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+++ "b/JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl"
@@ -1,37 +1,37 @@
-{"year": "2008", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "problem": "If for the real numbers $x, y, z, k$ the following conditions are valid, $x \\neq y \\neq z \\neq x$ and $x^{3}+y^{3}+k\\left(x^{2}+y^{2}\\right)=y^{3}+z^{3}+k\\left(y^{2}+z^{2}\\right)=z^{3}+x^{3}+k\\left(z^{2}+x^{2}\\right)=2008$, find the product $x y z$.", "solution": "$x^{3}+y^{3}+k\\left(x^{2}+y^{2}\\right)=y^{3}+z^{3}+k\\left(y^{2}+z^{2}\\right) \\Rightarrow x^{2}+x z+z^{2}=-k(x+z):(1)$ and $y^{3}+z^{3}+k\\left(y^{2}+z^{2}\\right)=z^{3}+x^{3}+k\\left(z^{2}+x^{2}\\right) \\Rightarrow y^{2}+y x+x^{2}=-k(y+x):(2)$\n\n- From (1) $-(2) \\Rightarrow x+y+z=-k:(*)$\n- If $x+z=0$, then from $(1) \\Rightarrow x^{2}+x z+z^{2}=0 \\Rightarrow(x+z)^{2}=x z \\Rightarrow x z=0$\n\nSo $x=z=0$, contradiction since $x \\neq z$ and therefore $(1) \\Rightarrow-k=\\frac{x^{2}+x z+z^{2}}{x+z}$\n\nSimilarly we have: $-k=\\frac{y^{2}+y x+x^{2}}{y+x}$.\n\nSo $\\frac{x^{2}+x z+z^{2}}{x+z}=\\frac{y^{2}+x y+x^{2}}{x+y}$ from which $x y+y z+z x=0:(* *)$.\n\nWe substitute $k$ in $x^{3}+y^{3}+k\\left(x^{2}+y^{2}\\right)=2008$ from the relation $(*)$ and using the $(* *)$, we finally obtain that $2 x y z=2008$ and therefore $x y z=1004$.\n\nRemark: $x, y, z$ must be the distinct real solutions of the equation $t^{3}+k t^{2}-1004=0$. Such solutions exist if (and only if) $k>3 \\sqrt[3]{251}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nA1 ", "solution_tag": "\nSolution"}}
-{"year": "2008", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "problem": "Find all real numbers $a, b, c, d$ such that $a+b+c+d=20$ and $a b+a c+a d+b c+b d+c d=$ 150 .", "solution": "$400=(a+b+c+d)^{2}=a^{2}+b^{2}+c^{2}+d^{2}+2 \\cdot 150$, so $a^{2}+b^{2}+c^{2}+d^{2}=100$. Now $(a-b)^{2}+(a-c)^{2}+(a-d)^{2}+(b-c)^{2}+(b-d)^{2}+(c-d)^{2}=3\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)-2(a b+$ $a c+a d+b c+b d+c d)=300-300=0$. Thus $a=b=c=d=5$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nA2 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "problem": "Let the real parameter $p$ be such that the system\n\n$$\n\\left\\{\\begin{array}{l}\np\\left(x^{2}-y^{2}\\right)=\\left(p^{2}-1\\right) x y \\\\\n|x-1|+|y|=1\n\\end{array}\\right.\n$$\n\nhas at least three different real solutions. Find $p$ and solve the system for that $p$.", "solution": "The second equation is invariant when $y$ is replaced by $-y$, so let us assume $y \\geq 0$. It is also invariant when $x-1$ is replaced by $-(x-1)$, so let us assume $x \\geq 1$. Under these conditions the equation becomes $x+y=2$, which defines a line on the coordinate plane. The set of points on it that satisfy the inequalities is a segment with endpoints $(1,1)$ and $(2,0)$. Now taking into account the invariance under the mentioned replacements, we conclude that the set of points satisfying the second equation is the square $\\diamond$ with vertices $(1,1),(2,0),(1,-1)$ and $(0,0)$.\n\nThe first equation is equivalent to\n\n$p x^{2}-p^{2} x y+x y-p y^{2}=0$\n\n$p x(x-p y)+y(x-p y)=0$\n\n$(p x+y)(x-p y)=0$.\n\nThus $y=-p x$ or $x=p y$. These are equations of two perpendicular lines passing through the origin, which is also a vertex of $\\diamond$. If one of them passes through an interior point of the square, the other cannot have any common points with $\\diamond$ other than $(0,0)$, so the system has two solutions. Since we have at least three different real solutions, the lines must contain some sides of $\\diamond$, i.e. the slopes of the lines have to be 1 and -1 . This happens if $p=1$ or $p=-1$. In either case $x^{2}=y^{2},|x|=|y|$, so the second equation becomes $|1-x|+|x|=1$. It is true exactly when $0 \\leq x \\leq 1$ and $y= \\pm x$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nA3 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "problem": "Find all triples $(x, y, z)$ of real numbers that satisfy the system\n\n$$\n\\left\\{\\begin{array}{l}\nx+y+z=2008 \\\\\nx^{2}+y^{2}+z^{2}=6024^{2} \\\\\n\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=\\frac{1}{2008}\n\\end{array}\\right.\n$$", "solution": "The last equation implies $x y z=2008(x y+y z+z x)$, therefore $x y z-2008(x y+y z+z x)+$ $2008^{2}(x+y+z)-2008^{3}=0$.\n\n$(x-2008)(y-2008)(z-2008)=0$.\n\nThus one of the variable equals 2008. Let this be $x$. Then the first equation implies $y=-z$. From the second one it now follows that $2 y^{2}=6024^{2}-2008^{2}=2008^{2}(9-1)=$ $2 \\cdot 4016^{2}$. Thus $(x, y, z)$ is the triple $(2008,4016,-4016)$ or any of its rearrangements.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nA4 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "A5", "problem_type": "Algebra", "problem": "Find all triples $(x, y, z)$ of real positive numbers, which satisfy the system\n\n$$\n\\left\\{\\begin{array}{l}\n\\frac{1}{x}+\\frac{4}{y}+\\frac{9}{z}=3 \\\\\nx+y+z \\leq 12\n\\end{array}\\right.\n$$", "solution": "If we multiply the given equation and inequality $(x>0, y>0, z>0)$, we have\n\n$$\n\\left(\\frac{4 x}{y}+\\frac{y}{x}\\right)+\\left(\\frac{z}{x}+\\frac{9 x}{z}\\right)+\\left(\\frac{4 z}{y}+\\frac{9 y}{z}\\right) \\leq 22\n$$\n\nFrom AM-GM we have\n\n$$\n\\frac{4 x}{y}+\\frac{y}{x} \\geq 4, \\quad \\frac{z}{x}+\\frac{9 x}{z} \\geq 6, \\quad \\frac{4 z}{y}+\\frac{9 y}{z} \\geq 12\n$$\n\nTherefore\n\n$$\n22 \\leq\\left(\\frac{4 x}{y}+\\frac{y}{x}\\right)+\\left(\\frac{z}{x}+\\frac{9 x}{z}\\right)+\\left(\\frac{4 z}{y}+\\frac{9 y}{z}\\right)\n$$\n\nNow from (1) and (3) we get\n\n$$\n\\left(\\frac{4 x}{y}+\\frac{y}{x}\\right)+\\left(\\frac{z}{x}+\\frac{9 x}{z}\\right)+\\left(\\frac{4 z}{y}+\\frac{9 y}{z}\\right)=22\n$$\n\nwhich means that in (2), everywhere equality holds i.e. we have equality between means, also $x+y+z=12$.\n\nTherefore $\\frac{4 x}{y}=\\frac{y}{x}, \\frac{z}{x}=\\frac{9 x}{z}$ and, as $x>0, y>0, z>0$, we get $y=2 x, z=3 x$. Finally if we substitute for $y$ and $z$, in $x+y+z=12$, we get $x=2$, therefore $y=2 \\cdot 2=4$ and $z=3 \\cdot 2=6$.\n\nThus the unique solution is $(x, y, z)=(2,4,6)$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nA5 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "A6", "problem_type": "Algebra", "problem": "If the real numbers $a, b, c, d$ are such that $0<a, b, c, d<1$, show that\n\n$$\n1+a b+b c+c d+d a+a c+b d>a+b+c+d\n$$", "solution": "If $1 \\geq a+b+c$ then we write the given inequality equivalently as\n\n$$\n\\begin{gathered}\n1-(a+b+c)+d[(a+b+c)-1]+a b+b c+c a>0 \\\\\n\\Leftrightarrow[1-(a+b+c)](1-d)+a b+b c+c a>0\n\\end{gathered}\n$$\n\nwhich is of course true.\n\nIf instead $a+b+c>1$, then $d(a+b+c)>d$ i.e.\n\n$$\nd a+d b+d c>d\n$$\n\nWe are going to prove that also\n\n$$\n1+a b+b c+c a>a+b+c\n$$\n\nthus adding (1) and (2) together we'll get the desired result in this case too.\n\nFor the truth of $(2)$ :\n\nIf $1 \\geq a+b$, then we rewrite (2) equivalently as\n\n$$\n\\begin{gathered}\n1-(a+b)+c[(a+b)-1]+a b>0 \\\\\n\\quad \\Leftrightarrow[1-(a+b)](1-c)+a b>0\n\\end{gathered}\n$$\n\nwhich is of course true.\n\nIf instead $a+b>1$, then $c(a+b)>c$, i.e.\n\n$$\nc a+c b>c\n$$\n\nBut it is also true that\n\n$$\n1+a b>a+b\n$$\n\nbecause this is equivalent to $(1-a)+b(a-1)>0$, i.e. to $(1-a)(1-b)>0$ which holds. Adding (3) and (4) together we get the truth of (2) in this case too and we are done. You can instead consider the following generalization:\n\nExercise. If for the real numbers $x_{1}, x_{2}, \\ldots, x_{n}$ it is $0<x_{i}<1$, for any $i$, show that\n\n$$\n1+\\sum_{1 \\leq i<j \\leq n} x_{i} x_{j}>\\sum_{i=1}^{n} x_{i}\n$$\n\n## Solution\n\nWe'll prove it by induction.\n\nFor $n=1$ the desired result becomes $1>x_{1}$ which is true.\n\nLet the result be true for some natural number $n \\geq 1$.\n\nWe'll prove it to be true for $n+1$ as well, and we'll be done.\n\nSo let $x_{1}, x_{2}, \\ldots, x_{n}, x_{n+1}$ be $n+1$ given real numbers with $0<x_{i}<1$, for any $i$. We wish to show that\n\n$$\n1+\\sum_{1 \\leq i<j \\leq n+1} x_{i} x_{j}>x_{1}+x_{2}+\\ldots+x_{n}+x_{n+1}\n$$\n\nIf $1 \\geq x_{1}+x_{2}+\\ldots+x_{n}$ then we rewrite (5) equivalently as\n\n$$\n1-\\left(x_{1}+x_{2}+\\ldots+x_{n}\\right)+x_{n+1}\\left(x_{1}+x_{2}+\\ldots+x_{n}-1\\right)+\\sum_{1 \\leq i<j \\leq n} x_{i} x_{j}>0\n$$\n\nThis is also written as\n\n$$\n\\left(1-x_{n+1}\\right)\\left[1-\\left(x_{1}+x_{2}+\\ldots+x_{n}\\right)\\right]+\\sum_{1 \\leq i<j \\leq n} x_{i} x_{j}>0\n$$\n\nwhich is clearly true.\n\nIf instead $x_{1}+x_{2}+\\ldots+x_{n}>1$ then $x_{n+1}\\left(x_{1}+x_{2}+\\ldots+x_{n}\\right)>x_{n+1}$, i.e.\n\n$$\nx_{n+1} x_{1}+x_{n+1} x_{2}+\\ldots+x_{n+1} x_{n}>x_{n+1}\n$$\n\nBy the induction hypothesis applied to the $n$ real numbers $x_{1}, x_{2}, \\ldots, x_{n}$ we also know that\n\n$$\n1+\\sum_{1 \\leq i<j \\leq n} x_{i} x_{j}>\\sum_{i=1}^{n} x_{i}\n$$\n\nAdding (6) and (7) together we get the validity of (5) in this case too, and we are done.\n\nYou can even consider the following variation:\n\nExercise. If the real numbers $x_{1}, x_{2}, \\ldots, x_{2008}$ are such that $0<x_{i}<1$, for any $i$, show that\n\n$$\n1+\\sum_{1 \\leq i<j \\leq 2008} x_{i} x_{j}>\\sum_{i=1}^{2008} x_{i}\n$$\n\nRemark: Inequality (2) follows directly from $(1-a)(1-b)(1-c)>0 \\Leftrightarrow 1-a-b-c+$ $a b+b c+c a>a b c>0$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nA6 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "A7", "problem_type": "Algebra", "problem": "Let $a, b$ and $c$ be a positive real numbers such that $a b c=1$. Prove the inequality\n\n$$\n\\left(a b+b c+\\frac{1}{c a}\\right)\\left(b c+c a+\\frac{1}{a b}\\right)\\left(c a+a b+\\frac{1}{b c}\\right) \\geq(1+2 a)(1+2 b)(1+2 c)\n$$", "solution": "By Cauchy-Schwarz inequality and $a b c=1$ we get\n\n$$\n\\begin{gathered}\n\\sqrt{\\left(b c+c a+\\frac{1}{a b}\\right)\\left(a b+b c+\\frac{1}{c a}\\right)}=\\sqrt{\\left(b c+c a+\\frac{1}{a b}\\right)\\left(\\frac{1}{c a}+a b+b c\\right)} \\geq \\\\\n\\left(\\sqrt{a b} \\cdot \\sqrt{\\frac{1}{a b}}+\\sqrt{b c} \\cdot \\sqrt{b c}+\\sqrt{\\frac{1}{c a}} \\cdot \\sqrt{c a}\\right)=(2+b c)=(2 a b c+b c)=b c(1+2 a)\n\\end{gathered}\n$$\n\nAnalogously we get $\\sqrt{\\left(b c+c a+\\frac{1}{a b}\\right)\\left(c a+a b+\\frac{1}{b c}\\right)} \\geq c a(1+2 b)$ and\n\n$\\sqrt{\\left(c a+a b+\\frac{1}{b c}\\right)\\left(a b+b c+\\frac{1}{c a}\\right)} \\geq a b(1+2 a)$.\n\nMultiplying these three inequalities we get:\n\n$$\n\\left(a b+b c+\\frac{1}{c a}\\right)\\left(b c+c a+\\frac{1}{a b}\\right)\\left(c a+a b+\\frac{1}{b c}\\right) \\geq a^{2} b^{2} c^{2}(1+2 a)(1+2 b)(1+2 c)=\n$$\n\n$(1+2 a)(1+2 b)(1+2 c)$ because $a b c=1$.\n\nEquality holds if and only if $a=b=c=1$.\n\n## Solution 2\n\nUsing $a b c=1$ we get\n\n$$\n\\begin{gathered}\n\\left(a b+b c+\\frac{1}{c a}\\right)\\left(b c+c a+\\frac{1}{a b}\\right)\\left(c a+a b+\\frac{1}{b c}\\right)= \\\\\n=\\left(\\frac{1}{c}+\\frac{1}{a}+b\\right)\\left(\\frac{1}{a}+\\frac{1}{b}+c\\right)\\left(\\frac{1}{b}+\\frac{1}{c}+a\\right)= \\\\\n=\\frac{(a+c+a b c)}{a c} \\cdot \\frac{(b+a+a b c)}{a b} \\cdot \\frac{(b+c+a b c)}{b c}=(a+b+1)(b+c+1)(c+a+1)\n\\end{gathered}\n$$\n\nThus, we need to prove\n\n$$\n(a+b+1)(b+c+1)(c+a+1) \\geq(1+2 a)(1+2 b)(1+2 c)\n$$\n\nAfter multiplication and using the fact $a b c=1$ we have to prove\n\n$$\n\\begin{gathered}\na^{2} b+a^{2} c+b^{2} c+b^{2} a+c^{2} a+c^{2} b+3(a b+b c+c a)+2(a+b+c)+a^{2}+b^{2}+c^{2}+3 \\geq \\\\\n\\geq 4(a b+b c+c a)+2(a+b+c)+9\n\\end{gathered}\n$$\n\nSo we need to prove\n\n$$\na^{2} b+a^{2} c+b^{2} c+b^{2} a+c^{2} a+c^{2} b+a^{2}+b^{2}+c^{2} \\geq a b+b c+c a+6\n$$\n\nThis follows from the well-known (AM-GM inequality) inequalities\n\n$$\na^{2}+b^{2}+c^{2} \\geq a b+b c+c a\n$$\n\nand\n\n$$\na^{2} b+a^{2} c+b^{2} c+b^{2} a+c^{2} a+c^{2} b \\geq 6 a b c=6\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nA7 ", "solution_tag": "## Solution 1"}}
-{"year": "2008", "tier": "T3", "problem_label": "A8", "problem_type": "Algebra", "problem": "Show that\n\n$$\n(x+y+z)\\left(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\right) \\geq 4\\left(\\frac{x}{x y+1}+\\frac{y}{y z+1}+\\frac{z}{z x+1}\\right)^{2}\n$$\n\nfor any real positive numbers $x, y$ and $z$.", "solution": "The idea is to split the inequality in two, showing that\n\n$$\n\\left(\\sqrt{\\frac{x}{y}}+\\sqrt{\\frac{y}{z}}+\\sqrt{\\frac{z}{x}}\\right)^{2}\n$$\n\ncan be intercalated between the left-hand side and the right-hand side. Indeed, using the Cauchy-Schwarz inequality one has\n\n$$\n(x+y+z)\\left(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\right) \\geq\\left(\\sqrt{\\frac{x}{y}}+\\sqrt{\\frac{y}{z}}+\\sqrt{\\frac{z}{x}}\\right)^{2}\n$$\n\nOn the other hand, as\n\n$$\n\\sqrt{\\frac{x}{y}} \\geq \\frac{2 x}{x y+1} \\Leftrightarrow(\\sqrt{x y}-1)^{2} \\geq 0\n$$\n\nby summation one has\n\n$$\n\\sqrt{\\frac{x}{y}}+\\sqrt{\\frac{y}{z}}+\\sqrt{\\frac{z}{x}} \\geq \\frac{2 x}{x y+1}+\\frac{2 y}{y z+1}+\\frac{2 z}{z x+1}\n$$\n\nThe rest is obvious.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nA8 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "A9", "problem_type": "Algebra", "problem": "Consider an integer $n \\geq 4$ and a sequence of real numbers $x_{1}, x_{2}, x_{3}, \\ldots, x_{n}$. An operation consists in eliminating all numbers not having the rank of the form $4 k+3$, thus leaving only the numbers $x_{3}, x_{7}, x_{11}, \\ldots$ (for example, the sequence $4,5,9,3,6,6,1,8$ produces the sequence 9,1 . Upon the sequence $1,2,3, \\ldots, 1024$ the operation is performed successively for 5 times. Show that at the end only 1 number remains and find this number.", "solution": "After the first operation 256 number remain; after the second one, 64 are left, then 16, next 4 and ultimately only one number.\n\nNotice that the 256 numbers left after the first operation are $3,7, \\ldots, 1023$, hence they are in arithmetical progression of common difference 4. Successively, the 64 numbers left after the second operation are in arithmetical progression of ratio 16 and so on.\n\nLet $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ be the first term in the 5 sequences obtained after each of the 5 operations. Thus $a_{1}=3$ and $a_{5}$ is the requested number. The sequence before the fifth operation has 4 numbers, namely\n\n$$\na_{4}, a_{4}+256, a_{4}+512, a_{4}+768\n$$\n\nand $a_{5}=a_{4}+512$. Similarly, $a_{4}=a_{3}+128, a_{3}=a_{2}+32, a_{2}=a_{1}+8$.\n\nSumming up yields $a_{5}=a_{1}+8+32+128+512=3+680=683$.\n\n### 2.2 Combinatorics", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nA9 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "problem": "On a $5 \\times 5$ board, $n$ white markers are positioned, each marker in a distinct $1 \\times 1$ square. A smart child got an assignment to recolor in black as many markers as possible, in the following manner: a white marker is taken from the board; it is colored in black, and then put back on the board on an empty square such that none of the neighboring squares contains a white marker (two squares are called neighboring if they contain a common side). If it is possible for the child to succeed in coloring all the markers black, we say that the initial positioning of the markers was good.\n\na) Prove that if $n=20$, then a good initial positioning exists.\n\nb) Prove that if $n=21$, then a good initial positioning does not exist.", "solution": "a) Position 20 white markers on the board such that the left-most column is empty. This\npositioning is good because the coloring can be realized column by column, starting with the second (from left), then the third, and so on, so that the white marker on position $(i, j)$ after the coloring is put on position $(i, j-1)$.\n\nb) Suppose there exists a good positioning with 21 white markers on the board i.e. there exists a re-coloring of them all, one by one. In any moment when there are 21 markers on the board, there must be at least one column completely filled with markers, and there must be at least one row completely filled with markers. So, there exists a \"cross\" of markers on the board. At the initial position, each such cross is completely white, at the final position each such cross is completely black, and at every moment when there are 21 markers on the board, each such cross is monochromatic. But this cannot be, since every two crosses have at least two common squares and therefore it is not possible for a white cross to vanish and for a black cross to appear by re-coloring of only one marker. Contradiction!", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nC1 ", "solution_tag": "\nSolution"}}
-{"year": "2008", "tier": "T3", "problem_label": "C2", "problem_type": "Combinatorics", "problem": "Kostas and Helene have the following dialogue:\n\nKostas: I have in my mind three positive real numbers with product 1 and sum equal to the sum of all their pairwise products.\n\nHelene: I think that I know the numbers you have in mind. They are all equal to 1.\n\nKostas: In fact, the numbers you mentioned satisfy my conditions, but I did not think of these numbers. The numbers you mentioned have the minimal sum between all possible solutions of the problem.\n\nCan you decide if Kostas is right? (Explain your answer).", "solution": "Kostas is right according to the following analysis:\n\nIf $x, y, z$ are the three positive real numbers Kostas thought about, then they satisfy the following equations:\n\n$$\n\\begin{gathered}\nx y+y z+z x=x+y+z \\\\\nx y z=1\n\\end{gathered}\n$$\n\nSubtracting (1) from (2) by parts we obtain\n\n$$\n\\begin{gathered}\nx y z-(x y+y z+z x)=1-(x+y+z) \\\\\n\\Leftrightarrow x y z-x y-y z-z x+x+y+z-1=0 \\\\\n\\Leftrightarrow x y(z-1)-x(z-1)-y(z-1)+(z-1)=0 \\\\\n\\Leftrightarrow(z-1)(x y-x-y+1)=0 \\\\\n(z-1)(x-1)(y-1)=0 \\\\\n\\Leftrightarrow x=1 \\text { or } y=1 \\text { or } z=1 .\n\\end{gathered}\n$$\n\nFor $x=1$, from (1) and (2) we have the equation $y z=1$, which has the solutions\n\n$$\n(y, z)=\\left(a, \\frac{1}{a}\\right), a>0\n$$\n\nAnd therefore the solutions of the problem are the triples\n\n$$\n(x, y, z)=\\left(1, a, \\frac{1}{a}\\right), a>0\n$$\n\nSimilarly, considering $y=1$ or $z=1$ we get the solutions\n\n$$\n(x, y, z)=\\left(a, 1, \\frac{1}{a}\\right) \\text { or }(x, y, z)=\\left(a, \\frac{1}{a}, 1\\right), a>0\n$$\n\nSince for each $a>0$ we have\n\n$$\nx+y+z=1+a+\\frac{1}{a} \\geq 1+2=3\n$$\n\nand equality is valid only for $a=1$, we conclude that among the solutions of the problem, the triple $(x, y, z)=(1,1,1)$ is the one whose sum $x+y+z$ is minimal.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nC2 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "problem": "Integers $1,2, \\ldots, 2 n$ are arbitrarily assigned to boxes labeled with numbers $1,2, \\ldots, 2 n$. Now, we add the number assigned to the box to the number on the box label. Show that two such sums give the same remainder modulo $2 n$.", "solution": "Let us assume that all sums give different remainder modulo $2 n$, and let $S$ denote the value of their sum.\n\nFor our assumption,\n\n$$\nS \\equiv 0+1+\\ldots+2 n-1=\\frac{(2 n-1) 2 n}{2}=(2 n-1) n \\equiv n \\quad(\\bmod 2 n)\n$$\n\nBut, if we sum, breaking all sums into its components, we derive\n\n$$\nS \\equiv 2(1+\\ldots+2 n)=2 \\cdot \\frac{2 n(2 n+1)}{2}=2 n(2 n+1) \\equiv 0 \\quad(\\bmod 2 n)\n$$\n\nFrom the last two conclusions we derive $n \\equiv 0(\\bmod 2 n)$. Contradiction.\n\nTherefore, there are two sums with the same remainder modulo $2 n$.\n\nRemark: The result is no longer true if one replaces $2 n$ by $2 n+1$. Indeed, one could assign the number $k$ to the box labeled $k$, thus obtaining the sums $2 k, k=\\overline{1,2 n+1}$. Two such numbers give different remainders when divided by $2 n+1$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nC3 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "C4", "problem_type": "Combinatorics", "problem": "Every cell of table $4 \\times 4$ is colored into white. It is permitted to place the cross (pictured below) on the table such that its center lies on the table (the whole figure does not need to lie on the table) and change colors of every cell which is covered into opposite (white and black). Find all $n$ such that after $n$ steps it is possible to get the table with every cell colored black.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_cef994f1dc1cf35663f9g-17.jpg?height=182&width=181&top_left_y=451&top_left_x=945)", "solution": "The cross covers at most five cells so we need at least 4 steps to change the color of every cell. If we place the cross 4 times such that its center lies in the cells marked below, we see that we can turn the whole square black in $n=4$ moves.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_cef994f1dc1cf35663f9g-17.jpg?height=198&width=201&top_left_y=1018&top_left_x=932)\n\nFurthermore, applying the same operation twice (,,do and undo\"), we get that is possible to turn all the cells black in $n$ steps for every even $n \\geq 4$.\n\nWe shall prove that for odd $n$ it is not possible to do that. Look at the picture below.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_cef994f1dc1cf35663f9g-17.jpg?height=196&width=206&top_left_y=1550&top_left_x=930)\n\nLet $k$ be a difference between white and black cells in the green area in picture. Every figure placed on the table covers an odd number of green cells, so after every step $k$ is changed by a number $\\equiv 2(\\bmod 4)$. At the beginning $k=10$, at the end $k=-10$. From this it is clear that we need an even number of steps.\n\nSolution for $n$ is: every even number except 2 .\n\n### 2.3 Geometry", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nC4 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "problem": "Two perpendicular chords of a circle, $A M, B N$, which intersect at point $K$, define on the circle four arcs with pairwise different length, with $A B$ being the smallest of them.\n\nWe draw the chords $A D, B C$ with $A D \\| B C$ and $C, D$ different from $N, M$. If $L$ is the point of intersection of $D N, M C$ and $T$ the point of intersection of $D C, K L$, prove that $\\angle K T C=\\angle K N L$.", "solution": "First we prove that $N L \\perp M C$. The arguments depend slightly on the position of $D$. The other cases are similar.\n\nFrom the cyclic quadrilaterals $A D C M$ and $D N B C$ we have:\n\n$$\n\\varangle D C L=\\varangle D A M \\text { and } \\varangle C D L=\\varangle C B N \\text {. }\n$$\n\nSo we obtain\n\n$$\n\\varangle D C L+\\varangle C D L=\\varangle D A M+\\varangle C B N .\n$$\n\nAnd because $A D \\| B C$, if $Z$ the point of intersection of $A M, B C$ then $\\varangle D A M=\\varangle B Z A$, and we have\n\n$$\n\\varangle D C L+\\varangle C D L=\\varangle B Z A+\\varangle C B N=90^{\\circ}\n$$\n\nLet $P$ the point of intersection of $K L, A C$, then $N P \\perp A C$, because the line $K P L$ is a Simson line of the point $N$ with respect to the triangle $A C M$.\n\nFrom the cyclic quadrilaterals $N P C L$ and $A N D C$ we obtain:\n\n$$\n\\varangle C P L=\\varangle C N L \\text { and } \\varangle C N L=\\varangle C A D \\text {, }\n$$\n\nso $\\varangle C P L=\\varangle C A D$, that is $K L\\|A D\\| B C$ therefore $\\varangle K T C=\\varangle A D C$ (1).\n\nBut $\\varangle A D C=\\varangle A N C=\\varangle A N K+\\varangle K N C=\\varangle C N L+\\varangle K N C$, so\n\n$$\n\\varangle A D C=\\varangle K N L\n$$\n\nFrom (1) and (2) we obtain the result.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_cef994f1dc1cf35663f9g-18.jpg?height=542&width=518&top_left_y=1710&top_left_x=782)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nG1 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "G2", "problem_type": "Geometry", "problem": "For a fixed triangle $A B C$ we choose a point $M$ on the ray $C A$ (after $A$ ), a point $N$ on the ray $A B$ (after $B$ ) and a point $P$ on the ray $B C$ (after $C$ ) in a way such that $A M-B C=B N-A C=C P-A B$. Prove that the angles of triangle $M N P$ do not depend on the choice of $M, N, P$.", "solution": "Consider the points $M^{\\prime}$ on the ray $B A$ (after $A$ ), $N^{\\prime}$ on the ray $C B$ (after $B$ ) and $P^{\\prime}$ on the ray $A C$ (after $C$ ), so that $A M=A M^{\\prime}, B N=B N^{\\prime}, C P=C P^{\\prime}$. Since $A M-B C=B N-A C=B N^{\\prime}-A C$, we get $C M=A C+A M=B C+B N^{\\prime}=C N^{\\prime}$. Thus triangle $M C N^{\\prime}$ is isosceles, so the perpendicular bisector of $\\left[M N^{\\prime}\\right]$ bisects angle $A C B$ and hence passes through the incenter $I$ of triangle $A B C$. Arguing similarly, we may conclude that $I$ lies also on the perpendicular bisectors of $\\left[N P^{\\prime}\\right]$ and $\\left[P M^{\\prime}\\right]$. On the other side, $I$ clearly lies on the perpendicular bisectors of $\\left[M M^{\\prime}\\right],\\left[N N^{\\prime}\\right]$ and $\\left[P P^{\\prime}\\right]$. Thus the hexagon $M^{\\prime} M N^{\\prime} N P^{\\prime} P$ is cyclic. Then angle $P M N$ equals angle $P N^{\\prime} N$, which measures $90^{\\circ}-\\frac{\\beta}{2}$ (the angles of triangle $A B C$ are $\\alpha, \\beta, \\gamma$ ). In the same way angle $M N P$ measures $90^{\\circ}-\\frac{\\gamma}{2}$ and angle $M P N$ measures $90^{\\circ}-\\frac{\\alpha}{2}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nG2 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "G3", "problem_type": "Geometry", "problem": "The vertices $A$ and $B$ of an equilateral $\\triangle A B C$ lie on a circle $k$ of radius 1 , and the vertex $C$ is inside $k$. The point $D \\neq B$ lies on $k, A D=A B$ and the line $D C$ intersects $k$ for the second time in point $E$. Find the length of the segment $C E$.", "solution": "As $A D=A C, \\triangle C D A$ is isosceles. If $\\varangle A D C=\\varangle A C D=\\alpha$ and $\\varangle B C E=\\beta$, then $\\beta=120^{\\circ}-\\alpha$. The quadrilateral $A B E D$ is cyclic, so $\\varangle A B E=180^{\\circ}-\\alpha$. Then $\\varangle C B E=$ $120^{\\circ}-\\alpha$ so $\\varangle C B E=\\beta$. Thus $\\triangle C B E$ is isosceles, so $A E$ is the perpendicular bisector of $B C$, so it bisects $\\varangle B A C$. Now the arc $B E$ is intercepted by a $30^{\\circ}$ inscribed angle, so it measures $60^{\\circ}$. Then $B E$ equals the radius of $k$, namely 1 . Hence $C E=B E=1$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_cef994f1dc1cf35663f9g-19.jpg?height=458&width=485&top_left_y=1614&top_left_x=798)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nG3 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "problem": "Let $A B C$ be a triangle, $(B C<A B)$. The line $\\ell$ passing trough the vertices $C$ and orthogonal to the angle bisector $B E$ of $\\angle B$, meets $B E$ and the median $B D$ of the side $A C$ at points $F$ and $G$, respectively. Prove that segment $D F$ bisect the segment $E G$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_cef994f1dc1cf35663f9g-20.jpg?height=434&width=314&top_left_y=474&top_left_x=884)", "solution": "Let $C F \\cap A B=\\{K\\}$ and $D F \\cap B C=\\{M\\}$. Since $B F \\perp K C$ and $B F$ is angle bisector of $\\varangle K B C$, we have that $\\triangle K B C$ is isosceles i.e. $B K=B C$, also $F$ is midpoint of $K C$. Hence $D F$ is midline for $\\triangle A C K$ i.e. $D F \\| A K$, from where it is clear that $M$ is a midpoint of $B C$.\n\nWe will prove that $G E \\| B C$. It is sufficient to show $\\frac{B G}{G D}=\\frac{C E}{E D}$. From $D F \\| A K$ and $D F=\\frac{A K}{2}$ we have\n\n$$\n\\frac{B G}{G D}=\\frac{B K}{D F}=\\frac{2 B K}{A K}\n$$\n\nAlso\n\n$$\n\\begin{gathered}\n\\frac{C E}{D E}=\\frac{C D-D E}{D E}=\\frac{C D}{D E}-1=\\frac{A D}{D E}-1=\\frac{A E-D E}{D E}-1=\\frac{A E}{D E}-2= \\\\\n=\\frac{A B}{D F}-2=\\frac{A K+B K}{\\frac{A K}{2}}-2=2+2 \\frac{B K}{A K}-2=\\frac{2 B K}{A K}\n\\end{gathered}\n$$\n\nFrom (1) and (2) we have $\\frac{B G}{G D}=\\frac{C E}{E D}$, so $G E \\| B C$, as $M$ is the midpoint of $B C$, it follows that the segment $D F$, bisects the segment $G E$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nG4 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "G5", "problem_type": "Geometry", "problem": "Is it possible to cover a given square with a few congruent right-angled triangles with acute angle equal to $30^{\\circ}$ ? (The triangles may not overlap and may not exceed the margins of the square.)", "solution": "We will prove that desired covering is impossible.\n\nLet assume the opposite i.e. a square with side length $a$, can be tiled with $k$ congruent right angled triangles, whose sides are of lengths $b, b \\sqrt{3}$ and $2 b$.\n\nThen the area of such a triangle is $\\frac{b^{2} \\sqrt{3}}{2}$.\n\nAnd the area of the square is\n\n$$\nS_{s q}=k b^{2} \\frac{\\sqrt{3}}{2}\n$$\n\nFurthermore, the length of the side of the square, $a$, is obtained by the contribution of an integer number of length $b, 2 b$ and $b \\sqrt{3}$, hence\n\n$$\na=m b \\sqrt{3}+n b\n$$\n\nwhere $m, n \\in \\mathbb{N} \\cup\\{0\\}$, and at least one of the numbers $m$ and $n$ is different from zero. So the area of the square is\n\n$$\nS_{s q}=a^{2}=(m b \\sqrt{3}+n b)^{2}=b^{2}\\left(3 m^{2}+n^{2}+2 \\sqrt{3} m n\\right)\n$$\n\nNow because of (1) and (2) it follows $3 m^{2}+n^{2}+2 \\sqrt{3} m n=k \\frac{\\sqrt{3}}{2}$ i.e.\n\n$$\n6 m^{2}+2 n^{2}=(k-4 m n) \\sqrt{3}\n$$\n\nBecause of $3 m^{2}+n^{2} \\neq 0$ and from the equality (3) it follows $4 m n \\neq k$. Using once more (3), we get\n\n$$\n\\sqrt{3}=\\frac{6 m^{2}+2 n^{2}}{k-4 m n}\n$$\n\nwhich contradicts at the fact that $\\sqrt{3}$ is irrational, because $\\frac{6 m^{2}+2 n^{2}}{k-4 m n}$ is a rational number.\n\nFinally, we have obtained a contradiction, which proves that the desired covering is impossible.\n\n## Remark.\n\nThis problem has been given in Russian Mathematical Olympiad 1993 - 1995 for 9-th Grade.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nG5 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "G6", "problem_type": "Geometry", "problem": "Let $A B C$ be a triangle with $A<90^{\\circ}$. Outside of a triangle we consider isosceles triangles $A B E$ and $A C Z$ with bases $A B$ and $A C$, respectively. If the midpoint $D$ of the side $B C$ is such that $D E \\perp D Z$ and $E Z=2 \\cdot E D$, prove that $\\widehat{A E B}=2 \\cdot \\widehat{A Z C}$.", "solution": "Since $D$ is the midpoint of the side $B C$, in the extension of the line segment $Z D$ we take a point $H$ such that $Z D=D H$. Then the quadrilateral $B H C Z$ is parallelogram and therefore we have\n\n$$\nB H=Z C=Z A\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_cef994f1dc1cf35663f9g-22.jpg?height=647&width=596&top_left_y=468&top_left_x=748)\n\nAlso from the isosceles triangle $A B E$ we get\n\n$$\nB E=A E\n$$\n\nSince $D E \\perp D Z, E D$ is altitude and median of the triangle $E Z H$ and so this triangle is isosceles with\n\n$$\nE H=E Z\n$$\n\nFrom (1), (2) and (3) we conclude that the triangles $B E H$ and $A E Z$ are equal. Therefore they have also\n\n$$\n\\widehat{B E H}=\\widehat{A E Z}, \\widehat{E B H}=\\widehat{E A Z} \\text { and } \\widehat{E H B}=\\widehat{A Z E}\n$$\n\nPutting $\\widehat{E B A}=\\widehat{E A B}=\\omega, \\widehat{Z A C}=\\widehat{Z C A}=\\varphi$, then we have $\\widehat{C B H}=\\widehat{B C Z}=\\widehat{C}+\\varphi$, and therefore from the equality $\\widehat{E B H}=\\widehat{E A Z}$ we receive:\n\n$$\n\\begin{gathered}\n360^{\\circ}-\\widehat{E B A}-\\widehat{B}-\\widehat{C B H}=\\widehat{E A B}+\\widehat{A}+\\widehat{Z A C} \\\\\n\\Rightarrow 360^{\\circ}-\\widehat{B}-\\omega-\\varphi-\\widehat{C}=\\omega+\\widehat{A}+\\varphi \\\\\n\\Rightarrow 2(\\omega+\\varphi)=360^{\\circ}-(\\widehat{A}+\\widehat{B}+\\widehat{C}) \\\\\n\\Rightarrow \\omega+\\varphi=90^{\\circ} \\\\\n\\Rightarrow \\frac{180^{\\circ}-\\widehat{A E B}}{2}+\\frac{180^{\\circ}-\\widehat{A Z C}}{2}=90^{\\circ} \\\\\n\\Rightarrow \\widehat{A E B}+\\widehat{A Z C}=180^{\\circ}\n\\end{gathered}\n$$\n\nFrom the supposition $E Z=2 \\cdot E D$, we get that the right triangle $Z E H$ has $\\widehat{E Z D}=30^{\\circ}$ and $\\widehat{Z E D}=60^{\\circ}$. Thus we have $\\widehat{Z E H}=120^{\\circ}$.\n\nHowever, since we have proved that $\\widehat{B E H}=\\widehat{A E Z}$, we get that\n\n$$\n\\widehat{A E B}=\\widehat{A E Z}+\\widehat{Z E B}=\\widehat{Z E B}+\\widehat{B E H}=\\widehat{Z E H}=120^{\\circ}\n$$\n\nFrom (5) and (6) we obtain that $\\widehat{A Z C}=60^{\\circ}$ and thus $\\widehat{A E B}=2 \\cdot \\widehat{A Z C}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nG6 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "G7", "problem_type": "Geometry", "problem": "Let $A B C$ be an isosceles triangle with $A C=B C$. The point $D$ lies on the side $A B$ such that the semicircle with diameter $[B D]$ and center $O$ is tangent to the side $A C$ in the point $P$ and intersects the side $B C$ at the point $Q$. The radius $O P$ intersects the chord $D Q$ at the point $E$ such that $5 \\cdot P E=3 \\cdot D E$. Find the ratio $\\frac{A B}{B C}$.", "solution": "We denote $O P=O D=O B=R, A C=B C=b$ and $A B=2 a$. Because $O P \\perp A C$ and $D Q \\perp B C$, then the right triangles $A P O$ and $B Q D$ are similar and $\\varangle B D Q=\\varangle A O P$. So, the triangle $D E O$ is isosceles with $D E=O E$. It follows that\n\n$$\n\\frac{P E}{D E}=\\frac{P E}{O E}=\\frac{3}{5}\n$$\n\nLet $F$ and $G$ are the orthogonal projections of the points $E$ and $P$ respectively on the side $A B$ and $M$ is the midpoint of the side $[A B]$. The triangles $O F E, O G P, O P A$ and $C M A$ are similar. We obtain the following relations\n\n$$\n\\frac{O F}{O E}=\\frac{O G}{O P}=\\frac{C M}{A C}=\\frac{O P}{O A}\n$$\n\nBut $C M=\\sqrt{b^{2}-a^{2}}$ and we have $O G=\\frac{R}{b} \\cdot \\sqrt{b^{2}-a^{2}}$. In isosceles triangle $D E O$ the point $F$ is the midpoint of the radius $D O$. So, $O F=R / 2$. By using Thales' theorem we obtain\n\n$$\n\\frac{3}{5}=\\frac{P E}{O E}=\\frac{G F}{O F}=\\frac{O G-O F}{O F}=\\frac{O G}{O F}-1=2 \\cdot \\sqrt{1-\\left(\\frac{a}{b}\\right)^{2}}-1\n$$\n\nFrom the last relations it is easy to obtain that $\\frac{a}{b}=\\frac{3}{5}$ and $\\frac{A B}{B C}=\\frac{6}{5}$.\n\nThe problem is solved.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nG7 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "G8", "problem_type": "Geometry", "problem": "The side lengths of a parallelogram are $a, b$ and diagonals have lengths $x$ and $y$, Knowing that $a b=\\frac{x y}{2}$, show that\n\n$$\na=\\frac{x}{\\sqrt{2}}, b=\\frac{y}{\\sqrt{2}} \\text { or } a=\\frac{y}{\\sqrt{2}}, b=\\frac{x}{\\sqrt{2}}\n$$", "solution": "## Solution 2.\n\nLet us consider a parallelogram $A B C D$, with $A B=a, B C=b, A C=x, B D=y$, $\\widehat{B O C}=\\theta$, and let us produce the line $A D$ towards $D$ and consider $M \\in(A D$ so that $A D=D M$. Then $B C M D$ is a parallelogram, so $C M=B D=y$.\n\nObserve also that $(A B C D)=2(A C D)=(A C M)$ which is written equivalently as\n\n$$\nC B \\cdot C D \\cdot \\sin C=\\frac{A C \\cdot C M \\cdot \\sin \\theta}{2} \\text { i.e. } a b \\sin C=\\frac{x y \\sin \\theta}{2}\n$$\n\nBecause of the given relation $a b=\\frac{x y}{2}$ the last relation becomes $\\sin C=\\sin \\theta$, i.e.\n\n$$\n\\theta=\\widehat{C} \\text { or } \\theta=180^{\\circ}-\\widehat{C}=\\widehat{B}\n$$\n\nIf $\\theta=\\widehat{C}$, then the triangles $A C M$ and $B C D$ are similar because their angles at $C$ are equal, as well as their angles at $B, M$ (remember $B C M D$ is a parallelogram).\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_cef994f1dc1cf35663f9g-25.jpg?height=496&width=553&top_left_y=462&top_left_x=775)\n\nThen\n\n$$\n\\frac{b}{y}=\\frac{a}{x}=\\frac{y}{2 b} \\Rightarrow\\left(b=\\frac{y}{2}, a=\\frac{x}{2}\\right)\n$$\n\nIf $\\theta=\\widehat{B}$, then similarly we prove that the triangles $A C M$ and $A C D$ are similar, which then implies\n\n$$\n\\frac{a}{y}=\\frac{b}{x}=\\frac{x}{2 b} \\Rightarrow\\left(a=\\frac{y}{2}, b=\\frac{x}{2}\\right)\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nG8 ", "solution_tag": "## Solution 1."}}
-{"year": "2008", "tier": "T3", "problem_label": "G8", "problem_type": "Geometry", "problem": "The side lengths of a parallelogram are $a, b$ and diagonals have lengths $x$ and $y$, Knowing that $a b=\\frac{x y}{2}$, show that\n\n$$\na=\\frac{x}{\\sqrt{2}}, b=\\frac{y}{\\sqrt{2}} \\text { or } a=\\frac{y}{\\sqrt{2}}, b=\\frac{x}{\\sqrt{2}}\n$$", "solution": "## Solution 3 .\n\nThe Parallelogram Law states that, in any parallelogram, the sum of the squares of its diagonals is equal to the sum of the squares of its sides.\n\nIn our case, this translates to $x^{2}+y^{2}=2\\left(a^{2}+b^{2}\\right)$. First adding $2 x y=4 a b$, then subtracting the same equality, yields $(x+y)^{2}=2(a+b)^{2}$ and $(x-y)^{2}=2(a-b)^{2}$. It follows that $x+y=a \\sqrt{2}+b \\sqrt{2}$ and either $x-y=a \\sqrt{2}-b \\sqrt{2}$, or $x-y=b \\sqrt{2}-a \\sqrt{2}$. In the first case one obtains $x=a \\sqrt{2}, y=b \\sqrt{2}$, in the latter case, $x=b \\sqrt{2}, y=a \\sqrt{2}$.\n\nFor the proof of the Parallelogram Law, simply apply the Law of cosines in triangles $A B C$ and $A B D$ and use the fact that $\\cos (\\varangle A B C)=-\\cos (\\varangle B A D)$. Adding the two relations gives the desired condition.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nG8 ", "solution_tag": "## Solution 1."}}
-{"year": "2008", "tier": "T3", "problem_label": "G9", "problem_type": "Geometry", "problem": "Let $O$ be a point inside the parallelogram $A B C D$ such that\n\n$$\n\\angle A O B+\\angle C O D=\\angle B O C+\\angle C O D\n$$\n\nProve that there exists a circle $k$ tangent to the circumscribed circles of the triangles $\\triangle A O B, \\triangle B O C, \\triangle C O D$ and $\\triangle D O A$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_cef994f1dc1cf35663f9g-26.jpg?height=610&width=615&top_left_y=438&top_left_x=728)", "solution": "From given condition it is clear that $\\varangle A O B+\\varangle C O D=\\varangle B O C+\\varangle A O D=180^{\\circ}$.\n\nLet $E$ be a point such that $A E=D O$ and $B E=C E$. Clearly, $\\triangle A E B \\equiv \\triangle D O C$ and from that $A E \\| D O$ and $B E \\| C O$. Also, $\\varangle A E B=\\varangle C O D$ so $\\varangle A O B+\\varangle A E B=$ $\\varangle A O B+\\varangle C O D=180^{\\circ}$. Thus, the quadrilateral $A O B E$ is cyclic.\n\nSo $\\triangle A O B$ and $\\triangle A E B$ the same circumcircle, therefor the circumcircles of the triangles $\\triangle A O B$ and $\\triangle C O D$ have the same radius.\n\nAlso, $A E \\| D O$ and $A E=D O$ gives $A E O D$ is parallelogram and $\\triangle A O D \\equiv \\triangle O A E$. So $\\triangle A O B, \\triangle C O D$ and $\\triangle D O A$ has the same radius of their circumcircle (the radius of the cyclic quadrilateral $A E B O)$. Analogously, triangles $\\triangle A O B, \\triangle B O C, \\triangle C O D$ and $\\triangle D O A$ has same radius $R$.\n\nObviously, the circle with center $O$ and radius $2 R$ is externally tangent to each of these circles, so this will be the circle $k$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nG9 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "G10", "problem_type": "Geometry", "problem": "Let $\\Gamma$ be a circle of center $O$, and $\\delta$ be a line in the plane of $\\Gamma$, not intersecting it. Denote by $A$ the foot of the perpendicular from $O$ onto $\\delta$, and let $M$ be a (variable) point on $\\Gamma$. Denote by $\\gamma$ the circle of diameter $A M$, by $X$ the (other than $M$ ) intersection point of $\\gamma$ and $\\Gamma$, and by $Y$ the (other than $A$ ) intersection point of $\\gamma$ and $\\delta$. Prove that the line $X Y$ passes through a fixed point.", "solution": "Consider the line $\\rho$ tangent to $\\gamma$ at $A$, and take the points $\\{K\\}=A M \\cap X Y,\\{L\\}=$ $\\rho \\cap X M$, and $\\{F\\}=O A \\cap X Y$.\n\n(Remark: Moving $M$ into its reflection with respect to the line $O A$ will move $X Y$ into its reflection with respect to $O A$. These old and the new $X Y$ meet on $O A$, hence it should be clear that the fixed point mult be $F$.)\n\nSince $\\varangle L M A=\\varangle F Y A$ and $\\varangle Y A F=\\varangle L A M=90^{\\circ}$, it follows that triangles $F A Y$ and $L A M$ are similar, therefore $\\varangle A F Y=\\varangle A L M$, hence the quadrilateral $A L X F$ is cyclic. But then $\\varangle A F L=\\varangle A X L=90^{\\circ}$, so $L F \\perp A F$, hence $L F \\| \\delta$.\n\nNow, $\\rho$ is the radical axis of circles $\\gamma$ and $A$ (consider $A$ as a circle of center $A$ and radius 0 ), while $X M$ is the radical axis of circles $\\gamma$ and $\\Gamma$, so $L$ is the radical center of the three circle, which means that $L$ lies on the radical axis of circles $\\Gamma$ and $A$. From $L F \\perp O A$, where $O A$ is the line of the centers of the circles $A$ and $\\Gamma$, and $F \\in X Y$, it follows that $F$ is (the) fixed point of $X Y$.\n\n(The degenerate two cases when $M \\in O A$, where $X \\equiv M$ and $Y \\equiv A$, also trivially satisfy the conclusion, as then $F \\in A M)$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_cef994f1dc1cf35663f9g-27.jpg?height=436&width=466&top_left_y=861&top_left_x=800)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nG10 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "G11", "problem_type": "Geometry", "problem": "Consider $A B C$ an acute-angled triangle with $A B \\neq A C$. Denote by $M$ the midpoint of $B C$, by $D, E$ the feet of the altitudes from $B, C$ respectively and let $P$ be the intersection point of the lines $D E$ and $B C$. The perpendicular from $M$ to $A C$ meets the perpendicular from $C$ to $B C$ at point $R$. Prove that lines $P R$ and $A M$ are perpendicular.", "solution": "Let $F$ be the foot of the altitude from $A$ and let $S$ be the intersection point of $A M$ and $R C$. As $P C$ is an altitude of the triangle $P R S$, the claim is equivalent to $R M \\perp P S$, since the latter implies that $M$ is the orthocenter of $P R S$. Due to $R M \\perp A C$, we need to prove that $A C \\| P S$, in other words\n\n$$\n\\frac{M C}{M P}=\\frac{M A}{M S}\n$$\n\nNotice that $A F \\| C S$, so $\\frac{M A}{M S}=\\frac{M F}{M C}$. Now the claim is reduced to proving $M C^{2}=$ $M F \\cdot M P$, a well-known result considering that $A F$ is the polar line of $P$ with respect to circle of radius $M C$ centered at $M$.\n\nThe \"elementary proof\" on the latter result may be obtained as follows: $\\frac{P B}{P C}=\\frac{F B}{F C}$, using, for instance, Menelaus and Ceva theorems with respect to $A B C$. Cross-multiplying one gets $(P M-x)(F M+x)=(x-F M)(P M+x)$\n\n- $x$ stands for the length of $M C$ - and then $P M \\cdot F M=x^{2}$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_cef994f1dc1cf35663f9g-28.jpg?height=604&width=393&top_left_y=473&top_left_x=844)\n\nComment. The proof above holds for both cases $A B<A C$ and $A B>A C$; it is for the committee to decide if a contestant is supposed to (even) mention this.\n\n### 2.4 Number Theory", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nG11 ", "solution_tag": "\nSolution"}}
-{"year": "2008", "tier": "T3", "problem_label": "NT1", "problem_type": "Number Theory", "problem": "Find all the positive integers $x$ and $y$ that satisfy the equation\n\n$$\nx(x-y)=8 y-7\n$$", "solution": "The given equation can be written as:\n\n$$\n\\begin{aligned}\n& x(x-y)=8 y-7 \\\\\n& x^{2}+7=y(x+8)\n\\end{aligned}\n$$\n\nLet $x+8=m, m \\in \\mathbb{N}$. Then we have: $x^{2}+7 \\equiv 0(\\bmod m)$, and $x^{2}+8 x \\equiv 0(\\bmod m)$. So we obtain that $8 x-7 \\equiv 0(\\bmod m) \\quad(1)$.\n\nAlso we obtain $8 x+8^{2}=8(x+8) \\equiv 0(\\bmod m) \\quad(2)$.\n\nFrom (1) and $(2)$ we obtain $(8 x+64)-(8 x-7)=71 \\equiv 0(\\bmod m)$, therefore $m \\mid 71$, since 71 is a prime number, we have:\n\n$x+8=1$ or $x+8=71$. The only accepted solution is $x=63$, and from the initial equation we obtain $y=56$.\n\nTherefore the equation has a unique solution, namely $(x, y)=(63,56)$.\n\nSolution 2:\n\nThe given equation is $x^{2}-x y+7-8 y=0$.\n\nDiscriminant is $\\Delta=y^{2}+32 y-28=(y+16)^{2}-284$ and must be perfect square. So $(y+16)^{2}-284=m^{2}$, and its follow $(y+16)^{2}-m^{2}=284$, and after some casework, $y+16-m=2$ and $y+16+m=142$, hence $y=56, x=63$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nNT1 ", "solution_tag": "## Solution 1:"}}
-{"year": "2008", "tier": "T3", "problem_label": "NT2", "problem_type": "Number Theory", "problem": "Let $n \\geq 2$ be a fixed positive integer. An integer will be called \" $n$-free\" if it is not a multiple of an $n$-th power of a prime. Let $M$ be an infinite set of rational numbers, such that the product of every $n$ elements of $M$ is an $n$-free integer. Prove that $M$ contains only integers.", "solution": "We first prove that $M$ can contain only a finite number of non-integers. Suppose that there are infinitely many of them: $\\frac{p_{1}}{q_{1}}, \\frac{p_{2}}{q_{2}}, \\ldots, \\frac{p_{k}}{q_{k}}, \\ldots$, with $\\left(p_{k}, q_{k}\\right)=1$ and $q_{k}>1$ for each $k$. Let $\\frac{p}{q}=\\frac{p_{1} p_{2} \\ldots p_{n-1}}{q_{1} q_{2} \\ldots q_{n-1}}$, where $(p, q)=1$. For each $i \\geq n$, the number $\\frac{p}{q} \\cdot \\frac{p_{i}}{q_{i}}$ is an integer, so $q_{i}$ is a divisor of $p$ (as $q_{i}$ and $p_{i}$ are coprime). But $p$ has a finite set of divisors, so there are $n$ numbers of $M$ with equal denominators. Their product cannot be an integer, a contradiction.\n\nNow suppose that $M$ contains a fraction $\\frac{a}{b}$ in lowest terms with $b>1$. Take a prime divisor $p$ of $b$. If we take any $n-1$ integers from $M$, their product with $\\frac{a}{b}$ is an integer, so some of them is a multiple of $p$. Therefore there are infinitely many multiples of $p$ in $M$, and the product of $n$ of them is not $n$-free, a contradiction.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nNT2 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "NT3", "problem_type": "Number Theory", "problem": "Let $s(a)$ denote the sum of digits of a given positive integer $a$. The sequence $a_{1}, a_{2}, \\ldots a_{n}, \\ldots$ of positive integers is such that $a_{n+1}=a_{n}+s\\left(a_{n}\\right)$ for each positive integer $n$. Find the greatest possible $n$ for which it is possible to have $a_{n}=2008$.", "solution": "Since $a_{n-1} \\equiv s\\left(a_{n-1}\\right)$ (all congruences are modulo 9 ), we have $2 a_{n-1} \\equiv a_{n} \\equiv 2008 \\equiv 10$, so $a_{n-1} \\equiv 5$. But $a_{n-1}<2008$, so $s\\left(a_{n-1}\\right) \\leq 28$ and thus $s\\left(a_{n-1}\\right)$ can equal 5,14 or 23 . We check $s(2008-5)=s(2003)=5, s(2008-14)=s(1994)=23, s(2008-23)=s(1985)=$ 23. Thus $a_{n-1}$ can equal 1985 or 2003 . As above $2 a_{n-2} \\equiv a_{n-1} \\equiv 5 \\equiv 14$, so $a_{n-2} \\equiv 7$. But $a_{n-2}<2003$, so $s\\left(a_{n-2}\\right) \\leq 28$ and thus $s\\left(a_{n-2}\\right)$ can equal 16 or 25 . Checking as above we see that the only possibility is $s(2003-25)=s(1978)=25$. Thus $a_{n-2}$ can be only 1978. Now $2 a_{n-3} \\equiv a_{n-2} \\equiv 7 \\equiv 16$ and $a_{n-3} \\equiv 8$. But $s\\left(a_{n-3}\\right) \\leq 27$ and thus $s\\left(a_{n-3}\\right)$ can equal 17 or 26 . The check works only for $s(1978-17)=s(1961)=17$. Thus $a_{n-3}=1961$ and similarly $a_{n-4}=1939 \\equiv 4, a_{n-5}=1919 \\equiv 2$ (if they exist). The search for $a_{n-6}$ requires a residue of 1 . But $a_{n-6}<1919$, so $s\\left(a_{n-6}\\right) \\leq 27$ and thus $s\\left(a_{n-6}\\right)$ can be equal only to 10 or 19 . The check fails for both $s(1919-10)=s(1909)=19$ and $s(1919-19)=s(1900)=10$. Thus $n \\leq 6$ and the case $n=6$ is constructed above (1919, 1939, 1961, 1978, 2003, 2008).", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nNT3 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "NT4", "problem_type": "Number Theory", "problem": "Find all integers $n$ such that $n^{4}+8 n+11$ is a product of two or more consecutive integers.", "solution": "We will prove that $n^{4}+8 n+11$ is never a multiple of 3 . This is clear if $n$ is a multiple of 3 . If\n$n$ is not a multiple of 3 , then $n^{4}+8 n+11=\\left(n^{4}-1\\right)+12+8 n=(n-1)(n+1)\\left(n^{2}+1\\right)+12+8 n$, where $8 n$ is the only term not divisible by 3 . Thus $n^{4}+8 n+11$ is never the product of three or more integers.\n\nIt remains to discuss the case when $n^{4}+8 n+11=y(y+1)$ for some integer $y$. We write this as $4\\left(n^{4}+8 n+11\\right)=4 y(y+1)$ or $4 n^{4}+32 n+45=(2 y+1)^{2}$. A check shows that among $n= \\pm 1$ and $n=0$ only $n=1$ satisfies the requirement, as $1^{4}+8 \\cdot 1+11=20=4 \\cdot 5$. Now let $|n| \\geq 2$. The identities $4 n^{2}+32 n+45=\\left(2 n^{2}-2\\right)^{2}+8(n+2)^{2}+9$ and $4 n^{4}+32 n+45=$ $\\left(2 n^{2}+8\\right)^{2}-32 n(n-1)-19$ indicate that for $|n| \\geq 2,2 n^{2}-2<2 y+1<2 n^{2}+8$. But $2 y+1$ is odd, so it can equal $2 n^{2} \\pm 1 ; 2 n^{2}+3 ; 2 n^{2}+5$ or $2 n^{2}+7$. We investigate them one by one.\n\nIf $4 n^{4}+32 n+45=\\left(2 n^{2}-1\\right)^{2} \\Rightarrow n^{2}+8 n+11=0 \\Rightarrow(n+4)^{2}=5$, which is impossible, as 5 is not a perfect square.\n\nIf $4 n^{4}+32 n+45=\\left(2 n^{2}+1\\right)^{2} \\Rightarrow n^{2}-8 n-11=0 \\Rightarrow(n-4)^{2}=27$ which also fails.\n\nAlso $4 n^{4}+32 n+45=\\left(2 n^{2}+3\\right)^{2} \\Rightarrow 3 n^{2}-8 n-9=0 \\Rightarrow 9 n^{2}-24 n-27=0 \\Rightarrow(3 n-4)^{2}=43$ fails.\n\nIf $4 n^{4}+32 n+45=\\left(2 n^{2}+5\\right)^{2} \\Rightarrow 5 n^{2}-8 n=5 \\Rightarrow 25 n^{2}-40 n=25 \\Rightarrow(5 n-4)^{2}=41$ which also fails.\n\nFinally, if $4 n^{4}+32 n+45=\\left(2 n^{2}+7\\right)^{2}$, then $28 n^{2}-32 n+4=0 \\Rightarrow 4(n-1)(7 n-1)=0$, whence $n=1$ that we already found. Thus the only solution is $n=1$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nNT4 ", "solution_tag": "\nSolution"}}
-{"year": "2008", "tier": "T3", "problem_label": "NT5", "problem_type": "Number Theory", "problem": "Is it possible to arrange the numbers $1^{1}, 2^{2}, \\ldots, 2008^{2008}$ one after the other, in such a way that the obtained number is a perfect square? (Explain your answer.)", "solution": "We will use the following lemmas.\n\nLemma 1. If $x \\in \\mathbb{N}$, then $x^{2} \\equiv 0$ or $1(\\bmod 3)$.\n\nProof: Let $x \\in \\mathbb{N}$, then $x=3 k, x=3 k+1$ or $x=3 k+2$, hence\n\n$$\n\\begin{aligned}\n& x^{2}=9 k^{2} \\equiv 0(\\bmod 3) \\\\\n& x^{2}=9 k^{2}+6 k+1 \\equiv 1(\\bmod 3), \\\\\n& x^{2}=9 k^{2}+12 k+4 \\equiv 1(\\bmod 3), \\text { respectively. }\n\\end{aligned}\n$$\n\nHence $x^{2} \\equiv 0$ or $1(\\bmod 3)$, for every positive integer $x$.\n\nWithout proof we will give the following lemma.\n\nLemma 2. If $a$ is a positive integer then $a \\equiv S(a)(\\bmod 3)$, where $S(a)$ is the sum of the digits of the number $a$.\n\nFurther we have\n\n$$\n\\begin{aligned}\n& (6 k+1)^{6 k+1}=\\left[(6 k+1)^{k}\\right]^{6} \\cdot(6 k+1) \\equiv 1(\\bmod 3) \\\\\n& (6 k+2)^{6 k+2}=\\left[(6 k+2)^{3 k+1}\\right]^{2} \\equiv 1(\\bmod 3) \\\\\n& (6 k+3)^{6 k+3} \\equiv 0(\\bmod 3) \\\\\n& (6 k+4)^{6 k+4}=\\left[(6 k+1)^{3 k+2}\\right]^{2} \\equiv 1(\\bmod 3) \\\\\n& (6 k+5)^{6 k+5}=\\left[(6 k+5)^{3 k+2}\\right]^{2} \\cdot(6 k+5) \\equiv 2(\\bmod 3) \\\\\n& (6 k+6)^{6 k+6} \\equiv 0(\\bmod 3)\n\\end{aligned}\n$$\n\nfor every $k=1,2,3, \\ldots$.\n\nLet us separate the numbers $1^{1}, 2^{2}, \\ldots, 2008^{2008}$ into the following six classes: $(6 k+1)^{6 k+1}$, $(6 k+2)^{6 k+2},(6 k+3)^{6 k+3},(6 k+4)^{6 k+4},(6 k+5)^{6 k+5},(6 k+6)^{6 k+6}, k=1,2, \\ldots$.\n\nFor $k=1,2,3, \\ldots$ let us denote by\n\n$s_{k}=(6 k+1)^{6 k+1}+(6 k+2)^{6 k+2}+(6 k+3)^{6 k+3}+(6 k+4)^{6 k+4}+(6 k+5)^{6 k+5}+(6 k+6)^{6 k+6}$.\n\nFrom (3) we have\n\n$$\ns_{k} \\equiv 1+1+0+1+2+0 \\equiv 2(\\bmod 3)\n$$\n\nfor every $k=1,2,3, \\ldots$.\n\nLet $A$ be the number obtained by writing one after the other (in some order) the numbers $1^{1}, 2^{2}, \\ldots, 2008^{2008}$.\n\nThe sum of the digits, $S(A)$, of the number $A$ is equal to the sum of the sums of digits, $S\\left(i^{i}\\right)$, of the numbers $i^{i}, i=1,2, \\ldots, 2008$, and so, from Lemma 2, it follows that\n\n$$\nA \\equiv S(A)=S\\left(1^{1}\\right)+S\\left(2^{2}\\right)+\\ldots+S\\left(2008^{2008}\\right) \\equiv 1^{1}+2^{2}+\\ldots+2008^{2008}(\\bmod 3)\n$$\n\nFurther on $2008=334 \\cdot 6+4$ and if we use (3) and (4) we get\n\n$$\n\\begin{aligned}\nA & \\equiv 1^{1}+2^{2}+\\ldots+2008^{2008} \\\\\n& \\equiv s_{1}+s_{2}+\\ldots+s_{334}+2005^{2005}+2006^{2006}+2007^{2007}+2008^{2008}(\\bmod 3) \\\\\n& \\equiv 334 \\cdot 2+1+1+0+1=671 \\equiv 2(\\bmod 3)\n\\end{aligned}\n$$\n\nFinally, from Lemma 1, it follows that $A$ can not be a perfect square.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nNT5 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "NT6", "problem_type": "Number Theory", "problem": "Let $f: \\mathbb{N} \\rightarrow \\mathbb{R}$ be a function, satisfying the following condition:\n\nfor every integer $n>1$, there exists a prime divisor $p$ of $n$ such that $f(n)=f\\left(\\frac{n}{p}\\right)-f(p)$. If\n\n$$\nf\\left(2^{2007}\\right)+f\\left(3^{2008}\\right)+f\\left(5^{2009}\\right)=2006\n$$\n\ndetermine the value of\n\n$$\nf\\left(2007^{2}\\right)+f\\left(2008^{3}\\right)+f\\left(2009^{5}\\right)\n$$", "solution": "If $n=p$ is prime number, we have\n\n$$\nf(p)=f\\left(\\frac{p}{p}\\right)-f(p)=f(1)-f(p)\n$$\n\ni.e.\n\n$$\nf(p)=\\frac{f(1)}{2}\n$$\n\nIf $n=p q$, where $p$ and $q$ are prime numbers, then\n\n$$\nf(n)=f\\left(\\frac{n}{p}\\right)-f(p)=f(q)-f(p)=\\frac{f(1)}{2}-\\frac{f(1)}{2}=0\n$$\n\nIf $n$ is a product of three prime numbers, we have\n\n$$\nf(n)=f\\left(\\frac{n}{p}\\right)-f(p)=0-f(p)=-f(p)=-\\frac{f(1)}{2}\n$$\n\nWith mathematical induction by a number of prime multipliers we shell prove that: if $n$ is a product of $k$ prime numbers then\n\n$$\nf(n)=(2-k) \\frac{f(1)}{2}\n$$\n\nFor $k=1$, clearly the statement (2), holds.\n\nLet statement (2) holds for all integers $n$, where $n$ is a product of $k$ prime numbers.\n\nNow let $n$ be a product of $k+1$ prime numbers. Then we have $n=n_{1} p$, where $n_{1}$ is a product of $k$ prime numbers.\n\nSo\n\n$$\nf(n)=f\\left(\\frac{n}{p}\\right)-f(p)=f\\left(n_{1}\\right)-f(p)=(2-k) \\frac{f(1)}{2}-\\frac{f(1)}{2}=(2-(k+1)) \\frac{f(1)}{2}\n$$\n\nSo (2) holds for every integer $n>1$.\n\nNow from $f\\left(2^{2007}\\right)+f\\left(3^{2008}\\right)+f\\left(5^{2009}\\right)=2006$ and because of (2) we have\n\n$$\n\\begin{aligned}\n2006 & =f\\left(2^{2007}\\right)+f\\left(3^{2008}\\right)+f\\left(5^{2009}\\right) \\\\\n& =\\frac{2-2007}{2} f(1)+\\frac{2-2008}{2} f(1)+\\frac{2-2009}{2} f(1)=-\\frac{3 \\cdot 2006}{2} f(1)\n\\end{aligned}\n$$\n\ni.e.\n\n$$\nf(1)=-\\frac{2}{3}\n$$\n\nSince\n\n$$\n2007=3^{2} \\cdot 223,2008=2^{3} \\cdot 251,2009=7^{2} \\cdot 41\n$$\n\nand because of (2) and (3), we get\n\n$$\n\\begin{aligned}\nf\\left(2007^{2}\\right)+f\\left(2008^{3}\\right)+f\\left(2009^{5}\\right) & =\\frac{2-6}{2} f(1)+\\frac{2-12}{2} f(1)+\\frac{2-15}{2} f(1) \\\\\n& =-\\frac{27}{2} f(1)=-\\frac{27}{2} \\cdot\\left(-\\frac{2}{3}\\right)=9\n\\end{aligned}\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nNT6 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "NT7", "problem_type": "Number Theory", "problem": "Determine the minimal prime number $p>3$ for which no natural number $n$ satisfies\n\n$$\n2^{n}+3^{n} \\equiv 0(\\bmod p)\n$$", "solution": "We put $A(n)=2^{n}+3^{n}$. From Fermat's little theorem, we have $2^{p-1} \\equiv 1(\\bmod p)$ and $3^{p-1} \\equiv 1(\\bmod p)$ from which we conclude $A(n) \\equiv 2(\\bmod p)$. Therefore, after $p-1$ steps\nat most, we will have repetition of the power. It means that in order to determine the minimal prime number $p$ we seek, it is enough to determine a complete set of remainders $S(p)=\\{0,1, \\ldots, p-1\\}$ such that $2^{n}+3^{n} \\not \\equiv 0(\\bmod p)$, for every $n \\in S(p)$.\n\nFor $p=5$ and $n=1$ we have $A(1) \\equiv 0(\\bmod 5)$.\n\nFor $p=7$ and $n=3$ we have $A(3) \\equiv 0(\\bmod 7)$.\n\nFor $p=11$ and $n=5$ we have $A(5) \\equiv 0(\\bmod 11)$.\n\nFor $p=13$ and $n=2$ we have $A(2) \\equiv 0(\\bmod 13)$.\n\nFor $p=17$ and $n=8$ we have $A(8) \\equiv 0(\\bmod 17)$.\n\nFor $p=19$ we have $A(n) \\not \\equiv 0(\\bmod 19)$, for all $n \\in S(19)$.\n\nHence the minimal value of $p$ is 19 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nNT7 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "NT8", "problem_type": "Number Theory", "problem": "Let $a, b, c, d, e, f$ are nonzero digits such that the natural numbers $\\overline{a b c}, \\overline{d e f}$ and $\\overline{a b c d e f}$ are squares.\n\na) Prove that $\\overline{a b c d e f}$ can be represented in two different ways as a sum of three squares of natural numbers.\n\nb) Give an example of such a number.", "solution": "a) Let $\\overline{a b c}=m^{2}, \\overline{d e f}=n^{2}$ and $\\overline{a b c d e f}=p^{2}$, where $11 \\leq m \\leq 31,11 \\leq n \\leq 31$ are natural numbers. So, $p^{2}=1000 \\cdot m^{2}+n^{2}$. But $1000=30^{2}+10^{2}=18^{2}+26^{2}$. We obtain the following relations\n\n$$\n\\begin{gathered}\np^{2}=\\left(30^{2}+10^{2}\\right) \\cdot m^{2}+n^{2}=\\left(18^{2}+26^{2}\\right) \\cdot m^{2}+n^{2}= \\\\\n=(30 m)^{2}+(10 m)^{2}+n^{2}=(18 m)^{2}+(26 m)^{2}+n^{2}\n\\end{gathered}\n$$\n\nThe assertion a) is proved.\n\nb) We write the equality $p^{2}=1000 \\cdot m^{2}+n^{2}$ in the equivalent form $(p+n)(p-n)=1000 \\cdot m^{2}$, where $349 \\leq p \\leq 979$. If $1000 \\cdot m^{2}=p_{1} \\cdot p_{2}$, such that $p+n=p_{1}$ and $p-n=p_{2}$, then $p_{1}$ and $p_{2}$ are even natural numbers with $p_{1}>p_{2} \\geq 318$ and $22 \\leq p_{1}-p_{2} \\leq 62$. For $m=15$ we obtain $p_{1}=500, p_{2}=450$. So, $n=25$ and $p=475$. We have\n\n$$\n225625=475^{2}=450^{2}+150^{2}+25^{2}=270^{2}+390^{2}+25^{2}\n$$\n\nThe problem is solved.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nNT8 ", "solution_tag": "\nSolution"}}
-{"year": "2008", "tier": "T3", "problem_label": "NT9", "problem_type": "Number Theory", "problem": "Let $p$ be a prime number. Find all positive integers $a$ and $b$ such that:\n\n$$\n\\frac{4 a+p}{b}+\\frac{4 b+p}{a}\n$$\n\nand\n\n$$\n\\frac{a^{2}}{b}+\\frac{b^{2}}{a}\n$$\n\nare integers.", "solution": "Since $a$ and $b$ are symmetric we can assume that $a \\leq b$. Let $d=(a, b), a=d u, b=d v$ and $(u, v)=1$. Then we have:\n\n$$\n\\frac{a^{2}}{b}+\\frac{b^{2}}{a}=\\frac{d\\left(u^{3}+v^{3}\\right)}{u v}\n$$\n\nSince,\n\n$$\n\\left(u^{3}+v^{3}, u\\right)=\\left(u^{3}+v^{3}, v\\right)=1\n$$\n\nwe deduce that $u \\mid d$ and $v \\mid d$. But as $(u, v)=1$, it follows that $u v \\mid d$.\n\nNow, let $d=u v t$. Furthermore,\n\n$$\n\\frac{4 a+p}{b}+\\frac{4 b+p}{a}=\\frac{4\\left(a^{2}+b^{2}\\right)+p(a+b)}{a b}=\\frac{4 u v t\\left(u^{2}+v^{2}\\right)+p(u+v)}{u^{2} v^{2} t}\n$$\n\nThis implies,\n\n$$\nu v \\mid p(u+v)\n$$\n\nBut from our assumption $1=(u, v)=(u, u+v)=(v, u+v)$ we conclude $u v \\mid p$. Therefore, we have three cases $\\{u=v=1\\},\\{u=1, v=p\\},\\{u=p, v=1\\}$. We assumed that $a \\leq b$, and this implies $u \\leq v$.\n\nIf $a=b$, we need $\\frac{4 a+p}{a}+\\frac{4 a+p}{a} \\in \\mathbb{N}$, i.e. $a \\mid 2 p$. Then $a \\in\\{1,2, p, 2 p\\}$. The other condition being fulfilled, we obtain the solutions $(1,1),(2,2),(p, p)$ and $(2 p, 2 p)$.\n\nNow, we have only one case to investigate, $u=1, v=p$. The last equation is transformed into:\n\n$$\n\\frac{4 a+p}{b}+\\frac{4 b+p}{a}=\\frac{4 p t\\left(1+p^{2}\\right)+p(p+1)}{p^{2} t}=\\frac{4 t+1+p(1+4 p t)}{p t}\n$$\n\nFrom the last equation we derive\n\n$$\np \\mid(4 t+1)\n$$\n\nLet $4 t+1=p q$. From here we derive\n\n$$\n\\frac{4 t+1+p(1+4 p t)}{p t}=\\frac{q+1+4 p t}{t}\n$$\n\nNow, we have\n\n$$\nt \\mid(q+1)\n$$\n\nor\n\n$$\nq+1=\\text { st. }\n$$\n\nTherefore,\n\n$$\np=\\frac{4 t+1}{q}=\\frac{4 t+1}{s t-1}\n$$\n\nSince $p$ is a prime number, we deduce\n\n$$\n\\frac{4 t+1}{s t-1} \\geq 2\n$$\n\nor\n\n$$\ns \\leq \\frac{4 t+3}{2 t}=2+\\frac{3}{2 t}<4\n$$\n\nCase 1: $s=1, p=\\frac{4 t+1}{t-1}=4+\\frac{5}{t-1}$. We conclude $t=2$ or $t=6$. But when $t=2$, we have $p=9$, not a prime. When $t=6, p=5, a=30, b=150$.\n\nCase 2: $s=2, p=\\frac{4 t+1}{2 t-1}=2+\\frac{3}{2 t-1}$. We conclude $t=1, p=5, a=5, b=25$ or $t=2, p=3, a=6, b=18$.\n\nCase 3: $s=3, p=\\frac{4 t+1}{3 t-1}$ or $3 p=4+\\frac{7}{3 t-1}$. As 7 does not have any divisors of the form $3 t-1$, in this case we have no solutions.\n\nSo, the solutions are\n\n$$\n(a, b)=\\{(1,1),(2,2),(p, p),(2 p, 2 p),(5,25),(6,18),(18,6),(25,5),(30,150),(150,30)\\}\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nNT9 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "NT10", "problem_type": "Number Theory", "problem": "Prove that $2^{n}+3^{n}$ is not a perfect cube for any positive integer $n$.", "solution": "If $n=1$ then $2^{1}+3^{1}=5$ is not perfect cube.\n\nPerfect cube gives residues $-1,0$ and 1 modulo 9 . If $2^{n}+3^{n}$ is a perfect cube, then $n$ must be divisible with 3 (congruence $2^{n}+3^{n}=x^{3}$ modulo 9 ).\n\nIf $n=3 k$ then $2^{3 k}+3^{2 k}>\\left(3^{k}\\right)^{3}$. Also, $\\left(3^{k}+1\\right)^{3}=3^{3 k}+3 \\cdot 3^{2 k}+3 \\cdot 3^{k}+1>3^{3 k}+3^{2 k}=$ $3^{3 k}+9^{k}>3^{3 k}+8^{k}=3^{3 k}+2^{3 k}$. But, $3^{k}$ and $3^{k}+1$ are two consecutive integers so $2^{3 k}+3^{3 k}$ is not a perfect cube.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nNT10 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "NT11", "problem_type": "Number Theory", "problem": "Determine the greatest number with $n$ digits in the decimal representation which is divisible by 429 and has the sum of all digits less than or equal to 11 .", "solution": "Let $A=\\overline{a_{n} a_{n-1} \\ldots a_{1}}$ and notice that $429=3 \\cdot 11 \\cdot 13$.\n\nSince the sum of the digits $\\sum a_{i} \\leq 11$ and $\\sum a_{i}$ is divisible by 3 , we get $\\sum a_{i}=3,6$ or 9. As 11 divides $A$, we have\n\n$$\n11 \\mid a_{n}-a_{n-1}+a_{n-2}-a_{n-3}+\\ldots\n$$\n\nin other words $11 \\mid \\sum_{i \\text { odd }} a_{i}-\\sum_{i \\text { even }} a_{i}$. But\n\n$$\n-9 \\leq-\\sum a_{i} \\leq \\sum_{i \\text { odd }} a_{i}-\\sum_{i \\text { even }} a_{i} \\leq \\sum a_{i} \\leq 9\n$$\n\nso $\\sum_{i \\text { odd }} a_{i}-\\sum_{i \\text { even }} a_{i}=0$. It follows that $\\sum a_{i}$ is even, so $\\sum a_{i}=6$ and $\\sum_{i \\text { odd }} a_{i}=\\sum_{i \\text { even }} a_{i}=3$.\n\nThe number 13 is a divisor of 1001 , hence\n\n$$\n13 \\mid \\overline{a_{3} a_{2} a_{1}}-\\overline{a_{6} a_{5} a_{4}}+\\overline{a_{9} a_{8} a_{7}}-\\overline{a_{12} a_{11} a_{10}}+\\ldots\n$$\n\nFor each $k=1,2,3,4,5,6$, let $s_{k}$ be the sum of the digits $a_{k+6 m}, m \\geq 0$; that is\n\n$$\ns_{1}=a_{1}+a_{7}+a_{13}+\\ldots \\text { and so on. }\n$$\n\nWith this notation, (1) rewrites as\n\n$$\n13 \\mid 100\\left(s_{3}-s_{6}\\right)+10\\left(s_{2}-s_{5}\\right)+\\left(s_{1}-s_{4}\\right), \\text { or } 13 \\mid 4\\left(s_{6}-s_{3}\\right)+3\\left(s_{5}-s_{2}\\right)+\\left(s_{1}-s_{4}\\right)\n$$\n\nLet $S_{3}=s_{3}-s_{6}, S_{2}=s_{2}-s_{5}$, and $S_{1}=s_{1}-s_{4}$. Recall that $\\sum_{i \\text { odd }} a_{i}=\\sum_{i \\text { even }} a_{i}$, which implies $S_{2}=S_{1}+S_{3}$. Then\n\n$$\n13\\left|4 S_{3}+3 S_{2}-S_{1}=7 S_{3}+2 S_{1} \\Rightarrow 13\\right| 49 S_{3}+14 S_{1} \\Rightarrow 13 \\mid S_{1}-3 S_{3}\n$$\n\nObserve that $\\left|S_{1}\\right| \\leq s_{1}=\\sum_{i \\text { odd }} a_{i}=3$ and likewise $\\left|S_{2}\\right|,\\left|S_{3}\\right| \\leq 3$. Then $-13<S_{1}-3 S_{3}<13$ and consequently $S_{1}=3 S_{3}$. Thus $S_{2}=4 S_{3}$ and $\\left|S_{2}\\right| \\leq 3$ yields $S_{2}=0$ and then $S_{1}=S_{3}=0$. We have $s_{1}=s_{4}, s_{2}=s_{5}, s_{3}=s_{6}$ and $s_{1}+s_{2}+s_{3}=3$, so the greatest number $A$ is $30030000 \\ldots$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nNT11 ", "solution_tag": "## Solution"}}
-{"year": "2008", "tier": "T3", "problem_label": "NT12", "problem_type": "Number Theory", "problem": "Solve the equation $\\frac{p}{q}-\\frac{4}{r+1}=1$ in prime numbers.", "solution": "We can rewrite the equation in the form\n\n$$\n\\begin{gathered}\n\\frac{p r+p-4 q}{q(r+1)}=1 \\Rightarrow p r+p-4 q=q r+q \\\\\np r-q r=5 q-p \\Rightarrow r(p-q)=5 q-p\n\\end{gathered}\n$$\n\nIt follows that $p \\neq q$ and\n\n$$\n\\begin{gathered}\nr=\\frac{5 q-p}{p-q}=\\frac{4 q+q-p}{p-q} \\\\\nr=\\frac{4 q}{p-q}-1\n\\end{gathered}\n$$\n\nAs $p$ is prime, $p-q \\neq q, p-q \\neq 2 q, p-q \\neq 4 q$.\n\nWe have $p-q=1$ or $p-q=2$ or $p-q=4$\n\ni) If $p-q=1$ then\n\n$$\nq=2, p=3, r=7\n$$\n\nii) If $p-q=2$ then $p=q+2, r=2 q-1$\n\nIf $q=1(\\bmod 3)$ then $q+2 \\equiv 0(\\bmod 3)$\n\n$$\nq+2=3 \\Rightarrow q=1\n$$\n\ncontradiction.\n\nIf $q \\equiv-1(\\bmod 3)$ then $r \\equiv-2-1 \\equiv 0(\\bmod 3)$\n\n$$\n\\begin{gathered}\nr=3 \\\\\nr=2 q-1=3 \\\\\nq=2 \\\\\np=4\n\\end{gathered}\n$$\n\ncontradiction.\n\nHence $q=3, p=5, r=5$.\n\niii) If $p-q=4$ then $p=q+4$.\n\n$r=q-1$\n\nHence $q=3, p=7, r=2$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_tag": "\nNT12 ", "solution_tag": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "problem": "If for the real numbers $x, y, z, k$ the following conditions are valid, $x \\neq y \\neq z \\neq x$ and $x^{3}+y^{3}+k\\left(x^{2}+y^{2}\\right)=y^{3}+z^{3}+k\\left(y^{2}+z^{2}\\right)=z^{3}+x^{3}+k\\left(z^{2}+x^{2}\\right)=2008$, find the product $x y z$.", "solution": "$x^{3}+y^{3}+k\\left(x^{2}+y^{2}\\right)=y^{3}+z^{3}+k\\left(y^{2}+z^{2}\\right) \\Rightarrow x^{2}+x z+z^{2}=-k(x+z):(1)$ and $y^{3}+z^{3}+k\\left(y^{2}+z^{2}\\right)=z^{3}+x^{3}+k\\left(z^{2}+x^{2}\\right) \\Rightarrow y^{2}+y x+x^{2}=-k(y+x):(2)$\n\n- From (1) $-(2) \\Rightarrow x+y+z=-k:(*)$\n- If $x+z=0$, then from $(1) \\Rightarrow x^{2}+x z+z^{2}=0 \\Rightarrow(x+z)^{2}=x z \\Rightarrow x z=0$\n\nSo $x=z=0$, contradiction since $x \\neq z$ and therefore $(1) \\Rightarrow-k=\\frac{x^{2}+x z+z^{2}}{x+z}$\n\nSimilarly we have: $-k=\\frac{y^{2}+y x+x^{2}}{y+x}$.\n\nSo $\\frac{x^{2}+x z+z^{2}}{x+z}=\\frac{y^{2}+x y+x^{2}}{x+y}$ from which $x y+y z+z x=0:(* *)$.\n\nWe substitute $k$ in $x^{3}+y^{3}+k\\left(x^{2}+y^{2}\\right)=2008$ from the relation $(*)$ and using the $(* *)$, we finally obtain that $2 x y z=2008$ and therefore $x y z=1004$.\n\nRemark: $x, y, z$ must be the distinct real solutions of the equation $t^{3}+k t^{2}-1004=0$. Such solutions exist if (and only if) $k>3 \\sqrt[3]{251}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nA1 ", "solution_match": "\nSolution"}}
+{"year": "2008", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "problem": "Find all real numbers $a, b, c, d$ such that $a+b+c+d=20$ and $a b+a c+a d+b c+b d+c d=$ 150 .", "solution": "$400=(a+b+c+d)^{2}=a^{2}+b^{2}+c^{2}+d^{2}+2 \\cdot 150$, so $a^{2}+b^{2}+c^{2}+d^{2}=100$. Now $(a-b)^{2}+(a-c)^{2}+(a-d)^{2}+(b-c)^{2}+(b-d)^{2}+(c-d)^{2}=3\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)-2(a b+$ $a c+a d+b c+b d+c d)=300-300=0$. Thus $a=b=c=d=5$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nA2 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "problem": "Let the real parameter $p$ be such that the system\n\n$$\n\\left\\{\\begin{array}{l}\np\\left(x^{2}-y^{2}\\right)=\\left(p^{2}-1\\right) x y \\\\\n|x-1|+|y|=1\n\\end{array}\\right.\n$$\n\nhas at least three different real solutions. Find $p$ and solve the system for that $p$.", "solution": "The second equation is invariant when $y$ is replaced by $-y$, so let us assume $y \\geq 0$. It is also invariant when $x-1$ is replaced by $-(x-1)$, so let us assume $x \\geq 1$. Under these conditions the equation becomes $x+y=2$, which defines a line on the coordinate plane. The set of points on it that satisfy the inequalities is a segment with endpoints $(1,1)$ and $(2,0)$. Now taking into account the invariance under the mentioned replacements, we conclude that the set of points satisfying the second equation is the square $\\diamond$ with vertices $(1,1),(2,0),(1,-1)$ and $(0,0)$.\n\nThe first equation is equivalent to\n\n$p x^{2}-p^{2} x y+x y-p y^{2}=0$\n\n$p x(x-p y)+y(x-p y)=0$\n\n$(p x+y)(x-p y)=0$.\n\nThus $y=-p x$ or $x=p y$. These are equations of two perpendicular lines passing through the origin, which is also a vertex of $\\diamond$. If one of them passes through an interior point of the square, the other cannot have any common points with $\\diamond$ other than $(0,0)$, so the system has two solutions. Since we have at least three different real solutions, the lines must contain some sides of $\\diamond$, i.e. the slopes of the lines have to be 1 and -1 . This happens if $p=1$ or $p=-1$. In either case $x^{2}=y^{2},|x|=|y|$, so the second equation becomes $|1-x|+|x|=1$. It is true exactly when $0 \\leq x \\leq 1$ and $y= \\pm x$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nA3 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "problem": "Find all triples $(x, y, z)$ of real numbers that satisfy the system\n\n$$\n\\left\\{\\begin{array}{l}\nx+y+z=2008 \\\\\nx^{2}+y^{2}+z^{2}=6024^{2} \\\\\n\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=\\frac{1}{2008}\n\\end{array}\\right.\n$$", "solution": "The last equation implies $x y z=2008(x y+y z+z x)$, therefore $x y z-2008(x y+y z+z x)+$ $2008^{2}(x+y+z)-2008^{3}=0$.\n\n$(x-2008)(y-2008)(z-2008)=0$.\n\nThus one of the variable equals 2008. Let this be $x$. Then the first equation implies $y=-z$. From the second one it now follows that $2 y^{2}=6024^{2}-2008^{2}=2008^{2}(9-1)=$ $2 \\cdot 4016^{2}$. Thus $(x, y, z)$ is the triple $(2008,4016,-4016)$ or any of its rearrangements.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nA4 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "A5", "problem_type": "Algebra", "problem": "Find all triples $(x, y, z)$ of real positive numbers, which satisfy the system\n\n$$\n\\left\\{\\begin{array}{l}\n\\frac{1}{x}+\\frac{4}{y}+\\frac{9}{z}=3 \\\\\nx+y+z \\leq 12\n\\end{array}\\right.\n$$", "solution": "If we multiply the given equation and inequality $(x>0, y>0, z>0)$, we have\n\n$$\n\\left(\\frac{4 x}{y}+\\frac{y}{x}\\right)+\\left(\\frac{z}{x}+\\frac{9 x}{z}\\right)+\\left(\\frac{4 z}{y}+\\frac{9 y}{z}\\right) \\leq 22\n$$\n\nFrom AM-GM we have\n\n$$\n\\frac{4 x}{y}+\\frac{y}{x} \\geq 4, \\quad \\frac{z}{x}+\\frac{9 x}{z} \\geq 6, \\quad \\frac{4 z}{y}+\\frac{9 y}{z} \\geq 12\n$$\n\nTherefore\n\n$$\n22 \\leq\\left(\\frac{4 x}{y}+\\frac{y}{x}\\right)+\\left(\\frac{z}{x}+\\frac{9 x}{z}\\right)+\\left(\\frac{4 z}{y}+\\frac{9 y}{z}\\right)\n$$\n\nNow from (1) and (3) we get\n\n$$\n\\left(\\frac{4 x}{y}+\\frac{y}{x}\\right)+\\left(\\frac{z}{x}+\\frac{9 x}{z}\\right)+\\left(\\frac{4 z}{y}+\\frac{9 y}{z}\\right)=22\n$$\n\nwhich means that in (2), everywhere equality holds i.e. we have equality between means, also $x+y+z=12$.\n\nTherefore $\\frac{4 x}{y}=\\frac{y}{x}, \\frac{z}{x}=\\frac{9 x}{z}$ and, as $x>0, y>0, z>0$, we get $y=2 x, z=3 x$. Finally if we substitute for $y$ and $z$, in $x+y+z=12$, we get $x=2$, therefore $y=2 \\cdot 2=4$ and $z=3 \\cdot 2=6$.\n\nThus the unique solution is $(x, y, z)=(2,4,6)$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nA5 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "A6", "problem_type": "Algebra", "problem": "If the real numbers $a, b, c, d$ are such that $0<a, b, c, d<1$, show that\n\n$$\n1+a b+b c+c d+d a+a c+b d>a+b+c+d\n$$", "solution": "If $1 \\geq a+b+c$ then we write the given inequality equivalently as\n\n$$\n\\begin{gathered}\n1-(a+b+c)+d[(a+b+c)-1]+a b+b c+c a>0 \\\\\n\\Leftrightarrow[1-(a+b+c)](1-d)+a b+b c+c a>0\n\\end{gathered}\n$$\n\nwhich is of course true.\n\nIf instead $a+b+c>1$, then $d(a+b+c)>d$ i.e.\n\n$$\nd a+d b+d c>d\n$$\n\nWe are going to prove that also\n\n$$\n1+a b+b c+c a>a+b+c\n$$\n\nthus adding (1) and (2) together we'll get the desired result in this case too.\n\nFor the truth of $(2)$ :\n\nIf $1 \\geq a+b$, then we rewrite (2) equivalently as\n\n$$\n\\begin{gathered}\n1-(a+b)+c[(a+b)-1]+a b>0 \\\\\n\\quad \\Leftrightarrow[1-(a+b)](1-c)+a b>0\n\\end{gathered}\n$$\n\nwhich is of course true.\n\nIf instead $a+b>1$, then $c(a+b)>c$, i.e.\n\n$$\nc a+c b>c\n$$\n\nBut it is also true that\n\n$$\n1+a b>a+b\n$$\n\nbecause this is equivalent to $(1-a)+b(a-1)>0$, i.e. to $(1-a)(1-b)>0$ which holds. Adding (3) and (4) together we get the truth of (2) in this case too and we are done. You can instead consider the following generalization:\n\nExercise. If for the real numbers $x_{1}, x_{2}, \\ldots, x_{n}$ it is $0<x_{i}<1$, for any $i$, show that\n\n$$\n1+\\sum_{1 \\leq i<j \\leq n} x_{i} x_{j}>\\sum_{i=1}^{n} x_{i}\n$$\n\n## Solution\n\nWe'll prove it by induction.\n\nFor $n=1$ the desired result becomes $1>x_{1}$ which is true.\n\nLet the result be true for some natural number $n \\geq 1$.\n\nWe'll prove it to be true for $n+1$ as well, and we'll be done.\n\nSo let $x_{1}, x_{2}, \\ldots, x_{n}, x_{n+1}$ be $n+1$ given real numbers with $0<x_{i}<1$, for any $i$. We wish to show that\n\n$$\n1+\\sum_{1 \\leq i<j \\leq n+1} x_{i} x_{j}>x_{1}+x_{2}+\\ldots+x_{n}+x_{n+1}\n$$\n\nIf $1 \\geq x_{1}+x_{2}+\\ldots+x_{n}$ then we rewrite (5) equivalently as\n\n$$\n1-\\left(x_{1}+x_{2}+\\ldots+x_{n}\\right)+x_{n+1}\\left(x_{1}+x_{2}+\\ldots+x_{n}-1\\right)+\\sum_{1 \\leq i<j \\leq n} x_{i} x_{j}>0\n$$\n\nThis is also written as\n\n$$\n\\left(1-x_{n+1}\\right)\\left[1-\\left(x_{1}+x_{2}+\\ldots+x_{n}\\right)\\right]+\\sum_{1 \\leq i<j \\leq n} x_{i} x_{j}>0\n$$\n\nwhich is clearly true.\n\nIf instead $x_{1}+x_{2}+\\ldots+x_{n}>1$ then $x_{n+1}\\left(x_{1}+x_{2}+\\ldots+x_{n}\\right)>x_{n+1}$, i.e.\n\n$$\nx_{n+1} x_{1}+x_{n+1} x_{2}+\\ldots+x_{n+1} x_{n}>x_{n+1}\n$$\n\nBy the induction hypothesis applied to the $n$ real numbers $x_{1}, x_{2}, \\ldots, x_{n}$ we also know that\n\n$$\n1+\\sum_{1 \\leq i<j \\leq n} x_{i} x_{j}>\\sum_{i=1}^{n} x_{i}\n$$\n\nAdding (6) and (7) together we get the validity of (5) in this case too, and we are done.\n\nYou can even consider the following variation:\n\nExercise. If the real numbers $x_{1}, x_{2}, \\ldots, x_{2008}$ are such that $0<x_{i}<1$, for any $i$, show that\n\n$$\n1+\\sum_{1 \\leq i<j \\leq 2008} x_{i} x_{j}>\\sum_{i=1}^{2008} x_{i}\n$$\n\nRemark: Inequality (2) follows directly from $(1-a)(1-b)(1-c)>0 \\Leftrightarrow 1-a-b-c+$ $a b+b c+c a>a b c>0$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nA6 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "A7", "problem_type": "Algebra", "problem": "Let $a, b$ and $c$ be a positive real numbers such that $a b c=1$. Prove the inequality\n\n$$\n\\left(a b+b c+\\frac{1}{c a}\\right)\\left(b c+c a+\\frac{1}{a b}\\right)\\left(c a+a b+\\frac{1}{b c}\\right) \\geq(1+2 a)(1+2 b)(1+2 c)\n$$", "solution": "By Cauchy-Schwarz inequality and $a b c=1$ we get\n\n$$\n\\begin{gathered}\n\\sqrt{\\left(b c+c a+\\frac{1}{a b}\\right)\\left(a b+b c+\\frac{1}{c a}\\right)}=\\sqrt{\\left(b c+c a+\\frac{1}{a b}\\right)\\left(\\frac{1}{c a}+a b+b c\\right)} \\geq \\\\\n\\left(\\sqrt{a b} \\cdot \\sqrt{\\frac{1}{a b}}+\\sqrt{b c} \\cdot \\sqrt{b c}+\\sqrt{\\frac{1}{c a}} \\cdot \\sqrt{c a}\\right)=(2+b c)=(2 a b c+b c)=b c(1+2 a)\n\\end{gathered}\n$$\n\nAnalogously we get $\\sqrt{\\left(b c+c a+\\frac{1}{a b}\\right)\\left(c a+a b+\\frac{1}{b c}\\right)} \\geq c a(1+2 b)$ and\n\n$\\sqrt{\\left(c a+a b+\\frac{1}{b c}\\right)\\left(a b+b c+\\frac{1}{c a}\\right)} \\geq a b(1+2 a)$.\n\nMultiplying these three inequalities we get:\n\n$$\n\\left(a b+b c+\\frac{1}{c a}\\right)\\left(b c+c a+\\frac{1}{a b}\\right)\\left(c a+a b+\\frac{1}{b c}\\right) \\geq a^{2} b^{2} c^{2}(1+2 a)(1+2 b)(1+2 c)=\n$$\n\n$(1+2 a)(1+2 b)(1+2 c)$ because $a b c=1$.\n\nEquality holds if and only if $a=b=c=1$.\n\n## Solution 2\n\nUsing $a b c=1$ we get\n\n$$\n\\begin{gathered}\n\\left(a b+b c+\\frac{1}{c a}\\right)\\left(b c+c a+\\frac{1}{a b}\\right)\\left(c a+a b+\\frac{1}{b c}\\right)= \\\\\n=\\left(\\frac{1}{c}+\\frac{1}{a}+b\\right)\\left(\\frac{1}{a}+\\frac{1}{b}+c\\right)\\left(\\frac{1}{b}+\\frac{1}{c}+a\\right)= \\\\\n=\\frac{(a+c+a b c)}{a c} \\cdot \\frac{(b+a+a b c)}{a b} \\cdot \\frac{(b+c+a b c)}{b c}=(a+b+1)(b+c+1)(c+a+1)\n\\end{gathered}\n$$\n\nThus, we need to prove\n\n$$\n(a+b+1)(b+c+1)(c+a+1) \\geq(1+2 a)(1+2 b)(1+2 c)\n$$\n\nAfter multiplication and using the fact $a b c=1$ we have to prove\n\n$$\n\\begin{gathered}\na^{2} b+a^{2} c+b^{2} c+b^{2} a+c^{2} a+c^{2} b+3(a b+b c+c a)+2(a+b+c)+a^{2}+b^{2}+c^{2}+3 \\geq \\\\\n\\geq 4(a b+b c+c a)+2(a+b+c)+9\n\\end{gathered}\n$$\n\nSo we need to prove\n\n$$\na^{2} b+a^{2} c+b^{2} c+b^{2} a+c^{2} a+c^{2} b+a^{2}+b^{2}+c^{2} \\geq a b+b c+c a+6\n$$\n\nThis follows from the well-known (AM-GM inequality) inequalities\n\n$$\na^{2}+b^{2}+c^{2} \\geq a b+b c+c a\n$$\n\nand\n\n$$\na^{2} b+a^{2} c+b^{2} c+b^{2} a+c^{2} a+c^{2} b \\geq 6 a b c=6\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nA7 ", "solution_match": "## Solution 1"}}
+{"year": "2008", "tier": "T3", "problem_label": "A8", "problem_type": "Algebra", "problem": "Show that\n\n$$\n(x+y+z)\\left(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\right) \\geq 4\\left(\\frac{x}{x y+1}+\\frac{y}{y z+1}+\\frac{z}{z x+1}\\right)^{2}\n$$\n\nfor any real positive numbers $x, y$ and $z$.", "solution": "The idea is to split the inequality in two, showing that\n\n$$\n\\left(\\sqrt{\\frac{x}{y}}+\\sqrt{\\frac{y}{z}}+\\sqrt{\\frac{z}{x}}\\right)^{2}\n$$\n\ncan be intercalated between the left-hand side and the right-hand side. Indeed, using the Cauchy-Schwarz inequality one has\n\n$$\n(x+y+z)\\left(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\right) \\geq\\left(\\sqrt{\\frac{x}{y}}+\\sqrt{\\frac{y}{z}}+\\sqrt{\\frac{z}{x}}\\right)^{2}\n$$\n\nOn the other hand, as\n\n$$\n\\sqrt{\\frac{x}{y}} \\geq \\frac{2 x}{x y+1} \\Leftrightarrow(\\sqrt{x y}-1)^{2} \\geq 0\n$$\n\nby summation one has\n\n$$\n\\sqrt{\\frac{x}{y}}+\\sqrt{\\frac{y}{z}}+\\sqrt{\\frac{z}{x}} \\geq \\frac{2 x}{x y+1}+\\frac{2 y}{y z+1}+\\frac{2 z}{z x+1}\n$$\n\nThe rest is obvious.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nA8 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "A9", "problem_type": "Algebra", "problem": "Consider an integer $n \\geq 4$ and a sequence of real numbers $x_{1}, x_{2}, x_{3}, \\ldots, x_{n}$. An operation consists in eliminating all numbers not having the rank of the form $4 k+3$, thus leaving only the numbers $x_{3}, x_{7}, x_{11}, \\ldots$ (for example, the sequence $4,5,9,3,6,6,1,8$ produces the sequence 9,1 . Upon the sequence $1,2,3, \\ldots, 1024$ the operation is performed successively for 5 times. Show that at the end only 1 number remains and find this number.", "solution": "After the first operation 256 number remain; after the second one, 64 are left, then 16, next 4 and ultimately only one number.\n\nNotice that the 256 numbers left after the first operation are $3,7, \\ldots, 1023$, hence they are in arithmetical progression of common difference 4. Successively, the 64 numbers left after the second operation are in arithmetical progression of ratio 16 and so on.\n\nLet $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ be the first term in the 5 sequences obtained after each of the 5 operations. Thus $a_{1}=3$ and $a_{5}$ is the requested number. The sequence before the fifth operation has 4 numbers, namely\n\n$$\na_{4}, a_{4}+256, a_{4}+512, a_{4}+768\n$$\n\nand $a_{5}=a_{4}+512$. Similarly, $a_{4}=a_{3}+128, a_{3}=a_{2}+32, a_{2}=a_{1}+8$.\n\nSumming up yields $a_{5}=a_{1}+8+32+128+512=3+680=683$.\n\n### 2.2 Combinatorics", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nA9 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "problem": "On a $5 \\times 5$ board, $n$ white markers are positioned, each marker in a distinct $1 \\times 1$ square. A smart child got an assignment to recolor in black as many markers as possible, in the following manner: a white marker is taken from the board; it is colored in black, and then put back on the board on an empty square such that none of the neighboring squares contains a white marker (two squares are called neighboring if they contain a common side). If it is possible for the child to succeed in coloring all the markers black, we say that the initial positioning of the markers was good.\n\na) Prove that if $n=20$, then a good initial positioning exists.\n\nb) Prove that if $n=21$, then a good initial positioning does not exist.", "solution": "a) Position 20 white markers on the board such that the left-most column is empty. This\npositioning is good because the coloring can be realized column by column, starting with the second (from left), then the third, and so on, so that the white marker on position $(i, j)$ after the coloring is put on position $(i, j-1)$.\n\nb) Suppose there exists a good positioning with 21 white markers on the board i.e. there exists a re-coloring of them all, one by one. In any moment when there are 21 markers on the board, there must be at least one column completely filled with markers, and there must be at least one row completely filled with markers. So, there exists a \"cross\" of markers on the board. At the initial position, each such cross is completely white, at the final position each such cross is completely black, and at every moment when there are 21 markers on the board, each such cross is monochromatic. But this cannot be, since every two crosses have at least two common squares and therefore it is not possible for a white cross to vanish and for a black cross to appear by re-coloring of only one marker. Contradiction!", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nC1 ", "solution_match": "\nSolution"}}
+{"year": "2008", "tier": "T3", "problem_label": "C2", "problem_type": "Combinatorics", "problem": "Kostas and Helene have the following dialogue:\n\nKostas: I have in my mind three positive real numbers with product 1 and sum equal to the sum of all their pairwise products.\n\nHelene: I think that I know the numbers you have in mind. They are all equal to 1.\n\nKostas: In fact, the numbers you mentioned satisfy my conditions, but I did not think of these numbers. The numbers you mentioned have the minimal sum between all possible solutions of the problem.\n\nCan you decide if Kostas is right? (Explain your answer).", "solution": "Kostas is right according to the following analysis:\n\nIf $x, y, z$ are the three positive real numbers Kostas thought about, then they satisfy the following equations:\n\n$$\n\\begin{gathered}\nx y+y z+z x=x+y+z \\\\\nx y z=1\n\\end{gathered}\n$$\n\nSubtracting (1) from (2) by parts we obtain\n\n$$\n\\begin{gathered}\nx y z-(x y+y z+z x)=1-(x+y+z) \\\\\n\\Leftrightarrow x y z-x y-y z-z x+x+y+z-1=0 \\\\\n\\Leftrightarrow x y(z-1)-x(z-1)-y(z-1)+(z-1)=0 \\\\\n\\Leftrightarrow(z-1)(x y-x-y+1)=0 \\\\\n(z-1)(x-1)(y-1)=0 \\\\\n\\Leftrightarrow x=1 \\text { or } y=1 \\text { or } z=1 .\n\\end{gathered}\n$$\n\nFor $x=1$, from (1) and (2) we have the equation $y z=1$, which has the solutions\n\n$$\n(y, z)=\\left(a, \\frac{1}{a}\\right), a>0\n$$\n\nAnd therefore the solutions of the problem are the triples\n\n$$\n(x, y, z)=\\left(1, a, \\frac{1}{a}\\right), a>0\n$$\n\nSimilarly, considering $y=1$ or $z=1$ we get the solutions\n\n$$\n(x, y, z)=\\left(a, 1, \\frac{1}{a}\\right) \\text { or }(x, y, z)=\\left(a, \\frac{1}{a}, 1\\right), a>0\n$$\n\nSince for each $a>0$ we have\n\n$$\nx+y+z=1+a+\\frac{1}{a} \\geq 1+2=3\n$$\n\nand equality is valid only for $a=1$, we conclude that among the solutions of the problem, the triple $(x, y, z)=(1,1,1)$ is the one whose sum $x+y+z$ is minimal.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nC2 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "problem": "Integers $1,2, \\ldots, 2 n$ are arbitrarily assigned to boxes labeled with numbers $1,2, \\ldots, 2 n$. Now, we add the number assigned to the box to the number on the box label. Show that two such sums give the same remainder modulo $2 n$.", "solution": "Let us assume that all sums give different remainder modulo $2 n$, and let $S$ denote the value of their sum.\n\nFor our assumption,\n\n$$\nS \\equiv 0+1+\\ldots+2 n-1=\\frac{(2 n-1) 2 n}{2}=(2 n-1) n \\equiv n \\quad(\\bmod 2 n)\n$$\n\nBut, if we sum, breaking all sums into its components, we derive\n\n$$\nS \\equiv 2(1+\\ldots+2 n)=2 \\cdot \\frac{2 n(2 n+1)}{2}=2 n(2 n+1) \\equiv 0 \\quad(\\bmod 2 n)\n$$\n\nFrom the last two conclusions we derive $n \\equiv 0(\\bmod 2 n)$. Contradiction.\n\nTherefore, there are two sums with the same remainder modulo $2 n$.\n\nRemark: The result is no longer true if one replaces $2 n$ by $2 n+1$. Indeed, one could assign the number $k$ to the box labeled $k$, thus obtaining the sums $2 k, k=\\overline{1,2 n+1}$. Two such numbers give different remainders when divided by $2 n+1$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nC3 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "C4", "problem_type": "Combinatorics", "problem": "Every cell of table $4 \\times 4$ is colored into white. It is permitted to place the cross (pictured below) on the table such that its center lies on the table (the whole figure does not need to lie on the table) and change colors of every cell which is covered into opposite (white and black). Find all $n$ such that after $n$ steps it is possible to get the table with every cell colored black.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_cef994f1dc1cf35663f9g-17.jpg?height=182&width=181&top_left_y=451&top_left_x=945)", "solution": "The cross covers at most five cells so we need at least 4 steps to change the color of every cell. If we place the cross 4 times such that its center lies in the cells marked below, we see that we can turn the whole square black in $n=4$ moves.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_cef994f1dc1cf35663f9g-17.jpg?height=198&width=201&top_left_y=1018&top_left_x=932)\n\nFurthermore, applying the same operation twice (,,do and undo\"), we get that is possible to turn all the cells black in $n$ steps for every even $n \\geq 4$.\n\nWe shall prove that for odd $n$ it is not possible to do that. Look at the picture below.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_cef994f1dc1cf35663f9g-17.jpg?height=196&width=206&top_left_y=1550&top_left_x=930)\n\nLet $k$ be a difference between white and black cells in the green area in picture. Every figure placed on the table covers an odd number of green cells, so after every step $k$ is changed by a number $\\equiv 2(\\bmod 4)$. At the beginning $k=10$, at the end $k=-10$. From this it is clear that we need an even number of steps.\n\nSolution for $n$ is: every even number except 2 .\n\n### 2.3 Geometry", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nC4 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "problem": "Two perpendicular chords of a circle, $A M, B N$, which intersect at point $K$, define on the circle four arcs with pairwise different length, with $A B$ being the smallest of them.\n\nWe draw the chords $A D, B C$ with $A D \\| B C$ and $C, D$ different from $N, M$. If $L$ is the point of intersection of $D N, M C$ and $T$ the point of intersection of $D C, K L$, prove that $\\angle K T C=\\angle K N L$.", "solution": "First we prove that $N L \\perp M C$. The arguments depend slightly on the position of $D$. The other cases are similar.\n\nFrom the cyclic quadrilaterals $A D C M$ and $D N B C$ we have:\n\n$$\n\\varangle D C L=\\varangle D A M \\text { and } \\varangle C D L=\\varangle C B N \\text {. }\n$$\n\nSo we obtain\n\n$$\n\\varangle D C L+\\varangle C D L=\\varangle D A M+\\varangle C B N .\n$$\n\nAnd because $A D \\| B C$, if $Z$ the point of intersection of $A M, B C$ then $\\varangle D A M=\\varangle B Z A$, and we have\n\n$$\n\\varangle D C L+\\varangle C D L=\\varangle B Z A+\\varangle C B N=90^{\\circ}\n$$\n\nLet $P$ the point of intersection of $K L, A C$, then $N P \\perp A C$, because the line $K P L$ is a Simson line of the point $N$ with respect to the triangle $A C M$.\n\nFrom the cyclic quadrilaterals $N P C L$ and $A N D C$ we obtain:\n\n$$\n\\varangle C P L=\\varangle C N L \\text { and } \\varangle C N L=\\varangle C A D \\text {, }\n$$\n\nso $\\varangle C P L=\\varangle C A D$, that is $K L\\|A D\\| B C$ therefore $\\varangle K T C=\\varangle A D C$ (1).\n\nBut $\\varangle A D C=\\varangle A N C=\\varangle A N K+\\varangle K N C=\\varangle C N L+\\varangle K N C$, so\n\n$$\n\\varangle A D C=\\varangle K N L\n$$\n\nFrom (1) and (2) we obtain the result.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_cef994f1dc1cf35663f9g-18.jpg?height=542&width=518&top_left_y=1710&top_left_x=782)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nG1 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "G2", "problem_type": "Geometry", "problem": "For a fixed triangle $A B C$ we choose a point $M$ on the ray $C A$ (after $A$ ), a point $N$ on the ray $A B$ (after $B$ ) and a point $P$ on the ray $B C$ (after $C$ ) in a way such that $A M-B C=B N-A C=C P-A B$. Prove that the angles of triangle $M N P$ do not depend on the choice of $M, N, P$.", "solution": "Consider the points $M^{\\prime}$ on the ray $B A$ (after $A$ ), $N^{\\prime}$ on the ray $C B$ (after $B$ ) and $P^{\\prime}$ on the ray $A C$ (after $C$ ), so that $A M=A M^{\\prime}, B N=B N^{\\prime}, C P=C P^{\\prime}$. Since $A M-B C=B N-A C=B N^{\\prime}-A C$, we get $C M=A C+A M=B C+B N^{\\prime}=C N^{\\prime}$. Thus triangle $M C N^{\\prime}$ is isosceles, so the perpendicular bisector of $\\left[M N^{\\prime}\\right]$ bisects angle $A C B$ and hence passes through the incenter $I$ of triangle $A B C$. Arguing similarly, we may conclude that $I$ lies also on the perpendicular bisectors of $\\left[N P^{\\prime}\\right]$ and $\\left[P M^{\\prime}\\right]$. On the other side, $I$ clearly lies on the perpendicular bisectors of $\\left[M M^{\\prime}\\right],\\left[N N^{\\prime}\\right]$ and $\\left[P P^{\\prime}\\right]$. Thus the hexagon $M^{\\prime} M N^{\\prime} N P^{\\prime} P$ is cyclic. Then angle $P M N$ equals angle $P N^{\\prime} N$, which measures $90^{\\circ}-\\frac{\\beta}{2}$ (the angles of triangle $A B C$ are $\\alpha, \\beta, \\gamma$ ). In the same way angle $M N P$ measures $90^{\\circ}-\\frac{\\gamma}{2}$ and angle $M P N$ measures $90^{\\circ}-\\frac{\\alpha}{2}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nG2 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "G3", "problem_type": "Geometry", "problem": "The vertices $A$ and $B$ of an equilateral $\\triangle A B C$ lie on a circle $k$ of radius 1 , and the vertex $C$ is inside $k$. The point $D \\neq B$ lies on $k, A D=A B$ and the line $D C$ intersects $k$ for the second time in point $E$. Find the length of the segment $C E$.", "solution": "As $A D=A C, \\triangle C D A$ is isosceles. If $\\varangle A D C=\\varangle A C D=\\alpha$ and $\\varangle B C E=\\beta$, then $\\beta=120^{\\circ}-\\alpha$. The quadrilateral $A B E D$ is cyclic, so $\\varangle A B E=180^{\\circ}-\\alpha$. Then $\\varangle C B E=$ $120^{\\circ}-\\alpha$ so $\\varangle C B E=\\beta$. Thus $\\triangle C B E$ is isosceles, so $A E$ is the perpendicular bisector of $B C$, so it bisects $\\varangle B A C$. Now the arc $B E$ is intercepted by a $30^{\\circ}$ inscribed angle, so it measures $60^{\\circ}$. Then $B E$ equals the radius of $k$, namely 1 . Hence $C E=B E=1$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_cef994f1dc1cf35663f9g-19.jpg?height=458&width=485&top_left_y=1614&top_left_x=798)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nG3 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "problem": "Let $A B C$ be a triangle, $(B C<A B)$. The line $\\ell$ passing trough the vertices $C$ and orthogonal to the angle bisector $B E$ of $\\angle B$, meets $B E$ and the median $B D$ of the side $A C$ at points $F$ and $G$, respectively. Prove that segment $D F$ bisect the segment $E G$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_cef994f1dc1cf35663f9g-20.jpg?height=434&width=314&top_left_y=474&top_left_x=884)", "solution": "Let $C F \\cap A B=\\{K\\}$ and $D F \\cap B C=\\{M\\}$. Since $B F \\perp K C$ and $B F$ is angle bisector of $\\varangle K B C$, we have that $\\triangle K B C$ is isosceles i.e. $B K=B C$, also $F$ is midpoint of $K C$. Hence $D F$ is midline for $\\triangle A C K$ i.e. $D F \\| A K$, from where it is clear that $M$ is a midpoint of $B C$.\n\nWe will prove that $G E \\| B C$. It is sufficient to show $\\frac{B G}{G D}=\\frac{C E}{E D}$. From $D F \\| A K$ and $D F=\\frac{A K}{2}$ we have\n\n$$\n\\frac{B G}{G D}=\\frac{B K}{D F}=\\frac{2 B K}{A K}\n$$\n\nAlso\n\n$$\n\\begin{gathered}\n\\frac{C E}{D E}=\\frac{C D-D E}{D E}=\\frac{C D}{D E}-1=\\frac{A D}{D E}-1=\\frac{A E-D E}{D E}-1=\\frac{A E}{D E}-2= \\\\\n=\\frac{A B}{D F}-2=\\frac{A K+B K}{\\frac{A K}{2}}-2=2+2 \\frac{B K}{A K}-2=\\frac{2 B K}{A K}\n\\end{gathered}\n$$\n\nFrom (1) and (2) we have $\\frac{B G}{G D}=\\frac{C E}{E D}$, so $G E \\| B C$, as $M$ is the midpoint of $B C$, it follows that the segment $D F$, bisects the segment $G E$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nG4 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "G5", "problem_type": "Geometry", "problem": "Is it possible to cover a given square with a few congruent right-angled triangles with acute angle equal to $30^{\\circ}$ ? (The triangles may not overlap and may not exceed the margins of the square.)", "solution": "We will prove that desired covering is impossible.\n\nLet assume the opposite i.e. a square with side length $a$, can be tiled with $k$ congruent right angled triangles, whose sides are of lengths $b, b \\sqrt{3}$ and $2 b$.\n\nThen the area of such a triangle is $\\frac{b^{2} \\sqrt{3}}{2}$.\n\nAnd the area of the square is\n\n$$\nS_{s q}=k b^{2} \\frac{\\sqrt{3}}{2}\n$$\n\nFurthermore, the length of the side of the square, $a$, is obtained by the contribution of an integer number of length $b, 2 b$ and $b \\sqrt{3}$, hence\n\n$$\na=m b \\sqrt{3}+n b\n$$\n\nwhere $m, n \\in \\mathbb{N} \\cup\\{0\\}$, and at least one of the numbers $m$ and $n$ is different from zero. So the area of the square is\n\n$$\nS_{s q}=a^{2}=(m b \\sqrt{3}+n b)^{2}=b^{2}\\left(3 m^{2}+n^{2}+2 \\sqrt{3} m n\\right)\n$$\n\nNow because of (1) and (2) it follows $3 m^{2}+n^{2}+2 \\sqrt{3} m n=k \\frac{\\sqrt{3}}{2}$ i.e.\n\n$$\n6 m^{2}+2 n^{2}=(k-4 m n) \\sqrt{3}\n$$\n\nBecause of $3 m^{2}+n^{2} \\neq 0$ and from the equality (3) it follows $4 m n \\neq k$. Using once more (3), we get\n\n$$\n\\sqrt{3}=\\frac{6 m^{2}+2 n^{2}}{k-4 m n}\n$$\n\nwhich contradicts at the fact that $\\sqrt{3}$ is irrational, because $\\frac{6 m^{2}+2 n^{2}}{k-4 m n}$ is a rational number.\n\nFinally, we have obtained a contradiction, which proves that the desired covering is impossible.\n\n## Remark.\n\nThis problem has been given in Russian Mathematical Olympiad 1993 - 1995 for 9-th Grade.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nG5 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "G6", "problem_type": "Geometry", "problem": "Let $A B C$ be a triangle with $A<90^{\\circ}$. Outside of a triangle we consider isosceles triangles $A B E$ and $A C Z$ with bases $A B$ and $A C$, respectively. If the midpoint $D$ of the side $B C$ is such that $D E \\perp D Z$ and $E Z=2 \\cdot E D$, prove that $\\widehat{A E B}=2 \\cdot \\widehat{A Z C}$.", "solution": "Since $D$ is the midpoint of the side $B C$, in the extension of the line segment $Z D$ we take a point $H$ such that $Z D=D H$. Then the quadrilateral $B H C Z$ is parallelogram and therefore we have\n\n$$\nB H=Z C=Z A\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_cef994f1dc1cf35663f9g-22.jpg?height=647&width=596&top_left_y=468&top_left_x=748)\n\nAlso from the isosceles triangle $A B E$ we get\n\n$$\nB E=A E\n$$\n\nSince $D E \\perp D Z, E D$ is altitude and median of the triangle $E Z H$ and so this triangle is isosceles with\n\n$$\nE H=E Z\n$$\n\nFrom (1), (2) and (3) we conclude that the triangles $B E H$ and $A E Z$ are equal. Therefore they have also\n\n$$\n\\widehat{B E H}=\\widehat{A E Z}, \\widehat{E B H}=\\widehat{E A Z} \\text { and } \\widehat{E H B}=\\widehat{A Z E}\n$$\n\nPutting $\\widehat{E B A}=\\widehat{E A B}=\\omega, \\widehat{Z A C}=\\widehat{Z C A}=\\varphi$, then we have $\\widehat{C B H}=\\widehat{B C Z}=\\widehat{C}+\\varphi$, and therefore from the equality $\\widehat{E B H}=\\widehat{E A Z}$ we receive:\n\n$$\n\\begin{gathered}\n360^{\\circ}-\\widehat{E B A}-\\widehat{B}-\\widehat{C B H}=\\widehat{E A B}+\\widehat{A}+\\widehat{Z A C} \\\\\n\\Rightarrow 360^{\\circ}-\\widehat{B}-\\omega-\\varphi-\\widehat{C}=\\omega+\\widehat{A}+\\varphi \\\\\n\\Rightarrow 2(\\omega+\\varphi)=360^{\\circ}-(\\widehat{A}+\\widehat{B}+\\widehat{C}) \\\\\n\\Rightarrow \\omega+\\varphi=90^{\\circ} \\\\\n\\Rightarrow \\frac{180^{\\circ}-\\widehat{A E B}}{2}+\\frac{180^{\\circ}-\\widehat{A Z C}}{2}=90^{\\circ} \\\\\n\\Rightarrow \\widehat{A E B}+\\widehat{A Z C}=180^{\\circ}\n\\end{gathered}\n$$\n\nFrom the supposition $E Z=2 \\cdot E D$, we get that the right triangle $Z E H$ has $\\widehat{E Z D}=30^{\\circ}$ and $\\widehat{Z E D}=60^{\\circ}$. Thus we have $\\widehat{Z E H}=120^{\\circ}$.\n\nHowever, since we have proved that $\\widehat{B E H}=\\widehat{A E Z}$, we get that\n\n$$\n\\widehat{A E B}=\\widehat{A E Z}+\\widehat{Z E B}=\\widehat{Z E B}+\\widehat{B E H}=\\widehat{Z E H}=120^{\\circ}\n$$\n\nFrom (5) and (6) we obtain that $\\widehat{A Z C}=60^{\\circ}$ and thus $\\widehat{A E B}=2 \\cdot \\widehat{A Z C}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nG6 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "G7", "problem_type": "Geometry", "problem": "Let $A B C$ be an isosceles triangle with $A C=B C$. The point $D$ lies on the side $A B$ such that the semicircle with diameter $[B D]$ and center $O$ is tangent to the side $A C$ in the point $P$ and intersects the side $B C$ at the point $Q$. The radius $O P$ intersects the chord $D Q$ at the point $E$ such that $5 \\cdot P E=3 \\cdot D E$. Find the ratio $\\frac{A B}{B C}$.", "solution": "We denote $O P=O D=O B=R, A C=B C=b$ and $A B=2 a$. Because $O P \\perp A C$ and $D Q \\perp B C$, then the right triangles $A P O$ and $B Q D$ are similar and $\\varangle B D Q=\\varangle A O P$. So, the triangle $D E O$ is isosceles with $D E=O E$. It follows that\n\n$$\n\\frac{P E}{D E}=\\frac{P E}{O E}=\\frac{3}{5}\n$$\n\nLet $F$ and $G$ are the orthogonal projections of the points $E$ and $P$ respectively on the side $A B$ and $M$ is the midpoint of the side $[A B]$. The triangles $O F E, O G P, O P A$ and $C M A$ are similar. We obtain the following relations\n\n$$\n\\frac{O F}{O E}=\\frac{O G}{O P}=\\frac{C M}{A C}=\\frac{O P}{O A}\n$$\n\nBut $C M=\\sqrt{b^{2}-a^{2}}$ and we have $O G=\\frac{R}{b} \\cdot \\sqrt{b^{2}-a^{2}}$. In isosceles triangle $D E O$ the point $F$ is the midpoint of the radius $D O$. So, $O F=R / 2$. By using Thales' theorem we obtain\n\n$$\n\\frac{3}{5}=\\frac{P E}{O E}=\\frac{G F}{O F}=\\frac{O G-O F}{O F}=\\frac{O G}{O F}-1=2 \\cdot \\sqrt{1-\\left(\\frac{a}{b}\\right)^{2}}-1\n$$\n\nFrom the last relations it is easy to obtain that $\\frac{a}{b}=\\frac{3}{5}$ and $\\frac{A B}{B C}=\\frac{6}{5}$.\n\nThe problem is solved.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nG7 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "G8", "problem_type": "Geometry", "problem": "The side lengths of a parallelogram are $a, b$ and diagonals have lengths $x$ and $y$, Knowing that $a b=\\frac{x y}{2}$, show that\n\n$$\na=\\frac{x}{\\sqrt{2}}, b=\\frac{y}{\\sqrt{2}} \\text { or } a=\\frac{y}{\\sqrt{2}}, b=\\frac{x}{\\sqrt{2}}\n$$", "solution": "## Solution 2.\n\nLet us consider a parallelogram $A B C D$, with $A B=a, B C=b, A C=x, B D=y$, $\\widehat{B O C}=\\theta$, and let us produce the line $A D$ towards $D$ and consider $M \\in(A D$ so that $A D=D M$. Then $B C M D$ is a parallelogram, so $C M=B D=y$.\n\nObserve also that $(A B C D)=2(A C D)=(A C M)$ which is written equivalently as\n\n$$\nC B \\cdot C D \\cdot \\sin C=\\frac{A C \\cdot C M \\cdot \\sin \\theta}{2} \\text { i.e. } a b \\sin C=\\frac{x y \\sin \\theta}{2}\n$$\n\nBecause of the given relation $a b=\\frac{x y}{2}$ the last relation becomes $\\sin C=\\sin \\theta$, i.e.\n\n$$\n\\theta=\\widehat{C} \\text { or } \\theta=180^{\\circ}-\\widehat{C}=\\widehat{B}\n$$\n\nIf $\\theta=\\widehat{C}$, then the triangles $A C M$ and $B C D$ are similar because their angles at $C$ are equal, as well as their angles at $B, M$ (remember $B C M D$ is a parallelogram).\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_cef994f1dc1cf35663f9g-25.jpg?height=496&width=553&top_left_y=462&top_left_x=775)\n\nThen\n\n$$\n\\frac{b}{y}=\\frac{a}{x}=\\frac{y}{2 b} \\Rightarrow\\left(b=\\frac{y}{2}, a=\\frac{x}{2}\\right)\n$$\n\nIf $\\theta=\\widehat{B}$, then similarly we prove that the triangles $A C M$ and $A C D$ are similar, which then implies\n\n$$\n\\frac{a}{y}=\\frac{b}{x}=\\frac{x}{2 b} \\Rightarrow\\left(a=\\frac{y}{2}, b=\\frac{x}{2}\\right)\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nG8 ", "solution_match": "## Solution 1."}}
+{"year": "2008", "tier": "T3", "problem_label": "G8", "problem_type": "Geometry", "problem": "The side lengths of a parallelogram are $a, b$ and diagonals have lengths $x$ and $y$, Knowing that $a b=\\frac{x y}{2}$, show that\n\n$$\na=\\frac{x}{\\sqrt{2}}, b=\\frac{y}{\\sqrt{2}} \\text { or } a=\\frac{y}{\\sqrt{2}}, b=\\frac{x}{\\sqrt{2}}\n$$", "solution": "## Solution 3 .\n\nThe Parallelogram Law states that, in any parallelogram, the sum of the squares of its diagonals is equal to the sum of the squares of its sides.\n\nIn our case, this translates to $x^{2}+y^{2}=2\\left(a^{2}+b^{2}\\right)$. First adding $2 x y=4 a b$, then subtracting the same equality, yields $(x+y)^{2}=2(a+b)^{2}$ and $(x-y)^{2}=2(a-b)^{2}$. It follows that $x+y=a \\sqrt{2}+b \\sqrt{2}$ and either $x-y=a \\sqrt{2}-b \\sqrt{2}$, or $x-y=b \\sqrt{2}-a \\sqrt{2}$. In the first case one obtains $x=a \\sqrt{2}, y=b \\sqrt{2}$, in the latter case, $x=b \\sqrt{2}, y=a \\sqrt{2}$.\n\nFor the proof of the Parallelogram Law, simply apply the Law of cosines in triangles $A B C$ and $A B D$ and use the fact that $\\cos (\\varangle A B C)=-\\cos (\\varangle B A D)$. Adding the two relations gives the desired condition.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nG8 ", "solution_match": "## Solution 1."}}
+{"year": "2008", "tier": "T3", "problem_label": "G9", "problem_type": "Geometry", "problem": "Let $O$ be a point inside the parallelogram $A B C D$ such that\n\n$$\n\\angle A O B+\\angle C O D=\\angle B O C+\\angle C O D\n$$\n\nProve that there exists a circle $k$ tangent to the circumscribed circles of the triangles $\\triangle A O B, \\triangle B O C, \\triangle C O D$ and $\\triangle D O A$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_cef994f1dc1cf35663f9g-26.jpg?height=610&width=615&top_left_y=438&top_left_x=728)", "solution": "From given condition it is clear that $\\varangle A O B+\\varangle C O D=\\varangle B O C+\\varangle A O D=180^{\\circ}$.\n\nLet $E$ be a point such that $A E=D O$ and $B E=C E$. Clearly, $\\triangle A E B \\equiv \\triangle D O C$ and from that $A E \\| D O$ and $B E \\| C O$. Also, $\\varangle A E B=\\varangle C O D$ so $\\varangle A O B+\\varangle A E B=$ $\\varangle A O B+\\varangle C O D=180^{\\circ}$. Thus, the quadrilateral $A O B E$ is cyclic.\n\nSo $\\triangle A O B$ and $\\triangle A E B$ the same circumcircle, therefor the circumcircles of the triangles $\\triangle A O B$ and $\\triangle C O D$ have the same radius.\n\nAlso, $A E \\| D O$ and $A E=D O$ gives $A E O D$ is parallelogram and $\\triangle A O D \\equiv \\triangle O A E$. So $\\triangle A O B, \\triangle C O D$ and $\\triangle D O A$ has the same radius of their circumcircle (the radius of the cyclic quadrilateral $A E B O)$. Analogously, triangles $\\triangle A O B, \\triangle B O C, \\triangle C O D$ and $\\triangle D O A$ has same radius $R$.\n\nObviously, the circle with center $O$ and radius $2 R$ is externally tangent to each of these circles, so this will be the circle $k$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nG9 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "G10", "problem_type": "Geometry", "problem": "Let $\\Gamma$ be a circle of center $O$, and $\\delta$ be a line in the plane of $\\Gamma$, not intersecting it. Denote by $A$ the foot of the perpendicular from $O$ onto $\\delta$, and let $M$ be a (variable) point on $\\Gamma$. Denote by $\\gamma$ the circle of diameter $A M$, by $X$ the (other than $M$ ) intersection point of $\\gamma$ and $\\Gamma$, and by $Y$ the (other than $A$ ) intersection point of $\\gamma$ and $\\delta$. Prove that the line $X Y$ passes through a fixed point.", "solution": "Consider the line $\\rho$ tangent to $\\gamma$ at $A$, and take the points $\\{K\\}=A M \\cap X Y,\\{L\\}=$ $\\rho \\cap X M$, and $\\{F\\}=O A \\cap X Y$.\n\n(Remark: Moving $M$ into its reflection with respect to the line $O A$ will move $X Y$ into its reflection with respect to $O A$. These old and the new $X Y$ meet on $O A$, hence it should be clear that the fixed point mult be $F$.)\n\nSince $\\varangle L M A=\\varangle F Y A$ and $\\varangle Y A F=\\varangle L A M=90^{\\circ}$, it follows that triangles $F A Y$ and $L A M$ are similar, therefore $\\varangle A F Y=\\varangle A L M$, hence the quadrilateral $A L X F$ is cyclic. But then $\\varangle A F L=\\varangle A X L=90^{\\circ}$, so $L F \\perp A F$, hence $L F \\| \\delta$.\n\nNow, $\\rho$ is the radical axis of circles $\\gamma$ and $A$ (consider $A$ as a circle of center $A$ and radius 0 ), while $X M$ is the radical axis of circles $\\gamma$ and $\\Gamma$, so $L$ is the radical center of the three circle, which means that $L$ lies on the radical axis of circles $\\Gamma$ and $A$. From $L F \\perp O A$, where $O A$ is the line of the centers of the circles $A$ and $\\Gamma$, and $F \\in X Y$, it follows that $F$ is (the) fixed point of $X Y$.\n\n(The degenerate two cases when $M \\in O A$, where $X \\equiv M$ and $Y \\equiv A$, also trivially satisfy the conclusion, as then $F \\in A M)$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_cef994f1dc1cf35663f9g-27.jpg?height=436&width=466&top_left_y=861&top_left_x=800)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nG10 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "G11", "problem_type": "Geometry", "problem": "Consider $A B C$ an acute-angled triangle with $A B \\neq A C$. Denote by $M$ the midpoint of $B C$, by $D, E$ the feet of the altitudes from $B, C$ respectively and let $P$ be the intersection point of the lines $D E$ and $B C$. The perpendicular from $M$ to $A C$ meets the perpendicular from $C$ to $B C$ at point $R$. Prove that lines $P R$ and $A M$ are perpendicular.", "solution": "Let $F$ be the foot of the altitude from $A$ and let $S$ be the intersection point of $A M$ and $R C$. As $P C$ is an altitude of the triangle $P R S$, the claim is equivalent to $R M \\perp P S$, since the latter implies that $M$ is the orthocenter of $P R S$. Due to $R M \\perp A C$, we need to prove that $A C \\| P S$, in other words\n\n$$\n\\frac{M C}{M P}=\\frac{M A}{M S}\n$$\n\nNotice that $A F \\| C S$, so $\\frac{M A}{M S}=\\frac{M F}{M C}$. Now the claim is reduced to proving $M C^{2}=$ $M F \\cdot M P$, a well-known result considering that $A F$ is the polar line of $P$ with respect to circle of radius $M C$ centered at $M$.\n\nThe \"elementary proof\" on the latter result may be obtained as follows: $\\frac{P B}{P C}=\\frac{F B}{F C}$, using, for instance, Menelaus and Ceva theorems with respect to $A B C$. Cross-multiplying one gets $(P M-x)(F M+x)=(x-F M)(P M+x)$\n\n- $x$ stands for the length of $M C$ - and then $P M \\cdot F M=x^{2}$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_cef994f1dc1cf35663f9g-28.jpg?height=604&width=393&top_left_y=473&top_left_x=844)\n\nComment. The proof above holds for both cases $A B<A C$ and $A B>A C$; it is for the committee to decide if a contestant is supposed to (even) mention this.\n\n### 2.4 Number Theory", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nG11 ", "solution_match": "\nSolution"}}
+{"year": "2008", "tier": "T3", "problem_label": "NT1", "problem_type": "Number Theory", "problem": "Find all the positive integers $x$ and $y$ that satisfy the equation\n\n$$\nx(x-y)=8 y-7\n$$", "solution": "The given equation can be written as:\n\n$$\n\\begin{aligned}\n& x(x-y)=8 y-7 \\\\\n& x^{2}+7=y(x+8)\n\\end{aligned}\n$$\n\nLet $x+8=m, m \\in \\mathbb{N}$. Then we have: $x^{2}+7 \\equiv 0(\\bmod m)$, and $x^{2}+8 x \\equiv 0(\\bmod m)$. So we obtain that $8 x-7 \\equiv 0(\\bmod m) \\quad(1)$.\n\nAlso we obtain $8 x+8^{2}=8(x+8) \\equiv 0(\\bmod m) \\quad(2)$.\n\nFrom (1) and $(2)$ we obtain $(8 x+64)-(8 x-7)=71 \\equiv 0(\\bmod m)$, therefore $m \\mid 71$, since 71 is a prime number, we have:\n\n$x+8=1$ or $x+8=71$. The only accepted solution is $x=63$, and from the initial equation we obtain $y=56$.\n\nTherefore the equation has a unique solution, namely $(x, y)=(63,56)$.\n\nSolution 2:\n\nThe given equation is $x^{2}-x y+7-8 y=0$.\n\nDiscriminant is $\\Delta=y^{2}+32 y-28=(y+16)^{2}-284$ and must be perfect square. So $(y+16)^{2}-284=m^{2}$, and its follow $(y+16)^{2}-m^{2}=284$, and after some casework, $y+16-m=2$ and $y+16+m=142$, hence $y=56, x=63$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nNT1 ", "solution_match": "## Solution 1:"}}
+{"year": "2008", "tier": "T3", "problem_label": "NT2", "problem_type": "Number Theory", "problem": "Let $n \\geq 2$ be a fixed positive integer. An integer will be called \" $n$-free\" if it is not a multiple of an $n$-th power of a prime. Let $M$ be an infinite set of rational numbers, such that the product of every $n$ elements of $M$ is an $n$-free integer. Prove that $M$ contains only integers.", "solution": "We first prove that $M$ can contain only a finite number of non-integers. Suppose that there are infinitely many of them: $\\frac{p_{1}}{q_{1}}, \\frac{p_{2}}{q_{2}}, \\ldots, \\frac{p_{k}}{q_{k}}, \\ldots$, with $\\left(p_{k}, q_{k}\\right)=1$ and $q_{k}>1$ for each $k$. Let $\\frac{p}{q}=\\frac{p_{1} p_{2} \\ldots p_{n-1}}{q_{1} q_{2} \\ldots q_{n-1}}$, where $(p, q)=1$. For each $i \\geq n$, the number $\\frac{p}{q} \\cdot \\frac{p_{i}}{q_{i}}$ is an integer, so $q_{i}$ is a divisor of $p$ (as $q_{i}$ and $p_{i}$ are coprime). But $p$ has a finite set of divisors, so there are $n$ numbers of $M$ with equal denominators. Their product cannot be an integer, a contradiction.\n\nNow suppose that $M$ contains a fraction $\\frac{a}{b}$ in lowest terms with $b>1$. Take a prime divisor $p$ of $b$. If we take any $n-1$ integers from $M$, their product with $\\frac{a}{b}$ is an integer, so some of them is a multiple of $p$. Therefore there are infinitely many multiples of $p$ in $M$, and the product of $n$ of them is not $n$-free, a contradiction.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nNT2 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "NT3", "problem_type": "Number Theory", "problem": "Let $s(a)$ denote the sum of digits of a given positive integer $a$. The sequence $a_{1}, a_{2}, \\ldots a_{n}, \\ldots$ of positive integers is such that $a_{n+1}=a_{n}+s\\left(a_{n}\\right)$ for each positive integer $n$. Find the greatest possible $n$ for which it is possible to have $a_{n}=2008$.", "solution": "Since $a_{n-1} \\equiv s\\left(a_{n-1}\\right)$ (all congruences are modulo 9 ), we have $2 a_{n-1} \\equiv a_{n} \\equiv 2008 \\equiv 10$, so $a_{n-1} \\equiv 5$. But $a_{n-1}<2008$, so $s\\left(a_{n-1}\\right) \\leq 28$ and thus $s\\left(a_{n-1}\\right)$ can equal 5,14 or 23 . We check $s(2008-5)=s(2003)=5, s(2008-14)=s(1994)=23, s(2008-23)=s(1985)=$ 23. Thus $a_{n-1}$ can equal 1985 or 2003 . As above $2 a_{n-2} \\equiv a_{n-1} \\equiv 5 \\equiv 14$, so $a_{n-2} \\equiv 7$. But $a_{n-2}<2003$, so $s\\left(a_{n-2}\\right) \\leq 28$ and thus $s\\left(a_{n-2}\\right)$ can equal 16 or 25 . Checking as above we see that the only possibility is $s(2003-25)=s(1978)=25$. Thus $a_{n-2}$ can be only 1978. Now $2 a_{n-3} \\equiv a_{n-2} \\equiv 7 \\equiv 16$ and $a_{n-3} \\equiv 8$. But $s\\left(a_{n-3}\\right) \\leq 27$ and thus $s\\left(a_{n-3}\\right)$ can equal 17 or 26 . The check works only for $s(1978-17)=s(1961)=17$. Thus $a_{n-3}=1961$ and similarly $a_{n-4}=1939 \\equiv 4, a_{n-5}=1919 \\equiv 2$ (if they exist). The search for $a_{n-6}$ requires a residue of 1 . But $a_{n-6}<1919$, so $s\\left(a_{n-6}\\right) \\leq 27$ and thus $s\\left(a_{n-6}\\right)$ can be equal only to 10 or 19 . The check fails for both $s(1919-10)=s(1909)=19$ and $s(1919-19)=s(1900)=10$. Thus $n \\leq 6$ and the case $n=6$ is constructed above (1919, 1939, 1961, 1978, 2003, 2008).", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nNT3 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "NT4", "problem_type": "Number Theory", "problem": "Find all integers $n$ such that $n^{4}+8 n+11$ is a product of two or more consecutive integers.", "solution": "We will prove that $n^{4}+8 n+11$ is never a multiple of 3 . This is clear if $n$ is a multiple of 3 . If\n$n$ is not a multiple of 3 , then $n^{4}+8 n+11=\\left(n^{4}-1\\right)+12+8 n=(n-1)(n+1)\\left(n^{2}+1\\right)+12+8 n$, where $8 n$ is the only term not divisible by 3 . Thus $n^{4}+8 n+11$ is never the product of three or more integers.\n\nIt remains to discuss the case when $n^{4}+8 n+11=y(y+1)$ for some integer $y$. We write this as $4\\left(n^{4}+8 n+11\\right)=4 y(y+1)$ or $4 n^{4}+32 n+45=(2 y+1)^{2}$. A check shows that among $n= \\pm 1$ and $n=0$ only $n=1$ satisfies the requirement, as $1^{4}+8 \\cdot 1+11=20=4 \\cdot 5$. Now let $|n| \\geq 2$. The identities $4 n^{2}+32 n+45=\\left(2 n^{2}-2\\right)^{2}+8(n+2)^{2}+9$ and $4 n^{4}+32 n+45=$ $\\left(2 n^{2}+8\\right)^{2}-32 n(n-1)-19$ indicate that for $|n| \\geq 2,2 n^{2}-2<2 y+1<2 n^{2}+8$. But $2 y+1$ is odd, so it can equal $2 n^{2} \\pm 1 ; 2 n^{2}+3 ; 2 n^{2}+5$ or $2 n^{2}+7$. We investigate them one by one.\n\nIf $4 n^{4}+32 n+45=\\left(2 n^{2}-1\\right)^{2} \\Rightarrow n^{2}+8 n+11=0 \\Rightarrow(n+4)^{2}=5$, which is impossible, as 5 is not a perfect square.\n\nIf $4 n^{4}+32 n+45=\\left(2 n^{2}+1\\right)^{2} \\Rightarrow n^{2}-8 n-11=0 \\Rightarrow(n-4)^{2}=27$ which also fails.\n\nAlso $4 n^{4}+32 n+45=\\left(2 n^{2}+3\\right)^{2} \\Rightarrow 3 n^{2}-8 n-9=0 \\Rightarrow 9 n^{2}-24 n-27=0 \\Rightarrow(3 n-4)^{2}=43$ fails.\n\nIf $4 n^{4}+32 n+45=\\left(2 n^{2}+5\\right)^{2} \\Rightarrow 5 n^{2}-8 n=5 \\Rightarrow 25 n^{2}-40 n=25 \\Rightarrow(5 n-4)^{2}=41$ which also fails.\n\nFinally, if $4 n^{4}+32 n+45=\\left(2 n^{2}+7\\right)^{2}$, then $28 n^{2}-32 n+4=0 \\Rightarrow 4(n-1)(7 n-1)=0$, whence $n=1$ that we already found. Thus the only solution is $n=1$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nNT4 ", "solution_match": "\nSolution"}}
+{"year": "2008", "tier": "T3", "problem_label": "NT5", "problem_type": "Number Theory", "problem": "Is it possible to arrange the numbers $1^{1}, 2^{2}, \\ldots, 2008^{2008}$ one after the other, in such a way that the obtained number is a perfect square? (Explain your answer.)", "solution": "We will use the following lemmas.\n\nLemma 1. If $x \\in \\mathbb{N}$, then $x^{2} \\equiv 0$ or $1(\\bmod 3)$.\n\nProof: Let $x \\in \\mathbb{N}$, then $x=3 k, x=3 k+1$ or $x=3 k+2$, hence\n\n$$\n\\begin{aligned}\n& x^{2}=9 k^{2} \\equiv 0(\\bmod 3) \\\\\n& x^{2}=9 k^{2}+6 k+1 \\equiv 1(\\bmod 3), \\\\\n& x^{2}=9 k^{2}+12 k+4 \\equiv 1(\\bmod 3), \\text { respectively. }\n\\end{aligned}\n$$\n\nHence $x^{2} \\equiv 0$ or $1(\\bmod 3)$, for every positive integer $x$.\n\nWithout proof we will give the following lemma.\n\nLemma 2. If $a$ is a positive integer then $a \\equiv S(a)(\\bmod 3)$, where $S(a)$ is the sum of the digits of the number $a$.\n\nFurther we have\n\n$$\n\\begin{aligned}\n& (6 k+1)^{6 k+1}=\\left[(6 k+1)^{k}\\right]^{6} \\cdot(6 k+1) \\equiv 1(\\bmod 3) \\\\\n& (6 k+2)^{6 k+2}=\\left[(6 k+2)^{3 k+1}\\right]^{2} \\equiv 1(\\bmod 3) \\\\\n& (6 k+3)^{6 k+3} \\equiv 0(\\bmod 3) \\\\\n& (6 k+4)^{6 k+4}=\\left[(6 k+1)^{3 k+2}\\right]^{2} \\equiv 1(\\bmod 3) \\\\\n& (6 k+5)^{6 k+5}=\\left[(6 k+5)^{3 k+2}\\right]^{2} \\cdot(6 k+5) \\equiv 2(\\bmod 3) \\\\\n& (6 k+6)^{6 k+6} \\equiv 0(\\bmod 3)\n\\end{aligned}\n$$\n\nfor every $k=1,2,3, \\ldots$.\n\nLet us separate the numbers $1^{1}, 2^{2}, \\ldots, 2008^{2008}$ into the following six classes: $(6 k+1)^{6 k+1}$, $(6 k+2)^{6 k+2},(6 k+3)^{6 k+3},(6 k+4)^{6 k+4},(6 k+5)^{6 k+5},(6 k+6)^{6 k+6}, k=1,2, \\ldots$.\n\nFor $k=1,2,3, \\ldots$ let us denote by\n\n$s_{k}=(6 k+1)^{6 k+1}+(6 k+2)^{6 k+2}+(6 k+3)^{6 k+3}+(6 k+4)^{6 k+4}+(6 k+5)^{6 k+5}+(6 k+6)^{6 k+6}$.\n\nFrom (3) we have\n\n$$\ns_{k} \\equiv 1+1+0+1+2+0 \\equiv 2(\\bmod 3)\n$$\n\nfor every $k=1,2,3, \\ldots$.\n\nLet $A$ be the number obtained by writing one after the other (in some order) the numbers $1^{1}, 2^{2}, \\ldots, 2008^{2008}$.\n\nThe sum of the digits, $S(A)$, of the number $A$ is equal to the sum of the sums of digits, $S\\left(i^{i}\\right)$, of the numbers $i^{i}, i=1,2, \\ldots, 2008$, and so, from Lemma 2, it follows that\n\n$$\nA \\equiv S(A)=S\\left(1^{1}\\right)+S\\left(2^{2}\\right)+\\ldots+S\\left(2008^{2008}\\right) \\equiv 1^{1}+2^{2}+\\ldots+2008^{2008}(\\bmod 3)\n$$\n\nFurther on $2008=334 \\cdot 6+4$ and if we use (3) and (4) we get\n\n$$\n\\begin{aligned}\nA & \\equiv 1^{1}+2^{2}+\\ldots+2008^{2008} \\\\\n& \\equiv s_{1}+s_{2}+\\ldots+s_{334}+2005^{2005}+2006^{2006}+2007^{2007}+2008^{2008}(\\bmod 3) \\\\\n& \\equiv 334 \\cdot 2+1+1+0+1=671 \\equiv 2(\\bmod 3)\n\\end{aligned}\n$$\n\nFinally, from Lemma 1, it follows that $A$ can not be a perfect square.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nNT5 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "NT6", "problem_type": "Number Theory", "problem": "Let $f: \\mathbb{N} \\rightarrow \\mathbb{R}$ be a function, satisfying the following condition:\n\nfor every integer $n>1$, there exists a prime divisor $p$ of $n$ such that $f(n)=f\\left(\\frac{n}{p}\\right)-f(p)$. If\n\n$$\nf\\left(2^{2007}\\right)+f\\left(3^{2008}\\right)+f\\left(5^{2009}\\right)=2006\n$$\n\ndetermine the value of\n\n$$\nf\\left(2007^{2}\\right)+f\\left(2008^{3}\\right)+f\\left(2009^{5}\\right)\n$$", "solution": "If $n=p$ is prime number, we have\n\n$$\nf(p)=f\\left(\\frac{p}{p}\\right)-f(p)=f(1)-f(p)\n$$\n\ni.e.\n\n$$\nf(p)=\\frac{f(1)}{2}\n$$\n\nIf $n=p q$, where $p$ and $q$ are prime numbers, then\n\n$$\nf(n)=f\\left(\\frac{n}{p}\\right)-f(p)=f(q)-f(p)=\\frac{f(1)}{2}-\\frac{f(1)}{2}=0\n$$\n\nIf $n$ is a product of three prime numbers, we have\n\n$$\nf(n)=f\\left(\\frac{n}{p}\\right)-f(p)=0-f(p)=-f(p)=-\\frac{f(1)}{2}\n$$\n\nWith mathematical induction by a number of prime multipliers we shell prove that: if $n$ is a product of $k$ prime numbers then\n\n$$\nf(n)=(2-k) \\frac{f(1)}{2}\n$$\n\nFor $k=1$, clearly the statement (2), holds.\n\nLet statement (2) holds for all integers $n$, where $n$ is a product of $k$ prime numbers.\n\nNow let $n$ be a product of $k+1$ prime numbers. Then we have $n=n_{1} p$, where $n_{1}$ is a product of $k$ prime numbers.\n\nSo\n\n$$\nf(n)=f\\left(\\frac{n}{p}\\right)-f(p)=f\\left(n_{1}\\right)-f(p)=(2-k) \\frac{f(1)}{2}-\\frac{f(1)}{2}=(2-(k+1)) \\frac{f(1)}{2}\n$$\n\nSo (2) holds for every integer $n>1$.\n\nNow from $f\\left(2^{2007}\\right)+f\\left(3^{2008}\\right)+f\\left(5^{2009}\\right)=2006$ and because of (2) we have\n\n$$\n\\begin{aligned}\n2006 & =f\\left(2^{2007}\\right)+f\\left(3^{2008}\\right)+f\\left(5^{2009}\\right) \\\\\n& =\\frac{2-2007}{2} f(1)+\\frac{2-2008}{2} f(1)+\\frac{2-2009}{2} f(1)=-\\frac{3 \\cdot 2006}{2} f(1)\n\\end{aligned}\n$$\n\ni.e.\n\n$$\nf(1)=-\\frac{2}{3}\n$$\n\nSince\n\n$$\n2007=3^{2} \\cdot 223,2008=2^{3} \\cdot 251,2009=7^{2} \\cdot 41\n$$\n\nand because of (2) and (3), we get\n\n$$\n\\begin{aligned}\nf\\left(2007^{2}\\right)+f\\left(2008^{3}\\right)+f\\left(2009^{5}\\right) & =\\frac{2-6}{2} f(1)+\\frac{2-12}{2} f(1)+\\frac{2-15}{2} f(1) \\\\\n& =-\\frac{27}{2} f(1)=-\\frac{27}{2} \\cdot\\left(-\\frac{2}{3}\\right)=9\n\\end{aligned}\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nNT6 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "NT7", "problem_type": "Number Theory", "problem": "Determine the minimal prime number $p>3$ for which no natural number $n$ satisfies\n\n$$\n2^{n}+3^{n} \\equiv 0(\\bmod p)\n$$", "solution": "We put $A(n)=2^{n}+3^{n}$. From Fermat's little theorem, we have $2^{p-1} \\equiv 1(\\bmod p)$ and $3^{p-1} \\equiv 1(\\bmod p)$ from which we conclude $A(n) \\equiv 2(\\bmod p)$. Therefore, after $p-1$ steps\nat most, we will have repetition of the power. It means that in order to determine the minimal prime number $p$ we seek, it is enough to determine a complete set of remainders $S(p)=\\{0,1, \\ldots, p-1\\}$ such that $2^{n}+3^{n} \\not \\equiv 0(\\bmod p)$, for every $n \\in S(p)$.\n\nFor $p=5$ and $n=1$ we have $A(1) \\equiv 0(\\bmod 5)$.\n\nFor $p=7$ and $n=3$ we have $A(3) \\equiv 0(\\bmod 7)$.\n\nFor $p=11$ and $n=5$ we have $A(5) \\equiv 0(\\bmod 11)$.\n\nFor $p=13$ and $n=2$ we have $A(2) \\equiv 0(\\bmod 13)$.\n\nFor $p=17$ and $n=8$ we have $A(8) \\equiv 0(\\bmod 17)$.\n\nFor $p=19$ we have $A(n) \\not \\equiv 0(\\bmod 19)$, for all $n \\in S(19)$.\n\nHence the minimal value of $p$ is 19 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nNT7 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "NT8", "problem_type": "Number Theory", "problem": "Let $a, b, c, d, e, f$ are nonzero digits such that the natural numbers $\\overline{a b c}, \\overline{d e f}$ and $\\overline{a b c d e f}$ are squares.\n\na) Prove that $\\overline{a b c d e f}$ can be represented in two different ways as a sum of three squares of natural numbers.\n\nb) Give an example of such a number.", "solution": "a) Let $\\overline{a b c}=m^{2}, \\overline{d e f}=n^{2}$ and $\\overline{a b c d e f}=p^{2}$, where $11 \\leq m \\leq 31,11 \\leq n \\leq 31$ are natural numbers. So, $p^{2}=1000 \\cdot m^{2}+n^{2}$. But $1000=30^{2}+10^{2}=18^{2}+26^{2}$. We obtain the following relations\n\n$$\n\\begin{gathered}\np^{2}=\\left(30^{2}+10^{2}\\right) \\cdot m^{2}+n^{2}=\\left(18^{2}+26^{2}\\right) \\cdot m^{2}+n^{2}= \\\\\n=(30 m)^{2}+(10 m)^{2}+n^{2}=(18 m)^{2}+(26 m)^{2}+n^{2}\n\\end{gathered}\n$$\n\nThe assertion a) is proved.\n\nb) We write the equality $p^{2}=1000 \\cdot m^{2}+n^{2}$ in the equivalent form $(p+n)(p-n)=1000 \\cdot m^{2}$, where $349 \\leq p \\leq 979$. If $1000 \\cdot m^{2}=p_{1} \\cdot p_{2}$, such that $p+n=p_{1}$ and $p-n=p_{2}$, then $p_{1}$ and $p_{2}$ are even natural numbers with $p_{1}>p_{2} \\geq 318$ and $22 \\leq p_{1}-p_{2} \\leq 62$. For $m=15$ we obtain $p_{1}=500, p_{2}=450$. So, $n=25$ and $p=475$. We have\n\n$$\n225625=475^{2}=450^{2}+150^{2}+25^{2}=270^{2}+390^{2}+25^{2}\n$$\n\nThe problem is solved.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nNT8 ", "solution_match": "\nSolution"}}
+{"year": "2008", "tier": "T3", "problem_label": "NT9", "problem_type": "Number Theory", "problem": "Let $p$ be a prime number. Find all positive integers $a$ and $b$ such that:\n\n$$\n\\frac{4 a+p}{b}+\\frac{4 b+p}{a}\n$$\n\nand\n\n$$\n\\frac{a^{2}}{b}+\\frac{b^{2}}{a}\n$$\n\nare integers.", "solution": "Since $a$ and $b$ are symmetric we can assume that $a \\leq b$. Let $d=(a, b), a=d u, b=d v$ and $(u, v)=1$. Then we have:\n\n$$\n\\frac{a^{2}}{b}+\\frac{b^{2}}{a}=\\frac{d\\left(u^{3}+v^{3}\\right)}{u v}\n$$\n\nSince,\n\n$$\n\\left(u^{3}+v^{3}, u\\right)=\\left(u^{3}+v^{3}, v\\right)=1\n$$\n\nwe deduce that $u \\mid d$ and $v \\mid d$. But as $(u, v)=1$, it follows that $u v \\mid d$.\n\nNow, let $d=u v t$. Furthermore,\n\n$$\n\\frac{4 a+p}{b}+\\frac{4 b+p}{a}=\\frac{4\\left(a^{2}+b^{2}\\right)+p(a+b)}{a b}=\\frac{4 u v t\\left(u^{2}+v^{2}\\right)+p(u+v)}{u^{2} v^{2} t}\n$$\n\nThis implies,\n\n$$\nu v \\mid p(u+v)\n$$\n\nBut from our assumption $1=(u, v)=(u, u+v)=(v, u+v)$ we conclude $u v \\mid p$. Therefore, we have three cases $\\{u=v=1\\},\\{u=1, v=p\\},\\{u=p, v=1\\}$. We assumed that $a \\leq b$, and this implies $u \\leq v$.\n\nIf $a=b$, we need $\\frac{4 a+p}{a}+\\frac{4 a+p}{a} \\in \\mathbb{N}$, i.e. $a \\mid 2 p$. Then $a \\in\\{1,2, p, 2 p\\}$. The other condition being fulfilled, we obtain the solutions $(1,1),(2,2),(p, p)$ and $(2 p, 2 p)$.\n\nNow, we have only one case to investigate, $u=1, v=p$. The last equation is transformed into:\n\n$$\n\\frac{4 a+p}{b}+\\frac{4 b+p}{a}=\\frac{4 p t\\left(1+p^{2}\\right)+p(p+1)}{p^{2} t}=\\frac{4 t+1+p(1+4 p t)}{p t}\n$$\n\nFrom the last equation we derive\n\n$$\np \\mid(4 t+1)\n$$\n\nLet $4 t+1=p q$. From here we derive\n\n$$\n\\frac{4 t+1+p(1+4 p t)}{p t}=\\frac{q+1+4 p t}{t}\n$$\n\nNow, we have\n\n$$\nt \\mid(q+1)\n$$\n\nor\n\n$$\nq+1=\\text { st. }\n$$\n\nTherefore,\n\n$$\np=\\frac{4 t+1}{q}=\\frac{4 t+1}{s t-1}\n$$\n\nSince $p$ is a prime number, we deduce\n\n$$\n\\frac{4 t+1}{s t-1} \\geq 2\n$$\n\nor\n\n$$\ns \\leq \\frac{4 t+3}{2 t}=2+\\frac{3}{2 t}<4\n$$\n\nCase 1: $s=1, p=\\frac{4 t+1}{t-1}=4+\\frac{5}{t-1}$. We conclude $t=2$ or $t=6$. But when $t=2$, we have $p=9$, not a prime. When $t=6, p=5, a=30, b=150$.\n\nCase 2: $s=2, p=\\frac{4 t+1}{2 t-1}=2+\\frac{3}{2 t-1}$. We conclude $t=1, p=5, a=5, b=25$ or $t=2, p=3, a=6, b=18$.\n\nCase 3: $s=3, p=\\frac{4 t+1}{3 t-1}$ or $3 p=4+\\frac{7}{3 t-1}$. As 7 does not have any divisors of the form $3 t-1$, in this case we have no solutions.\n\nSo, the solutions are\n\n$$\n(a, b)=\\{(1,1),(2,2),(p, p),(2 p, 2 p),(5,25),(6,18),(18,6),(25,5),(30,150),(150,30)\\}\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nNT9 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "NT10", "problem_type": "Number Theory", "problem": "Prove that $2^{n}+3^{n}$ is not a perfect cube for any positive integer $n$.", "solution": "If $n=1$ then $2^{1}+3^{1}=5$ is not perfect cube.\n\nPerfect cube gives residues $-1,0$ and 1 modulo 9 . If $2^{n}+3^{n}$ is a perfect cube, then $n$ must be divisible with 3 (congruence $2^{n}+3^{n}=x^{3}$ modulo 9 ).\n\nIf $n=3 k$ then $2^{3 k}+3^{2 k}>\\left(3^{k}\\right)^{3}$. Also, $\\left(3^{k}+1\\right)^{3}=3^{3 k}+3 \\cdot 3^{2 k}+3 \\cdot 3^{k}+1>3^{3 k}+3^{2 k}=$ $3^{3 k}+9^{k}>3^{3 k}+8^{k}=3^{3 k}+2^{3 k}$. But, $3^{k}$ and $3^{k}+1$ are two consecutive integers so $2^{3 k}+3^{3 k}$ is not a perfect cube.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nNT10 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "NT11", "problem_type": "Number Theory", "problem": "Determine the greatest number with $n$ digits in the decimal representation which is divisible by 429 and has the sum of all digits less than or equal to 11 .", "solution": "Let $A=\\overline{a_{n} a_{n-1} \\ldots a_{1}}$ and notice that $429=3 \\cdot 11 \\cdot 13$.\n\nSince the sum of the digits $\\sum a_{i} \\leq 11$ and $\\sum a_{i}$ is divisible by 3 , we get $\\sum a_{i}=3,6$ or 9. As 11 divides $A$, we have\n\n$$\n11 \\mid a_{n}-a_{n-1}+a_{n-2}-a_{n-3}+\\ldots\n$$\n\nin other words $11 \\mid \\sum_{i \\text { odd }} a_{i}-\\sum_{i \\text { even }} a_{i}$. But\n\n$$\n-9 \\leq-\\sum a_{i} \\leq \\sum_{i \\text { odd }} a_{i}-\\sum_{i \\text { even }} a_{i} \\leq \\sum a_{i} \\leq 9\n$$\n\nso $\\sum_{i \\text { odd }} a_{i}-\\sum_{i \\text { even }} a_{i}=0$. It follows that $\\sum a_{i}$ is even, so $\\sum a_{i}=6$ and $\\sum_{i \\text { odd }} a_{i}=\\sum_{i \\text { even }} a_{i}=3$.\n\nThe number 13 is a divisor of 1001 , hence\n\n$$\n13 \\mid \\overline{a_{3} a_{2} a_{1}}-\\overline{a_{6} a_{5} a_{4}}+\\overline{a_{9} a_{8} a_{7}}-\\overline{a_{12} a_{11} a_{10}}+\\ldots\n$$\n\nFor each $k=1,2,3,4,5,6$, let $s_{k}$ be the sum of the digits $a_{k+6 m}, m \\geq 0$; that is\n\n$$\ns_{1}=a_{1}+a_{7}+a_{13}+\\ldots \\text { and so on. }\n$$\n\nWith this notation, (1) rewrites as\n\n$$\n13 \\mid 100\\left(s_{3}-s_{6}\\right)+10\\left(s_{2}-s_{5}\\right)+\\left(s_{1}-s_{4}\\right), \\text { or } 13 \\mid 4\\left(s_{6}-s_{3}\\right)+3\\left(s_{5}-s_{2}\\right)+\\left(s_{1}-s_{4}\\right)\n$$\n\nLet $S_{3}=s_{3}-s_{6}, S_{2}=s_{2}-s_{5}$, and $S_{1}=s_{1}-s_{4}$. Recall that $\\sum_{i \\text { odd }} a_{i}=\\sum_{i \\text { even }} a_{i}$, which implies $S_{2}=S_{1}+S_{3}$. Then\n\n$$\n13\\left|4 S_{3}+3 S_{2}-S_{1}=7 S_{3}+2 S_{1} \\Rightarrow 13\\right| 49 S_{3}+14 S_{1} \\Rightarrow 13 \\mid S_{1}-3 S_{3}\n$$\n\nObserve that $\\left|S_{1}\\right| \\leq s_{1}=\\sum_{i \\text { odd }} a_{i}=3$ and likewise $\\left|S_{2}\\right|,\\left|S_{3}\\right| \\leq 3$. Then $-13<S_{1}-3 S_{3}<13$ and consequently $S_{1}=3 S_{3}$. Thus $S_{2}=4 S_{3}$ and $\\left|S_{2}\\right| \\leq 3$ yields $S_{2}=0$ and then $S_{1}=S_{3}=0$. We have $s_{1}=s_{4}, s_{2}=s_{5}, s_{3}=s_{6}$ and $s_{1}+s_{2}+s_{3}=3$, so the greatest number $A$ is $30030000 \\ldots$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nNT11 ", "solution_match": "## Solution"}}
+{"year": "2008", "tier": "T3", "problem_label": "NT12", "problem_type": "Number Theory", "problem": "Solve the equation $\\frac{p}{q}-\\frac{4}{r+1}=1$ in prime numbers.", "solution": "We can rewrite the equation in the form\n\n$$\n\\begin{gathered}\n\\frac{p r+p-4 q}{q(r+1)}=1 \\Rightarrow p r+p-4 q=q r+q \\\\\np r-q r=5 q-p \\Rightarrow r(p-q)=5 q-p\n\\end{gathered}\n$$\n\nIt follows that $p \\neq q$ and\n\n$$\n\\begin{gathered}\nr=\\frac{5 q-p}{p-q}=\\frac{4 q+q-p}{p-q} \\\\\nr=\\frac{4 q}{p-q}-1\n\\end{gathered}\n$$\n\nAs $p$ is prime, $p-q \\neq q, p-q \\neq 2 q, p-q \\neq 4 q$.\n\nWe have $p-q=1$ or $p-q=2$ or $p-q=4$\n\ni) If $p-q=1$ then\n\n$$\nq=2, p=3, r=7\n$$\n\nii) If $p-q=2$ then $p=q+2, r=2 q-1$\n\nIf $q=1(\\bmod 3)$ then $q+2 \\equiv 0(\\bmod 3)$\n\n$$\nq+2=3 \\Rightarrow q=1\n$$\n\ncontradiction.\n\nIf $q \\equiv-1(\\bmod 3)$ then $r \\equiv-2-1 \\equiv 0(\\bmod 3)$\n\n$$\n\\begin{gathered}\nr=3 \\\\\nr=2 q-1=3 \\\\\nq=2 \\\\\np=4\n\\end{gathered}\n$$\n\ncontradiction.\n\nHence $q=3, p=5, r=5$.\n\niii) If $p-q=4$ then $p=q+4$.\n\n$r=q-1$\n\nHence $q=3, p=7, r=2$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl", "problem_match": "\nNT12 ", "solution_match": "## Solution"}}