g(x) = \\sqrt{f(x)} \$.\n\nThen, taking \$ a = u^2 \$ and \$ b = v^2 \$, the equation rewrites as \$ g(a+b) = g(a) + g(b) \$.\n\nThis leads us to \$ g \$ being additive, and therefore, there exists \$ m \\in \\mathbb{N} \$ such that \$ g(x) = mx \$ for all \$ x > 0 \$. Since \$ g(1) = \\sqrt{f(1)} = 1 \$, we have \$ m = 1 \$.\n\nWe will prove that \$ f \$ is increasing on \$ [0, \\infty) \$. Given \$ a > b \\geq 0 \$, we express \$ a = u^2 + v^2 \$ and \$ b = u^2 \$ for \$ u, v \\in \\mathbb{R} \$.\n\nThen, \$ f(u^2 + v^2) = (f(u) + f(v))^2 = f(u^2) + 2f(uv) + f(v^2) > f(u^2) \$, implying \$ f(a) > f(b) \$, since \$ f \$ is multiplicative.\n\nTherefore, the only solutions are \$\\boxed{f(x) = 0}, \\boxed{ f(x) = 1/2 },\$ and \$\\boxed{f(x) = x^2}\$, which can be easily verified in the original equation." ]
IMO-2002-6	https://artofproblemsolving.com/wiki/index.php/2002_IMO_Problems/Problem_6	Let $n \ge 3$ be a positive integer. Let $C_1,C_2,...,C_n$ be unit circles in the plane, with centers $O_1,O_2,...,O_n$ respectively. If no line meets more than two of the circles, prove that \[ \sum_{1\le i< j \le n}^{}\frac{1}{O_iO_j}\le\frac{(n-1)\pi}{4} \]	[]
IMO-2003-1	https://artofproblemsolving.com/wiki/index.php/2003_IMO_Problems/Problem_1	$S$ is the set $\{1, 2, 3, \dots ,1000000\}$. Show that for any subset $A$ of $S$ with $101$ elements we can find $100$ distinct elements $x_i$ of $S$, such that the sets $\{a + x_i \mid a \in A\}$ are all pairwise disjoint.	[ "Consider the set \$D=\\{x-y \\mid x,y \\in A\\}\$. There are at most \$101 \\times 100 + 1 = 10101\$ elements in \$D\$. Two sets \$A + t_i\$ and \$A + t_j\$ have nonempty intersection if and only if \$t_i - t_j\$ is in \$D\$. So we need to choose the \$100\$ elements in such a way that we do not use a difference from \$D\$. Now select these elements by induction. Choose one element arbitrarily. Assume that \$k\$ elements, \$k \\leq 99\$, are already chosen. An element \$x\$ that is already chosen prevents us from selecting any element from the set \$x + D\$. Thus after \$k\$ elements are chosen, at most \$10101k \\leq 999999\$ elements are forbidden. Hence we can select one more element." ]
IMO-2003-2	https://artofproblemsolving.com/wiki/index.php/2003_IMO_Problems/Problem_2	Determine all pairs of positive integers $(a,b)$ such that \[ \frac{a^2}{2ab^2-b^3+1} \] is a positive integer.	[ "The only solutions are of the form \$(a,b) = (2n,1)\$, \$(a,b) = (n,2n)\$, and \$(8n^4-n,2n)\$ for any positive integer \$n\$.\n\nFirst, we note that when \$b=1\$, the given expression is equivalent to \$a/2\$, which is an integer if and only if \$a\$ is even.\n\nNow, suppose that \$(a,b)\$ is a solution not of the form \$(2n,1)\$. We have already given all solutions for \$b=1\$; then for this new solution, we must have \$b>1\$. Let us denote\n\n\\[\n\\frac{a^2}{2ab^2-b^3+1} = k .\n\\]\n\nDenote\n\n\\[\nP(t) = t^2 - 2kb^2 t + k(b^3-1) .\n\\]\n\nSince \$k(b^3-1) >0\$, and \$a\$ is a positive integer root of \$P\$, there must be some other root \$a'\$ of \$P\$.\n\nWithout loss of generality, let \$a' \\ge a\$. Then \$a^2 \\le aa' = k(b^3-1)\$, so\n\n\\[\nk = \\frac{a^2}{2ab^2-b^3+1} \\le k \\frac{b^3-1}{2ab^2-b^3+1},\n\\]\n\nor\n\n\\[\n2ab^2 - (b^3-1) \\le b^3-1,\n\\]\n\nwhich reduces to\n\n\\[\na \\le b - \\frac{1}{b^2} < b .\n\\]\n\nIt follows that\n\n\\[\n0 < 2ab^2 -b^3 + 1 \\le a^2 < b^2 ,\n\\]\n\nor\n\n\\[\n0 < (2a-b)b^2 + 1 < b^2 .\n\\]\n\nSince \$a\$ and \$b\$ are integers, this can only happen when \$2a-b=0\$, so \$(a,b)\$ can be written as \$(n,2n)\$, and \$k = n^2\$. It follows that\n\n\\[\na' = \\frac{k(b^3-1)}{a} = 8n^4-n.\n\\]\n\nSince \$a'\$ is the other root of \$P\$, it follows that \$(a',b)\$ also satisfies the problem's condition. Therefore the solutions are exactly the ones given at the solution's start. \$\\blacksquare\$\n\n## Resources\n\n- <url>Forum/viewtopic.php?p=262#262 Discussion on AoPS/MathLinks</url>" ]
IMO-2003-3	https://artofproblemsolving.com/wiki/index.php/2003_IMO_Problems/Problem_3	Each pair of opposite sides of convex hexagon has the property that the distance between their midpoints is $\frac{\sqrt{3}}{2}$ times the sum of their lengths. Prove that the hexagon is equiangular.	[]
IMO-2003-4	https://artofproblemsolving.com/wiki/index.php/2003_IMO_Problems/Problem_4	Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, and $R$ be the feet of perpendiculars from $D$ to lines $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively. Show that $PQ=QR$ if and only if the bisectors of angles $ABC$ and $ADC$ meet on segment $\overline{AC}$.	[ "Clearly \$PQR\$ is the Simson Line and \$APDQ\$, \$BPDR\$, \$CQDR\$ is cyclic. By angle chasing we have \$\\triangle DPQ\\sim\\triangle DBC\$, \$\\triangle DQR\\sim\\triangle DAB\$. Then by \$PQ=QR\$ we have \$\\frac{DC}{CB}=\\frac{DQ}{QP}=\\frac{DQ}{QR}=\\frac{DA}{AB}\$. Rearranging and using the angle bisector theorem we are done." ]
IMO-2003-5	https://artofproblemsolving.com/wiki/index.php/2003_IMO_Problems/Problem_5	Let $n$ be a positive integer and let $x_1 \le x_2 \le \cdots \le x_n$ be real numbers. Prove that \[ \left( \sum_{i=1}^{n}\sum_{j=i}^{n} \|x_i-x_j\|\right)^2 \le \frac{2(n^2-1)}{3}\sum_{i=1}^{n}\sum_{j=i}^{n}(x_i-x_j)^2 \] with equality if and only if $x_1, x_2, ..., x_n$ form an arithmetic sequence.	[ "We have \\begin{align}\\left(\\sum_{i,j=1}^{n}\|x_i-x_j\|\\right)^2 &=\\left(2\\sum_{1\\le i\\le j\\le n}(x_j-x_i)\\right)^2 \\\\ &= \\left((2n-2)x_n+(2n-6)x_{n-1}+\\dots +(2-2n)x_1\\right)^2 \\\\ &\\le ((2n-2)^2+(2n-6)^2+(2n-10)^2+\\dots + (2-2n)^2)(x_1^2+x_2^2+\\dots + x_n^2) \\\\ &= \\frac{4(n-1)(n)(n+1)}{3}(x_1^2+x_2^2+\\dots + x_n^2) \\\\ &= \\frac{2(n^2-1)}{3}\\cdot 2(nx_1^2 + nx_2^2 + \\dots + nx_n^2) \\\\ &= \\frac{2(n^2-1)}{3}\\cdot 2\\left((n-1)\\left(\\sum_{i=1}^{n}{x_i^2}\\right) + \\left(\\sum_{i=1}^{n}{x_i}\\right)^2 - 2\\sum_{1\\le i<j\\le n}x_ix_j\\right) \\\\ &= \\frac{2(n^2-1)}{3}\\cdot 2\\left(\\sum_{1\\le i<j\\le n}(x_i-x_j)^2\\right) \\\\ &= \\frac{2(n^2-1)}{3}\\sum_{i,j=1}^{n}(x_i-x_j)^2 \\end{align}" ]
IMO-2003-6	https://artofproblemsolving.com/wiki/index.php/2003_IMO_Problems/Problem_6	Let $p$ be a prime number. Prove that there exists a prime number $q$ such that for every integer $n$, the number $n^p-p$ is not divisible by $q$.	[ "Let N be \$1 + p + p^2 + ... + p^{p-1}\$ which equals \$\\frac{p^p-1}{p-1}\$ \$N\\equiv{p+1}\\pmod{p^2}\$ Which means there exists q which is a prime factor of n that doesn't satisfy \$q\\equiv{1}\\pmod{p^2}\$. \\\\unfinished", "For \$p\$ prime and \$gcd(n, p) = 1\$, \$n^{p}\\equiv{n}\\pmod{p}\$ simply by Fermat's Little Theorem. Therefore, \$n^{p} - p \\equiv{n}\\pmod{p}\$. We are going to prove by contradiction. Let \$q\$ be a prime that divides \$n^{p} - p\$. If \$q\$ divides this quantity, then it must also divide \$pk + n\$. But on the other hand, clearly \$p\$ doesn't divide this quantity. \$p\$ is a prime and so is \$q\$. If \$p\$ doesn't divide this, then no other prime should divide \$n^{p} - p\$ so we have proved the problem statement is true for \$gcd(n, any prime) = 1\$. Now, consider the case where \$gcd(n, p) = p\$. Then, \$n^{p} - p = p^{p}k^{p} - p\$ which is divisible by \$p\$. Notice \$n^{p} = p^{p}k^{p}\$. This has \$(p + 1)^{2}\$ factors if \$k\$ is prime. We assumed that \$q\$ also divided this original quantity. If \$q\$ divided this quantity, then the expression must have at most \$4\$ factors: \$1, p , q, pq\$. In fact, since \$p\$ is prime, \$(p + 1)^{2}\$ is at least \$9\$ meaning this tends towards the claim that we must have at least \$9\$ factors. But this is absurd! So we have proved the problem statement is true if \$k\$ is prime. If \$k\$ isn't prime, then by inspection, we will have way more than \$9\$ factors, and by similar logic, \$q\$ cannot divide the above quantity. Therefore, we are done.\n\n(This is my first IMO problem 6 and the solution is probably all wrong. If someone can correct my solution mistake(s), feel free to edit my solution or put down comments.)" ]
IMO-2004-1	https://artofproblemsolving.com/wiki/index.php/2004_IMO_Problems/Problem_1	Let $ABC$ be an acute-angled triangle with $AB\neq AC$. The circle with diameter $BC$ intersects the sides $AB$ and $AC$ at $M$ and $N$ respectively. Denote by $O$ the midpoint of the side $BC$. The bisectors of the angles $\angle BAC$ and $\angle MON$ intersect at $R$. Prove that the circumcircles of the triangles $BMR$ and $CNR$ have a common point lying on the side $BC$.	[ "\\[\n[asy] unitsize(2.5cm); size(60); pair A,B,C,O,K,M,N,R,D; C=(0,0); B=(3.2,0); A=(1.1,3.35); O=(1.6,0); K=(1.51,0); R=(1.42,0.73); M=(2.3,1.44); N=(0.31,0.95); D=(1.1, 2.02); draw(arc((1.6,0),1.6,0,180)); draw(A--B--C--cycle); draw(M--N); draw(A--K); draw(O--D); draw(R--M); draw(C--R); draw(B--R); draw(N--R); draw(N--O); draw(O--M); draw(N--D); draw(M--D); draw(A--D); draw(circumcircle(A,M,N)); draw(circumcircle(C,R,N)); draw(circumcircle(B,M,R)); dot(A); dot(B); dot(C); dot(O); dot(K); dot(M); dot(N); dot(R); dot(D); label(\"$A$\",A,N); label(\"$B$\",B,SE); label(\"$O$\",O,SE); label(\"$K$\",(1.38,-0.1)); label(\"$C$\",C,SW); label(\"$M$\",M,NE); label(\"$N$\",(0.24,1.13)); label(\"$R$\",(1.3,0.59)); markscalefactor=0.025; draw(anglemark(R,N,M),blue); draw(anglemark(N,M,R),blue); draw(anglemark(N,A,R),blue); draw(anglemark(R,A,M),blue); draw(anglemark(O,C,N),green); draw(anglemark(C,N,O),green); draw(anglemark(A,M,N),green); draw(anglemark(M,N,A),red); draw(anglemark(M,B,O),red); draw(anglemark(O,M,B),red); draw(anglemark(O,N,R),yellow); draw(anglemark(R,M,O),yellow); markscalefactor=0.01; draw(rightanglemark(N,(1.305,1.195),O)); [/asy]\n\\]\n\nLet \$\\angle{ACB} = a\$, \$\\angle{CBA} = b\$, and \$\\angle{ONR} = k\$. Call \$\\omega\$ the circle with diameter \$BC\$ and \$\\odot{AMN}\$ the circumcircle of \$\\triangle{AMN}\$.\n\nOur ultimate goal is to show that \$\\angle{CNR} + \\angle{RMB} = 180^\\circ\$. To show why this solves the problem, assume this statement holds true. Call \$K\$ the intersection point of the circumcircle of \$\\triangle{CNR}\$ with side \$BC\$. Then, \$\\angle{RKC} = 180^\\circ - \\angle{CNR}\$, and \$\\angle{RKB} = \\angle{CNR}\$. Since \$\\angle{RMB} = 180^\\circ - \\angle{CNR}\$, \$\\angle{RKB} + \\angle{RMB} = 180^\\circ\$, implying \$K\$ also lies on the circumcircle of \$\\triangle{BMR}\$, thereby solving the problem.\n\nWe now prove that \$\\angle{CNR} + \\angle{RMB} = 180^\\circ\$. Note that \$ON\$ and \$OM\$ are radii of \$\\omega\$, so \$\\triangle{MON}\$ is isosceles. The bisector of \$\\angle{MON}\$ is thus the perpendicular bisector of \$MN\$. Since \$R\$ lies on the bisector of \$\\angle{MON}\$, \$\\angle{ONR} = \\angle{RMO} = k\$. Angle computations yield that\n\n\\[\n\\angle{RMN} = 180^\\circ - \\angle{BMO} - \\angle{RMO} - \\angle{AMN}\n\\]\n\n\\[\n= 180^\\circ - b - k - a,\n\\]\n\nfrom \$\\angle{AMN} = 180^\\circ - \\angle{BMN} = 180^\\circ - (180^\\circ - \\angle{ACB}) = \\angle{ACB}\$ and from \$\\angle{BMO} = \\angle{MBO} =\$ \$\\angle{ABC}\$.\n\nThe bisector of \$\\angle{BAC}\$ hits \$\\odot{AMN}\$ at the midpoint of the arc \$MN\$ not containing \$A\$. This point must lie on the perpendicular bisector of segment \$MN\$, which is the bisector of \$\\angle{MON}\$. It follows that \$R\$ is indeed the midpoint of arc \$MN\$, so \$A\$, \$M\$, \$R\$, \$N\$, are concyclic. Since \$\\angle{RAN}\$ and \$\\angle{RMN}\$ subtend the same arc \$NR\$, \$\\angle{RAN}\$ = \$\\angle{RMN}\$. With \$AR\$ being the bisector of \$\\angle{BAC}\$, we have\n\n\\[\n\\angle{RAN} = \\angle{RMN} = \\tfrac{\\angle{CAB}}{2} = \\tfrac{180^\\circ - a - b}{2} = 90^\\circ - \\tfrac{a}{2} - \\tfrac{b}{2}.\n\\]\n\nWe know that \$\\angle{RMN} = 180^\\circ - b - k - a\$. so we have \$180^\\circ - b - k - a = 90^\\circ - \\tfrac{a}{2} - \\tfrac{b}{2}\$. Since \$\\angle{CNR} = \\angle{CNO} + \\angle{ONR} = a + k\$, and \$\\angle{RMB} = \\angle{OMB} + \\angle{RMO} = b + k\$, we have\n\n\\[\n\\angle{CNR} + \\angle{RMB} = a + k + b + k = a + b + 180^\\circ - a - b = 180^\\circ.\n\\]\n\nThe problem is solved.\n\n\$\\textbf{NOTE:}\$ We have \$\\angle{RKB} + \\angle{KBA} + \\angle{BAK} = 180^\\circ \\implies \\angle{BAK} = 180^\\circ - (180^\\circ - \\angle{RMB}) - \\angle{KBA}\$. Noting that \$180^\\circ - \\angle{RMB} = 180^\\circ - b - k = 180^\\circ - (90^\\circ - \\tfrac{a}{2} + \\tfrac{b}{2}) = 90^\\circ + \\tfrac{a}{2} - \\tfrac{b}{2}\$, we then have\n\n\\[\n\\angle{BAK} = 180^\\circ - (90^\\circ + \\tfrac{a}{2} - \\tfrac{b}{2}) - b = 90^\\circ - \\tfrac{a}{2} - \\tfrac{b}{2}\n\\]\n\nwhich is indeed the measure of \$\\angle{BAR}\$. This implies that \$K\$ lies on the bisector of \$\\angle{BAC}\$, and from this, it is clear that \$K\$ must lie on the interior of segment \$BC\$. Not proving that \$K\$ had to lie in the interior of \$BC\$ was a reason that a large portion of students who submitted a solution received a 1-point deduction." ]
IMO-2004-2	https://artofproblemsolving.com/wiki/index.php/2004_IMO_Problems/Problem_2	Find all polynomials $f$ with real coefficients such that for all reals $a,b,c$ such that $ab + bc + ca = 0$ we have the following relations \[ f(a - b) + f(b - c) + f(c - a) = 2f(a + b + c). \]	[ "From \$b=c=0\$, we have \$f(a) + f(-a) = 2f(a) \\Longrightarrow f(a) = f(-a)\$, so \$f\$ is even, and all the degrees all of its terms are even. Let \$\\text{deg}\\, f(x) = n\$\n\nLet \$(a,b,c) = (6x, 3x, -2x)\$*; then we have \$f(3x) + f(5x) + f(8x) = 2f(7x)\$. Comparing lead coefficients, we have \$3^n + 5^n + 8^n = 2 \\cdot 7^n\$, which cannot be true for \$n \\ge 5\$. Hence, we have \$f(x) = c_1 x^4 + c_2 x^2\$. We can easily verify by expanding that all such polynomials work.\n\n- The substitution arises from writing \$a = \\frac{-bc}{b+c}\$.", "Let \$a = (1 - \\sqrt {3})x\$, \$b = x\$, and \$c = (1 + \\sqrt {3})x\$. Then it is easy to check that \$ab + bc + ca = 0\$, so\n\n\\[\nP( - \\sqrt {3}x) + P( - \\sqrt {3}x) + P(2 \\sqrt {3} x) = 2P(3x)\n\\]\n\nfor all \$x\$. Hence, for the coefficient of \$x^n\$ to be nonzero, we must have \$2( - \\sqrt {3})^n + (2 \\sqrt {3})^n = 2 \\cdot 3^n\$.\n\nThis does not hold for \$n = 1\$, and if \$n\$ is odd and \$n \\ge 3\$, then the LHS is irrational and the RHS is a positive integer, so \$n\$ must be even.\n\nLet \$n = 2m\$. Then \$2 \\cdot 3^m + 12^m = 2 \\cdot 3^{2m}\$, so \$2 + 4^m = 2 \\cdot 3^m\$. This holds for \$m = 1\$ and \$m = 2\$, and \$(4/3)^3 = 64/27 > 2\$, so \$2 + 4^m > 4^m > 2 \\cdot 3^m\$ for \$m \\ge 3\$. Therefore, \$P(x)\$ must be of the form \$a_2 x^2 + a_4 x^4\$." ]
IMO-2004-3	https://artofproblemsolving.com/wiki/index.php/2004_IMO_Problems/Problem_3	Define a "hook" to be a figure made up of six unit squares as shown below in the picture, or any of the figures obtained by applying rotations and reflections to this figure. \[ [asy] unitsize(0.5 cm); draw((0,0)--(1,0)); draw((0,1)--(1,1)); draw((2,1)--(3,1)); draw((0,2)--(3,2)); draw((0,3)--(3,3)); draw((0,0)--(0,3)); draw((1,0)--(1,3)); draw((2,1)--(2,3)); draw((3,1)--(3,3)); [/asy] \] Determine all $m \times n$ rectangles that can be covered without gaps and without overlaps with hooks such that; (a) the rectangle is covered without gaps and without overlaps, (b) no part of a hook covers area outside the rectangle.	[]
IMO-2004-4	https://artofproblemsolving.com/wiki/index.php/2004_IMO_Problems/Problem_4	Let $n \geq 3$ be an integer. Let $t_1, t_2, \dots , t_n$ be positive real numbers such that \[ n^2 + 1 > \left( t_1 + t_2 + ... + t_n \right) \left( \frac {1}{t_1} + \frac {1}{t_2} + ... + \frac {1}{t_n} \right). \] Show that $t_i$, $t_j$, $t_k$ are side lengths of a triangle for all $i$, $j$, $k$ with $1 \leq i < j < k \leq n$.	[ "For \$n=3\$, suppose (for sake of contradiction) that \$t_3 = t_2 + t_1 + k\$ for \$k \\ge 0\$; then (by Cauchy-Schwarz Inequality)\n\n\\[\n\\begin{align}10 &> [2(t_1 + t_2) + k]\\left(\\frac {1}{t_1} + \\frac {1}{t_2} + \\frac 1{t_1 + t_2 + k}\\right) = 2(t_1+t_2)\\left(\\frac 1{t_1} + \\frac{1}{t_2}\\right) + \\left(\\frac{k}{t_1} + \\frac k{t_2}\\right) + \\frac{2t_1 + 2t_2 + k}{t_1 + t_2 + k}\\\\ &\\ge 8 + \\left(\\frac{k}{t_1} + \\frac k{t_2}\\right) + 2 - \\frac{k}{t_1 + t_2 + k}\\\\ &= 10 + k\\left(\\frac{1}{t_1} + \\frac {1}{t_2} - \\frac{1}{t_1 + t_2+k}\\right) \\ge 10\\end{align}\n\\]\n\nso it is true for \$n=3\$. We now claim the result by induction; for \$n \\ge 4\$, we have\n\n\\[\n\\begin{align}f(n) &:= \\left( t_1 + t_2 + ... + t_n \\right) \\left( \\frac {1}{t_1} + \\frac {1}{t_2} + ... + \\frac {1}{t_n} \\right) \\\\ &= \\left( t_1 + t_2 + ... + t_{n-1} \\right) \\left( \\frac {1}{t_1} + \\frac {1}{t_2} + ... + \\frac {1}{t_{n-1}} \\right) + t_n \\sum_{i=1}^{n-1} \\frac 1{t_i} + \\frac{1}{t_n} \\sum_{i=1}^{n-1} t_i + 1\\end{align}\n\\]\n\nBy AM-GM, \$\\frac{t_n}{t_i} + \\frac{t_i}{t_n} \\ge 2\$, so \$f(n) \\ge f(n-1) + 2(n-1) + 1 = f(n-1) + 2n - 1\$. Then the problem is reduced to proving the statement true for \$n-1\$ numbers, as desired.\$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\square\$" ]
IMO-2004-5	https://artofproblemsolving.com/wiki/index.php/2004_IMO_Problems/Problem_5	In a convex quadrilateral $ABCD$, the diagonal $BD$ bisects neither the angle $ABC$ nor the angle $CDA$. The point $P$ lies inside $ABCD$ and satisfies \[ \angle PBC = \angle DBA \text{ and } \angle PDC = \angle BDA. \] Prove that $ABCD$ is a cyclic quadrilateral if and only if $AP = CP.$	[ "Assume \$ABCD\$ is cyclic, let \$K\$ be the intersection of \$AC\$ and \$BE\$, let \$L\$ be the intersection of \$AC\$ and \$DF\$,\n\n\\[\n[asy] size(6cm); draw(circle((0,0),7.07)); draw((-3.7,-6)-- (3.7,-6)); draw((-6.8,-2)-- (6.8,-2)); draw((-5,5)-- (5,5)); draw((-5,5)-- (-3.7,-6)); draw((-5,5)-- (3.7,-6)); draw((-5,5)-- (-6.8,-2)); draw((-5,5)-- (6.8,-2)); draw((5,5)-- (-3.7,-6)); draw((5,5)-- (3.7,-6)); draw((5,5)-- (-6.8,-2)); draw((5,5)-- (6.8,-2)); draw((-3.7,-6)-- (-6.8,-2)); draw((-3.7,-6)-- (6.8,-2)); draw((3.7,-6)-- (-6.8,-2)); draw((3.7,-6)-- (6.8,-2)); label(\"$A$\", (-6.8,-2), SW); label(\"$B$\", (-3.7,-6), SW); label(\"$F$\", (3.7,-6), SE); label(\"$C$\", (6.8,-2), E); label(\"$E$\", (5,5), E); label(\"$D$\", (-5,5), W); label(\"$P$\", (0,-1.3), N); label(\"$K$\", (-1.6,-1.5), E); label(\"$L$\", (0.8,-1.5) ); [/asy]\n\\]\n\n\$\\angle PBC=\\angle DBA\$, so \$AD=CE\$, and \$DE//AC\$. \$\\angle PDC=\\angle BDA\$, so \$AB=CF\$, and \$AC//BF\$.\n\n~szhangmath" ]
IMO-2004-6	https://artofproblemsolving.com/wiki/index.php/2004_IMO_Problems/Problem_6	We call a positive integer alternating if every two consecutive digits in its decimal representation are of different parity. Find all positive integers $n$ which have an alternating multiple.	[ "We claim that all positive integers \$n\$ except multiples of \$20\$ have a multiple that is alternating. If \$n\$ is a multiple of \$20\$, then the units digit is \$0\$ and the tens digit is a multiple of \$2\$, so both digits are even. Now, we will prove that if \$20\\nmid n\$, then there is a multiple of \$n\$ that is alternating.\n\nWe claim that there exists a \$2k\$ digit alternating integer that is a multiple of \$2^{2k+1}\$ and there exists an alternating integer with at most \$k\$ digits that is a multiple of \$2\\cdot5^k\$. We will prove this by induction.\n\nBase Case: \$k=1\$ We can let the number be \$16\$.\n\nInductive Step: Let \$a\$ be a \$2k-2\$ digit alternating number such that \$2^{2k-1}\\mid a\$. Since \$a\$ is even, this means that the first digit of \$a\$ is odd. Let \$b\\cdot2^{2k-1}\\equiv a\\pmod{2^{2k+1}}\$, where \$b\\in\\{0,1,2,3\\}\$. Let \$x=16-2b\$. We see that \$x\$ is alternating, so the number \$x\\cdot10^{2k-2}+a\$ is also alternating. We also have \$x\\equiv-2b\\pmod8\$. Since \$a\\equiv b\\cdot2^{2k-1}\\pmod{2^{2k+1}}\$, this means that we have \\begin{align} x\\cdot10^{2k-2}+a&\\equiv x\\cdot10^{2k-2}+b\\cdot2^{2k-1}\\pmod{2^{2k+1}}\\\\ &\\equiv2^{2k-2}\\left(x\\cdot5^{2k-2}+2b\\right)\\pmod{2^{2k+1}}\\\\ &\\equiv2^{2k-2}\\left(-2b\\cdot5^{2k-2}+2b\\right)\\pmod{2^{2k+1}}\\\\ &\\equiv -b\\cdot2^{2k-1}\\left(5^{2k-2}-1\\right)\\pmod{2^{2k+1}}.\\end{align} Since \$4\\mid5^{2k-2}-1\$, this means that \$x\\cdot10^{2k-2}+a\\equiv0\\pmod{2^{2k+1}}\$, so \$x\\cdot10^{2k-2}+a\$ is a \$2k\$ digit alternating number that is a multiple of \$2^{2k+1}\$.\n\nNow, we will prove that there exists an alternating integer with at most \$k\$ digits that is a multiple of \$2\\cdot5^k\$.\n\nBase Case: \$k=1\$ We can let the number be \$0\$.\n\nInductive Step: Let \$a\$ be a \$k-1\$ digit alternating number such that \$2\\cdot5^{k-1}\\mid a\$. Let \$S\$ be the set of digits such that \$x\\cdot10^{k-1}+a\$ is alternating. We see that either \$S=\\{0,2,4,6,8\\}\$ or \$S=\\{1,3,5,7,9\\}\$. In each possible set, each \$\\mod5\$ residue appears exactly once. Let \$a=2\\cdot5^{k-1}\\cdot b\$. Then, \$x\\cdot10^{k-1}+a=2\\cdot5^{k-1}\\left(x\\cdot2^{k-2}+b\\right)\$. Therefore, there exists an \$x\$ such that \$2\\cdot5^k\\mid x\\cdot10^{k-1}+a\$, so there exists an alternating integer with at most \$k\$ digits that is a multiple of \$2\\cdot5^k\$. If \$k\$ is even, then \$0\\not\\in S\$, so it has exactly \$k\$ digits.\n\nNow, let \$n=2^a\\cdot5^b\\cdot m\$. Then, if \$20\\nmid n\$, then \$a<2\$ or \$b=0\$. If \$b=0\$, then there exists a \$2a\$ digit alternating integer \$x\$ that is a multiple of \$2^{2a+1}\$. Consider the sequence \$a_i=x\\cdot\\frac{10^{2ai}-1}{10^{2a}-1}\$. Then, the decimal representation of \$a_i\$ is the result when \$x\$ is written \$i\$ times. For any prime \$p\$ that divides \$n\$, if \$p\\mid10^{2a}-1\$, then \$\\nu_p\\left(10^{2ai}-1\\right)=\\nu_p\\left(10^{2a}-1\\right)+\\nu_p(i)\$, so \$\\nu_p(a_i)=\\nu_p(x)+\\nu_p(i)\\geq\\nu_p(i)\$. Therefore, there exists an \$i\$ such that \$p^{\\nu_p(n)}\\mid a_{ci}\$ for all \$c\$. If \$p\\nmid10^{2a}-1\$, then if \$i=\\phi(n)\$, then \$p^{\\nu_p(n)}\\mid10^{2ai}-1\$ if \$p\\neq2,5\$. Since \$5\\nmid n\$, this means that we only need to make sure \$2^a\\mid x\\cdot\\frac{10^{2ai}-1}{10^{2a}-1}\$. Since \$2^{2a+1}\\mid x\$, this means that this is true, so if \$5\\nmid n\$, then there exists a multiple of \$n\$ that is alternating.\n\nIf \$a<2\$, then there exists an alternating integer \$x\$ with exactly \$2b\$ digits that is a multiple of \$2\\cdot5^{2b}\$. Then, the first digit of \$x\$ is odd. Let \$a_i=x\\cdot\\frac{10^{2bi}-1}{10^{2b}-1}\$. Then, if \$p\\mid10^{2b}-1\$, then \$\\nu_p(a_i)\\geq\\nu_p(i)\$, so there exists an \$i\$ such that \$p^{\\nu_p(n)}\\mid a_i\$. If \$p\\nmid10^{2a}-1\$, then we can let \$i=\\phi(n)\$, so \$p^{\\nu_p(n)}\\mid a_i\$. If \$p=5\$, then \$5^b\\mid5^{2b}\\mid x\$, and if \$p=2\$, then \$a=0,1\$, so \$p^a\\mid x\$. Therefore, \$n\\mid a_i\$ for some \$i\$.\n\nTherefore, all positive integers except multiples of \$20\$ have an alternating multiple." ]
IMO-2005-1	https://artofproblemsolving.com/wiki/index.php/2005_IMO_Problems/Problem_1	Six points are chosen on the sides of an equilateral triangle $ABC$: $A_1, A_2$ on $BC$, $B_1$, $B_2$ on $CA$ and $C_1$, $C_2$ on $AB$, such that they are the vertices of a convex hexagon $A_1A_2B_1B_2C_1C_2$ with equal side lengths. Prove that the lines $A_1B_2, B_1C_2$ and $C_1A_2$ are concurrent.	[ "Let \$D = C_2A_1 \\cap A_2B_1\$, and similarly define \$E\$ and \$F\$. We claim that \$DEF\$ is equilateral.\n\nProof: Note that the vectors \$\\overrightarrow{A_1A_2}\$, \$\\overrightarrow{A_2B_1}\$, \$\\overrightarrow{B_1B_2}\$, \$\\overrightarrow{B_2C_1}\$, \$\\overrightarrow{C_1C_2}\$, \$\\overrightarrow{C_2A_1}\$ add to 0. But we also have:\n\n\\[\n\\overrightarrow{A_1A_2}+\\overrightarrow{B_1B_2}+\\overrightarrow{C_1C_2}=0\n\\]\n\nsince these are at the sides of an equilateral triangle. This means that\n\n\\[\n\\overrightarrow{A_2B_1}+\\overrightarrow{B_2C_1}+\\overrightarrow{C_2A_1}=0.\n\\]\n\nThese three unit vectors thus form an equilateral triangle.\n\nNow \$\\angle A_1DB_1 = 60^{\\circ} = \\angle A_1CB_1\$, so quadrilateral \$A_1DCB_1\$ is cyclic. So \$\\angle CA_1D = \\angle CB_1D\$ and \$\\angle C_2A_1A_2 = \\angle A_2B_1B_2\$. This means that triangles \$C_2A_1A_2\$ and \$A_2B_1B_2\$ are congruent, so \$C_2A_2 = A_2B_2\$. But also, \$C_2C_1 = C_1B_2\$, so \$C_1A_2\$ is the perpendicular bisector of \$A_1B_1C_1\$. Similarly, \$A_1B_2\$ and \$B_1C_2\$ are also perpendicular bisectors. Therefore, they concur." ]
IMO-2005-2	https://artofproblemsolving.com/wiki/index.php/2005_IMO_Problems/Problem_2	Let $a_1, a_2, \dots$ be a sequence of integers with infinitely many positive and negative terms. Suppose that for every positive integer $n$ the numbers $a_1, a_2, \dots, a_n$ leave $n$ different remainders upon division by $n$. Prove that every integer occurs exactly once in the sequence.	[ "\${a_n}\$ satisfies the conditions if and only if \${a_n-a_1}\$ does. Therefore we can assume that \$a_1=0\$\n\nFirst of all, \$\|a_n\|<n\$ Otherwise, \$a_n = 0\$ mod \$a_{a_n}\$\n\nClaim: \$a_{k+1}\$ is either the smallest positive number not in \${a_1, a_2, ..., a_k}\$ or the largest negative number not in this set.\n\n- Proof by induction: the induction hypothesis implies that \$a_1, a_2, ..., a_k\$ are consecutive. Let m and M be the smallest and largest numbers in this consecutive set, respectively. It is clear that \$a_{k+1}=m-1=M+1\$ mod \$k+1\$. But since \$\|a_n\|<n\$, \$a_{k+1}=m-1\$ or \$M+1\$\n\nThis proves that if \${a_n}\$ contains an infinite number of positive and negative numbers, it must contain each integer exactly once.\n\n~Kscv" ]
IMO-2005-3	https://artofproblemsolving.com/wiki/index.php/2005_IMO_Problems/Problem_3	Let $x, y, z > 0$ satisfy $xyz\ge 1$. Prove that \[ \frac{x^5-x^2}{x^5+y^2+z^2} + \frac{y^5-y^2}{x^2+y^5+z^2} + \frac{z^5-z^2}{x^2+y^2+z^5} \ge 0. \]	[]
IMO-2005-4	https://artofproblemsolving.com/wiki/index.php/2005_IMO_Problems/Problem_4	Determine all positive integers relatively prime to all the terms of the infinite sequence \[ a_n=2^n+3^n+6^n -1,\ n\geq 1. \]	[ "Let \$k\$ be a positive integer that satisfies the given condition.\n\nFor all primes \$p>3\$, by Fermat's Little Theorem, \$n^{p-1} \\equiv 1\\pmod p\$ if \$n\$ and \$p\$ are relatively prime. This means that \$n^{p-3} \\equiv \\frac{1}{n^2} \\pmod p\$. Plugging \$n = p-3\$ back into the equation, we see that the value\$\\mod p\$ is simply \$\\frac{2}{9} + \\frac{3}{4} + \\frac{1}{36} - 1 = 0\$. Thus, the expression is divisible by all primes \$p>3.\$ Since \$a_2 = 48 = 2^4 \\cdot 3,\$ we can conclude that \$k\$ cannot have any prime divisors. Therefore, our answer is only \$1.\$" ]
IMO-2005-5	https://artofproblemsolving.com/wiki/index.php/2005_IMO_Problems/Problem_5	Let $ABCD$ be a fixed convex quadrilateral with $BC = DA$ and $BC \nparallel DA$. Let two variable points $E$ and $F$ lie of the sides $BC$ and $DA$, respectively, and satisfy $BE = DF$. The lines $AC$ and $BD$ meet at $P$, the lines $BD$ and $EF$ meet at $Q$, the lines $EF$ and $AC$ meet at $R$. Prove that the circumcircles of the triangles $PQR$, as $E$ and $F$ vary, have a common point other than $P$.	[]
IMO-2005-6	https://artofproblemsolving.com/wiki/index.php/2005_IMO_Problems/Problem_6	In a mathematical competition, in which 6 problems were posed to the participants, every two of these problems were solved by more than 2/5 of the contestants. Moreover, no contestant solved all the 6 problems. Show that there are at least 2 contestants who solved exactly 5 problems each.	[]
IMO-2006-1	https://artofproblemsolving.com/wiki/index.php/2006_IMO_Problems/Problem_1	Let $ABC$ be triangle with incenter $I$. A point $P$ in the interior of the triangle satisfies $\angle PBA+\angle PCA = \angle PBC+\angle PCB$. Show that $AP \geq AI$, and that equality holds if and only if $P=I.$	[ "We have\n\n\\[\n\\angle IBP = \\angle IBC - \\angle PBC = \\frac{1}{2} \\angle ABC - \\angle PBC = \\frac{1}{2}(\\angle PCB - \\angle PCA).\n\\]\n\nand similarly\n\n\\[\n\\angle ICP = \\angle PCB - \\angle ICB = \\angle PCB - \\frac{1}{2} \\angle ACB = \\frac{1}{2}(\\angle PBA - \\angle PBC).\n\\]\n\nSince \$\\angle PBA + \\angle PCA = \\angle PBC + \\angle PCB\$, we have \$\\angle PCB - \\angle PCA = \\angle PBA - \\angle PBC.\$\n\nIt follows that\n\n\\[\n\\angle IBP = \\frac{1}{2} (\\angle PCB - \\angle PCA) = \\frac{1}{2} (\\angle PBA - \\angle PBC) = \\angle ICP.\n\\]\n\nHence, \$B,P,I,\$ and \$C\$ are concyclic.\n\nLet ray \$AI\$ meet the circumcircle of \$\\triangle ABC\\,\$ at point \$J\$. Then, by the Incenter-Excenter Lemma, \$JB=JC=JI=JP\$.\n\nFinally, \$AP+JP \\geq AJ = AI+IJ\$ (since triangle APJ can be degenerate, which happens only when \$P=I\$), but \$JI=JP\$; hence \$AP \\geq AI\$ and we are done.\n\nBy Mengsay LOEM , Cambodia IMO Team 2015\n\nlatexed by tluo5458 :)\n\nminor edits by lpieleanu" ]
IMO-2006-2	https://artofproblemsolving.com/wiki/index.php/2006_IMO_Problems/Problem_2	Let $P$ be a regular 2006-gon. A diagonal of $P$ is called good if its endpoints divide the boundary of $P$ into two parts, each composed of an odd number of sides of $P$. The sides of $P$ are also called good. Suppose $P$ has been dissected into triangles by 2003 diagonals, no two of which have a common point in the interior of $P$. Find the maximum number of isosceles triangles having two good sides that could appear in such a configuration.	[ "Call an isosceles triangle odd if it has two odd sides. Suppose we are given a dissection as in the problem statement. A triangle in the dissection which is odd and isosceles will be called iso-odd for brevity. Let ABC be an iso-odd triangle, with AB and BC odd sides. This means that there are an odd number of sides of the 2006-gon between A and B and also between B and C. We say that these sides belong to the iso-odd triangle ABC. At least one side in each of these groups does not belong to any other iso-odd triangle. This is so because any odd triangle whose vertices are among the points between A and B has two sides of equal length and therefore has an even number of sides belonging to it in total. Eliminating all sides belonging to any other iso-odd triangle in this area must therefore leave one side that belongs to no other iso-odd triangle. Let us assign these two sides (one in each group) to the triangle ABC. To each iso-odd triangle we have thus assigned a pair of sides, with no two triangles sharing an assigned side. It follows that at most 1003 iso-odd triangles can appear in the dissection. This value can be attained, as shows the example from the ﬁrst solution." ]
IMO-2006-3	https://artofproblemsolving.com/wiki/index.php/2006_IMO_Problems/Problem_3	Determine the least real number $M$ such that the inequality \[ \left\| ab(a^{2}-b^{2})+bc(b^{2}-c^{2})+ca(c^{2}-a^{2})\right\|\leq M(a^{2}+b^{2}+c^{2})^{2} \] holds for all real numbers $a,b$ and $c$.	[]
IMO-2006-4	https://artofproblemsolving.com/wiki/index.php/2006_IMO_Problems/Problem_4	Determine all pairs $(x, y)$ of integers such that \[ 1+2^{x}+2^{2x+1}= y^{2}. \]	[ "If \$(x,y)\$ is a solution then obviously \$x\\geq 0\$ and \$(x,-y)\$ is a solution too. For \$x=0\$ we get the two solutions \$(0,2)\$ and \$(0,-2)\$.\n\nNow let \$(x,y)\$ be a solution with \$x > 0\$; without loss of generality confine attention to \$y > 0\$. The equation rewritten as\n\n\\[\n2^x(1+2^{x+1}) = (y-1)(y+1)\n\\]\n\nshows that the factors \$y-1\$ and \$y+1\$ are even, exactly one of them divisible by \$4\$. Hence \$x\\geq 3\$ and one of these factors is divisible by \$2^{x-1}\$ but not by \$2^x\$. So\n\n\\[\ny = 2^{x-1}m + \\varepsilon, \\qquad m\\text{ odd},\\qquad \\varepsilon = \\pm 1.\\qquad\\qquad()\n\\]\n\nPlugging this into the original equation we obtain\n\n\\[\n2^x(1+2^{x+1}) = (2^{x+1}m + \\varepsilon)^2 - 1 = 2^{2x-2}m^2 + 2^xm\\varepsilon,\n\\]\n\nor, equivalently\n\n\\[\n1 + 2^{x+1} = 2^{x-2}m^2 + m\\varepsilon.\n\\]\n\nTherefore\n\n\\[\n1 - \\varepsilon m = 2^{x-2}(m^2 - 8).\\qquad(\\dagger)\n\\]\n\nFor \$\\varepsilon = 1\$ this yields \$m^2 - 8 < 0\$, i.e. \$m=1\$, which fails to satisfy \$(\\dagger)\$. For \$\\varepsilon = -1\$ equation \$(\\dagger)\$ gives us\n\n\\[\n1 + m = 2^{x-2}(m^2 - 8) \\geq 2(m^2 - 8),\n\\]\n\nimplying \$2m^2 - m - 17 \\leq 0\$. Hence \$m\\leq 3\$; on the other hand \$m\$ cannot be \$1\$ by \$(\\dagger)\$. Because \$m\$ is odd, we obtain \$m=3\$, leading to \$x=4\$. From \$()\$ we get \$y=23\$. These values indeed satisfy the given equation. Recall that then \$y = -23\$ is also good. Thus we have the complete list of solutions \$(x,y)\$: \$(0,2)\$, \$(0,-2)\$, \$(4,23)\$, \$(4,-23)\$." ]
IMO-2006-5	https://artofproblemsolving.com/wiki/index.php/2006_IMO_Problems/Problem_5	Let $P(x)$ be a polynomial of degree $n>1$ with integer coefficients, and let $k$ be a positive integer. Consider the polynomial $Q(x) = P( P ( \ldots P(P(x)) \ldots ))$, where $P$ occurs $k$ times. Prove that there are at most $n$ integers $t$ such that $Q(t)=t$.	[ "We use the notation \$P^k(x)\$ for \$Q(x)\$.\n\nLemma 1. The problem statement holds for \$k=2\$.\n\nProof. Suppose that \$a_1, \\dotsc, a_k\$, \$b_1, \\dotsc, b_k\$ are integers such that \$P(a_j) = b_j\$ and \$P(b_j) = a_j\$ for all indices \$j\$. Let the set \$\\{ a_1, \\dotsc, a_k, b_1, \\dotsc, b_k \\}\$ have \$m\$ distinct elements. It suffices to show that \$\\deg (P) \\ge m\$.\n\nIf \$a_j = b_j\$ for all indices \$j\$, then the polynomial \$P(x)-x\$ has at least \$m\$ roots; since \$P\$ is not linear, it follows that \$\\deg P \\ge m\$ by the division algorithm.\n\nSuppose on the other hand that \$a_i \\neq b_i\$, for some index \$i\$. In this case, we claim that \$a_j + b_j\$ is constant for every index \$j\$. Indeed, we note that\n\n\\[\na_j - a_i \\mid P(a_j) - P(a_i) = b_j - b_i \\mid P(b_j) - P(b_i) = a_j - a_i ,\n\\]\n\nso \$\\lvert a_j - a_i \\rvert = \\lvert b_j - b_i \\rvert\$. Similarly,\n\n\\[\na_j - b_i \\mid P(a_j) - P(b_i) = b_j - a_i \\mid P(b_j) - P(a_i) = a_j - b_i,\n\\]\n\nso \$\\lvert a_j - b_i \\rvert = \\lvert b_j - a_i \\rvert\$. It follows that \$a_j + b_j = a_i + b_i\$.\n\nThis proves our claim. It follows that the polynomial\n\n\\[\nP(x) - \\left( a_i + b_i - x \\right)\n\\]\n\nhas at least \$m\$ roots. Since \$P\$ is not linear it follows again that \$\\deg P \\ge m\$, as desired. Thus the lemma is proven. \$\\blacksquare\$\n\nLemma 2. If \$a\$ is a positive integer such that \$P^r(a)\$ for some positive integer \$r\$, then \$P^2(a) = a\$.\n\nProof. Let us denote \$a_0 = a\$, and \$a_j = P^j(a)\$, for positive integers \$j\$. Then \$a_0 = a_r\$, and\n\n\\[\n\\begin{align} a_1 - a_0 &\\mid P(a_1)- P(a_0) = a_2-a_1 \\\\ &\\mid P(a_2) - P(a_1) = a_3 - a_2 \\\\ &\\vdots \\\\ &\\mid P(a_r)- P(a_{r-1}) = a_1 - a_0 . \\end{align}\n\\]\n\nIt follows that \$\\lvert a_{j+1} - a_j \\rvert\$ is constant for all indices \$j\$; let us abbreviate this quantity \$d\$. Now, since\n\n\\[\n(a_1-a_0) + (a_2-a_1) + \\dotsb + (a_r-a_{r-1}) = a_r-a_0 = 0,\n\\]\n\nit follows that for some index \$j\$,\n\n\\[\na_j - a_{j+1} = -(a_{j+1} - a_{j+2}),\n\\]\n\nor \$a_j = a_{j+2} = P^2(a_j)\$. Since \$a = a_r = P^{r-j}(a_j)\$, it then follows that \$P^2(a) = a\$, as desired. \$\\blacksquare\$\n\nNow, if there are more than \$n\$ integers \$t\$ for which \$Q(t) = t\$, then by Lemma 2, there are more than \$n\$ integers \$t\$ such that \$P^2(t) = t\$, which is a contradiction by Lemma 1. Thus the problem is solved. \$\\blacksquare\$\n\n## Resources\n\n- 1974 USAMO Problems/Problem 1, which implies a special case of this problem" ]
IMO-2006-6	https://artofproblemsolving.com/wiki/index.php/2006_IMO_Problems/Problem_6	Assign to each side $b$ of a convex polygon $P$ the maximum area of a triangle that has $b$ as a side and is contained in $P$. Show that the sum of the areas assigned to the sides of $P$ is at least twice the area of $P$.	[]
IMO-2007-1	https://artofproblemsolving.com/wiki/index.php/2007_IMO_Problems/Problem_1	Real numbers $a_1, a_2, \dots , a_n$ are given. For each $i$ ($1\le i\le n$) define \[ d_i=\max\{a_j:1\le j\le i\}-\min\{a_j:i\le j\le n\} \] and let \[ d=\max\{d_i:1\le i\le n\}. \] (a) Prove that, for any real numbers $x_1\le x_2\le \cdots\le x_n$, \[ \max\{\|x_i-a_i\|:1\le i\le n\}\ge \dfrac{d}{2} () \] (b) Show that there are real numbers $x_1\le x_2\le x_n$ such that equality holds in ()	[ "Since \$d_i=\\max\\{a_j:1\\le j\\le i\\}-\\min\\{a_j:i\\le j\\le n\\}\$, all \$d_i\$ can be expressed as \$a_p-a_q\$ where \$1\\le p\\le i\\le q \\le n\$. Thus \$d\$ can be expressed as \$a_p-a_q\$ for some \$p\$ and \$q\$ with \$1\\le p\\le q\\le n\$\n\nLemma: \$d\\ge 0\$.\n\nSince \$a_p \\geq a_i\$ and \$a_i \\geq a_q\$ for some \$i\$ satisfying \$p \\leq i \\leq q\$ it follows \$d \\geq 0\$.\n\n(a):\n\nCase \$d=0:\$\n\nIf \$d=0\$, \$\\max\\{\|x_i-a_i\|:1\\le i\\le n\\}\$ is the maximum of a set of non-negative number, which must be at least \$0\$.\n\nCase \$d>0:\$\n\nAssume for the sake of contradiction that \$\\max\\{\|x_i-a_i\|:1\\le i\\le n\\}<\\dfrac{d}{2}\$.\n\nThen \$\\forall i\$, \$\|x_i-a_i\|<\\dfrac{d}{2}\$. So \$\|x_p-a_p\|<\\dfrac{d}{2}\$ and \$\|x_q-a_q\|<\\dfrac{d}{2}\$.\n\nThus \$x_p>a_p-\\dfrac{d}{2}\$ and \$x_q<a_q+\\dfrac{d}{2}\$. Subtracting the two inequalities we obtain\n\n\\[\nx_p-x_q>a_p-a_q-d=a_p-a_q-a_p+a_q=0.\n\\]\n\ni.e. \$x_p>x_q\$ which contradicts \$x_p\\le x_q\$ (since \$p \\leq q\$).\n\nThus \$\\max\\{\|x_i-a_i\|:1\\le i\\le n\\}\\ge\\dfrac{d}{2}\$.\n\n(b):\n\nA set of \$x_i\$ where equality holds in () is:\n\n\\[\nx_i=\\max\\{a_j:1\\le j\\le i\\}-\\frac{d}{2}\n\\]\n\nfor all \$i\$. Since \$\\max\\{a_j:1\\le j\\le i\\}\$ is a non-decreasing function, \$x_i\$ is non-decreasing.\n\n\$\\forall i\$ let \$a_{m_i}=\\max\\{a_j:1\\le j\\le i\\}\$. Then \$a_{m_i}-a_i<a_{m_i}-\\min\\{a_j:i\\le j\\le n\\}=d_i\$.\n\nThus \$0\\le a_{m_i}-a_i \\le d\$. \$(0\\le a_{m_i}-a_i\$ because \$a_m\$ is the max of a set including \$a_i\$).\n\nTherefore one has\n\n\\[\n-\\dfrac{d}{2} \\le a_{m_i}-a_i - \\dfrac{d}{2} \\le \\dfrac{d}{2}\n\\]\n\nhence \$\|x_i-a_i\| = \\left\|a_{m_i}- \\frac{d}{2} -a_i \\right\| \\le \\frac{d}{2}\$. Lastly since \$\\max\\{\|x_i-a_i\|:1\\le i\\le n\\}\\ge\\dfrac{d}{2}\$ and \$\|x_i-a_i\|\\le \\frac{d}{2} \\ \\forall i\$, it follows \$\\max\\{\|x_i-a_i\|:1\\le i\\le n\\}=\\dfrac{d}{2}\$.\n\nThis is written by Mo Lam--- who is a horrible proof writer, so please fix the proof for me. Thank you. O, also the formatting.\n\n(edited by not_detriti)", "(a): Let \$d_j\$ satisfy \$d_j = d\$. Then \$\\exists p,q\$ with \$p \\leq j \\leq q\$ such that \$d_j = a_p - a_q\$ (note \$a_p \\geq a_j \\geq a_q\$. Note this reasoning also implies \$d_i \\geq 0\$ for all \$i\$). And for any \$x_p \\leq x_q\$ if \$x_p > a_p - d/2 = a_q + d/2\$ then \$\|x_q-a_q\| > d/2\$ in which case we're done. So we may assume \$x_p \\leq a_p-d/2\$. But then \$\|x_p-a_p\| \\geq d/2\$ and we're done.\n\n(b): Let \$f(k) = \\max\\{a_i : 1 \\leq i \\leq k\\} - \\max\\{a_i: 1 \\leq i \\leq k-1\\}\$ for \$k=2,...,n\$. Then let \$\\{a_{i_1},...,a_{i_m}\\} = f^{-1}(0,\\infty)\$ such that \$i_1 < \\cdots < i_m\$. Let \$i_0 = 1\$. Then let \$j_l\$ be such that \$i_l \\leq j_l\$ and \$d_{i_l} = a_{i_l} - a_{j_l}\$ for \$l=0,...,m\$. Note that \$a_{i_0} < a_{i_1} < \\cdots < a_{i_m}\$ and \$a_{j_0} \\leq a_{j_1} \\leq \\cdots \\leq a_{j_m}\$ since \$a_{j_l} = \\min\\{a_i: i_l \\leq i \\leq n\\}\$.\n\nTherefore if we set\n\n\\[\nx_i = \\frac{1}{2}\\big(a_{i_l}+a_{j_l} \\big)\\ \\text{if } i_l \\leq i < i_{l+1}\n\\]\n\nand\n\n\\[\nx_i = \\frac{1}{2}\\big(a_{i_m} + a_{j_m}\\big)\\ \\text{if } i_m \\leq i\n\\]\n\none will have \$x_1 \\leq x_2 \\leq \\cdots \\leq x_n\$. Furthermore if \$i_l \\leq i < i_{l+1}\$ then \$a_{i_l} \\geq a_i \\geq a_{j_l}\$ so that\n\n\\[\na_{i_l}-x_i = \\frac{1}{2}\\big(a_{i_l}-a_{j_l}\\big) \\geq a_i - x_i \\geq a_{j_l}-x_i = \\frac{1}{2}\\big(-a_{i_l}+a_{j_l} \\big)\n\\]\n\nhence\n\n\\[\n\|a_i-x_i\| \\leq \\frac{1}{2}\\Big(a_{i_l}-a_{j_l}\\Big) = \\frac{d_{i_l}}{2} \\leq \\frac{d}{2}.\n\\]\n\nby definition of \$d\$. A similar argument shows the above equation also holds if \$i \\geq i_l\$. Since \$i\$ was arbitrary, combined with (a) it follows equality holds in () for this choice of \$x_1 \\leq \\cdots \\leq x_n\$.\n\n~not_detriti\n\n## Resources\n\n- <url>viewtopic.php?p=894656#p894656 Discussion on AoPS/MathLinks</url>" ]
IMO-2007-2	https://artofproblemsolving.com/wiki/index.php/2007_IMO_Problems/Problem_2	Consider five points $A,B,C,D$, and $E$ such that $ABCD$ is a parallelogram and $BCED$ is a cyclic quadrilateral. Let $\ell$ be a line passing through $A$. Suppose that $\ell$ intersects the interior of the segment $DC$ at $F$ and intersects line $BC$ at $G$. Suppose also that $EF=EG=EC$. Prove that $\ell$ is the bisector of $\angle DAB$.	[ "\\[\n[asy] import cse5; import graph; import olympiad; dotfactor = 3; unitsize(1.5inch); path circle = Circle(origin, 1); draw(circle); pair D = (-sqrt(3)/2, -0.5), C = (sqrt(3)/2, -0.5); //G = bisectorpoint(C, B, D); pair Ee = rotate(38,C)D; pair E = IP(C--Ee, circle,1); pair Gg = rotate(76,C)D; path circle2 = Circle(E, length(C-E)); pair G = IP(C--Gg, circle2, 1); pair F = IP(C--D, circle2, 1); pair Bb = rotate(-104,C)*D; pair B = IP(C--Bb, circle, 1); pair A = extension((-1,B.y),(1,B.y),G,F); draw(circle2, dashed); draw(A--G); draw(C--D--A--B); draw(G--B); draw(E--F); draw(E--C); draw(E--G); dot(\"$C$\", C, dir(30)); dot(\"$D$\", D, SW); dot(\"$G$\", G, SE); dot(\"$E$\", E, SW); dot(\"$F$\", F, SW); dot(\"$A$\", A, SW); dot(\"$B$\", B, SE); draw(D--E,dashed); draw(B--E,dashed); [/asy]\n\\]\n\nSince \$\\angle{DAF}=\\angle{CGF}\$, \$\\angle{BAF}=\\angle{CFG}\$, it suffices to prove \$CF=CG\$.\n\nLet \$\\angle{FCE}=\\alpha\$, \$\\angle{GCE}=\\beta\$, \$\\angle{CDE}=\\gamma\$. We have:\n\n\\[\nCF=2CE\\cos{\\alpha}, CG=2CE\\cos{\\beta}\n\\]\n\nso,\n\n\\[\n\\dfrac{CF}{CG}=\\dfrac{\\cos{\\alpha}}{\\cos{\\beta}}\n\\]\n\nMeantime, using Law of Sines on \$\\triangle{DEC}\$, we have,\n\n\\[\n\\dfrac{CE}{\\sin{\\gamma}}=\\dfrac{DC}{\\sin{(180-\\alpha-\\gamma)}}=\\dfrac{DC}{\\sin{(\\alpha+\\gamma)}}\n\\]\n\nUsing Law of Sines on \$\\triangle{BEG}\$, and notice that \$\\angle{CBE}=\\angle{CDE}=\\gamma\$, we have,\n\n\\[\n\\dfrac{CE}{\\sin{\\gamma}}=\\dfrac{BG}{\\sin{(180-\\beta-\\gamma)}}=\\dfrac{BG}{\\sin{(\\beta+\\gamma)}}\n\\]\n\nso,\n\n\\[\n\\dfrac{DC}{BG}=\\dfrac{\\sin{(\\alpha+\\gamma)}}{\\sin{(\\beta+\\gamma)}}\n\\]\n\nSince \$\\triangle{GFC} \\sim \\triangle{GAB}\$, and \$DC=AB\$, we have, \$\\dfrac{DC}{BG}=\\dfrac{AB}{BG}=\\dfrac{CF}{CG}\$. Hence,\n\n\\[\n\\dfrac{\\sin{(\\alpha+\\gamma)}}{\\sin{(\\beta+\\gamma)}}=\\dfrac{\\cos{\\alpha}}{\\cos{\\beta}}\n\\]\n\nor,\n\n\\[\n\\sin{(\\alpha+\\gamma)}\\cos{\\beta}=\\sin{(\\beta+\\gamma)}\\cos{\\alpha}\n\\]\n\n\\[\n\\dfrac{1}{2}(\\sin{(\\alpha+\\gamma+\\beta)}-\\sin{(\\alpha+\\gamma-\\beta)})=\\dfrac{1}{2}(\\sin{(\\alpha+\\gamma+\\beta)}-\\sin{(\\beta+\\gamma-\\alpha)})\n\\]\n\n\\[\n\\sin{(\\alpha+\\gamma-\\beta)}=\\sin{(\\beta+\\gamma-\\alpha)}\n\\]\n\nThere are two possibilities: (1) \$\\alpha+\\gamma-\\beta = \\beta+\\gamma-\\alpha\$, or (2) \$\\alpha+\\gamma-\\beta = 180 - (\\beta+\\gamma-\\alpha)\$. However, (2) would mean \$\\gamma=90\$, then \$EC\$ would be a diameter, and \$EF < EC\$ because \$F\$ is inside the circle, so (2) is not valid. From condition (1), we have \$\\alpha=\\beta\$, therefore \$CF=CG\$. \$\\square\$\n\nSolution by Mathdummy" ]
IMO-2007-3	https://artofproblemsolving.com/wiki/index.php/2007_IMO_Problems/Problem_3	In a mathematical competition some competitors are friends. Friendship is always mutual. Call a group of competitors a clique if each two of them are friends. (In particular, any group of fewer than two competitors is a clique.) The number of members of a clique is called its size. Given that, in this competition, the largest size of a clique is even, prove that the competitors can be arranged in two rooms such that the largest size of a clique contained in one room is the same as the largest size of a clique contained in the other room.	[]
IMO-2007-4	https://artofproblemsolving.com/wiki/index.php/2007_IMO_Problems/Problem_4	In $\triangle ABC$ the bisector of $\angle{BCA}$ intersects the circumcircle again at $R$, the perpendicular bisector of $BC$ at $P$, and the perpendicular bisector of $AC$ at $Q$. The midpoint of $BC$ is $K$ and the midpoint of $AC$ is $L$. Prove that the triangles $RPK$ and $RQL$ have the same area.	[ "\$\\angle{RQL} = 90+\\angle{QCL} = 90+\\dfrac{C}{2}\$, and similarly \$\\angle{RPK} = 90+\\angle{PCK} = 90+\\dfrac{C}{2}\$. Therefore, \$\\angle{RQL} = \\angle{RPK}\$. Using the triangle area formula \$A = \\dfrac{1}{2}bc\\sin{\\angle{A}}\$ yields \$RQ \\cdot QL = RP \\cdot PK = \\dfrac{PK}{QL} = \\dfrac{RQ}{RP}\$ after cancelling the sines and constant. Draw line \$QD\$ perpendicular to \$BC\$ that intersects \$BC\$ at \$D\$, then \$QD=QL\$ because the perpendicular bisectors are congruent, (or alternatively \$\\triangle QDC\\cong\\triangle QLC\$). This presents us \$\\dfrac{PK}{QL}=\\dfrac{PK}{QD}=\\dfrac{PC}{QC}\$ by similar triangles; now, we have only to prove \$\\dfrac{PC}{QC}=\\dfrac{RQ}{RP}\$, or \$RQ \\cdot QC=RP \\cdot PC\$.\n\nSince \$\\angle{OPQ} =1 80-\\angle{RPK} =1 80-\\angle{RQL} = \\angle{OQP}\$, we have \$\\triangle OPQ\$ is isosceles. Draw the perpendicular from \$O\$ to \$RC\$, intersecting at \$E\$. Then \$PE = QE = x\$ for a real \$x\$, now because the perpendicular from the center of a circle to a chord bisects that chord, \$RE = CE\$. Let \$y = RE\$, and then \$RQ \\cdot QC = (y+x) \\cdot (y-x) = PC \\cdot RP\$, proving our claim.", "Since \$\\angle{OPQ}=180-\\angle{RPK}=180-\\angle{RQL}=\\angle{OQP}\$, we have \$OQ=OP=x\$. Let the radius of the circumcircle be \$r\$, then the diameter through \$P\$ is divided by point \$P\$ into lengths of \$r+x\$ and \$r-x\$. By power of point, \$RPPC=(r+x)(r-x)\$. Similarly, \$RQQC=(r+x)(r-x)\$. Therefore \$RPPC=RQQC\$. \$\\square\$\n\nSolution by ~KingRavi\n\nAlternate Solution by ~mathdummy\n\nEdifying edits made by ~TheGrandioseGeometrician", "The area of \$\\triangle{RQL}\$ is given by \$\\dfrac{1}{2}QLRQ\\sin{\\angle{RQL}}\$ and the area of \$\\triangle{RPK}\$ is \$\\dfrac{1}{2}RPPK\\sin{\\angle{RPK}}\$. Let \$\\angle{BCA}=C\$, \$\\angle{BAC}=A\$, and \$\\angle{ABC}=B\$. Now \$\\angle{KCP}=\\angle{QCL}=\\dfrac{C}{2}\$ and \$\\angle{PKC}=\\angle{QLC}=90\$, thus \$\\angle{RPK}=\\angle{RQL}=90+\\dfrac{C}{2}\$. \$\\triangle{PKC} \\sim \\triangle{QLC}\$, so \$\\dfrac{PK}{QL}=\\dfrac{KC}{LC}\$, or \$\\dfrac{PK}{QL}=\\dfrac{BC}{AB}\$. The ratio of the areas is \$\\dfrac{[RPK]}{[RQL]}=\\dfrac{BCRP}{ACRQ}\$. The two areas are only equal when the ratio is 1, therefore it suffices to show \$\\dfrac{RP}{RQ}=\\dfrac{AC}{BC}\$. Let \$O\$ be the center of the circle. Then \$\\angle{ROK}=A+C\$, and \$\\angle{ROP}=180-(A+C)=B\$. Using law of sines on \$\\triangle{RPO}\$ we have: \$\\dfrac{RP}{\\sin{B}}=\\dfrac{OR}{\\sin{(90+\\dfrac{C}{2})}}\$ so \$RP\\sin{(90+\\dfrac{C}{2})}=OR\\sin{B}\$. \$OR\\sin{B}=\\dfrac{1}{2}AC\$ by law of sines, and \$\\sin{(90+\\dfrac{C}{2})}=\\cos{\\dfrac{C}{2}}\$, thus 1) \$2RP\\cos{\\dfrac{C}{2}}=AC\$. Similarly, law of sines on \$\\triangle{ROQ}\$ results in \$\\dfrac{RQ}{\\sin{(180-A)}}=\\dfrac{OR}{\\sin{(90-\\dfrac{C}{2})}}\$ or \$\\dfrac{RQ}{\\sin{A}}=\\dfrac{OR}{\\cos{\\dfrac{C}{2}}}\$. Cross multiplying we have \$RQ\\cos{\\dfrac{C}{2}}=OR\\sin{A}\$ or 2) \$2RQ\\cos{\\dfrac{C}{2}}=BC\$. Dividing 1) by 2) we have \$\\dfrac{RP}{RQ}=\\dfrac{AC}{BC}\$ \$\\square\$\n\n\\[\n(tkhalid)\n\\]", "(Image Link) https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYi85LzQ5OTJlZGNhYTQ0YjJjODcxMTBmZGNmMTdiZDdkMGRjZGUyOWQ5LnBuZw==&rn=U2NyZWVuIFNob3QgMjAxOS0wOC0wOCBhdCAxMi4yMC4zOCBQTS5wbmc=\n\nWLOG, let the diameter of \$(ACBR)\$ be \$1.\$\n\nWe see that \$PK = \\dfrac{1}{2}a \\tan \\dfrac{1}{2}C\$ and \$QL = \\dfrac{1}{2}b \\tan \\dfrac{1}{2}C\$ from right triangles \$\\triangle PKC\$ and \$\\triangle QLC.\$\n\nWe now look at \$AR.\$ By the Extended Law of Sines on \$\\triangle ACR,\$ we get that \$AR = \\sin\\frac{1}{2}C.\$ Similarly, \$BR = \\sin \\frac{1}{2}C.\$\n\nWe now look at \$CR.\$ By Ptolemy's Theorem, we have\n\n\\[\nAR \\cdot BC + BR \\cdot AC = AB \\cdot CR,\n\\]\n\nwhich gives us\n\n\\[\n\\sin \\frac{1}{2}C (a + b) = c(CR).\n\\]\n\nThis means that\n\n\\[\nCR = \\dfrac{\\sin \\frac{1}{2}C (a + b)}{c}.\n\\]\n\nWe now seek to relate the lengths computed with the areas.\n\nTo do this, we consider the altitude from \$R\$ to \$PK.\$ This is to find the area of \$RPK.\$ Finding the area of \$\\triangle RQL\$ is similar.\n\nWe claim that \$RF = \\dfrac{1}{2}b.\$ In order to prove this, we will prove that \$\\triangle RFP \\cong \\triangle QLC.\$ In other words, we wish to prove that \$PR = QC.\$ This is equivalent to proving that \$PC + QC = CR.\$\n\nNote that \$PC = \\dfrac{PK}{\\sin \\frac{1}{2}C}\$ and \$QC = \\dfrac{QL}{\\sin \\frac{1}{2}C}.\$ Therefore, we get that\n\n\\[\nPC + QC = \\dfrac{PK}{\\sin \\frac{1}{2}C} + \\dfrac{QL}{\\sin\\frac{1}{2}C}\n\\]\n\n\\[\n= \\dfrac{PK + QL}{\\sin\\frac{1}{2}C}\n\\]\n\n\\[\n= \\dfrac{PK(1 + \\frac{b}{a})}{\\sin\\frac{1}{2}C}\n\\]\n\n\\[\n= \\dfrac{PK(\\frac{a + b}{a})}{\\sin\\frac{1}{2}C}\n\\]\n\n\\[\n= \\dfrac{\\frac{1}{2}a\\tan\\frac{1}{2}C \\cdot (a + b)}{a\\sin\\frac{1}{2}C}\n\\]\n\n\\[\n= \\dfrac{\\frac{1}{2}a\\sin\\frac{1}{2}C \\cdot (a + b)}{a\\sin\\frac{1}{2}C\\cos\\frac{1}{2}C}\n\\]\n\n\\[\n= \\dfrac{\\sin\\frac{1}{2}C \\cdot (a + b)}{2\\sin \\frac{1}{2}C\\cos\\frac{1}{2}C}\n\\]\n\n\\[\n= \\dfrac{\\sin\\frac{1}{2}C \\cdot (a + b)}{\\sin C}\n\\]\n\n\\[\n= \\dfrac{\\sin\\frac{1}{2}C \\cdot (a + b)}{c}\n\\]\n\n\\[\n= CR.\n\\]\n\nThus, \$RF = \\dfrac{1}{2}b.\$ In this way, we get that the altidude from \$R\$ to \$QL\$ has length \$\\dfrac{1}{2}a.\$ Therefore, we see that \$[RPK] = \\dfrac{1}{8}ab \\tan \\frac{1}{2}C\$ and \$[RQL] = \\dfrac{1}{8}ab \\tan \\frac{1}{2}C,\$ so the two areas are equal.\n\nSolution by Ilikeapos", "\\[\n[\\triangle{RPK}]=[\\triangle{RQL}], LQRQ\\sin\\angle{LQR}=KPPR\\sin\\angle{RPK}\n\\]\n\nSince \$CR\$ bisects \$\\angle{ACB},\\angle{QCL}=\\angle{PCK}\$, \$OL,OK\$ are perpendicular to sides \$AC,BC\$ separately, \$\\angle{QLC}=\\angle{PKC}=90^{\\circ}, \\angle{CQL}=\\angle{CPK}, \\angle{LQR}=\\angle{RPK}\$\n\nSo now, we only have to prove \$RPPK=LQQR\$, which is \$RP(CQ+QP)\\cos\\angle{CPK}=CQ(QP+PR)\\cos\\angle{CQL}\$, as mentioned above, the two angles are the same, we have to prove that \$RP(CQ+QP)=CQ(QP+PR)\$, which is equivalent to \$RPQP=QPCQ\$, we have to prove \$RP=CQ\$\n\nNow notice that \$\\triangle{OQP}\$ is isosceles. \$\\angle{CQL}=\\angle{OQP}=\\angle{OPQ}\$, construct \$OJ \\bot CR\$, \$CJ=JR,JQ=JP,CQ=PR\$ as desired\n\n~bluesoul" ]
IMO-2007-5	https://artofproblemsolving.com/wiki/index.php/2007_IMO_Problems/Problem_5	Let $a$ and $b$ be positive integers. Show that if $4ab-1$ divides $(4a^2-1)^2$, then $a=b$.	[ "Lemma. If there is a counterexample for some value of \$a\$, then there is a counterexample \$(a,b)\$ for this value of \$a\$ such that \$b<a\$.\n\nProof. Suppose that \$b >a\$. Note that \$4ab -1 \\equiv -1 \\pmod{4a}\$, but \$(4a^2-1)^2 \\equiv 1 \\pmod{4a}\$. It follows that \$(4a^2-1)^2/(4ab-1) \\equiv -1 \\pmod{4a}\$. Since\n\n\\[\n0<(4a^2-1)^2/(4ab-1) < (4a^2-1)^2/(4a^2-1) = 4a^2-1,\n\\]\n\nit follows that \$(4a^2-1)^2/(4ab-1)\$ can be written as \$4ab'-1\$, with \$0<b'<a\$. Then \$(a,b')\$ is a counterexample for which \$b'<a\$. \$\\blacksquare\$\n\nNow, suppose a counterexample exists. Let \$(a,b)\$ be a counterexample for which \$a\$ is minimal and \$b<a\$. We note that\n\n\\[\n\\gcd(4ab-1,2a-1) \\mid 4ab-1 - 2b(2a-1) = 2b-1,\n\\]\n\nand\n\n\\[\n\\gcd(4ab-1,2a+1) \\mid 2b(2a+1) - (4ab-1) = 2b+1 .\n\\]\n\nNow,\n\n\\[\n\\begin{align} 4ab-1 &\\mid (4a^2-1)^2 = (2a-1)^2(2a+1)^2 \\\\ &\\mid \\gcd(4ab-1,2a-1)^2 \\cdot \\gcd(4ab-1,2a+1)^2 \\\\ &\\mid (2b-1)^2(2b+1)^2 = (4b^2-1)^2 . \\end{align}\n\\]\n\nThus \$(b,a)\$ is a counterexample. But \$b<a\$, which contradicts the minimality of \$a\$. Therefore no counterexample exists. \$\\blacksquare\$\n\n(Sean Yu, US)", "As \$(4a^2-1)^2 = (4a^2-4ab + 4ab-1)^2 \\equiv (4a)^2(a-b)^2 \\pmod {4ab-1}\$, since \$gcd((4a)^2, 4ab -1 ) = 1\$, we only need to show that \$4ab -1 \\mid (a-b)^2 \\Rightarrow a=b\$. We will use the Vieta Jumping technique to solve it.\n\nSuppose there exist counterexamples, which means the set \$S = \\{(x, y): \\mbox {x and y are positive integers}, x \\ne y, 4xy-1 \\mid (x-y)^2\\}\$ is not empty. Then there exists a pair \$(a, b)\\in S\$ such that \$a+b\$ has the minimum sum among all pairs in \$S\$. Without loss of generality, we assume \$a > b\$. Let \$k=\\frac{(a-b)^2}{4ab-1}\$. Then \$k\$ is a positive integer and \$a^2-(2+4k)ba + b^2+k = 0\$. So \$a\$ is a root for the quadratic equation \$x^2-(2+4k)bx + b^2+k = 0\$. Using the Vieta Lemma, we know the equation has another root \$a' = \\frac{b^2 + k}{a} > 0\$ and from \$a' = (2+4k)b -a\$ we know it is an integer. Note that since \$4ab -1 \\mid(a-b)^2\$ and \$a>b\$, we have \$4ab -1 \\le (a-b)^2 \\le a^2-1\$, which implies \$b\\le a/4\$. Also \$k=\\frac{(a-b)^2}{4ab-1} \\le \\frac{a^2}{3ab} \\le a/3\$, thus \$a' = \\frac{b^2 + k}{a} \\le \\frac{(a/4)^2 + a/3}{a} = a/16 + 1/3 < a\$, and we get another pair \$(a', b)\\in S\$, but \$a' + b < a+b\$, which contradicts the assumption that \$a+b\$ has the minimum sum. \$\\hfill\\blacksquare\$\n\n## Resources\n\n- <url>viewtopic.php?p=894656#p894656 Discussion on AoPS/MathLinks</url>" ]
IMO-2007-6	https://artofproblemsolving.com/wiki/index.php/2007_IMO_Problems/Problem_6	Let $n$ be a positive integer. Consider \[ S=\{(x,y,z)~:~x,y,z\in \{0,1,\ldots,n \},~x+y+z>0\} \] as a set of $(n+1)^3-1$ points in three-dimensional space. Determine the smallest possible number of planes, the union of which contain $S$ but does not include $(0,0,0)$.	[ "We will prove the result using the following Lemma, which has an easy proof by induction.\n\nLemma Let \$S_1 = \\{0, 1, \\ldots, n_1\\}\$, \$S_2 = \\{0, 1, \\ldots, n_2\\}\$ and \$S_3 = \\{0, 1, \\ldots, n_3\\}\$. If \$P\$ is a polynomial in \$\\mathbb{R}[x, y, z]\$ that vanishes on all points of the grid \$S = S_1 \\times S_2 \\times S_3\$ except at the origin, then\n\n\\[\n\\deg P \\geq n_1 + n_2 + n_3.\n\\]\n\nProof. We will prove this by induction on \$n = n_1 + n_2 + n_3\$. If \$n = 0\$, then the result follows trivially. Say \$n > 0\$. WLOG, we can assume that \$n_1 > 0\$. By polynomial division over \$\\mathbb{R}[y, z][x]\$ we can write\n\n\\[\nP = (x - n_1)Q + R.\n\\]\n\nSince \$x-n_1\$ is a monomial, the remainder \$R\$ must be a constant in \$\\mathbb{R}[y, z]\$, i.e., \$R\$ is a polynomial in two variables \$y\$, \$z\$. Pick an element of \$S\$ of the form \$(n_1, y, z)\$ and substitute it in the equation. Since \$P\$ vanishes on all such points, we get that \$R(y, z) = 0\$ for all \$(y, z) \\in S_2 \\times S_3\$. Let \$S_1' = \\{0, 1, \\ldots, n_1 - 1\\}\$ and \$S' = S_1' \\times S_2 \\times S_3\$. For every point \$(x, y, z)\$ in \$S'\$ we have\n\n\\[\nP(x, y, z) = (x - n_1)Q(x, y, z) + R(y, z) = (x-n_1)Q(x, y, z),\n\\]\n\nwhere \$x - n_1 \\neq 0\$. Therefore, the polynomial \$Q\$ vanishes on all points of \$S'\$ except the origin. By induction hypothesis, we must have \$\\deg Q \\geq n_1 - 1 + n_2 + n_3\$. But, \$\\deg P \\geq \\deg Q + 1\$ and hence we have \$\\deg P \\geq n_1 + n_2 + n_3\$.\n\nNow, to solve the problem let \$H_1, \\ldots, H_m\$ be \$m\$ planes that cover all points of \$S\$ except the origin. Since these planes don't pass through origin, each \$H_i\$ can be written as \$a_i x + b_i y + c_i z = 1\$. Define \$P\$ to be the polynomial \$\\prod (a_ix + b_iy + c_iz - 1)\$. Then \$P\$ vanishes at all points of \$S\$ except at the origin, and hence \$\\deg P = m \\geq 3n\$." ]
IMO-2008-1	https://artofproblemsolving.com/wiki/index.php/2008_IMO_Problems/Problem_1	An acute-angled triangle $ABC$ has orthocentre $H$. The circle passing through $H$ with centre the midpoint of $BC$ intersects the line $BC$ at $A_1$ and $A_2$. Similarly, the circle passing through $H$ with centre the midpoint of $CA$ intersects the line $CA$ at $B_1$ and $B_2$, and the circle passing through $H$ with centre the midpoint of $AB$ intersects the line $AB$ at $C_1$ and $C_2$. Show that $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$ lie on a circle.	[ "Let \$M_A\$, \$M_B\$, and \$M_C\$ be the midpoints of sides \$BC\$, \$CA\$, and \$AB\$, respectively. It's not hard to see that \$M_BM_C\\parallel BC\$. We also have that \$AH\\perp BC\$, so \$AH \\perp M_BM_C\$. Now note that the radical axis of two circles is perpendicular to the line connecting their centers. We know that \$H\$ is on the radical axis of the circles centered at \$M_B\$ and \$M_C\$, so \$A\$ is too. We then have \$AC_1\\cdot AC_2=AB_2\\cdot AB_1\\Rightarrow \\frac{AB_2}{AC_1}=\\frac{AC_2}{AB_1}\$. This implies that \$\\triangle AB_2C_1\\sim \\triangle AC_2B_1\$, so \$\\angle AB_2C_1=\\angle AC_2B_1\$. Therefore \$\\angle C_1B_2B_1=180^{\\circ}-\\angle AB_2C_1=180^{\\circ}-\\angle AC_2B_1\$. This shows that quadrilateral \$C_1C_2B_1B_2\$ is cyclic. Note that the center of its circumcircle is at the intersection of the perpendicular bisectors of the segments \$C_1C_2\$ and \$B_1B_2\$. However, these are just the perpendicular bisectors of \$AB\$ and \$CA\$, which meet at the circumcenter of \$ABC\$, so the circumcenter of \$C_1C_2B_1B_2\$ is the circumcenter of triangle \$ABC\$. Similarly, the circumcenters of \$A_1A_2B_1B_2\$ and \$C_1C_2A_1A_2\$ are coincident with the circumcenter of \$ABC\$. The desired result follows." ]
IMO-2008-2	https://artofproblemsolving.com/wiki/index.php/2008_IMO_Problems/Problem_2	(i) If $x$, $y$ and $z$ are three real numbers, all different from $1$, such that $xyz = 1$, then prove that $\frac {x^{2}}{\left(x - 1\right)^{2}} + \frac {y^{2}}{\left(y - 1\right)^{2}} + \frac {z^{2}}{\left(z - 1\right)^{2}} \geq 1$. (With the $\sum$ sign for cyclic summation, this inequality could be rewritten as $\sum \frac {x^{2}}{\left(x - 1\right)^{2}} \geq 1$.) (ii) Prove that equality is achieved for infinitely many triples of rational numbers $x$, $y$ and $z$.	[ "Consider the transormation \$f:\\mathbb{R}/\\{1\\} \\rightarrow \\mathbb{R}/\\{-1\\}\$ defined by \$f(u) = \\frac{u}{1-u}\$ and put \$\\alpha = f(x), \\beta = f(y), \\gamma = f(z)\$. Since \$f\$ is also one-to one from \$\\mathbb{Q}/\\{1\\}\$ to \$\\mathbb{Q}/\\{-1\\}\$, the problem is equivalent to showing that\n\n\\[\n\\alpha^2+\\beta^2+\\gamma^2 \\ge 1 \\quad (1)\n\\]\n\nsubject to\n\n\\[\n\\left(\\frac{\\alpha}{\\alpha+1}\\right) \\left(\\frac{\\beta}{\\beta+1}\\right) \\left(\\frac{\\gamma}{\\gamma+1}\\right) = 1 \\quad (2)\n\\]\n\nand that equallity holds for infinitely many triplets of rational \$\\alpha,\\beta,\\gamma\$.\n\nNow, rewrite (2) as \$\\alpha\\beta\\gamma = (1+\\alpha)(1+\\beta)(1+\\gamma)\$ and express it as\n\n\\[\n0 = 1 + p + q\n\\]\n\nwhere \$p=\\alpha+\\beta+\\gamma\$ and \$q = \\alpha\\beta+\\beta\\gamma+\\gamma\\alpha\$. Notice that (1) can be written as\n\n\\[\np^2-2q \\ge 1.\n\\]\n\nBut from \$p = -1-q\$, we get\n\n\\[\np^2-2q = (1+q)^2-2q = 1 + q^2 \\ge 1,\n\\]\n\nwith equality holding iff \$q = 0\$. That proves part (i) and points us in the direction of looking for rational \$\\alpha,\\beta,\\gamma\$ for which \$q=0\$ and (hence) \$p=-1\$, that is:\n\n\\[\n\\begin{align} \\alpha+\\beta+\\gamma & = -1\\\\ \\alpha\\beta+\\beta\\gamma+\\gamma\\alpha & = 0 \\end{align}\n\\]\n\nExpressing \$\\alpha\$ from the first equation and substituting into the second, we get\n\n\\[\n\\beta\\gamma + ( \\beta+\\gamma ) ( -1 - \\beta -\\gamma ) = 0\n\\]\n\nas the sole condition we need to satisfy in rational numbers.\n\nIf \$\\beta = \\frac{b}{m}\$ and \$\\gamma = \\frac{c}{m}\$ for some integers \$b\$,\$c\$,and \$m\$, they would need to satisfy\n\n\\[\nbc = m(b+c)+(b+c)^2 \\Leftrightarrow m = \\frac{bc}{b+c} - (b+c).\n\\]\n\nFor \$m\$ to be integer, we would like \$b+c\$ to divide \$bc\$. Consider the example\n\n\\[\nb=t, c=t^2-t, m = t-1-t^2,\n\\]\n\nwhere \$b+c = t^2\$ divides \$bc = t(t^2-t)\$ for any integer \$t \\ne 0\$. Substituting back, that gives us\n\n\\[\n\\beta = \\frac{t}{t-1-t^2},\\quad \\gamma = \\frac{t^2-t}{t-1-t^2},\\quad \\alpha = \\frac{1-t}{t-1-t^2}.\n\\]\n\nA simple check shows that \$\\alpha,\\beta,\\gamma\$ are rational and well defined and that \$p=-1\$ and \$q=0\$ for any integer \$t\$ (even for \$t=0\$).\n\nMoreover, from \$\\lim_{t\\rightarrow +\\infty} \\beta = 0\$ and \$\\beta < 0\$ for large \$t\$, we see that infinitely many \$t\$ generate infinitely many different triplets of \$\\alpha\$, \$\\beta\$, and \$\\gamma\$. That completes the proof of part (ii). --Vbarzov 03:03, 5 September 2008 (UTC)" ]
IMO-2008-3	https://artofproblemsolving.com/wiki/index.php/2008_IMO_Problems/Problem_3	Prove that there are infinitely many positive integers $n$ such that $n^{2} + 1$ has a prime divisor greater than $2n + \sqrt {2n}$.	[ "Solution 1\n\nThe main idea is to take a gaussian prime \$a+bi\$ and multiply it by a \"twice as small\" \$c+di\$ to get \$n+i\$. The rest is just making up the little details.\n\nFor each sufficiently large prime \$p\$ of the form \$4k+1\$, we shall find a corresponding \$n\$ such that \$p\$ divides \$n^2+1\$ and \$p>2n+\\sqrt{2n}\$. Since there exist infinitely many such primes and, for each of them, \$n \\ge \\sqrt{p-1}\$, we will have found infinitely many distinct \$n\$ satisfying the hypothesis.\n\nTake a prime \$p\$ of the form \$4k+1\$ and consider its \"sum-of-two squares\" representation \$p=a^2+b^2\$, which we know to exist for all such primes. As \$a\\ne b\$, assume without loss of generality that \$b>a\$. If \$a=1\$, then \$n=b\$ is what we are looking for, and \$p=n^2+1 > 2n+\\sqrt{2n}\$ as long as \$p\$ (and hence \$n\$) is large enough. Assume from now on that \$b>a>1\$.\n\nSince \$a\$ and \$b\$ are (obviously) co-prime, there must exist integers \$c\$ and \$d\$ such that\n\n\\[\nad+bc=1. \\quad(1)\n\\]\n\nIn fact, if \$c\$ and \$d\$ are such numbers, then \$c\\pm ma\$ and \$d\\mp mb\$ work as well for any integer \$m\$, so we can assume that \$c \\in \\left[-\\frac{a}{2}, \\frac{a}{2}\\right]\$.\n\nDefine \$n=\|ac-bd\|\$ and let's see why this is a good choice. For starters, notice that \$(a^2+b^2)(c^2+d^2)=n^2+1\$.\n\nIf \$c=\\pm\\frac{a}{2}\$, from (1) we see that \$a\$ must divide \$2\$ and hence \$a=2\$. This implies, \$d=-\\frac{b-1}{2}\$ and \$n=\\frac{b(b-1)}{2}-2\$. Therefore, \$\\left(b-\\frac{1}{2}\\right)^2 = 1/4 + 2(n+2) > 2n\$, so \$b > \\sqrt{2n}+\\frac{1}{2}\$. Finally, \$p=b^2+2^2 > 2n+\\sqrt{2n}\$ and the case \$c=\\pm\\frac{a}{2}\$ is cleared.\n\nWe can safely assume now that\n\n\\[\n\|c\| \\le \\frac{a-1}{2}.\n\\]\n\nAs \$b>a>1\$ implies \$b>2\$, we have\n\n\\[\n\|d\| = \\left\|\\frac{1-bc}{a}\\right\| \\le \\frac{b(a-1)+2}{2a} < \\frac{ba}{2a} = \\frac{b}{2},\n\\]\n\nso\n\n\\[\n\|d\| \\le \\frac{b-1}{2}.\n\\]\n\nTherefore,\n\n\\[\nn^2+1 = (a^2+b^2)(c^2+d^2) \\le p\\left( \\frac{(a-1)^2}{4}+\\frac{(b-1)^2}{4} \\right). \\quad (2)\n\\]\n\nBefore we proceed, we would like to show first that \$a+b-1 > \\sqrt{p}\$. Observe that the function \$x+\\sqrt{p-x^2}\$ over \$x\\in(2,\\sqrt{p-4})\$ reaches its minima on the ends, so \$a+b\$ given \$a^2+b^2=p\$ is minimized for \$a = 2\$, where it equals \$2+\\sqrt{p-2^2}\$. So we want to show that\n\n\\[\n2+\\sqrt{p-4} > \\sqrt{p} + 1,\n\\]\n\nwhich is not hard to show for large enough \$p\$.\n\nNow armed with \$a+b-1>\\sqrt{p}\$ and (2), we get\n\n\\[\n4(n^2+1)\\le p( a^2+b^2 - 2(a+b-1) )< p( p-2\\sqrt{p} )< u^2(u-1)^2,\n\\]\n\nwhere \$u=\\sqrt{p}.\$\n\nFinally,\n\n\\[\nu^2(u-1)^2 > 4n^2+4 > 4n^2\\Rightarrow u(u-1) > 2n \\Rightarrow (u-\\frac{1}{2})^2 > 2n+\\frac{1}{4} > 2n \\Rightarrow u > \\sqrt{2n} + \\frac{1}{2} \\Rightarrow p = u^2 > 2n + \\sqrt{2n}.\n\\]\n\nThe proof is complete.\n\nSolution 2\n\nWe begin with a lemma.\n\nLemma: For every prime \$p \\equiv 1 \\mod{4}\$, there exists a positive integer \$n \\le \\dfrac{p - \\sqrt{p-4}}{2}\$ such that \$p\\mid n^2 + 1\$.\n\nProof: We know that there must be a solution \$x\$ to the equation\n\n\\[\nx^2+1 \\equiv 0 \\mod{p}\n\\]\n\nwith \$1 \\le x \\le p-1\$. However, if \$x\$ is a solution then so is \$p-x\$, so there must be a solution \$x \\le p-1\$. Let \$n\$ denote this solution and suppose that \$n = \\frac{p-k}{2}\$. Then, we have\n\n\\[\n\\left(\\dfrac{p-k}{2}\\right)^2 \\equiv -1 \\mod{p}\n\\]\n\n\\[\nk^2 \\equiv -4 \\mod{p}.\n\\]\n\nSince \$k\$ is a positive integer, it follows that \$k^2 \\ge p-4\$, so we have\n\n\\[\nn \\le \\dfrac{p - \\sqrt{p-4}}{2},\n\\]\n\nas desired. The lemma is proven.\n\nSuppose for sake of contradiction that there are only finitely many integers \$n_1,n_2,\\dots,n_k\$ that work. Let \$p \\equiv 1 \\mod{4}\$ be a prime with \$p>20\$ and such that \$p\$ is relatively prime to \$n_i^2+1\$ for all \$i\$ (the existence of \$p\$ is guaranteed by the existence of infinitely many primes of the form \$4k+1\$). Then, we know that there exists an \$N\$ such that \$N\\le \\dfrac{p - \\sqrt{p-4}}{2}\$ and \$p \| N^2 + 1\$ (this last condition shows that \$N\$ is not among \$n_1,n_2,\\dots,n_k\$. We want to show that\n\n\\[\np > 2N + \\sqrt{2N}\n\\]\n\n\\[\np - 2N > \\sqrt{2N}.\n\\]\n\nBut, we know that \$p-2N \\ge \\sqrt{p-4}\$, so it suffices to show that\n\n\\[\np-4 > 2N\n\\]\n\n\\[\np - 2N > 4.\n\\]\n\nOnce again, it suffices to show that\n\n\\[\n\\sqrt{p-4} > 4,\n\\]\n\nwhich follows from \$p>20\$.\n\nThus, \$N\$ satisfies the required condition and it follows that there exist infinitely many values of \$n\$ that satisfy the given condition, as desired." ]
IMO-2008-4	https://artofproblemsolving.com/wiki/index.php/2008_IMO_Problems/Problem_4	Find all functions $f: (0, \infty) \mapsto (0, \infty)$ (so $f$ is a function from the positive real numbers) such that \[ \frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2} \] for all positive real numbers $w,x,y,z,$ satisfying $wx = yz.$	[ "Considering \$w=1\$ and \$z=y=\\sqrt{x}\$ which satisfy the constraint \$wx=yz\$ we get the following equation:\n\n\\[\n\\frac{(f(1))^2 + (f(x))^2}{f(x) + f(x)} = \\frac{1+x^2}{x+x} \\Leftrightarrow x((f(1))^2 + (f(x))^2) = (1+x^2)f(x)\n\\]\n\nAt once considering \$x=1\$ we get \$(f(1))^2 = f(1)\$ and knowing that \$f : \\mathbb{R}^+ \\rightarrow \\mathbb{R}^+\$ the only possible solution is \$f(1)=1\$ since \$f(1)=0\$ is impossible.\n\nSo we get the quadratic equation:\n\n\\[\nx(f(x))^2 - (1+x^2)f(x) + x = 0\n\\]\n\nSolving for \$f(x)\$ as a function of \$x\$ we get:\n\n\\[\nf(x) = \\frac{ 1+x^2 \\pm \\sqrt{(1+x^2)^2-4x^2} }{2x} = \\frac{ 1+x^2 \\pm \\sqrt{(1-x^2)^2} }{2x}\n\\]\n\nAt once we see that for one value of \$x\$, \$f(x)\$ can only take one of 2 possible values:\n\n\\[\nf(x) = x \\vee f(x) = \\frac{1}{x}\n\\]\n\n.\n\nTake into consideration that \$f(2) = 2\$ but \$f(3) = \\frac{1}{3}\$ verifies the quadratic equation and thus so far we can't say that \$f(x)=x \\, \\forall_{x \\in \\mathbb{R}^+}\$ or alternatively \$f(x)=\\frac{1}{x} \\, \\forall_{x \\in \\mathbb{R}^+}\$. This is indeed the case but we haven't proved it yet.\n\nTo prove the previous assertion consider 2 values \$a,b \\in \\mathbb{R}^+\$ such that \$a\\ne 1 \\wedge b \\ne 1 \\wedge a \\ne b\$ while having \$f(a) = a \\wedge f(b)=\\frac{1}{b}\$\n\nConsider now the original functional equation with \$w=a,\\ x=b,\\ y=z=\\sqrt{ab}\$ which verifies the constraint. Substituting we have:\n\n\\[\n\\frac{ (f(a))^2 + (f(b))^2}{ f(ab) + f(ab)} = \\frac{a^2 + b^2}{ab + ab} \\Leftrightarrow f(ab) = ab\\frac{a^2 + \\frac{1}{b^2}}{a^2 + b^2}\n\\]\n\nNow either \$f(ab)=ab\$ or \$f(ab)=\\frac{1}{ab}\$. (notice that \$ab \\ne b \\wedge ab\\ne b\$ by hypothesis)\n\nIf \$f(ab)=ab\$ then we have \$ab = ab\\frac{a^2 + \\frac{1}{b^2}}{a^2 + b^2} \\Leftrightarrow b^4=1\$ and since \$b>0\$ the only solution is \$b=1\$.\n\nIf \$f(ab)=\\frac{1}{ab}\$ then we have \$\\frac{1}{ab} = ab\\frac{a^2 + \\frac{1}{b^2}}{a^2 + b^2} \\Leftrightarrow a^2+b^2 = a^4b^2 +a^2 \\Leftrightarrow a^4=1\$ and since \$a>0\$ the only solution is \$a=1\$.\n\nSo the only solutions are \$a=1\$ or \$b=1\$ in which case both alternatives imply \$f(1)=1\$. Thus we conclude that solutions to the functional equation are a subset of \$\\left\\{f(x)=x \\ \\forall_{x \\in \\mathbb{R}^+},\\ f(x)=\\frac{1}{x}\\ \\forall_{x \\in \\mathbb{R}^+} \\right\\}\$.\n\nFinally, plug each of these 2 functions into the functional equation and verify that they indeed are solutions.\n\nThis is trivial since \$f(x)=x\$ is an obvious solution and for \$f(x)=\\frac{1}{x}\$ we have:\n\n\\[\n\\frac{ \\frac{1}{w^2} + \\frac{1}{x^2} }{ \\frac{1}{y^2} + \\frac{1}{z^2}} = \\frac{ \\frac{w^2+x^2}{(wx)^2} }{ \\frac{y^2+z^2}{(yz)^2} } = \\frac{w^2+x^2}{y^2+z^2}\n\\]\n\nprovided that \$(wx)^2 = (yz)^2\$ which verifies the original constraint.\n\nSo the functional equation has 2 solutions:\n\n\\[\nf(x) = x\\ \\forall_{x \\in \\mathbb{R}^+}\\ \\vee\\ f(x)=\\frac{1}{x}\\ \\forall_{x \\in \\mathbb{R}^+}\n\\]" ]
IMO-2008-5	https://artofproblemsolving.com/wiki/index.php/2008_IMO_Problems/Problem_5	Let $n$ and $k$ be positive integers with $k \geq n$ and $k - n$ an even number. Let $2n$ lamps labelled $1$, $2$, ..., $2n$ be given, each of which can be either on or off. Initially all the lamps are off. We consider sequences of steps: at each step one of the lamps is switched (from on to off or from off to on). Let $N$ be the number of such sequences consisting of $k$ steps and resulting in the state where lamps $1$ through $n$ are all on, and lamps $n + 1$ through $2n$ are all off. Let $M$ be number of such sequences consisting of $k$ steps, resulting in the state where lamps $1$ through $n$ are all on, and lamps $n + 1$ through $2n$ are all off, but where none of the lamps $n + 1$ through $2n$ is ever switched on. Determine $\frac {N}{M}$.	[ "For convenience, let \$A\$ denote the set \$(1,2,\\ldots n)\$ and \$B\$ the set \$(n+1,n+2,\\ldots,2n)\$.\n\nWe can describe each sequences of switching the lamps as a \$k\$-dimensional vector \$(a_1, a_2, \\ldots, a_k)\$, where \$a_i \\in A \\cup B\$ signifies which lamp was switched on the \$i\$-th move for \$i=1,2,\\ldots k\$.\n\nLet \$\\cal{N}\$ consist of those sequences that contain each of the numbers in \$A\$ an odd number of times and each of the numbers in \$B\$ an even number of times. Similarly, let \$\\cal{M}\$ denote the set of those sequences that contain no numbers from \$B\$ and each of the numbers in \$A\$ an odd number of times. By definition, \$M=\|\\cal{M}\|\$ and \$N=\|\\cal{N}\|\$.\n\nDefine the mapping \$f:\\cal{N} \\rightarrow \\cal{M}\$ as\n\n\\[\nf(a_1, a_2, \\ldots, a_k) = (b_1,b_2,\\ldots b_k) : b_i = \\begin{cases} a_i, & \\mbox{ if } a_i \\in A \\\\ a_i-n, & \\mbox{ if } a_i \\in B \\end{cases}\n\\]\n\nWhat we want to show now is that each element of \$\\cal{M}\$ is an image of exactly \$2^{k-n}\$ elements from \$\\cal{N}\$, which would imply \$N = 2^{k-n}M\$ and solve the problem.\n\nConsider an arbitrary element \$y\$ of \$\\cal{M}\$ and let \$l_i\$ be the number of appearances of the number \$i\$ in \$y\$ for \$i=1,2,\\ldots n\$. Now consider the set of pre-images of \$y\$, that is \$X_y = \\{ x \| f(x) = y \\}\$.\n\nIt is easy to see that each element \$x\\in X_y\$ is derived from \$y\$ by flipping an even number of its \$1\$-s, \$2\$-s, and so on, where flipping means changing the number \$j\\in A\$ to \$j+n\\in B\$. Since each such set of flippings results in a unique \$x\$, all we want to count is the number of flippings. We can flip exactly \$0, 2, 4,\\ldots\$ of the \$1\$-s, so that results in\n\n\\[\n\\binom{l_1}{0} + \\binom{l_1}{2}+\\binom{l_1}{4}+\\cdots = 2^{l_1-1}\n\\]\n\nflippings. Combine each of them with the \$2^{l_2-1}\$, \$2^{l_3-1}\$, etc. ways of flipping the \$2\$-s, \$3\$-s etc. respectively to get the total number of flippings:\n\n\\[\n2^{l_1-1}2^{l_2-1}\\cdots2^{l_n-1} = 2^{l_1+l_2+\\cdots+l_n-n} = 2^{k-n}.\n\\]\n\nThis shows that \$\|X_y\| = 2^{k-n}\$ and the proof is complete. --Vbarzov 03:04, 5 September 2008 (UTC)" ]
IMO-2008-6	https://artofproblemsolving.com/wiki/index.php/2008_IMO_Problems/Problem_6	Let $ABCD$ be a convex quadrilateral with $BA \ne BC$. Denote the incircles of triangles $ABC$ and $ADC$ by $\omega _1$ and $\omega _2$ respectively. Suppose that there exists a circle $\omega$ tangent to ray $BA$ beyond $A$ and to the ray $BC$ beyond $C$, which is also tangent to the lines $AD$ and $CD$. Prove that the common external tangents to $\omega _1$ and $\omega _2$ intersect on $\omega$	[ "Here are some hints:\n\nLet B be the top vertex of triangle ABC, O and K are the centers of the incircles of triangles ABC and ADC with radii R and r, respectively. S is the center of the circumcircle tangential to the extensions of AB, AC and DC. And let\n\nE be the foot of the projection of O to AB. U be the foot of the projection of O to AC. V be the foot of the projection of K to AC. M be the foot of the projection of S to DC. L be the foot of the projection of K to DC. L be the intercept of DC and AB.\n\nWe have\n\n/_EOB = /_LSC /_AOU = /_SOC /_ASO = /_KSC /_ASK = /_OSC /_LSB = /_KCO UV = BC – AB AU = VC OA. AK. cos(/_OAK) = OC. KC. cos(/_OCK) OK2 = (R + r)2 + UV2 SK2 = (R' + r)2 + ML2\n\nUse sin (90-x) = cos x and cos(90-x) = sinx and characteristic of triangle a2 = b2 + c**2 - 2.b.c.cosine(angle) to solve.\n\nVo Duc Dien" ]
IMO-2009-1	https://artofproblemsolving.com/wiki/index.php/2009_IMO_Problems/Problem_1	Let $n$ be a positive integer and let $a_1,\ldots,a_k (k\ge2)$ be distinct integers in the set $\{1,\ldots,n\}$ such that $n$ divides $a_i(a_{i+1}-1)$ for $i=1,\ldots,k-1$. Prove that $n$ doesn't divide $a_k(a_1-1)$.	[ "Let \$n=pq\$ such that \$p\\mid a_1\$ and \$q\\mid a_2-1\$. Suppose \$n\$ divides \$a_k(a_1-1)\$. Note \$q\\mid a_2-1\$ implies \$(q,a_2)=1\$ and hence \$q\\mid a_3-1\$. Similarly one has \$q\\mid a_i-1\$ for all \$i\$'s, in particular, \$p\\mid a_1\$ and \$q\\mid a_1-1\$ force \$(p,q)=1\$. Now \$(p,a_1-1)=1\$ gives \$p\\mid a_k\$, similarly one has \$p\\mid a_i\$ for all \$i\$'s, that is \$a_i\$'s satisfy \$p\\mid a_i\$ and \$q\\mid a_i-1\$, but there should be at most one such integer satisfies them within the range of \$1,2,\\ldots,n\$ for \$n=pq\$ and \$(p,q)=1\$. A contradiction!!!", "Let \$n = p_1^{b_1}p_2^{b_2} \\cdots p_s^{b_s}\$. Then after toying around with the \$p_i^{b_i}\$ and what they divide, we have that \$p_i^{b_i} \\nmid a_k\$, and so in particular, \$n \\nmid a_k\$.\n\nAssume by way of contradiction that \$n \\mid a_k(a_1 - 1)\$. Then \$n \\mid a_1 - 1\$. Now we shift our view towards the \$a_i(a_{i + 1} - 1)\$. Here each \$p_i^{b_i}\$ divides \$a_i(a_{i + 1} - 1) \\implies a_ia_{i + 1} \\equiv a_i \\pmod{p_i^{b_i}}\$. Hence we have the chain of equivalences \$a_1a_2 \\equiv a_1 \\pmod{p_i^{b_i}}, a_2a_3 \\equiv a_2 \\pmod{p_i^{b_i}}, \\dots, a_{k - 1}a_k \\equiv a_{k - 1} \\pmod {p_i^{b_i}}\$. Now we also have that \$p_i^{b_i} \\mid n \\mid a_1 - 1\$. Thus \$a_1 \\equiv 1 \\pmod{p_i^{b_i}}\$. Now plugging this value of \$a_1\$ modulo \$p_i^{b_i}\$, we obtain that \$a_1 \\equiv a_2 \\equiv a_3 \\equiv \\cdots a_k \\equiv 1 \\pmod{p_i^{b_i}}\$. Hence this chain of congruences is also true for \$n\$ as \$p_i\$ was arbitrary. However as all the \$a_i \\in \\{1, 2, \\dots, n\\}\$ we have that not all the \$a_i\$ are distinct, and so this is a contradiction.", "Suppose by contradiction that \$n=pq\$ and \$n \| a_k(a_1-1)\$ such that \$p \| a_1 -1\$ and \$q \| a_k\$. Since \$(p,a_1) = 1\$, but \$pq \| a_1(a_2-1)\$ one has \$p\|a_2-1\$. Furthermore by induction \$p\|a_k-1\$. Similarly since \$(q,a_k-1)=1\$, but \$pq \| a_{k-1}(a_k-1)\$ one has \$q \| a_{k-1}\$. By induction \$q\| a_1\$. We've shown\n\n\\[\nn=pq \| a_1(a_k-1)\n\\]\n\nTherefore \$n \| a_1(a_k-1) - a_k(a_1-1) = a_k-a_1\$ which is impossible since \$0< \|a_k-a_1\| < n\$.\n\n~not_detriti" ]
IMO-2009-2	https://artofproblemsolving.com/wiki/index.php/2009_IMO_Problems/Problem_2	Let $ABC$ be a triangle with circumcentre $O$. The points $P$ and $Q$ are interior points of the sides $CA$ and $AB$ respectively. Let $K,L$ and $M$ be the midpoints of the segments $BP,CQ$ and $PQ$, respectively, and let $\Gamma$ be the circle passing through $K,L$ and $M$. Suppose that the line $PQ$ is tangent to the circle $\Gamma$. Prove that $OP=OQ$.	[ "## Diagram\n\n\\[\n[asy] dot(\"O\", (50, 38), N); dot(\"A\", (40, 100), N); dot(\"B\", (0, 0), S); dot(\"C\", (100, 0), S); dot(\"Q\", (24, 60), W); dot(\"P\", (52, 80), E); dot(\"L\", (62, 30), SE); dot(\"M\", (38, 70), N); dot(\"K\", (27, 42), W); draw((100, 0)--(24, 60), dotted); draw((0, 0)--(52, 80), dashed); draw((0, 0)--(100, 0)--(40, 100)--cycle); draw((24, 60)--(52, 80)); draw((27, 42)--(38, 70)--(62, 30)--cycle); draw(circle((49, 49), 23)); label(\"$\\Gamma$\", (72, 49), E); draw(circle((50, 38), 63)); label(\"$\\omega$\", (-13, 38), NW); [/asy]\n\\]\n\nDiagram by qwertysri987\n\nBy parallel lines and the tangency condition,\n\n\\[\n\\angle APM\\cong \\angle LMP \\cong \\angle LKM.\n\\]\n\nSimilarly,\n\n\\[\n\\angle AQP\\cong \\angle KLM,\n\\]\n\nso AA similarity implies\n\n\\[\n\\triangle APQ\\sim \\triangle MKL.\n\\]\n\nLet \$\\omega\$ denote the circumcircle of \$\\triangle ABC,\$ and \$R\$ its circumradius. As both \$P\$ and \$Q\$ are inside \$\\omega,\$\n\n\\[\n\\begin{align} R^2-QO^2&=\\text{Pow}_{\\omega}(Q)\\\\ &=QB\\cdot AQ \\\\ &=2AQ\\cdot MK\\\\ &=2AP\\cdot ML\\\\ &=AP\\cdot PC\\\\ &=\\text{Pow}_{\\omega}(P)\\\\ &=R^2-PO^2. \\end{align}\n\\]\n\nIt follows that \$OP=OQ.\$ \$\\blacksquare\$" ]
IMO-2009-3	https://artofproblemsolving.com/wiki/index.php/2009_IMO_Problems/Problem_3	Suppose that $s_1,s_2,s_3,\ldots$ is a strictly increasing sequence of positive integers such that the subsequences \[ s_{s_1},s_{s_2},s_{s_3},\ldots \] and \[ s_{s_1+1},s_{s_2+1},s_{s_3+1},\ldots \] are both arithmetic progressions. Prove that the sequence $s_1,s_2,s_3,\ldots$ is itself an arithmetic progression.	[ "then\n\ni.s.w.\n\nput S(3) = b, S(2) = a, S(1) = k\n\ntherefore, every S(n) is an arithmetic sequence. Q.E.D." ]
IMO-2009-4	https://artofproblemsolving.com/wiki/index.php/2009_IMO_Problems/Problem_4	Let $ABC$ be a triangle with $AB=AC$. The angle bisectors of $\angle CAB$ and $\angle ABC$ meet the sides $BC$ and $CA$ at $D$ and $E$, respectively. Let $K$ be the incenter of triangle $ADC$. Suppose that $\angle BEK=45^\circ$. Find all possible values of $\angle CAB$.	[ "Extend \$CK\$ to meet \$BE\$ at \$I\$. Then, we can see that \$I\$ is the incenter of \$\\triangle ABC\$, so \$IM=ID\$, where \$M\$ is the intersection of the incircle with \$\\overline{AC}\$.\n\nSince \$CI\$ bisects \$\\angle ACB\$, we have \$\\triangle IDC \\cong \\triangle IMC\$, so \$\\angle IMK = \\angle IDK = 45^\\circ\$.\n\nFrom here, there are two possibilities: either \$M\$ and \$E\$ coincide or they don't. If \$M\$ and \$E\$ coincide, then \$BM\$ is the median and the altitude from \$B\$, so \$BC = AB\$, and therefore \$\\triangle ABC\$ is equilateral, so \$\\angle BAC = 60^\\circ\$.\n\nOtherwise, we have \$MIKE\$ is cyclic, and \$\\angle IME = 90^\\circ\$, so \$IE\$ is the diameter of \$(MIKE)\$, so \$\\angle IKE = 90^\\circ\$ and \$\\angle KIE = 45^\\circ\$. Also, \$\\angle BIC = 180^\\circ - \\angle KIE = 135^\\circ\$, so \$2x = 45^\\circ\$, so \$\\angle CAB = 180^\\circ = 4x = 90^\\circ\$, which is the second possible value of \$\\angle CAB\$.\n\n\\[\n2009 IMO P4 (1).png\n\\]", "We will be using the trig bash solution given in EGMO.\n\nLet \$I\$ be the incenter, and set \$\\angle DAC = 2x\$ with \$0 < x < 45\$. From \$\\angle AIE = \\angle DIC\$, it is easy to compute\n\n\\[\n\\angle KIE = 90^\\circ - 2x, \\quad \\angle ECI = 45^\\circ - x, \\quad \\angle IEK = 45^\\circ, \\quad \\angle KEC = 3x.\n\\]\n\nThen, by LoS, we have\n\n\\[\n\\dfrac{IK}{KC} = \\dfrac{\\sin 45^\\circ \\cdot \\frac{EK}{\\sin(90^\\circ - 2x)}}{\\sin(3x) \\cdot \\frac{EK}{\\sin(45^\\circ-x)}} = \\frac{\\sin 45^\\circ \\sin(45^\\circ - x)}{\\sin(3x) \\sin(90^\\circ-2x)}.\n\\]\n\nAlso, by the angle bisector theorem on \$\\triangle IDC\$, we get\n\n\\[\n\\dfrac{IK}{KC} = \\dfrac{ID}{DC} = \\frac{\\sin(45^\\circ-x)}{\\sin(45^\\circ+x)}.\n\\]\n\nEquating the above two equations and cancelling \$\\sin(45^\\circ-x)\$, we get\n\n\\[\n\\sin 45^\\circ \\sin(45^\\circ+x)=\\sin 3x \\sin(90^\\circ - 2x).\n\\]\n\nApplying product to sum, we get\n\n\\[\n\\cos(x)-\\cos(90^\\circ+x) = \\cos(5x-90^\\circ)-\\cos(90^\\circ+x),\n\\]\n\nor simply \$\\cos x = \\cos(5x-90^\\circ)\$. Then, applying difference to product, we get\n\n\\[\n0 = \\cos(5x-90^\\circ)-\\cos x = 2 \\sin(3x-45^\\circ) \\sin(2x-45^\\circ).\n\\]\n\nThen, we get two cases: \$\\sin(3x-45^\\circ) = 0\$ or \$\\sin(2x-45^\\circ) = 0\$. Note that these hold iff the expressions inside the sines are multiples of \$180^\\circ\$. Using the bound \$0 < x < 45\$, we can easily get that the only possible values are \$x = 15^\\circ\$ and \$x = \\frac{45}{2}^\\circ\$, so \$\\angle A = 60^\\circ, 90^\\circ\$.\n\n\\[\n2009 IMO P4 (2).png\n\\]" ]

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