distinguish JBMO and JBMO-SL

JBMO/segment_script/core.py

@@ -168,6 +168,7 @@ def write_pairs(filename, pairs):
                 len(problem) > prob_tag.end() - prob_tag.start() + min_length
                 and len(solution) > sol_tag.end() - sol_tag.start() + min_length
             ):
+                resource_path = Path(filename).relative_to(project_root).as_posix()
                 f.write(
                     json.dumps(
                         {
@@ -175,11 +176,11 @@ def write_pairs(filename, pairs):
                             "tier": "T3",
                             "problem_label": prob_label,
                             "problem_type": problem_type_mapping.get(prob_label[0]),
-                            "exam": "JBMO",
+                            "exam": "JBMO-SL" if "en-shortlist" in resource_path else "JBMO",
                             "problem": problem,
                             "solution": solution,
                             "metadata": {
-                                "resource_path": Path(filename).relative_to(project_root).as_posix(),
+                                "resource_path": resource_path,
                                 "problem_match": prob_tag.group(0),
                                 "solution_match": sol_tag.group(0),
                             },

JBMO/segmented/en-shortlist/en-alg-20111.jsonl

@@ -1,9 +1,9 @@
-{"year": "2011", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that:\n\n$\\left(a^{5}+a^{4}+a^{3}+a^{2}+a+1\\right)\\left(b^{5}+b^{4}+b^{3}+b^{2}+b+1\\right)\\left(c^{5}+c^{4}+c^{3}+c^{2}+c+1\\right) \\geq 8\\left(a^{2}+a+1\\right)\\left(b^{2}+b+1\\right)\\left(c^{2}+c+1\\right)$.", "solution": "We have $x^{5}+x^{4}+x^{3}+x^{2}+x+1=\\left(x^{3}+1\\right)\\left(x^{2}+x+1\\right)$ for all $x \\in \\mathbb{R}_{+}$.\n\nTake $S=\\left(a^{2}+a+1\\right)\\left(b^{2}+b+1\\right)\\left(c^{2}+c+1\\right)$.\n\nThe inequality becomes $S\\left(a^{3}+1\\right)\\left(b^{3}+1\\right)\\left(c^{3}+1\\right) \\geq 8 S$.\n\nIt remains to prove that $\\left(a^{3}+1\\right)\\left(b^{3}+1\\right)\\left(c^{3}+1\\right) \\geq 8$.\n\nBy $A M-G M$ we have $x^{3}+1 \\geq 2 \\sqrt{x^{3}}$ for all $x \\in \\mathbb{R}_{+}$.\n\nSo $\\left(a^{3}+1\\right)\\left(b^{3}+1\\right)\\left(c^{3}+1\\right) \\geq 2^{3} \\cdot \\sqrt{a^{3} b^{3} c^{3}}=8$ and we are done.\n\nEquality holds when $a=b=c=1$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-alg-20111.jsonl", "problem_match": "\nA1 ", "solution_match": "## Solution"}}
-{"year": "2011", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $x, y, z$ be positive real numbers. Prove that:\n\n$$\n\\frac{x+2 y}{z+2 x+3 y}+\\frac{y+2 z}{x+2 y+3 z}+\\frac{z+2 x}{y+2 z+3 x} \\leq \\frac{3}{2}\n$$", "solution": "Notice that $\\sum_{c y c} \\frac{x+2 y}{z+2 x+3 y}=\\sum_{c y c}\\left(1-\\frac{x+y+z}{z+2 x+3 y}\\right)=3-(x+y+z) \\sum_{c y c} \\frac{1}{z+2 x+3 y}$.\n\nWe have to proof that $3-(x+y+z) \\sum_{c y c} \\frac{1}{z+2 x+3 y} \\leq \\frac{3}{2}$ or $\\frac{3}{2(x+y+z)} \\leq \\sum_{c y c} \\frac{1}{z+2 x+3 y}$.\n\nBy Cauchy-Schwarz we obtain $\\sum_{\\text {cyc }} \\frac{1}{z+2 x+3 y} \\geq \\frac{(1+1+1)^{2}}{\\sum_{\\text {cyc }}(z+2 x+3 y)}=\\frac{3}{2(x+y+z)}$.\n\n## Solution 2\n\nBecause the inequality is homogenous, we can take $x+y+z=1$.\n\nDenote $x+2 y=a, y+2 z=b, z+2 x=c$. Hence, $a+b+c=3(x+y+z)=3$.\n\nWe have $(k-1)^{2} \\geq 0 \\Leftrightarrow(k+1)^{2} \\geq 4 k \\Leftrightarrow \\frac{k+1}{4} \\geq \\frac{k}{k+1}$ for all $k>0$.\n\nHence $\\sum_{\\text {cyc }} \\frac{x+2 y}{z+2 x+3 y}=\\sum \\frac{a}{1+a} \\leq \\sum \\frac{a+1}{4}=\\frac{a+b+c+3}{4}=\\frac{3}{2}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-alg-20111.jsonl", "problem_match": "\nA2 ", "solution_match": "## Solution 1"}}
-{"year": "2011", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $a, b$ be positive real numbers. Prove that $\\sqrt{\\frac{a^{2}+a b+b^{2}}{3}}+\\sqrt{a b} \\leq a+b$.", "solution": "Applying $x+y \\leq \\sqrt{2\\left(x^{2}+y^{2}\\right)}$ for $x=\\sqrt{\\frac{a^{2}+a b+b^{2}}{3}}$ and $y=\\sqrt{a b}$, we will obtain $\\sqrt{\\frac{a^{2}+a b+b^{2}}{3}}+\\sqrt{a b} \\leq \\sqrt{\\frac{2 a^{2}+2 a b+2 b^{2}+6 a b}{3}} \\leq \\sqrt{\\frac{3\\left(a^{2}+b^{2}+2 a b\\right)}{3}}=a+b$.\n\n## Solution 2\n\nThe inequality is equivalent to $\\frac{a^{2}+a b+b^{2}}{3}+\\frac{3 a b}{3}+2 \\sqrt{\\frac{a b\\left(a^{2}+a b+b^{2}\\right)}{3}} \\leq \\frac{3 a^{2}+6 a b+3 b^{2}}{3}$. This can be rewritten as $2 \\sqrt{\\frac{a b\\left(a^{2}+a b+b^{2}\\right)}{3}} \\leq \\frac{2\\left(a^{2}+a b+b^{2}\\right)}{3}$ or $\\sqrt{a b} \\leq \\sqrt{\\frac{a^{2}+a b+b^{2}}{3}}$ which is obviously true since $a^{2}+b^{2}+a b \\geq 2 a b+a b=3 a b$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-alg-20111.jsonl", "problem_match": "\nA3 ", "solution_match": "\nSolution 1"}}
-{"year": "2011", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $x, y$ be positive real numbers such that $x^{3}+y^{3} \\leq x^{2}+y^{2}$. Find the greatest possible value of the product $x y$.", "solution": "We have $(x+y)\\left(x^{2}+y^{2}\\right) \\geq(x+y)\\left(x^{3}+y^{3}\\right) \\geq\\left(x^{2}+y^{2}\\right)^{2}$, hence $x+y \\geq x^{2}+y^{2}$. Now $2(x+y) \\geq(1+1)\\left(x^{2}+y^{2}\\right) \\geq(x+y)^{2}$, thus $2 \\geq x+y$. Because $x+y \\geq 2 \\sqrt{x y}$, we will obtain $1 \\geq x y$. Equality holds when $x=y=1$.\n\nSo the greatest possible value of the product $x y$ is 1 .\n\n## Solution 2\n\nBy $A M-G M$ we have $x^{3}+y^{3} \\geq \\sqrt{x y} \\cdot\\left(x^{2}+y^{2}\\right)$, hence $1 \\geq \\sqrt{x y}$ since $x^{2}+y^{2} \\geq x^{3}+y^{3}$. Equality holds when $x=y=1$. So the greatest possible value of the product $x y$ is 1 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-alg-20111.jsonl", "problem_match": "\nA4 ", "solution_match": "## Solution 1"}}
-{"year": "2011", "tier": "T3", "problem_label": "A5", "problem_type": "Algebra", "exam": "JBMO", "problem": "Determine the positive integers $a, b$ such that $a^{2} b^{2}+208=4\\{l c m[a ; b]+g c d(a ; b)\\}^{2}$.", "solution": "Let $d=\\operatorname{gcd}(a, b)$ and $x, y \\in \\mathbb{Z}_{+}$such that $a=d x, b=d y$. Obviously, $(x, y)=1$. The equation is equivalent to $d^{4} x^{2} y^{2}+208=4 d^{2}(x y+1)^{2}$. Hence $d^{2} \\mid 208$ or $d^{2} \\mid 13 \\cdot 4^{2}$, so $d \\in\\{1,2,4\\}$. Take $t=x y$ with $t \\in \\mathbb{Z}_{+}$.\n\nCase I. If $d=1$, then $(x y)^{2}+208=4(x y+1)^{2}$ or $3 t^{2}+8 t-204=0$, without solutions.\n\nCase II. If $d=2$, then $16 x^{2} y^{2}+208=16(x y+1)^{2}$ or $t^{2}+13=t^{2}+2 t+1 \\Rightarrow t=6$, so $(x, y) \\in\\{(1,6) ;(2,3) ;(3,2) ;(6,1)\\} \\Rightarrow(a, b) \\in\\{(2,12) ;(4,6) ;(6,4) ;(12 ; 2)\\}$.\n\nCase III. If $d=4$, then $16^{2} x^{2} y^{2}+208=4 \\cdot 16(x y+1)^{2}$ or $16 t^{2}+13=4(t+1)^{2}$ and if $t \\in \\mathbb{Z}$, then 13 must be even, contradiction!\n\nFinally, the solutions are $(a, b) \\in\\{(2,12) ;(4,6) ;(6,4) ;(12 ; 2)\\}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-alg-20111.jsonl", "problem_match": "\nA5 ", "solution_match": "## Solution"}}
-{"year": "2011", "tier": "T3", "problem_label": "A6", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $x_{i}>1$, for all $i \\in\\{1,2,3, \\ldots, 2011\\}$. Prove the inequality $\\sum_{i=1}^{2011} \\frac{x_{i}^{2}}{x_{i+1}-1} \\geq 8044$ where $x_{2012}=x_{1}$. When does equality hold?", "solution": "Realize that $\\left(x_{i}-2\\right)^{2} \\geq 0 \\Leftrightarrow x_{i}^{2} \\geq 4\\left(x_{i}-1\\right)$. So we get:\n\n$\\frac{x_{1}^{2}}{x_{2}-1}+\\frac{x_{2}^{2}}{x_{3}-1}+\\ldots+\\frac{x_{2011}^{2}}{x_{1}-1} \\geq 4\\left(\\frac{x_{1}-1}{x_{2}-1}+\\frac{x_{2}-1}{x_{3}-1}+\\ldots+\\frac{x_{2011}-1}{x_{1}-1}\\right)$. By $A M-G M$ :\n$\\frac{x_{1}-1}{x_{2}-1}+\\frac{x_{2}-1}{x_{3}-1}+\\ldots+\\frac{x_{2011}-1}{x_{1}-1} \\geq 2011 \\cdot \\sqrt[2011]{\\frac{x_{1}-1}{x_{2}-1} \\cdot \\frac{x_{2}-1}{x_{3}-1} \\cdot \\ldots \\cdot \\frac{x_{2011}-1}{x_{1}-1}}=2011$\n\nFinally, we obtain that $\\frac{x_{1}^{2}}{x_{2}-1}+\\frac{x_{2}^{2}}{x_{3}-1}+\\ldots+\\frac{x_{2011}^{2}}{x_{1}-1} \\geq 8044$.\n\nEquality holds when $\\left(x_{i}-2\\right)^{2}=0,(\\forall) i=\\overline{1,2011}$, or $x_{1}=x_{2}=\\ldots=x_{2011}=2$.\n\n## Solution 2\n\nAll the denominators are greater than 0 , so by Cauchy - Schwar $z$ we have:\n\n$\\frac{x_{1}^{2}}{x_{2}-1}+\\frac{x_{2}^{2}}{x_{3}-1}+\\ldots+\\frac{x_{2011}^{2}}{x_{1}-1} \\geq \\frac{\\left(x_{1}+x_{2}+\\ldots+x_{2011}\\right)^{2}}{x_{1}+x_{2}+\\ldots+x_{2011}-2011}$. It remains to prove that $\\frac{\\left(x_{1}+x_{2}+\\ldots+x_{2011}\\right)^{2}}{x_{1}+x_{2}+\\ldots+x_{2011}-2011} \\geq 8044$ or $\\left(\\sum_{i=1}^{2011} x_{i}\\right)^{2}+4 \\cdot 2011^{2} \\geq 4 \\cdot 2011 \\cdot \\sum_{i=1}^{2011} x_{i}$ which is obviously true by $A M-G M$ for $\\left(\\sum_{i=1}^{2011} x_{i}\\right)^{2}$ and $4 \\cdot 2011^{2}$.\n\nEquality holds when $x_{1}+x_{2}+\\ldots+x_{2011}=4022$ and $\\frac{x_{1}}{x_{2}-1}=\\frac{x_{2}}{x_{3}-1}=\\ldots=\\frac{x_{2011}}{x_{1}-1}$ or $x_{i}^{2}-x_{i}=x_{i-1} x_{i+1}-x_{i-1},(\\forall) i=\\overline{1,2011} \\Rightarrow \\sum_{i=1}^{2011} x_{i}^{2}=\\sum_{i=1}^{2011} x_{i} x_{i+2}$ where $x_{2012}=x_{1}$ and $x_{2013}=x_{2}$. This means that $x_{1}=x_{2}=\\ldots=x_{2011}$.\n\nSo equality holds when $x_{1}=x_{2}=\\ldots=x_{2011}=2$ since $x_{1}+x_{2}+\\ldots+x_{2011}=4022$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-alg-20111.jsonl", "problem_match": "\nA6 ", "solution_match": "## Solution 1"}}
-{"year": "2011", "tier": "T3", "problem_label": "A7", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $a, b, c$ be positive real numbers with $a b c=1$. Prove the inequality:\n\n$$\n\\frac{2 a^{2}+\\frac{1}{a}}{b+\\frac{1}{a}+1}+\\frac{2 b^{2}+\\frac{1}{b}}{c+\\frac{1}{b}+1}+\\frac{2 c^{2}+\\frac{1}{c}}{a+\\frac{1}{c}+1} \\geq 3\n$$", "solution": "By $A M-G M$ we have $2 x^{2}+\\frac{1}{x}=x^{2}+x^{2}+\\frac{1}{x} \\geq 3 \\sqrt[3]{\\frac{x^{4}}{x}}=3 x$ for all $x>0$, so we have:\n\n$\\sum_{\\text {cyc }} \\frac{2 a^{2}+\\frac{1}{a}}{b+\\frac{1}{a}+1} \\geq \\sum_{c y c} \\frac{3 a}{1+b+b c}=3\\left(\\sum_{c y c} \\frac{a^{2}}{1+a+a b}\\right) \\geq \\frac{3(a+b+c)^{2}}{3+a+b+c+a b+b c+c a}$.\n\nBy $A M-G M$ we have $a b+b c+c a \\geq 3$ and $a+b+c \\geq 3$. But $3\\left(a^{2}+b^{2}+c^{2}\\right) \\geq(a+b+c)^{2} \\geq$ $3(a+b+c)$. So $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a \\geq 3+a+b+c+a b+b c+c a$. Hence $\\sum_{c y c} \\frac{2 a^{2}+\\frac{1}{a}}{b+\\frac{1}{a}+1} \\geq \\frac{3(a+b+c)^{2}}{3+a+b+c+a b+b c+c a} \\geq \\frac{3(a+b+c)^{2}}{(a+b+c)^{2}}=3$.\n\nSolution 2\n\nDenote $a=\\frac{y}{x}, b=\\frac{z}{y}$ and $c=\\frac{x}{z}$. We have $\\frac{2 a^{2}+\\frac{1}{a}}{b+\\frac{1}{a}+1}=\\frac{\\frac{2 y^{2}}{x^{2}}+\\frac{x}{y}}{\\frac{z}{y}+\\frac{x}{y}+1}=\\frac{2 y^{3}+x^{3}}{x^{2}(x+y+z)}$.\n\nHence $\\sum_{c y c} \\frac{2 a^{2}+\\frac{1}{a}}{b+\\frac{1}{a}+1}=\\frac{1}{x+y+z} \\cdot \\sum_{c y c} \\frac{2 y^{3}+x^{3}}{x^{2}}=\\frac{1}{x+y+z} \\cdot\\left(x+y+z+2 \\sum_{c y c} \\frac{y^{3}}{x^{2}}\\right)$.\n\nBy Rearrangements Inequality we get $\\sum_{\\text {cyc }} \\frac{y^{3}}{x^{2}} \\geq x+y+z$.\n\nSo $\\sum_{\\text {cyc }} \\frac{2 a^{2}+\\frac{1}{a}}{b+\\frac{1}{a}+1} \\geq \\frac{1}{x+y+z} \\cdot(3 x+3 y+3 z)=3$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-alg-20111.jsonl", "problem_match": "\nA7 ", "solution_match": "## Solution 1"}}
-{"year": "2011", "tier": "T3", "problem_label": "A8", "problem_type": "Algebra", "exam": "JBMO", "problem": "Decipher the equality $(\\overline{L A R N}-\\overline{A C A}):(\\overline{C Y P}+\\overline{R U S})=C^{Y^{P}} \\cdot R^{U^{S}}$ where different symbols correspond to different digits and equal symbols correspond to equal digits. It is also supposed that all these digits are different from 0 .", "solution": "Denote $x=\\overline{L A R N}-\\overline{A C A}, y=\\overline{C Y P}+\\overline{R U S}$ and $z=C^{Y^{P}} \\cdot R^{U^{S}}$. It is obvious that $1823-898 \\leq x \\leq 9187-121,135+246 \\leq y \\leq 975+864$, that is $925 \\leq x \\leq 9075$ and $381 \\leq y \\leq 1839$, whence it follows that $\\frac{925}{1839} \\leq \\frac{x}{y} \\leq \\frac{9075}{381}$, or $0,502 \\ldots \\leq \\frac{x}{y} \\leq 23,81 \\ldots$ Since $\\frac{x}{y}=z$ is an integer, it follows that $1 \\leq \\frac{x}{y} \\leq 23$, hence $1 \\leq C^{Y^{P}} \\cdot R^{U^{S}} \\leq 23$. So both values $C^{Y^{P}}$ and $R^{U^{S}}$ are $\\leq 23$. From this and the fact that $2^{2^{3}}>23$ it follows that at least one of the symbols in the expression $C^{Y^{P}}$ and at least one of the symbols in the expression $R^{U^{S}}$ correspond to the digit 1. This is impossible because of the assumption that all the symbols in the set $\\{C, Y, P, R, U, S\\}$ correspond to different digits.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-alg-20111.jsonl", "problem_match": "\nA8 ", "solution_match": "## Solution"}}
-{"year": "2011", "tier": "T3", "problem_label": "A9", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $x_{1}, x_{2}, \\ldots, x_{n}$ be real numbers satisfying $\\sum_{k=1}^{n-1} \\min \\left(x_{k} ; x_{k+1}\\right)=\\min \\left(x_{1}, x_{n}\\right)$.\n\nProve that $\\sum_{k=2}^{n-1} x_{k} \\geq 0$.", "solution": "Case I. If $\\min \\left(x_{1}, x_{n}\\right)=x_{1}$, we know that $x_{k} \\geq \\min \\left(x_{k} ; x_{k+1}\\right)$ for all $k \\in\\{1,2,3, \\ldots, n-1\\}$. So $x_{1}+x_{2}+\\ldots+x_{n-1} \\geq \\sum_{k=1}^{n-1} \\min \\left(x_{k} ; x_{k+1}\\right)=\\min \\left(x_{1}, x_{n}\\right)=x_{1}$, hence $\\sum_{k=2}^{n-1} x_{k} \\geq 0$.\n\nCase II. If $\\min \\left(x_{1}, x_{n}\\right)=x_{n}$, we know that $x_{k} \\geq \\min \\left(x_{k-1} ; x_{k}\\right)$ for all $k \\in\\{2,3,4, \\ldots, n\\}$. So $x_{2}+x_{3}+\\ldots+x_{n} \\geq \\sum_{k=1}^{n-1} \\min \\left(x_{k} ; x_{k+1}\\right)=\\min \\left(x_{1}, x_{n}\\right)=x_{n}$, hence $\\sum_{k=2}^{n-1} x_{k} \\geq 0$.\n\n## Solution 2\n\nSince $\\min (a, b)=\\frac{1}{2}(a+b-|a-b|)$, after substitutions, we will have:\n\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n-1} \\frac{1}{2}\\left(x_{k}+x_{k+1}-\\left|x_{k}-x_{k+1}\\right|\\right) & =\\frac{1}{2}\\left(x_{1}+x_{n}-\\left|x_{1}-x_{n}\\right|\\right) \\Leftrightarrow \\ldots \\\\\n2\\left(x_{2}+x_{3}+\\ldots+x_{n-1}\\right)+\\left|x_{1}-x_{n}\\right| & =\\left|x_{1}-x_{2}\\right|+\\left|x_{2}-x_{3}\\right|+\\ldots+\\left|x_{n-1}-x_{n}\\right|\n\\end{aligned}\n$$\n\nAs $\\left|x_{1}-x_{2}\\right|+\\left|x_{2}-x_{3}\\right|+\\ldots+\\left|x_{n-1}-x_{n}\\right| \\geq\\left|x_{1}-x_{2}+x_{2}-x_{3}+\\ldots+x_{n-1}-x_{n}\\right|=\\left|x_{1}-x_{n}\\right|$, we obtain the desired result.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-alg-20111.jsonl", "problem_match": "\nA9 ", "solution_match": "## Solution 1"}}
+{"year": "2011", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that:\n\n$\\left(a^{5}+a^{4}+a^{3}+a^{2}+a+1\\right)\\left(b^{5}+b^{4}+b^{3}+b^{2}+b+1\\right)\\left(c^{5}+c^{4}+c^{3}+c^{2}+c+1\\right) \\geq 8\\left(a^{2}+a+1\\right)\\left(b^{2}+b+1\\right)\\left(c^{2}+c+1\\right)$.", "solution": "We have $x^{5}+x^{4}+x^{3}+x^{2}+x+1=\\left(x^{3}+1\\right)\\left(x^{2}+x+1\\right)$ for all $x \\in \\mathbb{R}_{+}$.\n\nTake $S=\\left(a^{2}+a+1\\right)\\left(b^{2}+b+1\\right)\\left(c^{2}+c+1\\right)$.\n\nThe inequality becomes $S\\left(a^{3}+1\\right)\\left(b^{3}+1\\right)\\left(c^{3}+1\\right) \\geq 8 S$.\n\nIt remains to prove that $\\left(a^{3}+1\\right)\\left(b^{3}+1\\right)\\left(c^{3}+1\\right) \\geq 8$.\n\nBy $A M-G M$ we have $x^{3}+1 \\geq 2 \\sqrt{x^{3}}$ for all $x \\in \\mathbb{R}_{+}$.\n\nSo $\\left(a^{3}+1\\right)\\left(b^{3}+1\\right)\\left(c^{3}+1\\right) \\geq 2^{3} \\cdot \\sqrt{a^{3} b^{3} c^{3}}=8$ and we are done.\n\nEquality holds when $a=b=c=1$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-alg-20111.jsonl", "problem_match": "\nA1 ", "solution_match": "## Solution"}}
+{"year": "2011", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $x, y, z$ be positive real numbers. Prove that:\n\n$$\n\\frac{x+2 y}{z+2 x+3 y}+\\frac{y+2 z}{x+2 y+3 z}+\\frac{z+2 x}{y+2 z+3 x} \\leq \\frac{3}{2}\n$$", "solution": "Notice that $\\sum_{c y c} \\frac{x+2 y}{z+2 x+3 y}=\\sum_{c y c}\\left(1-\\frac{x+y+z}{z+2 x+3 y}\\right)=3-(x+y+z) \\sum_{c y c} \\frac{1}{z+2 x+3 y}$.\n\nWe have to proof that $3-(x+y+z) \\sum_{c y c} \\frac{1}{z+2 x+3 y} \\leq \\frac{3}{2}$ or $\\frac{3}{2(x+y+z)} \\leq \\sum_{c y c} \\frac{1}{z+2 x+3 y}$.\n\nBy Cauchy-Schwarz we obtain $\\sum_{\\text {cyc }} \\frac{1}{z+2 x+3 y} \\geq \\frac{(1+1+1)^{2}}{\\sum_{\\text {cyc }}(z+2 x+3 y)}=\\frac{3}{2(x+y+z)}$.\n\n## Solution 2\n\nBecause the inequality is homogenous, we can take $x+y+z=1$.\n\nDenote $x+2 y=a, y+2 z=b, z+2 x=c$. Hence, $a+b+c=3(x+y+z)=3$.\n\nWe have $(k-1)^{2} \\geq 0 \\Leftrightarrow(k+1)^{2} \\geq 4 k \\Leftrightarrow \\frac{k+1}{4} \\geq \\frac{k}{k+1}$ for all $k>0$.\n\nHence $\\sum_{\\text {cyc }} \\frac{x+2 y}{z+2 x+3 y}=\\sum \\frac{a}{1+a} \\leq \\sum \\frac{a+1}{4}=\\frac{a+b+c+3}{4}=\\frac{3}{2}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-alg-20111.jsonl", "problem_match": "\nA2 ", "solution_match": "## Solution 1"}}
+{"year": "2011", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $a, b$ be positive real numbers. Prove that $\\sqrt{\\frac{a^{2}+a b+b^{2}}{3}}+\\sqrt{a b} \\leq a+b$.", "solution": "Applying $x+y \\leq \\sqrt{2\\left(x^{2}+y^{2}\\right)}$ for $x=\\sqrt{\\frac{a^{2}+a b+b^{2}}{3}}$ and $y=\\sqrt{a b}$, we will obtain $\\sqrt{\\frac{a^{2}+a b+b^{2}}{3}}+\\sqrt{a b} \\leq \\sqrt{\\frac{2 a^{2}+2 a b+2 b^{2}+6 a b}{3}} \\leq \\sqrt{\\frac{3\\left(a^{2}+b^{2}+2 a b\\right)}{3}}=a+b$.\n\n## Solution 2\n\nThe inequality is equivalent to $\\frac{a^{2}+a b+b^{2}}{3}+\\frac{3 a b}{3}+2 \\sqrt{\\frac{a b\\left(a^{2}+a b+b^{2}\\right)}{3}} \\leq \\frac{3 a^{2}+6 a b+3 b^{2}}{3}$. This can be rewritten as $2 \\sqrt{\\frac{a b\\left(a^{2}+a b+b^{2}\\right)}{3}} \\leq \\frac{2\\left(a^{2}+a b+b^{2}\\right)}{3}$ or $\\sqrt{a b} \\leq \\sqrt{\\frac{a^{2}+a b+b^{2}}{3}}$ which is obviously true since $a^{2}+b^{2}+a b \\geq 2 a b+a b=3 a b$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-alg-20111.jsonl", "problem_match": "\nA3 ", "solution_match": "\nSolution 1"}}
+{"year": "2011", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $x, y$ be positive real numbers such that $x^{3}+y^{3} \\leq x^{2}+y^{2}$. Find the greatest possible value of the product $x y$.", "solution": "We have $(x+y)\\left(x^{2}+y^{2}\\right) \\geq(x+y)\\left(x^{3}+y^{3}\\right) \\geq\\left(x^{2}+y^{2}\\right)^{2}$, hence $x+y \\geq x^{2}+y^{2}$. Now $2(x+y) \\geq(1+1)\\left(x^{2}+y^{2}\\right) \\geq(x+y)^{2}$, thus $2 \\geq x+y$. Because $x+y \\geq 2 \\sqrt{x y}$, we will obtain $1 \\geq x y$. Equality holds when $x=y=1$.\n\nSo the greatest possible value of the product $x y$ is 1 .\n\n## Solution 2\n\nBy $A M-G M$ we have $x^{3}+y^{3} \\geq \\sqrt{x y} \\cdot\\left(x^{2}+y^{2}\\right)$, hence $1 \\geq \\sqrt{x y}$ since $x^{2}+y^{2} \\geq x^{3}+y^{3}$. Equality holds when $x=y=1$. So the greatest possible value of the product $x y$ is 1 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-alg-20111.jsonl", "problem_match": "\nA4 ", "solution_match": "## Solution 1"}}
+{"year": "2011", "tier": "T3", "problem_label": "A5", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Determine the positive integers $a, b$ such that $a^{2} b^{2}+208=4\\{l c m[a ; b]+g c d(a ; b)\\}^{2}$.", "solution": "Let $d=\\operatorname{gcd}(a, b)$ and $x, y \\in \\mathbb{Z}_{+}$such that $a=d x, b=d y$. Obviously, $(x, y)=1$. The equation is equivalent to $d^{4} x^{2} y^{2}+208=4 d^{2}(x y+1)^{2}$. Hence $d^{2} \\mid 208$ or $d^{2} \\mid 13 \\cdot 4^{2}$, so $d \\in\\{1,2,4\\}$. Take $t=x y$ with $t \\in \\mathbb{Z}_{+}$.\n\nCase I. If $d=1$, then $(x y)^{2}+208=4(x y+1)^{2}$ or $3 t^{2}+8 t-204=0$, without solutions.\n\nCase II. If $d=2$, then $16 x^{2} y^{2}+208=16(x y+1)^{2}$ or $t^{2}+13=t^{2}+2 t+1 \\Rightarrow t=6$, so $(x, y) \\in\\{(1,6) ;(2,3) ;(3,2) ;(6,1)\\} \\Rightarrow(a, b) \\in\\{(2,12) ;(4,6) ;(6,4) ;(12 ; 2)\\}$.\n\nCase III. If $d=4$, then $16^{2} x^{2} y^{2}+208=4 \\cdot 16(x y+1)^{2}$ or $16 t^{2}+13=4(t+1)^{2}$ and if $t \\in \\mathbb{Z}$, then 13 must be even, contradiction!\n\nFinally, the solutions are $(a, b) \\in\\{(2,12) ;(4,6) ;(6,4) ;(12 ; 2)\\}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-alg-20111.jsonl", "problem_match": "\nA5 ", "solution_match": "## Solution"}}
+{"year": "2011", "tier": "T3", "problem_label": "A6", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $x_{i}>1$, for all $i \\in\\{1,2,3, \\ldots, 2011\\}$. Prove the inequality $\\sum_{i=1}^{2011} \\frac{x_{i}^{2}}{x_{i+1}-1} \\geq 8044$ where $x_{2012}=x_{1}$. When does equality hold?", "solution": "Realize that $\\left(x_{i}-2\\right)^{2} \\geq 0 \\Leftrightarrow x_{i}^{2} \\geq 4\\left(x_{i}-1\\right)$. So we get:\n\n$\\frac{x_{1}^{2}}{x_{2}-1}+\\frac{x_{2}^{2}}{x_{3}-1}+\\ldots+\\frac{x_{2011}^{2}}{x_{1}-1} \\geq 4\\left(\\frac{x_{1}-1}{x_{2}-1}+\\frac{x_{2}-1}{x_{3}-1}+\\ldots+\\frac{x_{2011}-1}{x_{1}-1}\\right)$. By $A M-G M$ :\n$\\frac{x_{1}-1}{x_{2}-1}+\\frac{x_{2}-1}{x_{3}-1}+\\ldots+\\frac{x_{2011}-1}{x_{1}-1} \\geq 2011 \\cdot \\sqrt[2011]{\\frac{x_{1}-1}{x_{2}-1} \\cdot \\frac{x_{2}-1}{x_{3}-1} \\cdot \\ldots \\cdot \\frac{x_{2011}-1}{x_{1}-1}}=2011$\n\nFinally, we obtain that $\\frac{x_{1}^{2}}{x_{2}-1}+\\frac{x_{2}^{2}}{x_{3}-1}+\\ldots+\\frac{x_{2011}^{2}}{x_{1}-1} \\geq 8044$.\n\nEquality holds when $\\left(x_{i}-2\\right)^{2}=0,(\\forall) i=\\overline{1,2011}$, or $x_{1}=x_{2}=\\ldots=x_{2011}=2$.\n\n## Solution 2\n\nAll the denominators are greater than 0 , so by Cauchy - Schwar $z$ we have:\n\n$\\frac{x_{1}^{2}}{x_{2}-1}+\\frac{x_{2}^{2}}{x_{3}-1}+\\ldots+\\frac{x_{2011}^{2}}{x_{1}-1} \\geq \\frac{\\left(x_{1}+x_{2}+\\ldots+x_{2011}\\right)^{2}}{x_{1}+x_{2}+\\ldots+x_{2011}-2011}$. It remains to prove that $\\frac{\\left(x_{1}+x_{2}+\\ldots+x_{2011}\\right)^{2}}{x_{1}+x_{2}+\\ldots+x_{2011}-2011} \\geq 8044$ or $\\left(\\sum_{i=1}^{2011} x_{i}\\right)^{2}+4 \\cdot 2011^{2} \\geq 4 \\cdot 2011 \\cdot \\sum_{i=1}^{2011} x_{i}$ which is obviously true by $A M-G M$ for $\\left(\\sum_{i=1}^{2011} x_{i}\\right)^{2}$ and $4 \\cdot 2011^{2}$.\n\nEquality holds when $x_{1}+x_{2}+\\ldots+x_{2011}=4022$ and $\\frac{x_{1}}{x_{2}-1}=\\frac{x_{2}}{x_{3}-1}=\\ldots=\\frac{x_{2011}}{x_{1}-1}$ or $x_{i}^{2}-x_{i}=x_{i-1} x_{i+1}-x_{i-1},(\\forall) i=\\overline{1,2011} \\Rightarrow \\sum_{i=1}^{2011} x_{i}^{2}=\\sum_{i=1}^{2011} x_{i} x_{i+2}$ where $x_{2012}=x_{1}$ and $x_{2013}=x_{2}$. This means that $x_{1}=x_{2}=\\ldots=x_{2011}$.\n\nSo equality holds when $x_{1}=x_{2}=\\ldots=x_{2011}=2$ since $x_{1}+x_{2}+\\ldots+x_{2011}=4022$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-alg-20111.jsonl", "problem_match": "\nA6 ", "solution_match": "## Solution 1"}}
+{"year": "2011", "tier": "T3", "problem_label": "A7", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $a, b, c$ be positive real numbers with $a b c=1$. Prove the inequality:\n\n$$\n\\frac{2 a^{2}+\\frac{1}{a}}{b+\\frac{1}{a}+1}+\\frac{2 b^{2}+\\frac{1}{b}}{c+\\frac{1}{b}+1}+\\frac{2 c^{2}+\\frac{1}{c}}{a+\\frac{1}{c}+1} \\geq 3\n$$", "solution": "By $A M-G M$ we have $2 x^{2}+\\frac{1}{x}=x^{2}+x^{2}+\\frac{1}{x} \\geq 3 \\sqrt[3]{\\frac{x^{4}}{x}}=3 x$ for all $x>0$, so we have:\n\n$\\sum_{\\text {cyc }} \\frac{2 a^{2}+\\frac{1}{a}}{b+\\frac{1}{a}+1} \\geq \\sum_{c y c} \\frac{3 a}{1+b+b c}=3\\left(\\sum_{c y c} \\frac{a^{2}}{1+a+a b}\\right) \\geq \\frac{3(a+b+c)^{2}}{3+a+b+c+a b+b c+c a}$.\n\nBy $A M-G M$ we have $a b+b c+c a \\geq 3$ and $a+b+c \\geq 3$. But $3\\left(a^{2}+b^{2}+c^{2}\\right) \\geq(a+b+c)^{2} \\geq$ $3(a+b+c)$. So $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a \\geq 3+a+b+c+a b+b c+c a$. Hence $\\sum_{c y c} \\frac{2 a^{2}+\\frac{1}{a}}{b+\\frac{1}{a}+1} \\geq \\frac{3(a+b+c)^{2}}{3+a+b+c+a b+b c+c a} \\geq \\frac{3(a+b+c)^{2}}{(a+b+c)^{2}}=3$.\n\nSolution 2\n\nDenote $a=\\frac{y}{x}, b=\\frac{z}{y}$ and $c=\\frac{x}{z}$. We have $\\frac{2 a^{2}+\\frac{1}{a}}{b+\\frac{1}{a}+1}=\\frac{\\frac{2 y^{2}}{x^{2}}+\\frac{x}{y}}{\\frac{z}{y}+\\frac{x}{y}+1}=\\frac{2 y^{3}+x^{3}}{x^{2}(x+y+z)}$.\n\nHence $\\sum_{c y c} \\frac{2 a^{2}+\\frac{1}{a}}{b+\\frac{1}{a}+1}=\\frac{1}{x+y+z} \\cdot \\sum_{c y c} \\frac{2 y^{3}+x^{3}}{x^{2}}=\\frac{1}{x+y+z} \\cdot\\left(x+y+z+2 \\sum_{c y c} \\frac{y^{3}}{x^{2}}\\right)$.\n\nBy Rearrangements Inequality we get $\\sum_{\\text {cyc }} \\frac{y^{3}}{x^{2}} \\geq x+y+z$.\n\nSo $\\sum_{\\text {cyc }} \\frac{2 a^{2}+\\frac{1}{a}}{b+\\frac{1}{a}+1} \\geq \\frac{1}{x+y+z} \\cdot(3 x+3 y+3 z)=3$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-alg-20111.jsonl", "problem_match": "\nA7 ", "solution_match": "## Solution 1"}}
+{"year": "2011", "tier": "T3", "problem_label": "A8", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Decipher the equality $(\\overline{L A R N}-\\overline{A C A}):(\\overline{C Y P}+\\overline{R U S})=C^{Y^{P}} \\cdot R^{U^{S}}$ where different symbols correspond to different digits and equal symbols correspond to equal digits. It is also supposed that all these digits are different from 0 .", "solution": "Denote $x=\\overline{L A R N}-\\overline{A C A}, y=\\overline{C Y P}+\\overline{R U S}$ and $z=C^{Y^{P}} \\cdot R^{U^{S}}$. It is obvious that $1823-898 \\leq x \\leq 9187-121,135+246 \\leq y \\leq 975+864$, that is $925 \\leq x \\leq 9075$ and $381 \\leq y \\leq 1839$, whence it follows that $\\frac{925}{1839} \\leq \\frac{x}{y} \\leq \\frac{9075}{381}$, or $0,502 \\ldots \\leq \\frac{x}{y} \\leq 23,81 \\ldots$ Since $\\frac{x}{y}=z$ is an integer, it follows that $1 \\leq \\frac{x}{y} \\leq 23$, hence $1 \\leq C^{Y^{P}} \\cdot R^{U^{S}} \\leq 23$. So both values $C^{Y^{P}}$ and $R^{U^{S}}$ are $\\leq 23$. From this and the fact that $2^{2^{3}}>23$ it follows that at least one of the symbols in the expression $C^{Y^{P}}$ and at least one of the symbols in the expression $R^{U^{S}}$ correspond to the digit 1. This is impossible because of the assumption that all the symbols in the set $\\{C, Y, P, R, U, S\\}$ correspond to different digits.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-alg-20111.jsonl", "problem_match": "\nA8 ", "solution_match": "## Solution"}}
+{"year": "2011", "tier": "T3", "problem_label": "A9", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $x_{1}, x_{2}, \\ldots, x_{n}$ be real numbers satisfying $\\sum_{k=1}^{n-1} \\min \\left(x_{k} ; x_{k+1}\\right)=\\min \\left(x_{1}, x_{n}\\right)$.\n\nProve that $\\sum_{k=2}^{n-1} x_{k} \\geq 0$.", "solution": "Case I. If $\\min \\left(x_{1}, x_{n}\\right)=x_{1}$, we know that $x_{k} \\geq \\min \\left(x_{k} ; x_{k+1}\\right)$ for all $k \\in\\{1,2,3, \\ldots, n-1\\}$. So $x_{1}+x_{2}+\\ldots+x_{n-1} \\geq \\sum_{k=1}^{n-1} \\min \\left(x_{k} ; x_{k+1}\\right)=\\min \\left(x_{1}, x_{n}\\right)=x_{1}$, hence $\\sum_{k=2}^{n-1} x_{k} \\geq 0$.\n\nCase II. If $\\min \\left(x_{1}, x_{n}\\right)=x_{n}$, we know that $x_{k} \\geq \\min \\left(x_{k-1} ; x_{k}\\right)$ for all $k \\in\\{2,3,4, \\ldots, n\\}$. So $x_{2}+x_{3}+\\ldots+x_{n} \\geq \\sum_{k=1}^{n-1} \\min \\left(x_{k} ; x_{k+1}\\right)=\\min \\left(x_{1}, x_{n}\\right)=x_{n}$, hence $\\sum_{k=2}^{n-1} x_{k} \\geq 0$.\n\n## Solution 2\n\nSince $\\min (a, b)=\\frac{1}{2}(a+b-|a-b|)$, after substitutions, we will have:\n\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n-1} \\frac{1}{2}\\left(x_{k}+x_{k+1}-\\left|x_{k}-x_{k+1}\\right|\\right) & =\\frac{1}{2}\\left(x_{1}+x_{n}-\\left|x_{1}-x_{n}\\right|\\right) \\Leftrightarrow \\ldots \\\\\n2\\left(x_{2}+x_{3}+\\ldots+x_{n-1}\\right)+\\left|x_{1}-x_{n}\\right| & =\\left|x_{1}-x_{2}\\right|+\\left|x_{2}-x_{3}\\right|+\\ldots+\\left|x_{n-1}-x_{n}\\right|\n\\end{aligned}\n$$\n\nAs $\\left|x_{1}-x_{2}\\right|+\\left|x_{2}-x_{3}\\right|+\\ldots+\\left|x_{n-1}-x_{n}\\right| \\geq\\left|x_{1}-x_{2}+x_{2}-x_{3}+\\ldots+x_{n-1}-x_{n}\\right|=\\left|x_{1}-x_{n}\\right|$, we obtain the desired result.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-alg-20111.jsonl", "problem_match": "\nA9 ", "solution_match": "## Solution 1"}}

JBMO/segmented/en-shortlist/en-combi-2011.jsonl

@@ -1,9 +1,9 @@
-{"year": "2011", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Inside of a square whose side length is 1 there are a few circles such that the sum of their circumferences is equal to 10 . Show that there exists a line that meets alt least four of these circles.", "solution": "Find projections of all given circles on one of the sides of the square. The projection of each circle is a segment whose length is equal to the length of a diameter of this circle. Since the sum of the lengths of all circles' diameters is equal to $10 / \\pi$, it follows that the sum of the lengths of all mentioned projections is equal to $10 / \\pi>3$. Because the side of the square is equal to 1 , we conclude that at least one point is covered with at least four of these projections. Hence, a perpendicular line to the projection side passing through this point meets at least four of the given circles, so this is a line with the desired property.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-combi-2011.jsonl", "problem_match": "\nC1 ", "solution_match": "\nSolution"}}
-{"year": "2011", "tier": "T3", "problem_label": "C2", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Can we divide an equilateral triangle $\\triangle A B C$ into 2011 small triangles using 122 straight lines? (there should be 2011 triangles that are not themselves divided into smaller parts and there should be no polygons which are not triangles)", "solution": "Firstly, for each side of the triangle, we draw 37 equidistant, parallel lines to it. In this way we get $38^{2}=1444$ triangles. Then we erase 11 lines which are closest to the vertex $A$ and parallel to the side $B C$ and we draw 21 lines perpendicular to $B C$, the first starting from the vertex $A$ and 10 on each of the two sides, the lines which are closest to the vertex $A$, distributed symmetrically. In this way we get $26 \\cdot 21+10=$ 556 new triangles. Therefore we obtain a total of 2000 triangles and we have used $37 \\cdot 3-11+21=121$ lines. Let $D$ be the $12^{\\text {th }}$ point on side $A B$, starting from $B$ (including it). The perpendicular to $B C$ passing through $D$ will be the last line we draw. In this way we obtain the required configuration.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-combi-2011.jsonl", "problem_match": "\nC2 ", "solution_match": "## Solution"}}
-{"year": "2011", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "We can change a natural number $n$ in three ways:\n\na) If the number $n$ has at least two digits, we erase the last digit and we subtract that digit from the remaining number (for example, from 123 we get $12-3=9$ );\n\nb) If the last digit is different from 0 , we can change the order of the digits in the opposite one (for example, from 123 we get 321 );\n\nc) We can multiply the number $n$ by a number from the set $\\{1,2,3, \\ldots, 2010\\}$.\n\nCan we get the number 21062011 from the number 1012011?", "solution": "The answer is NO. We will prove that if the first number is divisible by 11, then all the numbers which we can get from $n$, are divisible by 11 . When we use $a$ ), from the number $10 a+b$, we will get the number $m=a-b=11 a-n$, so $11 \\mid m$ since $11 \\mid n$. It's well-known that a number is divisible by 11 if and only if the difference between sum of digits on even places and sum of digits on odd places is divisible by 11 . Hence, when we use $b$ ), from a number which is divisible by 11, we will get a number which is also divisible by 11 . When we use $c$ ), the obtained number remains divisible by 11 . So, the answer is NO since 1012011 is divisible by 11 and 21062011 is not.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-combi-2011.jsonl", "problem_match": "\nC3 ", "solution_match": "\nSolution"}}
-{"year": "2011", "tier": "T3", "problem_label": "C4", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "In a group of $n$ people, each one had a different ball. They performed a sequence of swaps; in each swap, two people swapped the ball they had at that moment. Each pair of people performed at least one swap. In the end each person had the ball he/she had at the start. Find the least possible number of swaps, if: $a$ ) $n=5$; b) $n=6$.", "solution": "We will denote the people by $A, B, C, \\ldots$ and their initial balls by the corresponding small letters. Thus the initial state is $A a, B b, C c, D d, E e(, F f)$. A swap is denoted by the (capital) letters of the people involved.\n\na) Five people form 10 pairs, so at least 10 swaps are necessary.\n\nIn fact, 10 swaps are sufficient:\n\nSwap $A B$, then $B C$, then $C A$; the state is now $A a, B c, C b, D d, E e$.\n\nSwap $A D$, then $D E$, then $E A$; the state is now $A a, B c, C b, D e, E d$.\n\nSwap $B E$, then $C D$; the state is now $A a, B d, C e, D b, E c$.\n\nSwap $B D$, then $C E$; the state is now $A a, B b, C c, D d, E e$.\n\nAll requirements are fulfilled now, so the answer is 10 .\n\nb) Six people form 15 pairs, so at least 15 swaps are necessary. We will prove that the final number of swaps must be even. Call a pair formed by a ball and a person inverted if letter of the ball lies after letter of the person in the alphabet. Let $T$ be the number of inverted pairs; at the start we have $T=0$. Each swap changes $T$ by 1 , so it changes the parity of $T$. Since in the end $T=0$, the total number of swaps must be even. Hence, at least 16 swaps are necessary. In fact 16 swaps are sufficient:\n\nSwap $A B$, then $B C$, then $C A$; the state is now $A a, B c, C b, D d, E e, F f$. Swap $A D$, then $D E$, then $E A$; the state is now $A a, B c, C b, D e, E d, F f$. Swap $F B$, then $B E$, then $E F$; the state is now $A a, B d, C b, D e, E c, F f$. Swap $F C$, then $C D$, then $D F$; the state is now $A a, B d, C e, D b, E c, F f$. Swap $B D$, then $C E$, then twice $A F$, the state is now $A a, B b, C c, D d, E e, F f$. All requirements are fulfilled now, so the answer is 16 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-combi-2011.jsonl", "problem_match": "\nC4 ", "solution_match": "## Solution"}}
-{"year": "2011", "tier": "T3", "problem_label": "C5", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "A set $S$ of natural numbers is called good, if for each element $x \\in S, x$ does not divide the sum of the remaining numbers in $S$. Find the maximal possible number of elements of a good set which is a subset of the set $A=\\{1,2,3, \\ldots, 63\\}$.", "solution": "Let set $B$ be the good subset of $A$ which have the maximum number of elements. We can easily see that the number 1 does not belong to $B$ since 1 divides all natural numbers. Based on the property of divisibility, we know that $x$ divides the sum of the remaining numbers if and only if $x$ divides the sum of all numbers in the set $B$. If $B$ has exactly 62 elements, than $B=\\{2,3,4, \\ldots, 62\\}$, but this set can't be good since the sum of its elements is 2015 which is divisible by 5 . Therefore $B$ has at most 61 elements. Now we are looking for the set, whose elements does not divide their sum, so the best way to do that is making a sum of elements be a prime number. $2+3+4+\\ldots+63=2015$ and if we remove the number 4, we will obtain the prime number 2011. Hence the set $B=\\{2,3,5,6,7, \\ldots, 63\\}$ is a good one. We conclude that our number is 61 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-combi-2011.jsonl", "problem_match": "\nC5 ", "solution_match": "\nSolution"}}
-{"year": "2011", "tier": "T3", "problem_label": "C6", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Let $n>3$ be a positive integer. An equilateral triangle $\\triangle A B C$ is divided into $n^{2}$ smaller congruent equilateral triangles (with sides parallel to its sides).\n\nLet $m$ be the number of rhombuses that contain two small equilateral triangles and $d$ the number of rhombuses that contain eight small equilateral triangles. Find the difference $m-d$ in terms of $n$.", "solution": "We will count the rhombuses having their large diagonal perpendicular to one side of the large triangle, then we will multiply by 3 . So consider the horizontal side, and then the other $n-1$ horizontal dividing lines, plus a horizontal line through the apex. Index them by $1,2, \\ldots, n$, from top down (the basis remains unindexed). Then each indexed $k$ line contains exactly $k$ nodes of this triangular lattice.\n\nThe top vertex of a rhombus containing two small triangles, and with its large diagonal vertical, may (and has to) a node of this triangular lattice found on the top $n-1$ lines, so there are $m=3[1+2+\\ldots+(n-1)]=\\frac{3 n(n-1)}{2}$ such rhombuses.\n\nThe top vertex of a rhombus containing eight small triangles, and with its large diagonal vertical, may (and has to) be a node of this triangular lattice found on the top $n-3$ lines, so there are $d=3[1+2+\\cdots+(n-3)]=\\frac{3(n-3)(n-2)}{2}$ such rhombuses. Finally we have $m-d=\\frac{3}{2} \\cdot\\left(n^{2}-n-n^{2}+5 n-6\\right)=6 n-9$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-combi-2011.jsonl", "problem_match": "\nC6 ", "solution_match": "## Solution"}}
-{"year": "2011", "tier": "T3", "problem_label": "C7", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Consider a rectangle whose lengths of sides are natural numbers. If someone places as many squares as possible, each with area 3 , inside of the given rectangle, such that\nthe sides of the squares are parallel to the rectangle sides, then the maximal number of these squares fill exactly half of the area of the rectangle. Determine the dimensions of all rectangles with this property.", "solution": "Let $A B C D$ be a rectangle with $A B=m$ and $A D=n$ where $m, n$ are natural numbers such that $m \\geq n \\geq 2$. Suppose that inside of the rectangle $A B C D$ is placed a rectangular lattice consisting of some identical squares whose areas are equals to 3 , where $k$ of them are placed along the side $A B$ and $l$ of them along the side $A D$.\n\nThe sum of areas of all of this squares is equal to $3 k l$. Besides of the obvious conditions $k \\sqrt{3}<m$ and $l \\sqrt{3} \\leq n \\mathbf{( 1 )}$, by the assumption of the maximality of the lattice consisting of these squares, we must have $(k+1) \\sqrt{3}>m$ and $(l+1) \\sqrt{3}>n$ (2).\n\nThe proposed problem is to determine all pairs $(m, n) \\in \\mathbb{N} \\times \\mathbb{N}$ with $m \\geq n \\geq 2$, for which the ratio $R_{m, n}=\\frac{3 k l}{m n}$ is equal to 0,5 where $k, l$ are natural numbers determined by the conditions (1) and (2).\n\nObserve that for $n \\geq 6$, using (2), we get $R_{m, n}=\\frac{k \\sqrt{3} \\cdot l \\sqrt{3}}{m n}>\\frac{(m-\\sqrt{3})(n-\\sqrt{3})}{m n}=$ $\\left(1-\\frac{\\sqrt{3}}{m}\\right)\\left(1-\\frac{\\sqrt{3}}{n}\\right) \\geq\\left(1-\\frac{\\sqrt{3}}{6}\\right)^{2}=\\frac{1}{2}+\\frac{7}{12}-\\frac{\\sqrt{3}}{3}>\\frac{1}{2}+\\frac{\\sqrt{48}}{12}-\\frac{\\sqrt{3}}{3}=0,5$\n\nSo, the condition $R_{m, n}=0,5$ yields $n \\leq 5$ or $n \\in\\{2,3,4,5\\}$. We have 4 possible cases:\n\nCase 1: $n=2$. Then $l=1$ and thus as above we get $R_{m, 2}=\\frac{3 k}{2 m}>\\frac{\\sqrt{3} \\cdot(m-\\sqrt{3})}{2 m}=$ $\\frac{\\sqrt{3}}{2} \\cdot\\left(1-\\frac{\\sqrt{3}}{m}\\right)$, which is greater than 0,5 for each $m>\\frac{\\sqrt{27}+3}{2}>\\frac{5+3}{2}=4$, hence $m \\in\\{2,3,4\\}$. Direct calculations give $R_{2,2}=R_{2,4}=0,75$ and $R_{2,3}=0,5$.\n\nCase 2: $n=3$. Then $l=1$ and thus as above we get $R_{m, 3}=\\frac{3 k}{3 m}>\\frac{\\sqrt{3} \\cdot(m-\\sqrt{3})}{3 m}=$ $\\frac{\\sqrt{3}}{3} \\cdot\\left(1-\\frac{\\sqrt{3}}{m}\\right)$, which is greater than 0,5 for each $m>4 \\sqrt{3}+6>12$, hence $m \\in$ $\\{3,4, \\ldots, 12\\}$. Direct calculations give $R_{3,3}=0,(3), R_{3,5}=0,4, R_{3,7}=4 / 7, R_{3,9}=$ $5 / 9, R_{3,11}=6 / 11$ and $R_{3,4}=R_{3,6}=R_{3,8}=R_{3,10}=R_{3,12}=0,5$.\n\nCase 3: $n=4$. Then $l=2$ and thus as above we get $R_{m, 4}=\\frac{6 k}{4 m}>\\frac{\\sqrt{3} \\cdot(m-\\sqrt{3})}{2 m}=$ $\\frac{\\sqrt{3}}{2} \\cdot\\left(1-\\frac{\\sqrt{3}}{m}\\right)$, which is greater than 0,5 for each $m>\\frac{\\sqrt{27}+3}{2}>\\frac{5+3}{2}=4$.\n\nHence $m=4$ and a calculation gives $R_{4,4}=0,75$.\n\nCase 4: $n=5$. Then $l=2$ and thus as above we get $R_{m, 5}=\\frac{6 k}{5 m}>\\frac{2 \\sqrt{3} \\cdot(m-\\sqrt{3})}{5 m}=$\n$\\frac{2 \\sqrt{3}}{5} \\cdot\\left(1-\\frac{\\sqrt{3}}{m}\\right)$, which is greater than 0,5 for each $m>\\frac{12(4 \\sqrt{3}+5}{23}>\\frac{12 \\cdot 11}{23}>6$, hence $m \\in\\{5,6\\}$. Direct calculations give $R_{5,5}=0,48$ and $R_{5,6}=0,6$.\n\nWe conclude that: $R_{i, j}=0,5$ for $(i, j) \\in\\{(2,3) ;(3,4) ;(, 3,6) ;(3,8) ;(3,10) ;(3,12)\\}$.\n\nThese pairs are the dimensions of all rectangles with desired property.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-combi-2011.jsonl", "problem_match": "\nC7 ", "solution_match": "## Solution"}}
-{"year": "2011", "tier": "T3", "problem_label": "C8", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Determine the polygons with $n$ sides $(n \\geq 4)$, not necessarily convex, which satisfy the property that the reflection of every vertex of polygon with respect to every diagonal of the polygon does not fall outside the polygon.\n\nNote: Each segment joining two non-neighboring vertices of the polygon is a diagonal. The reflection is considered with respect to the support line of the diagonal.", "solution": "A polygon with this property has to be convex, otherwise we consider an edge of the convex hull of this set of vertices which is not an edge of this polygon. All the others vertices are situated in one of the half-planes determined by the support-line of this edge, therefore the reflections of the others vertices falls outside the polygon.\n\nNow we choose a diagonal. It divides the polygon into two parts, $P 1$ and $P 2$. The reflection of $P 1$ falls into the interior of $P 2$ and viceversa. As a consequence, the diagonal is a symmetry axis for the polygon. Then every diagonal of the polygon bisects the angles of the polygon and this means that there are 4 vertices and the polygon is a rhombus. Each rhombus satisfies the desired condition.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-combi-2011.jsonl", "problem_match": "\nC8 ", "solution_match": "## Solution"}}
-{"year": "2011", "tier": "T3", "problem_label": "C9", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Decide if it is possible to consider 2011 points in a plane such that the distance between every two of these points is different from 1 and each unit circle centered at one of these points leaves exactly 1005 points outside the circle.", "solution": "NO. If such a configuration existed, the number of segments starting from each of the 2011 points towards the other one and having length less than 1 would be 1005 . Since each segment is counted twice, their total number would be $1005 \\cdot 2011 / 2$ which is not an integer, contradiction!", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-combi-2011.jsonl", "problem_match": "\nC9 ", "solution_match": "## Solution"}}
+{"year": "2011", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Inside of a square whose side length is 1 there are a few circles such that the sum of their circumferences is equal to 10 . Show that there exists a line that meets alt least four of these circles.", "solution": "Find projections of all given circles on one of the sides of the square. The projection of each circle is a segment whose length is equal to the length of a diameter of this circle. Since the sum of the lengths of all circles' diameters is equal to $10 / \\pi$, it follows that the sum of the lengths of all mentioned projections is equal to $10 / \\pi>3$. Because the side of the square is equal to 1 , we conclude that at least one point is covered with at least four of these projections. Hence, a perpendicular line to the projection side passing through this point meets at least four of the given circles, so this is a line with the desired property.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-combi-2011.jsonl", "problem_match": "\nC1 ", "solution_match": "\nSolution"}}
+{"year": "2011", "tier": "T3", "problem_label": "C2", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Can we divide an equilateral triangle $\\triangle A B C$ into 2011 small triangles using 122 straight lines? (there should be 2011 triangles that are not themselves divided into smaller parts and there should be no polygons which are not triangles)", "solution": "Firstly, for each side of the triangle, we draw 37 equidistant, parallel lines to it. In this way we get $38^{2}=1444$ triangles. Then we erase 11 lines which are closest to the vertex $A$ and parallel to the side $B C$ and we draw 21 lines perpendicular to $B C$, the first starting from the vertex $A$ and 10 on each of the two sides, the lines which are closest to the vertex $A$, distributed symmetrically. In this way we get $26 \\cdot 21+10=$ 556 new triangles. Therefore we obtain a total of 2000 triangles and we have used $37 \\cdot 3-11+21=121$ lines. Let $D$ be the $12^{\\text {th }}$ point on side $A B$, starting from $B$ (including it). The perpendicular to $B C$ passing through $D$ will be the last line we draw. In this way we obtain the required configuration.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-combi-2011.jsonl", "problem_match": "\nC2 ", "solution_match": "## Solution"}}
+{"year": "2011", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "We can change a natural number $n$ in three ways:\n\na) If the number $n$ has at least two digits, we erase the last digit and we subtract that digit from the remaining number (for example, from 123 we get $12-3=9$ );\n\nb) If the last digit is different from 0 , we can change the order of the digits in the opposite one (for example, from 123 we get 321 );\n\nc) We can multiply the number $n$ by a number from the set $\\{1,2,3, \\ldots, 2010\\}$.\n\nCan we get the number 21062011 from the number 1012011?", "solution": "The answer is NO. We will prove that if the first number is divisible by 11, then all the numbers which we can get from $n$, are divisible by 11 . When we use $a$ ), from the number $10 a+b$, we will get the number $m=a-b=11 a-n$, so $11 \\mid m$ since $11 \\mid n$. It's well-known that a number is divisible by 11 if and only if the difference between sum of digits on even places and sum of digits on odd places is divisible by 11 . Hence, when we use $b$ ), from a number which is divisible by 11, we will get a number which is also divisible by 11 . When we use $c$ ), the obtained number remains divisible by 11 . So, the answer is NO since 1012011 is divisible by 11 and 21062011 is not.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-combi-2011.jsonl", "problem_match": "\nC3 ", "solution_match": "\nSolution"}}
+{"year": "2011", "tier": "T3", "problem_label": "C4", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "In a group of $n$ people, each one had a different ball. They performed a sequence of swaps; in each swap, two people swapped the ball they had at that moment. Each pair of people performed at least one swap. In the end each person had the ball he/she had at the start. Find the least possible number of swaps, if: $a$ ) $n=5$; b) $n=6$.", "solution": "We will denote the people by $A, B, C, \\ldots$ and their initial balls by the corresponding small letters. Thus the initial state is $A a, B b, C c, D d, E e(, F f)$. A swap is denoted by the (capital) letters of the people involved.\n\na) Five people form 10 pairs, so at least 10 swaps are necessary.\n\nIn fact, 10 swaps are sufficient:\n\nSwap $A B$, then $B C$, then $C A$; the state is now $A a, B c, C b, D d, E e$.\n\nSwap $A D$, then $D E$, then $E A$; the state is now $A a, B c, C b, D e, E d$.\n\nSwap $B E$, then $C D$; the state is now $A a, B d, C e, D b, E c$.\n\nSwap $B D$, then $C E$; the state is now $A a, B b, C c, D d, E e$.\n\nAll requirements are fulfilled now, so the answer is 10 .\n\nb) Six people form 15 pairs, so at least 15 swaps are necessary. We will prove that the final number of swaps must be even. Call a pair formed by a ball and a person inverted if letter of the ball lies after letter of the person in the alphabet. Let $T$ be the number of inverted pairs; at the start we have $T=0$. Each swap changes $T$ by 1 , so it changes the parity of $T$. Since in the end $T=0$, the total number of swaps must be even. Hence, at least 16 swaps are necessary. In fact 16 swaps are sufficient:\n\nSwap $A B$, then $B C$, then $C A$; the state is now $A a, B c, C b, D d, E e, F f$. Swap $A D$, then $D E$, then $E A$; the state is now $A a, B c, C b, D e, E d, F f$. Swap $F B$, then $B E$, then $E F$; the state is now $A a, B d, C b, D e, E c, F f$. Swap $F C$, then $C D$, then $D F$; the state is now $A a, B d, C e, D b, E c, F f$. Swap $B D$, then $C E$, then twice $A F$, the state is now $A a, B b, C c, D d, E e, F f$. All requirements are fulfilled now, so the answer is 16 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-combi-2011.jsonl", "problem_match": "\nC4 ", "solution_match": "## Solution"}}
+{"year": "2011", "tier": "T3", "problem_label": "C5", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "A set $S$ of natural numbers is called good, if for each element $x \\in S, x$ does not divide the sum of the remaining numbers in $S$. Find the maximal possible number of elements of a good set which is a subset of the set $A=\\{1,2,3, \\ldots, 63\\}$.", "solution": "Let set $B$ be the good subset of $A$ which have the maximum number of elements. We can easily see that the number 1 does not belong to $B$ since 1 divides all natural numbers. Based on the property of divisibility, we know that $x$ divides the sum of the remaining numbers if and only if $x$ divides the sum of all numbers in the set $B$. If $B$ has exactly 62 elements, than $B=\\{2,3,4, \\ldots, 62\\}$, but this set can't be good since the sum of its elements is 2015 which is divisible by 5 . Therefore $B$ has at most 61 elements. Now we are looking for the set, whose elements does not divide their sum, so the best way to do that is making a sum of elements be a prime number. $2+3+4+\\ldots+63=2015$ and if we remove the number 4, we will obtain the prime number 2011. Hence the set $B=\\{2,3,5,6,7, \\ldots, 63\\}$ is a good one. We conclude that our number is 61 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-combi-2011.jsonl", "problem_match": "\nC5 ", "solution_match": "\nSolution"}}
+{"year": "2011", "tier": "T3", "problem_label": "C6", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Let $n>3$ be a positive integer. An equilateral triangle $\\triangle A B C$ is divided into $n^{2}$ smaller congruent equilateral triangles (with sides parallel to its sides).\n\nLet $m$ be the number of rhombuses that contain two small equilateral triangles and $d$ the number of rhombuses that contain eight small equilateral triangles. Find the difference $m-d$ in terms of $n$.", "solution": "We will count the rhombuses having their large diagonal perpendicular to one side of the large triangle, then we will multiply by 3 . So consider the horizontal side, and then the other $n-1$ horizontal dividing lines, plus a horizontal line through the apex. Index them by $1,2, \\ldots, n$, from top down (the basis remains unindexed). Then each indexed $k$ line contains exactly $k$ nodes of this triangular lattice.\n\nThe top vertex of a rhombus containing two small triangles, and with its large diagonal vertical, may (and has to) a node of this triangular lattice found on the top $n-1$ lines, so there are $m=3[1+2+\\ldots+(n-1)]=\\frac{3 n(n-1)}{2}$ such rhombuses.\n\nThe top vertex of a rhombus containing eight small triangles, and with its large diagonal vertical, may (and has to) be a node of this triangular lattice found on the top $n-3$ lines, so there are $d=3[1+2+\\cdots+(n-3)]=\\frac{3(n-3)(n-2)}{2}$ such rhombuses. Finally we have $m-d=\\frac{3}{2} \\cdot\\left(n^{2}-n-n^{2}+5 n-6\\right)=6 n-9$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-combi-2011.jsonl", "problem_match": "\nC6 ", "solution_match": "## Solution"}}
+{"year": "2011", "tier": "T3", "problem_label": "C7", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Consider a rectangle whose lengths of sides are natural numbers. If someone places as many squares as possible, each with area 3 , inside of the given rectangle, such that\nthe sides of the squares are parallel to the rectangle sides, then the maximal number of these squares fill exactly half of the area of the rectangle. Determine the dimensions of all rectangles with this property.", "solution": "Let $A B C D$ be a rectangle with $A B=m$ and $A D=n$ where $m, n$ are natural numbers such that $m \\geq n \\geq 2$. Suppose that inside of the rectangle $A B C D$ is placed a rectangular lattice consisting of some identical squares whose areas are equals to 3 , where $k$ of them are placed along the side $A B$ and $l$ of them along the side $A D$.\n\nThe sum of areas of all of this squares is equal to $3 k l$. Besides of the obvious conditions $k \\sqrt{3}<m$ and $l \\sqrt{3} \\leq n \\mathbf{( 1 )}$, by the assumption of the maximality of the lattice consisting of these squares, we must have $(k+1) \\sqrt{3}>m$ and $(l+1) \\sqrt{3}>n$ (2).\n\nThe proposed problem is to determine all pairs $(m, n) \\in \\mathbb{N} \\times \\mathbb{N}$ with $m \\geq n \\geq 2$, for which the ratio $R_{m, n}=\\frac{3 k l}{m n}$ is equal to 0,5 where $k, l$ are natural numbers determined by the conditions (1) and (2).\n\nObserve that for $n \\geq 6$, using (2), we get $R_{m, n}=\\frac{k \\sqrt{3} \\cdot l \\sqrt{3}}{m n}>\\frac{(m-\\sqrt{3})(n-\\sqrt{3})}{m n}=$ $\\left(1-\\frac{\\sqrt{3}}{m}\\right)\\left(1-\\frac{\\sqrt{3}}{n}\\right) \\geq\\left(1-\\frac{\\sqrt{3}}{6}\\right)^{2}=\\frac{1}{2}+\\frac{7}{12}-\\frac{\\sqrt{3}}{3}>\\frac{1}{2}+\\frac{\\sqrt{48}}{12}-\\frac{\\sqrt{3}}{3}=0,5$\n\nSo, the condition $R_{m, n}=0,5$ yields $n \\leq 5$ or $n \\in\\{2,3,4,5\\}$. We have 4 possible cases:\n\nCase 1: $n=2$. Then $l=1$ and thus as above we get $R_{m, 2}=\\frac{3 k}{2 m}>\\frac{\\sqrt{3} \\cdot(m-\\sqrt{3})}{2 m}=$ $\\frac{\\sqrt{3}}{2} \\cdot\\left(1-\\frac{\\sqrt{3}}{m}\\right)$, which is greater than 0,5 for each $m>\\frac{\\sqrt{27}+3}{2}>\\frac{5+3}{2}=4$, hence $m \\in\\{2,3,4\\}$. Direct calculations give $R_{2,2}=R_{2,4}=0,75$ and $R_{2,3}=0,5$.\n\nCase 2: $n=3$. Then $l=1$ and thus as above we get $R_{m, 3}=\\frac{3 k}{3 m}>\\frac{\\sqrt{3} \\cdot(m-\\sqrt{3})}{3 m}=$ $\\frac{\\sqrt{3}}{3} \\cdot\\left(1-\\frac{\\sqrt{3}}{m}\\right)$, which is greater than 0,5 for each $m>4 \\sqrt{3}+6>12$, hence $m \\in$ $\\{3,4, \\ldots, 12\\}$. Direct calculations give $R_{3,3}=0,(3), R_{3,5}=0,4, R_{3,7}=4 / 7, R_{3,9}=$ $5 / 9, R_{3,11}=6 / 11$ and $R_{3,4}=R_{3,6}=R_{3,8}=R_{3,10}=R_{3,12}=0,5$.\n\nCase 3: $n=4$. Then $l=2$ and thus as above we get $R_{m, 4}=\\frac{6 k}{4 m}>\\frac{\\sqrt{3} \\cdot(m-\\sqrt{3})}{2 m}=$ $\\frac{\\sqrt{3}}{2} \\cdot\\left(1-\\frac{\\sqrt{3}}{m}\\right)$, which is greater than 0,5 for each $m>\\frac{\\sqrt{27}+3}{2}>\\frac{5+3}{2}=4$.\n\nHence $m=4$ and a calculation gives $R_{4,4}=0,75$.\n\nCase 4: $n=5$. Then $l=2$ and thus as above we get $R_{m, 5}=\\frac{6 k}{5 m}>\\frac{2 \\sqrt{3} \\cdot(m-\\sqrt{3})}{5 m}=$\n$\\frac{2 \\sqrt{3}}{5} \\cdot\\left(1-\\frac{\\sqrt{3}}{m}\\right)$, which is greater than 0,5 for each $m>\\frac{12(4 \\sqrt{3}+5}{23}>\\frac{12 \\cdot 11}{23}>6$, hence $m \\in\\{5,6\\}$. Direct calculations give $R_{5,5}=0,48$ and $R_{5,6}=0,6$.\n\nWe conclude that: $R_{i, j}=0,5$ for $(i, j) \\in\\{(2,3) ;(3,4) ;(, 3,6) ;(3,8) ;(3,10) ;(3,12)\\}$.\n\nThese pairs are the dimensions of all rectangles with desired property.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-combi-2011.jsonl", "problem_match": "\nC7 ", "solution_match": "## Solution"}}
+{"year": "2011", "tier": "T3", "problem_label": "C8", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Determine the polygons with $n$ sides $(n \\geq 4)$, not necessarily convex, which satisfy the property that the reflection of every vertex of polygon with respect to every diagonal of the polygon does not fall outside the polygon.\n\nNote: Each segment joining two non-neighboring vertices of the polygon is a diagonal. The reflection is considered with respect to the support line of the diagonal.", "solution": "A polygon with this property has to be convex, otherwise we consider an edge of the convex hull of this set of vertices which is not an edge of this polygon. All the others vertices are situated in one of the half-planes determined by the support-line of this edge, therefore the reflections of the others vertices falls outside the polygon.\n\nNow we choose a diagonal. It divides the polygon into two parts, $P 1$ and $P 2$. The reflection of $P 1$ falls into the interior of $P 2$ and viceversa. As a consequence, the diagonal is a symmetry axis for the polygon. Then every diagonal of the polygon bisects the angles of the polygon and this means that there are 4 vertices and the polygon is a rhombus. Each rhombus satisfies the desired condition.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-combi-2011.jsonl", "problem_match": "\nC8 ", "solution_match": "## Solution"}}
+{"year": "2011", "tier": "T3", "problem_label": "C9", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Decide if it is possible to consider 2011 points in a plane such that the distance between every two of these points is different from 1 and each unit circle centered at one of these points leaves exactly 1005 points outside the circle.", "solution": "NO. If such a configuration existed, the number of segments starting from each of the 2011 points towards the other one and having length less than 1 would be 1005 . Since each segment is counted twice, their total number would be $1005 \\cdot 2011 / 2$ which is not an integer, contradiction!", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-combi-2011.jsonl", "problem_match": "\nC9 ", "solution_match": "## Solution"}}

JBMO/segmented/en-shortlist/en-geome-2011.jsonl

@@ -1,7 +1,7 @@
-{"year": "2011", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C$ be an isosceles triangle with $A B=A C$. On the extension of the side $[C A]$ we consider the point $D$ such that $A D<A C$. The perpendicular bisector of the segment $[B D]$ meets the internal and the external bisectors of the angle $\\widehat{B A C}$ at the points $E$ and $Z$, respectively. Prove that the points $A, E, D, Z$ are concyclic.", "solution": "## Solution 2\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_b2c1a517537edbece730g-1.jpg?height=494&width=746&top_left_y=1820&top_left_x=655)\n\nIn $\\triangle A B D$ the ray $[A Z$ bisects the angle $\\widehat{D A B}$ and the line $Z E$ is the perpendicular bisector of the side $[B D]$. Hence $Z$ belongs to the circumcircle of $\\triangle A B D$.\n\nTherefore the points $A, B, D, Z$ are concyclic.\n\nLet $M$ and $N$ be the midpoints of the sides $[B C]$ and $[D B]$, respectively. Then $N \\in Z E$ and $M \\in A E$. Next, $[M N]$ is a midline in $\\triangle B C D$, so $M N \\| C D \\Rightarrow \\widehat{N M B} \\equiv \\widehat{A C B}$.\n\nBut $[A Z$ is the external bisector of the angle $\\widehat{B A C}$ of $\\triangle A B C$, hence $\\widehat{B A Z} \\equiv \\widehat{A C B}$. Therefore, $\\widehat{N M B} \\equiv \\widehat{B A Z}$. In the quadrilateral $B M E N$ we have $\\widehat{B N E}=\\widehat{B M E}=90^{\\circ}$, so $B M E N$ is cyclic $\\Rightarrow \\widehat{N M B} \\equiv \\widehat{B E Z}$, hence $\\widehat{B A Z} \\equiv \\widehat{B E Z} \\Rightarrow A E B Z$ is cyclic.\n\nTherefore, since $A, B, D, Z$ are also concyclic, we conclude that $A E Z D$ is cyclic.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-geome-2011.jsonl", "problem_match": "\nG1 ", "solution_match": "\nSolution 1"}}
-{"year": "2011", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C$ be an isosceles triangle with $A B=A C$. On the extension of the side $[C A]$ we consider the point $D$ such that $A D<A C$. The perpendicular bisector of the segment $[B D]$ meets the internal and the external bisectors of the angle $\\widehat{B A C}$ at the points $E$ and $Z$, respectively. Prove that the points $A, E, D, Z$ are concyclic.", "solution": "## Solution 3\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_b2c1a517537edbece730g-2.jpg?height=568&width=671&top_left_y=904&top_left_x=687)\n\nLet $T$ be the symmetric of $B$ with respect to the axis $A Z$. Obviously $T \\in A D$. Since $A E$ and $B T$ are both perpendiculars to $A Z$, they are parallel, so $\\widehat{B A C} \\equiv \\widehat{B T A}$. (1)\n\nSince $Z B=Z T=Z D$, the point $Z$ is the circumcenter of $\\triangle B D T$.\n\nTherefore $\\widehat{B T A}=\\widehat{B Z D} / 2=\\widehat{E Z D}$.\n\nFrom (1) and (2) we conclude that $\\widehat{E A C} \\equiv \\widehat{E Z D}$, which gives that $A E D Z$ is cyclic.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-geome-2011.jsonl", "problem_match": "\nG1 ", "solution_match": "\nSolution 1"}}
-{"year": "2011", "tier": "T3", "problem_label": "G2", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A D, B F$ and $C E$ be the altitudes of $\\triangle A B C$. A line passing through $D$ and parallel to $A B$ intersects the line $E F$ at the point $G$. If $H$ is the orthocenter of $\\triangle A B C$, find the angle $\\widehat{C G H}$.", "solution": "We can see easily that points $C, D, H, F$ lies on a circle of diameter $[C H]$.\n\nTake $\\left\\{F, G^{\\prime}\\right\\}=\\odot(C H F) \\cap E F$. We have $\\widehat{E F H}=\\widehat{B A D}=\\widehat{B C E}=\\widehat{D F H}$ since the quadrilaterals $A E D C, A E H F, C D H F$ are cyclic. Hence $[F B$ is the bisector of $\\widehat{E F D}$, so $H$ is the midpoint of the arc $D G^{\\prime}$. It follows that $D G^{\\prime} \\perp C H$ since $[C H]$ is a diameter. Therefore $D G^{\\prime} \\| A B$ and $G \\equiv G^{\\prime}$. Finally $G$ lies on the circle $\\odot(C F H)$, so $\\widehat{H G C}=90^{\\circ}$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_b2c1a517537edbece730g-3.jpg?height=468&width=537&top_left_y=360&top_left_x=765)\n\n## Solution 2\n\nThe quadrilateral $A E H F$ is cyclic since $\\widehat{A E H}=\\widehat{A F H}=90^{\\circ}$, so $\\widehat{E A D} \\equiv \\widehat{G F H}$.\n\nBut $A B \\| G D$, hence $\\widehat{E A D} \\equiv \\widehat{G D H}$. Therefore $\\widehat{G F H} \\equiv \\widehat{G D H} \\Rightarrow D F G H$ is cyclic.\n\nBecause the quadrilateral $C D H F$ is cyclic since $\\widehat{C D H}=\\widehat{C F H}=90^{\\circ}$, we conclude that the quadrilateral $C F G H$ is cyclic, which gives that $\\widehat{C G H}=\\widehat{C F H}=90^{\\circ}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-geome-2011.jsonl", "problem_match": "\nG2 ", "solution_match": "## Solution 1"}}
-{"year": "2011", "tier": "T3", "problem_label": "G3", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C$ be a triangle in which ( $B L$ is the angle bisector of $\\widehat{A B C}(L \\in A C), A H$ is an altitude of $\\triangle A B C(H \\in B C)$ and $M$ is the midpoint of the side $[A B]$. It is known that the midpoints of the segments $[B L]$ and $[M H]$ coincides. Determine the internal angles of triangle $\\triangle A B C$.", "solution": "Let $N$ be the intersection of the segments $[B L]$ and $[M H]$. Because $N$ is the midpoint of both segments $[B L]$ and $[M H]$, it follows that $B M L H$ is a parallelogram. This implies that $M L \\| B C$ and $L H \\| A B$ and hence, since $M$ is the midpoint of $[A B]$, the angle bisector [ $B L$ and the altitude $A H$ are also medians of $\\triangle A B C$. This shows that $\\triangle A B C$ is an equilateral one with all internal angles measuring $60^{\\circ}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-geome-2011.jsonl", "problem_match": "\nG3 ", "solution_match": "## Solution"}}
-{"year": "2011", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO", "problem": "Point $D$ lies on the side $[B C]$ of $\\triangle A B C$. The circumcenters of $\\triangle A D C$ and $\\triangle B A D$ are $O_{1}$ and $O_{2}$, respectively and $O_{1} O_{2} \\| A B$. The orthocenter of $\\triangle A D C$ is $H$ and $A H=O_{1} O_{2}$. Find the angles of $\\triangle A B C$ if $2 m(<C)=3 m(<B)$.", "solution": "![](https://cdn.mathpix.com/cropped/2024_06_05_b2c1a517537edbece730g-3.jpg?height=471&width=768&top_left_y=2069&top_left_x=655)\n\nAs $A D$ is the radical axis of the circumcircles of $\\triangle A D C$ and $\\triangle B A D$, we have that $O_{1} O_{2} \\perp A D$, therefore $\\widehat{D A B}=90^{\\circ}$. Let $F$ be the midpoint of $[C D]$ and $[C E]$ be a diameter of the circumcircle of $\\triangle A D C$. Then $E D \\perp C D$ and $E A \\perp C A$, so $E D \\| A H$ and $E A \\| D H$ since $A H \\perp C D$ and $D H \\perp A C$ (H is the orthocenter of $\\triangle A D C$ ) and hence $E A H D$ is a parallelogram. Therefore $O_{1} O_{2}=A H=E D=2 O_{1} F$, so in $\\triangle F O_{1} O_{2}$ with $\\widehat{O_{1} F O_{2}}=90^{\\circ}$ we have $O_{1} O_{2}=2 F O_{1} \\Rightarrow \\widehat{A B C}=\\widehat{O_{1} O_{2} F}=30^{\\circ}$.\n\nThen we get $\\widehat{A C B}=45^{\\circ}$ and $\\widehat{B A C}=105^{\\circ}$.\n\n## Solution 2\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_b2c1a517537edbece730g-4.jpg?height=468&width=831&top_left_y=905&top_left_x=618)\n\nAs $A D$ is the radical axis of the circumcircles of $\\triangle A D C$ and $\\triangle B A D$, we have that $O_{1} O_{2} \\perp A D$, therefore $\\widehat{D A B}=90^{\\circ}$ and $O_{2}$ is the midpoint of $[B D]$.\n\nTake $\\{E\\}=D H \\cap A C,\\{F\\}=A H \\cap B C$ and $\\{M\\}=A D \\cap O_{1} O_{2}$.\n\nWe have $\\widehat{C E H}=\\widehat{C F H}=90^{\\circ} \\Rightarrow C E F H$ is cyclic, hence $\\widehat{A C D}=\\widehat{A H D}$.\n\nBut $\\widehat{A C D}=\\operatorname{arc} A D / 2=\\widehat{D O_{1} O_{2}}$, so $\\widehat{A H D}=\\widehat{D O_{1} O_{2}}$. We know that $A H=O_{1} O_{2}$.\n\nWe also have $\\widehat{D A H}=\\widehat{O_{1} O_{2} D}$ since $A O_{2} F M$ is cyclic with $\\widehat{A M O_{2}}=\\widehat{A F O_{2}}$.\n\nTherefore $\\triangle H D A \\equiv \\triangle O_{1} D O_{2} \\Rightarrow D A=D O_{2}=B D / 2$, so in right-angled $\\triangle A B D$ we have $m(\\widehat{A B D})=30^{\\circ}$. Then we get $\\widehat{A C B}=45^{\\circ}$ and $\\widehat{B A C}=105^{\\circ}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-geome-2011.jsonl", "problem_match": "\nG4 ", "solution_match": "## Solution 1"}}
-{"year": "2011", "tier": "T3", "problem_label": "G5", "problem_type": "Geometry", "exam": "JBMO", "problem": "Inside the square $A B C D$, the equilateral triangle $\\triangle A B E$ is constructed. Let $M$ be an interior point of the triangle $\\triangle A B E$ such that $M B=\\sqrt{2}, M C=\\sqrt{6}, M D=\\sqrt{5}$ and $M E=\\sqrt{3}$. Find the area of the square $A B C D$.", "solution": "Let $K, F, H, Z$ be the projections of point $M$ on the sides of the square.\n\nThen by Pythagorean Theorem we can prove that $M A^{2}+M C^{2}=M B^{2}+M D^{2}$.\n\nFrom the given condition we obtain $M A=1$.\n\nWith center $A$ and angle $60^{\\circ}$, we rotate $\\triangle A M E$, so we construct the triangle $A N B$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_b2c1a517537edbece730g-5.jpg?height=394&width=491&top_left_y=360&top_left_x=774)\n\nSince $A M=A N$ and $\\widehat{M A N}=60^{\\circ}$, it follows that $\\triangle A M N$ is equilateral and $M N=1$. Hence $\\triangle B M N$ is right-angled because $B M^{2}+M N^{2}=B N^{2}$.\n\nSo $m(\\widehat{B M A})=m(\\widehat{B M N})+m(\\widehat{A M N})=150^{\\circ}$.\n\nApplying Pythagorean Generalized Theorem in $\\triangle A M B$, we get:\n\n$A B^{2}=A M^{2}+B M^{2}-2 A M \\cdot B M \\cdot \\cos 150^{\\circ}=1+2+2 \\sqrt{2} \\cdot \\sqrt{3}: 2=3+\\sqrt{6}$.\n\nWe conclude that the area of the square $A B C D$ is $3+\\sqrt{6}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-geome-2011.jsonl", "problem_match": "\nG5 ", "solution_match": "## Solution"}}
-{"year": "2011", "tier": "T3", "problem_label": "G6", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C D$ be a convex quadrilateral, $E$ and $F$ points on the sides $A B$ and $C D$, respectively, such that $\\frac{A B}{A E}=\\frac{C D}{D F}=n$. Denote by $S$ the area of the quadrilateral $A E F D$. Prove that $S \\leq \\frac{A B \\cdot C D+n(n-1) \\cdot D A^{2}+n \\cdot A D \\cdot B C}{2 n^{2}}$.", "solution": "![](https://cdn.mathpix.com/cropped/2024_06_05_b2c1a517537edbece730g-5.jpg?height=383&width=560&top_left_y=1519&top_left_x=748)\n\nBy Ptolemy's Inequality in $A E F D$, we get $S=\\frac{A F \\cdot D E \\cdot \\sin (\\widehat{A F, D E})}{2} \\leq \\frac{A F \\cdot D E}{2} \\leq$ $\\frac{A E \\cdot D F+A D \\cdot E F}{2}=\\frac{A B \\cdot C D+n^{2} \\cdot D A \\cdot E F}{2 n^{2}}$.\n\nLet $G$ be a point on diagonal $B D$ such that $\\frac{D B}{D G}=n$. By Thales's Theorem we get $G E=\\frac{(n-1) A D}{n}$ and $G F=\\frac{B C}{n}$. Applying the inequality of triangle in $\\triangle E G F$ we get $E F \\leq E G+G F=\\frac{(n-1) A D+B C}{n}$. Now, we get:\n\n$S \\leq \\frac{A B \\cdot C D+n^{2} A D \\cdot E F}{2 n^{2}} \\leq \\frac{A B \\cdot C D+n(n-1) \\cdot D A^{2}+n \\cdot A D \\cdot B C}{2 n^{2}}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-geome-2011.jsonl", "problem_match": "\nG6 ", "solution_match": "## Solution"}}
+{"year": "2011", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C$ be an isosceles triangle with $A B=A C$. On the extension of the side $[C A]$ we consider the point $D$ such that $A D<A C$. The perpendicular bisector of the segment $[B D]$ meets the internal and the external bisectors of the angle $\\widehat{B A C}$ at the points $E$ and $Z$, respectively. Prove that the points $A, E, D, Z$ are concyclic.", "solution": "## Solution 2\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_b2c1a517537edbece730g-1.jpg?height=494&width=746&top_left_y=1820&top_left_x=655)\n\nIn $\\triangle A B D$ the ray $[A Z$ bisects the angle $\\widehat{D A B}$ and the line $Z E$ is the perpendicular bisector of the side $[B D]$. Hence $Z$ belongs to the circumcircle of $\\triangle A B D$.\n\nTherefore the points $A, B, D, Z$ are concyclic.\n\nLet $M$ and $N$ be the midpoints of the sides $[B C]$ and $[D B]$, respectively. Then $N \\in Z E$ and $M \\in A E$. Next, $[M N]$ is a midline in $\\triangle B C D$, so $M N \\| C D \\Rightarrow \\widehat{N M B} \\equiv \\widehat{A C B}$.\n\nBut $[A Z$ is the external bisector of the angle $\\widehat{B A C}$ of $\\triangle A B C$, hence $\\widehat{B A Z} \\equiv \\widehat{A C B}$. Therefore, $\\widehat{N M B} \\equiv \\widehat{B A Z}$. In the quadrilateral $B M E N$ we have $\\widehat{B N E}=\\widehat{B M E}=90^{\\circ}$, so $B M E N$ is cyclic $\\Rightarrow \\widehat{N M B} \\equiv \\widehat{B E Z}$, hence $\\widehat{B A Z} \\equiv \\widehat{B E Z} \\Rightarrow A E B Z$ is cyclic.\n\nTherefore, since $A, B, D, Z$ are also concyclic, we conclude that $A E Z D$ is cyclic.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-geome-2011.jsonl", "problem_match": "\nG1 ", "solution_match": "\nSolution 1"}}
+{"year": "2011", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C$ be an isosceles triangle with $A B=A C$. On the extension of the side $[C A]$ we consider the point $D$ such that $A D<A C$. The perpendicular bisector of the segment $[B D]$ meets the internal and the external bisectors of the angle $\\widehat{B A C}$ at the points $E$ and $Z$, respectively. Prove that the points $A, E, D, Z$ are concyclic.", "solution": "## Solution 3\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_b2c1a517537edbece730g-2.jpg?height=568&width=671&top_left_y=904&top_left_x=687)\n\nLet $T$ be the symmetric of $B$ with respect to the axis $A Z$. Obviously $T \\in A D$. Since $A E$ and $B T$ are both perpendiculars to $A Z$, they are parallel, so $\\widehat{B A C} \\equiv \\widehat{B T A}$. (1)\n\nSince $Z B=Z T=Z D$, the point $Z$ is the circumcenter of $\\triangle B D T$.\n\nTherefore $\\widehat{B T A}=\\widehat{B Z D} / 2=\\widehat{E Z D}$.\n\nFrom (1) and (2) we conclude that $\\widehat{E A C} \\equiv \\widehat{E Z D}$, which gives that $A E D Z$ is cyclic.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-geome-2011.jsonl", "problem_match": "\nG1 ", "solution_match": "\nSolution 1"}}
+{"year": "2011", "tier": "T3", "problem_label": "G2", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A D, B F$ and $C E$ be the altitudes of $\\triangle A B C$. A line passing through $D$ and parallel to $A B$ intersects the line $E F$ at the point $G$. If $H$ is the orthocenter of $\\triangle A B C$, find the angle $\\widehat{C G H}$.", "solution": "We can see easily that points $C, D, H, F$ lies on a circle of diameter $[C H]$.\n\nTake $\\left\\{F, G^{\\prime}\\right\\}=\\odot(C H F) \\cap E F$. We have $\\widehat{E F H}=\\widehat{B A D}=\\widehat{B C E}=\\widehat{D F H}$ since the quadrilaterals $A E D C, A E H F, C D H F$ are cyclic. Hence $[F B$ is the bisector of $\\widehat{E F D}$, so $H$ is the midpoint of the arc $D G^{\\prime}$. It follows that $D G^{\\prime} \\perp C H$ since $[C H]$ is a diameter. Therefore $D G^{\\prime} \\| A B$ and $G \\equiv G^{\\prime}$. Finally $G$ lies on the circle $\\odot(C F H)$, so $\\widehat{H G C}=90^{\\circ}$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_b2c1a517537edbece730g-3.jpg?height=468&width=537&top_left_y=360&top_left_x=765)\n\n## Solution 2\n\nThe quadrilateral $A E H F$ is cyclic since $\\widehat{A E H}=\\widehat{A F H}=90^{\\circ}$, so $\\widehat{E A D} \\equiv \\widehat{G F H}$.\n\nBut $A B \\| G D$, hence $\\widehat{E A D} \\equiv \\widehat{G D H}$. Therefore $\\widehat{G F H} \\equiv \\widehat{G D H} \\Rightarrow D F G H$ is cyclic.\n\nBecause the quadrilateral $C D H F$ is cyclic since $\\widehat{C D H}=\\widehat{C F H}=90^{\\circ}$, we conclude that the quadrilateral $C F G H$ is cyclic, which gives that $\\widehat{C G H}=\\widehat{C F H}=90^{\\circ}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-geome-2011.jsonl", "problem_match": "\nG2 ", "solution_match": "## Solution 1"}}
+{"year": "2011", "tier": "T3", "problem_label": "G3", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C$ be a triangle in which ( $B L$ is the angle bisector of $\\widehat{A B C}(L \\in A C), A H$ is an altitude of $\\triangle A B C(H \\in B C)$ and $M$ is the midpoint of the side $[A B]$. It is known that the midpoints of the segments $[B L]$ and $[M H]$ coincides. Determine the internal angles of triangle $\\triangle A B C$.", "solution": "Let $N$ be the intersection of the segments $[B L]$ and $[M H]$. Because $N$ is the midpoint of both segments $[B L]$ and $[M H]$, it follows that $B M L H$ is a parallelogram. This implies that $M L \\| B C$ and $L H \\| A B$ and hence, since $M$ is the midpoint of $[A B]$, the angle bisector [ $B L$ and the altitude $A H$ are also medians of $\\triangle A B C$. This shows that $\\triangle A B C$ is an equilateral one with all internal angles measuring $60^{\\circ}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-geome-2011.jsonl", "problem_match": "\nG3 ", "solution_match": "## Solution"}}
+{"year": "2011", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Point $D$ lies on the side $[B C]$ of $\\triangle A B C$. The circumcenters of $\\triangle A D C$ and $\\triangle B A D$ are $O_{1}$ and $O_{2}$, respectively and $O_{1} O_{2} \\| A B$. The orthocenter of $\\triangle A D C$ is $H$ and $A H=O_{1} O_{2}$. Find the angles of $\\triangle A B C$ if $2 m(<C)=3 m(<B)$.", "solution": "![](https://cdn.mathpix.com/cropped/2024_06_05_b2c1a517537edbece730g-3.jpg?height=471&width=768&top_left_y=2069&top_left_x=655)\n\nAs $A D$ is the radical axis of the circumcircles of $\\triangle A D C$ and $\\triangle B A D$, we have that $O_{1} O_{2} \\perp A D$, therefore $\\widehat{D A B}=90^{\\circ}$. Let $F$ be the midpoint of $[C D]$ and $[C E]$ be a diameter of the circumcircle of $\\triangle A D C$. Then $E D \\perp C D$ and $E A \\perp C A$, so $E D \\| A H$ and $E A \\| D H$ since $A H \\perp C D$ and $D H \\perp A C$ (H is the orthocenter of $\\triangle A D C$ ) and hence $E A H D$ is a parallelogram. Therefore $O_{1} O_{2}=A H=E D=2 O_{1} F$, so in $\\triangle F O_{1} O_{2}$ with $\\widehat{O_{1} F O_{2}}=90^{\\circ}$ we have $O_{1} O_{2}=2 F O_{1} \\Rightarrow \\widehat{A B C}=\\widehat{O_{1} O_{2} F}=30^{\\circ}$.\n\nThen we get $\\widehat{A C B}=45^{\\circ}$ and $\\widehat{B A C}=105^{\\circ}$.\n\n## Solution 2\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_b2c1a517537edbece730g-4.jpg?height=468&width=831&top_left_y=905&top_left_x=618)\n\nAs $A D$ is the radical axis of the circumcircles of $\\triangle A D C$ and $\\triangle B A D$, we have that $O_{1} O_{2} \\perp A D$, therefore $\\widehat{D A B}=90^{\\circ}$ and $O_{2}$ is the midpoint of $[B D]$.\n\nTake $\\{E\\}=D H \\cap A C,\\{F\\}=A H \\cap B C$ and $\\{M\\}=A D \\cap O_{1} O_{2}$.\n\nWe have $\\widehat{C E H}=\\widehat{C F H}=90^{\\circ} \\Rightarrow C E F H$ is cyclic, hence $\\widehat{A C D}=\\widehat{A H D}$.\n\nBut $\\widehat{A C D}=\\operatorname{arc} A D / 2=\\widehat{D O_{1} O_{2}}$, so $\\widehat{A H D}=\\widehat{D O_{1} O_{2}}$. We know that $A H=O_{1} O_{2}$.\n\nWe also have $\\widehat{D A H}=\\widehat{O_{1} O_{2} D}$ since $A O_{2} F M$ is cyclic with $\\widehat{A M O_{2}}=\\widehat{A F O_{2}}$.\n\nTherefore $\\triangle H D A \\equiv \\triangle O_{1} D O_{2} \\Rightarrow D A=D O_{2}=B D / 2$, so in right-angled $\\triangle A B D$ we have $m(\\widehat{A B D})=30^{\\circ}$. Then we get $\\widehat{A C B}=45^{\\circ}$ and $\\widehat{B A C}=105^{\\circ}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-geome-2011.jsonl", "problem_match": "\nG4 ", "solution_match": "## Solution 1"}}
+{"year": "2011", "tier": "T3", "problem_label": "G5", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Inside the square $A B C D$, the equilateral triangle $\\triangle A B E$ is constructed. Let $M$ be an interior point of the triangle $\\triangle A B E$ such that $M B=\\sqrt{2}, M C=\\sqrt{6}, M D=\\sqrt{5}$ and $M E=\\sqrt{3}$. Find the area of the square $A B C D$.", "solution": "Let $K, F, H, Z$ be the projections of point $M$ on the sides of the square.\n\nThen by Pythagorean Theorem we can prove that $M A^{2}+M C^{2}=M B^{2}+M D^{2}$.\n\nFrom the given condition we obtain $M A=1$.\n\nWith center $A$ and angle $60^{\\circ}$, we rotate $\\triangle A M E$, so we construct the triangle $A N B$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_b2c1a517537edbece730g-5.jpg?height=394&width=491&top_left_y=360&top_left_x=774)\n\nSince $A M=A N$ and $\\widehat{M A N}=60^{\\circ}$, it follows that $\\triangle A M N$ is equilateral and $M N=1$. Hence $\\triangle B M N$ is right-angled because $B M^{2}+M N^{2}=B N^{2}$.\n\nSo $m(\\widehat{B M A})=m(\\widehat{B M N})+m(\\widehat{A M N})=150^{\\circ}$.\n\nApplying Pythagorean Generalized Theorem in $\\triangle A M B$, we get:\n\n$A B^{2}=A M^{2}+B M^{2}-2 A M \\cdot B M \\cdot \\cos 150^{\\circ}=1+2+2 \\sqrt{2} \\cdot \\sqrt{3}: 2=3+\\sqrt{6}$.\n\nWe conclude that the area of the square $A B C D$ is $3+\\sqrt{6}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-geome-2011.jsonl", "problem_match": "\nG5 ", "solution_match": "## Solution"}}
+{"year": "2011", "tier": "T3", "problem_label": "G6", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C D$ be a convex quadrilateral, $E$ and $F$ points on the sides $A B$ and $C D$, respectively, such that $\\frac{A B}{A E}=\\frac{C D}{D F}=n$. Denote by $S$ the area of the quadrilateral $A E F D$. Prove that $S \\leq \\frac{A B \\cdot C D+n(n-1) \\cdot D A^{2}+n \\cdot A D \\cdot B C}{2 n^{2}}$.", "solution": "![](https://cdn.mathpix.com/cropped/2024_06_05_b2c1a517537edbece730g-5.jpg?height=383&width=560&top_left_y=1519&top_left_x=748)\n\nBy Ptolemy's Inequality in $A E F D$, we get $S=\\frac{A F \\cdot D E \\cdot \\sin (\\widehat{A F, D E})}{2} \\leq \\frac{A F \\cdot D E}{2} \\leq$ $\\frac{A E \\cdot D F+A D \\cdot E F}{2}=\\frac{A B \\cdot C D+n^{2} \\cdot D A \\cdot E F}{2 n^{2}}$.\n\nLet $G$ be a point on diagonal $B D$ such that $\\frac{D B}{D G}=n$. By Thales's Theorem we get $G E=\\frac{(n-1) A D}{n}$ and $G F=\\frac{B C}{n}$. Applying the inequality of triangle in $\\triangle E G F$ we get $E F \\leq E G+G F=\\frac{(n-1) A D+B C}{n}$. Now, we get:\n\n$S \\leq \\frac{A B \\cdot C D+n^{2} A D \\cdot E F}{2 n^{2}} \\leq \\frac{A B \\cdot C D+n(n-1) \\cdot D A^{2}+n \\cdot A D \\cdot B C}{2 n^{2}}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-geome-2011.jsonl", "problem_match": "\nG6 ", "solution_match": "## Solution"}}

JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl

@@ -1,13 +1,13 @@
-{"year": "2004", "tier": "T3", "problem_label": "NT1", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Let $a, b, p, q$ be positive integers such that $a$ and $b$ are relatively prime, $a b$ is even and $p, q \\geq 3$. Prove that\n\n$$\n2 a^{p} b-2 a b^{q}\n$$\n\n$$\na \\text { esche }\n$$\n\ncannot be a square of an integer number.", "solution": "Withou loss of\n\nLet $a=2 a^{\\prime}$. If $\\vdots$. Without loss of generality, assume that $a$ is even and consequently $b$ is odd.\n\n$$\n2 a^{p} b-2 a b^{q}=4 a^{\\prime} b\\left(a^{p-1}-b^{q-1}\\right)\n$$\n\nis a square, then $a^{\\prime}, b$ and $a^{p-1}-b^{q-1}$ are pairwise coprime.\n\nOn the other hand, $a^{p-1}$ is divisible by 4 and $b^{q-1}$ gives the remainder 1 when divided by 4. It follows that $a^{p-1}-b^{q-1}$ has the form $4 k+3$, a contradiction.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nNT1.", "solution_match": "\nSolution."}}
-{"year": "2004", "tier": "T3", "problem_label": "NT2", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all four digit numbers A such that\n\n$$\n\\frac{1}{3} A+2000=\\frac{2}{3} \\bar{A}\n$$\n\nwhere $\\bar{A}$ is the number with the same digits as $A$, but written in opposite order. (For example, $\\overline{1234}=4321$.)", "solution": "Let $A=1000 a+100 b+10 c+d$. Then we obtain the equality\n\n$$\n\\frac{1}{3}(1000 n+100 b+10 c+d)+2000=\\frac{2}{3}(1000 d+100 c+10 b+a)\n$$\n\nol\n\n$$\n1999 d+190 c=80 b+998 a+6000\n$$\n\nIt is clear that $d$ is even digit and $d>2$. So we have to investigate three cases: (i) $d=4$ : (ii) $d=6$; (iii) $d=8$.\n\n(i) If $d=4$. comparing the last digits in the upper equality we see that $a=2$ or $a=\\overline{1}$.\n\nIf $a=2$ then $19 c=80$, which is possible only when $40=c=0$. Hence the number $4=2004$ satisfies the condition.\n\nIf $a=\\overline{7}$ then $19 c-8 b=490$, which is impossible.\n\n(ii) If $d=6$ then $190 c+5994=80 b+998$ r. Comparing the last digits we obtain tilat. $a=3$ or $a=8$.\n\nIf $a=3$ then $80 b+998 a<80 \\cdot 9+1000 \\cdot 3<5994$.\n\nIf $a=8$ then $306+998 a \\geq 998 \\cdot \\delta=7984=5994+1990>599+190 c$.\n\n(iii) If $d=8$ then $190 c+9992=80 b+998 a$. Now $80 b+998 a \\leq 80 \\cdot 9+998 \\cdot 9=9702<$ $9992+190 c$\n\nHence we have the only solution $4=2004$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nNT2.", "solution_match": "\nSolution."}}
-{"year": "2004", "tier": "T3", "problem_label": "NT3", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all positive integers $n, n \\geq 3$, such that $n \\mid(n-2)$ !.", "solution": "For $n=3$ and $n=4$ we easily check that $n$ does not divide $(n-2)!$.\n\nIf $n$ is prime, $n \\geq 5$, then $(n-2)!=1 \\cdot 2 \\cdot 3 \\cdot \\ldots \\cdot(n-2)$ is not divided by $n$, since $n$ is a prime not included in the set of factors of ( $n-2)$ !.\n\nIf $n$ is composite, $n \\geq 6$, then $n=p m$, where $p$ is prime and $m>1$. Since $p \\geq 2$, we have $n=p m \\geq 2 m$, and consequently $m \\leq \\frac{n}{2}$. Moreover $m<n-2$, since $n>4$.\n\nTherefore, if $p \\neq m$, both $p$ and $m$ are distinct factors of ( $n-2)$ ! and so $n$ divides $(n-2)!$.\n\nIf $p=m$, then $n=p^{2}$ and $p>2$ (since $n>4$ ). Hence $2 p<p^{2}=n$. Moreover $2 p<n-1$ :(since $n>4$ ). Therefore, both $p$ and $2 p$ are distinct factors of $(n-2)$ ! and so $2 p^{2} \\mid(n-2)$ !. Hence $p^{2} \\mid(n-2)$ !, i.e. $n \\mid(n-2)$ !. So we conclude that $n \\mid(n-2)!$ if and only if $n$ is composite, $n \\geq 6$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nNT3.", "solution_match": "\nSolution."}}
-{"year": "2004", "tier": "T3", "problem_label": "NT4", "problem_type": "Number Theory", "exam": "JBMO", "problem": "If the positive integers $x$ and $y$ are such that both $3 x+4 y$ and $4 x+3 y$ are perfect squares, prove that both $x$ and $y$ are multiples of 7 .", "solution": "Let\n\n$$\n3 x+4 y=m^{2}, \\quad 4 x+3 y=n^{2}\n$$\n\nThen\n\n$$\n7(x+y)=m^{2}+n^{2} \\Rightarrow 7 \\mid m^{2}+n^{2}\n$$\n\nConsidering $m=7 k+r, \\quad r \\in\\{0,1,2,3,4,5,6\\}$ we find that $m^{2} \\equiv u(\\bmod 7), \\quad u \\in$ $\\{0,1,2,4\\}$ and similarly $n^{2} \\equiv v(\\bmod 7), \\quad v \\in\\{0,1,2,4\\}$. Therefore we have either $m^{2}+n^{2} \\equiv 0 \\quad(\\bmod 7)$, when $u=v=0$, or $m^{2}+n^{2} \\equiv w(\\bmod 7), w \\in\\{1,2,3,4,5,6\\}$. However, from (2) we have that $m^{2}+n^{2} \\equiv 0(\\bmod \\tau)$ and hence $u=v=0$ and\n\n$$\nm^{2}+n^{2} \\equiv 0 \\quad\\left(\\bmod \\tau^{2}\\right) \\Rightarrow 7(x+y) \\equiv 0 \\quad\\left(\\bmod \\tau^{2}\\right)\n$$\n\nand cousequently\n\n$$\nx+y \\equiv 0 \\quad(\\bmod 7)\n$$\n\nMoreover. from (1) we have $x-y=n^{2}-m^{2}$ and $n^{2}-m^{2} \\equiv 0\\left(\\bmod i^{2}\\right)$ (since $\\left.u=c=0\\right)$, so\n\n$$\nx-y \\equiv 0 \\quad(\\bmod 7) .\n$$\n\nFrom (3) and (4) we have tliat $x+y=7 k, x-y=7 l$, where $k$ and $l$ are positive incegers. Hence\n\n$$\n2 x=7(k+l), 2 y=7(k-l)\n$$\n\nwhere $k+l$ and $k-l$ are positive integers. It follows that $i \\mid 2 x$ and $i \\mid 2 y$, and finally $\\pi \\mid x$ and i|y.\n\n## ALGEBRA", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nNT4.", "solution_match": "\nSolution."}}
-{"year": "2004", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO", "problem": "Prove that\n\n$$\n(1+a b c)\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right) \\geq 3+a+b+c\n$$\n\nfor any real numbers $a, b, c \\geq 1$.", "solution": "The inequality rewrites as\n\n$$\n\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)+(b c+a c+a b) \\geq 3+a+b+c\n$$\n\nor\n\n$$\n\\left(\\frac{\\frac{1}{a}+\\frac{1}{b}}{2}+\\frac{\\frac{1}{b}+\\frac{1}{c}}{2}+\\frac{\\frac{1}{a}+\\frac{1}{c}}{2}\\right)+(b c+a c+a b) \\geq 3+a+b+c\n$$\n\nwhich is equivalent to\n\n$$\n\\frac{(2 a b-(a+b))(a b-1)}{2 a b}+\\frac{(2 b c-(b+c))(b c-1)}{2 b c}+\\frac{(2 c a-(c+a))(c a-1)}{2 c a} \\geq 0\n$$\n\nThe last inequality is true due to the obvious relations\n\n$$\n2 x y-(x+y)=x(y-1)+y(x-1) \\geq 0\n$$\n\nfor any two real numbers $x, y \\geq 1$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nA1.", "solution_match": "\nSolution."}}
-{"year": "2004", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO", "problem": "Prove that, for all real numbers $x, y, z$ :\n\n$$\n\\frac{x^{2}-y^{2}}{2 x^{2}+1}+\\frac{y^{2}-z^{2}}{2 y^{2}+1}+\\frac{z^{2}-x^{2}}{2 z^{2}+1} \\leq(x+y+z)^{2}\n$$\n\nWhen the equality holds?", "solution": "For $x=y=z=0$ the equality is valid.\n\nSince $(x+y+z)^{2} \\geq 0$ it is enongh to prove that\n\n$$\n\\frac{x^{2}-y^{2}}{2 x^{2}+1}+\\frac{y^{2}-z^{2}}{2 y^{2}+1}+\\frac{z^{2}-x^{2}}{2 z^{2}+1} \\leq 0\n$$\n\nwhich is equivalent to the inequality\n\n$$\n\\frac{x^{2}-y^{2}}{x^{2}+\\frac{1}{2}}+\\frac{y^{2}-z^{2}}{y^{2}+\\frac{1}{2}}+\\frac{z^{2}-x^{2}}{z^{2}+\\frac{1}{2}} \\leq 0\n$$\n\nDellote\n\n$$\na=x^{2}+\\frac{1}{2}, b=y^{2}+\\frac{1}{2}, c=z^{2}+\\frac{1}{2}\n$$\n\nThen (1) is equivalent to\n\n$$\n\\frac{a-b}{a}+\\frac{b-c}{b}+\\frac{c-a}{c} \\leq 0\n$$\n\nFrom very well known $A G$ inequality follows that\n\n$$\na^{2} b+b^{2} c+c^{2} a \\geq 3 a b c\n$$\n\nFron the equivalencies\n\n$$\na^{2} b+b^{2} c+c^{2} a \\geq 3 a b c \\Leftrightarrow \\frac{a}{c}+\\frac{b}{a}+\\frac{c}{b} \\geq-3 \\Leftrightarrow \\frac{a-b}{a}+\\frac{b-c}{b}+\\frac{c-a}{c} \\leq 0\n$$\n\nfollows thait the inequality (2) is valid, for positive real numbers $a, b, c$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nA2.", "solution_match": "\nSolution."}}
-{"year": "2004", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO", "problem": "Prove that for all real $x, y$\n\n$$\n\\frac{x+y}{x^{2}-x y+y^{2}} \\leq \\frac{2 \\sqrt{2}}{\\sqrt{x^{2}+y^{2}}}\n$$", "solution": "The inequality rewrites as\n\n$$\n\\frac{x+y}{x^{2}-x y+y^{2}} \\leq \\frac{\\sqrt{2\\left(x^{2}+y^{2}\\right)}}{\\frac{x^{2}+y^{2}}{2}}\n$$\n\nNow it is enough to prove the next two simple inequalities:\n\n$$\nx+y \\leq \\sqrt{2\\left(x^{2}+y^{2}\\right)}, \\quad x^{2}-x y+y^{2} \\geq \\frac{x^{2}+y^{2}}{2}\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nA3.", "solution_match": "\nSolution."}}
-{"year": "2004", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "exam": "JBMO", "problem": "Prove that if $0<\\frac{a}{b}<b<2 a$ then\n\n$$\n\\frac{2 a b-a^{2}}{7 a b-3 b^{2}-2 a^{2}}+\\frac{2 a b-b^{2}}{7 a b-3 a^{2}-2 b^{2}} \\geq 1+\\frac{1}{4}\\left(\\frac{a}{b}-\\frac{b}{a}\\right)^{2}\n$$", "solution": "If we denote\n\n$$\nu=2-\\frac{a}{b}, \\quad v=2-\\frac{b}{a}\n$$\n\nthen the inequality rewrites as\n\n$$\n\\begin{aligned}\n& \\frac{u}{v+u v}+\\frac{v}{u+u v} \\geq 1+\\frac{1}{4}(u-u)^{2} \\\\\n& \\frac{(u-v)^{2}+u v(1-u v)}{u v(u v+u+v+1)} \\geq \\frac{(u-u)^{2}}{4}\n\\end{aligned}\n$$\n\n$\\mathrm{Or}$\n\nSince $u>0, v>0, u+v \\leq 2, u v \\leq 1, u v(u+v+u v+1) \\leq 4$, the result is clear.\n\n## GEOMETRY", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nA4.", "solution_match": "\nSolution."}}
-{"year": "2004", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO", "problem": "Two circles $k_{1}$ and $k_{2}$ intersect a.t points $A$ and $B$. A circle $k_{3}$ centered at $A$ meet $k_{1}$ at $M$ and $P$ and $k_{2}$ at $N$ and $Q$, such that $N$ and $Q$ are on different sides of $M P$ and $A B>A M$.\n\nProve rhat the angles $\\angle M B Q$ and $\\angle N B P$ are equal.", "solution": "As $A M=A P$, we have\n\n$$\n\\angle M B A=\\frac{1}{2} \\operatorname{arcAM}=\\frac{1}{2} \\operatorname{arc} A P=\\angle A B P\n$$\n\nand likewise\n\n$$\n\\angle Q B A=\\frac{1}{2} \\operatorname{arc} A Q=\\frac{1}{2} \\operatorname{arc} c A N=\\angle A B N\n$$\n\nSumming these equalities yields $\\angle M B Q=\\angle N B P$ as needed.\n\nQ2. Let $E$ and $F$ be two distinct points inside of a parallelogram $A B C D$. Find the maximum number of triangles with the same area and having the vertices in three of the following five points: $A, B, C, D, E, F$.\n\nSolution. We shall use the following two well known results:\n\nLemma 1. Let $A, B, C, D$ be four points lying in the same plane such that the line $A B$ do not intersect the segment $C D$ (in particular $\\triangle B C D$ is a convex quadrilateral). If $[A B C]=[A B D]$, then $A B \\| C D$.\n\nLemma 2. Let $X$ be a point inside of a parallelogram $A B C D$. Then $[A C X]<[A B C]$ and $[B D X]<[4 B D]$. (Here and below the notation $[S]$ stands for the area of the surface of $S$.\n\nWith the points $A, B, C, D, E, F$ we can form 20 triangles. We will show that at most ten of them can have the same area. In that sense three cases may occur.\n\nGase 1. EF is parallel with one side of the parallelogram $\\triangle B C D$.\n\nWe can assume that $E F \\|: A D, E$ lies inside of the triangle $A B F$ and $F$ lies inside of thr hiangle $C D E$. With the points $A, B, C, D, E$ we can form ten pairs of triangles als follows:\n\n$(\\triangle .4 D E, \\triangle A E F) ;(\\triangle . A D F, \\triangle D E F) ;(\\triangle B C E, \\triangle B E F) ;(\\triangle B C F, \\triangle C E F) ;(\\triangle C D E, \\triangle C D F):$\n\n$(\\triangle A B E: \\triangle A B F) ;(\\triangle A D C ; \\triangle A C E) ;(\\triangle A B C, \\triangle . A C F) ;(\\triangle A B D, \\triangle B D E) ;(\\triangle C B D . \\triangle B D F)$.\n\nUsing Lemmas $1-2$ one can easily prove that any two triangles that belong to the same pair have distinct area, so there exists at most ten triangles having the same area.\n\nCase 2. $E F$ is parallel with a diagonal of the paralellogram $\\triangle B C D$.\n\nLet us assume $E F \\| A C$ and that $E, F$ lie inside of the triangle $A B C$. We consider the following pairs of triangles:\n\n$(\\triangle A B D, \\triangle B D E) ;(\\triangle A B C, \\triangle B C F) ;(\\triangle A C D, \\triangle B C E) ;(\\triangle A B F, \\triangle A B E) ;(\\triangle B E F, \\triangle D E F) ;$\n\n$(\\triangle A E F, \\triangle A C E) ;(\\triangle C E F, \\triangle A C F) ;(\\triangle A D E ; \\triangle A D F) ;(\\triangle D C E, \\triangle D C F) ;(\\triangle C B D, \\triangle B D F)$.\n\nWith the same idea as above we deduce that any two triangles that belong to the same pair have distinct area and the conclusion follows.\n\nWe also note that if $E$ and $F$ lie on $A C$ then only 16 of 20 triangles are nondegenerate. In this case we consider the following pairs:\n\n$$\n\\begin{aligned}\n& (\\triangle A B E, \\triangle A B F) ;(\\triangle A B C, \\triangle B C F) ;(\\triangle B C E, \\triangle B E F) ;(\\triangle A D E, \\triangle A D F) \\\\\n& (\\triangle A C D, \\triangle D C F) ;(\\triangle C D E, \\triangle E D F) ;(\\triangle B D E, \\triangle A B D) ;(\\triangle B D F, \\triangle B D C)\n\\end{aligned}\n$$\n\nCase 3. $E F$ is not parallel with any side or diagonal of $\\triangle B C D$.\n\nWe claim that at most two of the triangles $A E F, B E F, C E F, D E F$ can have the same area. Indeed, supposing the contrary, we may have $[A E F]=[B E F]=[C E F]$. We remark first that $A, B, C$ do not belong to $E F$ (elsewhere, exactly one of the above triangles is degenerate, contradiction!). Hence at least two of the points $A, B, C$ belong to the same side of the line $E F$. Using now Lemma 1 we get that $E F$ is parallel with $A B$ or $B C$ or $A C$. This is clearly a contradiction and our claim follows. With the remaining 16 triangles we form 8 pairs as follows:\n\n$(\\triangle A B D, \\triangle B D E) ;(\\triangle C D B, \\triangle B D F) ;(\\triangle A D C, \\triangle A C E) ;(\\triangle A B C, \\triangle A C F) ;$\n\n$(\\triangle A B E, \\triangle A B F) ;(\\triangle B C E, \\triangle B C F) ;(\\triangle A D E, \\triangle A D F) ;(\\triangle D C E, \\triangle D C F)$.\n\nWith the same arguments as above, we get at most ten triangles with the same area.\n\nTo conclude the proof, it remains only to give an example of points $E, F$ inside of the parallelogram $A B C D$ such that exactly ten of the triangles that can be formed with the vetices $A, B, C, D, E, F$ have the same area.\n\nDenote $A C \\cap B D=\\{O\\}$ and let $M, N$ be the midpoints of $A B$ and $C D$ respectively Consider $E$ and $F$ the midpoints of $M O$ and $N O$. Then $O, M, N, E, F$ are collinear and $M E=E O=F O=N F$. Since $E$ and $F$ are the centroids of the triangles $A B F$ and $C D E$ we get $[A B E]=[A E F]=[B E F]$ and $[C E F]=[D E F]=[C D F]$. On the other hand, taking into account that $A E C F$ and $B E D F$ are parallelograms we deduce $[4 E F]=[C E F]=[4 C E]=[A C F]$ and $[B E F]=[D E F]=[B D E]=[B D F]$. From the above equalities we conclude that the triangles\n\n## $\\triangle A B E, \\triangle C D F E, \\triangle A C E, \\triangle A C F, \\triangle B D E, \\triangle B D F, \\triangle A E F, \\triangle B E F, \\triangle C E F, \\triangle D E F$\n\nhave the same area. This finishes our proof.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nG1.", "solution_match": "\nSolution "}}
-{"year": "2004", "tier": "T3", "problem_label": "G3", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C$ be scalene triangle inscribed in the circle $k$. Circles $\\alpha, \\beta, \\gamma$ are internally tangent to $k$ at points $A_{1}, B_{1}, C_{1}$ respectively, and tangent to the sides $B C, C A, A B$ at points $A_{2}, B_{2}, C_{2}$ respectively, so that $A$ and $A_{1}$ are on opposite sides of $B C^{\\prime}, B$ and $B_{1}$ mre on opposite sides of $C A$, and $C$ and $C_{1}$ are on opposite sides of $A B$. Lines $A_{1} A_{2}, B_{1} B_{2}$ and $C_{1} C_{2}$ meet again the circle $k$ in the points $A^{\\prime}, B^{\\prime}, C^{\\prime \\prime}$ respectively. is right.\n\nProve that if $M$ is the intersection point of the lines $B B^{\\prime}$ and $C^{\\prime} C^{\\prime \\prime}$, the angle $\\angle M A .4^{\\prime}$", "solution": "The idea is to observe that $A^{\\prime}, B^{\\prime}, C^{\\prime \\prime}$ are the midpoints of the arcs $B C$, $C A$ and $A B$ of the circle $k$ which do not contain the points $A, B, C$ respectively. To prove this, consider the dilatation with the center $A_{1}$ taking' $\\alpha$ to $k$. The line $B C$, which touches $\\alpha$ at $A_{2}$, is taken to the line $t$ touching $l$ at $A^{\\prime}$. Since $t$ is parallel to $B C^{\\prime}$, it follows that $A^{\\prime}$ is the midpoint of thr arc $B C$ that do not touch $\\alpha$.\n\nConsequently; lines $B B^{\\prime}$ and $C C^{\\prime}$ are the exterior bisectors of the angles $B$ and $C$ of the scalene triangle $A B C$ and so $M$ is the excenter of $A B C$. Hence $A M$ and $4.4^{\\prime}$ are the interior and exterior bisectors of the angle $A$, implying $\\angle M A A^{i}=90^{\\circ}$.\n\nQ4. Let $A B C$ be isosceles triangle with $A C=B C, M$ be the midpoint of $A C, B H$ be the line through $C^{\\prime}$ perpendicular to $A B$. The circle through $B, C$ and $M$ intersects $C H$ in point $Q$. If $C Q=m$, find the radius of the circumcircle of $A B C$.\n\nSolution. Let $P$ be the center of circle $k_{1}$ through $B, C$ and $M, O$ be the center of the circumcircle of $A B C$, and $K E$ be the midpoint of $M C^{\\prime}$. Since $A C^{\\prime}=B C$, the center $O$ lies on $C H$. Let $K P$ intersects $C H$ in point $L$. Since $K P$ and $O M$ are perpendicular to AC, then $K P \\| O M$. From $M K=K C$ it follows that $O L=C L$. On the other hand $O P$ is perpendicular to $B C$, hence $\\angle L O P=\\angle C O P=90^{\\circ}-\\angle \\dot{B C H}$. Also we have $\\angle O L P=\\angle C L K^{\\circ}=90^{\\circ}-\\angle A C H$. Since $A B C$ is isosceles and $\\angle B C H=\\angle A C H$, then $\\angle L O P=\\angle O L P$ and $\\angle P=O P$. Since $C P=P Q$ we obtain that $\\angle C L P=\\angle Q O P$ and $C L=O Q$. Thus we have $C L=L O=O Q$, so $C O=\\frac{2}{3} C Q$. Finally for the radius $R$ of the circumcircle of $A B C$ we obtain $R=\\frac{2}{3} m$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_ed83a632d3567b652a08g-09.jpg?height=590&width=673&top_left_y=1947&top_left_x=745)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nG3.", "solution_match": "\nSolution."}}
-{"year": "2004", "tier": "T3", "problem_label": "G5", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C$ be a triangle with $\\angle C=90^{\\circ}$ and $D \\in C .4, E \\in C^{\\prime} B$, and $k_{1}, k_{2}, k_{3}, k_{4}$ semicircles with diameters $C A, C B, C D, C E$ respectively, which have common part with the triangle $A B C$. Let also,\n\n$$\nk_{1} \\cap k_{2}=\\{C, K\\}, k_{3} \\cap k_{4}=\\{C, M\\}, k_{2} \\cap k_{3}=\\{C, L\\}, k_{1} \\cap k_{4}=\\{C, N\\}\n$$\n\nProve that $K, L, M$ and $N$ are cocyclic points.", "solution": "The points $K, L, M, N$ belong to the segments $A B, B D, D E, E A$ respectively, where $C K \\perp A B, C L \\perp B D, C M \\perp D E, C N \\perp A E$. Then quadrilaterals $C D L M$ and $C E N M$ are inscribed. Let $\\angle C A E=\\varphi, \\angle D C L=\\theta$. Then $\\angle E M N=\\angle E C N=\\varphi$ and $\\angle D M L=\\angle D C L=\\theta$. So $\\angle D M L+\\angle E M N=\\varphi+\\theta$ and therefore $\\angle L M N=180^{\\circ}-\\varphi-\\theta$. The quadrilaterals $C B K L$ and $C A K N$ are also inscribed and hence $\\angle L K C=\\angle L B C=\\theta$, $\\angle C K N=\\angle C A N=\\varphi$, and so $\\angle L K N=\\varphi+\\theta$, while $\\angle L M N=180^{\\circ}-\\varphi-\\theta$, which means that $K L M N$ is inscribed.\n\nNote that $K L M N$ is convex because $L, N$ lie in the interior of the convex quadrilateral $A D E B$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_ed83a632d3567b652a08g-10.jpg?height=1024&width=725&top_left_y=1274&top_left_x=767)\n\n## COMBINATORICS", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nG5.", "solution_match": "\nSolution."}}
-{"year": "2004", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "A polygon having $n$ sides is arbitrarily decomposed in triangles having all the vertices among the vertices of the polygon. We paint in black the triangles that have two sides that are also sides of the polygon, in red if only one side of the triangle is side of the polygon and white those triangles that have in common with the poligon only vertices.\n\nProve that there are 2 more black triangles than white ones.", "solution": "Denote by $b, r, w$ the number of black, red white triangles respectively.\n\nIt is easy to prove that the polygon is divided into $n-2$ triangles, hence\n\n$$\nb+r+w=n-2\n$$\n\nEach side of the polygon is a side of exactly one triangle of the decomposition, and thus\n\n$$\n2 b+r=n\n$$\n\nSubtracting the two relations yields $w=b-2$, as needed. allowed:", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nC1.", "solution_match": "\nSolution."}}
-{"year": "2004", "tier": "T3", "problem_label": "C2", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Given $m \\times n$ table, each cell signed with \"-\". The following operations are\n\n(i) to change all the signs in entire row to the opposite, i. e. every \"-\" to \"+\", and every \"+\" to \"-\";\n\n(ii) to change all the signs in entire column to the opposite, i. e. every \"-\" to \"+\" and every \"+\" to \" -\".\n\n(a) Prove that if $m=n=100$, using the above operations one can not obtain 2004 signs \"t\".\n\n(b) If $m=1004$, find the least $n>100$ for which 2004 signs \" + \" can be obtained.", "solution": "If we apply (i) to $l$ rows and (ii) to $k$ columns we obtain $(m-k) l+(n-l) k$\n\n(a) We have equation $(100-k) l+(100-l) k=2004$, or $100 l+100 k-2 l k=2004$, le\n\n$$\n50 l+50 k-1 k=1002\n$$\n\nRewrite the lasc equation as\n\n$$\n(50-l)(50-h)=2.500-100.2=1498\n$$\n\nSince $1498=2 \\cdot 7 \\cdot 107$, this equation has no solitions in natural numbers.\n\n(b) Let $n=101$. Then we have\n\n$$\n(100-k) l+(101-l) k=2004\n$$\n\nOI\n\n$$\n100 l+101 k-2 l k=2004\n$$\n\nl.e.\n\n$$\n101 k=2004-100 l+2 l k \\div 2(1002-50 l+l k)\n$$\n\nHence $s=2 t$ and we have $101 t=501-25 l+2 l t$. From here we have\n\n$$\nt=\\frac{501-25 l}{101-2 l}=4+\\frac{97-17 l}{101-2 l}\n$$\n\nSince $t$ is natural number and $97-17 l<101-2 l$, this is a contradiction, Hence $n \\neq 101$. Let $n=1.02$. Then we have\n\n$$\n(100-k) l+(102-l) k=2004\n$$\n\nor\n\n$$\n100 l+102 k-2 l k=2004\n$$\n\n$$\n50 l+51 k-l k=1002\n$$\n\nRewrite the last equation as\n\n$$\n(51-l)(50-k)=25.50-1002=1.548\n$$\n\nSince $145 S=2 \\cdot 2 \\cdot 3 \\cdot 3 \\cdot 43$ we have $51-l=36$ and $50-k=43$. From here obtain $l=15$ and $k=7$. Indeed,\n\n$$\n(100-\\bar{\\imath}) \\cdot 15+(102-1.5) \\cdot \\overline{7}=93 \\cdot 15+87 \\cdot 7=1395+609=2004\n$$\n\nHence, the least $n$ is 102 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nC2.", "solution_match": "\nSolution."}}
+{"year": "2004", "tier": "T3", "problem_label": "NT1", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Let $a, b, p, q$ be positive integers such that $a$ and $b$ are relatively prime, $a b$ is even and $p, q \\geq 3$. Prove that\n\n$$\n2 a^{p} b-2 a b^{q}\n$$\n\n$$\na \\text { esche }\n$$\n\ncannot be a square of an integer number.", "solution": "Withou loss of\n\nLet $a=2 a^{\\prime}$. If $\\vdots$. Without loss of generality, assume that $a$ is even and consequently $b$ is odd.\n\n$$\n2 a^{p} b-2 a b^{q}=4 a^{\\prime} b\\left(a^{p-1}-b^{q-1}\\right)\n$$\n\nis a square, then $a^{\\prime}, b$ and $a^{p-1}-b^{q-1}$ are pairwise coprime.\n\nOn the other hand, $a^{p-1}$ is divisible by 4 and $b^{q-1}$ gives the remainder 1 when divided by 4. It follows that $a^{p-1}-b^{q-1}$ has the form $4 k+3$, a contradiction.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nNT1.", "solution_match": "\nSolution."}}
+{"year": "2004", "tier": "T3", "problem_label": "NT2", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all four digit numbers A such that\n\n$$\n\\frac{1}{3} A+2000=\\frac{2}{3} \\bar{A}\n$$\n\nwhere $\\bar{A}$ is the number with the same digits as $A$, but written in opposite order. (For example, $\\overline{1234}=4321$.)", "solution": "Let $A=1000 a+100 b+10 c+d$. Then we obtain the equality\n\n$$\n\\frac{1}{3}(1000 n+100 b+10 c+d)+2000=\\frac{2}{3}(1000 d+100 c+10 b+a)\n$$\n\nol\n\n$$\n1999 d+190 c=80 b+998 a+6000\n$$\n\nIt is clear that $d$ is even digit and $d>2$. So we have to investigate three cases: (i) $d=4$ : (ii) $d=6$; (iii) $d=8$.\n\n(i) If $d=4$. comparing the last digits in the upper equality we see that $a=2$ or $a=\\overline{1}$.\n\nIf $a=2$ then $19 c=80$, which is possible only when $40=c=0$. Hence the number $4=2004$ satisfies the condition.\n\nIf $a=\\overline{7}$ then $19 c-8 b=490$, which is impossible.\n\n(ii) If $d=6$ then $190 c+5994=80 b+998$ r. Comparing the last digits we obtain tilat. $a=3$ or $a=8$.\n\nIf $a=3$ then $80 b+998 a<80 \\cdot 9+1000 \\cdot 3<5994$.\n\nIf $a=8$ then $306+998 a \\geq 998 \\cdot \\delta=7984=5994+1990>599+190 c$.\n\n(iii) If $d=8$ then $190 c+9992=80 b+998 a$. Now $80 b+998 a \\leq 80 \\cdot 9+998 \\cdot 9=9702<$ $9992+190 c$\n\nHence we have the only solution $4=2004$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nNT2.", "solution_match": "\nSolution."}}
+{"year": "2004", "tier": "T3", "problem_label": "NT3", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all positive integers $n, n \\geq 3$, such that $n \\mid(n-2)$ !.", "solution": "For $n=3$ and $n=4$ we easily check that $n$ does not divide $(n-2)!$.\n\nIf $n$ is prime, $n \\geq 5$, then $(n-2)!=1 \\cdot 2 \\cdot 3 \\cdot \\ldots \\cdot(n-2)$ is not divided by $n$, since $n$ is a prime not included in the set of factors of ( $n-2)$ !.\n\nIf $n$ is composite, $n \\geq 6$, then $n=p m$, where $p$ is prime and $m>1$. Since $p \\geq 2$, we have $n=p m \\geq 2 m$, and consequently $m \\leq \\frac{n}{2}$. Moreover $m<n-2$, since $n>4$.\n\nTherefore, if $p \\neq m$, both $p$ and $m$ are distinct factors of ( $n-2)$ ! and so $n$ divides $(n-2)!$.\n\nIf $p=m$, then $n=p^{2}$ and $p>2$ (since $n>4$ ). Hence $2 p<p^{2}=n$. Moreover $2 p<n-1$ :(since $n>4$ ). Therefore, both $p$ and $2 p$ are distinct factors of $(n-2)$ ! and so $2 p^{2} \\mid(n-2)$ !. Hence $p^{2} \\mid(n-2)$ !, i.e. $n \\mid(n-2)$ !. So we conclude that $n \\mid(n-2)!$ if and only if $n$ is composite, $n \\geq 6$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nNT3.", "solution_match": "\nSolution."}}
+{"year": "2004", "tier": "T3", "problem_label": "NT4", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "If the positive integers $x$ and $y$ are such that both $3 x+4 y$ and $4 x+3 y$ are perfect squares, prove that both $x$ and $y$ are multiples of 7 .", "solution": "Let\n\n$$\n3 x+4 y=m^{2}, \\quad 4 x+3 y=n^{2}\n$$\n\nThen\n\n$$\n7(x+y)=m^{2}+n^{2} \\Rightarrow 7 \\mid m^{2}+n^{2}\n$$\n\nConsidering $m=7 k+r, \\quad r \\in\\{0,1,2,3,4,5,6\\}$ we find that $m^{2} \\equiv u(\\bmod 7), \\quad u \\in$ $\\{0,1,2,4\\}$ and similarly $n^{2} \\equiv v(\\bmod 7), \\quad v \\in\\{0,1,2,4\\}$. Therefore we have either $m^{2}+n^{2} \\equiv 0 \\quad(\\bmod 7)$, when $u=v=0$, or $m^{2}+n^{2} \\equiv w(\\bmod 7), w \\in\\{1,2,3,4,5,6\\}$. However, from (2) we have that $m^{2}+n^{2} \\equiv 0(\\bmod \\tau)$ and hence $u=v=0$ and\n\n$$\nm^{2}+n^{2} \\equiv 0 \\quad\\left(\\bmod \\tau^{2}\\right) \\Rightarrow 7(x+y) \\equiv 0 \\quad\\left(\\bmod \\tau^{2}\\right)\n$$\n\nand cousequently\n\n$$\nx+y \\equiv 0 \\quad(\\bmod 7)\n$$\n\nMoreover. from (1) we have $x-y=n^{2}-m^{2}$ and $n^{2}-m^{2} \\equiv 0\\left(\\bmod i^{2}\\right)$ (since $\\left.u=c=0\\right)$, so\n\n$$\nx-y \\equiv 0 \\quad(\\bmod 7) .\n$$\n\nFrom (3) and (4) we have tliat $x+y=7 k, x-y=7 l$, where $k$ and $l$ are positive incegers. Hence\n\n$$\n2 x=7(k+l), 2 y=7(k-l)\n$$\n\nwhere $k+l$ and $k-l$ are positive integers. It follows that $i \\mid 2 x$ and $i \\mid 2 y$, and finally $\\pi \\mid x$ and i|y.\n\n## ALGEBRA", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nNT4.", "solution_match": "\nSolution."}}
+{"year": "2004", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Prove that\n\n$$\n(1+a b c)\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right) \\geq 3+a+b+c\n$$\n\nfor any real numbers $a, b, c \\geq 1$.", "solution": "The inequality rewrites as\n\n$$\n\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)+(b c+a c+a b) \\geq 3+a+b+c\n$$\n\nor\n\n$$\n\\left(\\frac{\\frac{1}{a}+\\frac{1}{b}}{2}+\\frac{\\frac{1}{b}+\\frac{1}{c}}{2}+\\frac{\\frac{1}{a}+\\frac{1}{c}}{2}\\right)+(b c+a c+a b) \\geq 3+a+b+c\n$$\n\nwhich is equivalent to\n\n$$\n\\frac{(2 a b-(a+b))(a b-1)}{2 a b}+\\frac{(2 b c-(b+c))(b c-1)}{2 b c}+\\frac{(2 c a-(c+a))(c a-1)}{2 c a} \\geq 0\n$$\n\nThe last inequality is true due to the obvious relations\n\n$$\n2 x y-(x+y)=x(y-1)+y(x-1) \\geq 0\n$$\n\nfor any two real numbers $x, y \\geq 1$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nA1.", "solution_match": "\nSolution."}}
+{"year": "2004", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Prove that, for all real numbers $x, y, z$ :\n\n$$\n\\frac{x^{2}-y^{2}}{2 x^{2}+1}+\\frac{y^{2}-z^{2}}{2 y^{2}+1}+\\frac{z^{2}-x^{2}}{2 z^{2}+1} \\leq(x+y+z)^{2}\n$$\n\nWhen the equality holds?", "solution": "For $x=y=z=0$ the equality is valid.\n\nSince $(x+y+z)^{2} \\geq 0$ it is enongh to prove that\n\n$$\n\\frac{x^{2}-y^{2}}{2 x^{2}+1}+\\frac{y^{2}-z^{2}}{2 y^{2}+1}+\\frac{z^{2}-x^{2}}{2 z^{2}+1} \\leq 0\n$$\n\nwhich is equivalent to the inequality\n\n$$\n\\frac{x^{2}-y^{2}}{x^{2}+\\frac{1}{2}}+\\frac{y^{2}-z^{2}}{y^{2}+\\frac{1}{2}}+\\frac{z^{2}-x^{2}}{z^{2}+\\frac{1}{2}} \\leq 0\n$$\n\nDellote\n\n$$\na=x^{2}+\\frac{1}{2}, b=y^{2}+\\frac{1}{2}, c=z^{2}+\\frac{1}{2}\n$$\n\nThen (1) is equivalent to\n\n$$\n\\frac{a-b}{a}+\\frac{b-c}{b}+\\frac{c-a}{c} \\leq 0\n$$\n\nFrom very well known $A G$ inequality follows that\n\n$$\na^{2} b+b^{2} c+c^{2} a \\geq 3 a b c\n$$\n\nFron the equivalencies\n\n$$\na^{2} b+b^{2} c+c^{2} a \\geq 3 a b c \\Leftrightarrow \\frac{a}{c}+\\frac{b}{a}+\\frac{c}{b} \\geq-3 \\Leftrightarrow \\frac{a-b}{a}+\\frac{b-c}{b}+\\frac{c-a}{c} \\leq 0\n$$\n\nfollows thait the inequality (2) is valid, for positive real numbers $a, b, c$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nA2.", "solution_match": "\nSolution."}}
+{"year": "2004", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Prove that for all real $x, y$\n\n$$\n\\frac{x+y}{x^{2}-x y+y^{2}} \\leq \\frac{2 \\sqrt{2}}{\\sqrt{x^{2}+y^{2}}}\n$$", "solution": "The inequality rewrites as\n\n$$\n\\frac{x+y}{x^{2}-x y+y^{2}} \\leq \\frac{\\sqrt{2\\left(x^{2}+y^{2}\\right)}}{\\frac{x^{2}+y^{2}}{2}}\n$$\n\nNow it is enough to prove the next two simple inequalities:\n\n$$\nx+y \\leq \\sqrt{2\\left(x^{2}+y^{2}\\right)}, \\quad x^{2}-x y+y^{2} \\geq \\frac{x^{2}+y^{2}}{2}\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nA3.", "solution_match": "\nSolution."}}
+{"year": "2004", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Prove that if $0<\\frac{a}{b}<b<2 a$ then\n\n$$\n\\frac{2 a b-a^{2}}{7 a b-3 b^{2}-2 a^{2}}+\\frac{2 a b-b^{2}}{7 a b-3 a^{2}-2 b^{2}} \\geq 1+\\frac{1}{4}\\left(\\frac{a}{b}-\\frac{b}{a}\\right)^{2}\n$$", "solution": "If we denote\n\n$$\nu=2-\\frac{a}{b}, \\quad v=2-\\frac{b}{a}\n$$\n\nthen the inequality rewrites as\n\n$$\n\\begin{aligned}\n& \\frac{u}{v+u v}+\\frac{v}{u+u v} \\geq 1+\\frac{1}{4}(u-u)^{2} \\\\\n& \\frac{(u-v)^{2}+u v(1-u v)}{u v(u v+u+v+1)} \\geq \\frac{(u-u)^{2}}{4}\n\\end{aligned}\n$$\n\n$\\mathrm{Or}$\n\nSince $u>0, v>0, u+v \\leq 2, u v \\leq 1, u v(u+v+u v+1) \\leq 4$, the result is clear.\n\n## GEOMETRY", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nA4.", "solution_match": "\nSolution."}}
+{"year": "2004", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Two circles $k_{1}$ and $k_{2}$ intersect a.t points $A$ and $B$. A circle $k_{3}$ centered at $A$ meet $k_{1}$ at $M$ and $P$ and $k_{2}$ at $N$ and $Q$, such that $N$ and $Q$ are on different sides of $M P$ and $A B>A M$.\n\nProve rhat the angles $\\angle M B Q$ and $\\angle N B P$ are equal.", "solution": "As $A M=A P$, we have\n\n$$\n\\angle M B A=\\frac{1}{2} \\operatorname{arcAM}=\\frac{1}{2} \\operatorname{arc} A P=\\angle A B P\n$$\n\nand likewise\n\n$$\n\\angle Q B A=\\frac{1}{2} \\operatorname{arc} A Q=\\frac{1}{2} \\operatorname{arc} c A N=\\angle A B N\n$$\n\nSumming these equalities yields $\\angle M B Q=\\angle N B P$ as needed.\n\nQ2. Let $E$ and $F$ be two distinct points inside of a parallelogram $A B C D$. Find the maximum number of triangles with the same area and having the vertices in three of the following five points: $A, B, C, D, E, F$.\n\nSolution. We shall use the following two well known results:\n\nLemma 1. Let $A, B, C, D$ be four points lying in the same plane such that the line $A B$ do not intersect the segment $C D$ (in particular $\\triangle B C D$ is a convex quadrilateral). If $[A B C]=[A B D]$, then $A B \\| C D$.\n\nLemma 2. Let $X$ be a point inside of a parallelogram $A B C D$. Then $[A C X]<[A B C]$ and $[B D X]<[4 B D]$. (Here and below the notation $[S]$ stands for the area of the surface of $S$.\n\nWith the points $A, B, C, D, E, F$ we can form 20 triangles. We will show that at most ten of them can have the same area. In that sense three cases may occur.\n\nGase 1. EF is parallel with one side of the parallelogram $\\triangle B C D$.\n\nWe can assume that $E F \\|: A D, E$ lies inside of the triangle $A B F$ and $F$ lies inside of thr hiangle $C D E$. With the points $A, B, C, D, E$ we can form ten pairs of triangles als follows:\n\n$(\\triangle .4 D E, \\triangle A E F) ;(\\triangle . A D F, \\triangle D E F) ;(\\triangle B C E, \\triangle B E F) ;(\\triangle B C F, \\triangle C E F) ;(\\triangle C D E, \\triangle C D F):$\n\n$(\\triangle A B E: \\triangle A B F) ;(\\triangle A D C ; \\triangle A C E) ;(\\triangle A B C, \\triangle . A C F) ;(\\triangle A B D, \\triangle B D E) ;(\\triangle C B D . \\triangle B D F)$.\n\nUsing Lemmas $1-2$ one can easily prove that any two triangles that belong to the same pair have distinct area, so there exists at most ten triangles having the same area.\n\nCase 2. $E F$ is parallel with a diagonal of the paralellogram $\\triangle B C D$.\n\nLet us assume $E F \\| A C$ and that $E, F$ lie inside of the triangle $A B C$. We consider the following pairs of triangles:\n\n$(\\triangle A B D, \\triangle B D E) ;(\\triangle A B C, \\triangle B C F) ;(\\triangle A C D, \\triangle B C E) ;(\\triangle A B F, \\triangle A B E) ;(\\triangle B E F, \\triangle D E F) ;$\n\n$(\\triangle A E F, \\triangle A C E) ;(\\triangle C E F, \\triangle A C F) ;(\\triangle A D E ; \\triangle A D F) ;(\\triangle D C E, \\triangle D C F) ;(\\triangle C B D, \\triangle B D F)$.\n\nWith the same idea as above we deduce that any two triangles that belong to the same pair have distinct area and the conclusion follows.\n\nWe also note that if $E$ and $F$ lie on $A C$ then only 16 of 20 triangles are nondegenerate. In this case we consider the following pairs:\n\n$$\n\\begin{aligned}\n& (\\triangle A B E, \\triangle A B F) ;(\\triangle A B C, \\triangle B C F) ;(\\triangle B C E, \\triangle B E F) ;(\\triangle A D E, \\triangle A D F) \\\\\n& (\\triangle A C D, \\triangle D C F) ;(\\triangle C D E, \\triangle E D F) ;(\\triangle B D E, \\triangle A B D) ;(\\triangle B D F, \\triangle B D C)\n\\end{aligned}\n$$\n\nCase 3. $E F$ is not parallel with any side or diagonal of $\\triangle B C D$.\n\nWe claim that at most two of the triangles $A E F, B E F, C E F, D E F$ can have the same area. Indeed, supposing the contrary, we may have $[A E F]=[B E F]=[C E F]$. We remark first that $A, B, C$ do not belong to $E F$ (elsewhere, exactly one of the above triangles is degenerate, contradiction!). Hence at least two of the points $A, B, C$ belong to the same side of the line $E F$. Using now Lemma 1 we get that $E F$ is parallel with $A B$ or $B C$ or $A C$. This is clearly a contradiction and our claim follows. With the remaining 16 triangles we form 8 pairs as follows:\n\n$(\\triangle A B D, \\triangle B D E) ;(\\triangle C D B, \\triangle B D F) ;(\\triangle A D C, \\triangle A C E) ;(\\triangle A B C, \\triangle A C F) ;$\n\n$(\\triangle A B E, \\triangle A B F) ;(\\triangle B C E, \\triangle B C F) ;(\\triangle A D E, \\triangle A D F) ;(\\triangle D C E, \\triangle D C F)$.\n\nWith the same arguments as above, we get at most ten triangles with the same area.\n\nTo conclude the proof, it remains only to give an example of points $E, F$ inside of the parallelogram $A B C D$ such that exactly ten of the triangles that can be formed with the vetices $A, B, C, D, E, F$ have the same area.\n\nDenote $A C \\cap B D=\\{O\\}$ and let $M, N$ be the midpoints of $A B$ and $C D$ respectively Consider $E$ and $F$ the midpoints of $M O$ and $N O$. Then $O, M, N, E, F$ are collinear and $M E=E O=F O=N F$. Since $E$ and $F$ are the centroids of the triangles $A B F$ and $C D E$ we get $[A B E]=[A E F]=[B E F]$ and $[C E F]=[D E F]=[C D F]$. On the other hand, taking into account that $A E C F$ and $B E D F$ are parallelograms we deduce $[4 E F]=[C E F]=[4 C E]=[A C F]$ and $[B E F]=[D E F]=[B D E]=[B D F]$. From the above equalities we conclude that the triangles\n\n## $\\triangle A B E, \\triangle C D F E, \\triangle A C E, \\triangle A C F, \\triangle B D E, \\triangle B D F, \\triangle A E F, \\triangle B E F, \\triangle C E F, \\triangle D E F$\n\nhave the same area. This finishes our proof.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nG1.", "solution_match": "\nSolution "}}
+{"year": "2004", "tier": "T3", "problem_label": "G3", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C$ be scalene triangle inscribed in the circle $k$. Circles $\\alpha, \\beta, \\gamma$ are internally tangent to $k$ at points $A_{1}, B_{1}, C_{1}$ respectively, and tangent to the sides $B C, C A, A B$ at points $A_{2}, B_{2}, C_{2}$ respectively, so that $A$ and $A_{1}$ are on opposite sides of $B C^{\\prime}, B$ and $B_{1}$ mre on opposite sides of $C A$, and $C$ and $C_{1}$ are on opposite sides of $A B$. Lines $A_{1} A_{2}, B_{1} B_{2}$ and $C_{1} C_{2}$ meet again the circle $k$ in the points $A^{\\prime}, B^{\\prime}, C^{\\prime \\prime}$ respectively. is right.\n\nProve that if $M$ is the intersection point of the lines $B B^{\\prime}$ and $C^{\\prime} C^{\\prime \\prime}$, the angle $\\angle M A .4^{\\prime}$", "solution": "The idea is to observe that $A^{\\prime}, B^{\\prime}, C^{\\prime \\prime}$ are the midpoints of the arcs $B C$, $C A$ and $A B$ of the circle $k$ which do not contain the points $A, B, C$ respectively. To prove this, consider the dilatation with the center $A_{1}$ taking' $\\alpha$ to $k$. The line $B C$, which touches $\\alpha$ at $A_{2}$, is taken to the line $t$ touching $l$ at $A^{\\prime}$. Since $t$ is parallel to $B C^{\\prime}$, it follows that $A^{\\prime}$ is the midpoint of thr arc $B C$ that do not touch $\\alpha$.\n\nConsequently; lines $B B^{\\prime}$ and $C C^{\\prime}$ are the exterior bisectors of the angles $B$ and $C$ of the scalene triangle $A B C$ and so $M$ is the excenter of $A B C$. Hence $A M$ and $4.4^{\\prime}$ are the interior and exterior bisectors of the angle $A$, implying $\\angle M A A^{i}=90^{\\circ}$.\n\nQ4. Let $A B C$ be isosceles triangle with $A C=B C, M$ be the midpoint of $A C, B H$ be the line through $C^{\\prime}$ perpendicular to $A B$. The circle through $B, C$ and $M$ intersects $C H$ in point $Q$. If $C Q=m$, find the radius of the circumcircle of $A B C$.\n\nSolution. Let $P$ be the center of circle $k_{1}$ through $B, C$ and $M, O$ be the center of the circumcircle of $A B C$, and $K E$ be the midpoint of $M C^{\\prime}$. Since $A C^{\\prime}=B C$, the center $O$ lies on $C H$. Let $K P$ intersects $C H$ in point $L$. Since $K P$ and $O M$ are perpendicular to AC, then $K P \\| O M$. From $M K=K C$ it follows that $O L=C L$. On the other hand $O P$ is perpendicular to $B C$, hence $\\angle L O P=\\angle C O P=90^{\\circ}-\\angle \\dot{B C H}$. Also we have $\\angle O L P=\\angle C L K^{\\circ}=90^{\\circ}-\\angle A C H$. Since $A B C$ is isosceles and $\\angle B C H=\\angle A C H$, then $\\angle L O P=\\angle O L P$ and $\\angle P=O P$. Since $C P=P Q$ we obtain that $\\angle C L P=\\angle Q O P$ and $C L=O Q$. Thus we have $C L=L O=O Q$, so $C O=\\frac{2}{3} C Q$. Finally for the radius $R$ of the circumcircle of $A B C$ we obtain $R=\\frac{2}{3} m$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_ed83a632d3567b652a08g-09.jpg?height=590&width=673&top_left_y=1947&top_left_x=745)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nG3.", "solution_match": "\nSolution."}}
+{"year": "2004", "tier": "T3", "problem_label": "G5", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C$ be a triangle with $\\angle C=90^{\\circ}$ and $D \\in C .4, E \\in C^{\\prime} B$, and $k_{1}, k_{2}, k_{3}, k_{4}$ semicircles with diameters $C A, C B, C D, C E$ respectively, which have common part with the triangle $A B C$. Let also,\n\n$$\nk_{1} \\cap k_{2}=\\{C, K\\}, k_{3} \\cap k_{4}=\\{C, M\\}, k_{2} \\cap k_{3}=\\{C, L\\}, k_{1} \\cap k_{4}=\\{C, N\\}\n$$\n\nProve that $K, L, M$ and $N$ are cocyclic points.", "solution": "The points $K, L, M, N$ belong to the segments $A B, B D, D E, E A$ respectively, where $C K \\perp A B, C L \\perp B D, C M \\perp D E, C N \\perp A E$. Then quadrilaterals $C D L M$ and $C E N M$ are inscribed. Let $\\angle C A E=\\varphi, \\angle D C L=\\theta$. Then $\\angle E M N=\\angle E C N=\\varphi$ and $\\angle D M L=\\angle D C L=\\theta$. So $\\angle D M L+\\angle E M N=\\varphi+\\theta$ and therefore $\\angle L M N=180^{\\circ}-\\varphi-\\theta$. The quadrilaterals $C B K L$ and $C A K N$ are also inscribed and hence $\\angle L K C=\\angle L B C=\\theta$, $\\angle C K N=\\angle C A N=\\varphi$, and so $\\angle L K N=\\varphi+\\theta$, while $\\angle L M N=180^{\\circ}-\\varphi-\\theta$, which means that $K L M N$ is inscribed.\n\nNote that $K L M N$ is convex because $L, N$ lie in the interior of the convex quadrilateral $A D E B$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_ed83a632d3567b652a08g-10.jpg?height=1024&width=725&top_left_y=1274&top_left_x=767)\n\n## COMBINATORICS", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nG5.", "solution_match": "\nSolution."}}
+{"year": "2004", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "A polygon having $n$ sides is arbitrarily decomposed in triangles having all the vertices among the vertices of the polygon. We paint in black the triangles that have two sides that are also sides of the polygon, in red if only one side of the triangle is side of the polygon and white those triangles that have in common with the poligon only vertices.\n\nProve that there are 2 more black triangles than white ones.", "solution": "Denote by $b, r, w$ the number of black, red white triangles respectively.\n\nIt is easy to prove that the polygon is divided into $n-2$ triangles, hence\n\n$$\nb+r+w=n-2\n$$\n\nEach side of the polygon is a side of exactly one triangle of the decomposition, and thus\n\n$$\n2 b+r=n\n$$\n\nSubtracting the two relations yields $w=b-2$, as needed. allowed:", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nC1.", "solution_match": "\nSolution."}}
+{"year": "2004", "tier": "T3", "problem_label": "C2", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Given $m \\times n$ table, each cell signed with \"-\". The following operations are\n\n(i) to change all the signs in entire row to the opposite, i. e. every \"-\" to \"+\", and every \"+\" to \"-\";\n\n(ii) to change all the signs in entire column to the opposite, i. e. every \"-\" to \"+\" and every \"+\" to \" -\".\n\n(a) Prove that if $m=n=100$, using the above operations one can not obtain 2004 signs \"t\".\n\n(b) If $m=1004$, find the least $n>100$ for which 2004 signs \" + \" can be obtained.", "solution": "If we apply (i) to $l$ rows and (ii) to $k$ columns we obtain $(m-k) l+(n-l) k$\n\n(a) We have equation $(100-k) l+(100-l) k=2004$, or $100 l+100 k-2 l k=2004$, le\n\n$$\n50 l+50 k-1 k=1002\n$$\n\nRewrite the lasc equation as\n\n$$\n(50-l)(50-h)=2.500-100.2=1498\n$$\n\nSince $1498=2 \\cdot 7 \\cdot 107$, this equation has no solitions in natural numbers.\n\n(b) Let $n=101$. Then we have\n\n$$\n(100-k) l+(101-l) k=2004\n$$\n\nOI\n\n$$\n100 l+101 k-2 l k=2004\n$$\n\nl.e.\n\n$$\n101 k=2004-100 l+2 l k \\div 2(1002-50 l+l k)\n$$\n\nHence $s=2 t$ and we have $101 t=501-25 l+2 l t$. From here we have\n\n$$\nt=\\frac{501-25 l}{101-2 l}=4+\\frac{97-17 l}{101-2 l}\n$$\n\nSince $t$ is natural number and $97-17 l<101-2 l$, this is a contradiction, Hence $n \\neq 101$. Let $n=1.02$. Then we have\n\n$$\n(100-k) l+(102-l) k=2004\n$$\n\nor\n\n$$\n100 l+102 k-2 l k=2004\n$$\n\n$$\n50 l+51 k-l k=1002\n$$\n\nRewrite the last equation as\n\n$$\n(51-l)(50-k)=25.50-1002=1.548\n$$\n\nSince $145 S=2 \\cdot 2 \\cdot 3 \\cdot 3 \\cdot 43$ we have $51-l=36$ and $50-k=43$. From here obtain $l=15$ and $k=7$. Indeed,\n\n$$\n(100-\\bar{\\imath}) \\cdot 15+(102-1.5) \\cdot \\overline{7}=93 \\cdot 15+87 \\cdot 7=1395+609=2004\n$$\n\nHence, the least $n$ is 102 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl", "problem_match": "\nC2.", "solution_match": "\nSolution."}}

JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl

@@ -1,21 +1,21 @@
-{"year": "2005", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $a, b, c, d, e$ be real numbers such that $a+b+c+d+e=0$. Let, also $A=a b+b c+c d+d e+e a$ and $B=a c+c e+e b+b d+d a$.\n\nShow that\n\n$$\n2005 A+B \\leq 0 \\text { or } \\quad A+2005 B \\leq 0\n$$", "solution": "We have\n\n$$\n0=(a+b+c+d+e)^{2}=a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+2 A+2 B\n$$\n\nThis implies that\n\n$$\nA+B \\leq 0 \\text { or } 2006(\\dot{A}+B)=(2005 A+B)+(A+2005 B) \\leq 0\n$$\n\nThis implies the conclusion.\n\n## Alternative solution\n\nWe haye\n\n$$\n\\begin{aligned}\n2 A+2 B & =a(b+c+d+e)+b(c+d+e+a)+c(d+e+a+b) \\\\\n& +d(e+a+b+c)+e(a+b+c+d) \\\\\n& =-a^{2}-b^{2}-c^{2}-d^{2}-e^{2} \\leq 0\n\\end{aligned}\n$$\n\nTherefore we have $A+B \\leq 0$, etc.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nA1.", "solution_match": "## Solution"}}
-{"year": "2005", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO", "problem": "Find all positive integers $x, y$ satisfying the equation\n\n$$\n9\\left(x^{2}+y^{2}+1\\right)+2(3 x y+2)=2005\n$$", "solution": "The given equation can be written into the form\n\n$$\n2(x+y)^{2}+(x-y)^{2}=664\n$$\n\nTherefore, both numbers $x+y$ and $x-y$ are even.\n\nLet $x+y=2 m$ and $x-y=2 t, t \\in \\mathbb{Z}$.\n\nNow from (1) we have that $t$ and $t^{2}$ are even and $m$ is odd.\n\nSo, if $t=2 k, k \\in \\mathbb{Z}$ and $m=2 n+1, n \\in \\mathbb{N}$, then from (1) we get\n\n$$\nk^{2}=41-2 n(n+1)\n$$\n\nThus $41-2 n(n+1) \\geq 0$ or $2 n^{2}+2 n-41 \\leq 0$. The last inequality is satisfied for the positive integers $n=1,2,3,4$ and for $n=0$.\n\nHowever, only for $n=4$, equation (2) gives a perfect square $k^{2}=1 \\Leftrightarrow k= \\pm 1$. Therefore the solutions are $(x, y)=(11,7)$ or $(x, y)=(7,11)$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nA2.", "solution_match": "## Solution"}}
-{"year": "2005", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO", "problem": "Find the maximum value of the area of a triangle having side lengths $a, b, c$ with\n\n$$\na^{2}+b^{2}+c^{2}=a^{3}+b^{3}+c^{3}\n$$", "solution": "Without any loss of generality, we may assume that $a \\leq b \\leq c$.\n\nOn the one hand, Tchebyshev's inequality gives\n\n$$\n(a+b+c)\\left(a^{2}+b^{2}+c^{2}\\right) \\leq 3\\left(a^{3}+b^{3}+c^{3}\\right)\n$$\n\nTherefore using the given equation we get\n\n$$\na+b+c \\leq 3 \\text { or } p \\leq \\frac{3}{2}\n$$\n\nwhere $p$ denotes the semi perimeter of the triangle.\n\nOn the other hand,\n\n$$\np=(p-a)+(p-b)+(p-c) \\geq 3 \\sqrt[3]{(p-a)(p-b)(p-c)}\n$$\n\nHence\n\n$$\n\\begin{aligned}\np^{3} \\geq 27(p-a)(p-b)(p-c) & \\Leftrightarrow p^{4} \\geq 27 p(p-a)(p-b)(p-c) \\\\\n& \\Leftrightarrow p^{2} \\geq 3 \\sqrt{3} \\cdot S\n\\end{aligned}\n$$\n\nwhere $S$ is the area of the triangle.\n\nThus $S \\leq \\frac{\\sqrt{3}}{4}$ and equality holds whenever when $a=b=c=1$.\n\n## Comment\n\nCauchy's inequality implies the following two inequalities are true:\n\n$$\n\\frac{a+b+c}{3} \\leq \\frac{a^{2}+b^{2}+c^{2}}{a+b+c} \\leq \\frac{a^{3}+b^{3}+c^{3}}{a^{2}+b^{2}+c^{2}}\n$$\n\nNow note that\n\n$$\n\\frac{a+b+c}{3} \\leq \\frac{a^{2}+b^{2}+c^{2}}{a+b+c}\n$$\n\ngives\n\n$$\n(a+b+c)^{2} \\leq 3\\left(a^{2}+b^{2}+c^{2}\\right)\n$$\n\nwhereas $\\frac{a^{2}+b^{2}+c^{2}}{a+b+c} \\leq \\frac{a^{3}+b^{3}+c^{3}}{a^{2}+b^{2}+c^{2}}$, because of our assumptions, becomes $\\frac{a^{2}+b^{2}+c^{2}}{a+b+c} \\leq 1$, and so,\n\n$$\na^{2}+b^{2}+c^{2} \\leq a+b+c\n$$\n\nCombining (1) and (2) we get\n\n$(a+b+c)^{2} \\leq 3(a+b+c)$ and then $a+b+c \\leq 3$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nA3.", "solution_match": "## Solution"}}
-{"year": "2005", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "exam": "JBMO", "problem": "Find all the integer solutions of the equation\n\n$$\n9 x^{2} y^{2}+9 x y^{2}+6 x^{2} y+18 x y+x^{2}+2 y^{2}+5 x+7 y+6=0\n$$", "solution": "The equation is equivalent to the following one\n\n$$\n\\begin{aligned}\n& \\left(9 y^{2}+6 y+1\\right) x^{2}+\\left(9 y^{2}+18 y+5\\right) x+2 y^{2}+7 y++6=0 \\\\\n& \\Leftrightarrow(3 y+1)^{2}\\left(x^{2}+x\\right)+4(3 y+1) x+2 y^{2}+7 y+6=0\n\\end{aligned}\n$$\n\nTherefore $3 y+1$ must divide $2 y^{2}+7 y+6$ and so it must also divide\n\n$$\n9\\left(2 y^{2}+7 y+6\\right)=18 y^{2}+63 y+54=2(3 y+1)^{2}+17(3 y+1)+35\n$$\n\nfrom which it follows that it must divide 35 as well. Since $3 y+1 \\in \\mathbb{Z}$ we conclude that $y \\in\\{0,-2,2,-12\\}$ and it is easy now to get all the solutions $(-2,0),(-3,0),(0,-2),(-1,2)$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nA4.", "solution_match": "## Solution"}}
-{"year": "2005", "tier": "T3", "problem_label": "A5", "problem_type": "Algebra", "exam": "JBMO", "problem": "Solve the equation\n\n$$\n8 x^{3}+8 x^{2} y+8 x y^{2}+8 y^{3}=15\\left(x^{2}+y^{2}+x y+1\\right)\n$$\n\nin the set of integers.", "solution": "We transform the equation to the following one\n\n$$\n\\left(x^{2}+y^{2}\\right)(8 x+8 y-15)=15(x y+1)\n$$\n\nSince the right side is divisible by 3 , then $3 /\\left(x^{2}+y^{2}\\right)(8 x+8 y-15)$. But if $3 /\\left(x^{2}+y^{2}\\right)$, then $3 / x$ and $3 / y, 009$ will wive $15(x y+1)$ and $3 /(x y+1)$, which is impossible. Hence $3 /(x+y)$ and 3 does not divide $x$ or $y$. Without loss of generality we can assume that $x=3 a+1$ and $y=3 b+2$. Substituting in the equation, we obtain\n\n$$\n\\left(x^{2}+y^{2}\\right)(8(a+b)+3)=5(x y+1)\n$$\n\nSince $x y+1 \\equiv 0(\\bmod 3)$, we conclude that $3 /(a+b)$.\n\nNow we distinguish the following cases:\n\n- If $a+b=0$, then $x=3 a+1$ and $y=-3 a+2$ from which we get\n\n$$\n\\left(9 a^{2}+6 a+1+9 a^{2}-12 a+4\\right) \\cdot 3=5\\left(-9 a^{2}+3 a+3\\right) \\text { or } 3 a^{2}-a=0\n$$\n\nBut $a=\\frac{1}{3}$ is not an integer, so $a=0$ and $x=1, y=2$. Thus, by symmetry, we have two solutions $(x, y)=(1,2)$ and $(x, y)=(2,1)$.\n\n- If $a+b \\neq 0$, then $|8(a+b)+3| \\geq 21$. So we obtain\n\n$$\n\\left|\\left(x^{2}+y^{2}\\right)(8(a+b)+3)\\right| \\geq 21 x^{2}+21 y^{2} \\geq|5 x y+5|\n$$\n\nwhich means that the equation has no other solutions.\n\n## Geometry", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nA5.", "solution_match": "## Solution"}}
-{"year": "2005", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C D$ be an isosceles trapezoid with $A B=A D=B C, A B / / D C, A B>D C$. Let $E$ be the point of intersection of the diagonals $A C$ and $B D$ and $N$ be the symmetric point of $\\mathrm{B}$ with respect to the line $\\mathrm{AC}$. Prove that quadrilateral $A N D E$ is cyclic.", "solution": "Let $\\omega$ be a circle passing through the points $A, N, D$ and let $M$ the point where $\\omega$ intersects $B D$ for the second time. The quadrilateral $A N D M$ is cyclic and it follows that\n\n$$\n\\angle N D M+\\angle N A M=\\angle N D M+\\angle B D C=180^{\\circ}\n$$\n\nand\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-05.jpg?height=681&width=852&top_left_y=1043&top_left_x=658)\n\nFigure 1\n\n$$\n\\angle N A M=\\angle B D C\n$$\n\nNow we have\n\n$$\n\\angle B D C=\\angle A C D=\\angle N A C\n$$\n\nand\n\n$$\n\\angle N A M=\\angle N A C\n$$\n\nSo the points $A, M, C$ are collinear and $M \\equiv E$.\n\n## Alternative solution\n\nIn this solution we do not need the circle passing through the points $A, N$ and $D$.\n\nBecause of the given symmetry we have\n\n$$\n\\angle A N E=\\angle A B D\n$$\n\nand from the equality $\\mathrm{AD}=\\mathrm{AB}$ the triangle $\\mathrm{ABD}$ is isosceles with\n\n$$\n\\angle A B D=\\angle A D E\n$$\n\nFrom (1) and (2) we get that $\\angle A N E=\\angle A D E$, which means that the quadrilateral $A N D E$ is cyclic.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nG1.", "solution_match": "## Solution"}}
-{"year": "2005", "tier": "T3", "problem_label": "G2", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C$ be a triangle inscribed in a circle $K$. The tangent from $A$ to the circle meets the line $B C$ at point $P$. Let $M$ be the midpoint of the line segment $A P$ and let $R$ be the intersection point of the circle $K$ with the line $B M$. The line $P R$ meets again the circle $K$ at the point $S$. Prove that the lines $A P$ and $C S$ are parallel.", "solution": "![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-06.jpg?height=678&width=1500&top_left_y=531&top_left_x=311)\n\nFigure 2\n\nAssume that point $C$ lies on the line segment $B P$. By the Power of Point theorem we have $M A^{2}=M R \\cdot M B$ and so $M P^{2}=M R \\cdot M B$. The last equality implies that the triangles $M R$ and $M P B$ are similar. Hence $\\angle M P R=\\angle M B P$ and since $\\angle P S C=\\angle M B P$, the claim is proved.\n\nSlight changes are to be made if the point $B$ lies on the line segment $P C$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-06.jpg?height=687&width=1156&top_left_y=1659&top_left_x=329)\n\nFigure 3", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nG2.", "solution_match": "## Solution"}}
-{"year": "2005", "tier": "T3", "problem_label": "G3", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C D E F$ be a regular hexagon. The points $\\mathrm{M}$ and $\\mathrm{N}$ are internal points of the sides $\\mathrm{DE}$ and $\\mathrm{DC}$ respectively, such that $\\angle A M N=90^{\\circ}$ and $A N=\\sqrt{2} \\cdot C M$. Find the measure of the angle $\\angle B A M$.", "solution": "Since $A C \\perp C D$ and $A M \\perp M N$ the quadrilateral $A M N C$ is inscribed. So, we have\n\n$$\n\\angle M A N=\\angle M C N\n$$\n\nLet $P$ be the projection of the point $M$ on the line $C D$. The triangles $A M N$ and $C P M$ are similar implying\n\n$$\n\\frac{A M}{C P}=\\frac{M N}{P M}=\\frac{A N}{C M}=\\sqrt{2}\n$$\n\nSo, we have\n\n$$\n\\frac{M P}{M N}=\\frac{1}{\\sqrt{2}} \\Rightarrow \\angle M N P=45^{\\circ}\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-07.jpg?height=679&width=830&top_left_y=1110&top_left_x=606)\n\nFigure 4\n\nHence we have\n\n$$\n\\angle C A M=\\angle M N P=45^{\\circ}\n$$\n\nand finally, we obtain\n\n$$\n\\angle B A M=\\angle B A C+\\angle C A M=75^{\\circ}\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nG3.", "solution_match": "## Solution"}}
-{"year": "2005", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $\\mathrm{ABC}$ be an isosceles triangle such that $A B=A C$ and $\\angle \\frac{A}{2}<\\angle B$. On the extension of the altitude $\\mathrm{AM}$ we get the points $\\mathrm{D}$ and $\\mathrm{Z}$ such that $\\angle C B D=\\angle A$ and $\\angle Z B A=90^{\\circ}$. $\\mathrm{E}$ is the foot of the perpendicular from $\\mathrm{M}$ to the altitude $\\mathrm{BF}$ and $\\mathrm{K}$ is the foot of the perpendicular from $\\mathrm{Z}$ to $\\mathrm{AE}$. Prove that $\\angle K D Z=\\angle K B D=\\angle K Z B$.", "solution": "The points $A, B, K, Z$ and $C$ are co-cyclic.\n\nBecause ME//AC so we have\n\n$$\n\\angle K E M=\\angle E A C=\\angle M B K\n$$\n\nTherefore the points $B, K, M$ and $E$ are co-cyclic. Now, we have\n\n$$\n\\begin{aligned}\n& \\angle A B F=\\angle A B C-\\angle F B C \\\\\n& =\\angle A K C-\\angle E K M=\\angle M K C\n\\end{aligned}\n$$\n\nAlso, we have\n\n$$\n\\begin{aligned}\n& \\angle A B F=90^{\\circ}-\\angle B A F=90^{\\circ}-\\angle M B D \\\\\n& =\\angle B D M=\\angle M D C\n\\end{aligned}\n$$\n\nFrom (1) and (2) we get $\\angle M K C=\\angle M D C$ and so the points $M, K, D$ and $C$ are co-cyclic.\n\nConsequently,\n\n$$\n\\angle K D M=\\angle K C M=\\angle B A K=\\angle B Z K \\text {, }\n$$\n\nand because the line $\\mathrm{BD}$ is tangent to the circumcircle of triangle $A B C$, we have\n\n$$\n\\angle K B D=\\angle B A K\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-08.jpg?height=950&width=562&top_left_y=501&top_left_x=1219)\n\nFigure 5\n\nFinally, we have\n\n$$\n\\angle K D Z=\\angle K B D=\\angle K Z B\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nG4.", "solution_match": "## Solution"}}
-{"year": "2005", "tier": "T3", "problem_label": "G5", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A$ and $P$ are the points of intersection of the circles $k_{1}$ and $k_{2}$ with centers $O$ and $K$, respectively. Let also $B$ and $C$ be the symmetric points of $A$ with respect to $O$ and $K$, respectively. A line through $A$ intersects the circles $k_{1}$ and $k_{2}$ at the points $D$ and $E$, respectively. Prove that the centre of the circumcircle of the triangle $D E P$ lies on the circumcircle $O K P$.", "solution": "The points $B, P, C$ are collinear, and\n\n$$\n\\angle A P C=\\angle A P B=90^{\\circ}\n$$\n\nLet $N$ be the midpoint of $D P$.\n\nSo we have:\n\n$$\n\\begin{aligned}\n& \\angle N O P=\\angle D A P \\\\\n& =\\angle E C P=\\angle E C A+\\angle A C P\n\\end{aligned}\n$$\n\nSince $O K / / B C$ and $O K$ is the bisector of $\\angle A K P$ we get\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-09.jpg?height=579&width=836&top_left_y=578&top_left_x=923)\n\nFigure 6\n\n$$\n\\angle A C P=O K P\n$$\n\nAlso, since $A P \\perp O K$ and $M K \\perp P E$ we have that\n\n$$\n\\angle A P E=\\angle M K O\n$$\n\nThe points $A, E, C, P$ are co-cyclic, and so $\\angle E C A=\\angle A P E$.\n\nTherefore, from (1), (2) and (3) we have that $\\angle N O P=\\angle M K P$.\n\nThus $O, M, K$ and $P$ are co-cyclic.\n\n## Comment\n\nPoints B and C may not be included in the statement of the problem\n\n## Alternative solution\n\nIt is sufficient to prove that the quadrilateral $M O P K$ is circumscrible.\n\nSince $M O$ and $M K$ are perpendicular bisectors of the line segments $P D$ and $P E$, respectively, we have\n\nTherefore the quadrilateral $M O P K$ is circumscrible.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-10.jpg?height=1198&width=1196&top_left_y=716&top_left_x=292)\n\nFigure 7", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nG5.", "solution_match": "## Solution"}}
-{"year": "2005", "tier": "T3", "problem_label": "G6", "problem_type": "Geometry", "exam": "JBMO", "problem": "A point $O$ and the circles $k_{1}$ with center $O$ and radius $3, k_{2}$ with center $O$ and radius 5, are given. Let $A$ be a point on $k_{1}$ and $B$ be a point on $k_{2}$. If $A B C$ is equilateral triangle, find the maximum value of the distance $O C$.", "solution": "## Alternative solution 1 (by the proposer)\n\nLet $\\phi$ be a $60^{\\circ}$ rotation with center at $B$. Then $\\phi(A)=C, \\phi(O)=P$ and $P C=O A$, $O P=O$, etc.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nG6.", "solution_match": "## Solution"}}
-{"year": "2005", "tier": "T3", "problem_label": "G6", "problem_type": "Geometry", "exam": "JBMO", "problem": "A point $O$ and the circles $k_{1}$ with center $O$ and radius $3, k_{2}$ with center $O$ and radius 5, are given. Let $A$ be a point on $k_{1}$ and $B$ be a point on $k_{2}$. If $A B C$ is equilateral triangle, find the maximum value of the distance $O C$.", "solution": "## Alternative solution 2\n\nLet $O A=x, A B=B C=C A=a$. From the second theorem of Ptolemy we get\n\n$$\nO A \\cdot B C+O B \\cdot A C \\geq A B \\cdot O C \\Leftrightarrow 3 a+5 a=a x \\Leftrightarrow x \\leq 8\n$$\n\nThe value $x=8$ is attained when the quadrilateral OACB is circumscrible, i.e. when $\\angle A O B=120^{\\circ}$.\n\nThe point $\\mathrm{B}$ can be constructed as follows:\n\nIt is the point of intersection of the circle $(0,5)$ with the ray coming from the rotation of the ray $\\mathrm{OA}$ with center $\\mathrm{O}$ by an angle $\\theta=-120^{\\circ}$, (figure 10).\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-12.jpg?height=1047&width=807&top_left_y=914&top_left_x=652)\n\nFigure 10", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nG6.", "solution_match": "## Solution"}}
-{"year": "2005", "tier": "T3", "problem_label": "G7", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C D$ be a parallelogram, $\\mathrm{P}$ a point on $C D$, and $Q$ a point on $A B$. Let also $M=A P \\cap D Q, \\quad N=B P \\cap C Q, K=M N \\cap A D$, and $L=M N \\cap B C$. Show that $B L=D K$.", "solution": "Let $O$ be the intersection of the diagonals. Let $P_{1}$ be on $A B$ such that $P P_{1} / / A D$, and let $Q_{1}$ be on $C D$ such that $\\mathrm{Q} Q_{1} / / A D$. Let $\\sigma$ be the central symmetry with center $\\mathrm{O}$. Let $\\left.P^{\\prime}=\\sigma(P), Q^{\\prime}=\\sigma(Q), P_{1}^{\\prime}=\\sigma\\left(P_{1}\\right)\\right)$ and, (figure 1).\n\nLet $M_{1}=A Q_{1} \\cap D P_{1}, N_{1}=B Q_{1} \\cap C P_{1}, N^{\\prime}=A Q^{\\prime} \\cap D P^{\\prime}$ and $M^{\\prime}=B Q^{\\prime} \\cap C P^{\\prime}$.\n\nThen: $M^{\\prime}=\\sigma(M), N^{\\prime}=\\sigma(N), M_{1}^{\\prime}=\\sigma\\left(M_{1}\\right)$ and $N_{1}^{\\prime}=\\sigma\\left(N_{1}\\right)$.\n\nSince $A P$ and $D P_{1}$ are the diagonals of the parallelogram $A P_{1} P D, C P_{1}$ and $B P$ are the diagonals of the parallelogram $P_{1} B C P$, and $A Q_{1}$ and $D O$ are the diagonals of the parallelogram $A Q Q_{1} D$, it follows that the points $U, V, W$ (figure 2) are collinear and they lie on the line passing through the midpoints $R$ of $A D$ and $Z$ of $B C$. The diagonals AM and $D M_{1}$ the quadrilateral $A M_{1} M D$ intersect at $U$ and the diagonals $A M_{1}$ and - $D M$ intersect at $W$. Since the midpoint of $A D$ is on the line $U W$, it follows that the quadrilateral $A M_{1} M D$ is a trapezoid. Hence, $M M_{1}$ is parallel to $A D$ and the midpoint $S$ of $M M_{1}$ lies on the line $U W$, (figure 2).\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-13.jpg?height=724&width=1446&top_left_y=1344&top_left_x=315)\n\nFigure 11\n\nSimilarly $M^{\\prime} M_{1}^{\\prime}$ is parallel to $A D$ and its midpoint lies on $U W$. So $M_{1} M^{\\prime} M_{1}^{\\prime} M$ is a parallelogram whose diagonals intersect at $\\mathrm{O}$.\n\nSimilarly, $N_{1}^{\\prime} N N_{1} N^{\\prime}$ is a parallelogram whose diagonals intersect at $O$.\n\nAll these imply that $M, N, M^{\\prime}, N^{\\prime}$ and $O$ are collinear, i.e. $O$ lies on the line $K L$. This implies that $K=\\sigma(L)$, and since $D=\\sigma(B)$, the conclusion follows.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-14.jpg?height=556&width=1504&top_left_y=224&top_left_x=252)\n\nFigure 12\n\n## Alternative solution\n\nLet the line ( $\\varepsilon$ ) through the points $\\mathrm{M}, \\mathrm{N}$ intersect the lines $\\mathrm{DC}, \\mathrm{AB}$ at points $\\mathrm{T}_{1}, \\mathrm{~T}_{2}$ respectively. Applying Menelaus Theorem to the triangles DQC, APB with intersecting line $(\\varepsilon)$ in both cases we get:\n\n$$\n\\frac{M D}{M Q} \\frac{N Q}{N C} \\frac{T_{1} C}{T_{1} D}=1 \\text { and } \\frac{M P}{M A} \\frac{N B}{N P} \\frac{T_{2} A}{T_{2} B}=1\n$$\n\nBut it is true that $\\frac{\\mathrm{MD}}{\\mathrm{MQ}}=\\frac{\\mathrm{MP}}{\\mathrm{MA}}$ and $\\frac{\\mathrm{NQ}}{\\mathrm{NC}}=\\frac{\\mathrm{NB}}{\\mathrm{NP}}$.\n\nIt follows that $\\frac{T_{1} C}{T_{1} D}=\\frac{T_{2} A}{T_{2} B}$ i.e. $\\frac{T_{1} D+D C}{T_{1} D}=\\frac{T_{2} B+B A}{T_{2} B}$ hence $T_{1} D=T_{2} B$ (since\n\n$\\mathrm{DC}=\\mathrm{BA})$.\n\nThen of course by the similarity of the triangles $\\mathrm{T}_{1} \\mathrm{DK}$ and $\\mathrm{T}_{2} \\mathrm{BL}$ we get the desired equality $\\mathrm{DK} \\cdot \\mathrm{BL}$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-14.jpg?height=585&width=1106&top_left_y=1659&top_left_x=454)\n\nFigure 13\n\n## Number Theory", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nG7.", "solution_match": "## Solution"}}
-{"year": "2005", "tier": "T3", "problem_label": "NT1", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all the natural numbers $m$ and $n$, such that the square of $m$ minus the product of $n$ with $k$, is 2 , where the number $k$ is obtained from $n$ by writing 1 on the left of the decimal notation of $n$.", "solution": "Let $t$ be the number of digits of $n$. Then $k=10^{t}+n$. So\n\n$$\n\\mathrm{m}^{2}=n\\left(10^{t}+\\mathrm{n}\\right)+2, \\text { i.e. } \\mathrm{m}^{2}-\\mathrm{n}^{2}=10^{t} n+2\n$$\n\nThis implies that $\\mathrm{m}, \\mathrm{n}$ are even and both $\\mathrm{m}, \\mathrm{n}$ are odd.\n\nIf $t=1$, then, 4 is divisor of $10 n+2$, so, $n$ is odd. We check that the only solution in this case is $\\mathrm{m}=11$ and $\\mathrm{n}=7$.\n\nIf $t>1$, then 4 is divisor of $\\mathrm{m}^{2}-\\mathrm{n}^{2}$, but 4 is not divisor of $10^{t}+2$.\n\nHence the only solution is $\\mathrm{m}=11$ and $\\mathrm{n}=7$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nNT1.", "solution_match": "## Solution"}}
-{"year": "2005", "tier": "T3", "problem_label": "NT2", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all natural numbers $n$ such that $5^{n}+12^{n}$ is perfect square.", "solution": "By checking the cases $n=1,2,3$ we get the solution $n=2$ and $13^{2}=5^{2}+12^{2}$.\n\nIf $n=2 k+1$ is odd, we consider the equation modulo 5 and we obtain\n\n$$\n\\begin{aligned}\nx^{2} & \\equiv 5^{2 k+1}+12^{2 k+1}(\\bmod 5) \\equiv 2^{2 k} \\cdot 2(\\bmod 5) \\\\\n& \\equiv(-1)^{k} \\cdot 2(\\bmod 5) \\equiv \\pm 2(\\bmod 5)\n\\end{aligned}\n$$\n\nThis is not possible, because the square residue of any natural number module 5 is 0,1 or 4. Therefore $n$ is even and $x^{2}=5^{2 k}+12^{2 k}$. Rearrange this equation in the form\n\n$$\n5^{2 k}=\\left(x-12^{k}\\right)\\left(x+12^{k}\\right)\n$$\n\nIf 5 divides both factors on the right, it must also divide their difference, that is\n\n$$\n5 \\mid\\left(x+12^{k}\\right)-\\left(x-12^{k}\\right)=2 \\cdot 12^{k}\n$$\n\nwhich is not possible. Therefore we must have\n\n$$\nx-12^{k}=1 \\text { and } x+12^{k}=5^{2 k}\n$$\n\nBy adding the above equalities we get\n\n$$\n5^{2 k}-1=2 \\cdot 12^{k}\n$$\n\nFor $k \\geq 2$, we have the inequality\n\n$$\n25^{k}-1>24^{k}=2^{k} \\cdot 12^{k}>2 \\cdot 12^{k}\n$$\n\nThus we conclude that there exists a unique solution to our problem, namely $n=2$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nNT2.", "solution_match": "## Solution"}}
-{"year": "2005", "tier": "T3", "problem_label": "NT3", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Let $p$ be an odd prime. Prove that $p$ divides the integer\n\n$$\n\\frac{2^{p!}-1}{2^{k}-1}\n$$\n\nfor all integers $k=1,2, \\ldots, p$.", "solution": "At first, note that $\\frac{2^{p!}-1}{2^{k}-1}$ is indeed an integer.\n\nWe start with the case $\\mathrm{k}=\\mathrm{p}$. Since $p \\mid 2^{p}-2$, then $p / 22^{p}-1$ and so it suffices to prove that $p \\mid 2^{(p)!}-1$. This is obvious as $p \\mid 2^{p-1}-1$ and $\\left(2^{p-1}-1\\right) \\mid 2^{(p)!}-1$.\n\nIf $\\mathrm{k}=1,2, \\ldots, \\mathrm{p}-1$, let $m=\\frac{(p-1)!}{k} \\in \\mathbb{N}$ and observe that $p!=k m p$. Consider $a \\in \\mathbb{N}$ so that $p^{a} \\mid 2^{k}-1$ and observe that it suffices to prove $p^{a+1} \\mid 2^{p!}-1$. The case $a=0$ is solved as the case $k=p$. If else, write $2^{k}=1+p^{a} \\cdot l, l \\in \\mathbb{N}$ and rising at the power mp gives\n\n$$\n2^{p!}=\\left(1+p^{a} \\cdot l\\right)^{m p}=1+m p \\cdot p^{a} \\cdot l+M p^{2 a}\n$$\n\nwhere $M n$ stands for a multiply of $\\mathrm{n}$. Now it is clear that $p^{a+1} \\mid 2^{p!}-1$, as claimed.\n\nComment. The case $\\mathrm{k}=\\mathrm{p}$ can be included in the case $a=0$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nNT3.", "solution_match": "## Solution"}}
-{"year": "2005", "tier": "T3", "problem_label": "NT4", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all the three digit numbers $\\overline{a b c}$ such that\n\n$$\n\\overline{a b c}=a b c(a+b+c)\n$$", "solution": "We will show that the only solutions are 135 and 144 .\n\nWe have $a>0, b>0, c>0$ and\n\n$$\n9(11 a+b)=(a+b+c)(a b c-1)\n$$\n\n- If $a+b+c \\equiv 0(\\bmod 3)$ and $a b c-1 \\equiv 0(\\bmod 3)$, then $a \\equiv b \\equiv c \\equiv 1(\\bmod 3)$ and $11 a+b \\equiv 0(\\bmod 3)$. It follows now that\n\n$$\na+b+c \\equiv 0(\\bmod 9) ; \\text { or } a b c-1 \\equiv 0(\\bmod 9)\n$$\n\n- If . $a b c-1 \\equiv 0(\\bmod 9)$\n\nwe have $11 a+b=(a+b+c) k$, where $k$ is an integer\n\nand is easy to see that we must have $1<k \\leq 10$.\n\nSo we must deal only with the cases $k=2,3,4,5,6,7,8,9,10$\n\nIf $k=2.4,5,6,8,9,10$ then $9 k+1$ has prime divisors greater than 9 ,so we must see only the cases $\\quad k=3,7$.\n\n- If $k=3$ we have\n\n$$\n8 a=2 b+3 c \\text { and } \\quad a b c=28\n$$\n\nIt is clear that $\\mathrm{c}$ is even, $c=2 c_{1}$ and that both $b$ and $c_{1}$ are odd.\n\nFrem $a b c_{1}=14$ it follows that $a=2, b=7, c_{l}=1$ or $a=2, b=1, c_{i}=7$ and it is clear that there exists no solution in this case. - If $k=7$ we have $c$-even, $c=2 c_{k}, 2 a=3 b+7 c_{1}, a b c_{1}=32$, then both $b$ and $c_{I}$ are\neven.\n\nBut this is impossible, because they imply that $a>9$.\n\nNow we will deal with the case when $a+b+c \\equiv 0(\\bmod 9)$ or $a+b+c=9 l$, where $l$ is an integer.\n\n- If $l \\geq 2$ we have $a+b+c \\geq 18, \\max \\{a, b, c\\} \\geq 6$ and it is easy to see that $a b c \\geq 72$ and $a b c(a+b+c)>1000$,so the case $l \\geq 2$ is impossible.\n- If $l=1$ we have\n\n$$\n11 a+b=a b c-1 \\text { or } 11 a+b+1=a b c \\leq\\left(\\frac{a+b+c}{3}\\right)^{3}=27\n$$\n\nSo we have only two cases: $a=1$ or $a=2$.\n\n- If $a=1$, we have $b+c=8$ and $11+b=b c-1$ or $b+(c-1)=7$ and $b(c-1)=12$ and the solutions are $(a, b, c)=(1,3,5)$ and $(a, b, c)=(1,4,4)$, and the answer is 135 and 144.\n- If $a=2$ we have $b(2 c-1)=23$ and there is no solution for the problem.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nNT4.", "solution_match": "## Solution"}}
-{"year": "2005", "tier": "T3", "problem_label": "NT5", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Let $p$ be a prime number and let $a$ be an integer. Show that if $n^{2}-5$ is not divisible by $p$ for any integer $n$, there exist infinitely many integers $m$ so that $p$ divides $m^{5}+a$.", "solution": "We start with a simple fact:\n\nLemma: If $b$ is an integer not divisible by $p$ then there is an integer $s$ so that $s b$ has the remainder $l$ when divided by $p$.\n\nFor a proof, just note that numbers $b, 2 b, \\ldots,(p-1) b$ have distinct non-zero remainders when divided by $p$, and hence one of them is equal to 1 .\n\nWe prove that if $x, y=0,1,2, \\ldots, p-1$ and $\\mathrm{p}$ divides $x^{5}-y^{5}$, then $x=y$.\n\nIndeed, assume that $x \\neq y$. If $x=0$, then $p \\mid y^{5}$ and so $y=0$, a contradiction.\n\nTo this point we have $x, y \\neq 0$. Since\n\n$$\np \\mid(x-y)\\left(x^{4}+x^{3} y+x^{2} y^{2}+x y^{3}+y^{4}\\right) \\text { and } p /(x-y)\n$$\n\nwe have\n\n$$\n\\begin{aligned}\n& p l\\left(x^{2}+y^{2}\\right)^{2}+x y\\left(x^{2}+y^{2}\\right)-x^{2} y^{2} \\text {, and so } \\\\\n& p \\|\\left(2\\left(x^{2}+y^{2}\\right)+x y\\right)^{2}-5 x^{2} y^{2}\n\\end{aligned}\n$$\n\nAs $p / x y$, from the lemma we find an integer $s$ so that $s x y=k p+1, k \\in \\mathbb{N}$. Then\n\n$$\np \\mid\\left[s\\left(2 x^{2}+2 y^{2}+x y\\right)\\right]^{2}-5\\left(k^{2} p^{2}+2 k p+1\\right)\n$$\n\nand so $p \\mid z^{2}-5$, where $z=s\\left(2 x^{2}+2 y^{2}+x y\\right)$, a contradiction.\n\nConsequeatly $r=y$.\n\nSince we have proved that numbers $0^{5}, 1^{5}, \\ldots,(p-1)^{5}$ have distinct remainders when divided by $p$, the same goes for the numbers $0^{5}+a, 1^{5}+a, \\ldots,(p-1)^{5}+a$ and the conclusion can be reached easily.\n\n## Comments\n\n1. For beauty we may choose $a=-2$ or any other value.\n2. Moreover, we may ask only for one value of $m$, instead of \"infinitely many\".\n3. A simple version will be to ask for a proof that the numbers $0^{5}, 1^{5}, \\ldots,(p-1)^{5}$ have distinct remainders when divided by $p$.\n\n## Combinatorics", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nNT5.", "solution_match": "## Solution"}}
-{"year": "2005", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "A triangle with area 2003 is divided into non-overlapping small triangles. The number of all the vertices of all those triangles is 2005 . Show that at mest one of the smaller triangles has area less or equal to 1.", "solution": "Since all the vertices are 2005 , and the vertices of the big triangle are among them, it follows that the number of the small triangles is at least 2003. So, it follows that at least one of the small triangles has area at most 1", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nC1.", "solution_match": "## Solution"}}
-{"year": "2005", "tier": "T3", "problem_label": "C2", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Government of Greece wants to change the building. So they will move in building with square shape $2005 \\times 2005$. Only demand is that every room has exactly 2 doors and doors cannot lead outside the building. Is this possible? What about a building $2004 \\times 2005$ ?", "solution": "The key idea is to color the table $2005 \\times 2005$ in two colors, like chessboard. One door connects two adjacent squares, so they have different colors. We can count the number of doors in two ways. The number of doors is twice the number of white squares and it is also twice the number of black squares. This is, of course, Ipossible, because we have one more black square on the table. So, if $n$ is odd, we cannot achieve that every room has two doors.\n\nFor table $2004 \\times 2005$ this is possible, and one of many solutions is presented on the picture below.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-19.jpg?height=710&width=761&top_left_y=1662&top_left_x=612)\n\nFigure 14\n\n## Alternative statement of the problem\n\nIs it possible to open holes on two distinct sides of each cell of a $2005 \\times 2005$ square grid so that no hole is on the perimeter of the grid? What about a $2004 \\times 2005$ grid?\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=579&width=514&top_left_y=703&top_left_x=673)\n\n$$\n\\cdot \\varsigma 00 Z<8 \\downarrow 0 Z=(9 \\mathrm{I}) \\mathrm{X}=(\\varsigma 9) \\Psi\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=80&width=1236&top_left_y=1361&top_left_x=436)\n\n$$\n\\tau_{2} w 8=(\\mathfrak{I}-w \\tau) 8+{ }_{\\tau}(I-u) 8=\\tau \\cdot(I-w) \\nabla+(I-w) \\tau \\cdot \\nabla+8+(\\tau-u) X=(u) Y\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=60&width=251&top_left_y=1550&top_left_x=1432)\n\n$$\n\\dot{r}_{\\tau} \\varepsilon \\cdot 8=8 \\cdot 6=\\tau \\cdot 8+\\tau \\cdot \\tau \\cdot \\nabla+8+(\\tau) X=\\underset{\\partial \\Lambda E Y}{(\\varepsilon) Y}\n$$\n\n$$\n{ }^{2} \\tau \\cdot 8=8 \\cdot t=\\tau \\cdot \\downarrow+\\tau \\cdot \\downarrow+8+\\text { (I) } y=\\text { (乙) } y\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=77&width=1015&top_left_y=1746&top_left_x=665)\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=80&width=1426&top_left_y=1803&top_left_x=254)\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=94&width=1443&top_left_y=1846&top_left_x=240)\n\n$$\n8=(\\mathrm{L}) X\n$$\n\n:әлеч $\\partial M$ ঈ uo sұutod\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=94&width=1409&top_left_y=1988&top_left_x=277)\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=82&width=953&top_left_y=2037&top_left_x=730)\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=77&width=1472&top_left_y=2094&top_left_x=208)\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=105&width=1484&top_left_y=2131&top_left_x=208)\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=83&width=975&top_left_y=2179&top_left_x=719)\nuo!n [0S\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=73&width=859&top_left_y=2328&top_left_x=780)\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=63&width=739&top_left_y=2379&top_left_x=891)\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=57&width=368&top_left_y=2428&top_left_x=1254)\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=80&width=1420&top_left_y=2476&top_left_x=277)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nC2.", "solution_match": "## Solution"}}
-{"year": "2005", "tier": "T3", "problem_label": "C4", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Let $p_{1}, p_{2}, \\ldots, p_{2005}$ be different prime numbers. Let $\\mathrm{S}$ be a set of natural numbers which elements have the property that their simple divisors are some of the numbers $p_{1}, p_{2}, \\ldots, p_{2005}$ and product of any two elements from $\\mathrm{S}$ is not perfect square.\n\nWhat is the maximum number of elements in $\\mathrm{S}$ ?", "solution": "Let $a, b$ be two arbitrary numbers from $\\mathrm{S}$. They can be written as\n\n$$\na=p_{1}^{a_{1}} p_{2}^{a_{2}} \\cdots p_{2005}^{a_{2005}} \\text { and } b=p_{1}^{\\beta_{1}} p_{2}^{\\beta_{2}} \\cdots p_{2005}^{\\beta_{2005}}\n$$\n\nIn order for the product of the elements $a$ and $b$ to be a square all the sums of the corresponding exponents need to be even from where we can conclude that for every $i, a_{i}$ and $\\beta_{i}$ have the same parity. If we replace all exponents of $\\mathrm{a}$ and $b$ by their remainders modulo 2 , then we get two numbers $a^{\\prime}, b^{\\prime}$ whose product is a perfect square if and only if ab is a perfect square.\n\nIn order for the product $a^{\\prime} b^{\\prime}$ not to be a perfect square, at least one pair of the corresponding exponents modulo 2, need to be of opposite parity.\n\nSince we form 2005 such pairs modulo 2, and each number in these pairs is 1 or 2 , we conclude that we can obtain $2^{2005}$ distinct products none of which is a perfect square.\n\nNow if we are given $2^{2005}+1$ numbers, thanks to Dirichlet's principle, there are at least two with the same sequence of modulo 2 exponents, thus giving a product equal to a square.\n\nSo, the maximal number of the elements of $S$ is $2^{2005}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nC4.", "solution_match": "## Solution"}}
+{"year": "2005", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $a, b, c, d, e$ be real numbers such that $a+b+c+d+e=0$. Let, also $A=a b+b c+c d+d e+e a$ and $B=a c+c e+e b+b d+d a$.\n\nShow that\n\n$$\n2005 A+B \\leq 0 \\text { or } \\quad A+2005 B \\leq 0\n$$", "solution": "We have\n\n$$\n0=(a+b+c+d+e)^{2}=a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+2 A+2 B\n$$\n\nThis implies that\n\n$$\nA+B \\leq 0 \\text { or } 2006(\\dot{A}+B)=(2005 A+B)+(A+2005 B) \\leq 0\n$$\n\nThis implies the conclusion.\n\n## Alternative solution\n\nWe haye\n\n$$\n\\begin{aligned}\n2 A+2 B & =a(b+c+d+e)+b(c+d+e+a)+c(d+e+a+b) \\\\\n& +d(e+a+b+c)+e(a+b+c+d) \\\\\n& =-a^{2}-b^{2}-c^{2}-d^{2}-e^{2} \\leq 0\n\\end{aligned}\n$$\n\nTherefore we have $A+B \\leq 0$, etc.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nA1.", "solution_match": "## Solution"}}
+{"year": "2005", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Find all positive integers $x, y$ satisfying the equation\n\n$$\n9\\left(x^{2}+y^{2}+1\\right)+2(3 x y+2)=2005\n$$", "solution": "The given equation can be written into the form\n\n$$\n2(x+y)^{2}+(x-y)^{2}=664\n$$\n\nTherefore, both numbers $x+y$ and $x-y$ are even.\n\nLet $x+y=2 m$ and $x-y=2 t, t \\in \\mathbb{Z}$.\n\nNow from (1) we have that $t$ and $t^{2}$ are even and $m$ is odd.\n\nSo, if $t=2 k, k \\in \\mathbb{Z}$ and $m=2 n+1, n \\in \\mathbb{N}$, then from (1) we get\n\n$$\nk^{2}=41-2 n(n+1)\n$$\n\nThus $41-2 n(n+1) \\geq 0$ or $2 n^{2}+2 n-41 \\leq 0$. The last inequality is satisfied for the positive integers $n=1,2,3,4$ and for $n=0$.\n\nHowever, only for $n=4$, equation (2) gives a perfect square $k^{2}=1 \\Leftrightarrow k= \\pm 1$. Therefore the solutions are $(x, y)=(11,7)$ or $(x, y)=(7,11)$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nA2.", "solution_match": "## Solution"}}
+{"year": "2005", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Find the maximum value of the area of a triangle having side lengths $a, b, c$ with\n\n$$\na^{2}+b^{2}+c^{2}=a^{3}+b^{3}+c^{3}\n$$", "solution": "Without any loss of generality, we may assume that $a \\leq b \\leq c$.\n\nOn the one hand, Tchebyshev's inequality gives\n\n$$\n(a+b+c)\\left(a^{2}+b^{2}+c^{2}\\right) \\leq 3\\left(a^{3}+b^{3}+c^{3}\\right)\n$$\n\nTherefore using the given equation we get\n\n$$\na+b+c \\leq 3 \\text { or } p \\leq \\frac{3}{2}\n$$\n\nwhere $p$ denotes the semi perimeter of the triangle.\n\nOn the other hand,\n\n$$\np=(p-a)+(p-b)+(p-c) \\geq 3 \\sqrt[3]{(p-a)(p-b)(p-c)}\n$$\n\nHence\n\n$$\n\\begin{aligned}\np^{3} \\geq 27(p-a)(p-b)(p-c) & \\Leftrightarrow p^{4} \\geq 27 p(p-a)(p-b)(p-c) \\\\\n& \\Leftrightarrow p^{2} \\geq 3 \\sqrt{3} \\cdot S\n\\end{aligned}\n$$\n\nwhere $S$ is the area of the triangle.\n\nThus $S \\leq \\frac{\\sqrt{3}}{4}$ and equality holds whenever when $a=b=c=1$.\n\n## Comment\n\nCauchy's inequality implies the following two inequalities are true:\n\n$$\n\\frac{a+b+c}{3} \\leq \\frac{a^{2}+b^{2}+c^{2}}{a+b+c} \\leq \\frac{a^{3}+b^{3}+c^{3}}{a^{2}+b^{2}+c^{2}}\n$$\n\nNow note that\n\n$$\n\\frac{a+b+c}{3} \\leq \\frac{a^{2}+b^{2}+c^{2}}{a+b+c}\n$$\n\ngives\n\n$$\n(a+b+c)^{2} \\leq 3\\left(a^{2}+b^{2}+c^{2}\\right)\n$$\n\nwhereas $\\frac{a^{2}+b^{2}+c^{2}}{a+b+c} \\leq \\frac{a^{3}+b^{3}+c^{3}}{a^{2}+b^{2}+c^{2}}$, because of our assumptions, becomes $\\frac{a^{2}+b^{2}+c^{2}}{a+b+c} \\leq 1$, and so,\n\n$$\na^{2}+b^{2}+c^{2} \\leq a+b+c\n$$\n\nCombining (1) and (2) we get\n\n$(a+b+c)^{2} \\leq 3(a+b+c)$ and then $a+b+c \\leq 3$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nA3.", "solution_match": "## Solution"}}
+{"year": "2005", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Find all the integer solutions of the equation\n\n$$\n9 x^{2} y^{2}+9 x y^{2}+6 x^{2} y+18 x y+x^{2}+2 y^{2}+5 x+7 y+6=0\n$$", "solution": "The equation is equivalent to the following one\n\n$$\n\\begin{aligned}\n& \\left(9 y^{2}+6 y+1\\right) x^{2}+\\left(9 y^{2}+18 y+5\\right) x+2 y^{2}+7 y++6=0 \\\\\n& \\Leftrightarrow(3 y+1)^{2}\\left(x^{2}+x\\right)+4(3 y+1) x+2 y^{2}+7 y+6=0\n\\end{aligned}\n$$\n\nTherefore $3 y+1$ must divide $2 y^{2}+7 y+6$ and so it must also divide\n\n$$\n9\\left(2 y^{2}+7 y+6\\right)=18 y^{2}+63 y+54=2(3 y+1)^{2}+17(3 y+1)+35\n$$\n\nfrom which it follows that it must divide 35 as well. Since $3 y+1 \\in \\mathbb{Z}$ we conclude that $y \\in\\{0,-2,2,-12\\}$ and it is easy now to get all the solutions $(-2,0),(-3,0),(0,-2),(-1,2)$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nA4.", "solution_match": "## Solution"}}
+{"year": "2005", "tier": "T3", "problem_label": "A5", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Solve the equation\n\n$$\n8 x^{3}+8 x^{2} y+8 x y^{2}+8 y^{3}=15\\left(x^{2}+y^{2}+x y+1\\right)\n$$\n\nin the set of integers.", "solution": "We transform the equation to the following one\n\n$$\n\\left(x^{2}+y^{2}\\right)(8 x+8 y-15)=15(x y+1)\n$$\n\nSince the right side is divisible by 3 , then $3 /\\left(x^{2}+y^{2}\\right)(8 x+8 y-15)$. But if $3 /\\left(x^{2}+y^{2}\\right)$, then $3 / x$ and $3 / y, 009$ will wive $15(x y+1)$ and $3 /(x y+1)$, which is impossible. Hence $3 /(x+y)$ and 3 does not divide $x$ or $y$. Without loss of generality we can assume that $x=3 a+1$ and $y=3 b+2$. Substituting in the equation, we obtain\n\n$$\n\\left(x^{2}+y^{2}\\right)(8(a+b)+3)=5(x y+1)\n$$\n\nSince $x y+1 \\equiv 0(\\bmod 3)$, we conclude that $3 /(a+b)$.\n\nNow we distinguish the following cases:\n\n- If $a+b=0$, then $x=3 a+1$ and $y=-3 a+2$ from which we get\n\n$$\n\\left(9 a^{2}+6 a+1+9 a^{2}-12 a+4\\right) \\cdot 3=5\\left(-9 a^{2}+3 a+3\\right) \\text { or } 3 a^{2}-a=0\n$$\n\nBut $a=\\frac{1}{3}$ is not an integer, so $a=0$ and $x=1, y=2$. Thus, by symmetry, we have two solutions $(x, y)=(1,2)$ and $(x, y)=(2,1)$.\n\n- If $a+b \\neq 0$, then $|8(a+b)+3| \\geq 21$. So we obtain\n\n$$\n\\left|\\left(x^{2}+y^{2}\\right)(8(a+b)+3)\\right| \\geq 21 x^{2}+21 y^{2} \\geq|5 x y+5|\n$$\n\nwhich means that the equation has no other solutions.\n\n## Geometry", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nA5.", "solution_match": "## Solution"}}
+{"year": "2005", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C D$ be an isosceles trapezoid with $A B=A D=B C, A B / / D C, A B>D C$. Let $E$ be the point of intersection of the diagonals $A C$ and $B D$ and $N$ be the symmetric point of $\\mathrm{B}$ with respect to the line $\\mathrm{AC}$. Prove that quadrilateral $A N D E$ is cyclic.", "solution": "Let $\\omega$ be a circle passing through the points $A, N, D$ and let $M$ the point where $\\omega$ intersects $B D$ for the second time. The quadrilateral $A N D M$ is cyclic and it follows that\n\n$$\n\\angle N D M+\\angle N A M=\\angle N D M+\\angle B D C=180^{\\circ}\n$$\n\nand\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-05.jpg?height=681&width=852&top_left_y=1043&top_left_x=658)\n\nFigure 1\n\n$$\n\\angle N A M=\\angle B D C\n$$\n\nNow we have\n\n$$\n\\angle B D C=\\angle A C D=\\angle N A C\n$$\n\nand\n\n$$\n\\angle N A M=\\angle N A C\n$$\n\nSo the points $A, M, C$ are collinear and $M \\equiv E$.\n\n## Alternative solution\n\nIn this solution we do not need the circle passing through the points $A, N$ and $D$.\n\nBecause of the given symmetry we have\n\n$$\n\\angle A N E=\\angle A B D\n$$\n\nand from the equality $\\mathrm{AD}=\\mathrm{AB}$ the triangle $\\mathrm{ABD}$ is isosceles with\n\n$$\n\\angle A B D=\\angle A D E\n$$\n\nFrom (1) and (2) we get that $\\angle A N E=\\angle A D E$, which means that the quadrilateral $A N D E$ is cyclic.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nG1.", "solution_match": "## Solution"}}
+{"year": "2005", "tier": "T3", "problem_label": "G2", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C$ be a triangle inscribed in a circle $K$. The tangent from $A$ to the circle meets the line $B C$ at point $P$. Let $M$ be the midpoint of the line segment $A P$ and let $R$ be the intersection point of the circle $K$ with the line $B M$. The line $P R$ meets again the circle $K$ at the point $S$. Prove that the lines $A P$ and $C S$ are parallel.", "solution": "![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-06.jpg?height=678&width=1500&top_left_y=531&top_left_x=311)\n\nFigure 2\n\nAssume that point $C$ lies on the line segment $B P$. By the Power of Point theorem we have $M A^{2}=M R \\cdot M B$ and so $M P^{2}=M R \\cdot M B$. The last equality implies that the triangles $M R$ and $M P B$ are similar. Hence $\\angle M P R=\\angle M B P$ and since $\\angle P S C=\\angle M B P$, the claim is proved.\n\nSlight changes are to be made if the point $B$ lies on the line segment $P C$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-06.jpg?height=687&width=1156&top_left_y=1659&top_left_x=329)\n\nFigure 3", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nG2.", "solution_match": "## Solution"}}
+{"year": "2005", "tier": "T3", "problem_label": "G3", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C D E F$ be a regular hexagon. The points $\\mathrm{M}$ and $\\mathrm{N}$ are internal points of the sides $\\mathrm{DE}$ and $\\mathrm{DC}$ respectively, such that $\\angle A M N=90^{\\circ}$ and $A N=\\sqrt{2} \\cdot C M$. Find the measure of the angle $\\angle B A M$.", "solution": "Since $A C \\perp C D$ and $A M \\perp M N$ the quadrilateral $A M N C$ is inscribed. So, we have\n\n$$\n\\angle M A N=\\angle M C N\n$$\n\nLet $P$ be the projection of the point $M$ on the line $C D$. The triangles $A M N$ and $C P M$ are similar implying\n\n$$\n\\frac{A M}{C P}=\\frac{M N}{P M}=\\frac{A N}{C M}=\\sqrt{2}\n$$\n\nSo, we have\n\n$$\n\\frac{M P}{M N}=\\frac{1}{\\sqrt{2}} \\Rightarrow \\angle M N P=45^{\\circ}\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-07.jpg?height=679&width=830&top_left_y=1110&top_left_x=606)\n\nFigure 4\n\nHence we have\n\n$$\n\\angle C A M=\\angle M N P=45^{\\circ}\n$$\n\nand finally, we obtain\n\n$$\n\\angle B A M=\\angle B A C+\\angle C A M=75^{\\circ}\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nG3.", "solution_match": "## Solution"}}
+{"year": "2005", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $\\mathrm{ABC}$ be an isosceles triangle such that $A B=A C$ and $\\angle \\frac{A}{2}<\\angle B$. On the extension of the altitude $\\mathrm{AM}$ we get the points $\\mathrm{D}$ and $\\mathrm{Z}$ such that $\\angle C B D=\\angle A$ and $\\angle Z B A=90^{\\circ}$. $\\mathrm{E}$ is the foot of the perpendicular from $\\mathrm{M}$ to the altitude $\\mathrm{BF}$ and $\\mathrm{K}$ is the foot of the perpendicular from $\\mathrm{Z}$ to $\\mathrm{AE}$. Prove that $\\angle K D Z=\\angle K B D=\\angle K Z B$.", "solution": "The points $A, B, K, Z$ and $C$ are co-cyclic.\n\nBecause ME//AC so we have\n\n$$\n\\angle K E M=\\angle E A C=\\angle M B K\n$$\n\nTherefore the points $B, K, M$ and $E$ are co-cyclic. Now, we have\n\n$$\n\\begin{aligned}\n& \\angle A B F=\\angle A B C-\\angle F B C \\\\\n& =\\angle A K C-\\angle E K M=\\angle M K C\n\\end{aligned}\n$$\n\nAlso, we have\n\n$$\n\\begin{aligned}\n& \\angle A B F=90^{\\circ}-\\angle B A F=90^{\\circ}-\\angle M B D \\\\\n& =\\angle B D M=\\angle M D C\n\\end{aligned}\n$$\n\nFrom (1) and (2) we get $\\angle M K C=\\angle M D C$ and so the points $M, K, D$ and $C$ are co-cyclic.\n\nConsequently,\n\n$$\n\\angle K D M=\\angle K C M=\\angle B A K=\\angle B Z K \\text {, }\n$$\n\nand because the line $\\mathrm{BD}$ is tangent to the circumcircle of triangle $A B C$, we have\n\n$$\n\\angle K B D=\\angle B A K\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-08.jpg?height=950&width=562&top_left_y=501&top_left_x=1219)\n\nFigure 5\n\nFinally, we have\n\n$$\n\\angle K D Z=\\angle K B D=\\angle K Z B\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nG4.", "solution_match": "## Solution"}}
+{"year": "2005", "tier": "T3", "problem_label": "G5", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A$ and $P$ are the points of intersection of the circles $k_{1}$ and $k_{2}$ with centers $O$ and $K$, respectively. Let also $B$ and $C$ be the symmetric points of $A$ with respect to $O$ and $K$, respectively. A line through $A$ intersects the circles $k_{1}$ and $k_{2}$ at the points $D$ and $E$, respectively. Prove that the centre of the circumcircle of the triangle $D E P$ lies on the circumcircle $O K P$.", "solution": "The points $B, P, C$ are collinear, and\n\n$$\n\\angle A P C=\\angle A P B=90^{\\circ}\n$$\n\nLet $N$ be the midpoint of $D P$.\n\nSo we have:\n\n$$\n\\begin{aligned}\n& \\angle N O P=\\angle D A P \\\\\n& =\\angle E C P=\\angle E C A+\\angle A C P\n\\end{aligned}\n$$\n\nSince $O K / / B C$ and $O K$ is the bisector of $\\angle A K P$ we get\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-09.jpg?height=579&width=836&top_left_y=578&top_left_x=923)\n\nFigure 6\n\n$$\n\\angle A C P=O K P\n$$\n\nAlso, since $A P \\perp O K$ and $M K \\perp P E$ we have that\n\n$$\n\\angle A P E=\\angle M K O\n$$\n\nThe points $A, E, C, P$ are co-cyclic, and so $\\angle E C A=\\angle A P E$.\n\nTherefore, from (1), (2) and (3) we have that $\\angle N O P=\\angle M K P$.\n\nThus $O, M, K$ and $P$ are co-cyclic.\n\n## Comment\n\nPoints B and C may not be included in the statement of the problem\n\n## Alternative solution\n\nIt is sufficient to prove that the quadrilateral $M O P K$ is circumscrible.\n\nSince $M O$ and $M K$ are perpendicular bisectors of the line segments $P D$ and $P E$, respectively, we have\n\nTherefore the quadrilateral $M O P K$ is circumscrible.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-10.jpg?height=1198&width=1196&top_left_y=716&top_left_x=292)\n\nFigure 7", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nG5.", "solution_match": "## Solution"}}
+{"year": "2005", "tier": "T3", "problem_label": "G6", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "A point $O$ and the circles $k_{1}$ with center $O$ and radius $3, k_{2}$ with center $O$ and radius 5, are given. Let $A$ be a point on $k_{1}$ and $B$ be a point on $k_{2}$. If $A B C$ is equilateral triangle, find the maximum value of the distance $O C$.", "solution": "## Alternative solution 1 (by the proposer)\n\nLet $\\phi$ be a $60^{\\circ}$ rotation with center at $B$. Then $\\phi(A)=C, \\phi(O)=P$ and $P C=O A$, $O P=O$, etc.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nG6.", "solution_match": "## Solution"}}
+{"year": "2005", "tier": "T3", "problem_label": "G6", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "A point $O$ and the circles $k_{1}$ with center $O$ and radius $3, k_{2}$ with center $O$ and radius 5, are given. Let $A$ be a point on $k_{1}$ and $B$ be a point on $k_{2}$. If $A B C$ is equilateral triangle, find the maximum value of the distance $O C$.", "solution": "## Alternative solution 2\n\nLet $O A=x, A B=B C=C A=a$. From the second theorem of Ptolemy we get\n\n$$\nO A \\cdot B C+O B \\cdot A C \\geq A B \\cdot O C \\Leftrightarrow 3 a+5 a=a x \\Leftrightarrow x \\leq 8\n$$\n\nThe value $x=8$ is attained when the quadrilateral OACB is circumscrible, i.e. when $\\angle A O B=120^{\\circ}$.\n\nThe point $\\mathrm{B}$ can be constructed as follows:\n\nIt is the point of intersection of the circle $(0,5)$ with the ray coming from the rotation of the ray $\\mathrm{OA}$ with center $\\mathrm{O}$ by an angle $\\theta=-120^{\\circ}$, (figure 10).\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-12.jpg?height=1047&width=807&top_left_y=914&top_left_x=652)\n\nFigure 10", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nG6.", "solution_match": "## Solution"}}
+{"year": "2005", "tier": "T3", "problem_label": "G7", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C D$ be a parallelogram, $\\mathrm{P}$ a point on $C D$, and $Q$ a point on $A B$. Let also $M=A P \\cap D Q, \\quad N=B P \\cap C Q, K=M N \\cap A D$, and $L=M N \\cap B C$. Show that $B L=D K$.", "solution": "Let $O$ be the intersection of the diagonals. Let $P_{1}$ be on $A B$ such that $P P_{1} / / A D$, and let $Q_{1}$ be on $C D$ such that $\\mathrm{Q} Q_{1} / / A D$. Let $\\sigma$ be the central symmetry with center $\\mathrm{O}$. Let $\\left.P^{\\prime}=\\sigma(P), Q^{\\prime}=\\sigma(Q), P_{1}^{\\prime}=\\sigma\\left(P_{1}\\right)\\right)$ and, (figure 1).\n\nLet $M_{1}=A Q_{1} \\cap D P_{1}, N_{1}=B Q_{1} \\cap C P_{1}, N^{\\prime}=A Q^{\\prime} \\cap D P^{\\prime}$ and $M^{\\prime}=B Q^{\\prime} \\cap C P^{\\prime}$.\n\nThen: $M^{\\prime}=\\sigma(M), N^{\\prime}=\\sigma(N), M_{1}^{\\prime}=\\sigma\\left(M_{1}\\right)$ and $N_{1}^{\\prime}=\\sigma\\left(N_{1}\\right)$.\n\nSince $A P$ and $D P_{1}$ are the diagonals of the parallelogram $A P_{1} P D, C P_{1}$ and $B P$ are the diagonals of the parallelogram $P_{1} B C P$, and $A Q_{1}$ and $D O$ are the diagonals of the parallelogram $A Q Q_{1} D$, it follows that the points $U, V, W$ (figure 2) are collinear and they lie on the line passing through the midpoints $R$ of $A D$ and $Z$ of $B C$. The diagonals AM and $D M_{1}$ the quadrilateral $A M_{1} M D$ intersect at $U$ and the diagonals $A M_{1}$ and - $D M$ intersect at $W$. Since the midpoint of $A D$ is on the line $U W$, it follows that the quadrilateral $A M_{1} M D$ is a trapezoid. Hence, $M M_{1}$ is parallel to $A D$ and the midpoint $S$ of $M M_{1}$ lies on the line $U W$, (figure 2).\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-13.jpg?height=724&width=1446&top_left_y=1344&top_left_x=315)\n\nFigure 11\n\nSimilarly $M^{\\prime} M_{1}^{\\prime}$ is parallel to $A D$ and its midpoint lies on $U W$. So $M_{1} M^{\\prime} M_{1}^{\\prime} M$ is a parallelogram whose diagonals intersect at $\\mathrm{O}$.\n\nSimilarly, $N_{1}^{\\prime} N N_{1} N^{\\prime}$ is a parallelogram whose diagonals intersect at $O$.\n\nAll these imply that $M, N, M^{\\prime}, N^{\\prime}$ and $O$ are collinear, i.e. $O$ lies on the line $K L$. This implies that $K=\\sigma(L)$, and since $D=\\sigma(B)$, the conclusion follows.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-14.jpg?height=556&width=1504&top_left_y=224&top_left_x=252)\n\nFigure 12\n\n## Alternative solution\n\nLet the line ( $\\varepsilon$ ) through the points $\\mathrm{M}, \\mathrm{N}$ intersect the lines $\\mathrm{DC}, \\mathrm{AB}$ at points $\\mathrm{T}_{1}, \\mathrm{~T}_{2}$ respectively. Applying Menelaus Theorem to the triangles DQC, APB with intersecting line $(\\varepsilon)$ in both cases we get:\n\n$$\n\\frac{M D}{M Q} \\frac{N Q}{N C} \\frac{T_{1} C}{T_{1} D}=1 \\text { and } \\frac{M P}{M A} \\frac{N B}{N P} \\frac{T_{2} A}{T_{2} B}=1\n$$\n\nBut it is true that $\\frac{\\mathrm{MD}}{\\mathrm{MQ}}=\\frac{\\mathrm{MP}}{\\mathrm{MA}}$ and $\\frac{\\mathrm{NQ}}{\\mathrm{NC}}=\\frac{\\mathrm{NB}}{\\mathrm{NP}}$.\n\nIt follows that $\\frac{T_{1} C}{T_{1} D}=\\frac{T_{2} A}{T_{2} B}$ i.e. $\\frac{T_{1} D+D C}{T_{1} D}=\\frac{T_{2} B+B A}{T_{2} B}$ hence $T_{1} D=T_{2} B$ (since\n\n$\\mathrm{DC}=\\mathrm{BA})$.\n\nThen of course by the similarity of the triangles $\\mathrm{T}_{1} \\mathrm{DK}$ and $\\mathrm{T}_{2} \\mathrm{BL}$ we get the desired equality $\\mathrm{DK} \\cdot \\mathrm{BL}$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-14.jpg?height=585&width=1106&top_left_y=1659&top_left_x=454)\n\nFigure 13\n\n## Number Theory", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nG7.", "solution_match": "## Solution"}}
+{"year": "2005", "tier": "T3", "problem_label": "NT1", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all the natural numbers $m$ and $n$, such that the square of $m$ minus the product of $n$ with $k$, is 2 , where the number $k$ is obtained from $n$ by writing 1 on the left of the decimal notation of $n$.", "solution": "Let $t$ be the number of digits of $n$. Then $k=10^{t}+n$. So\n\n$$\n\\mathrm{m}^{2}=n\\left(10^{t}+\\mathrm{n}\\right)+2, \\text { i.e. } \\mathrm{m}^{2}-\\mathrm{n}^{2}=10^{t} n+2\n$$\n\nThis implies that $\\mathrm{m}, \\mathrm{n}$ are even and both $\\mathrm{m}, \\mathrm{n}$ are odd.\n\nIf $t=1$, then, 4 is divisor of $10 n+2$, so, $n$ is odd. We check that the only solution in this case is $\\mathrm{m}=11$ and $\\mathrm{n}=7$.\n\nIf $t>1$, then 4 is divisor of $\\mathrm{m}^{2}-\\mathrm{n}^{2}$, but 4 is not divisor of $10^{t}+2$.\n\nHence the only solution is $\\mathrm{m}=11$ and $\\mathrm{n}=7$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nNT1.", "solution_match": "## Solution"}}
+{"year": "2005", "tier": "T3", "problem_label": "NT2", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all natural numbers $n$ such that $5^{n}+12^{n}$ is perfect square.", "solution": "By checking the cases $n=1,2,3$ we get the solution $n=2$ and $13^{2}=5^{2}+12^{2}$.\n\nIf $n=2 k+1$ is odd, we consider the equation modulo 5 and we obtain\n\n$$\n\\begin{aligned}\nx^{2} & \\equiv 5^{2 k+1}+12^{2 k+1}(\\bmod 5) \\equiv 2^{2 k} \\cdot 2(\\bmod 5) \\\\\n& \\equiv(-1)^{k} \\cdot 2(\\bmod 5) \\equiv \\pm 2(\\bmod 5)\n\\end{aligned}\n$$\n\nThis is not possible, because the square residue of any natural number module 5 is 0,1 or 4. Therefore $n$ is even and $x^{2}=5^{2 k}+12^{2 k}$. Rearrange this equation in the form\n\n$$\n5^{2 k}=\\left(x-12^{k}\\right)\\left(x+12^{k}\\right)\n$$\n\nIf 5 divides both factors on the right, it must also divide their difference, that is\n\n$$\n5 \\mid\\left(x+12^{k}\\right)-\\left(x-12^{k}\\right)=2 \\cdot 12^{k}\n$$\n\nwhich is not possible. Therefore we must have\n\n$$\nx-12^{k}=1 \\text { and } x+12^{k}=5^{2 k}\n$$\n\nBy adding the above equalities we get\n\n$$\n5^{2 k}-1=2 \\cdot 12^{k}\n$$\n\nFor $k \\geq 2$, we have the inequality\n\n$$\n25^{k}-1>24^{k}=2^{k} \\cdot 12^{k}>2 \\cdot 12^{k}\n$$\n\nThus we conclude that there exists a unique solution to our problem, namely $n=2$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nNT2.", "solution_match": "## Solution"}}
+{"year": "2005", "tier": "T3", "problem_label": "NT3", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Let $p$ be an odd prime. Prove that $p$ divides the integer\n\n$$\n\\frac{2^{p!}-1}{2^{k}-1}\n$$\n\nfor all integers $k=1,2, \\ldots, p$.", "solution": "At first, note that $\\frac{2^{p!}-1}{2^{k}-1}$ is indeed an integer.\n\nWe start with the case $\\mathrm{k}=\\mathrm{p}$. Since $p \\mid 2^{p}-2$, then $p / 22^{p}-1$ and so it suffices to prove that $p \\mid 2^{(p)!}-1$. This is obvious as $p \\mid 2^{p-1}-1$ and $\\left(2^{p-1}-1\\right) \\mid 2^{(p)!}-1$.\n\nIf $\\mathrm{k}=1,2, \\ldots, \\mathrm{p}-1$, let $m=\\frac{(p-1)!}{k} \\in \\mathbb{N}$ and observe that $p!=k m p$. Consider $a \\in \\mathbb{N}$ so that $p^{a} \\mid 2^{k}-1$ and observe that it suffices to prove $p^{a+1} \\mid 2^{p!}-1$. The case $a=0$ is solved as the case $k=p$. If else, write $2^{k}=1+p^{a} \\cdot l, l \\in \\mathbb{N}$ and rising at the power mp gives\n\n$$\n2^{p!}=\\left(1+p^{a} \\cdot l\\right)^{m p}=1+m p \\cdot p^{a} \\cdot l+M p^{2 a}\n$$\n\nwhere $M n$ stands for a multiply of $\\mathrm{n}$. Now it is clear that $p^{a+1} \\mid 2^{p!}-1$, as claimed.\n\nComment. The case $\\mathrm{k}=\\mathrm{p}$ can be included in the case $a=0$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nNT3.", "solution_match": "## Solution"}}
+{"year": "2005", "tier": "T3", "problem_label": "NT4", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all the three digit numbers $\\overline{a b c}$ such that\n\n$$\n\\overline{a b c}=a b c(a+b+c)\n$$", "solution": "We will show that the only solutions are 135 and 144 .\n\nWe have $a>0, b>0, c>0$ and\n\n$$\n9(11 a+b)=(a+b+c)(a b c-1)\n$$\n\n- If $a+b+c \\equiv 0(\\bmod 3)$ and $a b c-1 \\equiv 0(\\bmod 3)$, then $a \\equiv b \\equiv c \\equiv 1(\\bmod 3)$ and $11 a+b \\equiv 0(\\bmod 3)$. It follows now that\n\n$$\na+b+c \\equiv 0(\\bmod 9) ; \\text { or } a b c-1 \\equiv 0(\\bmod 9)\n$$\n\n- If . $a b c-1 \\equiv 0(\\bmod 9)$\n\nwe have $11 a+b=(a+b+c) k$, where $k$ is an integer\n\nand is easy to see that we must have $1<k \\leq 10$.\n\nSo we must deal only with the cases $k=2,3,4,5,6,7,8,9,10$\n\nIf $k=2.4,5,6,8,9,10$ then $9 k+1$ has prime divisors greater than 9 ,so we must see only the cases $\\quad k=3,7$.\n\n- If $k=3$ we have\n\n$$\n8 a=2 b+3 c \\text { and } \\quad a b c=28\n$$\n\nIt is clear that $\\mathrm{c}$ is even, $c=2 c_{1}$ and that both $b$ and $c_{1}$ are odd.\n\nFrem $a b c_{1}=14$ it follows that $a=2, b=7, c_{l}=1$ or $a=2, b=1, c_{i}=7$ and it is clear that there exists no solution in this case. - If $k=7$ we have $c$-even, $c=2 c_{k}, 2 a=3 b+7 c_{1}, a b c_{1}=32$, then both $b$ and $c_{I}$ are\neven.\n\nBut this is impossible, because they imply that $a>9$.\n\nNow we will deal with the case when $a+b+c \\equiv 0(\\bmod 9)$ or $a+b+c=9 l$, where $l$ is an integer.\n\n- If $l \\geq 2$ we have $a+b+c \\geq 18, \\max \\{a, b, c\\} \\geq 6$ and it is easy to see that $a b c \\geq 72$ and $a b c(a+b+c)>1000$,so the case $l \\geq 2$ is impossible.\n- If $l=1$ we have\n\n$$\n11 a+b=a b c-1 \\text { or } 11 a+b+1=a b c \\leq\\left(\\frac{a+b+c}{3}\\right)^{3}=27\n$$\n\nSo we have only two cases: $a=1$ or $a=2$.\n\n- If $a=1$, we have $b+c=8$ and $11+b=b c-1$ or $b+(c-1)=7$ and $b(c-1)=12$ and the solutions are $(a, b, c)=(1,3,5)$ and $(a, b, c)=(1,4,4)$, and the answer is 135 and 144.\n- If $a=2$ we have $b(2 c-1)=23$ and there is no solution for the problem.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nNT4.", "solution_match": "## Solution"}}
+{"year": "2005", "tier": "T3", "problem_label": "NT5", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Let $p$ be a prime number and let $a$ be an integer. Show that if $n^{2}-5$ is not divisible by $p$ for any integer $n$, there exist infinitely many integers $m$ so that $p$ divides $m^{5}+a$.", "solution": "We start with a simple fact:\n\nLemma: If $b$ is an integer not divisible by $p$ then there is an integer $s$ so that $s b$ has the remainder $l$ when divided by $p$.\n\nFor a proof, just note that numbers $b, 2 b, \\ldots,(p-1) b$ have distinct non-zero remainders when divided by $p$, and hence one of them is equal to 1 .\n\nWe prove that if $x, y=0,1,2, \\ldots, p-1$ and $\\mathrm{p}$ divides $x^{5}-y^{5}$, then $x=y$.\n\nIndeed, assume that $x \\neq y$. If $x=0$, then $p \\mid y^{5}$ and so $y=0$, a contradiction.\n\nTo this point we have $x, y \\neq 0$. Since\n\n$$\np \\mid(x-y)\\left(x^{4}+x^{3} y+x^{2} y^{2}+x y^{3}+y^{4}\\right) \\text { and } p /(x-y)\n$$\n\nwe have\n\n$$\n\\begin{aligned}\n& p l\\left(x^{2}+y^{2}\\right)^{2}+x y\\left(x^{2}+y^{2}\\right)-x^{2} y^{2} \\text {, and so } \\\\\n& p \\|\\left(2\\left(x^{2}+y^{2}\\right)+x y\\right)^{2}-5 x^{2} y^{2}\n\\end{aligned}\n$$\n\nAs $p / x y$, from the lemma we find an integer $s$ so that $s x y=k p+1, k \\in \\mathbb{N}$. Then\n\n$$\np \\mid\\left[s\\left(2 x^{2}+2 y^{2}+x y\\right)\\right]^{2}-5\\left(k^{2} p^{2}+2 k p+1\\right)\n$$\n\nand so $p \\mid z^{2}-5$, where $z=s\\left(2 x^{2}+2 y^{2}+x y\\right)$, a contradiction.\n\nConsequeatly $r=y$.\n\nSince we have proved that numbers $0^{5}, 1^{5}, \\ldots,(p-1)^{5}$ have distinct remainders when divided by $p$, the same goes for the numbers $0^{5}+a, 1^{5}+a, \\ldots,(p-1)^{5}+a$ and the conclusion can be reached easily.\n\n## Comments\n\n1. For beauty we may choose $a=-2$ or any other value.\n2. Moreover, we may ask only for one value of $m$, instead of \"infinitely many\".\n3. A simple version will be to ask for a proof that the numbers $0^{5}, 1^{5}, \\ldots,(p-1)^{5}$ have distinct remainders when divided by $p$.\n\n## Combinatorics", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nNT5.", "solution_match": "## Solution"}}
+{"year": "2005", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "A triangle with area 2003 is divided into non-overlapping small triangles. The number of all the vertices of all those triangles is 2005 . Show that at mest one of the smaller triangles has area less or equal to 1.", "solution": "Since all the vertices are 2005 , and the vertices of the big triangle are among them, it follows that the number of the small triangles is at least 2003. So, it follows that at least one of the small triangles has area at most 1", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nC1.", "solution_match": "## Solution"}}
+{"year": "2005", "tier": "T3", "problem_label": "C2", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Government of Greece wants to change the building. So they will move in building with square shape $2005 \\times 2005$. Only demand is that every room has exactly 2 doors and doors cannot lead outside the building. Is this possible? What about a building $2004 \\times 2005$ ?", "solution": "The key idea is to color the table $2005 \\times 2005$ in two colors, like chessboard. One door connects two adjacent squares, so they have different colors. We can count the number of doors in two ways. The number of doors is twice the number of white squares and it is also twice the number of black squares. This is, of course, Ipossible, because we have one more black square on the table. So, if $n$ is odd, we cannot achieve that every room has two doors.\n\nFor table $2004 \\times 2005$ this is possible, and one of many solutions is presented on the picture below.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-19.jpg?height=710&width=761&top_left_y=1662&top_left_x=612)\n\nFigure 14\n\n## Alternative statement of the problem\n\nIs it possible to open holes on two distinct sides of each cell of a $2005 \\times 2005$ square grid so that no hole is on the perimeter of the grid? What about a $2004 \\times 2005$ grid?\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=579&width=514&top_left_y=703&top_left_x=673)\n\n$$\n\\cdot \\varsigma 00 Z<8 \\downarrow 0 Z=(9 \\mathrm{I}) \\mathrm{X}=(\\varsigma 9) \\Psi\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=80&width=1236&top_left_y=1361&top_left_x=436)\n\n$$\n\\tau_{2} w 8=(\\mathfrak{I}-w \\tau) 8+{ }_{\\tau}(I-u) 8=\\tau \\cdot(I-w) \\nabla+(I-w) \\tau \\cdot \\nabla+8+(\\tau-u) X=(u) Y\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=60&width=251&top_left_y=1550&top_left_x=1432)\n\n$$\n\\dot{r}_{\\tau} \\varepsilon \\cdot 8=8 \\cdot 6=\\tau \\cdot 8+\\tau \\cdot \\tau \\cdot \\nabla+8+(\\tau) X=\\underset{\\partial \\Lambda E Y}{(\\varepsilon) Y}\n$$\n\n$$\n{ }^{2} \\tau \\cdot 8=8 \\cdot t=\\tau \\cdot \\downarrow+\\tau \\cdot \\downarrow+8+\\text { (I) } y=\\text { (乙) } y\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=77&width=1015&top_left_y=1746&top_left_x=665)\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=80&width=1426&top_left_y=1803&top_left_x=254)\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=94&width=1443&top_left_y=1846&top_left_x=240)\n\n$$\n8=(\\mathrm{L}) X\n$$\n\n:әлеч $\\partial M$ ঈ uo sұutod\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=94&width=1409&top_left_y=1988&top_left_x=277)\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=82&width=953&top_left_y=2037&top_left_x=730)\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=77&width=1472&top_left_y=2094&top_left_x=208)\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=105&width=1484&top_left_y=2131&top_left_x=208)\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=83&width=975&top_left_y=2179&top_left_x=719)\nuo!n [0S\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=73&width=859&top_left_y=2328&top_left_x=780)\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=63&width=739&top_left_y=2379&top_left_x=891)\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=57&width=368&top_left_y=2428&top_left_x=1254)\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f0d06a71e884e5e6ec7bg-20.jpg?height=80&width=1420&top_left_y=2476&top_left_x=277)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nC2.", "solution_match": "## Solution"}}
+{"year": "2005", "tier": "T3", "problem_label": "C4", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Let $p_{1}, p_{2}, \\ldots, p_{2005}$ be different prime numbers. Let $\\mathrm{S}$ be a set of natural numbers which elements have the property that their simple divisors are some of the numbers $p_{1}, p_{2}, \\ldots, p_{2005}$ and product of any two elements from $\\mathrm{S}$ is not perfect square.\n\nWhat is the maximum number of elements in $\\mathrm{S}$ ?", "solution": "Let $a, b$ be two arbitrary numbers from $\\mathrm{S}$. They can be written as\n\n$$\na=p_{1}^{a_{1}} p_{2}^{a_{2}} \\cdots p_{2005}^{a_{2005}} \\text { and } b=p_{1}^{\\beta_{1}} p_{2}^{\\beta_{2}} \\cdots p_{2005}^{\\beta_{2005}}\n$$\n\nIn order for the product of the elements $a$ and $b$ to be a square all the sums of the corresponding exponents need to be even from where we can conclude that for every $i, a_{i}$ and $\\beta_{i}$ have the same parity. If we replace all exponents of $\\mathrm{a}$ and $b$ by their remainders modulo 2 , then we get two numbers $a^{\\prime}, b^{\\prime}$ whose product is a perfect square if and only if ab is a perfect square.\n\nIn order for the product $a^{\\prime} b^{\\prime}$ not to be a perfect square, at least one pair of the corresponding exponents modulo 2, need to be of opposite parity.\n\nSince we form 2005 such pairs modulo 2, and each number in these pairs is 1 or 2 , we conclude that we can obtain $2^{2005}$ distinct products none of which is a perfect square.\n\nNow if we are given $2^{2005}+1$ numbers, thanks to Dirichlet's principle, there are at least two with the same sequence of modulo 2 exponents, thus giving a product equal to a square.\n\nSo, the maximal number of the elements of $S$ is $2^{2005}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl", "problem_match": "\nC4.", "solution_match": "## Solution"}}

JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl

@@ -1,17 +1,17 @@
-{"year": "2007", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $a$ be a real positive number such that $a^{3}=6(a+1)$. Prove that the equation $x^{2}+a x+a^{2}-6=0$ has no solution in the set of the real number.", "solution": "The discriminant of the equation is $\\Delta=3\\left(8-a^{2}\\right)$. If we accept that $\\Delta \\geq 0$, then $a \\leq 2 \\sqrt{2}$ and $\\frac{1}{a} \\geq \\frac{\\sqrt{2}}{4}$, from where $a^{2} \\geq 6+6 \\cdot \\frac{\\sqrt{2}}{4}=6+\\frac{6}{a} \\geq 6+\\frac{3 \\sqrt{2}}{2}>8$ (contradiction).", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nA1 ", "solution_match": "## Solution"}}
-{"year": "2007", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO", "problem": "Prove that $\\frac{a^{2}-b c}{2 a^{2}+b c}+\\frac{b^{2}-c a}{2 b^{2}+c a}+\\frac{c^{2}-a b}{2 c^{2}+a b} \\leq 0$ for any real positive numbers $a, b, c$.", "solution": "The inequality rewrites as $\\sum \\frac{2 a^{2}+b c-3 b c}{2 a^{2}+b c} \\leq 0$, or $3-3 \\sum \\frac{b c}{2 a^{2}+b c} \\leq 0$ in other words $\\sum \\frac{b c}{2 a^{2}+b c} \\geq 1$.\n\nUsing Cauchy-Schwarz inequality we have\n\n$$\n\\sum \\frac{b c}{2 a^{2}+b c}=\\sum \\frac{b^{2} c^{2}}{2 a^{2} b c+b^{2} c^{2}} \\geq \\frac{\\left(\\sum b c\\right)^{2}}{2 a b c(a+b+c)+\\sum b^{2} c^{2}}=1\n$$\n\nas claimed.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nA2 ", "solution_match": "## Solution"}}
-{"year": "2007", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $A$ be a set of positive integers containing the number 1 and at least one more element. Given that for any two different elements $m, n$ of $A$ the number $\\frac{m+1}{(m+1, n+1)}$ is also an element of $A$, prove that $A$ coincides with the set of positive integers.", "solution": "Let $a>1$ be lowest number in $A \\backslash\\{1\\}$. For $m=a, n=1$ one gets $y=\\frac{a+1}{(2, a+1)} \\in A$. Since $(2, a+1)$ is either 1 or 2 , then $y=a+1$ or $y=\\frac{a+1}{2}$.\n\nBut $1<\\frac{a+1}{2}<a$, hence $y=a+1$. Applying the given property for $m=a+1, n=a$ one has $\\frac{a+2}{(a+2, a+1)}=a+2 \\in A$, and inductively $t \\in A$ for all integers $t \\geq a$.\n\nFurthermore, take $m=2 a-1, n=3 a-1$ (now in $A!$ ); as $(m+1, n+1)=(2 a, 3 a)=a$ one obtains $\\frac{2 a}{a}=2 \\in A$, so $a=2$, by the definition of $a$.\n\nThe conclusion follows immediately.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nA3 ", "solution_match": "## Solution"}}
-{"year": "2007", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $a$ and $b$ be positive integers bigger than 2. Prove that there exists a positive integer $k$ and a sequence $n_{1}, n_{2}, \\ldots, n_{k}$ consisting of positive integers, such that $n_{1}=a$, $n_{k}=b$, and $\\left(n_{i}+n_{i+1}\\right) \\mid n_{i} n_{i+1}$ for all $i=1,2, \\ldots, k-1$.", "solution": "We write $a \\Leftrightarrow b$ if the required sequence exists. It is clear that $\\Leftrightarrow$ is equivalence relation, i.e. $a \\Leftrightarrow a,(a \\Leftrightarrow b$ implies $b \\Rightarrow a)$ and $(a \\Leftrightarrow b, b \\Leftrightarrow c$ imply $a \\Leftrightarrow c$ ).\n\nWe shall prove that for every $a \\geq 3$, ( $a-$ an integer), $a \\Leftrightarrow 3$.\n\nIf $a=2^{s} t$, where $t>1$ is an odd number, we take the sequence\n\n$$\n2^{s} t, 2^{s}\\left(t^{2}-t\\right), 2^{s}\\left(t^{2}+t\\right), 2^{s}(t+1)=2^{s+1} \\cdot \\frac{t+1}{2}\n$$\n\nSince $\\frac{t+1}{2}<t$ after a finite number of steps we shall get a power of 2 . On the other side, if $s>1$ we have $2^{s}, 3 \\cdot 2^{s}, 3 \\cdot 2^{s-1}, 3 \\cdot 2^{s-2}, \\ldots, 3$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nA4 ", "solution_match": "## Solution"}}
-{"year": "2007", "tier": "T3", "problem_label": "A5", "problem_type": "Algebra", "exam": "JBMO", "problem": "The real numbers $x, y, z, m, n$ are positive, such that $m+n \\geq 2$. Prove that\n\n$$\n\\begin{gathered}\nx \\sqrt{y z(x+m y)(x+n z)}+y \\sqrt{x z(y+m x)(y+n z)}+z \\sqrt{x y(z+m x)(x+n y)} \\leq \\\\\n\\frac{3(m+n)}{8}(x+y)(y+z)(z+x) .\n\\end{gathered}\n$$", "solution": "Using the AM-GM inequality we have\n\n$$\n\\begin{aligned}\n& \\sqrt{y z(x+m y)(x+n z)}=\\sqrt{(x z+m y z)(x y+n y z)} \\leq \\frac{x y+x z+(m+n) y z}{2} \\\\\n& \\sqrt{x z(y+m x)(y+n z)}=\\sqrt{(y z+m x z)(x y+n x z)} \\leq \\frac{x y+y z+(m+n) x z}{2} \\\\\n& \\sqrt{x y(z+m x)(z+n y)}=\\sqrt{(y z+m x y)(x z+n x y)} \\leq \\frac{x z+y z+(m+n) x y}{2}\n\\end{aligned}\n$$\n\nThus it is enough to prove that\n\n$$\n\\begin{aligned}\nx[x y+x z+(m+n) y z] & +y[x y+y z+(m+n) x z]+z[x y+y z+(m+n) x z] \\leq \\\\\n\\leq & \\frac{3(m+n)}{4}(x+y)(y+z)(z+x),\n\\end{aligned}\n$$\n\nor\n\n$$\n4[A+3(m+n) B] \\leq 3(m+n)(A+2 B) \\Leftrightarrow 6(m+n) B \\leq[3(m+n)-4] A\n$$\n\nwhere $A=x^{2} y+x^{2} z+x y^{2}+y^{2} z+x z^{2}+y z^{2}, B=x y z$.\n\nBecause $m+n \\geq 2$ we obtain the inequality $m+n \\leq 3(m+n)-4$. From AMGM inequality it follows that $6 B \\leq A$. From the last two inequalities we deduce that $6(m+n) B \\leq[3(m+n)-4] A$. The inequality is proved.\n\nEquality holds when $m=n=1$ and $x=y=z$.\n\n### 2.2 Combinatorics", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nA5 ", "solution_match": "## Solution"}}
-{"year": "2007", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "We call a tiling of an $m \\times n$ rectangle with corners (see figure below) \"regular\" if there is no sub-rectangle which is tiled with corners. Prove that if for some $m$ and $n$ there exists a \"regular\" tiling of the $m \\times n$ rectangular then there exists a \"regular\" tiling also for the $2 m \\times 2 n$ rectangle.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_5f424aab603cc04b261bg-06.jpg?height=92&width=83&top_left_y=1259&top_left_x=992)", "solution": "A corner-shaped tile consists of 3 squares. Let us call \"center of the tile\" the square that has two neighboring squares. Notice that in a \"regular\" tiling, the squares situated in the corners of the rectangle have to be covered by the \"center\" of a tile, otherwise a $2 \\times 3$ (or $3 \\times 2$ ) rectangle tiled with two tiles would form.\n\nConsider a $2 m \\times 2 n$ rectangle, divide it into four $m \\times n$ rectangles by drawing its midlines, then do a \"regular\" tiling for each of these rectangles. In the center of the $2 m \\times 2 n$ rectangle we will necessarily obtain the following configuration:\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_5f424aab603cc04b261bg-06.jpg?height=200&width=209&top_left_y=1896&top_left_x=932)\n\nNow simply change the position of these four tiles into:\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_5f424aab603cc04b261bg-06.jpg?height=188&width=203&top_left_y=2299&top_left_x=938)\n\nIt is easy to see that this tiling is \"regular\".", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nC1 ", "solution_match": "## Solution"}}
-{"year": "2007", "tier": "T3", "problem_label": "C2", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Consider 50 points in the plane, no three of them belonging to the same line. The points have been colored into four colors. Prove that there are at least 130 scalene triangles whose vertices are colored in the same color.", "solution": "Since $50=4 \\cdot 12+2$, according to the pigeonhole principle we will have at least 13 points colored in the same color. We start with the:\n\nLemma. Given $n>8$ points in the plane, no three of them collinear, then there are at least $\\frac{n(n-1)(n-8)}{6}$ scalene triangles with vertices among the given points.\n\nProof. There are $\\frac{n(n-1)}{2}$ segments and $\\frac{n(n-1)(n-2)}{6}$ triangles with vertices among the given points. We shall prove that there are at most $n(n-1)$ isosceles triangles. Indeed, for every segment $A B$ we can construct at most two isosceles triangles (if we have three $A B C, A B D$ and $A B E$, than $C, D, E$ will be collinear). Hence we have at least\n\n$$\n\\frac{n(n-1)(n-2)}{6}-n(n-1)=\\frac{n(n-1)(n-8)}{6} \\text { scalene triangles. }\n$$\n\nFor $n=13$ we have $\\frac{13 \\cdot 12 \\cdot 5}{6}=130$, QED.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nC2 ", "solution_match": "## Solution"}}
-{"year": "2007", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "The nonnegative integer $n$ and $(2 n+1) \\times(2 n+1)$ chessboard with squares colored alternatively black and white are given. For every natural number $m$ with $1<m<2 n+1$, an $m \\times m$ square of the given chessboard that has more than half of its area colored in black, is called a $B$-square. If the given chessboard is a $B$-square, find in terms of $n$ the total number of $B$-squares of this chessboard.", "solution": "Every square with even side length will have an equal number of black and white $1 \\times 1$ squares, so it isn't a $B$-square. In a square with odd side length, there is one more $1 \\times 1$ black square than white squares, if it has black corner squares. So, a square with odd side length is a $B$-square either if it is a $1 \\times 1$ black square or it has black corners.\n\nLet the given $(2 n+1) \\times(2 n+1)$ chessboard be a $B$-square and denote by $b_{i}(i=$ $1,2, \\ldots, n+1)$ the lines of the chessboard, which have $n+1$ black $1 \\times 1$ squares, by $w_{i}(i=1,2, \\ldots, n)$ the lines of the chessboard, which have $n$ black $1 \\times 1$ squares and by $T_{m}(m=1,3,5, \\ldots, 2 n-1,2 n+1)$ the total number of $B$-squares of dimension $m \\times m$ of the given chessboard.\n\nFor $T_{1}$ we obtain $T_{1}=(n+1)(n+1)+n \\cdot n=(n+1)^{2}+n^{2}$.\n\nFor computing $T_{3}$ we observe that there are $n 3 \\times 3 B$-squares, which have the black corners on each pair of lines $\\left(b_{i}, b_{i+1}\\right)$ for $i=1,2, \\ldots, n$ and there are $n-13 \\times 3 B$-squares, which have the black corners on each pair of lines $\\left(w_{i}, w_{i+1}\\right)$ for $i=1,2, \\ldots, n-1$. So, we have\n\n$$\nT_{3}=n \\cdot n+(n-1)(n-1)=n^{2}+(n-1)^{2} \\text {. }\n$$\n\nBy using similar arguments for each pair of lines $\\left(b_{i}, b_{i+2}\\right)$ for $i=1,2, \\ldots, n-1$ and for each pair of lines $\\left(w_{i}, w_{i+2}\\right)$ for $i=1,2, \\ldots, n-2$ we compute\n\n$$\nT_{5}=(n-1)(n-1)+(n-2)(n-2)=(n-1)^{2}+(n-2)^{2}\n$$\n\nStep by step, we obtain\n\n$$\n\\begin{gathered}\nT_{7}=(n-2)(n-2)+(n-3)(n-3)=(n-2)^{2}+(n-3)^{2} \\\\\n\\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\\\\nT_{2 n-1}=2 \\cdot 2+1 \\cdot 1=2^{2}+1^{2} \\\\\nT_{2 n+1}=1 \\cdot 1=1^{2}\n\\end{gathered}\n$$\n\nThe total number of $B$-squares of the given chessboard equals to\n\n$$\n\\begin{gathered}\nT_{1}+T_{3}+T_{5}+\\ldots+T_{2 n+1}=2\\left(1^{2}+2^{2}+\\ldots+n^{2}\\right)+(n+1)^{2}= \\\\\n\\frac{n(n+1)(2 n+1)}{3}+(n+1)^{2}=\\frac{(n+1)\\left(2 n^{2}+4 n+3\\right)}{3}\n\\end{gathered}\n$$\n\nThe problem is solved.\n\n### 2.3 Geometry", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nC3 ", "solution_match": "## Solution"}}
-{"year": "2007", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $M$ be an interior point of the triangle $A B C$ with angles $\\varangle B A C=70^{\\circ}$ and $\\varangle A B C=80^{\\circ}$. If $\\varangle A C M=10^{\\circ}$ and $\\varangle C B M=20^{\\circ}$, prove that $A B=M C$.", "solution": "Let $O$ be the circumcenter of the triangle $A B C$. Because the triangle $A B C$ is acute, $O$ is in the interior of $\\triangle A B C$. Now we have that $\\varangle A O C=2 \\varangle A B C=160^{\\circ}$, so $\\varangle A C O=10^{\\circ}$ and $\\varangle B O C=2 \\varangle B A C=140^{\\circ}$, so $\\varangle C B O=20^{\\circ}$. Therefore $O \\equiv M$, thus $M A=$ $M B=M C$. Because $\\varangle A B O=80^{\\circ}-20^{\\circ}=60^{\\circ}$, the triangle $A B M$ is equilateral and so $A B=M B=M C$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_5f424aab603cc04b261bg-08.jpg?height=354&width=508&top_left_y=1999&top_left_x=791)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nG1 ", "solution_match": "## Solution"}}
-{"year": "2007", "tier": "T3", "problem_label": "G2", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C D$ be a convex quadrilateral with $\\varangle D A C=\\varangle B D C=36^{\\circ}, \\varangle C B D=18^{\\circ}$ and $\\varangle B A C=72^{\\circ}$. If $P$ is the point of intersection of the diagonals $A C$ and $B D$, find the measure of $\\varangle A P D$.", "solution": "Alternative solution. Let $X$ be the intersection point of the angle bisector of $\\varangle C A D$ and $P D$. As $\\varangle C A X=\\varangle C B X=18^{\\circ}, A B C X$ is cyclic, hence $\\varangle B X C=72^{\\circ}$. It follows that $C X D$ is isosceles. From the SSA criterion for triangles $A X C$ and $A X D$, it follows that either $\\varangle A C X=\\varangle A D X$, or $\\varangle A C X+\\varangle A D X=180^{\\circ}$. The latter being excluded, it follows that triangles $A X C$ and $A X D$ are congruent. Immediate angle chasing leads to the conclusion.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nG2 ", "solution_match": "## Solution"}}
-{"year": "2007", "tier": "T3", "problem_label": "G2", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C D$ be a convex quadrilateral with $\\varangle D A C=\\varangle B D C=36^{\\circ}, \\varangle C B D=18^{\\circ}$ and $\\varangle B A C=72^{\\circ}$. If $P$ is the point of intersection of the diagonals $A C$ and $B D$, find the measure of $\\varangle A P D$.", "solution": "Alternative solution. Let $S$ be the reflection of $D$ in the line $B C$. Triangle $B D S$ is isosceles, with $\\varangle D B S=36^{\\circ}$, hence $\\varangle S D B=\\varangle B S D=72^{\\circ}$. It follows that $A B S D$ is cyclic $\\left(\\varangle B S D+\\varangle B A D=180^{\\circ}\\right)$, hence $\\varangle B A S=\\varangle B D S=72^{\\circ}$ which means that $A, C, S$\nare collinear. $C$ is the incenter of $\\triangle B S D$, therefore $\\varangle P S B=\\varangle P B S=36^{\\circ}$, which leads to $\\varangle D P A=108^{\\circ}$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_5f424aab603cc04b261bg-10.jpg?height=671&width=534&top_left_y=590&top_left_x=767)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nG2 ", "solution_match": "## Solution"}}
-{"year": "2007", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $S$ be a point inside $\\varangle p O q$, and let $k$ be a circle which contains $S$ and touches the legs $O p$ and $O q$ in points $P$ and $Q$ respectively. Straight line $s$ parallel to $O p$ from $S$ intersects $O q$ in a point $R$. Let $T$ be the point of intersection of the ray $(P S$ and circumscribed circle of $\\triangle S Q R$ and $T \\neq S$. Prove that $O T \\| S Q$ and $O T$ is a tangent of the circumscribed circle of $\\triangle S Q R$.", "solution": "Let $\\varangle O P S=\\varphi_{1}$ and $\\varangle O Q S=\\varphi_{2}$. We have that $\\varangle O P S=\\varangle P Q S=\\varphi_{1}$ and $\\varangle O Q S=$ $\\varangle Q P S=\\varphi_{2}$ (tangents to circle $k$ ).\n\nBecause $R S \\| O P$ we have $\\varangle O P S=\\varangle R S T=\\varphi_{1}$ and $\\varangle R Q T=\\varangle R S T=\\varphi_{1}$ (cyclic quadrilateral $R S Q T$ ). So, we have as follows $\\varangle O P T=\\varphi_{1}=\\varangle R Q T=\\varangle O Q T$, which implies that the quadrilateral $O P Q T$ is cyclic. From that we directly obtain $\\varangle Q O T=$ $\\varangle Q P T=\\varphi_{2}=\\varangle O Q S$, so $O T \\| S Q$. From the cyclic quadrilateral $O P Q T$ by easy calculation we get\n\n$$\n\\varangle O T R=\\varangle O T P-\\varangle R T S=\\varangle O Q P-\\varangle R Q S=\\left(\\varphi_{1}+\\varphi_{2}\\right)-\\varphi_{2}=\\varphi_{1}=\\varangle R Q T\n$$\n\nThus, $O T$ is a tangent to the circumscribed circle of $\\triangle S Q R$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_5f424aab603cc04b261bg-11.jpg?height=543&width=803&top_left_y=1319&top_left_x=638)\n\n### 2.4 Number Theory", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nG4 ", "solution_match": "## Solution"}}
-{"year": "2007", "tier": "T3", "problem_label": "NT1", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all the pairs positive integers $(x, y)$ such that\n\n$$\n\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{[x, y]}+\\frac{1}{(x, y)}=\\frac{1}{2}\n$$\n\nwhere $(x, y)$ is the greatest common divisor of $x, y$ and $[x, y]$ is the least common multiple of $x, y$.", "solution": "We put $x=d u$ and $y=d v$ where $d=(x, y)$. So we have $(u, v)=1$. From the conclusion we obtain $2(u+1)(v+1)=d u v$. Because $(v, v+1)=1, v$ divides $2(u+1)$.\n\nCase 1. $u=v$. Hence $x=y=[x, y]=(x, y)$, which leads to the solution $x=8$ and $y=8$.\n\nCase 2. $u<v$. Then $u+1 \\leq v \\Leftrightarrow 2(u+1) \\leq 2 v \\Leftrightarrow \\frac{2(u+1)}{v} \\leq 2$, so $\\frac{2(u+1)}{v} \\in\\{1,2\\}$. But $\\frac{2(u+1)}{v}=\\frac{d u}{v+1}$.\n\nIf $\\frac{2(u+1)}{v}=1$ then we have $(d-2) u=3$. Therefore $(d, u)=(3,3)$ or $(d, u)=(5,1)$ so $(d, u, v)=(3,3,8)$ or $(d, u, v)=(5,1,4)$.\n\nThus we get $(x, y)=(9,24)$ or $(x, y)=(5,20)$.\n\nIf $\\frac{2(u+1)}{v}=2$ we similarly get $(d-2) u=4$ from where $(d, u)=(3,4)$, or $(d, u)=(4,2)$, or $(d, u)=(6,1)$. This leads $(x, y)=(12,15)$ or $(x, y)=(8,12)$ or $(x, y)=(6,12)$.\n\nCase 3. $u>v$. Because of the symmetry of $u, v$ and $x, y$ respectively we get exactly the symmetrical solutions of case 2 .\n\nFinally the pairs of $(x, y)$ which are solutions of the problem are:\n\n$(8,8),(9,24),(24,9),(5,20),(20,5),(12,15),(15,12),(8,12),(12,8),(6,12),(12,6)$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nNT1 ", "solution_match": "## Solution"}}
-{"year": "2007", "tier": "T3", "problem_label": "NT2", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Prove that the equation $x^{2006}-4 y^{2006}-2006=4 y^{2007}+2007 y$ has no solution in the set of the positive integers.", "solution": "We assume the contrary is true. So there are $x$ and $y$ that satisfy the equation. Hence we have\n\n$$\n\\begin{gathered}\nx^{2006}=4 y^{2007}+4 y^{2006}+2007 y+2006 \\\\\nx^{2006}+1=4 y^{2006}(y+1)+2007(y+1) \\\\\nx^{2006}+1=\\left(4 y^{2006}+2007\\right)(y+1)\n\\end{gathered}\n$$\n\nBut $4 y^{2006}+2007 \\equiv 3(\\bmod 4)$, so $x^{2006}+1$ will have at least one prime divisor of the type $4 k+3$. It is known (and easily obtainable by using Fermat's Little Theorem) that this is impossible.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nNT2 ", "solution_match": "## Solution"}}
-{"year": "2007", "tier": "T3", "problem_label": "NT3", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Let $n>1$ be a positive integer and $p$ a prime number such that $n \\mid(p-1)$ and $p \\mid\\left(n^{6}-1\\right)$. Prove that at least one of the numbers $p-n$ and $p+n$ is a perfect square.", "solution": "Since $n \\mid p-1$, then $p=1+n a$, where $a \\geq 1$ is an integer. From the condition $p \\mid n^{6}-1$, it follows that $p|n-1, p| n+1, p \\mid n^{2}+n+1$ or $p \\mid n^{2}-n+1$.\n\n- Let $p \\mid n-1$. Then $n \\geq p+1>n$ which is impossible.\n- Let $p \\mid n+1$. Then $n+1 \\geq p=1+n a$ which is possible only when $a=1$ and $p=n+1$, i.e. $p-n=1=1^{2}$.\n- Let $p \\mid n^{2}+n+1$, i.e. $n^{2}+n+1=p b$, where $b \\geq 1$ is an integer.\n\nThe equality $p=1+n a$ implies $n \\mid b-1$, from where $b=1+n c, c \\geq 0$ is an integer. We have\n\n$$\nn^{2}+n+1=p b=(1+n a)(1+n c)=1+(a+c) n+a c n^{2} \\text { or } n+1=a c n+a+c\n$$\n\nIf $a c \\geq 1$ then $a+c \\geq 2$, which is impossible. If $a c=0$ then $c=0$ and $a=n+1$. Thus we obtain $p=n^{2}+n+1$ from where $p+n=n^{2}+2 n+1=(n+1)^{2}$.\n\n- Let $p \\mid n^{2}-n+1$, i.e. $n^{2}-n+1=p b$ and analogously $b=1+n c$. So\n\n$$\nn^{2}-n+1=p b=(1+n a)(1+n c)=1+(a+c) n+a c n^{2} \\text { or } n-1=a c n+a+c\n$$\n\nSimilarly, we have $c=0, a=n-1$ and $p=n^{2}-n+1$ from where $p-n=n^{2}-2 n+1=$ $(n-1)^{2}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nNT3 ", "solution_match": "## Solution"}}
-{"year": "2007", "tier": "T3", "problem_label": "NT4", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Let $a, b$ be two co-prime positive integers. A number is called good if it can be written in the form $a x+b y$ for non-negative integers $x, y$. Define the function $f: \\mathbb{Z} \\rightarrow \\mathbb{Z}$ as $f(n)=n-n_{a}-n_{b}$, where $s_{t}$ represents the remainder of $s$ upon division by $t$. Show that an integer $n$ is good if and only if the infinite sequence $n, f(n), f(f(n)), \\ldots$ contains only non-negative integers.", "solution": "If $n$ is good then $n=a x+b y$ also $n_{a}=(b y)_{a}$ and $n_{b}=(a x)_{b}$ so\n\n$$\nf(n)=a x-(a x)_{b}+b y-(b y)_{a}=b y^{\\prime}+a x^{\\prime}\n$$\n\nis also good, thus the sequence contains only good numbers which are non-negative.\n\nNow we have to prove that if the sequence contains only non-negative integers then $n$ is good. Because the sequence is non-increasing then the sequence will become constant from some point onwards. But $f(k)=k$ implies that $k$ is a multiple of $a b$ thus some term of the sequence is good. We are done if we prove the following:\n\nLemma: $f(n)$ is good implies $n$ is good.\n\nProof of Lemma: $n=2 n-n_{a}-n_{b}-f(n)=a x^{\\prime}+b y^{\\prime}-a x-b y=a\\left(x^{\\prime}-x\\right)+b\\left(y^{\\prime}-y\\right)$ and $x^{\\prime} \\geq x$ because $n \\geq f(n) \\Rightarrow n-n_{a} \\geq f(n)-f(n)_{a} \\Rightarrow a x^{\\prime} \\geq a x+b y-(b y)_{a} \\geq a x$. Similarly $y^{\\prime} \\geq y$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nNT4 ", "solution_match": "## Solution"}}
-{"year": "2007", "tier": "T3", "problem_label": "NT5", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Let $p$ be a prime number. Show that $7 p+3^{p}-4$ is not a perfect square.", "solution": "Assume that for a prime number $p$ greater than $3, m=7 p+3^{p}-4$ is a perfect square. Let $m=n^{2}$ for some $n \\in \\mathbb{Z}$. By Fermat's Little Theorem,\n\n$$\nm=7 p+3^{p}-4 \\equiv 3-4 \\equiv-1 \\quad(\\bmod p)\n$$\n\nIf $p=4 k+3, k \\in \\mathbb{Z}$, then again by Fermat's Little Theorem\n\n$$\n-1 \\equiv m^{2 k+1} \\equiv n^{4 k+2} \\equiv n^{p-1} \\equiv 1 \\quad(\\bmod p), \\text { but } p>3\n$$\n\na contradiction. So $p \\equiv 1(\\bmod 4)$.\n\nTherefore $m=7 p+3^{p}-4 \\equiv 3-1 \\equiv 2(\\bmod 4)$. But this is a contradiction since 2 is not perfect square in $(\\bmod 4)$. For $p=2$ we have $m=19$ and for $p=3$ we have $m=44$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nNT5 ", "solution_match": "## Solution"}}
+{"year": "2007", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $a$ be a real positive number such that $a^{3}=6(a+1)$. Prove that the equation $x^{2}+a x+a^{2}-6=0$ has no solution in the set of the real number.", "solution": "The discriminant of the equation is $\\Delta=3\\left(8-a^{2}\\right)$. If we accept that $\\Delta \\geq 0$, then $a \\leq 2 \\sqrt{2}$ and $\\frac{1}{a} \\geq \\frac{\\sqrt{2}}{4}$, from where $a^{2} \\geq 6+6 \\cdot \\frac{\\sqrt{2}}{4}=6+\\frac{6}{a} \\geq 6+\\frac{3 \\sqrt{2}}{2}>8$ (contradiction).", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nA1 ", "solution_match": "## Solution"}}
+{"year": "2007", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Prove that $\\frac{a^{2}-b c}{2 a^{2}+b c}+\\frac{b^{2}-c a}{2 b^{2}+c a}+\\frac{c^{2}-a b}{2 c^{2}+a b} \\leq 0$ for any real positive numbers $a, b, c$.", "solution": "The inequality rewrites as $\\sum \\frac{2 a^{2}+b c-3 b c}{2 a^{2}+b c} \\leq 0$, or $3-3 \\sum \\frac{b c}{2 a^{2}+b c} \\leq 0$ in other words $\\sum \\frac{b c}{2 a^{2}+b c} \\geq 1$.\n\nUsing Cauchy-Schwarz inequality we have\n\n$$\n\\sum \\frac{b c}{2 a^{2}+b c}=\\sum \\frac{b^{2} c^{2}}{2 a^{2} b c+b^{2} c^{2}} \\geq \\frac{\\left(\\sum b c\\right)^{2}}{2 a b c(a+b+c)+\\sum b^{2} c^{2}}=1\n$$\n\nas claimed.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nA2 ", "solution_match": "## Solution"}}
+{"year": "2007", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $A$ be a set of positive integers containing the number 1 and at least one more element. Given that for any two different elements $m, n$ of $A$ the number $\\frac{m+1}{(m+1, n+1)}$ is also an element of $A$, prove that $A$ coincides with the set of positive integers.", "solution": "Let $a>1$ be lowest number in $A \\backslash\\{1\\}$. For $m=a, n=1$ one gets $y=\\frac{a+1}{(2, a+1)} \\in A$. Since $(2, a+1)$ is either 1 or 2 , then $y=a+1$ or $y=\\frac{a+1}{2}$.\n\nBut $1<\\frac{a+1}{2}<a$, hence $y=a+1$. Applying the given property for $m=a+1, n=a$ one has $\\frac{a+2}{(a+2, a+1)}=a+2 \\in A$, and inductively $t \\in A$ for all integers $t \\geq a$.\n\nFurthermore, take $m=2 a-1, n=3 a-1$ (now in $A!$ ); as $(m+1, n+1)=(2 a, 3 a)=a$ one obtains $\\frac{2 a}{a}=2 \\in A$, so $a=2$, by the definition of $a$.\n\nThe conclusion follows immediately.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nA3 ", "solution_match": "## Solution"}}
+{"year": "2007", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $a$ and $b$ be positive integers bigger than 2. Prove that there exists a positive integer $k$ and a sequence $n_{1}, n_{2}, \\ldots, n_{k}$ consisting of positive integers, such that $n_{1}=a$, $n_{k}=b$, and $\\left(n_{i}+n_{i+1}\\right) \\mid n_{i} n_{i+1}$ for all $i=1,2, \\ldots, k-1$.", "solution": "We write $a \\Leftrightarrow b$ if the required sequence exists. It is clear that $\\Leftrightarrow$ is equivalence relation, i.e. $a \\Leftrightarrow a,(a \\Leftrightarrow b$ implies $b \\Rightarrow a)$ and $(a \\Leftrightarrow b, b \\Leftrightarrow c$ imply $a \\Leftrightarrow c$ ).\n\nWe shall prove that for every $a \\geq 3$, ( $a-$ an integer), $a \\Leftrightarrow 3$.\n\nIf $a=2^{s} t$, where $t>1$ is an odd number, we take the sequence\n\n$$\n2^{s} t, 2^{s}\\left(t^{2}-t\\right), 2^{s}\\left(t^{2}+t\\right), 2^{s}(t+1)=2^{s+1} \\cdot \\frac{t+1}{2}\n$$\n\nSince $\\frac{t+1}{2}<t$ after a finite number of steps we shall get a power of 2 . On the other side, if $s>1$ we have $2^{s}, 3 \\cdot 2^{s}, 3 \\cdot 2^{s-1}, 3 \\cdot 2^{s-2}, \\ldots, 3$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nA4 ", "solution_match": "## Solution"}}
+{"year": "2007", "tier": "T3", "problem_label": "A5", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "The real numbers $x, y, z, m, n$ are positive, such that $m+n \\geq 2$. Prove that\n\n$$\n\\begin{gathered}\nx \\sqrt{y z(x+m y)(x+n z)}+y \\sqrt{x z(y+m x)(y+n z)}+z \\sqrt{x y(z+m x)(x+n y)} \\leq \\\\\n\\frac{3(m+n)}{8}(x+y)(y+z)(z+x) .\n\\end{gathered}\n$$", "solution": "Using the AM-GM inequality we have\n\n$$\n\\begin{aligned}\n& \\sqrt{y z(x+m y)(x+n z)}=\\sqrt{(x z+m y z)(x y+n y z)} \\leq \\frac{x y+x z+(m+n) y z}{2} \\\\\n& \\sqrt{x z(y+m x)(y+n z)}=\\sqrt{(y z+m x z)(x y+n x z)} \\leq \\frac{x y+y z+(m+n) x z}{2} \\\\\n& \\sqrt{x y(z+m x)(z+n y)}=\\sqrt{(y z+m x y)(x z+n x y)} \\leq \\frac{x z+y z+(m+n) x y}{2}\n\\end{aligned}\n$$\n\nThus it is enough to prove that\n\n$$\n\\begin{aligned}\nx[x y+x z+(m+n) y z] & +y[x y+y z+(m+n) x z]+z[x y+y z+(m+n) x z] \\leq \\\\\n\\leq & \\frac{3(m+n)}{4}(x+y)(y+z)(z+x),\n\\end{aligned}\n$$\n\nor\n\n$$\n4[A+3(m+n) B] \\leq 3(m+n)(A+2 B) \\Leftrightarrow 6(m+n) B \\leq[3(m+n)-4] A\n$$\n\nwhere $A=x^{2} y+x^{2} z+x y^{2}+y^{2} z+x z^{2}+y z^{2}, B=x y z$.\n\nBecause $m+n \\geq 2$ we obtain the inequality $m+n \\leq 3(m+n)-4$. From AMGM inequality it follows that $6 B \\leq A$. From the last two inequalities we deduce that $6(m+n) B \\leq[3(m+n)-4] A$. The inequality is proved.\n\nEquality holds when $m=n=1$ and $x=y=z$.\n\n### 2.2 Combinatorics", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nA5 ", "solution_match": "## Solution"}}
+{"year": "2007", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "We call a tiling of an $m \\times n$ rectangle with corners (see figure below) \"regular\" if there is no sub-rectangle which is tiled with corners. Prove that if for some $m$ and $n$ there exists a \"regular\" tiling of the $m \\times n$ rectangular then there exists a \"regular\" tiling also for the $2 m \\times 2 n$ rectangle.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_5f424aab603cc04b261bg-06.jpg?height=92&width=83&top_left_y=1259&top_left_x=992)", "solution": "A corner-shaped tile consists of 3 squares. Let us call \"center of the tile\" the square that has two neighboring squares. Notice that in a \"regular\" tiling, the squares situated in the corners of the rectangle have to be covered by the \"center\" of a tile, otherwise a $2 \\times 3$ (or $3 \\times 2$ ) rectangle tiled with two tiles would form.\n\nConsider a $2 m \\times 2 n$ rectangle, divide it into four $m \\times n$ rectangles by drawing its midlines, then do a \"regular\" tiling for each of these rectangles. In the center of the $2 m \\times 2 n$ rectangle we will necessarily obtain the following configuration:\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_5f424aab603cc04b261bg-06.jpg?height=200&width=209&top_left_y=1896&top_left_x=932)\n\nNow simply change the position of these four tiles into:\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_5f424aab603cc04b261bg-06.jpg?height=188&width=203&top_left_y=2299&top_left_x=938)\n\nIt is easy to see that this tiling is \"regular\".", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nC1 ", "solution_match": "## Solution"}}
+{"year": "2007", "tier": "T3", "problem_label": "C2", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Consider 50 points in the plane, no three of them belonging to the same line. The points have been colored into four colors. Prove that there are at least 130 scalene triangles whose vertices are colored in the same color.", "solution": "Since $50=4 \\cdot 12+2$, according to the pigeonhole principle we will have at least 13 points colored in the same color. We start with the:\n\nLemma. Given $n>8$ points in the plane, no three of them collinear, then there are at least $\\frac{n(n-1)(n-8)}{6}$ scalene triangles with vertices among the given points.\n\nProof. There are $\\frac{n(n-1)}{2}$ segments and $\\frac{n(n-1)(n-2)}{6}$ triangles with vertices among the given points. We shall prove that there are at most $n(n-1)$ isosceles triangles. Indeed, for every segment $A B$ we can construct at most two isosceles triangles (if we have three $A B C, A B D$ and $A B E$, than $C, D, E$ will be collinear). Hence we have at least\n\n$$\n\\frac{n(n-1)(n-2)}{6}-n(n-1)=\\frac{n(n-1)(n-8)}{6} \\text { scalene triangles. }\n$$\n\nFor $n=13$ we have $\\frac{13 \\cdot 12 \\cdot 5}{6}=130$, QED.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nC2 ", "solution_match": "## Solution"}}
+{"year": "2007", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "The nonnegative integer $n$ and $(2 n+1) \\times(2 n+1)$ chessboard with squares colored alternatively black and white are given. For every natural number $m$ with $1<m<2 n+1$, an $m \\times m$ square of the given chessboard that has more than half of its area colored in black, is called a $B$-square. If the given chessboard is a $B$-square, find in terms of $n$ the total number of $B$-squares of this chessboard.", "solution": "Every square with even side length will have an equal number of black and white $1 \\times 1$ squares, so it isn't a $B$-square. In a square with odd side length, there is one more $1 \\times 1$ black square than white squares, if it has black corner squares. So, a square with odd side length is a $B$-square either if it is a $1 \\times 1$ black square or it has black corners.\n\nLet the given $(2 n+1) \\times(2 n+1)$ chessboard be a $B$-square and denote by $b_{i}(i=$ $1,2, \\ldots, n+1)$ the lines of the chessboard, which have $n+1$ black $1 \\times 1$ squares, by $w_{i}(i=1,2, \\ldots, n)$ the lines of the chessboard, which have $n$ black $1 \\times 1$ squares and by $T_{m}(m=1,3,5, \\ldots, 2 n-1,2 n+1)$ the total number of $B$-squares of dimension $m \\times m$ of the given chessboard.\n\nFor $T_{1}$ we obtain $T_{1}=(n+1)(n+1)+n \\cdot n=(n+1)^{2}+n^{2}$.\n\nFor computing $T_{3}$ we observe that there are $n 3 \\times 3 B$-squares, which have the black corners on each pair of lines $\\left(b_{i}, b_{i+1}\\right)$ for $i=1,2, \\ldots, n$ and there are $n-13 \\times 3 B$-squares, which have the black corners on each pair of lines $\\left(w_{i}, w_{i+1}\\right)$ for $i=1,2, \\ldots, n-1$. So, we have\n\n$$\nT_{3}=n \\cdot n+(n-1)(n-1)=n^{2}+(n-1)^{2} \\text {. }\n$$\n\nBy using similar arguments for each pair of lines $\\left(b_{i}, b_{i+2}\\right)$ for $i=1,2, \\ldots, n-1$ and for each pair of lines $\\left(w_{i}, w_{i+2}\\right)$ for $i=1,2, \\ldots, n-2$ we compute\n\n$$\nT_{5}=(n-1)(n-1)+(n-2)(n-2)=(n-1)^{2}+(n-2)^{2}\n$$\n\nStep by step, we obtain\n\n$$\n\\begin{gathered}\nT_{7}=(n-2)(n-2)+(n-3)(n-3)=(n-2)^{2}+(n-3)^{2} \\\\\n\\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\\\\nT_{2 n-1}=2 \\cdot 2+1 \\cdot 1=2^{2}+1^{2} \\\\\nT_{2 n+1}=1 \\cdot 1=1^{2}\n\\end{gathered}\n$$\n\nThe total number of $B$-squares of the given chessboard equals to\n\n$$\n\\begin{gathered}\nT_{1}+T_{3}+T_{5}+\\ldots+T_{2 n+1}=2\\left(1^{2}+2^{2}+\\ldots+n^{2}\\right)+(n+1)^{2}= \\\\\n\\frac{n(n+1)(2 n+1)}{3}+(n+1)^{2}=\\frac{(n+1)\\left(2 n^{2}+4 n+3\\right)}{3}\n\\end{gathered}\n$$\n\nThe problem is solved.\n\n### 2.3 Geometry", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nC3 ", "solution_match": "## Solution"}}
+{"year": "2007", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $M$ be an interior point of the triangle $A B C$ with angles $\\varangle B A C=70^{\\circ}$ and $\\varangle A B C=80^{\\circ}$. If $\\varangle A C M=10^{\\circ}$ and $\\varangle C B M=20^{\\circ}$, prove that $A B=M C$.", "solution": "Let $O$ be the circumcenter of the triangle $A B C$. Because the triangle $A B C$ is acute, $O$ is in the interior of $\\triangle A B C$. Now we have that $\\varangle A O C=2 \\varangle A B C=160^{\\circ}$, so $\\varangle A C O=10^{\\circ}$ and $\\varangle B O C=2 \\varangle B A C=140^{\\circ}$, so $\\varangle C B O=20^{\\circ}$. Therefore $O \\equiv M$, thus $M A=$ $M B=M C$. Because $\\varangle A B O=80^{\\circ}-20^{\\circ}=60^{\\circ}$, the triangle $A B M$ is equilateral and so $A B=M B=M C$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_5f424aab603cc04b261bg-08.jpg?height=354&width=508&top_left_y=1999&top_left_x=791)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nG1 ", "solution_match": "## Solution"}}
+{"year": "2007", "tier": "T3", "problem_label": "G2", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C D$ be a convex quadrilateral with $\\varangle D A C=\\varangle B D C=36^{\\circ}, \\varangle C B D=18^{\\circ}$ and $\\varangle B A C=72^{\\circ}$. If $P$ is the point of intersection of the diagonals $A C$ and $B D$, find the measure of $\\varangle A P D$.", "solution": "Alternative solution. Let $X$ be the intersection point of the angle bisector of $\\varangle C A D$ and $P D$. As $\\varangle C A X=\\varangle C B X=18^{\\circ}, A B C X$ is cyclic, hence $\\varangle B X C=72^{\\circ}$. It follows that $C X D$ is isosceles. From the SSA criterion for triangles $A X C$ and $A X D$, it follows that either $\\varangle A C X=\\varangle A D X$, or $\\varangle A C X+\\varangle A D X=180^{\\circ}$. The latter being excluded, it follows that triangles $A X C$ and $A X D$ are congruent. Immediate angle chasing leads to the conclusion.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nG2 ", "solution_match": "## Solution"}}
+{"year": "2007", "tier": "T3", "problem_label": "G2", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C D$ be a convex quadrilateral with $\\varangle D A C=\\varangle B D C=36^{\\circ}, \\varangle C B D=18^{\\circ}$ and $\\varangle B A C=72^{\\circ}$. If $P$ is the point of intersection of the diagonals $A C$ and $B D$, find the measure of $\\varangle A P D$.", "solution": "Alternative solution. Let $S$ be the reflection of $D$ in the line $B C$. Triangle $B D S$ is isosceles, with $\\varangle D B S=36^{\\circ}$, hence $\\varangle S D B=\\varangle B S D=72^{\\circ}$. It follows that $A B S D$ is cyclic $\\left(\\varangle B S D+\\varangle B A D=180^{\\circ}\\right)$, hence $\\varangle B A S=\\varangle B D S=72^{\\circ}$ which means that $A, C, S$\nare collinear. $C$ is the incenter of $\\triangle B S D$, therefore $\\varangle P S B=\\varangle P B S=36^{\\circ}$, which leads to $\\varangle D P A=108^{\\circ}$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_5f424aab603cc04b261bg-10.jpg?height=671&width=534&top_left_y=590&top_left_x=767)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nG2 ", "solution_match": "## Solution"}}
+{"year": "2007", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $S$ be a point inside $\\varangle p O q$, and let $k$ be a circle which contains $S$ and touches the legs $O p$ and $O q$ in points $P$ and $Q$ respectively. Straight line $s$ parallel to $O p$ from $S$ intersects $O q$ in a point $R$. Let $T$ be the point of intersection of the ray $(P S$ and circumscribed circle of $\\triangle S Q R$ and $T \\neq S$. Prove that $O T \\| S Q$ and $O T$ is a tangent of the circumscribed circle of $\\triangle S Q R$.", "solution": "Let $\\varangle O P S=\\varphi_{1}$ and $\\varangle O Q S=\\varphi_{2}$. We have that $\\varangle O P S=\\varangle P Q S=\\varphi_{1}$ and $\\varangle O Q S=$ $\\varangle Q P S=\\varphi_{2}$ (tangents to circle $k$ ).\n\nBecause $R S \\| O P$ we have $\\varangle O P S=\\varangle R S T=\\varphi_{1}$ and $\\varangle R Q T=\\varangle R S T=\\varphi_{1}$ (cyclic quadrilateral $R S Q T$ ). So, we have as follows $\\varangle O P T=\\varphi_{1}=\\varangle R Q T=\\varangle O Q T$, which implies that the quadrilateral $O P Q T$ is cyclic. From that we directly obtain $\\varangle Q O T=$ $\\varangle Q P T=\\varphi_{2}=\\varangle O Q S$, so $O T \\| S Q$. From the cyclic quadrilateral $O P Q T$ by easy calculation we get\n\n$$\n\\varangle O T R=\\varangle O T P-\\varangle R T S=\\varangle O Q P-\\varangle R Q S=\\left(\\varphi_{1}+\\varphi_{2}\\right)-\\varphi_{2}=\\varphi_{1}=\\varangle R Q T\n$$\n\nThus, $O T$ is a tangent to the circumscribed circle of $\\triangle S Q R$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_5f424aab603cc04b261bg-11.jpg?height=543&width=803&top_left_y=1319&top_left_x=638)\n\n### 2.4 Number Theory", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nG4 ", "solution_match": "## Solution"}}
+{"year": "2007", "tier": "T3", "problem_label": "NT1", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all the pairs positive integers $(x, y)$ such that\n\n$$\n\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{[x, y]}+\\frac{1}{(x, y)}=\\frac{1}{2}\n$$\n\nwhere $(x, y)$ is the greatest common divisor of $x, y$ and $[x, y]$ is the least common multiple of $x, y$.", "solution": "We put $x=d u$ and $y=d v$ where $d=(x, y)$. So we have $(u, v)=1$. From the conclusion we obtain $2(u+1)(v+1)=d u v$. Because $(v, v+1)=1, v$ divides $2(u+1)$.\n\nCase 1. $u=v$. Hence $x=y=[x, y]=(x, y)$, which leads to the solution $x=8$ and $y=8$.\n\nCase 2. $u<v$. Then $u+1 \\leq v \\Leftrightarrow 2(u+1) \\leq 2 v \\Leftrightarrow \\frac{2(u+1)}{v} \\leq 2$, so $\\frac{2(u+1)}{v} \\in\\{1,2\\}$. But $\\frac{2(u+1)}{v}=\\frac{d u}{v+1}$.\n\nIf $\\frac{2(u+1)}{v}=1$ then we have $(d-2) u=3$. Therefore $(d, u)=(3,3)$ or $(d, u)=(5,1)$ so $(d, u, v)=(3,3,8)$ or $(d, u, v)=(5,1,4)$.\n\nThus we get $(x, y)=(9,24)$ or $(x, y)=(5,20)$.\n\nIf $\\frac{2(u+1)}{v}=2$ we similarly get $(d-2) u=4$ from where $(d, u)=(3,4)$, or $(d, u)=(4,2)$, or $(d, u)=(6,1)$. This leads $(x, y)=(12,15)$ or $(x, y)=(8,12)$ or $(x, y)=(6,12)$.\n\nCase 3. $u>v$. Because of the symmetry of $u, v$ and $x, y$ respectively we get exactly the symmetrical solutions of case 2 .\n\nFinally the pairs of $(x, y)$ which are solutions of the problem are:\n\n$(8,8),(9,24),(24,9),(5,20),(20,5),(12,15),(15,12),(8,12),(12,8),(6,12),(12,6)$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nNT1 ", "solution_match": "## Solution"}}
+{"year": "2007", "tier": "T3", "problem_label": "NT2", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Prove that the equation $x^{2006}-4 y^{2006}-2006=4 y^{2007}+2007 y$ has no solution in the set of the positive integers.", "solution": "We assume the contrary is true. So there are $x$ and $y$ that satisfy the equation. Hence we have\n\n$$\n\\begin{gathered}\nx^{2006}=4 y^{2007}+4 y^{2006}+2007 y+2006 \\\\\nx^{2006}+1=4 y^{2006}(y+1)+2007(y+1) \\\\\nx^{2006}+1=\\left(4 y^{2006}+2007\\right)(y+1)\n\\end{gathered}\n$$\n\nBut $4 y^{2006}+2007 \\equiv 3(\\bmod 4)$, so $x^{2006}+1$ will have at least one prime divisor of the type $4 k+3$. It is known (and easily obtainable by using Fermat's Little Theorem) that this is impossible.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nNT2 ", "solution_match": "## Solution"}}
+{"year": "2007", "tier": "T3", "problem_label": "NT3", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Let $n>1$ be a positive integer and $p$ a prime number such that $n \\mid(p-1)$ and $p \\mid\\left(n^{6}-1\\right)$. Prove that at least one of the numbers $p-n$ and $p+n$ is a perfect square.", "solution": "Since $n \\mid p-1$, then $p=1+n a$, where $a \\geq 1$ is an integer. From the condition $p \\mid n^{6}-1$, it follows that $p|n-1, p| n+1, p \\mid n^{2}+n+1$ or $p \\mid n^{2}-n+1$.\n\n- Let $p \\mid n-1$. Then $n \\geq p+1>n$ which is impossible.\n- Let $p \\mid n+1$. Then $n+1 \\geq p=1+n a$ which is possible only when $a=1$ and $p=n+1$, i.e. $p-n=1=1^{2}$.\n- Let $p \\mid n^{2}+n+1$, i.e. $n^{2}+n+1=p b$, where $b \\geq 1$ is an integer.\n\nThe equality $p=1+n a$ implies $n \\mid b-1$, from where $b=1+n c, c \\geq 0$ is an integer. We have\n\n$$\nn^{2}+n+1=p b=(1+n a)(1+n c)=1+(a+c) n+a c n^{2} \\text { or } n+1=a c n+a+c\n$$\n\nIf $a c \\geq 1$ then $a+c \\geq 2$, which is impossible. If $a c=0$ then $c=0$ and $a=n+1$. Thus we obtain $p=n^{2}+n+1$ from where $p+n=n^{2}+2 n+1=(n+1)^{2}$.\n\n- Let $p \\mid n^{2}-n+1$, i.e. $n^{2}-n+1=p b$ and analogously $b=1+n c$. So\n\n$$\nn^{2}-n+1=p b=(1+n a)(1+n c)=1+(a+c) n+a c n^{2} \\text { or } n-1=a c n+a+c\n$$\n\nSimilarly, we have $c=0, a=n-1$ and $p=n^{2}-n+1$ from where $p-n=n^{2}-2 n+1=$ $(n-1)^{2}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nNT3 ", "solution_match": "## Solution"}}
+{"year": "2007", "tier": "T3", "problem_label": "NT4", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Let $a, b$ be two co-prime positive integers. A number is called good if it can be written in the form $a x+b y$ for non-negative integers $x, y$. Define the function $f: \\mathbb{Z} \\rightarrow \\mathbb{Z}$ as $f(n)=n-n_{a}-n_{b}$, where $s_{t}$ represents the remainder of $s$ upon division by $t$. Show that an integer $n$ is good if and only if the infinite sequence $n, f(n), f(f(n)), \\ldots$ contains only non-negative integers.", "solution": "If $n$ is good then $n=a x+b y$ also $n_{a}=(b y)_{a}$ and $n_{b}=(a x)_{b}$ so\n\n$$\nf(n)=a x-(a x)_{b}+b y-(b y)_{a}=b y^{\\prime}+a x^{\\prime}\n$$\n\nis also good, thus the sequence contains only good numbers which are non-negative.\n\nNow we have to prove that if the sequence contains only non-negative integers then $n$ is good. Because the sequence is non-increasing then the sequence will become constant from some point onwards. But $f(k)=k$ implies that $k$ is a multiple of $a b$ thus some term of the sequence is good. We are done if we prove the following:\n\nLemma: $f(n)$ is good implies $n$ is good.\n\nProof of Lemma: $n=2 n-n_{a}-n_{b}-f(n)=a x^{\\prime}+b y^{\\prime}-a x-b y=a\\left(x^{\\prime}-x\\right)+b\\left(y^{\\prime}-y\\right)$ and $x^{\\prime} \\geq x$ because $n \\geq f(n) \\Rightarrow n-n_{a} \\geq f(n)-f(n)_{a} \\Rightarrow a x^{\\prime} \\geq a x+b y-(b y)_{a} \\geq a x$. Similarly $y^{\\prime} \\geq y$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nNT4 ", "solution_match": "## Solution"}}
+{"year": "2007", "tier": "T3", "problem_label": "NT5", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Let $p$ be a prime number. Show that $7 p+3^{p}-4$ is not a perfect square.", "solution": "Assume that for a prime number $p$ greater than $3, m=7 p+3^{p}-4$ is a perfect square. Let $m=n^{2}$ for some $n \\in \\mathbb{Z}$. By Fermat's Little Theorem,\n\n$$\nm=7 p+3^{p}-4 \\equiv 3-4 \\equiv-1 \\quad(\\bmod p)\n$$\n\nIf $p=4 k+3, k \\in \\mathbb{Z}$, then again by Fermat's Little Theorem\n\n$$\n-1 \\equiv m^{2 k+1} \\equiv n^{4 k+2} \\equiv n^{p-1} \\equiv 1 \\quad(\\bmod p), \\text { but } p>3\n$$\n\na contradiction. So $p \\equiv 1(\\bmod 4)$.\n\nTherefore $m=7 p+3^{p}-4 \\equiv 3-1 \\equiv 2(\\bmod 4)$. But this is a contradiction since 2 is not perfect square in $(\\bmod 4)$. For $p=2$ we have $m=19$ and for $p=3$ we have $m=44$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl", "problem_match": "\nNT5 ", "solution_match": "## Solution"}}

JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl

@@ -1,19 +1,19 @@
-{"year": "2009", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO", "problem": "Determine all integers $a, b, c$ satisfying the identities:\n\n$$\n\\begin{gathered}\na+b+c=15 \\\\\n(a-3)^{3}+(b-5)^{3}+(c-7)^{3}=540\n\\end{gathered}\n$$", "solution": "Solution II: We use the substitution $a-3=x, b-5=y, c-7=z$.\n\nNow, equations are transformed to:\n\n$$\n\\begin{aligned}\nx+y+z & =0 \\\\\nx^{3}+y^{3}+z^{3} & =540 .\n\\end{aligned}\n$$\n\nSubstituting $z=-x-y$ in second equation, we get:\n\n$$\n-3 x y^{2}-3 x^{2} y=540\n$$\n\nor\n\n$$\nx y(x+y)=-180\n$$\n\nor\n\n$$\nx y z=180 \\text {. }\n$$\n\nReturning to starting problem we have:\n\n$$\n(a-3)(b-5)(c-7)=180\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nA1 ", "solution_match": "\nSolution "}}
-{"year": "2009", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO", "problem": "Determine all integers $a, b, c$ satisfying the identities:\n\n$$\n\\begin{gathered}\na+b+c=15 \\\\\n(a-3)^{3}+(b-5)^{3}+(c-7)^{3}=540\n\\end{gathered}\n$$", "solution": "Solution proceeds as the previous one.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nA1 ", "solution_match": "\nSolution "}}
-{"year": "2009", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO", "problem": "Find the maximum value of $z+x$, if $(x, y, z, t)$ satisfies the conditions:\n\n$$\n\\left\\{\\begin{array}{l}\nx^{2}+y^{2}=4 \\\\\nz^{2}+t^{2}=9 \\\\\nx t+y z \\geq 6\n\\end{array}\\right.\n$$", "solution": "I: From the conditions we have\n\n$$\n36=\\left(x^{2}+y^{2}\\right)\\left(z^{2}+t^{2}\\right)=(x t+y z)^{2}+(x z-y t)^{2} \\geq 36+(x z-y t)^{2}\n$$\n\nand this implies $x z-y t=0$.\n\nNow it is clear that\n\n$$\nx^{2}+z^{2}+y^{2}+t^{2}=(x+z)^{2}+(y-t)^{2}=13\n$$\n\nand the maximum value of $z+x$ is $\\sqrt{13}$. It is achieved for $x=\\frac{4}{\\sqrt{13}}, y=t=\\frac{6}{\\sqrt{13}}$ and $z=\\frac{9}{\\sqrt{13}}$.\n\nSolution II: From inequality $x t+y z \\geq 6$ and problem conditions we have:\n\n$$\n\\begin{gathered}\n(x t+y z)^{2}-36 \\geq 0 \\Leftrightarrow \\\\\n(x t+y z)^{2}-\\left(x^{2}+y^{2}\\right)\\left(z^{2}+t^{2}\\right) \\geq 0 \\Leftrightarrow \\\\\n2 x y z t-x^{2} y^{2}-y^{2} t^{2} \\geq 0 \\Leftrightarrow \\\\\n-(x z-y t)^{2} \\geq 0\n\\end{gathered}\n$$\n\nFrom here we have $x z=y t$.\n\nFurthermore,\n\n$$\nx^{2}+y^{2}+z^{2}+t^{2}=(x+z)^{2}+(y-t)^{2}=13\n$$\n\nand it follows that\n\n$$\n(x+z)^{2} \\leq 13\n$$\n\nThus,\n\n$$\nx+z \\leq \\sqrt{13}\n$$\n\nEquality $x+z=\\sqrt{13}$ holds if we have $y=t$ and $z^{2}-x^{2}=5$, which leads to $z-x=\\frac{5}{\\sqrt{13}}$. Therefore, $x=\\frac{4}{\\sqrt{13}}, y=t=\\frac{6}{\\sqrt{13}}, z=\\frac{9}{\\sqrt{13}}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nA2 ", "solution_match": "\nSolution "}}
-{"year": "2009", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO", "problem": "Find all values of the real parameter $a$, for which the system\n\n$$\n\\left\\{\\begin{array}{c}\n(|x|+|y|-2)^{2}=1 \\\\\ny=a x+5\n\\end{array}\\right.\n$$\n\nhas exactly three solutions.", "solution": "The first equation is equivalent to\n\n$$\n|x|+|y|=1\n$$\n\nor\n\n$$\n|x|+|y|=3\n$$\n\nThe graph of the first equation is symmetric with respect to both axes. In the first quadrant it is reduced to $x+y=1$, whose graph is segment connecting points $(1,0)$ and $(0,1)$. Thus, the graph of\n\n$$\n|x|+|y|=1\n$$\n\nis square with vertices $(1,0),(0,1),(-1,0)$ and $(0,-1)$. Similarly, the graph of\n\n$$\n|x|+|y|=3\n$$\n\nis a square with vertices $(3,0),(0,3),(-3,0)$ and $(0,-3)$. The graph of the second equation of the system is a straight line with slope $a$ passing through (0,5). This line intersects the graph of the first equation in three points exactly, when passing through one of the points $(1,0)$ or $(-1,0)$. This happens if and only if $a=5$ or $a=-5$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nA3 ", "solution_match": "\nSolution:"}}
-{"year": "2009", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "exam": "JBMO", "problem": "Real numbers $x, y, z$ satisfy\n\n$$\n0<x, y, z<1\n$$\n\nand\n\n$$\nx y z=(1-x)(1-y)(1-z) .\n$$\n\nShow that\n\n$$\n\\frac{1}{4} \\leq \\max \\{(1-x) y,(1-y) z,(1-z) x\\}\n$$", "solution": "It is clear that $a(1-a) \\leq \\frac{1}{4}$ for any real numbers $a$ (equivalent to $0<(2 a-1)^{2}$ ). Thus,\n\n$$\n\\begin{gathered}\nx y z=(1-x)(1-y)(1-z) \\\\\n(x y z)^{2}=[x(1-x)][y(1-y)][z(1-z)] \\leq \\frac{1}{4} \\cdot \\frac{1}{4} \\cdot \\frac{1}{4}=\\frac{1}{4^{3}} \\\\\nx y z \\leq \\frac{1}{2^{3}}\n\\end{gathered}\n$$\n\nIt implies that at least one of $x, y, z$ is at less or equal to $\\frac{1}{2}$. Let us say that $x \\leq \\frac{1}{2}$, and notice that $1-x \\geq \\frac{1}{2}$.\n\nAssume contrary to required result, that we have\n\n$$\n\\frac{1}{4}>\\max \\{(1-x) y,(1-y) x,(1-z) x\\}\n$$\n\nNow\n\n$$\n(1-x) y<\\frac{1}{4}, \\quad(1-y) z<\\frac{1}{4}, \\quad(1-z) x<\\frac{1}{4}\n$$\n\nFrom here we deduce:\n\n$$\ny<\\frac{1}{4} \\cdot \\frac{1}{1-x} \\leq \\frac{1}{4} \\cdot 2=\\frac{1}{2}\n$$\n\nNotice that $1-y>\\frac{1}{2}$.\n\nUsing same reasoning we conclude:\n\n$$\nz<\\frac{1}{2}, \\quad 1-z>\\frac{1}{2}\n$$\n\nUsing these facts we derive:\n\n$$\n\\frac{1}{8}=\\frac{1}{2} \\cdot \\frac{1}{2} \\cdot \\frac{1}{2}>x y z=(1-x)(1-y)(1-z)>\\frac{1}{2} \\cdot \\frac{1}{2} \\cdot \\frac{1}{2}=\\frac{1}{8}\n$$\n\nContradiction!\n\nRemark: The exercise along with its proof generalizes for any given (finite) number of numbers, and you can consider this new form in place of the proposed one:\n\nExercise: If for the real numbers $x_{1}, x_{2}, \\ldots, x_{n}, 0<x_{i}<1$, for all indices $i$, and\n\n$$\nx_{1} x_{2} \\ldots x_{n}=\\left(1-x_{1}\\right)\\left(1-x_{2}\\right) \\ldots\\left(1-x_{n}\\right)\n$$\n\nshow that\n\n$$\n\\frac{1}{4} \\leq \\max _{1 \\leq i \\leq n}\\left(1-x_{i}\\right) x_{i+1}\n$$\n\n(where $x_{n+1}=x_{1}$ ).\n\nOr you can consider the following variation:\n\nExercise: If for the real numbers $x_{1}, x_{2}, \\ldots, x_{2009}, 0<x_{i}<1$, for all indices $i$, and\n\n$$\nx_{1} x_{2} \\ldots x_{2009}=\\left(1-x_{1}\\right)\\left(1-x_{2}\\right) \\ldots\\left(1-x_{2009}\\right)\n$$\n\nshow that\n\n$$\n\\frac{1}{4} \\leq \\max _{1 \\leq i \\leq 2009}\\left(1-x_{i}\\right) x_{i+1}\n$$\n\n(where $x_{2010}=x_{1}$ ).", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nA4 ", "solution_match": "\nSolution:"}}
-{"year": "2009", "tier": "T3", "problem_label": "A5", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $x, y, z$ be positive real numbers. Prove that:\n\n$$\n\\left(x^{2}+y+1\\right)\\left(x^{2}+z+1\\right)\\left(y^{2}+z+1\\right)\\left(y^{2}+x+1\\right)\\left(z^{2}+x+1\\right)\\left(z^{2}+y+1\\right) \\geq(x+y+z)^{6}\n$$", "solution": "I: Applying Cauchy-Schwarz's inequality:\n\n$$\n\\left(x^{2}+y+1\\right)\\left(z^{2}+y+1\\right)=\\left(x^{2}+y+1\\right)\\left(1+y+z^{2}\\right) \\geq(x+y+z)^{2}\n$$\n\nUsing the same reasoning we deduce:\n\n$$\n\\left(x^{2}+z+1\\right)\\left(y^{2}+z+1\\right) \\geq(x+y+z)^{2}\n$$\n\nand\n\n$$\n\\left(y^{2}+x+1\\right)\\left(z^{2}+x+1\\right) \\geq(x+y+z)^{2}\n$$\n\nMultiplying these three inequalities we get the desired result.\n\nSolution II: We have\n\n$$\n\\begin{gathered}\n\\left(x^{2}+y+1\\right)\\left(z^{2}+y+1\\right) \\geq(x+y+z)^{2} \\Leftrightarrow \\\\\nx^{2} z^{2}+x^{2} y+x^{2}+y z^{2}+y^{2}+y+z^{2}+y+1 \\geq x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x \\Leftrightarrow \\\\\n\\left(x^{2} z^{2}-2 z x+1\\right)+\\left(x^{2} y-2 x y+y\\right)+\\left(y z^{2}-2 y z+y\\right) \\geq 0 \\Leftrightarrow \\\\\n(x z-1)^{2}+y(x-1)^{2}+y(z-1)^{2} \\geq 0\n\\end{gathered}\n$$\n\nwhich is correct.\n\nUsing the same reasoning we get:\n\n$$\n\\begin{aligned}\n& \\left(x^{2}+z+1\\right)\\left(y^{2}+z+1\\right) \\geq(x+y+z)^{2} \\\\\n& \\left(y^{2}+x+1\\right)\\left(z^{2}+x+1\\right) \\geq(x+y+z)^{2}\n\\end{aligned}\n$$\n\nMultiplying these three inequalities we get the desired result. Equality is attained at $x=y=z=1$.\n\n### 2.2 Combinatorics", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nA5 ", "solution_match": "\nSolution "}}
-{"year": "2009", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Each one of 2009 distinct points in the plane is coloured in blue or red, so that on every blue-centered unit circle there are exactly two red points. Find the gratest possible number of blue points.", "solution": "Each pair of red points can belong to at most two blue-centered unit circles. As $n$ red points form $\\frac{n(n-1)}{2}$ pairs, we can have not more than twice that number of blue points, i.e. $n(n-1)$ blue points. Thus, the total number of points can not exceed\n\n$$\nn+n(n-1)=n^{2}\n$$\n\nAs $44^{2}<2009, n$ must be at least 45 . We can arrange 45 distinct red points on a segment of length 1 , and color blue all but $16\\left(=45^{2}-2009\\right)$ points on intersections of the redcentered unit circles (all points of intersection are distinct, as no blue-centered unit circle can intersect the segment more than twice). Thus, the greatest possible number of blue points is $2009-45=1964$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nC1 ", "solution_match": "\nSolution:"}}
-{"year": "2009", "tier": "T3", "problem_label": "C2", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Five players $(A, B, C, D, E)$ take part in a bridge tournament. Every two players must play (as partners) against every other two players. Any two given players can be partners not more than once per day. What is the least number of days needed for this tournament?", "solution": "A given pair must play with three other pairs and these plays must be in different days, so at three days are needed. Suppose that three days suffice. Let the pair $A B$ play against $C D$ on day $x$. Then $A B-D E$ and $C D-B E$ cannot play on day $x$. Then one of the other two plays of $D E$ (with $A C$ and $B C$ ) must be on day $x$. Similarly, one of the plays of $B E$ with $A C$ or $A D$ must be on day $x$. Thus, two of the plays in the chain $B C-D E-A C-B E-A D$ are on day $x$ (more than two among these cannot be on one day).\n\nConsider the chain $A B-C D-E A-B D-C E-A B$. At least three days are needed for playing all the matches within it. For each of these days we conclude (as above) that there are exactly two of the plays in the chain $B C-D E-A C-B E-A D-B C$ on that day. This is impossible, as this chain consists of five plays.\n\nIt remains to show that four days will suffice:\n\nDay 1: $A B-C D, A C-D E, A D-C E, A E-B C$\n\nDay 2: $A B-D E, A C-B D, A D-B C, B E-C D$\n\nDay 3: $A B-C E, A D-B E, A E-B D, B C-D E$\n\nDay 4: $A C-B E, A E-C D, B D-C E$.\n\nRemark: It is possible to have 5 games in one day (but not on each day).\n\n## Alternative solution:\n\nThere are 10 pairs. Each of them plays 3 games, so the tournament needs to last at least 3 days. Assume the tournament could finish in 3 days. Then every pair must play one game on each day. There are 15 games to be played, so you must have 5 games on each day. Call \"Day 1\" the day AB plays against CD, \"Day 2\" the day AB plays against DE\nand \"Day 3\" the day AB plays against CE. Let us examine the other possible games on Day 1. CE can't play $\\mathrm{AB}$, so it must play either $\\mathrm{AD}$ or $\\mathrm{BD}$. $\\mathrm{DE}$ can't play $\\mathrm{AB}$, so it must play AC or BC. Similarly, AE can't play CD, so it must play BC or BD, and BE must play either $\\mathrm{AC}$ or $\\mathrm{AD}$. We obtain the following circular diagram in which exact every other game has to take place on Day 1, either the red ones or the blue ones:\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_17e61ec1e039d21af3a0g-10.jpg?height=463&width=485&top_left_y=768&top_left_x=794)\n\nSimilar reasoning leads un to the following diagram for Day 2. Here again, either all the red matches have to take place, or all the blue matches have to take place on Day 2.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_17e61ec1e039d21af3a0g-10.jpg?height=463&width=483&top_left_y=1593&top_left_x=795)\n\nOne can't have the blue matches on both Day 1 and Day 2 because AD-BE would repeat itself.\n\nWe can't have the red mathes on both days as this would repeat the match BD-CE.\n\nWe can not have the blue matches on Day 1 and the red matches on Day 2 because this would repeat the game AC-BE. Finally, choosing the red matches on Day 1 and the blue ones on Day 2 won't work either as the game AE-BC would repeat itself.\n\nIn conclusion, the tournament has to last at least four days. An example of how it could\nbe organized in four days is given in the previous solution.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nC2 ", "solution_match": "\nSolution:"}}
-{"year": "2009", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "a) In how many ways can we read the word SARAJEVO from the table below, if it is allowed to jump from cell to an adjacent cell (by vertex or a side) cell?\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_17e61ec1e039d21af3a0g-11.jpg?height=497&width=596&top_left_y=631&top_left_x=730)\n\nb) After the letter in one cell was deleted, only 525 ways to read the word SARAJEVO remained. Find all possible positions of that cell.", "solution": "In the first of the tables below the number in each cell shows the number of ways to reach that cell from the start (which is the sum of the quantities in the cells, from which we can come), and in the second one are the number of ways to arrive from that cell to the end (which is the sum of the quantities in the cells, to which we can go).\n\na) The answer is 750 , as seen from the second table.\n\nb) If we delete the letter in a cell, the number of ways to read SARAJEVO will decrease by the product of the numbers in the corresponding cell in the two tables. As $750-525=225$, this product has to be 225. This happens only for two cells on the third row. Here is the table with the products:", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nC3 ", "solution_match": "\nSolution:"}}
-{"year": "2009", "tier": "T3", "problem_label": "C4", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Determine all pairs $(m, n)$ for which it is possible to tile the table $m \\times n$ with \"corners\" as in the figure below, with the condition that in the tiling there is no rectangle (except for the $m \\times n$ one) regularly covered with corners.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_17e61ec1e039d21af3a0g-11.jpg?height=85&width=83&top_left_y=1942&top_left_x=992)", "solution": "Every \"corner\" covers exactly 3 squares, so a necessary condition for the tiling to exists is $3 \\mid m n$.\n\nFirst, we shall prove that for a tiling with our condition to exist, it is necessary that both $m, n$ for $m, n>3$ to be even. Suppose the contrary, i.e. suppose that that $m>3$ is odd (without losing generality). Look at the \"corners\" that cover squares on the side of length $m$ of table $m \\times n$. Because $m$ is odd, there must be a \"corner\" which covers exactly one square of that side. But any placement of that corner forces existence of a $2 \\times 3$ rectangle in the tiling. Thus, $m$ and $n$ for $m, n>3$ must be even and at least one of them is divisible by 3 .\n\nNotice that in the corners of table $m \\times n$, the \"corner\" must be placed such that it covers the square in the corner of the rectangle and its two neighboring squares, otherwise, again, a $2 \\times 3$ rectangle would form.\n\nIf one of $m$ and $n$ is 2 then condition forces that the only convenient tables are $2 \\times 3$ and $3 \\times 2$. If we try to find the desired tiling when $m=4$, then we are forced to stop at table $4 \\times 6$ because of the conditions of problem.\n\nWe easily find an example of a desired tiling for the table $6 \\times 6$ and, more generally, a tiling for a $6 \\times 2 k$ table.\n\nThus, it will be helpful to prove that the desired tiling exists for tables $6 k \\times 4 \\ell$, for $k, \\ell \\geq 2$. Divide that table at rectangle $6 \\times 4$ and tile that rectangle as we described. Now, change placement of problematic \"corners\" as in figure.\n\nThus, we get desired tilling for this type of table.\n\nSimilarly, we prove existence in case $6 k \\times(4 k+2)$ where $m, \\ell \\geq 2$. But, we first divide table at two tables $6 k \\times 6$ and $6 k \\times 4(\\ell-1)$. Divide them at rectangles $6 \\times 6$ and $6 \\times 4$. Tile them as we described earlier, and arrange problematic \"corners\" as in previous case. So, $2 \\times 3,3 \\times 2,6 \\times 2 k, 2 k \\times 6, k \\geq 2$, and $6 k \\times 4 \\ell$ for $k, \\ell \\geq 2$ and $6 k \\times(4 \\ell+2)$ for $k, \\ell \\geq 2$ are the convenient pairs.\n\nRemark: The problem is inspired by a problem given at Romanian Selection Test 2000, but it is completely different.\n\nRemark: Alternatively, the problem can be relaxed by asking: \"Does such a tiling exist for some concrete values of $m$ and $n$ ?\" .\n\n### 2.3 Geometry", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nC4 ", "solution_match": "\nSolution:"}}
-{"year": "2009", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C D$ be a parallelogram with $A C>B D$, and let $O$ be the point of intersection of $A C$ and $B D$. The circle with center at $O$ and radius $O A$ intersects the extensions of $A D$ and $A B$ at points $G$ and $L$, respectively. Let $Z$ be intersection point of lines $B D$ and $G L$. Prove that $\\angle Z C A=90^{\\circ}$.", "solution": "From the point $L$ we draw a parallel line to $B D$ that intersects lines $A C$ and $A G$ at points $N$ and $R$ respectively. Since $D O=O B$, we have that $N R=N L$, and point $N$ is the midpoint of segment $L R$.\n\nLet $K$ be the midpoint of $G L$. Now, $N K \\| R G$, and\n\n$$\n\\angle A G L=\\angle N K L=\\angle A C L\n$$\n\nTherefore, from the cyclic quadrilateral $N K C L$ we deduce:\n\n$$\n\\angle K C N=\\angle K L N\n$$\n\nNow, since $L R \\| D Z$, we have\n\n$$\n\\angle K L N=\\angle K Z O\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_17e61ec1e039d21af3a0g-13.jpg?height=729&width=817&top_left_y=475&top_left_x=631)\n\nIt implies that quadrilateral $O K C Z$ is cyclic, and\n\n$$\n\\angle O K Z=\\angle O C Z\n$$\n\nSince $O K \\perp G L$, we derive that $\\angle Z C A=90^{\\circ}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nG1 ", "solution_match": "## Solution:"}}
-{"year": "2009", "tier": "T3", "problem_label": "G2", "problem_type": "Geometry", "exam": "JBMO", "problem": "In a right trapezoid $A B C D(A B \\| C D)$ the angle at vertex $B$ measures $75^{\\circ}$. Point $H$ is the foot of the perpendicular from point $A$ to the line $B C$. If $B H=D C$ and $A D+A H=8$, find the area of $A B C D$.", "solution": "Produce the legs of the trapezoid until they intersect at point $E$. The triangles $A B H$ and $E C D$ are congruent (ASA). The area of $A B C D$ is equal to area of triangle $E A H$ of hypotenuse\n\n$$\nA E=A D+D E=A D+A H=8\n$$\n\nLet $M$ be the midpoint of $A E$. Then\n\n$$\nM E=M A=M H=4\n$$\n\nand $\\angle A M H=30^{\\circ}$. Now, the altitude from $H$ to AM equals one half of $M H$, namely 2. Finally, the area is 8 .\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_17e61ec1e039d21af3a0g-13.jpg?height=802&width=300&top_left_y=1689&top_left_x=1272)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nG2 ", "solution_match": "\nSolution:"}}
-{"year": "2009", "tier": "T3", "problem_label": "G3", "problem_type": "Geometry", "exam": "JBMO", "problem": "A parallelogram $A B C D$ with obtuse angle $\\angle A B C$ is given. After rotating the triangle $A C D$ around the vertex $C$, we get a triangle $C D^{\\prime} A^{\\prime}$, such that points $B, C$ and $D^{\\prime}$ are collinear. The extension of the median of triangle $C D^{\\prime} A^{\\prime}$ that passes through $D^{\\prime}$ intersects the straight line $B D$ at point $P$. Prove that $P C$ is the bisector of the angle $\\angle B P D^{\\prime}$.", "solution": "Let $A C \\cap B D=\\{X\\}$ and $P D^{\\prime} \\cap C A^{\\prime}=\\{Y\\}$. Because $A X=C X$ and $C Y=Y A^{\\prime}$, we deduce:\n\n$$\n\\triangle A B C \\cong \\triangle C D A \\cong \\triangle C D^{\\prime} A^{\\prime} \\Rightarrow \\triangle A B X \\cong \\triangle C D^{\\prime} Y, \\triangle B C X \\cong \\triangle D^{\\prime} A^{\\prime} Y\n$$\n\nIt follows that\n\n$$\n\\angle A B X=\\angle C D^{\\prime} Y\n$$\n\nLet $M$ and $N$ be orthogonal projections of the point $C$ on the straight lines $P D^{\\prime}$ and $B P$, respectively, and $Q$ is the orthogonal projection of the point $A$ on the straight line $B P$. Because $C D^{\\prime}=A B$, we have that $\\triangle A B Q \\cong \\triangle C D^{\\prime} M$.\n\nWe conclude that $C M=A Q$. But, $A X=C X$ and $\\triangle A Q X \\cong \\triangle C N X$. So, $C M=C N$ and $P C$ is the bisector of the angle $\\angle B P D^{\\prime}$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_17e61ec1e039d21af3a0g-14.jpg?height=1003&width=1305&top_left_y=1286&top_left_x=387)\n\nMuch shortened: $\\triangle C D^{\\prime} Y \\equiv \\triangle C D X$ means their altitudes from $C$ are also equal, i.e. $C M=C N$ and the conclusion.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nG3 ", "solution_match": "\nSolution:"}}
-{"year": "2009", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C D E$ be a convex pentagon such that $A B+C D=B C+D E$ and let $k$ be a semicircle with center on side $A E$ that touches the sides $A B, B C, C D$ and $D E$ of the pentagon, respectively, at points $P, Q, R$ and $S$ (different from the vertices of the pentagon). Prove that $P S \\| A E$.", "solution": "Let $O$ be center of $k$. We deduce that $B P=B Q, C Q=C R, D R=D S$, since those are tangents to the circle $k$. Using the condition $A B+C D=B C+D E$, we derive:\n\n$$\nA P+B P+C R+D R=B Q+C Q+D S+E S\n$$\n\nFrom here we have $A P=E S$.\n\nThus,\n\n$$\n\\triangle A P O \\cong \\triangle E S O\\left(A P=E S, \\angle A P O=\\angle E S O=90^{\\circ}, P O=S O\\right)\n$$\n\nThis implies\n\n$$\n\\angle O P S=\\angle O S P\n$$\n\nTherefore,\n\n$$\n\\angle A P S=\\angle A P O+\\angle O P S=90^{\\circ}+\\angle O P S=90^{\\circ}+\\angle O S P=\\angle P S E\n$$\n\nNow, from quadrilateral $A P S E$ we deduce:\n\n$$\n2 \\angle E A P+2 \\angle A P S=\\angle E A P+\\angle A P S+\\angle P S E+\\angle S E A=360^{\\circ}\n$$\n\nSo,\n\n$$\n\\angle E A P+\\angle A P S=180^{\\circ}\n$$\n\nand $A P S E$ is isosceles trapezoid. Therefore, $A E \\| P S$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_17e61ec1e039d21af3a0g-15.jpg?height=571&width=939&top_left_y=1733&top_left_x=567)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nG4 ", "solution_match": "\nSolution:"}}
-{"year": "2009", "tier": "T3", "problem_label": "G5", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A, B, C$ and $O$ be four points in the plane, such that $\\angle A B C>90^{\\circ}$ and $O A=$ $O B=O C$. Define the point $D \\in A B$ and the line $\\ell$ such that $D \\in \\ell, A C \\perp D C$ and $\\ell \\perp A O$. Line $\\ell$ cuts $A C$ at $E$ and the circumcircle of $\\triangle A B C$ at $F$. Prove that the circumcircles of triangles $B E F$ and $C F D$ are tangent at $F$.", "solution": "Let $\\ell \\cap A C=\\{K\\}$ and define $G$ to be the mirror image of the point $A$ with respect to $O$. Then $A G$ is a diameter of the circumcircle of the triangle $A B C$, therefore $A C \\perp C G$. On the other hand we have $A C \\perp D C$, and it implies that points $D, C, G$ are collinear.\n\nMoreover, as $A E \\perp D G$ and $D E \\perp A G$, we obtain that $E$ is the orthocenter of triangle $A D G$ and $G E \\perp A D$. As $A G$ is a diameter, we have $A B \\perp B G$, and since $A D \\perp G E$, the points $E, G$, and $B$ are collinear.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_17e61ec1e039d21af3a0g-16.jpg?height=1083&width=851&top_left_y=972&top_left_x=634)\n\nNotice that\n\n$$\n\\angle C A G=90^{\\circ}-\\angle A G C=\\angle K D C\n$$\n\nand\n\n$$\n\\angle C A G=\\angle G F C\n$$\n\nsince both subtend the same arc.\n\nHence,\n\n$$\n\\angle F D G=\\angle G F C\n$$\n\nTherefore, $G F$ is tangent to the circumcircle of the triangle $C D F$ at point $F$.\n\nWe claim that line $G F$ is also tangent to the circumcircle of triangle $B E F$ at point $F$, which concludes the proof.\n\nThe claim is equivalent to $\\angle G B F=\\angle E F G$. Denote by $F^{\\prime}$ the second intersection point - other than $F$ - of line $\\ell$ with the circumcircle of triangle $A B C$. Observe that $\\angle G B F=\\angle G F^{\\prime} F$, because both angles subtend the same arc, and $\\angle F F^{\\prime} G=\\angle E F G$, since $A G$ is the perpendicular bisector of the chord $F F^{\\prime}$, and we are done.\n\n### 2.4 Number Theory", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nG5 ", "solution_match": "\nSolution:"}}
-{"year": "2009", "tier": "T3", "problem_label": "NT1", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Determine all positive integer numbers $k$ for which the numbers $k+9$ are perfect squares and the only prime factors of $k$ are 2 and 3 .", "solution": "We have an integer $x$ such that\n\n$$\nx^{2}=k+9\n$$\n\n$k=2^{a} 3^{b}, a, b \\geq 0, a, b \\in \\mathbb{N}$.\n\nTherefore,\n\n$$\n(x-3)(x+3)=k \\text {. }\n$$\n\nIf $b=0$ then we have $k=16$.\n\nIf $b>0$ then we have $3 \\mid k+9$. Hence, $3 \\mid x^{2}$ and $9 \\mid k$.\n\nTherefore, we have $b \\geq 2$. Let $x=3 y$.\n\n$$\n(y-1)(y+1)=2^{a} 3^{b-2}\n$$\n\nIf $a=0$ then $b=3$ and we have $k=27$.\n\nIf $a \\geq 1$, then the numbers $y-1$ and $y+1$ are even. Therefore, we have $a \\geq 2$, and\n\n$$\n\\frac{y-1}{2} \\cdot \\frac{y+1}{2}=2^{a-2} 3^{b-2}\n$$\n\nSince the numbers $\\frac{y-1}{2}, \\frac{y+1}{2}$ are consecutive numbers, these numbers have to be powers of 2 and 3 . Let $m=a-2, n=b-2$.\n\n- If $2^{m}-3^{n}=1$ then we have $m \\geq n$. For $n=0$ we have $m=1, a=3, b=2$ and $k=72$. For $n>0$ using $\\bmod 3$ we have that $m$ is even number. Let $m=2 t$. Therefore,\n\n$$\n\\left(2^{t}-1\\right)\\left(2^{t}+1\\right)=3^{n}\n$$\n\nHence, $t=1, m=2, n=1$ and $a=4, b=3, k=432$.\n\n- If $3^{n}-2^{m}=1$, then $m>0$. For $m=1$ we have $n=1, a=3, b=3, k=216$. For $m>1$ using $\\bmod 4$ we have that $n$ is even number. Let $n=2 t$.\n\n$$\n\\left(3^{t}-1\\right)\\left(3^{t}+1\\right)=2^{m}\n$$\n\nTherefore, $t=1, n=2, m=3, a=5, b=4, k=2592$.\n\nSet of solutions: $\\{16,27,72,216,432,2592\\}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nNT1 ", "solution_match": "\nSolution:"}}
-{"year": "2009", "tier": "T3", "problem_label": "NT2", "problem_type": "Number Theory", "exam": "JBMO", "problem": "A group of $n>1$ pirates of different age owned total of 2009 coins. Initially each pirate (except for the youngest one) had one coin more than the next younger.\n\na) Find all possible values of $n$.\n\nb) Every day a pirate was chosen. The chosen pirate gave a coin to each of the other pirates. If $n=7$, find the largest possible number of coins a pirate can have after several days.", "solution": "a) If $n$ is odd, then it is a divisor of $2009=7 \\times 7 \\times 41$. If $n>49$, then $n$ is at least $7 \\times 41$, while the average pirate has 7 coins, so the initial division is impossible. So, we can have $n=7, n=41$ or $n=49$. Each of these cases is possible (e.g. if $n=49$, the average pirate has 41 coins, so the initial amounts are from $41-24=17$ to $41+24=65$ ).\n\nIf $n$ is even, then 2009 is multiple of the sum $S$ of the oldest and the youngest pirate. If $S<7 \\times 41$, then $S$ is at most 39 and the pairs of pirates of sum $S$ is at least 41 , so we must have at least 82 pirates, a contradiction. So we can have just $S=7 \\times 41=287$ and $S=49 \\times 41=2009$; respectively, $n=2 \\times 7=14$ or $n=2 \\times 1=2$. Each of these cases is possible (e.g. if $n=14$, the initial amounts are from $144-7=137$ to $143+7=150$ ). In total, $n$ is one of the numbers $2,7,13,41$ and 49 .\n\nb) If $n=7$, the average pirate has $7 \\times 41=287$ coins, so the initial amounts are from 284 to 290; they have different residues modulo 7. The operation decreases one of the amounts by 6 and increases the other ones by 1 , so the residues will be different at all times. The largest possible amount in one pirate's possession will be achieved if all the others have as little as possible, namely $0,1,2,3,4$ and 5 coins (the residues modulo 7 have to be different). If this happens, the wealthiest pirate will have $2009-14=1994$ coins. Indeed, this can be achieved e.g. if every day (until that moment) the coins are given by the second wealthiest: while he has more than 5 coins, he can provide the 6 coins needed, and when he has no more than five, the coins at the poorest six pirates have to be $0,1,2,3,4,5$. Thus, $n=1994$ can be achieved.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nNT2 ", "solution_match": "## Solution:"}}
-{"year": "2009", "tier": "T3", "problem_label": "NT3", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all pairs $(x, y)$ of integers which satisfy the equation\n\n$$\n(x+y)^{2}\\left(x^{2}+y^{2}\\right)=2009^{2}\n$$", "solution": "Let $x+y=s, x y=p$ with $s \\in \\mathbb{Z}^{*}$ and $p \\in \\mathbb{Z}$. The given equation can be written in the form\n\n$$\ns^{2}\\left(s^{2}-2 p\\right)=2009^{2}\n$$\n\nor\n\n$$\ns^{2}-2 p=\\left(\\frac{2009}{s}\\right)^{2}\n$$\n\nSo, $s$ divides $2009=7^{2} \\times 41$ and it follows that $p \\neq 0$.\n\nIf $p>0$, then $2009^{2}=s^{2}\\left(s^{2}-2 p\\right)=s^{4}-2 p s^{2}<s^{4}$. We obtain that $s$ divides 2009 and $|s| \\geq 49$. Thus, $s \\in\\{ \\pm 49, \\pm 287, \\pm 2009\\}$.\n\n- For $s= \\pm 49$, we have $p=360$, and $(x, y)=\\{(40,9),(9,40),(-40,-9),(-9,-40)\\}$.\n- For $s \\in\\{ \\pm 287, \\pm 2009\\}$ the equation has no integer solutions.\n\nIf $p<0$, then $2009^{2}=s^{4}-2 p s^{2}>s^{4}$. We obtain that $s$ divides 2009 and $|s| \\leq 41$. Thus, $s \\in\\{ \\pm 1, \\pm 7, \\pm 41\\}$. For these values of $s$ the equation has no integer solutions.\n\nSo, the given equation has only the solutions $(40,9),(9,40),(-40,-9),(-9,-40)$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nNT3 ", "solution_match": "\nSolution:"}}
-{"year": "2009", "tier": "T3", "problem_label": "NT4", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Determine all prime numbers $p_{1}, p_{2}, \\ldots, p_{12}, p_{13}, p_{1} \\leq p_{2} \\leq \\ldots \\leq p_{12} \\leq p_{13}$, such that\n\n$$\np_{1}^{2}+p_{2}^{2}+\\ldots+p_{12}^{2}=p_{13}^{2}\n$$\n\nand one of them is equal to $2 p_{1}+p_{9}$.", "solution": "Obviously, $p_{13} \\neq 2$, because sum of squares of 12 prime numbers is greater or equal to $12 \\times 2^{2}=48$. Thus, $p_{13}$ is odd number and $p_{13} \\geq 7$.\n\nWe have that $n^{2} \\equiv 1(\\bmod 8)$, when $n$ is odd. Let $k$ be the number of prime numbers equal to 2 . Looking at equation modulo 8 we get:\n\n$$\n4 k+12-k \\equiv 1 \\quad(\\bmod 8)\n$$\n\nSo, $k \\equiv 7(\\bmod 8)$ and because $k \\leq 12$ we get $k=7$. Therefore, $p_{1}=p_{2}=\\ldots=p_{7}=2$. Furthermore, we are looking for solutions of equations:\n\n$$\n28+p_{8}^{2}+p_{9}^{2}+p_{10}^{2}+p_{11}^{2}+p_{12}^{2}=p_{13}^{2}\n$$\n\nwhere $p_{8}, p_{9}, \\ldots, p_{13}$ are odd prime numbers and one of them is equal to $p_{9}+4$.\n\nNow, we know that when $n$ is not divisible by $3, n^{2} \\equiv 1(\\bmod 3)$. Let $s$ be number of prime numbers equal to 3 . Looking at equation modulo 3 we get:\n\n$$\n28+5-s \\equiv 1 \\quad(\\bmod 3)\n$$\n\nThus, $s \\equiv 2(\\bmod 3)$ and because $s \\leq 5, s$ is either 2 or 5 . We will consider both cases. i. When $s=2$, we get $p_{8}=p_{9}=3$. Thus, we are looking for prime numbers $p_{10} \\leq p_{11} \\leq$ $p_{12} \\leq p_{13}$ greater than 3 and at least one of them is 7 (certainly $p_{13} \\neq 7$ ), that satisfy\n\n$$\n46+p_{10}^{2}+p_{11}^{2}+p_{12}^{2}=p_{13}^{2}\n$$\n\nWe know that $n^{2} \\equiv 1(\\bmod 5)$ or $n^{2}=4(\\bmod 5)$ when $n$ is not divisible by 5 . It is not possible that $p_{10}=p_{11}=5$, because in that case $p_{12}$ must be equal to 7 and the left-hand side would be divisible by 5 , which contradicts the fact that $p_{13} \\geq 7$. So, we proved that $p_{10}=5$ or $p_{10}=7$.\n\nIf $p_{10}=5$ then $p_{11}=7$ because $p_{11}$ is the least of remaining prime numbers. Thus, we are looking for solutions of equation\n\n$$\n120=p_{13}^{2}-p_{12}^{2}\n$$\n\nin prime numbers. Now, from\n\n$$\n2^{3} \\cdot 3 \\cdot 5=\\left(p_{12}-p_{12}\\right)\\left(p_{13}+p_{12}\\right)\n$$\n\nthat desired solutions are $p_{12}=7, p_{13}=13 ; p_{12}=13, p_{13}=17 ; p_{12}=29, p_{13}=31$.\n\nIf $p_{10}=7$ we are solving equation:\n\n$$\n95+p_{11}^{2}+p_{12}^{2}=p_{13}^{2}\n$$\n\nin prime numbers greater than 5 . But left side can give residues 0 or 3 modulo 5 , while right side can give only 1 or 4 modulo 5 . So, in this case we do not have solution.\n\nii. When $s=5$ we get equation:\n\n$$\n28+45=73=p_{13}^{2}\n$$\n\nbut 73 is not square or integer and we do not have solution in this case.\n\nFinally, only solutions are:\n\n$\\{(2,2,2,2,2,2,2,3,3,5,7,7,13),(2,2,2,2,2,2,2,3,3,5,7,13,17),(2,2,2,2,2,2,2,3,3,5,7,29,31)\\}$. NT5 Show that there are infinitely many positive integers $c$, such that the following equations both have solutions in positive integers:\n\n$$\n\\left(x^{2}-c\\right)\\left(y^{2}-c\\right)=z^{2}-c\n$$\n\nand\n\n$$\n\\left(x^{2}+c\\right)\\left(y^{2}-c\\right)=z^{2}-c\n$$\n\nSolution: The firs equation always has solutions, namely the triples $\\{x, x+1, x(x+1)-c\\}$ for all $x \\in \\mathbb{N}$. Indeed,\n\n$$\n\\left(x^{2}-c\\right)\\left((x+1)^{2}-c\\right)=x^{2}(x+1)^{2}-2 c\\left(x^{2}+(x+1)^{2}\\right)+c^{2}=(x(x+1)-c)^{2}-c .\n$$\n\nFor second equation, we try $z=|x y-c|$. We need\n\n$$\n\\left(x^{2}+c\\right)\\left(y^{2}-c\\right)=(x y-c)^{2}\n$$\n\nor\n\n$$\nx^{2} y^{2}+c\\left(y^{2}-x^{2}\\right)-c^{2}=x^{2} y^{2}-2 x y c+c^{2}\n$$\n\nCancelling the common terms we get\n\n$$\nc\\left(x^{2}-y^{2}+2 x y\\right)=2 c^{2}\n$$\n\nor\n\n$$\nc=\\frac{x^{2}-y^{2}+2 x y}{2}\n$$\n\nTherefore, all $c$ of this form will work. This expression is a positive integer if $x$ and $y$ have the same parity, and it clearly takes infinitely many positive values. We only need to check $z \\neq 0$, i.e. $c \\neq x y$, which is true for $x \\neq y$. For example, one can take\n\n$$\ny=x-2\n$$\n\nand\n\n$$\nz=\\frac{x^{2}-(x-2)^{2}+2 x(x-2)}{2}=x^{2}-2 \\text {. }\n$$\n\nThus, $\\{(x, x-2,2 x-2)\\}$ is a solution for $c=x^{2}-2$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nNT4 ", "solution_match": "\nSolution:"}}
+{"year": "2009", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Determine all integers $a, b, c$ satisfying the identities:\n\n$$\n\\begin{gathered}\na+b+c=15 \\\\\n(a-3)^{3}+(b-5)^{3}+(c-7)^{3}=540\n\\end{gathered}\n$$", "solution": "Solution II: We use the substitution $a-3=x, b-5=y, c-7=z$.\n\nNow, equations are transformed to:\n\n$$\n\\begin{aligned}\nx+y+z & =0 \\\\\nx^{3}+y^{3}+z^{3} & =540 .\n\\end{aligned}\n$$\n\nSubstituting $z=-x-y$ in second equation, we get:\n\n$$\n-3 x y^{2}-3 x^{2} y=540\n$$\n\nor\n\n$$\nx y(x+y)=-180\n$$\n\nor\n\n$$\nx y z=180 \\text {. }\n$$\n\nReturning to starting problem we have:\n\n$$\n(a-3)(b-5)(c-7)=180\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nA1 ", "solution_match": "\nSolution "}}
+{"year": "2009", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Determine all integers $a, b, c$ satisfying the identities:\n\n$$\n\\begin{gathered}\na+b+c=15 \\\\\n(a-3)^{3}+(b-5)^{3}+(c-7)^{3}=540\n\\end{gathered}\n$$", "solution": "Solution proceeds as the previous one.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nA1 ", "solution_match": "\nSolution "}}
+{"year": "2009", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Find the maximum value of $z+x$, if $(x, y, z, t)$ satisfies the conditions:\n\n$$\n\\left\\{\\begin{array}{l}\nx^{2}+y^{2}=4 \\\\\nz^{2}+t^{2}=9 \\\\\nx t+y z \\geq 6\n\\end{array}\\right.\n$$", "solution": "I: From the conditions we have\n\n$$\n36=\\left(x^{2}+y^{2}\\right)\\left(z^{2}+t^{2}\\right)=(x t+y z)^{2}+(x z-y t)^{2} \\geq 36+(x z-y t)^{2}\n$$\n\nand this implies $x z-y t=0$.\n\nNow it is clear that\n\n$$\nx^{2}+z^{2}+y^{2}+t^{2}=(x+z)^{2}+(y-t)^{2}=13\n$$\n\nand the maximum value of $z+x$ is $\\sqrt{13}$. It is achieved for $x=\\frac{4}{\\sqrt{13}}, y=t=\\frac{6}{\\sqrt{13}}$ and $z=\\frac{9}{\\sqrt{13}}$.\n\nSolution II: From inequality $x t+y z \\geq 6$ and problem conditions we have:\n\n$$\n\\begin{gathered}\n(x t+y z)^{2}-36 \\geq 0 \\Leftrightarrow \\\\\n(x t+y z)^{2}-\\left(x^{2}+y^{2}\\right)\\left(z^{2}+t^{2}\\right) \\geq 0 \\Leftrightarrow \\\\\n2 x y z t-x^{2} y^{2}-y^{2} t^{2} \\geq 0 \\Leftrightarrow \\\\\n-(x z-y t)^{2} \\geq 0\n\\end{gathered}\n$$\n\nFrom here we have $x z=y t$.\n\nFurthermore,\n\n$$\nx^{2}+y^{2}+z^{2}+t^{2}=(x+z)^{2}+(y-t)^{2}=13\n$$\n\nand it follows that\n\n$$\n(x+z)^{2} \\leq 13\n$$\n\nThus,\n\n$$\nx+z \\leq \\sqrt{13}\n$$\n\nEquality $x+z=\\sqrt{13}$ holds if we have $y=t$ and $z^{2}-x^{2}=5$, which leads to $z-x=\\frac{5}{\\sqrt{13}}$. Therefore, $x=\\frac{4}{\\sqrt{13}}, y=t=\\frac{6}{\\sqrt{13}}, z=\\frac{9}{\\sqrt{13}}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nA2 ", "solution_match": "\nSolution "}}
+{"year": "2009", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Find all values of the real parameter $a$, for which the system\n\n$$\n\\left\\{\\begin{array}{c}\n(|x|+|y|-2)^{2}=1 \\\\\ny=a x+5\n\\end{array}\\right.\n$$\n\nhas exactly three solutions.", "solution": "The first equation is equivalent to\n\n$$\n|x|+|y|=1\n$$\n\nor\n\n$$\n|x|+|y|=3\n$$\n\nThe graph of the first equation is symmetric with respect to both axes. In the first quadrant it is reduced to $x+y=1$, whose graph is segment connecting points $(1,0)$ and $(0,1)$. Thus, the graph of\n\n$$\n|x|+|y|=1\n$$\n\nis square with vertices $(1,0),(0,1),(-1,0)$ and $(0,-1)$. Similarly, the graph of\n\n$$\n|x|+|y|=3\n$$\n\nis a square with vertices $(3,0),(0,3),(-3,0)$ and $(0,-3)$. The graph of the second equation of the system is a straight line with slope $a$ passing through (0,5). This line intersects the graph of the first equation in three points exactly, when passing through one of the points $(1,0)$ or $(-1,0)$. This happens if and only if $a=5$ or $a=-5$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nA3 ", "solution_match": "\nSolution:"}}
+{"year": "2009", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Real numbers $x, y, z$ satisfy\n\n$$\n0<x, y, z<1\n$$\n\nand\n\n$$\nx y z=(1-x)(1-y)(1-z) .\n$$\n\nShow that\n\n$$\n\\frac{1}{4} \\leq \\max \\{(1-x) y,(1-y) z,(1-z) x\\}\n$$", "solution": "It is clear that $a(1-a) \\leq \\frac{1}{4}$ for any real numbers $a$ (equivalent to $0<(2 a-1)^{2}$ ). Thus,\n\n$$\n\\begin{gathered}\nx y z=(1-x)(1-y)(1-z) \\\\\n(x y z)^{2}=[x(1-x)][y(1-y)][z(1-z)] \\leq \\frac{1}{4} \\cdot \\frac{1}{4} \\cdot \\frac{1}{4}=\\frac{1}{4^{3}} \\\\\nx y z \\leq \\frac{1}{2^{3}}\n\\end{gathered}\n$$\n\nIt implies that at least one of $x, y, z$ is at less or equal to $\\frac{1}{2}$. Let us say that $x \\leq \\frac{1}{2}$, and notice that $1-x \\geq \\frac{1}{2}$.\n\nAssume contrary to required result, that we have\n\n$$\n\\frac{1}{4}>\\max \\{(1-x) y,(1-y) x,(1-z) x\\}\n$$\n\nNow\n\n$$\n(1-x) y<\\frac{1}{4}, \\quad(1-y) z<\\frac{1}{4}, \\quad(1-z) x<\\frac{1}{4}\n$$\n\nFrom here we deduce:\n\n$$\ny<\\frac{1}{4} \\cdot \\frac{1}{1-x} \\leq \\frac{1}{4} \\cdot 2=\\frac{1}{2}\n$$\n\nNotice that $1-y>\\frac{1}{2}$.\n\nUsing same reasoning we conclude:\n\n$$\nz<\\frac{1}{2}, \\quad 1-z>\\frac{1}{2}\n$$\n\nUsing these facts we derive:\n\n$$\n\\frac{1}{8}=\\frac{1}{2} \\cdot \\frac{1}{2} \\cdot \\frac{1}{2}>x y z=(1-x)(1-y)(1-z)>\\frac{1}{2} \\cdot \\frac{1}{2} \\cdot \\frac{1}{2}=\\frac{1}{8}\n$$\n\nContradiction!\n\nRemark: The exercise along with its proof generalizes for any given (finite) number of numbers, and you can consider this new form in place of the proposed one:\n\nExercise: If for the real numbers $x_{1}, x_{2}, \\ldots, x_{n}, 0<x_{i}<1$, for all indices $i$, and\n\n$$\nx_{1} x_{2} \\ldots x_{n}=\\left(1-x_{1}\\right)\\left(1-x_{2}\\right) \\ldots\\left(1-x_{n}\\right)\n$$\n\nshow that\n\n$$\n\\frac{1}{4} \\leq \\max _{1 \\leq i \\leq n}\\left(1-x_{i}\\right) x_{i+1}\n$$\n\n(where $x_{n+1}=x_{1}$ ).\n\nOr you can consider the following variation:\n\nExercise: If for the real numbers $x_{1}, x_{2}, \\ldots, x_{2009}, 0<x_{i}<1$, for all indices $i$, and\n\n$$\nx_{1} x_{2} \\ldots x_{2009}=\\left(1-x_{1}\\right)\\left(1-x_{2}\\right) \\ldots\\left(1-x_{2009}\\right)\n$$\n\nshow that\n\n$$\n\\frac{1}{4} \\leq \\max _{1 \\leq i \\leq 2009}\\left(1-x_{i}\\right) x_{i+1}\n$$\n\n(where $x_{2010}=x_{1}$ ).", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nA4 ", "solution_match": "\nSolution:"}}
+{"year": "2009", "tier": "T3", "problem_label": "A5", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $x, y, z$ be positive real numbers. Prove that:\n\n$$\n\\left(x^{2}+y+1\\right)\\left(x^{2}+z+1\\right)\\left(y^{2}+z+1\\right)\\left(y^{2}+x+1\\right)\\left(z^{2}+x+1\\right)\\left(z^{2}+y+1\\right) \\geq(x+y+z)^{6}\n$$", "solution": "I: Applying Cauchy-Schwarz's inequality:\n\n$$\n\\left(x^{2}+y+1\\right)\\left(z^{2}+y+1\\right)=\\left(x^{2}+y+1\\right)\\left(1+y+z^{2}\\right) \\geq(x+y+z)^{2}\n$$\n\nUsing the same reasoning we deduce:\n\n$$\n\\left(x^{2}+z+1\\right)\\left(y^{2}+z+1\\right) \\geq(x+y+z)^{2}\n$$\n\nand\n\n$$\n\\left(y^{2}+x+1\\right)\\left(z^{2}+x+1\\right) \\geq(x+y+z)^{2}\n$$\n\nMultiplying these three inequalities we get the desired result.\n\nSolution II: We have\n\n$$\n\\begin{gathered}\n\\left(x^{2}+y+1\\right)\\left(z^{2}+y+1\\right) \\geq(x+y+z)^{2} \\Leftrightarrow \\\\\nx^{2} z^{2}+x^{2} y+x^{2}+y z^{2}+y^{2}+y+z^{2}+y+1 \\geq x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x \\Leftrightarrow \\\\\n\\left(x^{2} z^{2}-2 z x+1\\right)+\\left(x^{2} y-2 x y+y\\right)+\\left(y z^{2}-2 y z+y\\right) \\geq 0 \\Leftrightarrow \\\\\n(x z-1)^{2}+y(x-1)^{2}+y(z-1)^{2} \\geq 0\n\\end{gathered}\n$$\n\nwhich is correct.\n\nUsing the same reasoning we get:\n\n$$\n\\begin{aligned}\n& \\left(x^{2}+z+1\\right)\\left(y^{2}+z+1\\right) \\geq(x+y+z)^{2} \\\\\n& \\left(y^{2}+x+1\\right)\\left(z^{2}+x+1\\right) \\geq(x+y+z)^{2}\n\\end{aligned}\n$$\n\nMultiplying these three inequalities we get the desired result. Equality is attained at $x=y=z=1$.\n\n### 2.2 Combinatorics", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nA5 ", "solution_match": "\nSolution "}}
+{"year": "2009", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Each one of 2009 distinct points in the plane is coloured in blue or red, so that on every blue-centered unit circle there are exactly two red points. Find the gratest possible number of blue points.", "solution": "Each pair of red points can belong to at most two blue-centered unit circles. As $n$ red points form $\\frac{n(n-1)}{2}$ pairs, we can have not more than twice that number of blue points, i.e. $n(n-1)$ blue points. Thus, the total number of points can not exceed\n\n$$\nn+n(n-1)=n^{2}\n$$\n\nAs $44^{2}<2009, n$ must be at least 45 . We can arrange 45 distinct red points on a segment of length 1 , and color blue all but $16\\left(=45^{2}-2009\\right)$ points on intersections of the redcentered unit circles (all points of intersection are distinct, as no blue-centered unit circle can intersect the segment more than twice). Thus, the greatest possible number of blue points is $2009-45=1964$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nC1 ", "solution_match": "\nSolution:"}}
+{"year": "2009", "tier": "T3", "problem_label": "C2", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Five players $(A, B, C, D, E)$ take part in a bridge tournament. Every two players must play (as partners) against every other two players. Any two given players can be partners not more than once per day. What is the least number of days needed for this tournament?", "solution": "A given pair must play with three other pairs and these plays must be in different days, so at three days are needed. Suppose that three days suffice. Let the pair $A B$ play against $C D$ on day $x$. Then $A B-D E$ and $C D-B E$ cannot play on day $x$. Then one of the other two plays of $D E$ (with $A C$ and $B C$ ) must be on day $x$. Similarly, one of the plays of $B E$ with $A C$ or $A D$ must be on day $x$. Thus, two of the plays in the chain $B C-D E-A C-B E-A D$ are on day $x$ (more than two among these cannot be on one day).\n\nConsider the chain $A B-C D-E A-B D-C E-A B$. At least three days are needed for playing all the matches within it. For each of these days we conclude (as above) that there are exactly two of the plays in the chain $B C-D E-A C-B E-A D-B C$ on that day. This is impossible, as this chain consists of five plays.\n\nIt remains to show that four days will suffice:\n\nDay 1: $A B-C D, A C-D E, A D-C E, A E-B C$\n\nDay 2: $A B-D E, A C-B D, A D-B C, B E-C D$\n\nDay 3: $A B-C E, A D-B E, A E-B D, B C-D E$\n\nDay 4: $A C-B E, A E-C D, B D-C E$.\n\nRemark: It is possible to have 5 games in one day (but not on each day).\n\n## Alternative solution:\n\nThere are 10 pairs. Each of them plays 3 games, so the tournament needs to last at least 3 days. Assume the tournament could finish in 3 days. Then every pair must play one game on each day. There are 15 games to be played, so you must have 5 games on each day. Call \"Day 1\" the day AB plays against CD, \"Day 2\" the day AB plays against DE\nand \"Day 3\" the day AB plays against CE. Let us examine the other possible games on Day 1. CE can't play $\\mathrm{AB}$, so it must play either $\\mathrm{AD}$ or $\\mathrm{BD}$. $\\mathrm{DE}$ can't play $\\mathrm{AB}$, so it must play AC or BC. Similarly, AE can't play CD, so it must play BC or BD, and BE must play either $\\mathrm{AC}$ or $\\mathrm{AD}$. We obtain the following circular diagram in which exact every other game has to take place on Day 1, either the red ones or the blue ones:\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_17e61ec1e039d21af3a0g-10.jpg?height=463&width=485&top_left_y=768&top_left_x=794)\n\nSimilar reasoning leads un to the following diagram for Day 2. Here again, either all the red matches have to take place, or all the blue matches have to take place on Day 2.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_17e61ec1e039d21af3a0g-10.jpg?height=463&width=483&top_left_y=1593&top_left_x=795)\n\nOne can't have the blue matches on both Day 1 and Day 2 because AD-BE would repeat itself.\n\nWe can't have the red mathes on both days as this would repeat the match BD-CE.\n\nWe can not have the blue matches on Day 1 and the red matches on Day 2 because this would repeat the game AC-BE. Finally, choosing the red matches on Day 1 and the blue ones on Day 2 won't work either as the game AE-BC would repeat itself.\n\nIn conclusion, the tournament has to last at least four days. An example of how it could\nbe organized in four days is given in the previous solution.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nC2 ", "solution_match": "\nSolution:"}}
+{"year": "2009", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "a) In how many ways can we read the word SARAJEVO from the table below, if it is allowed to jump from cell to an adjacent cell (by vertex or a side) cell?\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_17e61ec1e039d21af3a0g-11.jpg?height=497&width=596&top_left_y=631&top_left_x=730)\n\nb) After the letter in one cell was deleted, only 525 ways to read the word SARAJEVO remained. Find all possible positions of that cell.", "solution": "In the first of the tables below the number in each cell shows the number of ways to reach that cell from the start (which is the sum of the quantities in the cells, from which we can come), and in the second one are the number of ways to arrive from that cell to the end (which is the sum of the quantities in the cells, to which we can go).\n\na) The answer is 750 , as seen from the second table.\n\nb) If we delete the letter in a cell, the number of ways to read SARAJEVO will decrease by the product of the numbers in the corresponding cell in the two tables. As $750-525=225$, this product has to be 225. This happens only for two cells on the third row. Here is the table with the products:", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nC3 ", "solution_match": "\nSolution:"}}
+{"year": "2009", "tier": "T3", "problem_label": "C4", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Determine all pairs $(m, n)$ for which it is possible to tile the table $m \\times n$ with \"corners\" as in the figure below, with the condition that in the tiling there is no rectangle (except for the $m \\times n$ one) regularly covered with corners.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_17e61ec1e039d21af3a0g-11.jpg?height=85&width=83&top_left_y=1942&top_left_x=992)", "solution": "Every \"corner\" covers exactly 3 squares, so a necessary condition for the tiling to exists is $3 \\mid m n$.\n\nFirst, we shall prove that for a tiling with our condition to exist, it is necessary that both $m, n$ for $m, n>3$ to be even. Suppose the contrary, i.e. suppose that that $m>3$ is odd (without losing generality). Look at the \"corners\" that cover squares on the side of length $m$ of table $m \\times n$. Because $m$ is odd, there must be a \"corner\" which covers exactly one square of that side. But any placement of that corner forces existence of a $2 \\times 3$ rectangle in the tiling. Thus, $m$ and $n$ for $m, n>3$ must be even and at least one of them is divisible by 3 .\n\nNotice that in the corners of table $m \\times n$, the \"corner\" must be placed such that it covers the square in the corner of the rectangle and its two neighboring squares, otherwise, again, a $2 \\times 3$ rectangle would form.\n\nIf one of $m$ and $n$ is 2 then condition forces that the only convenient tables are $2 \\times 3$ and $3 \\times 2$. If we try to find the desired tiling when $m=4$, then we are forced to stop at table $4 \\times 6$ because of the conditions of problem.\n\nWe easily find an example of a desired tiling for the table $6 \\times 6$ and, more generally, a tiling for a $6 \\times 2 k$ table.\n\nThus, it will be helpful to prove that the desired tiling exists for tables $6 k \\times 4 \\ell$, for $k, \\ell \\geq 2$. Divide that table at rectangle $6 \\times 4$ and tile that rectangle as we described. Now, change placement of problematic \"corners\" as in figure.\n\nThus, we get desired tilling for this type of table.\n\nSimilarly, we prove existence in case $6 k \\times(4 k+2)$ where $m, \\ell \\geq 2$. But, we first divide table at two tables $6 k \\times 6$ and $6 k \\times 4(\\ell-1)$. Divide them at rectangles $6 \\times 6$ and $6 \\times 4$. Tile them as we described earlier, and arrange problematic \"corners\" as in previous case. So, $2 \\times 3,3 \\times 2,6 \\times 2 k, 2 k \\times 6, k \\geq 2$, and $6 k \\times 4 \\ell$ for $k, \\ell \\geq 2$ and $6 k \\times(4 \\ell+2)$ for $k, \\ell \\geq 2$ are the convenient pairs.\n\nRemark: The problem is inspired by a problem given at Romanian Selection Test 2000, but it is completely different.\n\nRemark: Alternatively, the problem can be relaxed by asking: \"Does such a tiling exist for some concrete values of $m$ and $n$ ?\" .\n\n### 2.3 Geometry", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nC4 ", "solution_match": "\nSolution:"}}
+{"year": "2009", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C D$ be a parallelogram with $A C>B D$, and let $O$ be the point of intersection of $A C$ and $B D$. The circle with center at $O$ and radius $O A$ intersects the extensions of $A D$ and $A B$ at points $G$ and $L$, respectively. Let $Z$ be intersection point of lines $B D$ and $G L$. Prove that $\\angle Z C A=90^{\\circ}$.", "solution": "From the point $L$ we draw a parallel line to $B D$ that intersects lines $A C$ and $A G$ at points $N$ and $R$ respectively. Since $D O=O B$, we have that $N R=N L$, and point $N$ is the midpoint of segment $L R$.\n\nLet $K$ be the midpoint of $G L$. Now, $N K \\| R G$, and\n\n$$\n\\angle A G L=\\angle N K L=\\angle A C L\n$$\n\nTherefore, from the cyclic quadrilateral $N K C L$ we deduce:\n\n$$\n\\angle K C N=\\angle K L N\n$$\n\nNow, since $L R \\| D Z$, we have\n\n$$\n\\angle K L N=\\angle K Z O\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_17e61ec1e039d21af3a0g-13.jpg?height=729&width=817&top_left_y=475&top_left_x=631)\n\nIt implies that quadrilateral $O K C Z$ is cyclic, and\n\n$$\n\\angle O K Z=\\angle O C Z\n$$\n\nSince $O K \\perp G L$, we derive that $\\angle Z C A=90^{\\circ}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nG1 ", "solution_match": "## Solution:"}}
+{"year": "2009", "tier": "T3", "problem_label": "G2", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "In a right trapezoid $A B C D(A B \\| C D)$ the angle at vertex $B$ measures $75^{\\circ}$. Point $H$ is the foot of the perpendicular from point $A$ to the line $B C$. If $B H=D C$ and $A D+A H=8$, find the area of $A B C D$.", "solution": "Produce the legs of the trapezoid until they intersect at point $E$. The triangles $A B H$ and $E C D$ are congruent (ASA). The area of $A B C D$ is equal to area of triangle $E A H$ of hypotenuse\n\n$$\nA E=A D+D E=A D+A H=8\n$$\n\nLet $M$ be the midpoint of $A E$. Then\n\n$$\nM E=M A=M H=4\n$$\n\nand $\\angle A M H=30^{\\circ}$. Now, the altitude from $H$ to AM equals one half of $M H$, namely 2. Finally, the area is 8 .\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_17e61ec1e039d21af3a0g-13.jpg?height=802&width=300&top_left_y=1689&top_left_x=1272)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nG2 ", "solution_match": "\nSolution:"}}
+{"year": "2009", "tier": "T3", "problem_label": "G3", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "A parallelogram $A B C D$ with obtuse angle $\\angle A B C$ is given. After rotating the triangle $A C D$ around the vertex $C$, we get a triangle $C D^{\\prime} A^{\\prime}$, such that points $B, C$ and $D^{\\prime}$ are collinear. The extension of the median of triangle $C D^{\\prime} A^{\\prime}$ that passes through $D^{\\prime}$ intersects the straight line $B D$ at point $P$. Prove that $P C$ is the bisector of the angle $\\angle B P D^{\\prime}$.", "solution": "Let $A C \\cap B D=\\{X\\}$ and $P D^{\\prime} \\cap C A^{\\prime}=\\{Y\\}$. Because $A X=C X$ and $C Y=Y A^{\\prime}$, we deduce:\n\n$$\n\\triangle A B C \\cong \\triangle C D A \\cong \\triangle C D^{\\prime} A^{\\prime} \\Rightarrow \\triangle A B X \\cong \\triangle C D^{\\prime} Y, \\triangle B C X \\cong \\triangle D^{\\prime} A^{\\prime} Y\n$$\n\nIt follows that\n\n$$\n\\angle A B X=\\angle C D^{\\prime} Y\n$$\n\nLet $M$ and $N$ be orthogonal projections of the point $C$ on the straight lines $P D^{\\prime}$ and $B P$, respectively, and $Q$ is the orthogonal projection of the point $A$ on the straight line $B P$. Because $C D^{\\prime}=A B$, we have that $\\triangle A B Q \\cong \\triangle C D^{\\prime} M$.\n\nWe conclude that $C M=A Q$. But, $A X=C X$ and $\\triangle A Q X \\cong \\triangle C N X$. So, $C M=C N$ and $P C$ is the bisector of the angle $\\angle B P D^{\\prime}$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_17e61ec1e039d21af3a0g-14.jpg?height=1003&width=1305&top_left_y=1286&top_left_x=387)\n\nMuch shortened: $\\triangle C D^{\\prime} Y \\equiv \\triangle C D X$ means their altitudes from $C$ are also equal, i.e. $C M=C N$ and the conclusion.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nG3 ", "solution_match": "\nSolution:"}}
+{"year": "2009", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C D E$ be a convex pentagon such that $A B+C D=B C+D E$ and let $k$ be a semicircle with center on side $A E$ that touches the sides $A B, B C, C D$ and $D E$ of the pentagon, respectively, at points $P, Q, R$ and $S$ (different from the vertices of the pentagon). Prove that $P S \\| A E$.", "solution": "Let $O$ be center of $k$. We deduce that $B P=B Q, C Q=C R, D R=D S$, since those are tangents to the circle $k$. Using the condition $A B+C D=B C+D E$, we derive:\n\n$$\nA P+B P+C R+D R=B Q+C Q+D S+E S\n$$\n\nFrom here we have $A P=E S$.\n\nThus,\n\n$$\n\\triangle A P O \\cong \\triangle E S O\\left(A P=E S, \\angle A P O=\\angle E S O=90^{\\circ}, P O=S O\\right)\n$$\n\nThis implies\n\n$$\n\\angle O P S=\\angle O S P\n$$\n\nTherefore,\n\n$$\n\\angle A P S=\\angle A P O+\\angle O P S=90^{\\circ}+\\angle O P S=90^{\\circ}+\\angle O S P=\\angle P S E\n$$\n\nNow, from quadrilateral $A P S E$ we deduce:\n\n$$\n2 \\angle E A P+2 \\angle A P S=\\angle E A P+\\angle A P S+\\angle P S E+\\angle S E A=360^{\\circ}\n$$\n\nSo,\n\n$$\n\\angle E A P+\\angle A P S=180^{\\circ}\n$$\n\nand $A P S E$ is isosceles trapezoid. Therefore, $A E \\| P S$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_17e61ec1e039d21af3a0g-15.jpg?height=571&width=939&top_left_y=1733&top_left_x=567)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nG4 ", "solution_match": "\nSolution:"}}
+{"year": "2009", "tier": "T3", "problem_label": "G5", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A, B, C$ and $O$ be four points in the plane, such that $\\angle A B C>90^{\\circ}$ and $O A=$ $O B=O C$. Define the point $D \\in A B$ and the line $\\ell$ such that $D \\in \\ell, A C \\perp D C$ and $\\ell \\perp A O$. Line $\\ell$ cuts $A C$ at $E$ and the circumcircle of $\\triangle A B C$ at $F$. Prove that the circumcircles of triangles $B E F$ and $C F D$ are tangent at $F$.", "solution": "Let $\\ell \\cap A C=\\{K\\}$ and define $G$ to be the mirror image of the point $A$ with respect to $O$. Then $A G$ is a diameter of the circumcircle of the triangle $A B C$, therefore $A C \\perp C G$. On the other hand we have $A C \\perp D C$, and it implies that points $D, C, G$ are collinear.\n\nMoreover, as $A E \\perp D G$ and $D E \\perp A G$, we obtain that $E$ is the orthocenter of triangle $A D G$ and $G E \\perp A D$. As $A G$ is a diameter, we have $A B \\perp B G$, and since $A D \\perp G E$, the points $E, G$, and $B$ are collinear.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_17e61ec1e039d21af3a0g-16.jpg?height=1083&width=851&top_left_y=972&top_left_x=634)\n\nNotice that\n\n$$\n\\angle C A G=90^{\\circ}-\\angle A G C=\\angle K D C\n$$\n\nand\n\n$$\n\\angle C A G=\\angle G F C\n$$\n\nsince both subtend the same arc.\n\nHence,\n\n$$\n\\angle F D G=\\angle G F C\n$$\n\nTherefore, $G F$ is tangent to the circumcircle of the triangle $C D F$ at point $F$.\n\nWe claim that line $G F$ is also tangent to the circumcircle of triangle $B E F$ at point $F$, which concludes the proof.\n\nThe claim is equivalent to $\\angle G B F=\\angle E F G$. Denote by $F^{\\prime}$ the second intersection point - other than $F$ - of line $\\ell$ with the circumcircle of triangle $A B C$. Observe that $\\angle G B F=\\angle G F^{\\prime} F$, because both angles subtend the same arc, and $\\angle F F^{\\prime} G=\\angle E F G$, since $A G$ is the perpendicular bisector of the chord $F F^{\\prime}$, and we are done.\n\n### 2.4 Number Theory", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nG5 ", "solution_match": "\nSolution:"}}
+{"year": "2009", "tier": "T3", "problem_label": "NT1", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Determine all positive integer numbers $k$ for which the numbers $k+9$ are perfect squares and the only prime factors of $k$ are 2 and 3 .", "solution": "We have an integer $x$ such that\n\n$$\nx^{2}=k+9\n$$\n\n$k=2^{a} 3^{b}, a, b \\geq 0, a, b \\in \\mathbb{N}$.\n\nTherefore,\n\n$$\n(x-3)(x+3)=k \\text {. }\n$$\n\nIf $b=0$ then we have $k=16$.\n\nIf $b>0$ then we have $3 \\mid k+9$. Hence, $3 \\mid x^{2}$ and $9 \\mid k$.\n\nTherefore, we have $b \\geq 2$. Let $x=3 y$.\n\n$$\n(y-1)(y+1)=2^{a} 3^{b-2}\n$$\n\nIf $a=0$ then $b=3$ and we have $k=27$.\n\nIf $a \\geq 1$, then the numbers $y-1$ and $y+1$ are even. Therefore, we have $a \\geq 2$, and\n\n$$\n\\frac{y-1}{2} \\cdot \\frac{y+1}{2}=2^{a-2} 3^{b-2}\n$$\n\nSince the numbers $\\frac{y-1}{2}, \\frac{y+1}{2}$ are consecutive numbers, these numbers have to be powers of 2 and 3 . Let $m=a-2, n=b-2$.\n\n- If $2^{m}-3^{n}=1$ then we have $m \\geq n$. For $n=0$ we have $m=1, a=3, b=2$ and $k=72$. For $n>0$ using $\\bmod 3$ we have that $m$ is even number. Let $m=2 t$. Therefore,\n\n$$\n\\left(2^{t}-1\\right)\\left(2^{t}+1\\right)=3^{n}\n$$\n\nHence, $t=1, m=2, n=1$ and $a=4, b=3, k=432$.\n\n- If $3^{n}-2^{m}=1$, then $m>0$. For $m=1$ we have $n=1, a=3, b=3, k=216$. For $m>1$ using $\\bmod 4$ we have that $n$ is even number. Let $n=2 t$.\n\n$$\n\\left(3^{t}-1\\right)\\left(3^{t}+1\\right)=2^{m}\n$$\n\nTherefore, $t=1, n=2, m=3, a=5, b=4, k=2592$.\n\nSet of solutions: $\\{16,27,72,216,432,2592\\}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nNT1 ", "solution_match": "\nSolution:"}}
+{"year": "2009", "tier": "T3", "problem_label": "NT2", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "A group of $n>1$ pirates of different age owned total of 2009 coins. Initially each pirate (except for the youngest one) had one coin more than the next younger.\n\na) Find all possible values of $n$.\n\nb) Every day a pirate was chosen. The chosen pirate gave a coin to each of the other pirates. If $n=7$, find the largest possible number of coins a pirate can have after several days.", "solution": "a) If $n$ is odd, then it is a divisor of $2009=7 \\times 7 \\times 41$. If $n>49$, then $n$ is at least $7 \\times 41$, while the average pirate has 7 coins, so the initial division is impossible. So, we can have $n=7, n=41$ or $n=49$. Each of these cases is possible (e.g. if $n=49$, the average pirate has 41 coins, so the initial amounts are from $41-24=17$ to $41+24=65$ ).\n\nIf $n$ is even, then 2009 is multiple of the sum $S$ of the oldest and the youngest pirate. If $S<7 \\times 41$, then $S$ is at most 39 and the pairs of pirates of sum $S$ is at least 41 , so we must have at least 82 pirates, a contradiction. So we can have just $S=7 \\times 41=287$ and $S=49 \\times 41=2009$; respectively, $n=2 \\times 7=14$ or $n=2 \\times 1=2$. Each of these cases is possible (e.g. if $n=14$, the initial amounts are from $144-7=137$ to $143+7=150$ ). In total, $n$ is one of the numbers $2,7,13,41$ and 49 .\n\nb) If $n=7$, the average pirate has $7 \\times 41=287$ coins, so the initial amounts are from 284 to 290; they have different residues modulo 7. The operation decreases one of the amounts by 6 and increases the other ones by 1 , so the residues will be different at all times. The largest possible amount in one pirate's possession will be achieved if all the others have as little as possible, namely $0,1,2,3,4$ and 5 coins (the residues modulo 7 have to be different). If this happens, the wealthiest pirate will have $2009-14=1994$ coins. Indeed, this can be achieved e.g. if every day (until that moment) the coins are given by the second wealthiest: while he has more than 5 coins, he can provide the 6 coins needed, and when he has no more than five, the coins at the poorest six pirates have to be $0,1,2,3,4,5$. Thus, $n=1994$ can be achieved.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nNT2 ", "solution_match": "## Solution:"}}
+{"year": "2009", "tier": "T3", "problem_label": "NT3", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all pairs $(x, y)$ of integers which satisfy the equation\n\n$$\n(x+y)^{2}\\left(x^{2}+y^{2}\\right)=2009^{2}\n$$", "solution": "Let $x+y=s, x y=p$ with $s \\in \\mathbb{Z}^{*}$ and $p \\in \\mathbb{Z}$. The given equation can be written in the form\n\n$$\ns^{2}\\left(s^{2}-2 p\\right)=2009^{2}\n$$\n\nor\n\n$$\ns^{2}-2 p=\\left(\\frac{2009}{s}\\right)^{2}\n$$\n\nSo, $s$ divides $2009=7^{2} \\times 41$ and it follows that $p \\neq 0$.\n\nIf $p>0$, then $2009^{2}=s^{2}\\left(s^{2}-2 p\\right)=s^{4}-2 p s^{2}<s^{4}$. We obtain that $s$ divides 2009 and $|s| \\geq 49$. Thus, $s \\in\\{ \\pm 49, \\pm 287, \\pm 2009\\}$.\n\n- For $s= \\pm 49$, we have $p=360$, and $(x, y)=\\{(40,9),(9,40),(-40,-9),(-9,-40)\\}$.\n- For $s \\in\\{ \\pm 287, \\pm 2009\\}$ the equation has no integer solutions.\n\nIf $p<0$, then $2009^{2}=s^{4}-2 p s^{2}>s^{4}$. We obtain that $s$ divides 2009 and $|s| \\leq 41$. Thus, $s \\in\\{ \\pm 1, \\pm 7, \\pm 41\\}$. For these values of $s$ the equation has no integer solutions.\n\nSo, the given equation has only the solutions $(40,9),(9,40),(-40,-9),(-9,-40)$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nNT3 ", "solution_match": "\nSolution:"}}
+{"year": "2009", "tier": "T3", "problem_label": "NT4", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Determine all prime numbers $p_{1}, p_{2}, \\ldots, p_{12}, p_{13}, p_{1} \\leq p_{2} \\leq \\ldots \\leq p_{12} \\leq p_{13}$, such that\n\n$$\np_{1}^{2}+p_{2}^{2}+\\ldots+p_{12}^{2}=p_{13}^{2}\n$$\n\nand one of them is equal to $2 p_{1}+p_{9}$.", "solution": "Obviously, $p_{13} \\neq 2$, because sum of squares of 12 prime numbers is greater or equal to $12 \\times 2^{2}=48$. Thus, $p_{13}$ is odd number and $p_{13} \\geq 7$.\n\nWe have that $n^{2} \\equiv 1(\\bmod 8)$, when $n$ is odd. Let $k$ be the number of prime numbers equal to 2 . Looking at equation modulo 8 we get:\n\n$$\n4 k+12-k \\equiv 1 \\quad(\\bmod 8)\n$$\n\nSo, $k \\equiv 7(\\bmod 8)$ and because $k \\leq 12$ we get $k=7$. Therefore, $p_{1}=p_{2}=\\ldots=p_{7}=2$. Furthermore, we are looking for solutions of equations:\n\n$$\n28+p_{8}^{2}+p_{9}^{2}+p_{10}^{2}+p_{11}^{2}+p_{12}^{2}=p_{13}^{2}\n$$\n\nwhere $p_{8}, p_{9}, \\ldots, p_{13}$ are odd prime numbers and one of them is equal to $p_{9}+4$.\n\nNow, we know that when $n$ is not divisible by $3, n^{2} \\equiv 1(\\bmod 3)$. Let $s$ be number of prime numbers equal to 3 . Looking at equation modulo 3 we get:\n\n$$\n28+5-s \\equiv 1 \\quad(\\bmod 3)\n$$\n\nThus, $s \\equiv 2(\\bmod 3)$ and because $s \\leq 5, s$ is either 2 or 5 . We will consider both cases. i. When $s=2$, we get $p_{8}=p_{9}=3$. Thus, we are looking for prime numbers $p_{10} \\leq p_{11} \\leq$ $p_{12} \\leq p_{13}$ greater than 3 and at least one of them is 7 (certainly $p_{13} \\neq 7$ ), that satisfy\n\n$$\n46+p_{10}^{2}+p_{11}^{2}+p_{12}^{2}=p_{13}^{2}\n$$\n\nWe know that $n^{2} \\equiv 1(\\bmod 5)$ or $n^{2}=4(\\bmod 5)$ when $n$ is not divisible by 5 . It is not possible that $p_{10}=p_{11}=5$, because in that case $p_{12}$ must be equal to 7 and the left-hand side would be divisible by 5 , which contradicts the fact that $p_{13} \\geq 7$. So, we proved that $p_{10}=5$ or $p_{10}=7$.\n\nIf $p_{10}=5$ then $p_{11}=7$ because $p_{11}$ is the least of remaining prime numbers. Thus, we are looking for solutions of equation\n\n$$\n120=p_{13}^{2}-p_{12}^{2}\n$$\n\nin prime numbers. Now, from\n\n$$\n2^{3} \\cdot 3 \\cdot 5=\\left(p_{12}-p_{12}\\right)\\left(p_{13}+p_{12}\\right)\n$$\n\nthat desired solutions are $p_{12}=7, p_{13}=13 ; p_{12}=13, p_{13}=17 ; p_{12}=29, p_{13}=31$.\n\nIf $p_{10}=7$ we are solving equation:\n\n$$\n95+p_{11}^{2}+p_{12}^{2}=p_{13}^{2}\n$$\n\nin prime numbers greater than 5 . But left side can give residues 0 or 3 modulo 5 , while right side can give only 1 or 4 modulo 5 . So, in this case we do not have solution.\n\nii. When $s=5$ we get equation:\n\n$$\n28+45=73=p_{13}^{2}\n$$\n\nbut 73 is not square or integer and we do not have solution in this case.\n\nFinally, only solutions are:\n\n$\\{(2,2,2,2,2,2,2,3,3,5,7,7,13),(2,2,2,2,2,2,2,3,3,5,7,13,17),(2,2,2,2,2,2,2,3,3,5,7,29,31)\\}$. NT5 Show that there are infinitely many positive integers $c$, such that the following equations both have solutions in positive integers:\n\n$$\n\\left(x^{2}-c\\right)\\left(y^{2}-c\\right)=z^{2}-c\n$$\n\nand\n\n$$\n\\left(x^{2}+c\\right)\\left(y^{2}-c\\right)=z^{2}-c\n$$\n\nSolution: The firs equation always has solutions, namely the triples $\\{x, x+1, x(x+1)-c\\}$ for all $x \\in \\mathbb{N}$. Indeed,\n\n$$\n\\left(x^{2}-c\\right)\\left((x+1)^{2}-c\\right)=x^{2}(x+1)^{2}-2 c\\left(x^{2}+(x+1)^{2}\\right)+c^{2}=(x(x+1)-c)^{2}-c .\n$$\n\nFor second equation, we try $z=|x y-c|$. We need\n\n$$\n\\left(x^{2}+c\\right)\\left(y^{2}-c\\right)=(x y-c)^{2}\n$$\n\nor\n\n$$\nx^{2} y^{2}+c\\left(y^{2}-x^{2}\\right)-c^{2}=x^{2} y^{2}-2 x y c+c^{2}\n$$\n\nCancelling the common terms we get\n\n$$\nc\\left(x^{2}-y^{2}+2 x y\\right)=2 c^{2}\n$$\n\nor\n\n$$\nc=\\frac{x^{2}-y^{2}+2 x y}{2}\n$$\n\nTherefore, all $c$ of this form will work. This expression is a positive integer if $x$ and $y$ have the same parity, and it clearly takes infinitely many positive values. We only need to check $z \\neq 0$, i.e. $c \\neq x y$, which is true for $x \\neq y$. For example, one can take\n\n$$\ny=x-2\n$$\n\nand\n\n$$\nz=\\frac{x^{2}-(x-2)^{2}+2 x(x-2)}{2}=x^{2}-2 \\text {. }\n$$\n\nThus, $\\{(x, x-2,2 x-2)\\}$ is a solution for $c=x^{2}-2$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl", "problem_match": "\nNT4 ", "solution_match": "\nSolution:"}}

JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl

@@ -1,13 +1,13 @@
-{"year": "2010", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO", "problem": "The real numbers $a, b, c, d$ satisfy simultaneously the equations\n\n$$\na b c-d=1, b c d-a=2, c d a-b=3, d a b-c=-6\n$$\n\nProve that $a+b+c+d \\neq 0$.", "solution": "Suppose that $a+b+c+d=0$. Then\n\n$$\na b c+b c d+c d a+d a b=0\n$$\n\nIf $a b c d=0$, then one of numbers, say $d$, must be 0 . In this case $a b c=0$, and so at least two of the numbers $a, b, c, d$ will be equal to 0 , making one of the given equations impossible. Hence $a b c d \\neq 0$ and, from (1),\n\n$$\n\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{d}=0\n$$\n\nimplying\n\n$$\n\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=\\frac{1}{a+b+c}\n$$\n\nIt follows that $(a+b)(b+c)(c+a)=0$, which is impossible (for instance, if $a+b=0$, then adding the second and third given equations would lead to $0=2+3$, a contradiction). Thus $a+b+c+d \\neq 0$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem A1.", "solution_match": "\nSolution."}}
-{"year": "2010", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO", "problem": "Determine all four digit numbers $\\overline{a b c d}$ such that\n\n$$\na(a+b+c+d)\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)\\left(a^{6}+2 b^{6}+3 c^{6}+4 d^{6}\\right)=\\overline{a b c d}\n$$", "solution": "From $\\overline{a b c d}<10000$ and\n\n$$\na^{10} \\leq a(a+b+c+d)\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)\\left(a^{6}+2 b^{6}+3 c^{6}+4 d^{6}\\right)=\\overline{a b c d}\n$$\n\nfollows that $a \\leq 2$. We thus have two cases:\n\nCase I: $a=1$.\n\nObviously $2000>\\overline{1 b c d}=(1+b+c+d)\\left(1+b^{2}+c^{2}+d^{2}\\right)\\left(1+2 b^{6}+3 c^{6}+4 d^{6}\\right) \\geq$ $(b+1)\\left(b^{2}+1\\right)\\left(2 b^{6}+1\\right)$, so $b \\leq 2$. Similarly one gets $c<2$ and $d<2$. By direct check there is no solution in this case.\n\nCase II: $a=2$.\n\nWe have $3000>\\overline{2 b c d}=2(2+b+c+d)\\left(4+b^{2}+c^{2}+d^{2}\\right)\\left(64+2 b^{6}+3 c^{6}+4 d^{6}\\right) \\geq$ $2(b+2)\\left(b^{2}+4\\right)\\left(2 b^{6}+64\\right)$, imposing $b \\leq 1$. In the same way one proves $c<2$ and $d<2$. By direct check, we find out that 2010 is the only solution.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem A2.", "solution_match": "\nSolution."}}
-{"year": "2010", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO", "problem": "Find all pairs $(x, y)$ of real numbers such that $|x|+|y|=1340$ and $x^{3}+y^{3}+2010 x y=670^{3}$.", "solution": "Answer: $(-670 ;-670),(1005 ;-335),(-335 ; 1005)$.\n\nTo prove this, let $z=-670$. We have\n\n$$\n0=x^{3}+y^{3}+z^{3}-3 x y z=\\frac{1}{2}(x+y+z)\\left((x-y)^{2}+(y-z)^{2}+(z-x)^{2}\\right)\n$$\n\nThus either $x+y+z=0$, or $x=y=z$. In the latter case we get $x=y=-670$, which satisfies both the equations. In the former case we get $x+y=670$. Then at least one of $x, y$ is positive, but not both, as from the second equation we would get $x+y=1340$. If $x>0 \\geq y$, we get $x-y=1340$, which together with $x+y=670$ yields $x=1005, y=-335$. If $y>0 \\geq x$ we get similarly $x=-335, y=1005$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem A3.", "solution_match": "\nSolution."}}
-{"year": "2010", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $a, b, c$ be positive real numbers such that $a b c(a+b+c)=3$. Prove the inequality\n\n$$\n(a+b)(b+c)(c+a) \\geq 8\n$$\n\nand determine all cases when equality holds.", "solution": "We have\n\n$A=(a+b)(b+c)(c+a)=\\left(a b+a c+b^{2}+b c\\right)(c+a)=(b(a+b+c)+a c)(c+a)$,\n\nso by the given condition\n\n$$\nA=\\left(\\frac{3}{a c}+a c\\right)(c+a)=\\left(\\frac{1}{a c}+\\frac{1}{a c}+\\frac{1}{a c}+a c\\right)(c+a)\n$$\n\nAplying the AM-GM inequality for four and two terms respectively, we get\n\n$$\nA \\geq 4 \\sqrt[4]{\\frac{a c}{(a c)^{3}}} \\cdot 2 \\sqrt{a c}=8\n$$\n\nFrom the last part, it is easy to see that inequality holds when $a=c$ and $\\frac{1}{a c}=a c$, i.e. $a=b=c=1$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem A4.", "solution_match": "\nSolution."}}
-{"year": "2010", "tier": "T3", "problem_label": "A5", "problem_type": "Algebra", "exam": "JBMO", "problem": "The real positive numbers $x, y, z$ satisfy the relations $x \\leq 2$, $y \\leq 3, x+y+z=11$. Prove that $\\sqrt{x y z} \\leq 6$.", "solution": "For $x=2, y=3$ and $z=6$ the equality holds.\n\nAfter the substitutions $x=2-u, y=3-v$ with $u \\in[0,2), v \\in[0,3)$, we obtain that $z=6+u+v$ and the required inequality becomes\n\n$$\n(2-u)(3-v)(6+u+v) \\leqslant 36\n$$\n\nWe shall need the following lemma.\n\nLemma. If real numbers $a$ and $b$ satisfy the relations $0<b \\leq a$, then for every real number $y \\in[0, b)$ the inequality\n\n$$\n\\frac{a}{a+y} \\geqslant \\frac{b-y}{b}\n$$\n\nholds.\n\nProof of the lemma. The inequality (2) is equivalent to\n\n$$\na b \\geq a b-a y+b y-y^{2} \\Leftrightarrow y^{2}+(a-b) y \\geq 0\n$$\n\nThe last inequality is true, because $a \\geq b>0$ and $y \\geq 0$.\n\nThe equality in (2) holds if $y=0$. The lemma is proved.\n\nBy using the lemma we can write the following inequalities:\n\n$$\n\\begin{gathered}\n\\frac{6}{6+u} \\geqslant \\frac{2-u}{2} \\\\\n\\frac{6}{6+v} \\geqslant \\frac{3-v}{3} \\\\\n\\frac{6+u}{6+u+v} \\geqslant \\frac{6}{6+v}\n\\end{gathered}\n$$\n\nBy multiplying the inequalities (3), (4) and (5) we obtain:\n\n$$\n\\begin{gathered}\n\\frac{6 \\cdot 6 \\cdot(6+u)}{(6+u)(6+v)(6+u+v)} \\geqslant \\frac{6(2-u)(3-v)}{2 \\cdot 3(6+v)} \\Leftrightarrow \\\\\n(2-u)(3-v)(6+u+v) \\leqslant 2 \\cdot 3 \\cdot 6=36 \\Leftrightarrow \\quad(1)\n\\end{gathered}\n$$\n\nBy virtue of lemma, the equality holds if and only if $u=v=0$.\n\nAlternative solution. With the same substitutions write the inequality as\n\n$$\n(6-u-v)(6+u+v)+(u v-2 u-v)(6+u+v) \\leq 36\n$$\n\nAs the first product on the lefthand side is $36-(u+v)^{2} \\leq 36$, it is enough to prove that the second product is nonpositive. This comes easily from $|u-1| \\leq 1$, $|v-2| \\leq 2$ and $u v-2 u-v=(u-1)(v-2)-2$, which implies $u v-v-2 u \\leq 0$.\n\n## Geometry", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem A5.", "solution_match": "\nSolution."}}
-{"year": "2010", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO", "problem": "Consider a triangle $A B C$ with $\\angle A C B=90^{\\circ}$. Let $F$ be the foot of the altitude from $C$. Circle $\\omega$ touches the line segment $F B$ at point $P$, the altitude $C F$ at point $Q$ and the circumcircle of $A B C$ at point $R$. Prove that points $A, Q, R$ are collinear and $A P=A C$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_5ab10fadba60f287b615g-04.jpg?height=539&width=515&top_left_y=793&top_left_x=816)", "solution": "Let $M$ be the midpoint of $A B$ and let $N$ be the center of $\\omega$. Then $M$ is the circumcenter of triangle $A B C$, so points $M, N$ and $R$ are collinear. From $Q N \\| A M$ we get $\\angle A M R=\\angle Q N R$. Besides that, triangles $A M R$ and $Q N R$ are isosceles, therefore $\\angle M R A=\\angle N R Q$; thus points $A, Q, R$ are collinear.\n\nRight angled triangles $A F Q$ and $A R B$ are similar, which implies $\\frac{A Q}{A B}=\\frac{A F}{A R}$, that is $A Q \\cdot A R=A F \\cdot A B$. The power of point $A$ with respect to $\\omega$ gives $A Q \\cdot A R=A P^{2}$. Also, from similar triangles $A B C$ and $A C F$ we get $A F \\cdot A B=$ $A C^{2}$. Now, the claim follows from $A C^{2}=A F \\cdot A B=A Q \\cdot A R=A P^{2}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem G1.", "solution_match": "\nSolution."}}
-{"year": "2010", "tier": "T3", "problem_label": "G2", "problem_type": "Geometry", "exam": "JBMO", "problem": "Consider a triangle $A B C$ and let $M$ be the midpoint of the side $B C$. Suppose $\\angle M A C=\\angle A B C$ and $\\angle B A M=105^{\\circ}$. Find the measure of $\\angle A B C$.", "solution": "The angle measure is $30^{\\circ}$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_5ab10fadba60f287b615g-05.jpg?height=510&width=965&top_left_y=721&top_left_x=583)\n\nLet $O$ be the circumcenter of the triangle $A B M$. From $\\angle B A M=105^{\\circ}$ follows $\\angle M B O=15^{\\circ}$. Let $M^{\\prime}, C^{\\prime}$ be the projections of points $M, C$ onto the line $B O$. Since $\\angle M B O=15^{\\circ}$, then $\\angle M O M^{\\prime}=30^{\\circ}$ and consequently $M M^{\\prime}=\\frac{M O}{2}$. On the other hand, $M M^{\\prime}$ joins the midpoints of two sides of the triangle $B C C^{\\prime}$, which implies $C C^{\\prime}=M O=A O$.\n\nThe relation $\\angle M A C=\\angle A B C$ implies $C A$ tangent to $\\omega$, hence $A O \\perp A C$. It follows that $\\triangle A C O \\equiv \\triangle O C C^{\\prime}$, and furthermore $O B \\| A C$.\n\nTherefore $\\angle A O M=\\angle A O M^{\\prime}-\\angle M O M^{\\prime}=90^{\\circ}-30^{\\circ}=60^{\\circ}$ and $\\angle A B M=$ $\\frac{\\angle A O M}{2}=30^{\\circ}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem G2.", "solution_match": "\nSolution."}}
-{"year": "2010", "tier": "T3", "problem_label": "G3", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C$ be an acute-angled triangle. A circle $\\omega_{1}\\left(O_{1}, R_{1}\\right)$ passes through points $B$ and $C$ and meets the sides $A B$ and $A C$ at points $D$ and $E$, respectively. Let $\\omega_{2}\\left(O_{2}, R_{2}\\right)$ be the circumcircle of the triangle $A D E$. Prove that $O_{1} O_{2}$ is equal to the circumradius of the triangle $A B C$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_5ab10fadba60f287b615g-06.jpg?height=455&width=398&top_left_y=710&top_left_x=858)", "solution": "Recall that, in every triangle, the altitude and the diameter of the circumcircle drawn from the same vertex are isogonal. The proof offers no difficulty, being a simple angle chasing around the circumcircle of the triangle.\n\nLet $O$ be the circumcenter of the triangle $A B C$. From the above, one has $\\angle O A E=90^{\\circ}-B$. On the other hand $\\angle D E A=B$, for $B C D E$ is cyclic. Thus $A O \\perp D E$, implying that in the triangle $A D E$ cevians $A O$ and $A O_{2}$ are isogonal. So, since $A O$ is a radius of the circumcircle of triangle $A B C$, one obtains that $A O_{2}$ is an altitude in this triangle.\n\nMoreover, since $O O_{1}$ is the perpendicular bisector of the line segment $B C$, one has $O O_{1} \\perp B C$, and furthermore $A O_{2} \\| O O_{1}$.\n\nChord $D E$ is common to $\\omega_{1}$ and $\\omega_{2}$, hence $O_{1} O_{2} \\perp D E$. It follows that $A O \\|$ $O_{1} O_{2}$, so $A O O_{1} O_{2}$ is a parallelogram. The conclusion is now obvious.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem G3.", "solution_match": "\nSolution."}}
-{"year": "2010", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A L$ and $B K$ be angle bisectors in the non-isosceles triangle $A B C(L \\in B C, K \\in A C)$. The perpendicular bisector of $B K$ intersects the line $A L$ at point $M$. Point $N$ lies on the line $B K$ such that $L N \\| M K$. Prove that $L N=N A$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_5ab10fadba60f287b615g-07.jpg?height=439&width=721&top_left_y=667&top_left_x=694)", "solution": "The point $M$ lies on the circumcircle of $\\triangle A B K$ (since both $A L$ and the perpendicular bisector of $B K$ bisect the arc $B K$ of this circle). Then $\\angle C B K=\\angle A B K=\\angle A M K=\\angle N L A$. Thus $A B L N$ is cyclic, whence $\\angle N A L=$ $\\angle N B L=\\angle C B K=\\angle N L A$. Now it follows that $L N=N A$.\n\n## Combinatorics", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem G4.", "solution_match": "\nSolution."}}
-{"year": "2010", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "There are two piles of coins, each containing 2010 pieces. Two players A and B play a game taking turns (A plays first). At each turn, the player on play has to take one or more coins from one pile or exactly one coin from each pile. Whoever takes the last coin is the winner. Which player will win if they both play in the best possible way?", "solution": "B wins.\n\nIn fact, we will show that A will lose if the total number of coins is a multiple of 3 and the two piles differ by not more than one coin (call this a balanced position). To this end, firstly notice that it is not possible to move from one balanced position to another. The winning strategy for B consists in returning A to a balanced position (notice that the initial position is a balanced position).\n\nThere are two types of balanced positions; for each of them consider the moves of $\\mathrm{A}$ and the replies of B.\n\nIf the number in each pile is a multiple of 3 and there is at least one coin:\n\n- if A takes $3 n$ coins from one pile, then $\\mathrm{B}$ takes $3 n$ coins from the other one.\n- if A takes $3 n+1$ coins from one pile, then $\\mathrm{B}$ takes $3 n+2$ coins from the other one.\n- if A takes $3 n+2$ coins from one pile, then $\\mathrm{B}$ takes $3 n+1$ coins from the other one.\n- if A takes a coin from each pile, then $\\mathrm{B}$ takes one coin from one pile.\n\nIf the numbers are not multiples of 3 , then we have $3 m+1$ coins in one pile and $3 m+2$ in the other one. Hence:\n\n- if A takes $3 n$ coins from one pile, then $\\mathrm{B}$ takes $3 n$ coins from the other one.\n- if A takes $3 n+1$ coins from the first pile $(n \\leq m)$, then $\\mathrm{B}$ takes $3 n+2$ coins from the second one.\n- if A takes $3 n+2$ coins from the second pile $(n \\leq m)$, then $\\mathrm{B}$ takes $3 n+1$ coins from the first one.\n- if A takes $3 n+2$ coins from the first pile $(n \\leq m-1)$, then $\\mathrm{B}$ takes $3 n+4$ coins from the second one.\n- if A takes $3 n+1$ coins from the second pile $(n \\leq m)$, then $\\mathrm{B}$ takes $3 n-1$ coins from the first one. This is impossible if A has taken only one coin from the second pile; in this case $\\mathrm{B}$ takes one coin from each pile.\n- if A takes a coin from each pile, then B takes one coin from the second pile.\n\nIn all these cases, the position after B's move is again a balanced position. Since the number of coins decreases and $(0 ; 0)$ is a balanced position, after a finite number of moves, there will be no coins left after B's move. Thus, B wins.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem C1.", "solution_match": "\nSolution."}}
-{"year": "2010", "tier": "T3", "problem_label": "C2", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "A $9 \\times 7$ rectangle is tiled with pieces of two types, shown in the picture below.\n![](https://cdn.mathpix.com/cropped/2024_06_05_5ab10fadba60f287b615g-09.jpg?height=124&width=504&top_left_y=588&top_left_x=801)\n\nFind the possible values of the number of the $2 \\times 2$ pieces which can be used in such a tiling.", "solution": "Answer: 0 or 3.\n\nDenote by $x$ the number of the pieces of the type 'corner' and by $y$ the number of the pieces of the type of $2 \\times 2$. Mark 20 squares of the rectangle as in the figure below.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_5ab10fadba60f287b615g-09.jpg?height=295&width=374&top_left_y=1105&top_left_x=867)\n\nObviously, each piece covers at most one marked square.\n\nThus, $x+y \\geq 20$ (1) and consequently $3 x+3 y \\geq 60$ (2). On the other hand $3 x+4 y=63$ (3). From (2) and (3) it follows $y \\leq 3$ and from (3), $3 \\mid y$.\n\nThe proof is finished if we produce tilings with 3 , respectively $0,2 \\times 2$ tiles:\n![](https://cdn.mathpix.com/cropped/2024_06_05_5ab10fadba60f287b615g-09.jpg?height=300&width=830&top_left_y=1682&top_left_x=642)\n\n## Number Theory", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem C2.", "solution_match": "\nSolution."}}
-{"year": "2010", "tier": "T3", "problem_label": "N1", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all positive integers $n$ such that $n 2^{n+1}+1$ is a perfect square.", "solution": "Answer: $n=0$ and $n=3$.\n\nClearly $n 2^{n+1}+1$ is odd, so, if this number is a perfect square, then $n 2^{n+1}+1=$ $(2 x+1)^{2}, x \\in \\mathbb{N}$, whence $n 2^{n-1}=x(x+1)$.\n\nThe integers $x$ and $x+1$ are coprime, so one of them must divisible by $2^{n-1}$, which means that the other must be at most $n$. This shows that $2^{n-1} \\leqslant n+1$.\n\nAn easy induction shows that the above inequality is false for all $n \\geqslant 4$, and a direct inspection confirms that the only convenient values in the case $n \\leqslant 3$ are 0 and 3 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem N1.", "solution_match": "\nSolution."}}
-{"year": "2010", "tier": "T3", "problem_label": "N2", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all positive integers $n$ such that $36^{n}-6$ is a product of two or more consecutive positive integers.", "solution": "Answer: $n=1$.\n\nAmong each four consecutive integers there is a multiple of 4 . As $36^{n}-6$ is not a multiple of 4 , it must be the product of two or three consecutive positive integers.\n\nCase I. If $36^{n}-6=x(x+1)$ (all letters here and below denote positive integers), then $4 \\cdot 36^{n}-23=(2 x+1)^{2}$, whence $\\left(2 \\cdot 6^{n}+2 x+1\\right)\\left(2 \\cdot 6^{n}-2 x-1\\right)=23$. As 23 is prime, this leads to $2 \\cdot 6^{n}+2 x+1=23,2 \\cdot 6^{n}-2 x-1=1$. Subtracting these yields $4 x+2=22, x=5, n=1$, which is a solution to the problem.\n\nCase II. If $36^{n}-6=(y-1) y(y+1)$, then\n\n$$\n36^{n}=y^{3}-y+6=\\left(y^{3}+8\\right)-(y+2)=(y+2)\\left(y^{2}-2 y+3\\right)\n$$\n\nThus each of $y+2$ and $y^{2}-2 y+3$ can have only 2 and 3 as prime factors, so the same is true for their GCD. This, combined with the identity $y^{2}-2 y+3=$ $(y+2)(y-4)+11$ yields $\\operatorname{GCD}\\left(y+2 ; y^{2}-2 y+3\\right)=1$. Now $y+2<y^{2}-2 y+3$ and the latter number is odd, so $y+2=4^{n}, y^{2}-2 y+3=9^{n}$. The former identity implies $y$ is even and now by the latter one $9^{n} \\equiv 3(\\bmod 4)$, while in fact $9^{n} \\equiv 1(\\bmod 4)$ - a contradiction. So, in this case there is no such $n$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem N2.", "solution_match": "\nSolution."}}
+{"year": "2010", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "The real numbers $a, b, c, d$ satisfy simultaneously the equations\n\n$$\na b c-d=1, b c d-a=2, c d a-b=3, d a b-c=-6\n$$\n\nProve that $a+b+c+d \\neq 0$.", "solution": "Suppose that $a+b+c+d=0$. Then\n\n$$\na b c+b c d+c d a+d a b=0\n$$\n\nIf $a b c d=0$, then one of numbers, say $d$, must be 0 . In this case $a b c=0$, and so at least two of the numbers $a, b, c, d$ will be equal to 0 , making one of the given equations impossible. Hence $a b c d \\neq 0$ and, from (1),\n\n$$\n\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{d}=0\n$$\n\nimplying\n\n$$\n\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=\\frac{1}{a+b+c}\n$$\n\nIt follows that $(a+b)(b+c)(c+a)=0$, which is impossible (for instance, if $a+b=0$, then adding the second and third given equations would lead to $0=2+3$, a contradiction). Thus $a+b+c+d \\neq 0$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem A1.", "solution_match": "\nSolution."}}
+{"year": "2010", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Determine all four digit numbers $\\overline{a b c d}$ such that\n\n$$\na(a+b+c+d)\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)\\left(a^{6}+2 b^{6}+3 c^{6}+4 d^{6}\\right)=\\overline{a b c d}\n$$", "solution": "From $\\overline{a b c d}<10000$ and\n\n$$\na^{10} \\leq a(a+b+c+d)\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)\\left(a^{6}+2 b^{6}+3 c^{6}+4 d^{6}\\right)=\\overline{a b c d}\n$$\n\nfollows that $a \\leq 2$. We thus have two cases:\n\nCase I: $a=1$.\n\nObviously $2000>\\overline{1 b c d}=(1+b+c+d)\\left(1+b^{2}+c^{2}+d^{2}\\right)\\left(1+2 b^{6}+3 c^{6}+4 d^{6}\\right) \\geq$ $(b+1)\\left(b^{2}+1\\right)\\left(2 b^{6}+1\\right)$, so $b \\leq 2$. Similarly one gets $c<2$ and $d<2$. By direct check there is no solution in this case.\n\nCase II: $a=2$.\n\nWe have $3000>\\overline{2 b c d}=2(2+b+c+d)\\left(4+b^{2}+c^{2}+d^{2}\\right)\\left(64+2 b^{6}+3 c^{6}+4 d^{6}\\right) \\geq$ $2(b+2)\\left(b^{2}+4\\right)\\left(2 b^{6}+64\\right)$, imposing $b \\leq 1$. In the same way one proves $c<2$ and $d<2$. By direct check, we find out that 2010 is the only solution.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem A2.", "solution_match": "\nSolution."}}
+{"year": "2010", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Find all pairs $(x, y)$ of real numbers such that $|x|+|y|=1340$ and $x^{3}+y^{3}+2010 x y=670^{3}$.", "solution": "Answer: $(-670 ;-670),(1005 ;-335),(-335 ; 1005)$.\n\nTo prove this, let $z=-670$. We have\n\n$$\n0=x^{3}+y^{3}+z^{3}-3 x y z=\\frac{1}{2}(x+y+z)\\left((x-y)^{2}+(y-z)^{2}+(z-x)^{2}\\right)\n$$\n\nThus either $x+y+z=0$, or $x=y=z$. In the latter case we get $x=y=-670$, which satisfies both the equations. In the former case we get $x+y=670$. Then at least one of $x, y$ is positive, but not both, as from the second equation we would get $x+y=1340$. If $x>0 \\geq y$, we get $x-y=1340$, which together with $x+y=670$ yields $x=1005, y=-335$. If $y>0 \\geq x$ we get similarly $x=-335, y=1005$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem A3.", "solution_match": "\nSolution."}}
+{"year": "2010", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $a, b, c$ be positive real numbers such that $a b c(a+b+c)=3$. Prove the inequality\n\n$$\n(a+b)(b+c)(c+a) \\geq 8\n$$\n\nand determine all cases when equality holds.", "solution": "We have\n\n$A=(a+b)(b+c)(c+a)=\\left(a b+a c+b^{2}+b c\\right)(c+a)=(b(a+b+c)+a c)(c+a)$,\n\nso by the given condition\n\n$$\nA=\\left(\\frac{3}{a c}+a c\\right)(c+a)=\\left(\\frac{1}{a c}+\\frac{1}{a c}+\\frac{1}{a c}+a c\\right)(c+a)\n$$\n\nAplying the AM-GM inequality for four and two terms respectively, we get\n\n$$\nA \\geq 4 \\sqrt[4]{\\frac{a c}{(a c)^{3}}} \\cdot 2 \\sqrt{a c}=8\n$$\n\nFrom the last part, it is easy to see that inequality holds when $a=c$ and $\\frac{1}{a c}=a c$, i.e. $a=b=c=1$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem A4.", "solution_match": "\nSolution."}}
+{"year": "2010", "tier": "T3", "problem_label": "A5", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "The real positive numbers $x, y, z$ satisfy the relations $x \\leq 2$, $y \\leq 3, x+y+z=11$. Prove that $\\sqrt{x y z} \\leq 6$.", "solution": "For $x=2, y=3$ and $z=6$ the equality holds.\n\nAfter the substitutions $x=2-u, y=3-v$ with $u \\in[0,2), v \\in[0,3)$, we obtain that $z=6+u+v$ and the required inequality becomes\n\n$$\n(2-u)(3-v)(6+u+v) \\leqslant 36\n$$\n\nWe shall need the following lemma.\n\nLemma. If real numbers $a$ and $b$ satisfy the relations $0<b \\leq a$, then for every real number $y \\in[0, b)$ the inequality\n\n$$\n\\frac{a}{a+y} \\geqslant \\frac{b-y}{b}\n$$\n\nholds.\n\nProof of the lemma. The inequality (2) is equivalent to\n\n$$\na b \\geq a b-a y+b y-y^{2} \\Leftrightarrow y^{2}+(a-b) y \\geq 0\n$$\n\nThe last inequality is true, because $a \\geq b>0$ and $y \\geq 0$.\n\nThe equality in (2) holds if $y=0$. The lemma is proved.\n\nBy using the lemma we can write the following inequalities:\n\n$$\n\\begin{gathered}\n\\frac{6}{6+u} \\geqslant \\frac{2-u}{2} \\\\\n\\frac{6}{6+v} \\geqslant \\frac{3-v}{3} \\\\\n\\frac{6+u}{6+u+v} \\geqslant \\frac{6}{6+v}\n\\end{gathered}\n$$\n\nBy multiplying the inequalities (3), (4) and (5) we obtain:\n\n$$\n\\begin{gathered}\n\\frac{6 \\cdot 6 \\cdot(6+u)}{(6+u)(6+v)(6+u+v)} \\geqslant \\frac{6(2-u)(3-v)}{2 \\cdot 3(6+v)} \\Leftrightarrow \\\\\n(2-u)(3-v)(6+u+v) \\leqslant 2 \\cdot 3 \\cdot 6=36 \\Leftrightarrow \\quad(1)\n\\end{gathered}\n$$\n\nBy virtue of lemma, the equality holds if and only if $u=v=0$.\n\nAlternative solution. With the same substitutions write the inequality as\n\n$$\n(6-u-v)(6+u+v)+(u v-2 u-v)(6+u+v) \\leq 36\n$$\n\nAs the first product on the lefthand side is $36-(u+v)^{2} \\leq 36$, it is enough to prove that the second product is nonpositive. This comes easily from $|u-1| \\leq 1$, $|v-2| \\leq 2$ and $u v-2 u-v=(u-1)(v-2)-2$, which implies $u v-v-2 u \\leq 0$.\n\n## Geometry", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem A5.", "solution_match": "\nSolution."}}
+{"year": "2010", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Consider a triangle $A B C$ with $\\angle A C B=90^{\\circ}$. Let $F$ be the foot of the altitude from $C$. Circle $\\omega$ touches the line segment $F B$ at point $P$, the altitude $C F$ at point $Q$ and the circumcircle of $A B C$ at point $R$. Prove that points $A, Q, R$ are collinear and $A P=A C$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_5ab10fadba60f287b615g-04.jpg?height=539&width=515&top_left_y=793&top_left_x=816)", "solution": "Let $M$ be the midpoint of $A B$ and let $N$ be the center of $\\omega$. Then $M$ is the circumcenter of triangle $A B C$, so points $M, N$ and $R$ are collinear. From $Q N \\| A M$ we get $\\angle A M R=\\angle Q N R$. Besides that, triangles $A M R$ and $Q N R$ are isosceles, therefore $\\angle M R A=\\angle N R Q$; thus points $A, Q, R$ are collinear.\n\nRight angled triangles $A F Q$ and $A R B$ are similar, which implies $\\frac{A Q}{A B}=\\frac{A F}{A R}$, that is $A Q \\cdot A R=A F \\cdot A B$. The power of point $A$ with respect to $\\omega$ gives $A Q \\cdot A R=A P^{2}$. Also, from similar triangles $A B C$ and $A C F$ we get $A F \\cdot A B=$ $A C^{2}$. Now, the claim follows from $A C^{2}=A F \\cdot A B=A Q \\cdot A R=A P^{2}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem G1.", "solution_match": "\nSolution."}}
+{"year": "2010", "tier": "T3", "problem_label": "G2", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Consider a triangle $A B C$ and let $M$ be the midpoint of the side $B C$. Suppose $\\angle M A C=\\angle A B C$ and $\\angle B A M=105^{\\circ}$. Find the measure of $\\angle A B C$.", "solution": "The angle measure is $30^{\\circ}$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_5ab10fadba60f287b615g-05.jpg?height=510&width=965&top_left_y=721&top_left_x=583)\n\nLet $O$ be the circumcenter of the triangle $A B M$. From $\\angle B A M=105^{\\circ}$ follows $\\angle M B O=15^{\\circ}$. Let $M^{\\prime}, C^{\\prime}$ be the projections of points $M, C$ onto the line $B O$. Since $\\angle M B O=15^{\\circ}$, then $\\angle M O M^{\\prime}=30^{\\circ}$ and consequently $M M^{\\prime}=\\frac{M O}{2}$. On the other hand, $M M^{\\prime}$ joins the midpoints of two sides of the triangle $B C C^{\\prime}$, which implies $C C^{\\prime}=M O=A O$.\n\nThe relation $\\angle M A C=\\angle A B C$ implies $C A$ tangent to $\\omega$, hence $A O \\perp A C$. It follows that $\\triangle A C O \\equiv \\triangle O C C^{\\prime}$, and furthermore $O B \\| A C$.\n\nTherefore $\\angle A O M=\\angle A O M^{\\prime}-\\angle M O M^{\\prime}=90^{\\circ}-30^{\\circ}=60^{\\circ}$ and $\\angle A B M=$ $\\frac{\\angle A O M}{2}=30^{\\circ}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem G2.", "solution_match": "\nSolution."}}
+{"year": "2010", "tier": "T3", "problem_label": "G3", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C$ be an acute-angled triangle. A circle $\\omega_{1}\\left(O_{1}, R_{1}\\right)$ passes through points $B$ and $C$ and meets the sides $A B$ and $A C$ at points $D$ and $E$, respectively. Let $\\omega_{2}\\left(O_{2}, R_{2}\\right)$ be the circumcircle of the triangle $A D E$. Prove that $O_{1} O_{2}$ is equal to the circumradius of the triangle $A B C$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_5ab10fadba60f287b615g-06.jpg?height=455&width=398&top_left_y=710&top_left_x=858)", "solution": "Recall that, in every triangle, the altitude and the diameter of the circumcircle drawn from the same vertex are isogonal. The proof offers no difficulty, being a simple angle chasing around the circumcircle of the triangle.\n\nLet $O$ be the circumcenter of the triangle $A B C$. From the above, one has $\\angle O A E=90^{\\circ}-B$. On the other hand $\\angle D E A=B$, for $B C D E$ is cyclic. Thus $A O \\perp D E$, implying that in the triangle $A D E$ cevians $A O$ and $A O_{2}$ are isogonal. So, since $A O$ is a radius of the circumcircle of triangle $A B C$, one obtains that $A O_{2}$ is an altitude in this triangle.\n\nMoreover, since $O O_{1}$ is the perpendicular bisector of the line segment $B C$, one has $O O_{1} \\perp B C$, and furthermore $A O_{2} \\| O O_{1}$.\n\nChord $D E$ is common to $\\omega_{1}$ and $\\omega_{2}$, hence $O_{1} O_{2} \\perp D E$. It follows that $A O \\|$ $O_{1} O_{2}$, so $A O O_{1} O_{2}$ is a parallelogram. The conclusion is now obvious.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem G3.", "solution_match": "\nSolution."}}
+{"year": "2010", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A L$ and $B K$ be angle bisectors in the non-isosceles triangle $A B C(L \\in B C, K \\in A C)$. The perpendicular bisector of $B K$ intersects the line $A L$ at point $M$. Point $N$ lies on the line $B K$ such that $L N \\| M K$. Prove that $L N=N A$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_5ab10fadba60f287b615g-07.jpg?height=439&width=721&top_left_y=667&top_left_x=694)", "solution": "The point $M$ lies on the circumcircle of $\\triangle A B K$ (since both $A L$ and the perpendicular bisector of $B K$ bisect the arc $B K$ of this circle). Then $\\angle C B K=\\angle A B K=\\angle A M K=\\angle N L A$. Thus $A B L N$ is cyclic, whence $\\angle N A L=$ $\\angle N B L=\\angle C B K=\\angle N L A$. Now it follows that $L N=N A$.\n\n## Combinatorics", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem G4.", "solution_match": "\nSolution."}}
+{"year": "2010", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "There are two piles of coins, each containing 2010 pieces. Two players A and B play a game taking turns (A plays first). At each turn, the player on play has to take one or more coins from one pile or exactly one coin from each pile. Whoever takes the last coin is the winner. Which player will win if they both play in the best possible way?", "solution": "B wins.\n\nIn fact, we will show that A will lose if the total number of coins is a multiple of 3 and the two piles differ by not more than one coin (call this a balanced position). To this end, firstly notice that it is not possible to move from one balanced position to another. The winning strategy for B consists in returning A to a balanced position (notice that the initial position is a balanced position).\n\nThere are two types of balanced positions; for each of them consider the moves of $\\mathrm{A}$ and the replies of B.\n\nIf the number in each pile is a multiple of 3 and there is at least one coin:\n\n- if A takes $3 n$ coins from one pile, then $\\mathrm{B}$ takes $3 n$ coins from the other one.\n- if A takes $3 n+1$ coins from one pile, then $\\mathrm{B}$ takes $3 n+2$ coins from the other one.\n- if A takes $3 n+2$ coins from one pile, then $\\mathrm{B}$ takes $3 n+1$ coins from the other one.\n- if A takes a coin from each pile, then $\\mathrm{B}$ takes one coin from one pile.\n\nIf the numbers are not multiples of 3 , then we have $3 m+1$ coins in one pile and $3 m+2$ in the other one. Hence:\n\n- if A takes $3 n$ coins from one pile, then $\\mathrm{B}$ takes $3 n$ coins from the other one.\n- if A takes $3 n+1$ coins from the first pile $(n \\leq m)$, then $\\mathrm{B}$ takes $3 n+2$ coins from the second one.\n- if A takes $3 n+2$ coins from the second pile $(n \\leq m)$, then $\\mathrm{B}$ takes $3 n+1$ coins from the first one.\n- if A takes $3 n+2$ coins from the first pile $(n \\leq m-1)$, then $\\mathrm{B}$ takes $3 n+4$ coins from the second one.\n- if A takes $3 n+1$ coins from the second pile $(n \\leq m)$, then $\\mathrm{B}$ takes $3 n-1$ coins from the first one. This is impossible if A has taken only one coin from the second pile; in this case $\\mathrm{B}$ takes one coin from each pile.\n- if A takes a coin from each pile, then B takes one coin from the second pile.\n\nIn all these cases, the position after B's move is again a balanced position. Since the number of coins decreases and $(0 ; 0)$ is a balanced position, after a finite number of moves, there will be no coins left after B's move. Thus, B wins.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem C1.", "solution_match": "\nSolution."}}
+{"year": "2010", "tier": "T3", "problem_label": "C2", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "A $9 \\times 7$ rectangle is tiled with pieces of two types, shown in the picture below.\n![](https://cdn.mathpix.com/cropped/2024_06_05_5ab10fadba60f287b615g-09.jpg?height=124&width=504&top_left_y=588&top_left_x=801)\n\nFind the possible values of the number of the $2 \\times 2$ pieces which can be used in such a tiling.", "solution": "Answer: 0 or 3.\n\nDenote by $x$ the number of the pieces of the type 'corner' and by $y$ the number of the pieces of the type of $2 \\times 2$. Mark 20 squares of the rectangle as in the figure below.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_5ab10fadba60f287b615g-09.jpg?height=295&width=374&top_left_y=1105&top_left_x=867)\n\nObviously, each piece covers at most one marked square.\n\nThus, $x+y \\geq 20$ (1) and consequently $3 x+3 y \\geq 60$ (2). On the other hand $3 x+4 y=63$ (3). From (2) and (3) it follows $y \\leq 3$ and from (3), $3 \\mid y$.\n\nThe proof is finished if we produce tilings with 3 , respectively $0,2 \\times 2$ tiles:\n![](https://cdn.mathpix.com/cropped/2024_06_05_5ab10fadba60f287b615g-09.jpg?height=300&width=830&top_left_y=1682&top_left_x=642)\n\n## Number Theory", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem C2.", "solution_match": "\nSolution."}}
+{"year": "2010", "tier": "T3", "problem_label": "N1", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all positive integers $n$ such that $n 2^{n+1}+1$ is a perfect square.", "solution": "Answer: $n=0$ and $n=3$.\n\nClearly $n 2^{n+1}+1$ is odd, so, if this number is a perfect square, then $n 2^{n+1}+1=$ $(2 x+1)^{2}, x \\in \\mathbb{N}$, whence $n 2^{n-1}=x(x+1)$.\n\nThe integers $x$ and $x+1$ are coprime, so one of them must divisible by $2^{n-1}$, which means that the other must be at most $n$. This shows that $2^{n-1} \\leqslant n+1$.\n\nAn easy induction shows that the above inequality is false for all $n \\geqslant 4$, and a direct inspection confirms that the only convenient values in the case $n \\leqslant 3$ are 0 and 3 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem N1.", "solution_match": "\nSolution."}}
+{"year": "2010", "tier": "T3", "problem_label": "N2", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all positive integers $n$ such that $36^{n}-6$ is a product of two or more consecutive positive integers.", "solution": "Answer: $n=1$.\n\nAmong each four consecutive integers there is a multiple of 4 . As $36^{n}-6$ is not a multiple of 4 , it must be the product of two or three consecutive positive integers.\n\nCase I. If $36^{n}-6=x(x+1)$ (all letters here and below denote positive integers), then $4 \\cdot 36^{n}-23=(2 x+1)^{2}$, whence $\\left(2 \\cdot 6^{n}+2 x+1\\right)\\left(2 \\cdot 6^{n}-2 x-1\\right)=23$. As 23 is prime, this leads to $2 \\cdot 6^{n}+2 x+1=23,2 \\cdot 6^{n}-2 x-1=1$. Subtracting these yields $4 x+2=22, x=5, n=1$, which is a solution to the problem.\n\nCase II. If $36^{n}-6=(y-1) y(y+1)$, then\n\n$$\n36^{n}=y^{3}-y+6=\\left(y^{3}+8\\right)-(y+2)=(y+2)\\left(y^{2}-2 y+3\\right)\n$$\n\nThus each of $y+2$ and $y^{2}-2 y+3$ can have only 2 and 3 as prime factors, so the same is true for their GCD. This, combined with the identity $y^{2}-2 y+3=$ $(y+2)(y-4)+11$ yields $\\operatorname{GCD}\\left(y+2 ; y^{2}-2 y+3\\right)=1$. Now $y+2<y^{2}-2 y+3$ and the latter number is odd, so $y+2=4^{n}, y^{2}-2 y+3=9^{n}$. The former identity implies $y$ is even and now by the latter one $9^{n} \\equiv 3(\\bmod 4)$, while in fact $9^{n} \\equiv 1(\\bmod 4)$ - a contradiction. So, in this case there is no such $n$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl", "problem_match": "\nProblem N2.", "solution_match": "\nSolution."}}

JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl

@@ -1,17 +1,17 @@
-{"year": "2013", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO", "problem": "Find all ordered triples $(x, y, z)$ of real numbers satisfying the following system of equations:\n\n$$\n\\begin{aligned}\nx^{3} & =\\frac{z}{y}-2 \\frac{y}{z} \\\\\ny^{3} & =\\frac{x}{z}-2 \\frac{z}{x} \\\\\nz^{3} & =\\frac{y}{x}-2 \\frac{x}{y}\n\\end{aligned}\n$$", "solution": "We have\n\n$$\n\\begin{aligned}\n& x^{3} y z=z^{2}-2 y^{2} \\\\\n& y^{3} z x=x^{2}-2 z^{2} \\\\\n& z^{3} x y=y^{2}-2 x^{2}\n\\end{aligned}\n$$\n\nwith $x y z \\neq 0$.\n\nAdding these up we obtain $\\left(x^{2}+y^{2}+z^{2}\\right)(x y z+1)=0$. Hence $x y z=-1$. Now the system of equations becomes:\n\n$$\n\\begin{aligned}\n& x^{2}=2 y^{2}-z^{2} \\\\\n& y^{2}=2 z^{2}-x^{2} \\\\\n& z^{2}=2 x^{2}-y^{2}\n\\end{aligned}\n$$\n\nThen the first two equations give $x^{2}=y^{2}=z^{2}$. As $x y z=-1$, we conclude that $(x, y, z)=$ $(1,1,-1),(1,-1,1),(-1,1,1)$ and $(-1,-1,-1)$ are the only solutions.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nA1.", "solution_match": "\nSolution."}}
-{"year": "2013", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO", "problem": "Find the largest possible value of the expression $\\left|\\sqrt{x^{2}+4 x+8}-\\sqrt{x^{2}+8 x+17}\\right|$ where $x$ is a real number.", "solution": "We observe that\n\n$$\n\\left|\\sqrt{x^{2}+4 x+8}-\\sqrt{x^{2}+8 x+17}\\right|=\\left|\\sqrt{(x-(-2))^{2}+(0-2)^{2}}-\\sqrt{\\left.(x-(-4))^{2}+(0-1)^{2}\\right)}\\right|\n$$\n\nis the absolute difference of the distances from the point $P(x, 0)$ in the $x y$-plane to the points $A(-2,2)$ and $B(-4,1)$.\n\nBy the Triangle Inequality, $|P A-P B| \\leq|A B|$ and the equality occurs exactly when $P$ lies on the line passing through $A$ and $B$, but not between them.\n\nIf $P, A, B$ are collinear, then $(x-(-2)) /(0-2)=((-4)-(-2)) /(1-2)$. This gives $x=-6$, and as $-6<-4<-2$,\n\n$$\n\\left|\\sqrt{(-6)^{2}+4(-6)+8}-\\sqrt{(-6)^{2}+8(-6)+17}\\right|=|\\sqrt{20}-\\sqrt{5}|=\\sqrt{5}\n$$\n\nis the largest possible value of the expression.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nA2.", "solution_match": "\nSolution."}}
-{"year": "2013", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO", "problem": "Show that\n\n$$\n\\left(a+2 b+\\frac{2}{a+1}\\right)\\left(b+2 a+\\frac{2}{b+1}\\right) \\geq 16\n$$\n\nfor all positive real numbers $a, b$ satisfying $a b \\geq 1$.", "solution": "Solution 2. Since $a b \\geq 1$, we have $a+b \\geq a+1 / a \\geq 2 \\sqrt{a \\cdot(1 / a)}=2$.\n\nThen\n\n$$\n\\begin{aligned}\na+2 b+\\frac{2}{a+1} & =b+(a+b)+\\frac{2}{a+1} \\\\\n& \\geq b+2+\\frac{2}{a+1} \\\\\n& =\\frac{b+1}{2}+\\frac{b+1}{2}+1+\\frac{2}{a+1} \\\\\n& \\geq 4 \\sqrt[4]{\\frac{(b+1)^{2}}{2(a+1)}}\n\\end{aligned}\n$$\n\nby the AM-GM Inequality. Similarly,\n\n$$\nb+2 a+\\frac{2}{b+1} \\geq 4 \\sqrt[4]{\\frac{(a+1)^{2}}{2(b+1)}}\n$$\n\nNow using these and applying the AM-GM Inequality another time we obtain:\n\n$$\n\\begin{aligned}\n\\left(a+2 b+\\frac{2}{a+1}\\right)\\left(b+2 a+\\frac{2}{b+1}\\right) & \\geq 16 \\sqrt[4]{\\frac{(a+1)(b+1)}{4}} \\\\\n& \\geq 16 \\sqrt[4]{\\frac{(2 \\sqrt{a})(2 \\sqrt{b})}{4}} \\\\\n& =16 \\sqrt[8]{a b} \\\\\n& \\geq 16\n\\end{aligned}\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nA3.", "solution_match": "\nSolution 1."}}
-{"year": "2013", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO", "problem": "Show that\n\n$$\n\\left(a+2 b+\\frac{2}{a+1}\\right)\\left(b+2 a+\\frac{2}{b+1}\\right) \\geq 16\n$$\n\nfor all positive real numbers $a, b$ satisfying $a b \\geq 1$.", "solution": "Solution 3. We have\n\n$$\n\\begin{aligned}\n\\left(a+2 b+\\frac{2}{a+1}\\right)\\left(b+2 a+\\frac{2}{b+1}\\right) & =\\left((a+b)+b+\\frac{2}{a+1}\\right)\\left((a+b)+a+\\frac{2}{b+1}\\right) \\\\\n& \\geq\\left(a+b+\\sqrt{a b}+\\frac{2}{\\sqrt{(a+1)(b+1)}}\\right)^{2}\n\\end{aligned}\n$$\n\nby the Cauchy-Schwarz Inequality.\n\nOn the other hand,\n\n$$\n\\frac{2}{\\sqrt{(a+1)(b+1)}} \\geq \\frac{4}{a+b+2}\n$$\n\nby the AM-GM Inequality and\n\n$$\na+b+\\sqrt{a b}+\\frac{2}{\\sqrt{(a+1)(b+1)}} \\geq a+b+1+\\frac{4}{a+b+2}=\\frac{(a+b+1)(a+b-2)}{a+b+2}+4 \\geq 4\n$$\n\nas $a+b \\geq 2 \\sqrt{a b} \\geq 2$, finishing the proof.\n\nQ1. Find the largest number of distinct integers that can be chosen from the set $\\{1,2, \\ldots, 2013\\}$ so that the difference of no two of them is equal to 17 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nA3.", "solution_match": "\nSolution 1."}}
-{"year": "2013", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO", "problem": "Show that\n\n$$\n\\left(a+2 b+\\frac{2}{a+1}\\right)\\left(b+2 a+\\frac{2}{b+1}\\right) \\geq 16\n$$\n\nfor all positive real numbers $a, b$ satisfying $a b \\geq 1$.", "solution": "Solution. Consider the sets $A_{m n}=\\{34 m+n-34,34 m+n-17\\}$ for $1 \\leq m \\leq 59$ and $1 \\leq n \\leq 17$, and $B_{k}=\\{2006+k\\}$ for $1 \\leq k \\leq 7$. As we cannot choose more than one number from each of these sets, we can choose at most $59 \\cdot 17+7=1010$ numbers. On the other hand, choosing the smaller element of each of these sets gives exactly 1010 numbers satisfying the condition.\n\nComment. The original problem proposal asks the question with the numbers 55 and 5 .\n\nQ2. On a billiards table in the shape of a rectangle $A B C D$ with $A B=2013$ and $A D=$ 1000 , a billiard ball is shot along the bisector of the angle $\\angle B A D$. Assuming that the ball is reflected from the sides at the same angle it comes in, determine whether it will ever go to the corner $B$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nA3.", "solution_match": "\nSolution 1."}}
-{"year": "2013", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO", "problem": "Show that\n\n$$\n\\left(a+2 b+\\frac{2}{a+1}\\right)\\left(b+2 a+\\frac{2}{b+1}\\right) \\geq 16\n$$\n\nfor all positive real numbers $a, b$ satisfying $a b \\geq 1$.", "solution": "Solution 1. The ball travels a horizontal distance of 1000 units between two bounces from the sides $A B$ and $C D$ as it always moves on a line making a $45^{\\circ}$ angle with the sides. Hence it is always at a distance of even number of units to the line $A D$ when it hits $A B$ or $C D$. Hence it can never hit $A B$ at $B$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nA3.", "solution_match": "\nSolution 1."}}
-{"year": "2013", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO", "problem": "Show that\n\n$$\n\\left(a+2 b+\\frac{2}{a+1}\\right)\\left(b+2 a+\\frac{2}{b+1}\\right) \\geq 16\n$$\n\nfor all positive real numbers $a, b$ satisfying $a b \\geq 1$.", "solution": "Solution 2. Consider a rectangle $A^{\\prime} B^{\\prime} C^{\\prime} D^{\\prime}$ which is wider $1 / 2$ units on all sides, divide it into unit squares, and color them black and white alternatingly with the vertex $A$ being the center of a black unit square. Then the ball always moves along the diagonals of the black unit squares. As $B$ lies at the center of a white unit square, the ball never reaches $B$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nA3.", "solution_match": "\nSolution 1."}}
-{"year": "2013", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO", "problem": "Show that\n\n$$\n\\left(a+2 b+\\frac{2}{a+1}\\right)\\left(b+2 a+\\frac{2}{b+1}\\right) \\geq 16\n$$\n\nfor all positive real numbers $a, b$ satisfying $a b \\geq 1$.", "solution": "Solution 3. The vertical lines $x=2013 m$ and the horizontal lines $y=1000 n$, where $m$ and $n$ are integers, divide the $x y$-plane into rectangles congruent to the rectangle $A B C D$. Let $A(0,0), B(2013,0), C(2013,1000), D(0,1000)$, and identify the other rectangles with $A B C D$ via reflections across these lines. Under this identification, the ball moves along the line $y=x$ and the coordinates of the points identified with $B$ have the form $(2013 k, 1000 l)$ where $k$ is an odd integer and $l$ is an even one. Hence the ball never goes to $B$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nA3.", "solution_match": "\nSolution 1."}}
-{"year": "2013", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "All possible pairs of $n$ apples are weighed and the results are given to us in an arbitrary order. Can we determine the weights of the apples if a. $n=4$, b. $n=5$, c. $n=6$ ?", "solution": "a. No. Four apples with weights $1,5,7,9$ and with weights $2,4,6,10$ both give the results $6,8,10,12,14,16$ when weighed in pairs.\n\nb. Yes. Let $a \\leq b \\leq c \\leq d \\leq e$ be the weights of the apples. As each apple is weighed 4 times, by adding all 10 pairwise weights and dividing the sum by 4 , we obtain $a+b+c+d+e$. Subtracting the smallest and the largest pairwise weights $a+b$ and $d+e$ from this we obtain c. Subtracting $c$ from the second largest pairwise weight $c+e$ we obtain $e$. Subtracting $e$ from the largest pairwise weight $d+e$ we obtain $d$. $a$ and $b$ are similarly determined.\n\nc. Yes. Let $a \\leq b \\leq c \\leq d \\leq e \\leq f$ be the weights of the apples. As each apple is weighed 5 times, by adding all 15 pairwise weights and dividing the sum by 5 , we obtain $a+b+c+d+e+f$. Subtracting the smallest and the largest pairwise weights $a+b$ and $e+f$ from this we obtain $c+d$.\n\nSubtracting the smallest and the second largest pairwise weights $a+b$ and $d+f$ from $a+b+c+d+e+f$ we obtain $c+e$. Similarly we obtain $b+d$. We use these to obtain $a+f$ and $b+e$.\n\nNow $a+d, a+e, b+c$ are the three smallest among the remaining six pairwise weights. If we add these up, subtract the known weights $c+d$ and $b+e$ form the sum and divide the difference by 2 , we obtain $a$. Then the rest follows.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nC3.", "solution_match": "\nSolution."}}
-{"year": "2013", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B$ be a diameter of a circle $\\omega$ with center $O$ and $O C$ be a radius of $\\omega$ which is perpendicular to $A B$. Let $M$ be a point on the line segment $O C$. Let $N$ be the second point of intersection of the line $A M$ with $\\omega$, and let $P$ be the point of intersection of the lines tangent to $\\omega$ at $N$ and at $B$. Show that the points $M, O, P, N$ are concyclic.", "solution": "Since the lines $P N$ and $B P$ are tangent to $\\omega, N P=P B$ and $O P$ is the bisector of $\\angle N O B$. Therefore the lines $O P$ and $N B$ are perpendicular. Since $\\angle A N B=90^{\\circ}$, it follows that the lines $A N$ and $O P$ are parallel. As $M O$ and $P B$ are also parallel and $A O=O B$, the triangles $A M O$ and $O P B$ are congruent and $M O=P B$. Hence $M O=N P$. Therefore $M O P N$ is an isosceles trapezoid and therefore cyclic. Hence the points $M, O, P, N$ are concyclic.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_66b0df6ec561017d2840g-16.jpg?height=1267&width=1249&top_left_y=779&top_left_x=382)\n\n8.2. $\\omega_{1}$ and $\\omega_{2}$ are two circles that are externally tangent to each other at the point $M$ and internally tangent to a circle $\\omega_{3}$ at the points $K$ and $L$, respectively. Let $A$ and $B$ be the two points where the common tangent line at $M$ to $\\omega_{1}$ and $\\omega_{2}$ intersects $\\omega_{3}$. Show that if $\\angle K A B=\\angle L A B$ then the line segment $A B$ is a diameter of $\\omega_{3}$.\n\nSolution. Let $C$ be the intersection point of the tangent lines to the circles $\\omega_{1}$ at $K$ and $\\omega_{2}$ at $L$. Point $C$ lies on the radical axis of circles $\\omega_{1}$ and $\\omega_{3}$, and also on the radical axis of the circles $\\omega_{2}$ and $\\omega_{3}$. Therefore $C$ lies on the radical axis of the circles $\\omega_{1}$ and $\\omega_{2}$ too. Therefore the points $A, B, C$ are collinear.\n\nSince $\\angle K A B=\\angle L A B$, the chords $K B$ and $B L$ have the same length. As we also have $C K=C L$, the triangles $K B C$ and $L B C$ are congruent. In particular, $\\angle K B A=\\angle L B A$. Therefore, $\\angle B K A=180^{\\circ}-(\\angle A B K+\\angle B A K)=180^{\\circ}-(\\angle L B K+\\angle L A K) / 2=180^{\\circ}-90^{\\circ}=$ $90^{\\circ}$, and $A B$ is a diameter.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_66b0df6ec561017d2840g-17.jpg?height=986&width=1290&top_left_y=885&top_left_x=447)\n\nComment. The original problem proposal gives $\\angle K A B=\\angle L A B=15^{\\circ}$ and asks the measures of the angles of the quadrilateral $A K B L$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nG1.", "solution_match": "\nSolution."}}
-{"year": "2013", "tier": "T3", "problem_label": "G3", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $D$ be a point on the side $B C$ of an acute triangle $A B C$ such that $\\angle B A D=\\angle C A O$ where $O$ is the center of the circumcircle $\\omega$ of the triangle $A B C$. Let $E$ be the second point of intersection of $\\omega$ and the line $A D$. Let $M, N, P$ be the midpoints of the line segments $B E, O D, A C$, respectively. Show that $M, N, P$ are collinear.", "solution": "We will show that $M O P D$ is a parallelogram. From this it follows that $M, N$, $P$ are collinear.\n\nSince $\\angle B A D=\\angle C A O=90^{\\circ}-\\angle A B C, D$ is the foot of the perpendicular from $A$ to side $B C$. Since $M$ is the midpoint of the line segment $B E$, we have $B M=M E=M D$ and hence $\\angle M D E=\\angle M E D=\\angle A C B$.\n\nLet the line $M D$ intersect the line $A C$ at $D_{1}$. Since $\\angle A D D_{1}=\\angle M D E=\\angle A C D, M D$ is perpendicular to $A C$. On the other hand, since $O$ is the center of the circumcircle of triangle $A B C$ and $P$ is the midpoint of the side $A C, O P$ is perpendicular to $A C$. Therefore $M D$ and $O P$ are parallel.\n\nSimilarly, since $P$ is the midpoint of the side $A C$, we have $A P=P C=D P$ and hence $\\angle P D C=\\angle A C B$. Let the line $P D$ intersect the line $B E$ at $D_{2}$. Since $\\angle B D D_{2}=\\angle P D C=$ $\\angle A C B=\\angle B E D$, we conclude that $P D$ is perpendicular to $B E$. Since $M$ is the midpoint of the line segment $B E, O M$ is perpendicular to $B E$ and hence $O M$ and $P D$ are parallel.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_66b0df6ec561017d2840g-18.jpg?height=1238&width=1335&top_left_y=1217&top_left_x=359)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nG3.", "solution_match": "\nSolution."}}
-{"year": "2013", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $I$ be the incenter and $A B$ the shortest side of a triangle $A B C$. The circle with center $I$ and passing through $C$ intersects the ray $A B$ at the point $P$ and the ray $B A$ at the point $Q$. Let $D$ be the point where the excircle of the triangle $A B C$ belonging to angle $A$ touches the side $B C$, and let $E$ be the symmetric of the point $C$ with respect to $D$. Show that the lines $P E$ and $C Q$ are perpendicular.", "solution": "First we will show that points $P$ and $Q$ are not on the line segment $A B$.\n\nAssume that $Q$ is on the line segment $A B$. Since $C I=Q I$ and $\\angle I B Q=\\angle I B C$, either the triangles $C B I$ and $Q B I$ are congruent or $\\angle I C B+\\angle I Q B=180^{\\circ}$. In the first case, we have $B C=B Q$ which contradicts $A B$ being the shortest side.\n\nIn the second case, we have $\\angle I Q A=\\angle I C B=\\angle I C A$ and the triangles $I A C$ and $I A Q$ are congruent. Hence this time we have $A C=A Q$, contradicting $A B$ being the shortest side.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_66b0df6ec561017d2840g-19.jpg?height=677&width=1058&top_left_y=874&top_left_x=520)\n\nCase 1\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_66b0df6ec561017d2840g-19.jpg?height=686&width=1058&top_left_y=1733&top_left_x=498)\n\nNow we will show that the lines $P E$ and $C Q$ are perpendicular.\n\nSince $\\angle I Q B \\leq \\angle I A B=(\\angle C A B) / 2<90^{\\circ}$ and $\\angle I C B=(\\angle A C B) / 2<90^{\\circ}$, the triangles $C B I$ and $Q B I$ are congruent. Hence $B C=B Q$ and $\\angle C Q P=\\angle C Q B=90^{\\circ}-(\\angle A B C) / 2$. Similarly, we have $A C=A P$ and hence $B P=A C-A B$.\n\nOn the other hand, as $D E=C D$ and $C D+A C=u$, where $u$ denotes the semiperimeter of the triangle $A B C$, we have $B E=B C-2(u-A C)=A C-A B$. Therefore $B P=B E$ and $\\angle Q P E=(\\angle A B C) / 2$.\n\nHence, $\\angle C Q P+\\angle Q P E=90^{\\circ}$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_66b0df6ec561017d2840g-20.jpg?height=1393&width=1427&top_left_y=773&top_left_x=333)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nG4.", "solution_match": "\nSolution."}}
-{"year": "2013", "tier": "T3", "problem_label": "G5", "problem_type": "Geometry", "exam": "JBMO", "problem": "A circle passing through the midpoint $M$ of the side $B C$ and the vertex $A$ of a triangle $A B C$ intersects the sides $A B$ and $A C$ for the second time at the points $P$ and $Q$, respectively. Show that if $\\angle B A C=60^{\\circ}$ then\n\n$$\nA P+A Q+P Q<A B+A C+\\frac{1}{2} B C\n$$", "solution": "Since the quadrilateral $A P M Q$ is cyclic, we have $\\angle P M Q=180^{\\circ}-\\angle P A Q=$ $180^{\\circ}-\\angle B A C=120^{\\circ}$. Therefore $\\angle P M B+\\angle Q M C=180^{\\circ}-\\angle P M Q=60^{\\circ}$.\n\nLet the point $B^{\\prime}$ be the symmetric of the point $B$ with respect to the line $P M$ and the point $C^{\\prime}$ be the symmetric of the point $C$ with respect to the line $Q M$. The triangles $B^{\\prime} M P$ and $B M P$ are congruent and the triangles $C^{\\prime} M Q$ and $C M Q$ are congruent. Hence $\\angle B^{\\prime} M C^{\\prime}=\\angle P M Q-\\angle B^{\\prime} M P-\\angle C^{\\prime} M Q=120^{\\circ}-\\angle B M P-\\angle C M Q=120^{\\circ}-60^{\\circ}=60^{\\circ}$. As we also have $B^{\\prime} M=B M=C M=C^{\\prime} M$, we conclude that the triangle $B^{\\prime} M C^{\\prime}$ is equilateral and $B^{\\prime} C^{\\prime}=B C / 2$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_66b0df6ec561017d2840g-21.jpg?height=898&width=1153&top_left_y=978&top_left_x=493)\n\nOn the other hand, we have $P B^{\\prime}+B^{\\prime} C^{\\prime}+C^{\\prime} Q \\geq P Q$ by the Triangle Inequality, and hence $P B+B C / 2+Q C \\geq P Q$. This gives the inequality $A B+B C / 2+A C \\geq A P+P Q+A Q$.\n\nWe get an equality only when the points $B^{\\prime}$ and $C^{\\prime}$ lie on the line segment $P Q$. If this is the case, then $\\angle P Q C+\\angle Q P B=2(\\angle P Q M+\\angle Q P M)=120^{\\circ}$ and therefore $\\angle A P Q+\\angle A Q P=$ $240^{\\circ} \\neq 120^{\\circ}$, a contradiction.\n$A B C D$. Let $K$ and $M$ be the points of intersection of the line $P D$ with $Q B$ and $Q A$, respectively, and let $N$ be the point of intersection of the lines $P A$ and $Q B$.\n\nLet $X, Y, Z$ be the midpoints of the line segments $A N, K N, A M$, respectively. Let $\\ell_{1}$ be the line passing through $X$ and perpendicular to $M K, \\ell_{2}$ be the line passing through $Y$ and perpendicular to $A M, \\ell_{3}$ be the line passing through $Z$ and perpendicular to $K N$. Show that $\\ell_{1}, \\ell_{2}, \\ell_{3}$ are concurrent.\n\nSolution. Let $R$ be the midpoint of the side $A D$. Then the lines $B R$ and $P D$ are parallel. Since $\\angle M A N=\\angle Q A P=\\angle Q B R=\\angle Q K M$, the points $A, N, K, M$ are concyclic.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_66b0df6ec561017d2840g-22.jpg?height=747&width=1266&top_left_y=650&top_left_x=408)\n\nLet $\\ell_{4}$ be the line passing through the midpoint $W$ of the line segment $M K$ and perpendicular to the line $A N$. Let $E_{1}$ be the point of intersection of $\\ell_{1}$ and $\\ell_{4}$, and $E_{2}$ be the point of intersection of $\\ell_{2}$ and $\\ell_{3}$. We will show that the points $E_{1}$ and $E_{2}$ coincide.\n\nLet $O$ be the circumcenter of the cyclic quadrilateral $A N K M$. $O W$ is perpendicular to the side $M K$ and $O X$ is perpendicular to the side $A N$. Hence $O W$ is parallel to $\\ell_{1}, O X$ is parallel to $\\ell_{4}$, and $X O W E_{1}$ is a parallelogram. Therefore the midpoints of the line segments $O E_{1}$ and $W X$ coincide. Similarly, the midpoints of the line segments $O E_{2}$ and $Y Z$ coincide.\n\nOn the other hand, as $X, Y, Z, W$ are midpoints of the sides of the quadrilateral $A N K M$, $X Y W Z$ is a parallelogram and therefore the midpoints of the line segments $W X$ and $Y Z$ coincide. Hence the midpoints of the line segments $O E_{1}$ and $O E_{2}$ coincide. In other words, $E_{1}$ and $E_{2}$ are the same point, and the lines $\\ell_{1}, \\ell_{2}, \\ell_{3}$ are concurrent.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_66b0df6ec561017d2840g-23.jpg?height=1304&width=1172&top_left_y=841&top_left_x=498)\n\nComment. The problem can be asked in the following form:\n\nLet $A N K M$ be a cyclic quadrilateral and let $X, Y, Z$ be the midpoints of the sides $A N$, $K N, A M$, respectively. Let $\\ell_{1}$ be the line passing through $X$ and perpendicular to $M K$, $\\ell_{2}$ be the line passing through $Y$ and perpendicular to $A M, \\ell_{3}$ be the line passing through $Z$ and perpendicular to $K N$. Show that $\\ell_{1}, \\ell_{2}, \\ell_{3}$ are concurrent.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nG5.", "solution_match": "\nSolution."}}
-{"year": "2013", "tier": "T3", "problem_label": "N1", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all positive integere on for othich $1^{2}+2^{2}+\\cdots+16^{2}+17$ is a perfort square", "solution": "We bave $1^{3}+2^{5}+\\cdots+16^{2}=(1+2+\\cdots+16)^{2}=5^{2} \\quad 17^{0}$ Herese. $11^{3}+2^{3}+$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_66b0df6ec561017d2840g-24.jpg?height=83&width=1607&top_left_y=309&top_left_x=263)\nmeeger t then $1 T^{\\prime}-(1+B)(t-3)$ A. $(1,5) \\quad(t-3)=16$ this can coly happen\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_66b0df6ec561017d2840g-24.jpg?height=74&width=1467&top_left_y=415&top_left_x=206)\n\nEirornd all ordered triples $(x, y, z)$ of integers satisfying $20^{x}+13^{y}=2013^{z}$.\n\nSolution. As $20 \\cdot 13=2^{2} \\cdot 5 \\cdot 13$ and $2013=3 \\cdot 11 \\cdot 61$ are relatively prime, $x, y$ and $z$ must be nonnegative.\n\nConsidering the equation modulo 3 , we observe that $x$ must be odd. Now considering the equation modulo 7 , we obtain $(-1)+(-1)^{y} \\equiv 4^{z}(\\bmod 7)$, which is impossible as the right hand side can only be 1,2 and 4 modulo 7 .\n\nThere are no solutions.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nN1.", "solution_match": "\nSolution."}}
-{"year": "2013", "tier": "T3", "problem_label": "N3", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all ordered pairs $(a, b)$ of positive integers for which the numbers $\\frac{a^{3} b-1}{a+1}$ and $\\frac{b^{3} a+1}{b-1}$ are positive integers.", "solution": "As $a^{3} b-1=b\\left(a^{3}+1\\right)-(b+1)$ and $a+1 \\mid a^{3}+1$, we have $a+1 \\mid b+1$.\n\nAs $b^{3} a+1=a\\left(b^{3}-1\\right)+(a+1)$ and $b-1 \\mid b^{3}-1$, we have $b-1 \\mid a+1$.\n\nSo $b-1 \\mid b+1$ and hence $b-1 \\mid 2$.\n\n- If $b=2$, then $a+1 \\mid b+1=3$ gives $a=2$. Hence $(a, b)=(2,2)$ is the only solution in this case.\n- If $b=3$, then $a+1 \\mid b+1=4$ gives $a=1$ or $a=3$. Hence $(a, b)=(1,3)$ and $(3,3)$ are the only solutions in this case.\n\nTo summarize, $(a, b)=(1,3),(2,2)$ and $(3,3)$ are the only solutions.\n$\\mathbf{N 4}$. A rectangle in the $x y$-plane is called latticed if all its vertices have integer coordinates.\n\na. Find a latticed rectangle with area 2013 whose sides are not parallel to the axes.\n\nb. Show that if a latticed rectangle has area 2011, then its sides are parallel to the axes.\n\nSolution. a. The rectangle $P Q R S$ with $P(0,0), Q(165,198), R(159,203), S(-6,5)$ has area 2013 .\n\nb. Suppose that the latticed rectangle $P Q R S$ has area 2011 and its sides are not parallel to the axes. Without loss of generality we may assume that its vertices are $P(0,0), Q(a, b)$, $S(c, d), R(a+c, b+d)$ where $a, b, c, d$ are integers and $a b c d \\neq 0$.\n\nThen $\\left(a^{2}+b^{2}\\right)\\left(c^{2}+d^{2}\\right)=2011^{2}$. As 2011 is a prime, $2011 \\mid a^{2}+b^{2}$ or $2011 \\mid c^{2}+d^{2}$. Assume that the first one is the case. Since $2011 \\equiv 3(\\bmod 4)$, this can happen only if $a=2011 a_{0}$ and $b=2011 b_{0}$ for some integers $a_{0}$ and $b_{0}$. Now we have $\\left(a_{0}^{2}+b_{0}^{2}\\right)\\left(c^{2}+d^{2}\\right)=1$. This means $c^{2}+d^{2}=1$ and hence $c=0$ or $d=0$, contradicting $c d \\neq 0$.\n\nComment. The original problem proposal has the numbers 13 and 11 instead.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nN3.", "solution_match": "\nSolution."}}
-{"year": "2013", "tier": "T3", "problem_label": "N5", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all ordered triples $(x, y, z)$ of positive integers satisfying the equation\n\n$$\n\\frac{1}{x^{2}}+\\frac{y}{x z}+\\frac{1}{z^{2}}=\\frac{1}{2013}\n$$", "solution": "We have $x^{2} z^{2}=2013\\left(x^{2}+x y z+z^{2}\\right)$. Let $d=\\operatorname{gcd}(x, z)$ and $x=d a, z=d b$. Then $a^{2} b^{2} d^{2}=2013\\left(a^{2}+a b y+b^{2}\\right)$.\n\nAs $\\operatorname{gcd}(a, b)=1$, we also have $\\operatorname{gcd}\\left(a^{2}, a^{2}+a b y+b^{2}\\right)=1$ and $\\operatorname{gcd}\\left(b^{2}, a^{2}+a b y+b^{2}\\right)=1$. Therefore $a^{2} \\mid 2013$ and $b^{2} \\mid 2013$. But $2013=3 \\cdot 11 \\cdot 61$ is squarefree and therefore $a=1=b$.\n\nNow we have $x=z=d$ and $d^{2}=2013(y+2)$. Once again as 2013 is squarefree, we must have $y+2=2013 n^{2}$ where $n$ is a positive integer.\n\nHence $(x, y, z)=\\left(2013 n, 2013 n^{2}-2,2013 n\\right)$ where $n$ is a positive integer.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nN5.", "solution_match": "\nSolution."}}
-{"year": "2013", "tier": "T3", "problem_label": "N6", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all ordered triples $(x, y, z)$ of integers satisfying the following system of equations:\n\n$$\n\\begin{aligned}\nx^{2}-y^{2} & =z \\\\\n3 x y+(x-y) z & =z^{2}\n\\end{aligned}\n$$", "solution": "If $z=0$, then $x=0$ and $y=0$, and $(x, y, z)=(0,0,0)$.\n\nLet us assume that $z \\neq 0$, and $x+y=a$ and $x-y=b$ where $a$ and $b$ are nonzero integers such that $z=a b$. Then $x=(a+b) / 2$ and $y=(a-b) / 2$, and the second equations gives $3 a^{2}-3 b^{2}+4 a b^{2}=4 a^{2} b^{2}$.\n\nHence\n\n$$\nb^{2}=\\frac{3 a^{2}}{4 a^{2}-4 a+3}\n$$\n\nand\n\n$$\n3 a^{2} \\geq 4 a^{2}-4 a+3\n$$\n\nwhich is satisfied only if $a=1,2$ or 3 .\n\n- If $a=1$, then $b^{2}=1 .(x, y, z)=(1,0,1)$ and $(0,1,-1)$ are the only solutions in this case.\n- If $a=2$, then $b^{2}=12 / 11$. There are no solutions in this case\n- If $a=3$, then $b^{2}=1 .(x, y, z)=(1,2,-3)$ and $(2,1,3)$ are the only solutions in this case.\n\nTo summarize, $(x, y, z)=(0,0,0),(1,0,1),(0,1,-1),(1,2,-3)$ and $(2,1,3)$ are the only solutions.\n\nComments. 1. The original problem proposal asks for the solutions when $z=p$ is a prime number.\n\n2. The problem can be asked with a single equation in the form:\n\n$$\n3 x y+(x-y)^{2}(x+y)=\\left(x^{2}-y^{2}\\right)^{2}\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nN6.", "solution_match": "\nSolution."}}
+{"year": "2013", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Find all ordered triples $(x, y, z)$ of real numbers satisfying the following system of equations:\n\n$$\n\\begin{aligned}\nx^{3} & =\\frac{z}{y}-2 \\frac{y}{z} \\\\\ny^{3} & =\\frac{x}{z}-2 \\frac{z}{x} \\\\\nz^{3} & =\\frac{y}{x}-2 \\frac{x}{y}\n\\end{aligned}\n$$", "solution": "We have\n\n$$\n\\begin{aligned}\n& x^{3} y z=z^{2}-2 y^{2} \\\\\n& y^{3} z x=x^{2}-2 z^{2} \\\\\n& z^{3} x y=y^{2}-2 x^{2}\n\\end{aligned}\n$$\n\nwith $x y z \\neq 0$.\n\nAdding these up we obtain $\\left(x^{2}+y^{2}+z^{2}\\right)(x y z+1)=0$. Hence $x y z=-1$. Now the system of equations becomes:\n\n$$\n\\begin{aligned}\n& x^{2}=2 y^{2}-z^{2} \\\\\n& y^{2}=2 z^{2}-x^{2} \\\\\n& z^{2}=2 x^{2}-y^{2}\n\\end{aligned}\n$$\n\nThen the first two equations give $x^{2}=y^{2}=z^{2}$. As $x y z=-1$, we conclude that $(x, y, z)=$ $(1,1,-1),(1,-1,1),(-1,1,1)$ and $(-1,-1,-1)$ are the only solutions.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nA1.", "solution_match": "\nSolution."}}
+{"year": "2013", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Find the largest possible value of the expression $\\left|\\sqrt{x^{2}+4 x+8}-\\sqrt{x^{2}+8 x+17}\\right|$ where $x$ is a real number.", "solution": "We observe that\n\n$$\n\\left|\\sqrt{x^{2}+4 x+8}-\\sqrt{x^{2}+8 x+17}\\right|=\\left|\\sqrt{(x-(-2))^{2}+(0-2)^{2}}-\\sqrt{\\left.(x-(-4))^{2}+(0-1)^{2}\\right)}\\right|\n$$\n\nis the absolute difference of the distances from the point $P(x, 0)$ in the $x y$-plane to the points $A(-2,2)$ and $B(-4,1)$.\n\nBy the Triangle Inequality, $|P A-P B| \\leq|A B|$ and the equality occurs exactly when $P$ lies on the line passing through $A$ and $B$, but not between them.\n\nIf $P, A, B$ are collinear, then $(x-(-2)) /(0-2)=((-4)-(-2)) /(1-2)$. This gives $x=-6$, and as $-6<-4<-2$,\n\n$$\n\\left|\\sqrt{(-6)^{2}+4(-6)+8}-\\sqrt{(-6)^{2}+8(-6)+17}\\right|=|\\sqrt{20}-\\sqrt{5}|=\\sqrt{5}\n$$\n\nis the largest possible value of the expression.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nA2.", "solution_match": "\nSolution."}}
+{"year": "2013", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Show that\n\n$$\n\\left(a+2 b+\\frac{2}{a+1}\\right)\\left(b+2 a+\\frac{2}{b+1}\\right) \\geq 16\n$$\n\nfor all positive real numbers $a, b$ satisfying $a b \\geq 1$.", "solution": "Solution 2. Since $a b \\geq 1$, we have $a+b \\geq a+1 / a \\geq 2 \\sqrt{a \\cdot(1 / a)}=2$.\n\nThen\n\n$$\n\\begin{aligned}\na+2 b+\\frac{2}{a+1} & =b+(a+b)+\\frac{2}{a+1} \\\\\n& \\geq b+2+\\frac{2}{a+1} \\\\\n& =\\frac{b+1}{2}+\\frac{b+1}{2}+1+\\frac{2}{a+1} \\\\\n& \\geq 4 \\sqrt[4]{\\frac{(b+1)^{2}}{2(a+1)}}\n\\end{aligned}\n$$\n\nby the AM-GM Inequality. Similarly,\n\n$$\nb+2 a+\\frac{2}{b+1} \\geq 4 \\sqrt[4]{\\frac{(a+1)^{2}}{2(b+1)}}\n$$\n\nNow using these and applying the AM-GM Inequality another time we obtain:\n\n$$\n\\begin{aligned}\n\\left(a+2 b+\\frac{2}{a+1}\\right)\\left(b+2 a+\\frac{2}{b+1}\\right) & \\geq 16 \\sqrt[4]{\\frac{(a+1)(b+1)}{4}} \\\\\n& \\geq 16 \\sqrt[4]{\\frac{(2 \\sqrt{a})(2 \\sqrt{b})}{4}} \\\\\n& =16 \\sqrt[8]{a b} \\\\\n& \\geq 16\n\\end{aligned}\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nA3.", "solution_match": "\nSolution 1."}}
+{"year": "2013", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Show that\n\n$$\n\\left(a+2 b+\\frac{2}{a+1}\\right)\\left(b+2 a+\\frac{2}{b+1}\\right) \\geq 16\n$$\n\nfor all positive real numbers $a, b$ satisfying $a b \\geq 1$.", "solution": "Solution 3. We have\n\n$$\n\\begin{aligned}\n\\left(a+2 b+\\frac{2}{a+1}\\right)\\left(b+2 a+\\frac{2}{b+1}\\right) & =\\left((a+b)+b+\\frac{2}{a+1}\\right)\\left((a+b)+a+\\frac{2}{b+1}\\right) \\\\\n& \\geq\\left(a+b+\\sqrt{a b}+\\frac{2}{\\sqrt{(a+1)(b+1)}}\\right)^{2}\n\\end{aligned}\n$$\n\nby the Cauchy-Schwarz Inequality.\n\nOn the other hand,\n\n$$\n\\frac{2}{\\sqrt{(a+1)(b+1)}} \\geq \\frac{4}{a+b+2}\n$$\n\nby the AM-GM Inequality and\n\n$$\na+b+\\sqrt{a b}+\\frac{2}{\\sqrt{(a+1)(b+1)}} \\geq a+b+1+\\frac{4}{a+b+2}=\\frac{(a+b+1)(a+b-2)}{a+b+2}+4 \\geq 4\n$$\n\nas $a+b \\geq 2 \\sqrt{a b} \\geq 2$, finishing the proof.\n\nQ1. Find the largest number of distinct integers that can be chosen from the set $\\{1,2, \\ldots, 2013\\}$ so that the difference of no two of them is equal to 17 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nA3.", "solution_match": "\nSolution 1."}}
+{"year": "2013", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Show that\n\n$$\n\\left(a+2 b+\\frac{2}{a+1}\\right)\\left(b+2 a+\\frac{2}{b+1}\\right) \\geq 16\n$$\n\nfor all positive real numbers $a, b$ satisfying $a b \\geq 1$.", "solution": "Solution. Consider the sets $A_{m n}=\\{34 m+n-34,34 m+n-17\\}$ for $1 \\leq m \\leq 59$ and $1 \\leq n \\leq 17$, and $B_{k}=\\{2006+k\\}$ for $1 \\leq k \\leq 7$. As we cannot choose more than one number from each of these sets, we can choose at most $59 \\cdot 17+7=1010$ numbers. On the other hand, choosing the smaller element of each of these sets gives exactly 1010 numbers satisfying the condition.\n\nComment. The original problem proposal asks the question with the numbers 55 and 5 .\n\nQ2. On a billiards table in the shape of a rectangle $A B C D$ with $A B=2013$ and $A D=$ 1000 , a billiard ball is shot along the bisector of the angle $\\angle B A D$. Assuming that the ball is reflected from the sides at the same angle it comes in, determine whether it will ever go to the corner $B$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nA3.", "solution_match": "\nSolution 1."}}
+{"year": "2013", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Show that\n\n$$\n\\left(a+2 b+\\frac{2}{a+1}\\right)\\left(b+2 a+\\frac{2}{b+1}\\right) \\geq 16\n$$\n\nfor all positive real numbers $a, b$ satisfying $a b \\geq 1$.", "solution": "Solution 1. The ball travels a horizontal distance of 1000 units between two bounces from the sides $A B$ and $C D$ as it always moves on a line making a $45^{\\circ}$ angle with the sides. Hence it is always at a distance of even number of units to the line $A D$ when it hits $A B$ or $C D$. Hence it can never hit $A B$ at $B$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nA3.", "solution_match": "\nSolution 1."}}
+{"year": "2013", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Show that\n\n$$\n\\left(a+2 b+\\frac{2}{a+1}\\right)\\left(b+2 a+\\frac{2}{b+1}\\right) \\geq 16\n$$\n\nfor all positive real numbers $a, b$ satisfying $a b \\geq 1$.", "solution": "Solution 2. Consider a rectangle $A^{\\prime} B^{\\prime} C^{\\prime} D^{\\prime}$ which is wider $1 / 2$ units on all sides, divide it into unit squares, and color them black and white alternatingly with the vertex $A$ being the center of a black unit square. Then the ball always moves along the diagonals of the black unit squares. As $B$ lies at the center of a white unit square, the ball never reaches $B$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nA3.", "solution_match": "\nSolution 1."}}
+{"year": "2013", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Show that\n\n$$\n\\left(a+2 b+\\frac{2}{a+1}\\right)\\left(b+2 a+\\frac{2}{b+1}\\right) \\geq 16\n$$\n\nfor all positive real numbers $a, b$ satisfying $a b \\geq 1$.", "solution": "Solution 3. The vertical lines $x=2013 m$ and the horizontal lines $y=1000 n$, where $m$ and $n$ are integers, divide the $x y$-plane into rectangles congruent to the rectangle $A B C D$. Let $A(0,0), B(2013,0), C(2013,1000), D(0,1000)$, and identify the other rectangles with $A B C D$ via reflections across these lines. Under this identification, the ball moves along the line $y=x$ and the coordinates of the points identified with $B$ have the form $(2013 k, 1000 l)$ where $k$ is an odd integer and $l$ is an even one. Hence the ball never goes to $B$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nA3.", "solution_match": "\nSolution 1."}}
+{"year": "2013", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "All possible pairs of $n$ apples are weighed and the results are given to us in an arbitrary order. Can we determine the weights of the apples if a. $n=4$, b. $n=5$, c. $n=6$ ?", "solution": "a. No. Four apples with weights $1,5,7,9$ and with weights $2,4,6,10$ both give the results $6,8,10,12,14,16$ when weighed in pairs.\n\nb. Yes. Let $a \\leq b \\leq c \\leq d \\leq e$ be the weights of the apples. As each apple is weighed 4 times, by adding all 10 pairwise weights and dividing the sum by 4 , we obtain $a+b+c+d+e$. Subtracting the smallest and the largest pairwise weights $a+b$ and $d+e$ from this we obtain c. Subtracting $c$ from the second largest pairwise weight $c+e$ we obtain $e$. Subtracting $e$ from the largest pairwise weight $d+e$ we obtain $d$. $a$ and $b$ are similarly determined.\n\nc. Yes. Let $a \\leq b \\leq c \\leq d \\leq e \\leq f$ be the weights of the apples. As each apple is weighed 5 times, by adding all 15 pairwise weights and dividing the sum by 5 , we obtain $a+b+c+d+e+f$. Subtracting the smallest and the largest pairwise weights $a+b$ and $e+f$ from this we obtain $c+d$.\n\nSubtracting the smallest and the second largest pairwise weights $a+b$ and $d+f$ from $a+b+c+d+e+f$ we obtain $c+e$. Similarly we obtain $b+d$. We use these to obtain $a+f$ and $b+e$.\n\nNow $a+d, a+e, b+c$ are the three smallest among the remaining six pairwise weights. If we add these up, subtract the known weights $c+d$ and $b+e$ form the sum and divide the difference by 2 , we obtain $a$. Then the rest follows.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nC3.", "solution_match": "\nSolution."}}
+{"year": "2013", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B$ be a diameter of a circle $\\omega$ with center $O$ and $O C$ be a radius of $\\omega$ which is perpendicular to $A B$. Let $M$ be a point on the line segment $O C$. Let $N$ be the second point of intersection of the line $A M$ with $\\omega$, and let $P$ be the point of intersection of the lines tangent to $\\omega$ at $N$ and at $B$. Show that the points $M, O, P, N$ are concyclic.", "solution": "Since the lines $P N$ and $B P$ are tangent to $\\omega, N P=P B$ and $O P$ is the bisector of $\\angle N O B$. Therefore the lines $O P$ and $N B$ are perpendicular. Since $\\angle A N B=90^{\\circ}$, it follows that the lines $A N$ and $O P$ are parallel. As $M O$ and $P B$ are also parallel and $A O=O B$, the triangles $A M O$ and $O P B$ are congruent and $M O=P B$. Hence $M O=N P$. Therefore $M O P N$ is an isosceles trapezoid and therefore cyclic. Hence the points $M, O, P, N$ are concyclic.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_66b0df6ec561017d2840g-16.jpg?height=1267&width=1249&top_left_y=779&top_left_x=382)\n\n8.2. $\\omega_{1}$ and $\\omega_{2}$ are two circles that are externally tangent to each other at the point $M$ and internally tangent to a circle $\\omega_{3}$ at the points $K$ and $L$, respectively. Let $A$ and $B$ be the two points where the common tangent line at $M$ to $\\omega_{1}$ and $\\omega_{2}$ intersects $\\omega_{3}$. Show that if $\\angle K A B=\\angle L A B$ then the line segment $A B$ is a diameter of $\\omega_{3}$.\n\nSolution. Let $C$ be the intersection point of the tangent lines to the circles $\\omega_{1}$ at $K$ and $\\omega_{2}$ at $L$. Point $C$ lies on the radical axis of circles $\\omega_{1}$ and $\\omega_{3}$, and also on the radical axis of the circles $\\omega_{2}$ and $\\omega_{3}$. Therefore $C$ lies on the radical axis of the circles $\\omega_{1}$ and $\\omega_{2}$ too. Therefore the points $A, B, C$ are collinear.\n\nSince $\\angle K A B=\\angle L A B$, the chords $K B$ and $B L$ have the same length. As we also have $C K=C L$, the triangles $K B C$ and $L B C$ are congruent. In particular, $\\angle K B A=\\angle L B A$. Therefore, $\\angle B K A=180^{\\circ}-(\\angle A B K+\\angle B A K)=180^{\\circ}-(\\angle L B K+\\angle L A K) / 2=180^{\\circ}-90^{\\circ}=$ $90^{\\circ}$, and $A B$ is a diameter.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_66b0df6ec561017d2840g-17.jpg?height=986&width=1290&top_left_y=885&top_left_x=447)\n\nComment. The original problem proposal gives $\\angle K A B=\\angle L A B=15^{\\circ}$ and asks the measures of the angles of the quadrilateral $A K B L$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nG1.", "solution_match": "\nSolution."}}
+{"year": "2013", "tier": "T3", "problem_label": "G3", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $D$ be a point on the side $B C$ of an acute triangle $A B C$ such that $\\angle B A D=\\angle C A O$ where $O$ is the center of the circumcircle $\\omega$ of the triangle $A B C$. Let $E$ be the second point of intersection of $\\omega$ and the line $A D$. Let $M, N, P$ be the midpoints of the line segments $B E, O D, A C$, respectively. Show that $M, N, P$ are collinear.", "solution": "We will show that $M O P D$ is a parallelogram. From this it follows that $M, N$, $P$ are collinear.\n\nSince $\\angle B A D=\\angle C A O=90^{\\circ}-\\angle A B C, D$ is the foot of the perpendicular from $A$ to side $B C$. Since $M$ is the midpoint of the line segment $B E$, we have $B M=M E=M D$ and hence $\\angle M D E=\\angle M E D=\\angle A C B$.\n\nLet the line $M D$ intersect the line $A C$ at $D_{1}$. Since $\\angle A D D_{1}=\\angle M D E=\\angle A C D, M D$ is perpendicular to $A C$. On the other hand, since $O$ is the center of the circumcircle of triangle $A B C$ and $P$ is the midpoint of the side $A C, O P$ is perpendicular to $A C$. Therefore $M D$ and $O P$ are parallel.\n\nSimilarly, since $P$ is the midpoint of the side $A C$, we have $A P=P C=D P$ and hence $\\angle P D C=\\angle A C B$. Let the line $P D$ intersect the line $B E$ at $D_{2}$. Since $\\angle B D D_{2}=\\angle P D C=$ $\\angle A C B=\\angle B E D$, we conclude that $P D$ is perpendicular to $B E$. Since $M$ is the midpoint of the line segment $B E, O M$ is perpendicular to $B E$ and hence $O M$ and $P D$ are parallel.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_66b0df6ec561017d2840g-18.jpg?height=1238&width=1335&top_left_y=1217&top_left_x=359)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nG3.", "solution_match": "\nSolution."}}
+{"year": "2013", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $I$ be the incenter and $A B$ the shortest side of a triangle $A B C$. The circle with center $I$ and passing through $C$ intersects the ray $A B$ at the point $P$ and the ray $B A$ at the point $Q$. Let $D$ be the point where the excircle of the triangle $A B C$ belonging to angle $A$ touches the side $B C$, and let $E$ be the symmetric of the point $C$ with respect to $D$. Show that the lines $P E$ and $C Q$ are perpendicular.", "solution": "First we will show that points $P$ and $Q$ are not on the line segment $A B$.\n\nAssume that $Q$ is on the line segment $A B$. Since $C I=Q I$ and $\\angle I B Q=\\angle I B C$, either the triangles $C B I$ and $Q B I$ are congruent or $\\angle I C B+\\angle I Q B=180^{\\circ}$. In the first case, we have $B C=B Q$ which contradicts $A B$ being the shortest side.\n\nIn the second case, we have $\\angle I Q A=\\angle I C B=\\angle I C A$ and the triangles $I A C$ and $I A Q$ are congruent. Hence this time we have $A C=A Q$, contradicting $A B$ being the shortest side.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_66b0df6ec561017d2840g-19.jpg?height=677&width=1058&top_left_y=874&top_left_x=520)\n\nCase 1\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_66b0df6ec561017d2840g-19.jpg?height=686&width=1058&top_left_y=1733&top_left_x=498)\n\nNow we will show that the lines $P E$ and $C Q$ are perpendicular.\n\nSince $\\angle I Q B \\leq \\angle I A B=(\\angle C A B) / 2<90^{\\circ}$ and $\\angle I C B=(\\angle A C B) / 2<90^{\\circ}$, the triangles $C B I$ and $Q B I$ are congruent. Hence $B C=B Q$ and $\\angle C Q P=\\angle C Q B=90^{\\circ}-(\\angle A B C) / 2$. Similarly, we have $A C=A P$ and hence $B P=A C-A B$.\n\nOn the other hand, as $D E=C D$ and $C D+A C=u$, where $u$ denotes the semiperimeter of the triangle $A B C$, we have $B E=B C-2(u-A C)=A C-A B$. Therefore $B P=B E$ and $\\angle Q P E=(\\angle A B C) / 2$.\n\nHence, $\\angle C Q P+\\angle Q P E=90^{\\circ}$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_66b0df6ec561017d2840g-20.jpg?height=1393&width=1427&top_left_y=773&top_left_x=333)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nG4.", "solution_match": "\nSolution."}}
+{"year": "2013", "tier": "T3", "problem_label": "G5", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "A circle passing through the midpoint $M$ of the side $B C$ and the vertex $A$ of a triangle $A B C$ intersects the sides $A B$ and $A C$ for the second time at the points $P$ and $Q$, respectively. Show that if $\\angle B A C=60^{\\circ}$ then\n\n$$\nA P+A Q+P Q<A B+A C+\\frac{1}{2} B C\n$$", "solution": "Since the quadrilateral $A P M Q$ is cyclic, we have $\\angle P M Q=180^{\\circ}-\\angle P A Q=$ $180^{\\circ}-\\angle B A C=120^{\\circ}$. Therefore $\\angle P M B+\\angle Q M C=180^{\\circ}-\\angle P M Q=60^{\\circ}$.\n\nLet the point $B^{\\prime}$ be the symmetric of the point $B$ with respect to the line $P M$ and the point $C^{\\prime}$ be the symmetric of the point $C$ with respect to the line $Q M$. The triangles $B^{\\prime} M P$ and $B M P$ are congruent and the triangles $C^{\\prime} M Q$ and $C M Q$ are congruent. Hence $\\angle B^{\\prime} M C^{\\prime}=\\angle P M Q-\\angle B^{\\prime} M P-\\angle C^{\\prime} M Q=120^{\\circ}-\\angle B M P-\\angle C M Q=120^{\\circ}-60^{\\circ}=60^{\\circ}$. As we also have $B^{\\prime} M=B M=C M=C^{\\prime} M$, we conclude that the triangle $B^{\\prime} M C^{\\prime}$ is equilateral and $B^{\\prime} C^{\\prime}=B C / 2$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_66b0df6ec561017d2840g-21.jpg?height=898&width=1153&top_left_y=978&top_left_x=493)\n\nOn the other hand, we have $P B^{\\prime}+B^{\\prime} C^{\\prime}+C^{\\prime} Q \\geq P Q$ by the Triangle Inequality, and hence $P B+B C / 2+Q C \\geq P Q$. This gives the inequality $A B+B C / 2+A C \\geq A P+P Q+A Q$.\n\nWe get an equality only when the points $B^{\\prime}$ and $C^{\\prime}$ lie on the line segment $P Q$. If this is the case, then $\\angle P Q C+\\angle Q P B=2(\\angle P Q M+\\angle Q P M)=120^{\\circ}$ and therefore $\\angle A P Q+\\angle A Q P=$ $240^{\\circ} \\neq 120^{\\circ}$, a contradiction.\n$A B C D$. Let $K$ and $M$ be the points of intersection of the line $P D$ with $Q B$ and $Q A$, respectively, and let $N$ be the point of intersection of the lines $P A$ and $Q B$.\n\nLet $X, Y, Z$ be the midpoints of the line segments $A N, K N, A M$, respectively. Let $\\ell_{1}$ be the line passing through $X$ and perpendicular to $M K, \\ell_{2}$ be the line passing through $Y$ and perpendicular to $A M, \\ell_{3}$ be the line passing through $Z$ and perpendicular to $K N$. Show that $\\ell_{1}, \\ell_{2}, \\ell_{3}$ are concurrent.\n\nSolution. Let $R$ be the midpoint of the side $A D$. Then the lines $B R$ and $P D$ are parallel. Since $\\angle M A N=\\angle Q A P=\\angle Q B R=\\angle Q K M$, the points $A, N, K, M$ are concyclic.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_66b0df6ec561017d2840g-22.jpg?height=747&width=1266&top_left_y=650&top_left_x=408)\n\nLet $\\ell_{4}$ be the line passing through the midpoint $W$ of the line segment $M K$ and perpendicular to the line $A N$. Let $E_{1}$ be the point of intersection of $\\ell_{1}$ and $\\ell_{4}$, and $E_{2}$ be the point of intersection of $\\ell_{2}$ and $\\ell_{3}$. We will show that the points $E_{1}$ and $E_{2}$ coincide.\n\nLet $O$ be the circumcenter of the cyclic quadrilateral $A N K M$. $O W$ is perpendicular to the side $M K$ and $O X$ is perpendicular to the side $A N$. Hence $O W$ is parallel to $\\ell_{1}, O X$ is parallel to $\\ell_{4}$, and $X O W E_{1}$ is a parallelogram. Therefore the midpoints of the line segments $O E_{1}$ and $W X$ coincide. Similarly, the midpoints of the line segments $O E_{2}$ and $Y Z$ coincide.\n\nOn the other hand, as $X, Y, Z, W$ are midpoints of the sides of the quadrilateral $A N K M$, $X Y W Z$ is a parallelogram and therefore the midpoints of the line segments $W X$ and $Y Z$ coincide. Hence the midpoints of the line segments $O E_{1}$ and $O E_{2}$ coincide. In other words, $E_{1}$ and $E_{2}$ are the same point, and the lines $\\ell_{1}, \\ell_{2}, \\ell_{3}$ are concurrent.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_66b0df6ec561017d2840g-23.jpg?height=1304&width=1172&top_left_y=841&top_left_x=498)\n\nComment. The problem can be asked in the following form:\n\nLet $A N K M$ be a cyclic quadrilateral and let $X, Y, Z$ be the midpoints of the sides $A N$, $K N, A M$, respectively. Let $\\ell_{1}$ be the line passing through $X$ and perpendicular to $M K$, $\\ell_{2}$ be the line passing through $Y$ and perpendicular to $A M, \\ell_{3}$ be the line passing through $Z$ and perpendicular to $K N$. Show that $\\ell_{1}, \\ell_{2}, \\ell_{3}$ are concurrent.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nG5.", "solution_match": "\nSolution."}}
+{"year": "2013", "tier": "T3", "problem_label": "N1", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all positive integere on for othich $1^{2}+2^{2}+\\cdots+16^{2}+17$ is a perfort square", "solution": "We bave $1^{3}+2^{5}+\\cdots+16^{2}=(1+2+\\cdots+16)^{2}=5^{2} \\quad 17^{0}$ Herese. $11^{3}+2^{3}+$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_66b0df6ec561017d2840g-24.jpg?height=83&width=1607&top_left_y=309&top_left_x=263)\nmeeger t then $1 T^{\\prime}-(1+B)(t-3)$ A. $(1,5) \\quad(t-3)=16$ this can coly happen\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_66b0df6ec561017d2840g-24.jpg?height=74&width=1467&top_left_y=415&top_left_x=206)\n\nEirornd all ordered triples $(x, y, z)$ of integers satisfying $20^{x}+13^{y}=2013^{z}$.\n\nSolution. As $20 \\cdot 13=2^{2} \\cdot 5 \\cdot 13$ and $2013=3 \\cdot 11 \\cdot 61$ are relatively prime, $x, y$ and $z$ must be nonnegative.\n\nConsidering the equation modulo 3 , we observe that $x$ must be odd. Now considering the equation modulo 7 , we obtain $(-1)+(-1)^{y} \\equiv 4^{z}(\\bmod 7)$, which is impossible as the right hand side can only be 1,2 and 4 modulo 7 .\n\nThere are no solutions.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nN1.", "solution_match": "\nSolution."}}
+{"year": "2013", "tier": "T3", "problem_label": "N3", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all ordered pairs $(a, b)$ of positive integers for which the numbers $\\frac{a^{3} b-1}{a+1}$ and $\\frac{b^{3} a+1}{b-1}$ are positive integers.", "solution": "As $a^{3} b-1=b\\left(a^{3}+1\\right)-(b+1)$ and $a+1 \\mid a^{3}+1$, we have $a+1 \\mid b+1$.\n\nAs $b^{3} a+1=a\\left(b^{3}-1\\right)+(a+1)$ and $b-1 \\mid b^{3}-1$, we have $b-1 \\mid a+1$.\n\nSo $b-1 \\mid b+1$ and hence $b-1 \\mid 2$.\n\n- If $b=2$, then $a+1 \\mid b+1=3$ gives $a=2$. Hence $(a, b)=(2,2)$ is the only solution in this case.\n- If $b=3$, then $a+1 \\mid b+1=4$ gives $a=1$ or $a=3$. Hence $(a, b)=(1,3)$ and $(3,3)$ are the only solutions in this case.\n\nTo summarize, $(a, b)=(1,3),(2,2)$ and $(3,3)$ are the only solutions.\n$\\mathbf{N 4}$. A rectangle in the $x y$-plane is called latticed if all its vertices have integer coordinates.\n\na. Find a latticed rectangle with area 2013 whose sides are not parallel to the axes.\n\nb. Show that if a latticed rectangle has area 2011, then its sides are parallel to the axes.\n\nSolution. a. The rectangle $P Q R S$ with $P(0,0), Q(165,198), R(159,203), S(-6,5)$ has area 2013 .\n\nb. Suppose that the latticed rectangle $P Q R S$ has area 2011 and its sides are not parallel to the axes. Without loss of generality we may assume that its vertices are $P(0,0), Q(a, b)$, $S(c, d), R(a+c, b+d)$ where $a, b, c, d$ are integers and $a b c d \\neq 0$.\n\nThen $\\left(a^{2}+b^{2}\\right)\\left(c^{2}+d^{2}\\right)=2011^{2}$. As 2011 is a prime, $2011 \\mid a^{2}+b^{2}$ or $2011 \\mid c^{2}+d^{2}$. Assume that the first one is the case. Since $2011 \\equiv 3(\\bmod 4)$, this can happen only if $a=2011 a_{0}$ and $b=2011 b_{0}$ for some integers $a_{0}$ and $b_{0}$. Now we have $\\left(a_{0}^{2}+b_{0}^{2}\\right)\\left(c^{2}+d^{2}\\right)=1$. This means $c^{2}+d^{2}=1$ and hence $c=0$ or $d=0$, contradicting $c d \\neq 0$.\n\nComment. The original problem proposal has the numbers 13 and 11 instead.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nN3.", "solution_match": "\nSolution."}}
+{"year": "2013", "tier": "T3", "problem_label": "N5", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all ordered triples $(x, y, z)$ of positive integers satisfying the equation\n\n$$\n\\frac{1}{x^{2}}+\\frac{y}{x z}+\\frac{1}{z^{2}}=\\frac{1}{2013}\n$$", "solution": "We have $x^{2} z^{2}=2013\\left(x^{2}+x y z+z^{2}\\right)$. Let $d=\\operatorname{gcd}(x, z)$ and $x=d a, z=d b$. Then $a^{2} b^{2} d^{2}=2013\\left(a^{2}+a b y+b^{2}\\right)$.\n\nAs $\\operatorname{gcd}(a, b)=1$, we also have $\\operatorname{gcd}\\left(a^{2}, a^{2}+a b y+b^{2}\\right)=1$ and $\\operatorname{gcd}\\left(b^{2}, a^{2}+a b y+b^{2}\\right)=1$. Therefore $a^{2} \\mid 2013$ and $b^{2} \\mid 2013$. But $2013=3 \\cdot 11 \\cdot 61$ is squarefree and therefore $a=1=b$.\n\nNow we have $x=z=d$ and $d^{2}=2013(y+2)$. Once again as 2013 is squarefree, we must have $y+2=2013 n^{2}$ where $n$ is a positive integer.\n\nHence $(x, y, z)=\\left(2013 n, 2013 n^{2}-2,2013 n\\right)$ where $n$ is a positive integer.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nN5.", "solution_match": "\nSolution."}}
+{"year": "2013", "tier": "T3", "problem_label": "N6", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all ordered triples $(x, y, z)$ of integers satisfying the following system of equations:\n\n$$\n\\begin{aligned}\nx^{2}-y^{2} & =z \\\\\n3 x y+(x-y) z & =z^{2}\n\\end{aligned}\n$$", "solution": "If $z=0$, then $x=0$ and $y=0$, and $(x, y, z)=(0,0,0)$.\n\nLet us assume that $z \\neq 0$, and $x+y=a$ and $x-y=b$ where $a$ and $b$ are nonzero integers such that $z=a b$. Then $x=(a+b) / 2$ and $y=(a-b) / 2$, and the second equations gives $3 a^{2}-3 b^{2}+4 a b^{2}=4 a^{2} b^{2}$.\n\nHence\n\n$$\nb^{2}=\\frac{3 a^{2}}{4 a^{2}-4 a+3}\n$$\n\nand\n\n$$\n3 a^{2} \\geq 4 a^{2}-4 a+3\n$$\n\nwhich is satisfied only if $a=1,2$ or 3 .\n\n- If $a=1$, then $b^{2}=1 .(x, y, z)=(1,0,1)$ and $(0,1,-1)$ are the only solutions in this case.\n- If $a=2$, then $b^{2}=12 / 11$. There are no solutions in this case\n- If $a=3$, then $b^{2}=1 .(x, y, z)=(1,2,-3)$ and $(2,1,3)$ are the only solutions in this case.\n\nTo summarize, $(x, y, z)=(0,0,0),(1,0,1),(0,1,-1),(1,2,-3)$ and $(2,1,3)$ are the only solutions.\n\nComments. 1. The original problem proposal asks for the solutions when $z=p$ is a prime number.\n\n2. The problem can be asked with a single equation in the form:\n\n$$\n3 x y+(x-y)^{2}(x+y)=\\left(x^{2}-y^{2}\\right)^{2}\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl", "problem_match": "\nN6.", "solution_match": "\nSolution."}}

JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl

@@ -1,21 +1,21 @@
-{"year": "2014", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO", "problem": "For any real number a, let $\\lfloor a\\rfloor$ denote the greatest integer not exceeding a. In positive real numbers solve the following equation\n\n$$\nn+\\lfloor\\sqrt{n}\\rfloor+\\lfloor\\sqrt[3]{n}\\rfloor=2014\n$$", "solution": "Obviously $n$ must be positive integer. Now note that $44^{2}=1936<2014<2025=45^{2}$ and $12^{3}<1900<2014<13^{3}$.\n\nIf $n<1950$ than $2014=n+\\lfloor\\sqrt{n}\\rfloor+\\lfloor\\sqrt[3]{n}\\rfloor<1950+44+12=2006$, a contradiction!\n\nSo $n \\geq 1950$. Also if $n>2000$ than $2014=n+\\lfloor\\sqrt{n}\\rfloor+\\lfloor\\sqrt[3]{n}\\rfloor>2000+44+12=2056$, a contradiction!\n\nSo $1950 \\leq n \\leq 2000$, therefore $\\lfloor\\sqrt{n}\\rfloor=44$ and $\\lfloor\\sqrt[3]{n}\\rfloor=12$. Plugging that into the original equation we get:\n\n$$\nn+\\lfloor\\sqrt{n}\\rfloor+\\lfloor\\sqrt[3]{n}\\rfloor=n+44+12=2014\n$$\n\nFrom which we get $n=1956$, which is the only solution.\n\nSolution2. Obviously $n$ must be positive integer. Since $n \\leq 2014, \\sqrt{n}<45$ and $\\sqrt[3]{n}<13$.\n\nForm $n=2014-\\lfloor\\sqrt{n}\\rfloor-\\lfloor\\sqrt[3]{n}\\rfloor>2014-45-13=1956, \\sqrt{n}>44$ and $\\sqrt[3]{n}>12$, thus $\\lfloor\\sqrt{n}\\rfloor=44$ and $\\lfloor\\sqrt[3]{n}\\rfloor=12$ and $n=2014-\\lfloor\\sqrt{n}\\rfloor-\\lfloor\\sqrt[3]{n}\\rfloor=2014-44-12=1958$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## A1", "solution_match": "\nSolution1."}}
-{"year": "2014", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $a, b$ and $c$ be positive real numbers such that abc $=\\frac{1}{8}$. Prove the inequality\n\n$$\na^{2}+b^{2}+c^{2}+a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \\geq \\frac{15}{16}\n$$\n\nWhen does equality hold?", "solution": "By using The Arithmetic-Geometric Mean Inequality for 15 positive numbers, we find that\n\n$$\n\\begin{aligned}\n& a^{2}+b^{2}+c^{2}+a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}= \\\\\n& \\quad=\\frac{a^{2}}{4}+\\frac{a^{2}}{4}+\\frac{a^{2}}{4}+\\frac{a^{2}}{4}+\\frac{b^{2}}{4}+\\frac{b^{2}}{4}+\\frac{b^{2}}{4}+\\frac{b^{2}}{4}+\\frac{c^{2}}{4}+\\frac{c^{2}}{4}+\\frac{c^{2}}{4}+\\frac{c^{2}}{4}+a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \\geq \\\\\n& \\quad \\geq 1515 \\sqrt{\\frac{a^{12} b^{12} c^{12}}{4^{12}}}=15 \\sqrt[5]{\\left(\\frac{a b c}{4}\\right)^{4}}=15 \\sqrt[5]{\\left(\\frac{1}{32}\\right)^{4}}=\\frac{15}{16}\n\\end{aligned}\n$$\n\nas desired. Equality holds if and only if $a=b=c=\\frac{1}{2}$.\n\nSolution2. By using AM-GM we obtain\n\n$$\n\\begin{aligned}\n& \\left(a^{2}+b^{2}+c^{2}\\right)+\\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\\right) \\geq 3 \\sqrt[3]{a^{2} b^{2} c^{2}}+3 \\sqrt[3]{a^{4} b^{4} c^{4}}= \\\\\n& =3 \\sqrt[3]{\\left(\\frac{1}{8}\\right)^{2}}+3 \\sqrt[3]{\\left(\\frac{1}{8}\\right)^{4}}=\\frac{3}{4}+\\frac{3}{16}=\\frac{15}{16}\n\\end{aligned}\n$$\n\nThe equality holds when $a^{2}=b^{2}=c^{2}$, i.e. $a=b=c=\\frac{1}{2}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## A2", "solution_match": "\nSolution1."}}
-{"year": "2014", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that:\n\n$$\n\\left(a+\\frac{1}{b}\\right)^{2}+\\left(b+\\frac{1}{c}\\right)^{2}+\\left(c+\\frac{1}{a}\\right)^{2} \\geq 3(a+b+c+1)\n$$\n\nWhen does equality hold?", "solution": "By using AM-GM $\\left(x^{2}+y^{2}+z^{2} \\geq x y+y z+z x\\right)$ we have\n\n$$\n\\begin{aligned}\n\\left(a+\\frac{1}{b}\\right)^{2}+\\left(b+\\frac{1}{c}\\right)^{2}+\\left(c+\\frac{1}{a}\\right)^{2} & \\geq\\left(a+\\frac{1}{b}\\right)\\left(b+\\frac{1}{c}\\right)+\\left(b+\\frac{1}{c}\\right)\\left(c+\\frac{1}{a}\\right)+\\left(c+\\frac{1}{a}\\right)\\left(a+\\frac{1}{b}\\right) \\\\\n& =\\left(a b+1+\\frac{a}{c}+a\\right)+\\left(b c+1+\\frac{b}{a}+b\\right)+\\left(c a+1+\\frac{c}{b}+c\\right) \\\\\n& =a b+b c+c a+\\frac{a}{c}+\\frac{c}{b}+\\frac{b}{a}+3+a+b+c\n\\end{aligned}\n$$\n\nNotice that by AM-GM we have $a b+\\frac{b}{a} \\geq 2 b, b c+\\frac{c}{b} \\geq 2 c$, and $c a+\\frac{a}{c} \\geq 2 a$.\n\nThus,\n\n$$\n\\left(a+\\frac{1}{b}\\right)^{2}+\\left(b+\\frac{1}{c}\\right)^{2}+\\left(c+\\frac{1}{a}\\right)^{2} \\geq\\left(a b+\\frac{b}{a}\\right)+\\left(b c+\\frac{c}{b}\\right)+\\left(c a+\\frac{a}{c}\\right)+3+a+b+c \\geq 3(a+b+c+1)\n$$\n\nThe equality holds if and only if $a=b=c=1$.\n\nSolution2. From QM-AM we obtain\n\n$$\n\\begin{aligned}\n& \\sqrt{\\frac{\\left(a+\\frac{1}{b}\\right)^{2}+\\left(b+\\frac{1}{c}\\right)^{2}+\\left(c+\\frac{1}{a}\\right)^{2}}{3}} \\geq \\frac{a+\\frac{1}{b}+b+\\frac{1}{c}+c+\\frac{1}{a}}{3} \\Leftrightarrow \\\\\n& \\left(a+\\frac{1}{b}\\right)^{2}+\\left(b+\\frac{1}{c}\\right)^{2}+\\left(c+\\frac{1}{a}\\right)^{2} \\geq \\frac{\\left(a+\\frac{1}{b}+b+\\frac{1}{c}+c+\\frac{1}{a}\\right)^{2}}{3}\n\\end{aligned}\n$$\n\nFrom AM-GM we have $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} \\geq 3 \\sqrt[3]{\\frac{1}{a b c}}=3$, and substituting in (1) we get\n\n$$\n\\begin{aligned}\n&\\left(a+\\frac{1}{b}\\right)^{2}+\\left(b+\\frac{1}{c}\\right)^{2}+\\left(c+\\frac{1}{a}\\right)^{2} \\geq \\frac{\\left(a+\\frac{1}{b}+b+\\frac{1}{c}+c+\\frac{1}{a}\\right)^{2}}{3} \\geq \\frac{(a+b+c+3)^{2}}{3}= \\\\\n&=\\frac{(a+b+c)(a+b+c)+6(a+b+c)+9}{3} \\geq \\frac{(a+b+c) 3 \\sqrt[3]{a b c}+6(a+b+c)+9}{3}= \\\\\n&=\\frac{9(a+b+c)+9}{3}=3(a+b+c+1)\n\\end{aligned}\n$$\n\nThe equality holds if and only if $a=b=c=1$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## A3", "solution_match": "\nSolution1."}}
-{"year": "2014", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $a, b, c$ be positive real numbers such that $a+b+c=1$. Prove that\n\n$$\n\\frac{7+2 b}{1+a}+\\frac{7+2 c}{1+b}+\\frac{7+2 a}{1+c} \\geq \\frac{69}{4}\n$$\n\nWhen does equality hold?", "solution": "Solution. Equality holds when $x=y=z=0$.\n\nApply AM-GM to $x+y+z=x y z$,\n\n$$\n\\begin{aligned}\n& x y z=x+y+z \\geq 3 \\sqrt[3]{x y z} \\Rightarrow(x y z)^{3} \\geq(3 \\sqrt[3]{x y z})^{3} \\\\\n& \\Rightarrow x^{3} y^{3} z^{3} \\geq 27 x y z \\\\\n& \\Rightarrow x^{2} y^{2} z^{2} \\geq 27 \\\\\n& \\Rightarrow \\sqrt[3]{x^{2} y^{2} z^{2}} \\geq 3\n\\end{aligned}\n$$\n\nAlso by AM-GM we have, $x^{2}+y^{2}+z^{2} \\geq 3 \\sqrt[3]{x^{2} y^{2} z^{2}} \\geq 9$.\n\nTherefore we get $x^{2}+y^{2}+z^{2} \\geq 9$.\n\nNow,\n\n$$\n\\begin{gathered}\n2\\left(x^{2}+y^{2}+z^{2}\\right) \\geq 3(x+y+z) \\Leftrightarrow \\frac{2\\left(x^{2}+y^{2}+z^{2}\\right)}{3} \\geq(x+y+z) \\\\\n\\Leftrightarrow 2 \\cdot \\frac{2\\left(x^{2}+y^{2}+z^{2}\\right)}{3} \\geq 2 \\cdot(x+y+z) \\\\\n\\Leftrightarrow \\frac{4\\left(x^{2}+y^{2}+z^{2}\\right)}{3} \\geq 2 \\cdot(x+y+z) \\\\\n\\Leftrightarrow x^{2}+y^{2}+z^{2}+\\frac{\\left(x^{2}+y^{2}+z^{2}\\right)}{3} \\geq 2 \\cdot(x+y+z) \\\\\n\\Leftrightarrow 3+\\frac{x^{2}}{3}+3+\\frac{y^{2}}{3}+3+\\frac{z^{2}}{3} \\geq 2(x+y+z) \\\\\n\\Leftrightarrow 3+\\frac{x^{2}}{3}+3+\\frac{y^{2}}{3}+3+\\frac{z^{2}}{3} \\geq 2(x+y+z) \\\\\n\\Leftrightarrow 2 \\sqrt{3 \\cdot \\frac{x^{2}}{3}}+2 \\sqrt{3 \\cdot \\frac{y^{2}}{3}}+2 \\sqrt{3 \\cdot \\frac{z^{2}}{3}} \\geq 2(x+y+z)\n\\end{gathered}\n$$\n\nEquality holds if $3=\\frac{x^{2}}{3}=\\frac{y^{2}}{3}=\\frac{z^{2}}{3}$, i.e. $x=y=z=3$, for which $x+y+z \\neq x y z$.\n\nRemark. The inequality can be improved: $x^{2}+y^{2}+z^{2} \\geq \\sqrt{3}(x+y+z)$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## A4", "solution_match": "\nSolution1."}}
-{"year": "2014", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $a, b, c$ be positive real numbers such that $a+b+c=1$. Prove that\n\n$$\n\\frac{7+2 b}{1+a}+\\frac{7+2 c}{1+b}+\\frac{7+2 a}{1+c} \\geq \\frac{69}{4}\n$$\n\nWhen does equality hold?", "solution": "Solution. If one of the numbers is zero, then from $x+y+z=x y z$ all three numbers are zero and the equality trivially holds.\n\nFrom AM-GM $x^{2}+y^{2}+z^{2} \\geq 3 \\sqrt[3]{x^{2} y^{2} z^{2}}=3 \\frac{x+y+z}{\\sqrt[3]{x y z}} \\geq 3 \\frac{x+y+z}{\\frac{x+y+z}{3}}=9$\n\nFrom QM-AM $\\frac{x^{2}+y^{2}+z^{2}}{3} \\geq\\left(\\frac{x+y+z}{3}\\right)^{2}$\n\nMultiplying (1) and (2) we get $\\frac{\\left(x^{2}+y^{2}+z^{2}\\right)^{2}}{3} \\geq 9 \\frac{(x+y+z)^{2}}{9}=(x+y+z)^{2}$. By taking square root on both sides we deduce the stated inequality.\n\nEquality holds only when $x=y=z=\\sqrt{3}$ or $x=y=z=0$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## A4", "solution_match": "\nSolution1."}}
-{"year": "2014", "tier": "T3", "problem_label": "A6", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $a, b, c$ be positive real numbers. Prove that\n\n$$\n\\left(\\left(3 a^{2}+1\\right)^{2}+2\\left(1+\\frac{3}{b}\\right)^{2}\\right)\\left(\\left(3 b^{2}+1\\right)^{2}+2\\left(1+\\frac{3}{c}\\right)^{2}\\right)\\left(\\left(3 c^{2}+1\\right)^{2}+2\\left(1+\\frac{3}{a}\\right)^{2}\\right) \\geq 48^{3}\n$$\n\nWhen does equality hold?", "solution": "Let $x$ be a positive real number. By AM-GM we have $\\frac{1+x+x+x}{4} \\geq x^{\\frac{3}{4}}$, or equivalently $1+3 x \\geq 4 x^{\\frac{3}{4}}$. Using this inequality we obtain:\n\n$$\n\\left(3 a^{2}+1\\right)^{2} \\geq 16 a^{3} \\text { and } 2\\left(1+\\frac{3}{b}\\right)^{2} \\geq 32 b^{-\\frac{3}{2}}\n$$\n\nMoreover, by inequality of arithmetic and geometric means we have\n\n$$\nf(a, b)=\\left(3 a^{2}+1\\right)^{2}+2\\left(1+\\frac{3}{b}\\right)^{2} \\geq 16 a^{3}+32 b^{-\\frac{3}{2}}=16\\left(a^{3}+b^{-\\frac{3}{2}}+b^{-\\frac{3}{2}}\\right) \\geq 48 \\frac{a}{b}\n$$\n\nTherefore, we obtain\n\n$$\nf(a, b) f(b, c) f(c, a) \\geq 48 \\cdot \\frac{a}{b} \\cdot 48 \\cdot \\frac{b}{c} \\cdot 48 \\cdot \\frac{c}{a}=48^{3}\n$$\n\nEquality holds only when $a=b=c=1$.\n\n## 1 IH $^{\\text {th J.M. }} 2014$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## A6", "solution_match": "\nSolution."}}
-{"year": "2014", "tier": "T3", "problem_label": "A8", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $x, y$ and $z$ be positive real numbers such that $x y z=1$. Prove the inequality\n\n$$\n\\frac{1}{x(a y+b)}+\\frac{1}{y(a z+b)}+\\frac{1}{z(a x+b)} \\geq 3 \\text {, if: }\n$$\n\na) $a=0$ and $b=1$;\nb) $a=1$ and $b=0$;\nc) $a+b=1$ for $a, b>0$\n\nWhen does the equality hold true?", "solution": "a) The inequality reduces to $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z} \\geq 3$, which follows directly from the AM-GM inequality.\n\nEquality holds only when $x=y=z=1$.\n\nb) Here the inequality reduces to $\\frac{1}{x y}+\\frac{1}{y z}+\\frac{1}{z x} \\geq 3$, i.e. $x+y+z \\geq 3$, which also follows from the AM-GM inequality.\n\nEquality holds only when $x=y=z=1$.\n\nc) Let $m, n$ and $p$ be such that $x=\\frac{m}{n}, y=\\frac{n}{p}$ и $z=\\frac{p}{m}$. The inequality reduces to\n\n$$\n\\frac{n p}{a m n+b m p}+\\frac{p m}{a n p+b n m}+\\frac{m n}{a p m+b p n} \\geq 3\n$$\n\nBy substituting $u=n p, v=p m$ and $w=m n$, (1) becomes\n\n$$\n\\frac{u}{a w+b v}+\\frac{v}{a u+b w}+\\frac{w}{a v+b u} \\geq 3\n$$\n\nThe last inequality is equivalent to\n\n$$\n\\frac{u^{2}}{a u w+b u v}+\\frac{v^{2}}{a u v+b v w}+\\frac{w^{2}}{a v w+b u w} \\geq 3\n$$\n\nCauchy-Schwarz Inequality implies\n\n$$\n\\frac{u^{2}}{a u w+b u v}+\\frac{v^{2}}{a u v+b v w}+\\frac{w^{2}}{a v w+b u w} \\geq \\frac{(u+v+w)^{2}}{a u w+b u v+a u v+b v w+a v w+b u w}=\\frac{(u+v+w)^{2}}{u w+v u+w v}\n$$\n\nThus, the problem simplifies to $(u+v+w)^{2} \\geq 3(u w+v u+w v)$, which is equivalent to $(u-v)^{2}+(v-w)^{2}+(w-u)^{2} \\geq 0$.\n\nEquality holds only when $u=v=w$, that is only for $x=y=z=1$.\n\nRemark. The problem can be reformulated:\n\nLet $a, b, x, y$ and $z$ be nonnegative real numbers such that $x y z=1$ and $a+b=1$. Prove the inequality\n\n$$\n\\frac{1}{x(a y+b)}+\\frac{1}{y(a z+b)}+\\frac{1}{z(a x+b)} \\geq 3\n$$\n\nWhen does the equality hold true?", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## A8", "solution_match": "\nSolution."}}
-{"year": "2014", "tier": "T3", "problem_label": "A9", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $n$ be a positive integer, and let $x_{1}, \\ldots, x_{n}, y_{1}, \\ldots, y_{n}$ be positive real numbers such that $x_{1}+\\ldots+x_{n}=y_{1}+\\ldots+y_{n}=1$. Show that\n\n$$\n\\left|x_{1}-y_{1}\\right|+\\ldots\\left|x_{n}-y_{n}\\right| \\leq 2-\\min _{1 \\leq i \\leq n} \\frac{x_{i}}{y_{i}}-\\min _{1 \\leq i \\leq n} \\frac{y_{i}}{x_{i}}\n$$", "solution": "Up to reordering the real numbers $x_{i}$ and $y_{i}$, we may assume that $\\frac{x_{1}}{y_{1}} \\leq \\ldots \\leq \\frac{x_{n}}{y_{n}}$. Let $A=\\frac{x_{1}}{y_{1}}$ and $B=\\frac{x_{n}}{y_{n}}$, and $\\mathrm{S}=\\left|x_{1}-y_{1}\\right|+\\ldots\\left|x_{n}-y_{n}\\right|$. Our aim is to prove that $S \\leq 2-A-\\frac{1}{B}$.\n\nFirst, note that we cannot have $A>1$, since that would imply $x_{i}>y_{i}$ for all $i \\leq n$, hence $x_{1}+\\ldots+x_{n}>y_{1}+\\ldots+y_{n}$. Similarly, we cannot have $B<1$, since that would imply $x_{i}<y_{i}$ for all $i \\leq n$, hence $x_{1}+\\ldots+x_{n}<y_{1}+\\ldots+y_{n}$.\n\nIf $n=1$, then $x_{1}=y_{1}=A=B=1$ and $S=0$, hence $S \\leq 2-A-\\frac{1}{B}$. For $n \\geq 2$ let $1 \\leq k<n$ be some integer such that $\\frac{x_{k}}{y_{k}} \\leq 1 \\leq \\frac{x_{k+1}}{y_{k+1}}$. We define the positive real numbers $X_{1}=x_{1}+\\ldots+x_{k}$, $X_{2}=x_{k+1}+\\ldots+x_{n}, Y_{1}=y_{1}+\\ldots+y_{k}, Y_{2}=y_{k+1}+\\ldots+y_{n}$. Note that $Y_{1} \\geq X_{1} \\geq A Y_{1}$ and $Y_{2} \\leq X_{2} \\leq B Y_{2}$. Thus, $A \\leq \\frac{X_{1}}{Y_{1}} \\leq 1 \\leq \\frac{X_{2}}{Y_{2}} \\leq B$. In addition, $S=Y_{1}-X_{1}+X_{2}-Y_{2}$.\n\nFrom $0<X_{2}, Y_{1} \\leq 1,0 \\leq Y_{1}-X_{1}$ and $0 \\leq X_{2}-Y_{2}$, follows\n\n$$\nS=Y_{1}-X_{1}+X_{2}-Y_{2}=\\frac{Y_{1}-X_{1}}{Y_{1}}+\\frac{X_{2}-Y_{2}}{X_{2}}=2-\\frac{X_{1}}{Y_{1}}-\\frac{Y_{2}}{X_{2}} \\leq 2-A-\\frac{1}{B}\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## A9", "solution_match": "\nSolution."}}
-{"year": "2014", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Several (at least two) segments are drawn on a board. Select two of them, and let $a$ and $b$ be their lengths. Delete the selected segments and draw a segment of length $\\frac{a b}{a+b}$. Continue this procedure until only one segment remains on the board. Prove:\n\na) the length of the last remaining segment does not depend on the order of the deletions.\n\nb) for every positive integer $n$, the initial segments on the board can be chosen with distinct integer lengths, such that the last remaining segment has length $n$.", "solution": "a) Observe that $\\frac{1}{\\frac{a b}{a+b}}=\\frac{1}{a}+\\frac{1}{b}$. Thus, if the lengths of the initial segments on the board were $a_{1}, a_{2}, \\ldots, a_{n}$, and $c$ is the length of the last remaining segment, then $\\frac{1}{c}=\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+\\ldots+\\frac{1}{a_{n}}$ , proving a).\n\nb) From a) and the equation $\\frac{1}{n}=\\frac{1}{2 n}+\\frac{1}{3 n}+\\frac{1}{6 n}$ it follows that if the lengths of the starting segments are $2 n, 3 n$ and $6 n$, then the length of the last remaining segment is $n$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## C1", "solution_match": "\nSolution."}}
-{"year": "2014", "tier": "T3", "problem_label": "C2", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "In a country with $n$ cities, all direct airlines are two-way. There are $r>2014$ routes between pairs of different cities that include no more than one intermediate stop (the direction of each route matters). Find the least possible $n$ and the least possible $r$ for that value of $n$.", "solution": "Denote by $X_{1}, X_{2}, \\ldots X_{n}$ the cities in the country and let $X_{i}$ be connected to exactly $m_{i}$ other cities by direct two-way airline. Then $X_{i}$ is a final destination of $m_{i}$ direct routes and an intermediate stop of $m_{i}\\left(m_{i}-1\\right)$ non-direct routes. Thus $r=m_{1}^{2}+\\ldots+m_{n}^{2}$. As each $m_{i}$ is at most $n-1$ and $13 \\cdot 12^{2}<2014$, we deduce $n \\geq 14$.\n\nConsider $n=14$. As each route appears in two opposite directions, $r$ is even, so $r \\geq 2016$. We can achieve $r=2016$ by arranging the 14 cities uniformly on a circle connect (by direct two-way airlines) all of them, except the diametrically opposite pairs. This way, there are exactly $14 \\cdot 12^{2}=2016$ routes.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## C2", "solution_match": "\nSolution."}}
-{"year": "2014", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "For a given positive integer n, two players $A$ and B play the following game: Given is pile of $\\boldsymbol{\\Omega}$ stones. The players take turn alternatively with A going first. On each turn the player is allowed to take one stone, a prime number of stones, or a multiple of $n$ stones. The winner is the one who takes the last stone. Assuming perfect play, find the number of values for $S_{\\infty}$, for which A cannot win.", "solution": "Denote by $k$ the sought number and let $\\left\\{a_{1}, a_{2}, \\ldots, a_{k}\\right\\}$ be the corresponding values for $a$. We will call each $a_{i}$ a losing number and every other positive integer a winning numbers. Clearly every multiple of $n$ is a winning number.\n\nSuppose there are two different losing numbers $a_{i}>a_{j}$, which are congruent modulo $n$. Then, on his first turn of play, the player $A$ may remove $a_{i}-a_{j}$ stones (since $n \\mid a_{i}-a_{j}$ ), leaving a pile with $a_{j}$ stones for B. This is in contradiction with both $a_{i}$ and $a_{j}$ being losing numbers. Therefore there are at most $n-1$ losing numbers, i.e. $k \\leq n-1$.\n\nSuppose there exists an integer $r \\in\\{1,2, \\ldots, n-1\\}$, such that $m n+r$ is a winning number for every $m \\in \\mathbb{N}_{0}$. Let us denote by $u$ the greatest losing number (if $k>0$ ) or 0 (if $k=0$ ), and let $s=\\operatorname{LCM}(2,3, \\ldots, u+n+1)$. Note that all the numbers $s+2, s+3, \\ldots, s+u+n+1$ are composite. Let $m^{\\prime} \\in \\mathbb{N}_{0}$, be such that $s+u+2 \\leq m^{\\prime} n+r \\leq s+u+n+1$. In order for $m^{\\prime} n+r$ to be a winning number, there must exist an integer $p$, which is either one, or prime, or a positive multiple of $n$, such that $m^{\\prime} n+r-p$ is a losing number or 0 , and hence lesser than or equal to $u$. Since $s+2 \\leq m^{\\prime} n+r-u \\leq p \\leq m^{\\prime} n+r \\leq s+u+n+1, p$ must be a composite, hence $p$ is a multiple of $n$ (say $p=q n$ ). But then $m^{\\prime} n+r-p=\\left(m^{\\prime}-q\\right) n+r$ must be a winning number, according to our assumption. This contradicts our assumption that all numbers $m n+r, m \\in \\mathbb{N}_{0}$ are winning.\n\nHence there are exactly $n-1$ losing numbers (one for each residue $r \\in\\{1,2, \\ldots, n-1\\}$ ).", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## C3", "solution_match": "\nSolution."}}
-{"year": "2014", "tier": "T3", "problem_label": "C4", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Let $A=1 \\cdot 4 \\cdot 7 \\cdot \\ldots \\cdot 2014$ be the product of the numbers less or equal to 2014 that give remainder 1 when divided by 3 . Find the last non-zero digit of $A$.", "solution": "Grouping the elements of the product by ten we get:\n\n$$\n\\begin{aligned}\n& (30 k+1)(30 k+4)(30 k+7)(30 k+10)(30 k+13)(30 k+16) \\\\\n& (30 k+19)(30 k+22)(30 k+25)(30 k+28)= \\\\\n& =(30 k+1)(15 k+2)(30 k+7)(120 k+40)(30 k+13)(15 k+8) \\\\\n& (30 k+19)(15 k+11)(120 k+100)(15 k+14)\n\\end{aligned}\n$$\n\n(We divide all even numbers not divisible by five, by two and multiply all numbers divisible by five with four.)\n\nWe denote $P_{k}=(30 k+1)(15 k+2)(30 k+7)(30 k+13)(15 k+8)(30 k+19)(15 k+11)(15 k+14)$. For all the numbers not divisible by five, only the last digit affects the solution, since the power of two in the numbers divisible by five is greater than the power of five. Considering this, for even $k, P_{k}$ ends with the same digit as $1 \\cdot 2 \\cdot 7 \\cdot 3 \\cdot 8 \\cdot 9 \\cdot 1 \\cdot 4$, i.e. six and for odd $k, P_{k}$ ends with the same digit as $1 \\cdot 7 \\cdot 7 \\cdot 3 \\cdot 3 \\cdot 9 \\cdot 6 \\cdot 9$, i.e. six. Thus $P_{0} P_{1} \\ldots P_{66}$ ends with six. If we remove one zero from the end of all numbers divisible with five, we get that the last nonzero digit of the given product is the same as the one from $6 \\cdot 2011 \\cdot 2014 \\cdot 4 \\cdot 10 \\cdot 16 \\cdot \\ldots .796 \\cdot 802$. Considering that $4 \\cdot 6 \\cdot 2 \\cdot 8$ ends with four and removing one zero from every fifth number we get that the last nonzero digit is the same as in $4 \\cdot 4^{26} \\cdot 784 \\cdot 796 \\cdot 802 \\cdot 1 \\cdot 4 \\cdot \\ldots \\cdot 76 \\cdot 79$. Repeating the process we did for the starting sequence we conclude that the last nonzero number will be the same as in $2 \\cdot 6 \\cdot 6 \\cdot 40 \\cdot 100 \\cdot 160 \\cdot 220 \\cdot 280 \\cdot 61 \\cdot 32 \\cdot 67 \\cdot 73 \\cdot 38 \\cdot 79$, which is two.\n\nLet $A B C$ be a triangle with $\\measuredangle B=\\measuredangle C=40^{\\circ}$. The bisector of the $\\measuredangle B$ meets $A C$ at the point $D$ . Prove that $\\overline{B D}+\\overline{D A}=\\overline{B C}$.\n\nSolution. Since $\\measuredangle B A C=100^{\\circ}$ and $\\measuredangle B D C=120^{\\circ}$ we have $\\overline{B D}<\\overline{B C}$. Let $E$ be the point on $\\overline{B C}$ such that $\\overline{B D}=\\overline{B E}$. Then $\\measuredangle D E C=100^{\\circ}$ and $\\measuredangle E D C=40^{\\circ}$, hence $\\overline{D E}=\\overline{E C}$, and $\\measuredangle B A C+\\measuredangle D E B=180^{\\circ}$. So $A, B, E$ and $D$ are concyclic, implying $\\overline{A D}=\\overline{D E}$ (since $\\measuredangle A B D=\\measuredangle D B C=20^{\\circ}$ ), which completes the proof.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## C4", "solution_match": "\nSolution."}}
-{"year": "2014", "tier": "T3", "problem_label": "G2", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C$ be an acute triangle with $\\overline{A B}<\\overline{A C}<\\overline{B C}$ and $c(O, R)$ be its circumcircle. Denote with $D$ and $E$ be the points diametrically opposite to the points $B$ and $C$, respectively. The circle $c_{1}(A, \\overline{A E})$ intersects $\\overline{A C}$ at point $K$, the circle $c_{2}(A, \\overline{A D})$ intersects $B A$ at point $L(A$ lies between $B$ and $L$ ). Prove that the lines $E K$ and $D L$ meet on the circle $c$.", "solution": "Let $\\mathrm{M}$ be the point of intersection of the line $D L$ with the circle $c(O, R)$ (we choose $M \\equiv D$ if $L D$ is tangent to $c$ and $M$ to be the second intersecting point otherwise). It is\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_eb51ab1a35cd98c2706cg-27.jpg?height=778&width=815&top_left_y=1069&top_left_x=359)\nsufficient to prove that the points $E, K$ and $M$ are collinear.\n\nWe have that $\\measuredangle E A C=90^{\\circ}$ (since $E C$ is diameter of the circle $c$ ). The triangle $A E K$ is right-angled and isosceles ( $\\overline{A E}$ and $\\overline{A K}$ are radii of the circle $c_{1}$ ). Therefore\n\n$$\n\\measuredangle \\mathrm{AEK}=\\measuredangle \\mathrm{AKE}=45^{\\circ} \\text {. }\n$$\n\nSimilarly, we obtain that $\\measuredangle B A D=90^{\\circ}=\\measuredangle D A L$. Since $\\overline{A D}=\\overline{A L}$ the triangle $A D L$ is right-angled and isosceles, we have\n\n$$\n\\measuredangle \\mathrm{ADL}=\\measuredangle \\mathrm{A} L D=45^{\\circ} .\n$$\n\nIf $M$ is between $D$ and $L$, then $\\measuredangle \\mathrm{ADM}=\\measuredangle A E M$, because they are inscribed in the circle $c(O, R)$ and they correspond to the same arch $\\overparen{A M}$. Hence $\\measuredangle \\mathrm{AEK}=\\measuredangle \\mathrm{AEM}=45^{\\circ}$ i.e. the points $E, K, M$ are collinear.\n\nIf $D$ is between $M$ and $L$, then $\\measuredangle \\mathrm{ADM}+\\measuredangle A E M=180^{\\circ}$ as opposite angles in cyclic quadrilateral. Hence $\\measuredangle \\mathrm{AEK}=\\measuredangle \\mathrm{AEM}=45^{\\circ}$ i.e. the points $E, K, M$ are collinear.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## G2", "solution_match": "\nSolution."}}
-{"year": "2014", "tier": "T3", "problem_label": "G3", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $C D \\perp A B(D \\in A B), D M \\perp A C(M \\in A C)$ and $D N \\perp B C(N \\in B C)$ for an acute triangle ABC with area $S$. If $H_{1}$ and $H_{2}$ are the orthocentres of the triangles $M N C$ and MND respectively. Evaluate the area of the quadrilateral $\\mathrm{AH}_{1} \\mathrm{BH}_{2}$.", "solution": "Let $O, P, K, R$ and $T$ be the midpoints of the segments $C D, M N, C N, C H_{1}$ and $M H_{1}$, respectively. From $\\triangle M N C$ we have that $\\overline{P K}=\\frac{1}{2} \\overline{M C}$ and $P K \\| M C$. Analogously, from $\\Delta M H_{1} C$ we have that $\\overline{T R}=\\frac{1}{2} \\overline{M C}$ and $T R \\| M C$. Consequently, $\\overline{P K}=\\overline{T R}$ and $P K \\| T R$. Also $O K \\| D N$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_eb51ab1a35cd98c2706cg-28.jpg?height=612&width=789&top_left_y=818&top_left_x=995)\n(from $\\triangle C D N$ ) and since $D N \\perp B C$ and $M H_{1} \\perp B C$, it follows that $T H_{1} \\| O K$. Since $O$ is the circumcenter of $\\triangle C M N, O P \\perp M N$. Thus, $C H_{1} \\perp M N$ implies $O P \\| C H_{1}$. We conclude $\\Delta T R H_{1} \\cong \\triangle K P O$ (they have parallel sides and $\\overline{T R}=\\overline{P K}$ ), hence $\\overline{R H_{1}}=\\overline{P O}$, i.e. $\\overline{C H_{1}}=2 \\overline{P O}$ and $\\mathrm{CH}_{1} \\| \\mathrm{PO}$.\n\nAnalogously, $\\overline{D H_{2}}=2 \\overline{P O}$ and $D H_{2} \\| P O$. From $\\overline{C H_{1}}=2 \\overline{P O}=\\overline{D H_{2}}$ and\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_eb51ab1a35cd98c2706cg-28.jpg?height=75&width=1581&top_left_y=1895&top_left_x=222)\n$H_{1} H_{2} \\| C D$. Therefore the area of the quadrilateral $A H_{1} B H_{2}$ is $\\frac{\\overline{A B} \\cdot \\overline{H_{1} H_{2}}}{2}=\\frac{\\overline{A B} \\cdot \\overline{C D}}{2}=S$.\n\nSolution2. Since $M H_{1} \\| D N$ and $N H_{1} \\| D M, M D N H_{1}$ is a parallelogram. Similarly, $\\mathrm{NH}_{2} \\| C M$ and $\\mathrm{MH}_{2} \\| \\mathrm{CN}$ imply $\\mathrm{MCNH}_{2}$ is a parallelogram. Let $P$ be the midpoint of the segment $\\overline{M N}$. Then $\\sigma_{P}(D)=H_{1}$ and $\\sigma_{P}(C)=H_{2}$, thus $C D \\| H_{1} H_{2}$ and $\\overline{C D}=\\overline{H_{1} H_{2}}$. From $C D \\perp A B$ we deduce $A_{A H_{1} B H_{2}}=\\frac{1}{2} \\overline{A B} \\cdot \\overline{C D}=S$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## G3", "solution_match": "\nSolution1."}}
-{"year": "2014", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C$ be a triangle such that $\\overline{A B} \\neq \\overline{A C}$. Let $M$ be a midpoint of $\\overline{B C}, H$ the orthocenter of $A B C, O_{1}$ the midpoint of $\\overline{A H}$ and $O_{2}$ the circumcenter of $B C H$. Prove that $O_{1} A M O_{2}$ is a parallelogram.", "solution": "Let $O_{2}^{\\prime}$ be the point such that $O_{1} A M O_{2}^{\\prime}$ is a parallelogram. Note that $\\overrightarrow{M O_{2}}=\\overrightarrow{A O_{1}}=\\overrightarrow{O_{1} H}$. Therefore, $O_{1} H O_{2}^{\\prime} M$ is a parallelogram and $\\overrightarrow{M O_{1}}=\\overrightarrow{O_{2} H}$.\n\nSince $M$ is the midpoint of $\\overline{B C}$ and $O_{1}$ is the midpoint of $\\overline{A H}$, it follows that $4 \\overrightarrow{M O_{1}}=\\overrightarrow{B A}+\\overrightarrow{B H}+\\overrightarrow{C A}+\\overrightarrow{C H}=2(\\overrightarrow{C A}+\\overrightarrow{B H})$. Moreover, let $B^{\\prime}$ be the midpoint of $\\overrightarrow{B H}$. Then,\n\n$$\n\\begin{aligned}\n2 \\overrightarrow{O_{2}^{\\prime B}} \\cdot \\overrightarrow{B H} & =\\left(\\overline{O_{2}^{\\prime H}}+\\overline{O_{2}^{\\prime B}}\\right) \\cdot \\overrightarrow{B H}=\\left(2 \\overline{O_{2}^{\\prime} H}+\\overrightarrow{H B}\\right) \\cdot \\overrightarrow{B H}= \\\\\n& =\\left(2 \\overline{M O_{1}}+\\overline{H B}\\right) \\cdot \\overline{B H}=(\\overline{C A}+\\overrightarrow{B H}+\\overrightarrow{H B}) \\cdot \\overrightarrow{B H}=\\overline{C A} \\cdot \\overrightarrow{B H}=0 .\n\\end{aligned}\n$$\n\nBy $\\vec{a} \\cdot \\vec{b}$ we denote the inner product of the vectors $\\vec{a}$ and $\\vec{b}$.\n\nTherefore, $O_{2}^{\\prime}$ lies on the perpendicular bisector of $\\overline{B H}$. Since $B$ and $C$ play symmetric roles, $\\mathrm{O}_{2}^{\\prime}$ also lies on the perpendicular bisector of $\\overline{\\mathrm{CH}}$, hence $\\mathrm{O}_{2}^{\\prime}$ is the circumcenter of $\\triangle B C H$ and $\\mathrm{O}_{2}=\\mathrm{O}_{2}^{\\prime}$.\n\nNote: The condition $\\overline{A B} \\neq \\overline{A C}$ just aims at ensuring that the parallelogram $O_{1} A N O_{2}$ is not degenerate, hence at helping students to focus on the \"general\" case.\n\nSolution2. We use the following two well-known facts:\n\n$\\sigma_{B C}(H)$ lies on the circumcircle of $\\triangle A B C$.\n\n$\\overrightarrow{A H}=-2 \\overrightarrow{M O}$, where $O$ is the circumcenter of $\\triangle A B C$.\n\nThe statement \" $\\mathrm{O}_{1} A M O_{2}$ is parallelogram\" is equivalent to \" $\\sigma_{B C}\\left(O_{2}\\right)=O$ \". The later is true because the circumcircles of $\\triangle A B C$ and $\\triangle B C H$ are symmetrical with respect to $B C$, from (1).", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## G4", "solution_match": "\nSolution1."}}
-{"year": "2014", "tier": "T3", "problem_label": "G5", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C$ be a triangle with $\\overline{A B} \\neq \\overline{B C}$, and let $B D$ be the internal bisector of $\\measuredangle A B C(D \\in A C)$. Denote the midpoint of the arc $A C$ which contains point BbyM. The circumcircle of the triangle $B D M$ intersects the segment $A B$ at point $K \\neq B$, and let $J$ be the reflection of $A$ with respect to $K$. If $D J \\cap A M=\\{O\\}$, prove that the points $J, B, M, O$ belong to the same circle.", "solution": "![](https://cdn.mathpix.com/cropped/2024_06_05_eb51ab1a35cd98c2706cg-30.jpg?height=1337&width=1232&top_left_y=1419&top_left_x=762)\n\nLet the circumcircle of the triangle $B D M$ intersect the line segment $B C$ at point $L \\neq B$. From $\\measuredangle C B D=\\measuredangle D B A$ we have $\\overline{D L}=\\overline{D K}$. Since $\\measuredangle L C M=\\measuredangle B C M=\\measuredangle B A M=\\measuredangle K A M, \\overline{M C}=\\overline{M A}$ and\n\n$$\n\\measuredangle L M C=\\measuredangle L M K-\\measuredangle C M K=\\measuredangle L B K-\\measuredangle C M K=\\measuredangle C B A-\\measuredangle C M K=\\measuredangle C M A-\\measuredangle C M K=\\measuredangle K M A,\n$$\n\nit follows that triangles $M L C$ and $M K A$ are congruent, which implies $\\overline{C L}=\\overline{A K}=\\overline{K J}$. Furthermore, $\\measuredangle C L D=180^{\\circ}-\\measuredangle B L D=\\measuredangle D K B=\\measuredangle D K J$ and $\\overline{D L}=\\overline{D K}$, it follows that triangles $D C L$ and $D J K$ are congruent. Hence, $\\angle D C L=\\angle D J K=\\measuredangle B J O$. Then\n\n$$\n\\measuredangle B J O+\\measuredangle B M O=\\angle D C L+\\angle B M A=\\angle B C A+180^{\\circ}-\\angle B C A=180^{\\circ}\n$$\n\nso the points $J, B, M, O$ belong to the same circle, q.e.d.\n\n## Solution2.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_eb51ab1a35cd98c2706cg-31.jpg?height=938&width=947&top_left_y=452&top_left_x=676)\n\nSince $\\overline{M C}=\\overline{M A}$ and $\\measuredangle C M A=\\measuredangle C B A$, we have $\\measuredangle A C M=\\measuredangle C A M=90^{\\circ}-\\frac{\\measuredangle C B A}{2}$. It follows that $\\measuredangle M B D=\\measuredangle M B A+\\measuredangle A B D=\\measuredangle A C M+\\measuredangle A B D=90^{\\circ}-\\frac{\\measuredangle C B A}{2}+\\frac{\\measuredangle C B A}{2}=90^{\\circ}$. Denote the midpoint of $\\overline{A C}$ by $N$. Since $\\measuredangle D N M=\\measuredangle C N M=90^{\\circ}, N$ belongs to the circumcircle of the triangle $B D M$. Since $N K$ is the midline of the triangle $A C J$ and $N K \\| C J$, we have\n\n$$\n\\measuredangle B J C=\\measuredangle B K N=180^{\\circ}-\\measuredangle N D B=\\measuredangle C D B\n$$\n\nHence, the quadrilateral $C D J B$ is cyclic (this can also be obtained from the power of a point theorem, because $\\overline{A N} \\cdot \\overline{A D}=\\overline{A K} \\cdot \\overline{A B}$ implies $\\overline{A C} \\cdot \\overline{A D}=\\overline{A J} \\cdot \\overline{A B}$ ), and\n\n$$\n\\measuredangle B J O=\\measuredangle 180^{\\circ}-\\measuredangle B J D=\\angle B C D=\\angle B C A=180^{\\circ}-\\angle B M A=180^{\\circ}-\\measuredangle B M O\n$$\n\nso the points $J, B, M, O$ belong to the same circle, q.e.d.\n\nRemark. If $J$ is between $A$ and $K$ the solution can be easily adapted.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## G5", "solution_match": "## Solution1."}}
-{"year": "2014", "tier": "T3", "problem_label": "G6", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C D$ be a quadrilateral whose sides $A B$ and $C D$ are not parallel, and let $O$ be the intersection of its diagonals. Denote with $H_{1}$ and $H_{2}$ the orthocenters of the triangles $O A B$ and OCD, respectively. If $M$ and $N$ are the midpoints of the segments $\\overline{A B}$ and $\\overline{C D}$, respectively, prove that the lines $M N$ and $\\mathrm{H}_{1} \\mathrm{H}_{2}$ are parallel if and only if $\\overline{A C}=\\overline{B D}$.", "solution": "![](https://cdn.mathpix.com/cropped/2024_06_05_eb51ab1a35cd98c2706cg-32.jpg?height=1378&width=1735&top_left_y=1403&top_left_x=461)\n\nLet $A^{\\prime}$ and $B^{\\prime}$ be the feet of the altitudes drawn from $A$ and $B$ respectively in the triangle $A O B$, and $C^{\\prime}$ and $D^{\\prime}$ are the feet of the altitudes drawn from $C$ and $D$ in the triangle $C O D$. Obviously, $A^{\\prime}$ and $D^{\\prime}$ belong to the circle $c_{1}$ of diameter $\\overline{A D}$, while $B^{\\prime}$ and $C^{\\prime}$ belong to the circle $c_{2}$ of diameter $\\overline{B C}$.\n\nIt is easy to see that triangles $H_{1} A B$ and $H_{1} B^{\\prime} A^{\\prime}$ are similar. It follows that $\\overline{H_{1} A} \\cdot \\overline{H_{1} A^{\\prime}}=\\overline{H_{1} B} \\cdot \\overline{H_{1} B^{\\prime}}$. (Alternatively, one could notice that the quadrilateral $A B A^{\\prime} B^{\\prime}$ is cyclic and obtain the previous relation by writing the power of $H_{1}$ with respect to its circumcircle.) It\nfollows that $H_{1}$ has the same power with respect to circles $c_{1}$ and $c_{2}$. Thus, $H_{1}$ (and similarly, $\\mathrm{H}_{2}$ ) is on the radical axis of the two circles.\n\nThe radical axis being perpendicular to the line joining the centers of the two circles, one concludes that $\\mathrm{H}_{1} \\mathrm{H}_{2}$ is perpendicular to $P Q$, where $P$ and $Q$ are the midpoints of the sides $\\overline{A D}$ and $\\overline{B C}$, respectively. ( $P$ and $Q$ are the centers of circles $c_{1}$ and $c_{2}$.)\n\nThe condition $H_{1} H_{2} \\| M N$ is equivalent to $M N \\perp P Q$. As $M P N Q$ is a parallelogram, we conclude that $H_{1} H_{2} \\| M N \\Leftrightarrow M N \\perp P Q \\Leftrightarrow M P N Q$ a rhombus $\\Leftrightarrow \\overline{M P}=\\overline{M Q} \\Leftrightarrow \\overline{A C}=\\overline{B D}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## G6", "solution_match": "## Solution."}}
-{"year": "2014", "tier": "T3", "problem_label": "N1", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Each letter of the word OHRID corresponds to a different digit belonging to the set $(1,2,3,4,5)$. Decipher the equality $(O+H+R+I+D)^{2}:(O-H-R+I+D)=O^{H^{R^{I_{D}^{D}}}}$.", "solution": "Since $O, H, R, I$ and $D$ are distinct numbers from $\\{1,2,3,4,5\\}$, we have $O+H+R+I+D=15$ and $O-H-R+I+D=O+H+R+I+D-2(H+R)<15$. From this $O^{H^{R^{I^{D}}}}=\\frac{(O+H+R+I+D)^{2}}{O-H-R+I+D}=\\frac{225}{15-2(H+R)}$, hence $O^{H^{R^{R^{D}}}}>15$ and divides 225 , which is only possible for $O^{H^{R^{I^{D}}}}=25$ (must be a power of three or five). This implies that $O=5, H=2$ and $R=1$. It's easy to check that both $I=3, D=4$ and $I=4, D=3$ satisfy the stated equation.\n\n## $\\mathbf{N} 2$\n\nFind all triples $(p, q, r)$ of distinct primes $p, q$ and $r$ such that\n\n$$\n3 p^{4}-5 q^{4}-4 r^{2}=26\n$$\n\nSolution. First notice that if both primes $q$ and $r$ differ from 3 , then $q^{2} \\equiv r^{2} \\equiv 1(\\bmod 3)$, hence the left hand side of the given equation is congruent to zero modulo 3 , which is impossible since 26 is not divisible by 3 . Thus, $q=3$ or $r=3$. We consider two cases.\n\nCase 1. $q=3$.\n\nThe equation reduces to $3 p^{4}-4 r^{2}=431$\n\nIf $p \\neq 5$, by Fermat's little theorem, $p^{4} \\equiv 1(\\bmod 5)$, which yields $3-4 r^{2} \\equiv 1(\\bmod 5)$, or equivalently, $r^{2}+2 \\equiv 0(\\bmod 5)$. The last congruence is impossible in view of the fact that a residue of a square of a positive integer belongs to the set $\\{0,1,4\\}$. Therefore $p=5$ and $r=19$.\n\nCase 2. $r=3$.\n\nThe equation becomes $3 p^{4}-5 q^{4}=62(2)$.\n\nObviously $p \\neq 5$. Hence, Fermat's little theorem gives $p^{4} \\equiv 1(\\bmod 5)$. But then $5 q^{4} \\equiv 1(\\bmod 5)$, which is impossible.\n\nHence, the only solution of the given equation is $p=5, q=3, r=19$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## N1", "solution_match": "\nSolution."}}
-{"year": "2014", "tier": "T3", "problem_label": "N3", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find the integer solutions of the equation\n\n$$\nx^{2}=y^{2}\\left(x+y^{4}+2 y^{2}\\right)\n$$", "solution": "If $x=0$, then $y=0$ and conversely, if $y=0$, then $x=0$. It follows that $(x, y)=(0,0)$ is a solution of the problem. Assume $x \\neq 0$ and $y \\neq 0$ satisfy the equation. The equation can be transformed in the form $x^{2}-x y^{2}=y^{6}+2 y^{4}$. Then $4 x^{2}-4 x y^{2}+y^{4}=4 y^{6}+9 y^{4}$ and consequently $\\left(\\frac{2 x}{y^{2}}-1\\right)^{2}=4 y^{2}+9 \\quad$ (1). Obviously $\\frac{2 x}{y^{2}}-1$ is integer. From (1), we get that the numbers $\\frac{2 x}{y^{2}}-1,2 y$ and 3 are Pythagorean triplets. It follows that $\\frac{2 x}{y^{2}}-1= \\pm 5$ and $2 y= \\pm 4$. Therefore, $x=3 y^{2}$ or $x=-2 y^{2}$ and $y= \\pm 2$. Hence $(x, y)=(12,-2),(x, y)=(12,2),(x, y)=(-8,-2)$ and $(x, y)=(-8,2)$ are the possible solutions. By substituting them in the initial equation we verify that all the 4 pairs are solution. Thus, together with the couple $(x, y)=(0,0)$ the problem has 5 solutions.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "\nN3", "solution_match": "\nSolution."}}
-{"year": "2014", "tier": "T3", "problem_label": "N4", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Prove there are no integers $a$ and $b$ satisfying the following conditions:\ni) $16 a-9 b$ is a prime number\n\nii) $\\quad a b$ is a perfect square\n\niii) $a+b$ is a perfect square", "solution": "Suppose $a$ and $b$ be integers satisfying the given conditions. Let $p$ be a prime number, $n$ and $m$ be integers. Then we can write the conditions as follows:\n\n$$\n\\begin{aligned}\n& 16 a-9 b=p \\\\\n& a b=n^{2} \\\\\n& a+b=m^{2}\n\\end{aligned}\n$$\n\nMoreover, let $d=g d c(a, b)$ and $a=d x, b=d y$ for some relatively prime integers $x$ and $y$. Obviously $a \\neq 0$ and $b \\neq 0, a$ and $b$ are positive (by (2) and (3)).\n\nFrom (2) follows that $x$ and $y$ are perfect squares, say $x=l^{2}$ and $y=s^{2}$.\n\nFrom (1), $d \\mid p$ and hence $d=p$ or $d=1$. If $d=p$, then $16 x-9 y=1$, and we obtain $x=9 k+4$, $y=16 k+7$ for some nonnegative integer $k$. But then $s^{2}=y \\equiv 3(\\bmod 4)$, which is a contradiction.\n\nIf $d=1$ then $16 l^{2}-9 s^{2}=p \\Rightarrow(4 l-3 s)(4 l+3 s)=p \\Rightarrow(4 l+3 s=p \\wedge 4 l-3 s=1)$.\n\nBy adding the last two equations we get $8 l=p+1$ and by subtracting them we get $6 s=p-1$. Therefore $p=24 t+7$ for some integer $t$ and $a=(3 t+1)^{2}$ and $b=(4 t+1)^{2}$ satisfy the conditions (1) and (2). By (3) we have $m^{2}=(3 t+1)^{2}+(4 t+1)^{2}=25 t^{2}+14 t+2$, or equivalently $25 m^{2}=(25 t+7)^{2}+1$.\n\nSince the difference between two nonzero perfect square cannot be 1 , we have a contradiction. As a result there is no solution.\n\n## NS\n\nFind all nonnegative integers $x, y, z$ such that\n\n$$\n2013^{x}+2014^{y}=2015^{z}\n$$\n\nSolution. Clearly, $y>0$, and $z>0$. If $x=0$ and $y=1$, then $z=1$ and $(x, y, z)=(0,1,1)$ is a solution. If $x=0$ and $y \\geq 2$, then modulo 4 we have $1+0 \\equiv(-1)^{z}$, hence $z$ is even $\\left(z=2 z_{1}\\right.$ for some integer $\\left.z_{1}\\right)$. Then $2^{y} 1007^{y}=\\left(2015^{z_{1}}-1\\right)\\left(2015^{z_{1}}+1\\right)$, and since $\\operatorname{gcd}\\left(1007,2015^{z_{1}}+1\\right)=1$ we obtain $2 \\cdot 1007^{y} \\mid 2015^{z_{1}}-1$ and $2015^{z_{1}}+1 \\mid 2^{y-1}$. From this we get $2015^{z_{1}}+1 \\leq 2^{y-1}<2 \\cdot 1007^{y} \\leq 2015^{z_{1}}-1$, which is impossible.\n\nNow for $x>0$, modulo 3 we get $0+1 \\equiv(-1)^{z}$, hence $z$ must be even $\\left(z=2 z_{1}\\right.$ for some integer $z_{1}$ ). Modulo 2014 we get $(-1)^{x}+0 \\equiv 1$, thus $x$ must be even $\\left(x=2 x_{1}\\right.$ for some integer $\\left.x_{1}\\right)$. We transform the equation to $2^{y} 1007^{y}=\\left(2015^{z_{1}}-2013^{x_{1}}\\right)\\left(2015^{z_{1}}+2013^{x_{1}}\\right)$ and since $\\operatorname{gcd}\\left(2015^{z_{1}}-2013^{x_{1}}, 2015^{z_{1}}+2013^{x_{1}}\\right)=2,1007^{y}$ divides $2015^{z_{1}}-2013^{x_{1}}$ or $2015^{z_{1}}+2013^{x_{1}}$ but not both. If $1007^{y} \\mid 2015^{z_{1}}-2013^{x_{1}}$, then $2015^{z_{1}}+2013^{x_{1}} \\leq 2^{y}<1007^{y} \\leq 2015^{z_{1}}-2013^{x_{1}}$, which is impossible. Hence $1007 \\mid 2015^{z_{1}}+2013^{x_{1}}$, and from $2015^{z_{1}}+2013^{x_{1}} \\equiv 1+(-1)^{x_{1}}(\\bmod 1007), x_{1}$ is odd $\\left(x=2 x_{1}=4 x_{2}+2\\right.$ for some integer $\\left.x_{2}\\right)$.\n\nNow modulo 5 we get $-1+(-1)^{y} \\equiv(-2)^{4 x_{2}+2}+(-1)^{y} \\equiv 0$, hence $y$ must be even $\\left(y=2 y_{1}\\right.$ for some integer $y_{1}$ ). Finally modulo 31 , we have $(-2)^{4 x_{2}+2}+(-1)^{2 y_{1}} \\equiv 0$ or $4^{2 x_{2}+1} \\equiv-1$. This is impossible since the reminders of the powers of 4 modulo 31 are 1, 2, 4, 8 and 16 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "\nN4", "solution_match": "\nSolution."}}
-{"year": "2014", "tier": "T3", "problem_label": "N6", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Vukasin, Dimitrije, Dusan, Stefan and Filip asked their professor to guess a three consecutive positive integer numbers after they had told him these (true) sentences:\n\nVukasin: \"Sum of the digits of one of them is a prime number. Sum of the digits of some of the other two is an even perfect number ( $n$ is perfect if $\\sigma(n)=2 n$ ). Sum of the digits of the remaining number is equal to the number of its positive divisors.\"\n\nDimitrije:\"Each of these three numbers has no more than two digits 1 in its decimal representation.\"\n\nDusan:\"If we add 11 to one of them, we obtain a square of an integer.\"\n\nStefan:\"Each of them has exactly one prime divisor less then 10.\"\n\nFilip:\"The 3 numbers are square-free.\"\n\nTheir professor gave the correct answer. Which numbers did he say?", "solution": "Let the middle number be $n$, so the numbers are $n-1, n$ and $n+1$. Since 4 does not divide any of them, $n \\equiv 2(\\bmod 4)$. Furthermore, neither 3,5 nor 7 divides $n$. Also $n+1+11 \\equiv 2(\\bmod 4)$ cannot be a square. Then 3 must divide $n-1$ or $n+1$. If $n-1+11$ is a square, then $3 \\mid n+1$ which implies $3 \\mid n+10$ (a square), so $9 \\mid n+10$ hence $9 \\mid n+1$, which is impossible. Thus must be $n+11=m^{2}$.\n\nFurther, 7 does not divide $n-1$, nor $n+1$, because $1+11 \\equiv 5(\\bmod 7)$ and $-1+11 \\equiv 3(\\bmod 7)$ are quadratic nonresidues modulo 7. This implies $5 \\mid n-1$ or $5 \\mid n+1$. Again, since $n+11$ is a square, it is impossible $5 \\mid n-1$, hence $5 \\mid n+1$ which implies $3 \\mid n-1$. This yields $n \\equiv 4(\\bmod 10)$ hence $S(n+1)=S(n)+1=S(n-1)+2(S(n)$ is sum of the digits of $n)$. Since the three numbers are square-free, their numbers of positive divisors are powers of 2 . Thus, we have two even sums of digits - they must be $S(n-1)$ and $S(n+1)$, so $S(n)$ is prime. From $3 \\mid n-1$, follows $S(n-1)$ is an even perfect number, and $S(n+1)=2^{p}$. Consequently $S(n)=2^{p}-1$ is a prime, so $p$ is a prime number. One easily verifies $p \\neq 2$, so $\\mathrm{p}$ is odd implying $3 \\mid 2^{p}-2$. Then\n$\\sigma\\left(2^{p}-2\\right) \\geq\\left(2^{p}-2\\right)\\left(1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{6}\\right)=2\\left(2^{p}-2\\right)$. Since this number is perfect, $\\frac{2^{p}-2}{6}$ must be one, i.e. $p=3$ and $S(n-1)=6, S(n)=7$ and $S(n+1)=8$.\n\nSince 4 does not divide $n$, the 2 -digit ending of $n$ must be 14 or 34 . But $n=34$ is impossible, since $n+11=45$ is not a square. Hence, $n=10^{a}+10^{b}+14$ with $a \\geq b \\geq 2$. If $a \\neq b$, then $n$ has three digits 1 in its decimal representation, which is impossible. Therefore $a=b$, and $n=2 \\cdot 10^{a}+14$. Now, $2 \\cdot 10^{a}+25=m^{2}$, hence $5 \\mid m$, say $m=5 t$, and $(t-1)(t+1)=2^{a+1} 5^{a-2}$. Because $\\operatorname{gcd}(t-1, t+1)=2$ there are three possibilities:\n\n1) $t-1=2, t+1=2^{a} 5^{a-2}$, which implies $a=2, t=3$;\n2) $t-1=2^{a}, t+1=2 \\cdot 5^{a-2}$, so $2^{a}+2=2 \\cdot 5^{a-2}$, which implies $a=3, t=9$;\n3) $t-1=2 \\cdot 5^{a-2}, t+1=2^{a}$, so $2 \\cdot 5^{a-2}+2=2^{a}$, which implies $a=2, t=3$, same as case 1 ).\n\nFrom the only two possibilities $(n-1, n, n+1)=(213,214,215)$ and $(n-1, n, n+1)=$ $(2013,2014,2015)$ the first one is not possible, because $S(215)=8$ and $\\tau(215)=4$. By checking the conditions, we conclude that the latter is a solution, so the professor said the numbers: 2013, 2014, 2015.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## N6", "solution_match": "\nSolution."}}
+{"year": "2014", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "For any real number a, let $\\lfloor a\\rfloor$ denote the greatest integer not exceeding a. In positive real numbers solve the following equation\n\n$$\nn+\\lfloor\\sqrt{n}\\rfloor+\\lfloor\\sqrt[3]{n}\\rfloor=2014\n$$", "solution": "Obviously $n$ must be positive integer. Now note that $44^{2}=1936<2014<2025=45^{2}$ and $12^{3}<1900<2014<13^{3}$.\n\nIf $n<1950$ than $2014=n+\\lfloor\\sqrt{n}\\rfloor+\\lfloor\\sqrt[3]{n}\\rfloor<1950+44+12=2006$, a contradiction!\n\nSo $n \\geq 1950$. Also if $n>2000$ than $2014=n+\\lfloor\\sqrt{n}\\rfloor+\\lfloor\\sqrt[3]{n}\\rfloor>2000+44+12=2056$, a contradiction!\n\nSo $1950 \\leq n \\leq 2000$, therefore $\\lfloor\\sqrt{n}\\rfloor=44$ and $\\lfloor\\sqrt[3]{n}\\rfloor=12$. Plugging that into the original equation we get:\n\n$$\nn+\\lfloor\\sqrt{n}\\rfloor+\\lfloor\\sqrt[3]{n}\\rfloor=n+44+12=2014\n$$\n\nFrom which we get $n=1956$, which is the only solution.\n\nSolution2. Obviously $n$ must be positive integer. Since $n \\leq 2014, \\sqrt{n}<45$ and $\\sqrt[3]{n}<13$.\n\nForm $n=2014-\\lfloor\\sqrt{n}\\rfloor-\\lfloor\\sqrt[3]{n}\\rfloor>2014-45-13=1956, \\sqrt{n}>44$ and $\\sqrt[3]{n}>12$, thus $\\lfloor\\sqrt{n}\\rfloor=44$ and $\\lfloor\\sqrt[3]{n}\\rfloor=12$ and $n=2014-\\lfloor\\sqrt{n}\\rfloor-\\lfloor\\sqrt[3]{n}\\rfloor=2014-44-12=1958$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## A1", "solution_match": "\nSolution1."}}
+{"year": "2014", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $a, b$ and $c$ be positive real numbers such that abc $=\\frac{1}{8}$. Prove the inequality\n\n$$\na^{2}+b^{2}+c^{2}+a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \\geq \\frac{15}{16}\n$$\n\nWhen does equality hold?", "solution": "By using The Arithmetic-Geometric Mean Inequality for 15 positive numbers, we find that\n\n$$\n\\begin{aligned}\n& a^{2}+b^{2}+c^{2}+a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}= \\\\\n& \\quad=\\frac{a^{2}}{4}+\\frac{a^{2}}{4}+\\frac{a^{2}}{4}+\\frac{a^{2}}{4}+\\frac{b^{2}}{4}+\\frac{b^{2}}{4}+\\frac{b^{2}}{4}+\\frac{b^{2}}{4}+\\frac{c^{2}}{4}+\\frac{c^{2}}{4}+\\frac{c^{2}}{4}+\\frac{c^{2}}{4}+a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \\geq \\\\\n& \\quad \\geq 1515 \\sqrt{\\frac{a^{12} b^{12} c^{12}}{4^{12}}}=15 \\sqrt[5]{\\left(\\frac{a b c}{4}\\right)^{4}}=15 \\sqrt[5]{\\left(\\frac{1}{32}\\right)^{4}}=\\frac{15}{16}\n\\end{aligned}\n$$\n\nas desired. Equality holds if and only if $a=b=c=\\frac{1}{2}$.\n\nSolution2. By using AM-GM we obtain\n\n$$\n\\begin{aligned}\n& \\left(a^{2}+b^{2}+c^{2}\\right)+\\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\\right) \\geq 3 \\sqrt[3]{a^{2} b^{2} c^{2}}+3 \\sqrt[3]{a^{4} b^{4} c^{4}}= \\\\\n& =3 \\sqrt[3]{\\left(\\frac{1}{8}\\right)^{2}}+3 \\sqrt[3]{\\left(\\frac{1}{8}\\right)^{4}}=\\frac{3}{4}+\\frac{3}{16}=\\frac{15}{16}\n\\end{aligned}\n$$\n\nThe equality holds when $a^{2}=b^{2}=c^{2}$, i.e. $a=b=c=\\frac{1}{2}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## A2", "solution_match": "\nSolution1."}}
+{"year": "2014", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that:\n\n$$\n\\left(a+\\frac{1}{b}\\right)^{2}+\\left(b+\\frac{1}{c}\\right)^{2}+\\left(c+\\frac{1}{a}\\right)^{2} \\geq 3(a+b+c+1)\n$$\n\nWhen does equality hold?", "solution": "By using AM-GM $\\left(x^{2}+y^{2}+z^{2} \\geq x y+y z+z x\\right)$ we have\n\n$$\n\\begin{aligned}\n\\left(a+\\frac{1}{b}\\right)^{2}+\\left(b+\\frac{1}{c}\\right)^{2}+\\left(c+\\frac{1}{a}\\right)^{2} & \\geq\\left(a+\\frac{1}{b}\\right)\\left(b+\\frac{1}{c}\\right)+\\left(b+\\frac{1}{c}\\right)\\left(c+\\frac{1}{a}\\right)+\\left(c+\\frac{1}{a}\\right)\\left(a+\\frac{1}{b}\\right) \\\\\n& =\\left(a b+1+\\frac{a}{c}+a\\right)+\\left(b c+1+\\frac{b}{a}+b\\right)+\\left(c a+1+\\frac{c}{b}+c\\right) \\\\\n& =a b+b c+c a+\\frac{a}{c}+\\frac{c}{b}+\\frac{b}{a}+3+a+b+c\n\\end{aligned}\n$$\n\nNotice that by AM-GM we have $a b+\\frac{b}{a} \\geq 2 b, b c+\\frac{c}{b} \\geq 2 c$, and $c a+\\frac{a}{c} \\geq 2 a$.\n\nThus,\n\n$$\n\\left(a+\\frac{1}{b}\\right)^{2}+\\left(b+\\frac{1}{c}\\right)^{2}+\\left(c+\\frac{1}{a}\\right)^{2} \\geq\\left(a b+\\frac{b}{a}\\right)+\\left(b c+\\frac{c}{b}\\right)+\\left(c a+\\frac{a}{c}\\right)+3+a+b+c \\geq 3(a+b+c+1)\n$$\n\nThe equality holds if and only if $a=b=c=1$.\n\nSolution2. From QM-AM we obtain\n\n$$\n\\begin{aligned}\n& \\sqrt{\\frac{\\left(a+\\frac{1}{b}\\right)^{2}+\\left(b+\\frac{1}{c}\\right)^{2}+\\left(c+\\frac{1}{a}\\right)^{2}}{3}} \\geq \\frac{a+\\frac{1}{b}+b+\\frac{1}{c}+c+\\frac{1}{a}}{3} \\Leftrightarrow \\\\\n& \\left(a+\\frac{1}{b}\\right)^{2}+\\left(b+\\frac{1}{c}\\right)^{2}+\\left(c+\\frac{1}{a}\\right)^{2} \\geq \\frac{\\left(a+\\frac{1}{b}+b+\\frac{1}{c}+c+\\frac{1}{a}\\right)^{2}}{3}\n\\end{aligned}\n$$\n\nFrom AM-GM we have $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} \\geq 3 \\sqrt[3]{\\frac{1}{a b c}}=3$, and substituting in (1) we get\n\n$$\n\\begin{aligned}\n&\\left(a+\\frac{1}{b}\\right)^{2}+\\left(b+\\frac{1}{c}\\right)^{2}+\\left(c+\\frac{1}{a}\\right)^{2} \\geq \\frac{\\left(a+\\frac{1}{b}+b+\\frac{1}{c}+c+\\frac{1}{a}\\right)^{2}}{3} \\geq \\frac{(a+b+c+3)^{2}}{3}= \\\\\n&=\\frac{(a+b+c)(a+b+c)+6(a+b+c)+9}{3} \\geq \\frac{(a+b+c) 3 \\sqrt[3]{a b c}+6(a+b+c)+9}{3}= \\\\\n&=\\frac{9(a+b+c)+9}{3}=3(a+b+c+1)\n\\end{aligned}\n$$\n\nThe equality holds if and only if $a=b=c=1$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## A3", "solution_match": "\nSolution1."}}
+{"year": "2014", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $a, b, c$ be positive real numbers such that $a+b+c=1$. Prove that\n\n$$\n\\frac{7+2 b}{1+a}+\\frac{7+2 c}{1+b}+\\frac{7+2 a}{1+c} \\geq \\frac{69}{4}\n$$\n\nWhen does equality hold?", "solution": "Solution. Equality holds when $x=y=z=0$.\n\nApply AM-GM to $x+y+z=x y z$,\n\n$$\n\\begin{aligned}\n& x y z=x+y+z \\geq 3 \\sqrt[3]{x y z} \\Rightarrow(x y z)^{3} \\geq(3 \\sqrt[3]{x y z})^{3} \\\\\n& \\Rightarrow x^{3} y^{3} z^{3} \\geq 27 x y z \\\\\n& \\Rightarrow x^{2} y^{2} z^{2} \\geq 27 \\\\\n& \\Rightarrow \\sqrt[3]{x^{2} y^{2} z^{2}} \\geq 3\n\\end{aligned}\n$$\n\nAlso by AM-GM we have, $x^{2}+y^{2}+z^{2} \\geq 3 \\sqrt[3]{x^{2} y^{2} z^{2}} \\geq 9$.\n\nTherefore we get $x^{2}+y^{2}+z^{2} \\geq 9$.\n\nNow,\n\n$$\n\\begin{gathered}\n2\\left(x^{2}+y^{2}+z^{2}\\right) \\geq 3(x+y+z) \\Leftrightarrow \\frac{2\\left(x^{2}+y^{2}+z^{2}\\right)}{3} \\geq(x+y+z) \\\\\n\\Leftrightarrow 2 \\cdot \\frac{2\\left(x^{2}+y^{2}+z^{2}\\right)}{3} \\geq 2 \\cdot(x+y+z) \\\\\n\\Leftrightarrow \\frac{4\\left(x^{2}+y^{2}+z^{2}\\right)}{3} \\geq 2 \\cdot(x+y+z) \\\\\n\\Leftrightarrow x^{2}+y^{2}+z^{2}+\\frac{\\left(x^{2}+y^{2}+z^{2}\\right)}{3} \\geq 2 \\cdot(x+y+z) \\\\\n\\Leftrightarrow 3+\\frac{x^{2}}{3}+3+\\frac{y^{2}}{3}+3+\\frac{z^{2}}{3} \\geq 2(x+y+z) \\\\\n\\Leftrightarrow 3+\\frac{x^{2}}{3}+3+\\frac{y^{2}}{3}+3+\\frac{z^{2}}{3} \\geq 2(x+y+z) \\\\\n\\Leftrightarrow 2 \\sqrt{3 \\cdot \\frac{x^{2}}{3}}+2 \\sqrt{3 \\cdot \\frac{y^{2}}{3}}+2 \\sqrt{3 \\cdot \\frac{z^{2}}{3}} \\geq 2(x+y+z)\n\\end{gathered}\n$$\n\nEquality holds if $3=\\frac{x^{2}}{3}=\\frac{y^{2}}{3}=\\frac{z^{2}}{3}$, i.e. $x=y=z=3$, for which $x+y+z \\neq x y z$.\n\nRemark. The inequality can be improved: $x^{2}+y^{2}+z^{2} \\geq \\sqrt{3}(x+y+z)$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## A4", "solution_match": "\nSolution1."}}
+{"year": "2014", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $a, b, c$ be positive real numbers such that $a+b+c=1$. Prove that\n\n$$\n\\frac{7+2 b}{1+a}+\\frac{7+2 c}{1+b}+\\frac{7+2 a}{1+c} \\geq \\frac{69}{4}\n$$\n\nWhen does equality hold?", "solution": "Solution. If one of the numbers is zero, then from $x+y+z=x y z$ all three numbers are zero and the equality trivially holds.\n\nFrom AM-GM $x^{2}+y^{2}+z^{2} \\geq 3 \\sqrt[3]{x^{2} y^{2} z^{2}}=3 \\frac{x+y+z}{\\sqrt[3]{x y z}} \\geq 3 \\frac{x+y+z}{\\frac{x+y+z}{3}}=9$\n\nFrom QM-AM $\\frac{x^{2}+y^{2}+z^{2}}{3} \\geq\\left(\\frac{x+y+z}{3}\\right)^{2}$\n\nMultiplying (1) and (2) we get $\\frac{\\left(x^{2}+y^{2}+z^{2}\\right)^{2}}{3} \\geq 9 \\frac{(x+y+z)^{2}}{9}=(x+y+z)^{2}$. By taking square root on both sides we deduce the stated inequality.\n\nEquality holds only when $x=y=z=\\sqrt{3}$ or $x=y=z=0$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## A4", "solution_match": "\nSolution1."}}
+{"year": "2014", "tier": "T3", "problem_label": "A6", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $a, b, c$ be positive real numbers. Prove that\n\n$$\n\\left(\\left(3 a^{2}+1\\right)^{2}+2\\left(1+\\frac{3}{b}\\right)^{2}\\right)\\left(\\left(3 b^{2}+1\\right)^{2}+2\\left(1+\\frac{3}{c}\\right)^{2}\\right)\\left(\\left(3 c^{2}+1\\right)^{2}+2\\left(1+\\frac{3}{a}\\right)^{2}\\right) \\geq 48^{3}\n$$\n\nWhen does equality hold?", "solution": "Let $x$ be a positive real number. By AM-GM we have $\\frac{1+x+x+x}{4} \\geq x^{\\frac{3}{4}}$, or equivalently $1+3 x \\geq 4 x^{\\frac{3}{4}}$. Using this inequality we obtain:\n\n$$\n\\left(3 a^{2}+1\\right)^{2} \\geq 16 a^{3} \\text { and } 2\\left(1+\\frac{3}{b}\\right)^{2} \\geq 32 b^{-\\frac{3}{2}}\n$$\n\nMoreover, by inequality of arithmetic and geometric means we have\n\n$$\nf(a, b)=\\left(3 a^{2}+1\\right)^{2}+2\\left(1+\\frac{3}{b}\\right)^{2} \\geq 16 a^{3}+32 b^{-\\frac{3}{2}}=16\\left(a^{3}+b^{-\\frac{3}{2}}+b^{-\\frac{3}{2}}\\right) \\geq 48 \\frac{a}{b}\n$$\n\nTherefore, we obtain\n\n$$\nf(a, b) f(b, c) f(c, a) \\geq 48 \\cdot \\frac{a}{b} \\cdot 48 \\cdot \\frac{b}{c} \\cdot 48 \\cdot \\frac{c}{a}=48^{3}\n$$\n\nEquality holds only when $a=b=c=1$.\n\n## 1 IH $^{\\text {th J.M. }} 2014$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## A6", "solution_match": "\nSolution."}}
+{"year": "2014", "tier": "T3", "problem_label": "A8", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $x, y$ and $z$ be positive real numbers such that $x y z=1$. Prove the inequality\n\n$$\n\\frac{1}{x(a y+b)}+\\frac{1}{y(a z+b)}+\\frac{1}{z(a x+b)} \\geq 3 \\text {, if: }\n$$\n\na) $a=0$ and $b=1$;\nb) $a=1$ and $b=0$;\nc) $a+b=1$ for $a, b>0$\n\nWhen does the equality hold true?", "solution": "a) The inequality reduces to $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z} \\geq 3$, which follows directly from the AM-GM inequality.\n\nEquality holds only when $x=y=z=1$.\n\nb) Here the inequality reduces to $\\frac{1}{x y}+\\frac{1}{y z}+\\frac{1}{z x} \\geq 3$, i.e. $x+y+z \\geq 3$, which also follows from the AM-GM inequality.\n\nEquality holds only when $x=y=z=1$.\n\nc) Let $m, n$ and $p$ be such that $x=\\frac{m}{n}, y=\\frac{n}{p}$ и $z=\\frac{p}{m}$. The inequality reduces to\n\n$$\n\\frac{n p}{a m n+b m p}+\\frac{p m}{a n p+b n m}+\\frac{m n}{a p m+b p n} \\geq 3\n$$\n\nBy substituting $u=n p, v=p m$ and $w=m n$, (1) becomes\n\n$$\n\\frac{u}{a w+b v}+\\frac{v}{a u+b w}+\\frac{w}{a v+b u} \\geq 3\n$$\n\nThe last inequality is equivalent to\n\n$$\n\\frac{u^{2}}{a u w+b u v}+\\frac{v^{2}}{a u v+b v w}+\\frac{w^{2}}{a v w+b u w} \\geq 3\n$$\n\nCauchy-Schwarz Inequality implies\n\n$$\n\\frac{u^{2}}{a u w+b u v}+\\frac{v^{2}}{a u v+b v w}+\\frac{w^{2}}{a v w+b u w} \\geq \\frac{(u+v+w)^{2}}{a u w+b u v+a u v+b v w+a v w+b u w}=\\frac{(u+v+w)^{2}}{u w+v u+w v}\n$$\n\nThus, the problem simplifies to $(u+v+w)^{2} \\geq 3(u w+v u+w v)$, which is equivalent to $(u-v)^{2}+(v-w)^{2}+(w-u)^{2} \\geq 0$.\n\nEquality holds only when $u=v=w$, that is only for $x=y=z=1$.\n\nRemark. The problem can be reformulated:\n\nLet $a, b, x, y$ and $z$ be nonnegative real numbers such that $x y z=1$ and $a+b=1$. Prove the inequality\n\n$$\n\\frac{1}{x(a y+b)}+\\frac{1}{y(a z+b)}+\\frac{1}{z(a x+b)} \\geq 3\n$$\n\nWhen does the equality hold true?", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## A8", "solution_match": "\nSolution."}}
+{"year": "2014", "tier": "T3", "problem_label": "A9", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $n$ be a positive integer, and let $x_{1}, \\ldots, x_{n}, y_{1}, \\ldots, y_{n}$ be positive real numbers such that $x_{1}+\\ldots+x_{n}=y_{1}+\\ldots+y_{n}=1$. Show that\n\n$$\n\\left|x_{1}-y_{1}\\right|+\\ldots\\left|x_{n}-y_{n}\\right| \\leq 2-\\min _{1 \\leq i \\leq n} \\frac{x_{i}}{y_{i}}-\\min _{1 \\leq i \\leq n} \\frac{y_{i}}{x_{i}}\n$$", "solution": "Up to reordering the real numbers $x_{i}$ and $y_{i}$, we may assume that $\\frac{x_{1}}{y_{1}} \\leq \\ldots \\leq \\frac{x_{n}}{y_{n}}$. Let $A=\\frac{x_{1}}{y_{1}}$ and $B=\\frac{x_{n}}{y_{n}}$, and $\\mathrm{S}=\\left|x_{1}-y_{1}\\right|+\\ldots\\left|x_{n}-y_{n}\\right|$. Our aim is to prove that $S \\leq 2-A-\\frac{1}{B}$.\n\nFirst, note that we cannot have $A>1$, since that would imply $x_{i}>y_{i}$ for all $i \\leq n$, hence $x_{1}+\\ldots+x_{n}>y_{1}+\\ldots+y_{n}$. Similarly, we cannot have $B<1$, since that would imply $x_{i}<y_{i}$ for all $i \\leq n$, hence $x_{1}+\\ldots+x_{n}<y_{1}+\\ldots+y_{n}$.\n\nIf $n=1$, then $x_{1}=y_{1}=A=B=1$ and $S=0$, hence $S \\leq 2-A-\\frac{1}{B}$. For $n \\geq 2$ let $1 \\leq k<n$ be some integer such that $\\frac{x_{k}}{y_{k}} \\leq 1 \\leq \\frac{x_{k+1}}{y_{k+1}}$. We define the positive real numbers $X_{1}=x_{1}+\\ldots+x_{k}$, $X_{2}=x_{k+1}+\\ldots+x_{n}, Y_{1}=y_{1}+\\ldots+y_{k}, Y_{2}=y_{k+1}+\\ldots+y_{n}$. Note that $Y_{1} \\geq X_{1} \\geq A Y_{1}$ and $Y_{2} \\leq X_{2} \\leq B Y_{2}$. Thus, $A \\leq \\frac{X_{1}}{Y_{1}} \\leq 1 \\leq \\frac{X_{2}}{Y_{2}} \\leq B$. In addition, $S=Y_{1}-X_{1}+X_{2}-Y_{2}$.\n\nFrom $0<X_{2}, Y_{1} \\leq 1,0 \\leq Y_{1}-X_{1}$ and $0 \\leq X_{2}-Y_{2}$, follows\n\n$$\nS=Y_{1}-X_{1}+X_{2}-Y_{2}=\\frac{Y_{1}-X_{1}}{Y_{1}}+\\frac{X_{2}-Y_{2}}{X_{2}}=2-\\frac{X_{1}}{Y_{1}}-\\frac{Y_{2}}{X_{2}} \\leq 2-A-\\frac{1}{B}\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## A9", "solution_match": "\nSolution."}}
+{"year": "2014", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Several (at least two) segments are drawn on a board. Select two of them, and let $a$ and $b$ be their lengths. Delete the selected segments and draw a segment of length $\\frac{a b}{a+b}$. Continue this procedure until only one segment remains on the board. Prove:\n\na) the length of the last remaining segment does not depend on the order of the deletions.\n\nb) for every positive integer $n$, the initial segments on the board can be chosen with distinct integer lengths, such that the last remaining segment has length $n$.", "solution": "a) Observe that $\\frac{1}{\\frac{a b}{a+b}}=\\frac{1}{a}+\\frac{1}{b}$. Thus, if the lengths of the initial segments on the board were $a_{1}, a_{2}, \\ldots, a_{n}$, and $c$ is the length of the last remaining segment, then $\\frac{1}{c}=\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+\\ldots+\\frac{1}{a_{n}}$ , proving a).\n\nb) From a) and the equation $\\frac{1}{n}=\\frac{1}{2 n}+\\frac{1}{3 n}+\\frac{1}{6 n}$ it follows that if the lengths of the starting segments are $2 n, 3 n$ and $6 n$, then the length of the last remaining segment is $n$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## C1", "solution_match": "\nSolution."}}
+{"year": "2014", "tier": "T3", "problem_label": "C2", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "In a country with $n$ cities, all direct airlines are two-way. There are $r>2014$ routes between pairs of different cities that include no more than one intermediate stop (the direction of each route matters). Find the least possible $n$ and the least possible $r$ for that value of $n$.", "solution": "Denote by $X_{1}, X_{2}, \\ldots X_{n}$ the cities in the country and let $X_{i}$ be connected to exactly $m_{i}$ other cities by direct two-way airline. Then $X_{i}$ is a final destination of $m_{i}$ direct routes and an intermediate stop of $m_{i}\\left(m_{i}-1\\right)$ non-direct routes. Thus $r=m_{1}^{2}+\\ldots+m_{n}^{2}$. As each $m_{i}$ is at most $n-1$ and $13 \\cdot 12^{2}<2014$, we deduce $n \\geq 14$.\n\nConsider $n=14$. As each route appears in two opposite directions, $r$ is even, so $r \\geq 2016$. We can achieve $r=2016$ by arranging the 14 cities uniformly on a circle connect (by direct two-way airlines) all of them, except the diametrically opposite pairs. This way, there are exactly $14 \\cdot 12^{2}=2016$ routes.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## C2", "solution_match": "\nSolution."}}
+{"year": "2014", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "For a given positive integer n, two players $A$ and B play the following game: Given is pile of $\\boldsymbol{\\Omega}$ stones. The players take turn alternatively with A going first. On each turn the player is allowed to take one stone, a prime number of stones, or a multiple of $n$ stones. The winner is the one who takes the last stone. Assuming perfect play, find the number of values for $S_{\\infty}$, for which A cannot win.", "solution": "Denote by $k$ the sought number and let $\\left\\{a_{1}, a_{2}, \\ldots, a_{k}\\right\\}$ be the corresponding values for $a$. We will call each $a_{i}$ a losing number and every other positive integer a winning numbers. Clearly every multiple of $n$ is a winning number.\n\nSuppose there are two different losing numbers $a_{i}>a_{j}$, which are congruent modulo $n$. Then, on his first turn of play, the player $A$ may remove $a_{i}-a_{j}$ stones (since $n \\mid a_{i}-a_{j}$ ), leaving a pile with $a_{j}$ stones for B. This is in contradiction with both $a_{i}$ and $a_{j}$ being losing numbers. Therefore there are at most $n-1$ losing numbers, i.e. $k \\leq n-1$.\n\nSuppose there exists an integer $r \\in\\{1,2, \\ldots, n-1\\}$, such that $m n+r$ is a winning number for every $m \\in \\mathbb{N}_{0}$. Let us denote by $u$ the greatest losing number (if $k>0$ ) or 0 (if $k=0$ ), and let $s=\\operatorname{LCM}(2,3, \\ldots, u+n+1)$. Note that all the numbers $s+2, s+3, \\ldots, s+u+n+1$ are composite. Let $m^{\\prime} \\in \\mathbb{N}_{0}$, be such that $s+u+2 \\leq m^{\\prime} n+r \\leq s+u+n+1$. In order for $m^{\\prime} n+r$ to be a winning number, there must exist an integer $p$, which is either one, or prime, or a positive multiple of $n$, such that $m^{\\prime} n+r-p$ is a losing number or 0 , and hence lesser than or equal to $u$. Since $s+2 \\leq m^{\\prime} n+r-u \\leq p \\leq m^{\\prime} n+r \\leq s+u+n+1, p$ must be a composite, hence $p$ is a multiple of $n$ (say $p=q n$ ). But then $m^{\\prime} n+r-p=\\left(m^{\\prime}-q\\right) n+r$ must be a winning number, according to our assumption. This contradicts our assumption that all numbers $m n+r, m \\in \\mathbb{N}_{0}$ are winning.\n\nHence there are exactly $n-1$ losing numbers (one for each residue $r \\in\\{1,2, \\ldots, n-1\\}$ ).", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## C3", "solution_match": "\nSolution."}}
+{"year": "2014", "tier": "T3", "problem_label": "C4", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Let $A=1 \\cdot 4 \\cdot 7 \\cdot \\ldots \\cdot 2014$ be the product of the numbers less or equal to 2014 that give remainder 1 when divided by 3 . Find the last non-zero digit of $A$.", "solution": "Grouping the elements of the product by ten we get:\n\n$$\n\\begin{aligned}\n& (30 k+1)(30 k+4)(30 k+7)(30 k+10)(30 k+13)(30 k+16) \\\\\n& (30 k+19)(30 k+22)(30 k+25)(30 k+28)= \\\\\n& =(30 k+1)(15 k+2)(30 k+7)(120 k+40)(30 k+13)(15 k+8) \\\\\n& (30 k+19)(15 k+11)(120 k+100)(15 k+14)\n\\end{aligned}\n$$\n\n(We divide all even numbers not divisible by five, by two and multiply all numbers divisible by five with four.)\n\nWe denote $P_{k}=(30 k+1)(15 k+2)(30 k+7)(30 k+13)(15 k+8)(30 k+19)(15 k+11)(15 k+14)$. For all the numbers not divisible by five, only the last digit affects the solution, since the power of two in the numbers divisible by five is greater than the power of five. Considering this, for even $k, P_{k}$ ends with the same digit as $1 \\cdot 2 \\cdot 7 \\cdot 3 \\cdot 8 \\cdot 9 \\cdot 1 \\cdot 4$, i.e. six and for odd $k, P_{k}$ ends with the same digit as $1 \\cdot 7 \\cdot 7 \\cdot 3 \\cdot 3 \\cdot 9 \\cdot 6 \\cdot 9$, i.e. six. Thus $P_{0} P_{1} \\ldots P_{66}$ ends with six. If we remove one zero from the end of all numbers divisible with five, we get that the last nonzero digit of the given product is the same as the one from $6 \\cdot 2011 \\cdot 2014 \\cdot 4 \\cdot 10 \\cdot 16 \\cdot \\ldots .796 \\cdot 802$. Considering that $4 \\cdot 6 \\cdot 2 \\cdot 8$ ends with four and removing one zero from every fifth number we get that the last nonzero digit is the same as in $4 \\cdot 4^{26} \\cdot 784 \\cdot 796 \\cdot 802 \\cdot 1 \\cdot 4 \\cdot \\ldots \\cdot 76 \\cdot 79$. Repeating the process we did for the starting sequence we conclude that the last nonzero number will be the same as in $2 \\cdot 6 \\cdot 6 \\cdot 40 \\cdot 100 \\cdot 160 \\cdot 220 \\cdot 280 \\cdot 61 \\cdot 32 \\cdot 67 \\cdot 73 \\cdot 38 \\cdot 79$, which is two.\n\nLet $A B C$ be a triangle with $\\measuredangle B=\\measuredangle C=40^{\\circ}$. The bisector of the $\\measuredangle B$ meets $A C$ at the point $D$ . Prove that $\\overline{B D}+\\overline{D A}=\\overline{B C}$.\n\nSolution. Since $\\measuredangle B A C=100^{\\circ}$ and $\\measuredangle B D C=120^{\\circ}$ we have $\\overline{B D}<\\overline{B C}$. Let $E$ be the point on $\\overline{B C}$ such that $\\overline{B D}=\\overline{B E}$. Then $\\measuredangle D E C=100^{\\circ}$ and $\\measuredangle E D C=40^{\\circ}$, hence $\\overline{D E}=\\overline{E C}$, and $\\measuredangle B A C+\\measuredangle D E B=180^{\\circ}$. So $A, B, E$ and $D$ are concyclic, implying $\\overline{A D}=\\overline{D E}$ (since $\\measuredangle A B D=\\measuredangle D B C=20^{\\circ}$ ), which completes the proof.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## C4", "solution_match": "\nSolution."}}
+{"year": "2014", "tier": "T3", "problem_label": "G2", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C$ be an acute triangle with $\\overline{A B}<\\overline{A C}<\\overline{B C}$ and $c(O, R)$ be its circumcircle. Denote with $D$ and $E$ be the points diametrically opposite to the points $B$ and $C$, respectively. The circle $c_{1}(A, \\overline{A E})$ intersects $\\overline{A C}$ at point $K$, the circle $c_{2}(A, \\overline{A D})$ intersects $B A$ at point $L(A$ lies between $B$ and $L$ ). Prove that the lines $E K$ and $D L$ meet on the circle $c$.", "solution": "Let $\\mathrm{M}$ be the point of intersection of the line $D L$ with the circle $c(O, R)$ (we choose $M \\equiv D$ if $L D$ is tangent to $c$ and $M$ to be the second intersecting point otherwise). It is\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_eb51ab1a35cd98c2706cg-27.jpg?height=778&width=815&top_left_y=1069&top_left_x=359)\nsufficient to prove that the points $E, K$ and $M$ are collinear.\n\nWe have that $\\measuredangle E A C=90^{\\circ}$ (since $E C$ is diameter of the circle $c$ ). The triangle $A E K$ is right-angled and isosceles ( $\\overline{A E}$ and $\\overline{A K}$ are radii of the circle $c_{1}$ ). Therefore\n\n$$\n\\measuredangle \\mathrm{AEK}=\\measuredangle \\mathrm{AKE}=45^{\\circ} \\text {. }\n$$\n\nSimilarly, we obtain that $\\measuredangle B A D=90^{\\circ}=\\measuredangle D A L$. Since $\\overline{A D}=\\overline{A L}$ the triangle $A D L$ is right-angled and isosceles, we have\n\n$$\n\\measuredangle \\mathrm{ADL}=\\measuredangle \\mathrm{A} L D=45^{\\circ} .\n$$\n\nIf $M$ is between $D$ and $L$, then $\\measuredangle \\mathrm{ADM}=\\measuredangle A E M$, because they are inscribed in the circle $c(O, R)$ and they correspond to the same arch $\\overparen{A M}$. Hence $\\measuredangle \\mathrm{AEK}=\\measuredangle \\mathrm{AEM}=45^{\\circ}$ i.e. the points $E, K, M$ are collinear.\n\nIf $D$ is between $M$ and $L$, then $\\measuredangle \\mathrm{ADM}+\\measuredangle A E M=180^{\\circ}$ as opposite angles in cyclic quadrilateral. Hence $\\measuredangle \\mathrm{AEK}=\\measuredangle \\mathrm{AEM}=45^{\\circ}$ i.e. the points $E, K, M$ are collinear.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## G2", "solution_match": "\nSolution."}}
+{"year": "2014", "tier": "T3", "problem_label": "G3", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $C D \\perp A B(D \\in A B), D M \\perp A C(M \\in A C)$ and $D N \\perp B C(N \\in B C)$ for an acute triangle ABC with area $S$. If $H_{1}$ and $H_{2}$ are the orthocentres of the triangles $M N C$ and MND respectively. Evaluate the area of the quadrilateral $\\mathrm{AH}_{1} \\mathrm{BH}_{2}$.", "solution": "Let $O, P, K, R$ and $T$ be the midpoints of the segments $C D, M N, C N, C H_{1}$ and $M H_{1}$, respectively. From $\\triangle M N C$ we have that $\\overline{P K}=\\frac{1}{2} \\overline{M C}$ and $P K \\| M C$. Analogously, from $\\Delta M H_{1} C$ we have that $\\overline{T R}=\\frac{1}{2} \\overline{M C}$ and $T R \\| M C$. Consequently, $\\overline{P K}=\\overline{T R}$ and $P K \\| T R$. Also $O K \\| D N$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_eb51ab1a35cd98c2706cg-28.jpg?height=612&width=789&top_left_y=818&top_left_x=995)\n(from $\\triangle C D N$ ) and since $D N \\perp B C$ and $M H_{1} \\perp B C$, it follows that $T H_{1} \\| O K$. Since $O$ is the circumcenter of $\\triangle C M N, O P \\perp M N$. Thus, $C H_{1} \\perp M N$ implies $O P \\| C H_{1}$. We conclude $\\Delta T R H_{1} \\cong \\triangle K P O$ (they have parallel sides and $\\overline{T R}=\\overline{P K}$ ), hence $\\overline{R H_{1}}=\\overline{P O}$, i.e. $\\overline{C H_{1}}=2 \\overline{P O}$ and $\\mathrm{CH}_{1} \\| \\mathrm{PO}$.\n\nAnalogously, $\\overline{D H_{2}}=2 \\overline{P O}$ and $D H_{2} \\| P O$. From $\\overline{C H_{1}}=2 \\overline{P O}=\\overline{D H_{2}}$ and\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_eb51ab1a35cd98c2706cg-28.jpg?height=75&width=1581&top_left_y=1895&top_left_x=222)\n$H_{1} H_{2} \\| C D$. Therefore the area of the quadrilateral $A H_{1} B H_{2}$ is $\\frac{\\overline{A B} \\cdot \\overline{H_{1} H_{2}}}{2}=\\frac{\\overline{A B} \\cdot \\overline{C D}}{2}=S$.\n\nSolution2. Since $M H_{1} \\| D N$ and $N H_{1} \\| D M, M D N H_{1}$ is a parallelogram. Similarly, $\\mathrm{NH}_{2} \\| C M$ and $\\mathrm{MH}_{2} \\| \\mathrm{CN}$ imply $\\mathrm{MCNH}_{2}$ is a parallelogram. Let $P$ be the midpoint of the segment $\\overline{M N}$. Then $\\sigma_{P}(D)=H_{1}$ and $\\sigma_{P}(C)=H_{2}$, thus $C D \\| H_{1} H_{2}$ and $\\overline{C D}=\\overline{H_{1} H_{2}}$. From $C D \\perp A B$ we deduce $A_{A H_{1} B H_{2}}=\\frac{1}{2} \\overline{A B} \\cdot \\overline{C D}=S$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## G3", "solution_match": "\nSolution1."}}
+{"year": "2014", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C$ be a triangle such that $\\overline{A B} \\neq \\overline{A C}$. Let $M$ be a midpoint of $\\overline{B C}, H$ the orthocenter of $A B C, O_{1}$ the midpoint of $\\overline{A H}$ and $O_{2}$ the circumcenter of $B C H$. Prove that $O_{1} A M O_{2}$ is a parallelogram.", "solution": "Let $O_{2}^{\\prime}$ be the point such that $O_{1} A M O_{2}^{\\prime}$ is a parallelogram. Note that $\\overrightarrow{M O_{2}}=\\overrightarrow{A O_{1}}=\\overrightarrow{O_{1} H}$. Therefore, $O_{1} H O_{2}^{\\prime} M$ is a parallelogram and $\\overrightarrow{M O_{1}}=\\overrightarrow{O_{2} H}$.\n\nSince $M$ is the midpoint of $\\overline{B C}$ and $O_{1}$ is the midpoint of $\\overline{A H}$, it follows that $4 \\overrightarrow{M O_{1}}=\\overrightarrow{B A}+\\overrightarrow{B H}+\\overrightarrow{C A}+\\overrightarrow{C H}=2(\\overrightarrow{C A}+\\overrightarrow{B H})$. Moreover, let $B^{\\prime}$ be the midpoint of $\\overrightarrow{B H}$. Then,\n\n$$\n\\begin{aligned}\n2 \\overrightarrow{O_{2}^{\\prime B}} \\cdot \\overrightarrow{B H} & =\\left(\\overline{O_{2}^{\\prime H}}+\\overline{O_{2}^{\\prime B}}\\right) \\cdot \\overrightarrow{B H}=\\left(2 \\overline{O_{2}^{\\prime} H}+\\overrightarrow{H B}\\right) \\cdot \\overrightarrow{B H}= \\\\\n& =\\left(2 \\overline{M O_{1}}+\\overline{H B}\\right) \\cdot \\overline{B H}=(\\overline{C A}+\\overrightarrow{B H}+\\overrightarrow{H B}) \\cdot \\overrightarrow{B H}=\\overline{C A} \\cdot \\overrightarrow{B H}=0 .\n\\end{aligned}\n$$\n\nBy $\\vec{a} \\cdot \\vec{b}$ we denote the inner product of the vectors $\\vec{a}$ and $\\vec{b}$.\n\nTherefore, $O_{2}^{\\prime}$ lies on the perpendicular bisector of $\\overline{B H}$. Since $B$ and $C$ play symmetric roles, $\\mathrm{O}_{2}^{\\prime}$ also lies on the perpendicular bisector of $\\overline{\\mathrm{CH}}$, hence $\\mathrm{O}_{2}^{\\prime}$ is the circumcenter of $\\triangle B C H$ and $\\mathrm{O}_{2}=\\mathrm{O}_{2}^{\\prime}$.\n\nNote: The condition $\\overline{A B} \\neq \\overline{A C}$ just aims at ensuring that the parallelogram $O_{1} A N O_{2}$ is not degenerate, hence at helping students to focus on the \"general\" case.\n\nSolution2. We use the following two well-known facts:\n\n$\\sigma_{B C}(H)$ lies on the circumcircle of $\\triangle A B C$.\n\n$\\overrightarrow{A H}=-2 \\overrightarrow{M O}$, where $O$ is the circumcenter of $\\triangle A B C$.\n\nThe statement \" $\\mathrm{O}_{1} A M O_{2}$ is parallelogram\" is equivalent to \" $\\sigma_{B C}\\left(O_{2}\\right)=O$ \". The later is true because the circumcircles of $\\triangle A B C$ and $\\triangle B C H$ are symmetrical with respect to $B C$, from (1).", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## G4", "solution_match": "\nSolution1."}}
+{"year": "2014", "tier": "T3", "problem_label": "G5", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C$ be a triangle with $\\overline{A B} \\neq \\overline{B C}$, and let $B D$ be the internal bisector of $\\measuredangle A B C(D \\in A C)$. Denote the midpoint of the arc $A C$ which contains point BbyM. The circumcircle of the triangle $B D M$ intersects the segment $A B$ at point $K \\neq B$, and let $J$ be the reflection of $A$ with respect to $K$. If $D J \\cap A M=\\{O\\}$, prove that the points $J, B, M, O$ belong to the same circle.", "solution": "![](https://cdn.mathpix.com/cropped/2024_06_05_eb51ab1a35cd98c2706cg-30.jpg?height=1337&width=1232&top_left_y=1419&top_left_x=762)\n\nLet the circumcircle of the triangle $B D M$ intersect the line segment $B C$ at point $L \\neq B$. From $\\measuredangle C B D=\\measuredangle D B A$ we have $\\overline{D L}=\\overline{D K}$. Since $\\measuredangle L C M=\\measuredangle B C M=\\measuredangle B A M=\\measuredangle K A M, \\overline{M C}=\\overline{M A}$ and\n\n$$\n\\measuredangle L M C=\\measuredangle L M K-\\measuredangle C M K=\\measuredangle L B K-\\measuredangle C M K=\\measuredangle C B A-\\measuredangle C M K=\\measuredangle C M A-\\measuredangle C M K=\\measuredangle K M A,\n$$\n\nit follows that triangles $M L C$ and $M K A$ are congruent, which implies $\\overline{C L}=\\overline{A K}=\\overline{K J}$. Furthermore, $\\measuredangle C L D=180^{\\circ}-\\measuredangle B L D=\\measuredangle D K B=\\measuredangle D K J$ and $\\overline{D L}=\\overline{D K}$, it follows that triangles $D C L$ and $D J K$ are congruent. Hence, $\\angle D C L=\\angle D J K=\\measuredangle B J O$. Then\n\n$$\n\\measuredangle B J O+\\measuredangle B M O=\\angle D C L+\\angle B M A=\\angle B C A+180^{\\circ}-\\angle B C A=180^{\\circ}\n$$\n\nso the points $J, B, M, O$ belong to the same circle, q.e.d.\n\n## Solution2.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_eb51ab1a35cd98c2706cg-31.jpg?height=938&width=947&top_left_y=452&top_left_x=676)\n\nSince $\\overline{M C}=\\overline{M A}$ and $\\measuredangle C M A=\\measuredangle C B A$, we have $\\measuredangle A C M=\\measuredangle C A M=90^{\\circ}-\\frac{\\measuredangle C B A}{2}$. It follows that $\\measuredangle M B D=\\measuredangle M B A+\\measuredangle A B D=\\measuredangle A C M+\\measuredangle A B D=90^{\\circ}-\\frac{\\measuredangle C B A}{2}+\\frac{\\measuredangle C B A}{2}=90^{\\circ}$. Denote the midpoint of $\\overline{A C}$ by $N$. Since $\\measuredangle D N M=\\measuredangle C N M=90^{\\circ}, N$ belongs to the circumcircle of the triangle $B D M$. Since $N K$ is the midline of the triangle $A C J$ and $N K \\| C J$, we have\n\n$$\n\\measuredangle B J C=\\measuredangle B K N=180^{\\circ}-\\measuredangle N D B=\\measuredangle C D B\n$$\n\nHence, the quadrilateral $C D J B$ is cyclic (this can also be obtained from the power of a point theorem, because $\\overline{A N} \\cdot \\overline{A D}=\\overline{A K} \\cdot \\overline{A B}$ implies $\\overline{A C} \\cdot \\overline{A D}=\\overline{A J} \\cdot \\overline{A B}$ ), and\n\n$$\n\\measuredangle B J O=\\measuredangle 180^{\\circ}-\\measuredangle B J D=\\angle B C D=\\angle B C A=180^{\\circ}-\\angle B M A=180^{\\circ}-\\measuredangle B M O\n$$\n\nso the points $J, B, M, O$ belong to the same circle, q.e.d.\n\nRemark. If $J$ is between $A$ and $K$ the solution can be easily adapted.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## G5", "solution_match": "## Solution1."}}
+{"year": "2014", "tier": "T3", "problem_label": "G6", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C D$ be a quadrilateral whose sides $A B$ and $C D$ are not parallel, and let $O$ be the intersection of its diagonals. Denote with $H_{1}$ and $H_{2}$ the orthocenters of the triangles $O A B$ and OCD, respectively. If $M$ and $N$ are the midpoints of the segments $\\overline{A B}$ and $\\overline{C D}$, respectively, prove that the lines $M N$ and $\\mathrm{H}_{1} \\mathrm{H}_{2}$ are parallel if and only if $\\overline{A C}=\\overline{B D}$.", "solution": "![](https://cdn.mathpix.com/cropped/2024_06_05_eb51ab1a35cd98c2706cg-32.jpg?height=1378&width=1735&top_left_y=1403&top_left_x=461)\n\nLet $A^{\\prime}$ and $B^{\\prime}$ be the feet of the altitudes drawn from $A$ and $B$ respectively in the triangle $A O B$, and $C^{\\prime}$ and $D^{\\prime}$ are the feet of the altitudes drawn from $C$ and $D$ in the triangle $C O D$. Obviously, $A^{\\prime}$ and $D^{\\prime}$ belong to the circle $c_{1}$ of diameter $\\overline{A D}$, while $B^{\\prime}$ and $C^{\\prime}$ belong to the circle $c_{2}$ of diameter $\\overline{B C}$.\n\nIt is easy to see that triangles $H_{1} A B$ and $H_{1} B^{\\prime} A^{\\prime}$ are similar. It follows that $\\overline{H_{1} A} \\cdot \\overline{H_{1} A^{\\prime}}=\\overline{H_{1} B} \\cdot \\overline{H_{1} B^{\\prime}}$. (Alternatively, one could notice that the quadrilateral $A B A^{\\prime} B^{\\prime}$ is cyclic and obtain the previous relation by writing the power of $H_{1}$ with respect to its circumcircle.) It\nfollows that $H_{1}$ has the same power with respect to circles $c_{1}$ and $c_{2}$. Thus, $H_{1}$ (and similarly, $\\mathrm{H}_{2}$ ) is on the radical axis of the two circles.\n\nThe radical axis being perpendicular to the line joining the centers of the two circles, one concludes that $\\mathrm{H}_{1} \\mathrm{H}_{2}$ is perpendicular to $P Q$, where $P$ and $Q$ are the midpoints of the sides $\\overline{A D}$ and $\\overline{B C}$, respectively. ( $P$ and $Q$ are the centers of circles $c_{1}$ and $c_{2}$.)\n\nThe condition $H_{1} H_{2} \\| M N$ is equivalent to $M N \\perp P Q$. As $M P N Q$ is a parallelogram, we conclude that $H_{1} H_{2} \\| M N \\Leftrightarrow M N \\perp P Q \\Leftrightarrow M P N Q$ a rhombus $\\Leftrightarrow \\overline{M P}=\\overline{M Q} \\Leftrightarrow \\overline{A C}=\\overline{B D}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## G6", "solution_match": "## Solution."}}
+{"year": "2014", "tier": "T3", "problem_label": "N1", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Each letter of the word OHRID corresponds to a different digit belonging to the set $(1,2,3,4,5)$. Decipher the equality $(O+H+R+I+D)^{2}:(O-H-R+I+D)=O^{H^{R^{I_{D}^{D}}}}$.", "solution": "Since $O, H, R, I$ and $D$ are distinct numbers from $\\{1,2,3,4,5\\}$, we have $O+H+R+I+D=15$ and $O-H-R+I+D=O+H+R+I+D-2(H+R)<15$. From this $O^{H^{R^{I^{D}}}}=\\frac{(O+H+R+I+D)^{2}}{O-H-R+I+D}=\\frac{225}{15-2(H+R)}$, hence $O^{H^{R^{R^{D}}}}>15$ and divides 225 , which is only possible for $O^{H^{R^{I^{D}}}}=25$ (must be a power of three or five). This implies that $O=5, H=2$ and $R=1$. It's easy to check that both $I=3, D=4$ and $I=4, D=3$ satisfy the stated equation.\n\n## $\\mathbf{N} 2$\n\nFind all triples $(p, q, r)$ of distinct primes $p, q$ and $r$ such that\n\n$$\n3 p^{4}-5 q^{4}-4 r^{2}=26\n$$\n\nSolution. First notice that if both primes $q$ and $r$ differ from 3 , then $q^{2} \\equiv r^{2} \\equiv 1(\\bmod 3)$, hence the left hand side of the given equation is congruent to zero modulo 3 , which is impossible since 26 is not divisible by 3 . Thus, $q=3$ or $r=3$. We consider two cases.\n\nCase 1. $q=3$.\n\nThe equation reduces to $3 p^{4}-4 r^{2}=431$\n\nIf $p \\neq 5$, by Fermat's little theorem, $p^{4} \\equiv 1(\\bmod 5)$, which yields $3-4 r^{2} \\equiv 1(\\bmod 5)$, or equivalently, $r^{2}+2 \\equiv 0(\\bmod 5)$. The last congruence is impossible in view of the fact that a residue of a square of a positive integer belongs to the set $\\{0,1,4\\}$. Therefore $p=5$ and $r=19$.\n\nCase 2. $r=3$.\n\nThe equation becomes $3 p^{4}-5 q^{4}=62(2)$.\n\nObviously $p \\neq 5$. Hence, Fermat's little theorem gives $p^{4} \\equiv 1(\\bmod 5)$. But then $5 q^{4} \\equiv 1(\\bmod 5)$, which is impossible.\n\nHence, the only solution of the given equation is $p=5, q=3, r=19$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## N1", "solution_match": "\nSolution."}}
+{"year": "2014", "tier": "T3", "problem_label": "N3", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find the integer solutions of the equation\n\n$$\nx^{2}=y^{2}\\left(x+y^{4}+2 y^{2}\\right)\n$$", "solution": "If $x=0$, then $y=0$ and conversely, if $y=0$, then $x=0$. It follows that $(x, y)=(0,0)$ is a solution of the problem. Assume $x \\neq 0$ and $y \\neq 0$ satisfy the equation. The equation can be transformed in the form $x^{2}-x y^{2}=y^{6}+2 y^{4}$. Then $4 x^{2}-4 x y^{2}+y^{4}=4 y^{6}+9 y^{4}$ and consequently $\\left(\\frac{2 x}{y^{2}}-1\\right)^{2}=4 y^{2}+9 \\quad$ (1). Obviously $\\frac{2 x}{y^{2}}-1$ is integer. From (1), we get that the numbers $\\frac{2 x}{y^{2}}-1,2 y$ and 3 are Pythagorean triplets. It follows that $\\frac{2 x}{y^{2}}-1= \\pm 5$ and $2 y= \\pm 4$. Therefore, $x=3 y^{2}$ or $x=-2 y^{2}$ and $y= \\pm 2$. Hence $(x, y)=(12,-2),(x, y)=(12,2),(x, y)=(-8,-2)$ and $(x, y)=(-8,2)$ are the possible solutions. By substituting them in the initial equation we verify that all the 4 pairs are solution. Thus, together with the couple $(x, y)=(0,0)$ the problem has 5 solutions.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "\nN3", "solution_match": "\nSolution."}}
+{"year": "2014", "tier": "T3", "problem_label": "N4", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Prove there are no integers $a$ and $b$ satisfying the following conditions:\ni) $16 a-9 b$ is a prime number\n\nii) $\\quad a b$ is a perfect square\n\niii) $a+b$ is a perfect square", "solution": "Suppose $a$ and $b$ be integers satisfying the given conditions. Let $p$ be a prime number, $n$ and $m$ be integers. Then we can write the conditions as follows:\n\n$$\n\\begin{aligned}\n& 16 a-9 b=p \\\\\n& a b=n^{2} \\\\\n& a+b=m^{2}\n\\end{aligned}\n$$\n\nMoreover, let $d=g d c(a, b)$ and $a=d x, b=d y$ for some relatively prime integers $x$ and $y$. Obviously $a \\neq 0$ and $b \\neq 0, a$ and $b$ are positive (by (2) and (3)).\n\nFrom (2) follows that $x$ and $y$ are perfect squares, say $x=l^{2}$ and $y=s^{2}$.\n\nFrom (1), $d \\mid p$ and hence $d=p$ or $d=1$. If $d=p$, then $16 x-9 y=1$, and we obtain $x=9 k+4$, $y=16 k+7$ for some nonnegative integer $k$. But then $s^{2}=y \\equiv 3(\\bmod 4)$, which is a contradiction.\n\nIf $d=1$ then $16 l^{2}-9 s^{2}=p \\Rightarrow(4 l-3 s)(4 l+3 s)=p \\Rightarrow(4 l+3 s=p \\wedge 4 l-3 s=1)$.\n\nBy adding the last two equations we get $8 l=p+1$ and by subtracting them we get $6 s=p-1$. Therefore $p=24 t+7$ for some integer $t$ and $a=(3 t+1)^{2}$ and $b=(4 t+1)^{2}$ satisfy the conditions (1) and (2). By (3) we have $m^{2}=(3 t+1)^{2}+(4 t+1)^{2}=25 t^{2}+14 t+2$, or equivalently $25 m^{2}=(25 t+7)^{2}+1$.\n\nSince the difference between two nonzero perfect square cannot be 1 , we have a contradiction. As a result there is no solution.\n\n## NS\n\nFind all nonnegative integers $x, y, z$ such that\n\n$$\n2013^{x}+2014^{y}=2015^{z}\n$$\n\nSolution. Clearly, $y>0$, and $z>0$. If $x=0$ and $y=1$, then $z=1$ and $(x, y, z)=(0,1,1)$ is a solution. If $x=0$ and $y \\geq 2$, then modulo 4 we have $1+0 \\equiv(-1)^{z}$, hence $z$ is even $\\left(z=2 z_{1}\\right.$ for some integer $\\left.z_{1}\\right)$. Then $2^{y} 1007^{y}=\\left(2015^{z_{1}}-1\\right)\\left(2015^{z_{1}}+1\\right)$, and since $\\operatorname{gcd}\\left(1007,2015^{z_{1}}+1\\right)=1$ we obtain $2 \\cdot 1007^{y} \\mid 2015^{z_{1}}-1$ and $2015^{z_{1}}+1 \\mid 2^{y-1}$. From this we get $2015^{z_{1}}+1 \\leq 2^{y-1}<2 \\cdot 1007^{y} \\leq 2015^{z_{1}}-1$, which is impossible.\n\nNow for $x>0$, modulo 3 we get $0+1 \\equiv(-1)^{z}$, hence $z$ must be even $\\left(z=2 z_{1}\\right.$ for some integer $z_{1}$ ). Modulo 2014 we get $(-1)^{x}+0 \\equiv 1$, thus $x$ must be even $\\left(x=2 x_{1}\\right.$ for some integer $\\left.x_{1}\\right)$. We transform the equation to $2^{y} 1007^{y}=\\left(2015^{z_{1}}-2013^{x_{1}}\\right)\\left(2015^{z_{1}}+2013^{x_{1}}\\right)$ and since $\\operatorname{gcd}\\left(2015^{z_{1}}-2013^{x_{1}}, 2015^{z_{1}}+2013^{x_{1}}\\right)=2,1007^{y}$ divides $2015^{z_{1}}-2013^{x_{1}}$ or $2015^{z_{1}}+2013^{x_{1}}$ but not both. If $1007^{y} \\mid 2015^{z_{1}}-2013^{x_{1}}$, then $2015^{z_{1}}+2013^{x_{1}} \\leq 2^{y}<1007^{y} \\leq 2015^{z_{1}}-2013^{x_{1}}$, which is impossible. Hence $1007 \\mid 2015^{z_{1}}+2013^{x_{1}}$, and from $2015^{z_{1}}+2013^{x_{1}} \\equiv 1+(-1)^{x_{1}}(\\bmod 1007), x_{1}$ is odd $\\left(x=2 x_{1}=4 x_{2}+2\\right.$ for some integer $\\left.x_{2}\\right)$.\n\nNow modulo 5 we get $-1+(-1)^{y} \\equiv(-2)^{4 x_{2}+2}+(-1)^{y} \\equiv 0$, hence $y$ must be even $\\left(y=2 y_{1}\\right.$ for some integer $y_{1}$ ). Finally modulo 31 , we have $(-2)^{4 x_{2}+2}+(-1)^{2 y_{1}} \\equiv 0$ or $4^{2 x_{2}+1} \\equiv-1$. This is impossible since the reminders of the powers of 4 modulo 31 are 1, 2, 4, 8 and 16 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "\nN4", "solution_match": "\nSolution."}}
+{"year": "2014", "tier": "T3", "problem_label": "N6", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Vukasin, Dimitrije, Dusan, Stefan and Filip asked their professor to guess a three consecutive positive integer numbers after they had told him these (true) sentences:\n\nVukasin: \"Sum of the digits of one of them is a prime number. Sum of the digits of some of the other two is an even perfect number ( $n$ is perfect if $\\sigma(n)=2 n$ ). Sum of the digits of the remaining number is equal to the number of its positive divisors.\"\n\nDimitrije:\"Each of these three numbers has no more than two digits 1 in its decimal representation.\"\n\nDusan:\"If we add 11 to one of them, we obtain a square of an integer.\"\n\nStefan:\"Each of them has exactly one prime divisor less then 10.\"\n\nFilip:\"The 3 numbers are square-free.\"\n\nTheir professor gave the correct answer. Which numbers did he say?", "solution": "Let the middle number be $n$, so the numbers are $n-1, n$ and $n+1$. Since 4 does not divide any of them, $n \\equiv 2(\\bmod 4)$. Furthermore, neither 3,5 nor 7 divides $n$. Also $n+1+11 \\equiv 2(\\bmod 4)$ cannot be a square. Then 3 must divide $n-1$ or $n+1$. If $n-1+11$ is a square, then $3 \\mid n+1$ which implies $3 \\mid n+10$ (a square), so $9 \\mid n+10$ hence $9 \\mid n+1$, which is impossible. Thus must be $n+11=m^{2}$.\n\nFurther, 7 does not divide $n-1$, nor $n+1$, because $1+11 \\equiv 5(\\bmod 7)$ and $-1+11 \\equiv 3(\\bmod 7)$ are quadratic nonresidues modulo 7. This implies $5 \\mid n-1$ or $5 \\mid n+1$. Again, since $n+11$ is a square, it is impossible $5 \\mid n-1$, hence $5 \\mid n+1$ which implies $3 \\mid n-1$. This yields $n \\equiv 4(\\bmod 10)$ hence $S(n+1)=S(n)+1=S(n-1)+2(S(n)$ is sum of the digits of $n)$. Since the three numbers are square-free, their numbers of positive divisors are powers of 2 . Thus, we have two even sums of digits - they must be $S(n-1)$ and $S(n+1)$, so $S(n)$ is prime. From $3 \\mid n-1$, follows $S(n-1)$ is an even perfect number, and $S(n+1)=2^{p}$. Consequently $S(n)=2^{p}-1$ is a prime, so $p$ is a prime number. One easily verifies $p \\neq 2$, so $\\mathrm{p}$ is odd implying $3 \\mid 2^{p}-2$. Then\n$\\sigma\\left(2^{p}-2\\right) \\geq\\left(2^{p}-2\\right)\\left(1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{6}\\right)=2\\left(2^{p}-2\\right)$. Since this number is perfect, $\\frac{2^{p}-2}{6}$ must be one, i.e. $p=3$ and $S(n-1)=6, S(n)=7$ and $S(n+1)=8$.\n\nSince 4 does not divide $n$, the 2 -digit ending of $n$ must be 14 or 34 . But $n=34$ is impossible, since $n+11=45$ is not a square. Hence, $n=10^{a}+10^{b}+14$ with $a \\geq b \\geq 2$. If $a \\neq b$, then $n$ has three digits 1 in its decimal representation, which is impossible. Therefore $a=b$, and $n=2 \\cdot 10^{a}+14$. Now, $2 \\cdot 10^{a}+25=m^{2}$, hence $5 \\mid m$, say $m=5 t$, and $(t-1)(t+1)=2^{a+1} 5^{a-2}$. Because $\\operatorname{gcd}(t-1, t+1)=2$ there are three possibilities:\n\n1) $t-1=2, t+1=2^{a} 5^{a-2}$, which implies $a=2, t=3$;\n2) $t-1=2^{a}, t+1=2 \\cdot 5^{a-2}$, so $2^{a}+2=2 \\cdot 5^{a-2}$, which implies $a=3, t=9$;\n3) $t-1=2 \\cdot 5^{a-2}, t+1=2^{a}$, so $2 \\cdot 5^{a-2}+2=2^{a}$, which implies $a=2, t=3$, same as case 1 ).\n\nFrom the only two possibilities $(n-1, n, n+1)=(213,214,215)$ and $(n-1, n, n+1)=$ $(2013,2014,2015)$ the first one is not possible, because $S(215)=8$ and $\\tau(215)=4$. By checking the conditions, we conclude that the latter is a solution, so the professor said the numbers: 2013, 2014, 2015.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl", "problem_match": "## N6", "solution_match": "\nSolution."}}

JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl

@@ -1,16 +1,16 @@
-{"year": "2017", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $a, b, c$ be positive real numbers such that $a+b+c+a b+b c+c a+a b c=7$. Prove that\n\n$$\n\\sqrt{a^{2}+b^{2}+2}+\\sqrt{b^{2}+c^{2}+2}+\\sqrt{c^{2}+a^{2}+2} \\geq 6\n$$", "solution": "First we see that $x^{2}+y^{2}+1 \\geq x y+x+y$. Indeed, this is equivalent to\n\n$$\n(x-y)^{2}+(x-1)^{2}+(y-1)^{2} \\geq 0\n$$\n\nTherefore\n\n$$\n\\begin{aligned}\n& \\sqrt{a^{2}+b^{2}+2}+\\sqrt{b^{2}+c^{2}+2}+\\sqrt{c^{2}+a^{2}+2} \\\\\n\\geq & \\sqrt{a b+a+b+1}+\\sqrt{b c+b+c+1}+\\sqrt{c a+c+a+1} \\\\\n= & \\sqrt{(a+1)(b+1)}+\\sqrt{(b+1)(a+1)}+\\sqrt{(c+1)(a+1)}\n\\end{aligned}\n$$\n\nIt follows from the AM-GM inequality that\n\n$$\n\\begin{aligned}\n& \\sqrt{(a+1)(b+1)}+\\sqrt{(b+1)(a+1)}+\\sqrt{(c+1)(a+1)} \\\\\n\\geq & 3 \\sqrt[3]{\\sqrt{(a+1)(b+1)} \\cdot \\sqrt{(b+1)(a+1)} \\cdot \\sqrt{(c+1)(a+1)}} \\\\\n= & 3 \\sqrt[3]{(a+1)(b+1)(c+1)}\n\\end{aligned}\n$$\n\nOn the other hand, the given condition is equivalent to $(a+1)(b+1)(c+1)=8$ and we get the desired inequality.\n\nObviously, equality is attained if and only if $a=b=c=1$.\n\nRemark. The condition of positivity of $a, b, c$ is superfluous and the equality $\\cdots=7$ can be replaced by the inequality $\\cdots \\geq 7$. Indeed, the above proof and the triangle inequality imply that\n\n$$\n\\begin{aligned}\n\\sqrt{a^{2}+b^{2}+2}+\\sqrt{b^{2}+c^{2}+2}+\\sqrt{c^{2}+a^{2}+2} & \\geq 3 \\sqrt[3]{(|a|+1)(|b|+1)(|c|+1)} \\\\\n& \\geq 3 \\sqrt[3]{|a+1| \\cdot|b+1| \\cdot|c+1|} \\geq 6\n\\end{aligned}\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nA1.", "solution_match": "\nSolution."}}
-{"year": "2017", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $a$ and $b$ be positive real numbers such that $3 a^{2}+2 b^{2}=3 a+2 b$. Find the minimum value of\n\n$$\nA=\\sqrt{\\frac{a}{b(3 a+2)}}+\\sqrt{\\frac{b}{a(2 b+3)}}\n$$", "solution": "By the Cauchy-Schwarz inequality we have that\n\n$$\n5\\left(3 a^{2}+2 b^{2}\\right)=5\\left(a^{2}+a^{2}+a^{2}+b^{2}+b^{2}\\right) \\geq(3 a+2 b)^{2}\n$$\n\n(or use that the last inequality is equivalent to $(a-b)^{2} \\geq 0$ ).\n\nSo, with the help of the given condition we get that $3 a+2 b \\leq 5$. Now, by the AM-GM inequality we have that\n\n$$\nA \\geq 2 \\sqrt{\\sqrt{\\frac{a}{b(3 a+2)}} \\cdot \\sqrt{\\frac{b}{a(2 b+3)}}}=\\frac{2}{\\sqrt[4]{(3 a+2)(2 b+3)}}\n$$\n\nFinally, using again the AM-GM inequality, we get that\n\n$$\n(3 a+2)(2 b+3) \\leq\\left(\\frac{3 a+2 b+5}{2}\\right)^{2} \\leq 25\n$$\n\nso $A \\geq 2 / \\sqrt{5}$ and the equality holds if and only if $a=b=1$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nA2.", "solution_match": "\nSolution."}}
-{"year": "2017", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $a, b, c, d$ be real numbers such that $0 \\leq a \\leq b \\leq c \\leq d$. Prove the inequality\n\n$$\na b^{3}+b c^{3}+c d^{3}+d a^{3} \\geq a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}\n$$", "solution": "The inequality is equivalent to\n\n$$\n\\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\\right)^{2} \\geq\\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}\\right)^{2}\n$$\n\nBy the Cauchy-Schwarz inequality,\n\n$$\n\\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\\right)\\left(a^{3} b+b^{3} c+c^{3} d+d^{3} a\\right) \\geq\\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}\\right)^{2}\n$$\n\nHence it is sufficient to prove that\n\n$$\n\\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\\right)^{2} \\geq\\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\\right)\\left(a^{3} b+b^{3} c+c^{3} d+d^{3} a\\right)\n$$\n\ni.e. to prove $a b^{3}+b c^{3}+c d^{3}+d a^{3} \\geq a^{3} b+b^{3} c+c^{3} d+d^{3} a$.\n\nThis inequality can be written successively\n\n$$\na\\left(b^{3}-d^{3}\\right)+b\\left(c^{3}-a^{3}\\right)+c\\left(d^{3}-b^{3}\\right)+d\\left(a^{3}-c^{3}\\right) \\geq 0\n$$\n\nor\n\n$$\n(a-c)\\left(b^{3}-d^{3}\\right)-(b-d)\\left(a^{3}-c^{3}\\right) \\geq 0\n$$\n\nwhich comes down to\n\n$$\n(a-c)(b-d)\\left(b^{2}+b d+d^{2}-a^{2}-a c-c^{2}\\right) \\geq 0\n$$\n\nThe last inequality is true because $a-c \\leq 0, b-d \\leq 0$, and $\\left(b^{2}-a^{2}\\right)+(b d-a c)+\\left(d^{2}-c^{2}\\right) \\geq 0$ as a sum of three non-negative numbers.\n\nThe last inequality is satisfied with equality whence $a=b$ and $c=d$. Combining this with the equality cases in the Cauchy-Schwarz inequality we obtain the equality cases for the initial inequality: $a=b=c=d$.\n\nRemark. Instead of using the Cauchy-Schwarz inequality, once the inequality $a b^{3}+b c^{3}+c d^{3}+$ $d a^{3} \\geq a^{3} b+b^{3} c+c^{3} d+d^{3} a$ is established, we have $2\\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\\right) \\geq\\left(a b^{3}+b c^{3}+c d^{3}+\\right.$ $\\left.d a^{3}\\right)+\\left(a^{3} b+b^{3} c+c^{3} d+d^{3} a\\right)=\\left(a b^{3}+a^{3} b\\right)+\\left(b c^{3}+b^{3} c\\right)+\\left(c d^{3}+c^{3} d\\right)+\\left(d a^{3}+d^{3} a\\right) \\stackrel{A M-G M}{\\geq}$ $2 a^{2} b^{2}+2 b^{2} c^{2}+2 c^{2} d^{2}+2 d^{2} a^{2}$ which gives the conclusion.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nA3.", "solution_match": "\nSolution."}}
-{"year": "2017", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $x, y, z$ be three distinct positive integers. Prove that\n\n$$\n(x+y+z)(x y+y z+z x-2) \\geq 9 x y z\n$$\n\nWhen does the equality hold?", "solution": "Since $x, y, z$ are distinct positive integers, the required inequality is symmetric and WLOG we can suppose that $x \\geq y+1 \\geq z+2$. We consider 2 possible cases:\n\nCase 1. $y \\geq z+2$. Since $x \\geq y+1 \\geq z+3$ it follows that\n\n$$\n(x-y)^{2} \\geq 1, \\quad(y-z)^{2} \\geq 4, \\quad(x-z)^{2} \\geq 9\n$$\n\nwhich are equivalent to\n\n$$\nx^{2}+y^{2} \\geq 2 x y+1, \\quad y^{2}+z^{2} \\geq 2 y z+4, \\quad x^{2}+z^{2} \\geq 2 x z+9\n$$\n\nor otherwise\n\n$$\nz x^{2}+z y^{2} \\geq 2 x y z+z, \\quad x y^{2}+x z^{2} \\geq 2 x y z+4 x, \\quad y x^{2}+y z^{2} \\geq 2 x y z+9 y\n$$\n\nAdding up the last three inequalities we have\n\n$$\nx y(x+y)+y z(y+z)+z x(z+x) \\geq 6 x y z+4 x+9 y+z\n$$\n\nwhich implies that $(x+y+z)(x y+y z+z x-2) \\geq 9 x y z+2 x+7 y-z$.\n\nSince $x \\geq z+3$ it follows that $2 x+7 y-z \\geq 0$ and our inequality follows.\n\nCase 2. $y=z+1$. Since $x \\geq y+1=z+2$ it follows that $x \\geq z+2$, and replacing $y=z+1$ in the required inequality we have to prove\n\n$$\n(x+z+1+z)(x(z+1)+(z+1) z+z x-2) \\geq 9 x(z+1) z\n$$\n\nwhich is equivalent to\n\n$$\n(x+2 z+1)\\left(z^{2}+2 z x+z+x-2\\right)-9 x(z+1) z \\geq 0\n$$\n\nDoing easy algebraic manipulations, this is equivalent to prove\n\n$$\n(x-z-2)(x-z+1)(2 z+1) \\geq 0\n$$\n\nwhich is satisfied since $x \\geq z+2$.\n\nThe equality is achieved only in the Case 2 for $x=z+2$, so we have equality when $(x, y, z)=$ $(k+2, k+1, k)$ and all the permutations for any positive integer $k$.\n\n## Combinatorics", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nA4.", "solution_match": "\nSolution."}}
-{"year": "2017", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Consider a regular $2 n+1$-gon $P$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$, if the line segment $S E$ contains no other points that lie on the sides of $P$ except $S$. We want to color the sides of $P$ in 3 colors, such that every side is colored in exactly one color, and each color must be used at least once. Moreover, from every point in the plane external to $P$, at most 2 different colors on $P$ can be seen (ignore the vertices of $P$, we consider them colorless). Find the largest positive integer for which such a coloring is possible.", "solution": "Answer: $n=1$ is clearly a solution, we can just color each side of the equilateral triangle in a different color, and the conditions are satisfied. We prove there is no larger $n$ that fulfills the requirements.\n\nLemma 1. Given a regular $2 n+1$-gon in the plane, and a sequence of $n+1$ consecutive sides $s_{1}, s_{2}, \\ldots, s_{n+1}$ there is an external point $Q$ in the plane, such that the color of each $s_{i}$ can be seen from $Q$, for $i=1,2, \\ldots, n+1$.\n\nProof. It is obvious that for a semi-circle $S$, there is a point $R$ in the plane far enough on the perpendicular bisector of the diameter of $S$ such that almost the entire semi-circle can be seen from $R$.\n\nNow, it is clear that looking at the circumscribed circle around the $2 n+1$-gon, there is a semi-circle $S$ such that each $s_{i}$ either has both endpoints on it, or has an endpoint that is on the semi-circle, and is not on the semicircle's end. So, take $Q$ to be a point in the plane from which almost all of $S$ can be seen, clearly, the color of each $s_{i}$ can be seen from $Q$. $\\diamond$ Take $n \\geq 2$, denote the sides $a_{1}, a_{2}, \\ldots, a_{2 n+1}$ in that order, and suppose we have a coloring that satisfies the condition of the problem. Let's call the 3 colors red, green and blue. We must have 2 adjacent sides of different colors, say $a_{1}$ is red and $a_{2}$ is green. Then, by Lemma 1 :\n\n(i) We cannot have a blue side among $a_{1}, a_{2}, \\ldots, a_{n+1}$.\n(ii) We cannot have a blue side among $a_{2}, a_{1}, a_{2 n+1}, \\ldots, a_{n+3}$.\n\nWe are required to have at least one blue side, and according to 1 ) and 2), that can only be $a_{n+2}$, so $a_{n+2}$ is blue. Now, applying Lemma 1 on the sequence of sides $a_{2}, a_{3}, \\ldots, a_{n+2}$ we get that $a_{2}, a_{3}, \\ldots, a_{n+1}$ are all green. Applying Lemma 1 on the sequence of sides $a_{1}, a_{2 n+1}, a_{2 n}, \\ldots, a_{n+2}$ we get that $a_{2 n+1}, a_{2 n}, \\ldots, a_{n+3}$ are all red.\n\nTherefore $a_{n+1}, a_{n+2}$ and $a_{n+3}$ are all of different colors, and for $n \\geq 2$ they can all be seen from the same point according to Lemma 1 , so we have a contradiction.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nC1.", "solution_match": "\nSolution."}}
-{"year": "2017", "tier": "T3", "problem_label": "C2", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Consider a regular $2 n$-gon $P$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$, if the line segment $S E$ contains no other points that lie on the sides of $P$ except $S$. We want to color the sides of $P$ in 3 colors, such that every side is colored in exactly one color, and each color must be used at least once. Moreover, from every point in the plane external to $P$, at most 2 different colors on $P$ can be seen (ignore the vertices of $P$, we consider them colorless). Find the number of distinct such colorings of $P$ (two colorings are considered distinct if at least one side is colored differently).", "solution": "Answer: For $n=2$, the answer is 36 ; for $n=3$, the answer is 30 and for $n \\geq 4$, the answer is $6 n$.\n\nLemma 1. Given a regular $2 n$-gon in the plane and a sequence of $n$ consecutive sides $s_{1}, s_{2}, \\ldots, s_{n}$ there is an external point $Q$ in the plane, such that the color of each $s_{i}$ can be seen from $Q$, for $i=1,2, \\ldots, n$.\n\nProof. It is obvious that for a semi-circle $S$, there is a point $R$ in the plane far enough on the bisector of its diameter such that almost the entire semi-circle can be seen from $R$.\n\nNow, it is clear that looking at the circumscribed circle around the $2 n$-gon, there is a semi-circle $S$ such that each $s_{i}$ either has both endpoints on it, or has an endpoint that's on the semi-circle, and is not on the semi-circle's end. So, take $Q$ to be a point in the plane from which almost all of $S$ can be seen, clearly, the color of each $s_{i}$ can be seen from $Q$.\n\nLemma 2. Given a regular $2 n$-gon in the plane, and a sequence of $n+1$ consecutive sides $s_{1}, s_{2}, \\ldots, s_{n+1}$ there is no external point $Q$ in the plane, such that the color of each $s_{i}$ can be seen from $Q$, for $i=1,2, \\ldots, n+1$.\n\nProof. Since $s_{1}$ and $s_{n+1}$ are parallel opposite sides of the $2 n$-gon, they cannot be seen at the same time from an external point.\n\nFor $n=2$, we have a square, so all we have to do is make sure each color is used. Two sides will be of the same color, and we have to choose which are these 2 sides, and then assign colors\naccording to this choice, so the answer is $\\binom{4}{2} \\cdot 3 \\cdot 2=36$.\n\nFor $n=3$, we have a hexagon. Denote the sides as $a_{1}, a_{2}, \\ldots q_{6}$, in that order. There must be 2 consecutive sides of different colors, say $a_{1}$ is red, $a_{2}$ is blue. We must have a green side, and only $a_{4}$ and $a_{5}$ can be green. We have 3 possibilities:\n\n1) $a_{4}$ is green, $a_{5}$ is not. So, $a_{3}$ must be blue and $a_{5}$ must be blue (by elimination) and $a_{6}$ must be blue, so we get a valid coloring.\n2) Both $a_{4}$ and $a_{5}$ are green, thus $a_{6}$ must be red and $a_{5}$ must be blue, and we get the coloring rbbggr.\n3) $a_{5}$ is green, $a_{4}$ is not. Then $a_{6}$ must be red. Subsequently, $a_{4}$ must be red (we assume it is not green). It remains that $a_{3}$ must be red, and the coloring is rbrrgr.\n\nThus, we have 2 kinds of configurations:\n\ni) 2 opposite sides have 2 opposite colors and all other sides are of the third color. This can happen in 3.(3.2.1) $=18$ ways (first choosing the pair of opposite sides, then assigning colors), ii) 3 pairs of consecutive sides, each pair in one of the 3 colors. This can happen in $2.6=12$ ways (2 partitioning into pairs of consecutive sides, for each partitioning, 6 ways to assign the colors).\n\nThus, for $n=3$, the answer is $18+12=30$.\n\nFinally, let's address the case $n \\geq 4$. The important thing now is that any 4 consecutive sides can be seen from an external point, by Lemma 1.\n\nDenote the sides as $a_{1}, a_{2}, \\ldots, a_{2 n}$. Again, there must be 2 adjacent sides that are of different colors, say $a_{1}$ is blue and $a_{2}$ is red. We must have a green side, and by Lemma 1, that can only be $a_{n+1}$ or $a_{n+2}$. So, we have 2 cases:\n\nCase 1: $a_{n+1}$ is green, so $a_{n}$ must be red (cannot be green due to Lemma 1 applied to $a_{1}, a_{2}, \\ldots, a_{n}$, cannot be blue for the sake of $a_{2}, \\ldots, a_{n+1}$. If $a_{n+2}$ is red, so are $a_{n+3}, \\ldots, a_{2 n}$, and we get a valid coloring: $a_{1}$ is blue, $a_{n+1}$ is green, and all the others are red.\n\nIf $a_{n+2}$ is green:\n\na) $a_{n+3}$ cannot be green, because of $a_{2}, a_{1}, a_{2 n} \\ldots, a_{n+3}$.\nb) $a_{n+3}$ cannot be blue, because the 4 adjacent sides $a_{n}, \\ldots, a_{n+3}$ can be seen (this is the case that makes the separate treatment of $n \\geq 4$ necessary)\n\nc) $a_{n+3}$ cannot be red, because of $a_{1}, a_{2 n}, \\ldots, a_{n+2}$.\n\nSo, in the case that $a_{n+2}$ is also green, we cannot get a valid coloring.\n\nCase 2: $a_{n+2}$ is green is treated the same way as Case 1.\n\nThis means that the only valid configuration for $n \\geq 4$ is having 2 opposite sides colored in 2 different colors, and all other sides colored in the third color. This can be done in $n \\cdot 3 \\cdot 2=6 n$ ways.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nC2.", "solution_match": "\nSolution."}}
-{"year": "2017", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "We have two piles with 2000 and 2017 coins respectively. Ann and Bob take alternate turns making the following moves: The player whose turn is to move picks a pile with at least two coins, removes from that pile $t$ coins for some $2 \\leqslant t \\leqslant 4$, and adds to the other pile 1 coin. The players can choose a different $t$ at each turn, and the player who cannot make a move loses. If Ann plays first determine which player has a winning strategy.", "solution": "Denote the number of coins in the two piles by $X$ and $Y$. We say that the pair $(X, Y)$ is losing if the player who begins the game loses and that the pair $(X, Y)$ is winning otherwise. We shall prove that $(X, Y)$ is loosing if $X-Y \\equiv 0,1,7 \\bmod 8$, and winning if $X-Y \\equiv 2,3,4,5,6 \\bmod 8$.\n\nLemma 1. If we have a winning pair $(X, Y)$ then we can always play in such a way that the other player is then faced with a losing pair.\n\nProof of Lemma 1. Assume $X \\geq Y$ and write $X=Y+8 k+\\ell$ for some non-negative integer $k$ and some $\\ell \\in\\{2,3,4,5,6\\}$. If $\\ell=2,3,4$ then we remove two coins from the first pile and add one coin to the second pile. If $\\ell=5,6$ then we remove four coins from the first pile and add one coin to the second pile. In each case we then obtain loosing pair\n\nLemma 2. If we are faced with a losing distribution then either we cannot play, or, however we play, the other player is faced with a winning distribution.\n\nProof of Lemma 2. Without loss of generality we may assume that we remove $k$ coins from the first pile. The following table show the new difference for all possible values of $k$ and all possible differences $X-Y$. So however we move, the other player will be faced with a winning distribution.\n\n| $k \\backslash X-Y$ | 0 | 1 | 7 |\n| :---: | :---: | :---: | :---: |\n| 2 | 5 | 6 | 4 |\n| 3 | 4 | 5 | 3 |\n| 4 | 3 | 4 | 2 |\n\nSince initially the coin difference is 1 mod 8 , by Lemmas 1 and 2 Bob has a winning strategy: He can play so that he is always faced with a winning distribution while Ann is always faced with a losing distribution. So Bob cannot lose. On the other hand the game finishes after at most 4017 moves, so Ann has to lose.\n\n## Geometry", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nC3.", "solution_match": "\nSolution."}}
-{"year": "2017", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO", "problem": "Given a parallelogram $A B C D$. The line perpendicular to $A C$ passing through $C$ and the line perpendicular to $B D$ passing through $A$ intersect at point $P$. The circle centered at point $P$ and radius $P C$ intersects the line $B C$ at point $X,(X \\neq C)$ and the line $D C$ at point $Y$, $(Y \\neq C)$. Prove that the line $A X$ passes through the point $Y$.", "solution": "Denote the feet of the perpendiculars from $P$ to the lines $B C$ and $D C$ by $M$ and $N$ respectively and let $O=A C \\cap B D$. Since the points $O, M$ and $N$ are midpoints of $C A, C X$ and $C Y$ respectively it suffices to prove that $M, N$ and $O$ are collinear. According to Menelaus's theorem for $\\triangle B C D$ and points $M, N$ and $O$ we have to prove that\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_7db2aacb858e7f253ee0g-11.jpg?height=575&width=616&top_left_y=905&top_left_x=1226)\n\n$$\n\\frac{B M}{M C} \\cdot \\frac{C N}{N D} \\cdot \\frac{D O}{O B}=1\n$$\n\nSince $D O=O B$ the above simplifies to $\\frac{B M}{C M}=\\frac{D N}{C N}$. It follows from $B M=B C+C M$ and $D N=D C-C N=A B-C N$ that the last equality is equivalent to:\n\n$$\n\\frac{B C}{C M}+2=\\frac{A B}{C N}\n$$\n\nDenote by $S$ the foot of the perpendicular from $B$ to $A C$. Since $\\Varangle B C S=\\Varangle C P M=\\varphi$ and $\\Varangle B A C=\\Varangle A C D=\\Varangle C P N=\\psi$ we conclude that $\\triangle C B S \\sim \\triangle P C M$ and $\\triangle A B S \\sim \\triangle P C N$. Therefore\n\n$$\n\\frac{C M}{B S}=\\frac{C P}{B C} \\text { and } \\frac{C N}{B S}=\\frac{C P}{A B}\n$$\n\nand thus,\n\n$$\nC M=\\frac{C P . B S}{B C} \\text { and } C N=\\frac{C P . B S}{A B}\n$$\n\nNow equality (1) becomes $A B^{2}-B C^{2}=2 C P . B S$. It follows from\n\n$$\nA B^{2}-B C^{2}=A S^{2}-C S^{2}=(A S-C S)(A S+C S)=2 O S . A C\n$$\n\nthat\n\n$$\nD C^{2}-B C^{2}=2 C P \\cdot B S \\Longleftrightarrow 2 O S \\cdot A C=2 C P . B S \\Longleftrightarrow O S \\cdot A C=C P \\cdot B S .\n$$\n\nSince $\\Varangle A C P=\\Varangle B S O=90^{\\circ}$ and $\\Varangle C A P=\\Varangle S B O$ we conclude that $\\triangle A C P \\sim \\triangle B S O$. This implies $O S . A C=C P . B S$, which completes the proof.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nG1.", "solution_match": "\nSolution."}}
-{"year": "2017", "tier": "T3", "problem_label": "G2", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C$ be an acute triangle such that $A B$ is the shortest side of the triangle. Let $D$ be the midpoint of the side $A B$ and $P$ be an interior point of the triangle such that\n\n$$\n\\Varangle C A P=\\Varangle C B P=\\Varangle A C B\n$$\n\nDenote by $M$ and $N$ the feet of the perpendiculars from $P$ to $B C$ and $A C$, respectively. Let $p$ be the line through $M$ parallel to $A C$ and $q$ be the line through $N$ parallel to $B C$. If $p$ and $q$ intersect at $K$ prove that $D$ is the circumcenter of triangle $M N K$.", "solution": "If $\\gamma=\\Varangle A C B$ then $\\Varangle C A P=\\Varangle C B P=\\Varangle A C B=\\gamma$. Let $E=K N \\cap A P$ and $F=K M \\cap B P$. We show that points $E$ and $F$ are midpoints of $A P$ and $B P$, respectively.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_7db2aacb858e7f253ee0g-13.jpg?height=545&width=1009&top_left_y=1137&top_left_x=539)\n\nIndeed, consider the triangle $A E N$. Since $K N \\| B C$, we have $\\Varangle E N A=\\Varangle B C A=\\gamma$. Moreover $\\Varangle E A N=\\gamma$ giving that triangle $A E N$ is isosceles, i.e. $A E=E N$. Next, consider the triangle $E N P$. Since $\\Varangle E N A=\\gamma$ we find that\n\n$$\n\\Varangle P N E=90^{\\circ}-\\Varangle E N A=90^{\\circ}-\\gamma\n$$\n\nNow $\\Varangle E P N=90^{\\circ}-\\gamma$ implies that the triangle $E N P$ is isosceles triangle, i.e. $E N=E P$.\n\nSince $A E=E N=E P$ point $E$ is the midpoint of $A P$ and analogously, $F$ is the midpoint of $B P$. Moreover, $D$ is also midpoint of $A B$ and we conclude that $D F P E$ is parallelogram.\n\nIt follows from $D E \\| A P$ and $K E \\| B C$ that $\\Varangle D E K=\\Varangle C B P=\\gamma$ and analogously $\\Varangle D F K=\\gamma$.\n\nWe conclude that $\\triangle E D N \\cong \\triangle F M D(E D=F P=F M, E N=E P=F D$ and $\\Varangle D E N=$ $\\left.\\Varangle M F D=180^{\\circ}-\\gamma\\right)$ and thus $N D=M D$. Therefore $D$ is a point on the perpendicular bisector of $M N$. Further,\n\n$$\n\\begin{aligned}\n\\Varangle F D E & =\\Varangle F P E=360^{\\circ}-\\Varangle B P M-\\Varangle M P N-\\Varangle N P A= \\\\\n& =360^{\\circ}-\\left(90^{\\circ}-\\gamma\\right)-\\left(180^{\\circ}-\\gamma\\right)-\\left(90^{\\circ}-\\gamma\\right)=3 \\gamma\n\\end{aligned}\n$$\n\nIt follows that\n\n$$\n\\begin{aligned}\n\\Varangle M D N & =\\Varangle F D E-\\Varangle F D M-\\Varangle E D N=\\Varangle F D E-\\Varangle E N D-\\Varangle E D N= \\\\\n& =\\Varangle F D E-(\\Varangle E N D+\\Varangle E D N)=3 \\gamma-\\gamma=2 \\gamma .\n\\end{aligned}\n$$\n\nFianlly, $K M C N$ is parallelogram, i.e. $\\Varangle M K N=\\Varangle M C N=\\gamma$. Therefore $D$ is a point on the perpendicular bisector of $M N$ and $\\Varangle M D N=2 \\Varangle M K N$, so $D$ is the circumcenter of $\\triangle M N K$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nG2.", "solution_match": "\nSolution."}}
-{"year": "2017", "tier": "T3", "problem_label": "G3", "problem_type": "Geometry", "exam": "JBMO", "problem": "Consider triangle $A B C$ such that $A B \\leq A C$. Point $D$ on the arc $B C$ of the circumcirle of $A B C$ not containing point $A$ and point $E$ on side $B C$ are such that\n\n$$\n\\Varangle B A D=\\Varangle C A E<\\frac{1}{2} \\Varangle B A C .\n$$\n\nLet $S$ be the midpoint of segment $A D$. If $\\Varangle A D E=\\Varangle A B C-\\Varangle A C B$ prove that\n\n$$\n\\Varangle B S C=2 \\Varangle B A C\n$$", "solution": "Let the tangent to the circumcircle of $\\triangle A B C$ at point $A$ intersect line $B C$ at $T$. Since $A B \\leq A C$ we get that $B$ lies between $T$ and $C$. Since $\\Varangle B A T=\\Varangle A C B$ and $\\Varangle A B T=\\Varangle 180^{\\circ}-\\Varangle A B C$ we get $\\Varangle E T A=\\Varangle B T A=\\Varangle A B C-\\Varangle A C B=\\Varangle A D E$ which gives that $A, E, D, T$ are concyclic. Since\n\n$$\n\\Varangle T D B+\\Varangle B C A=\\Varangle T D B+\\Varangle B D A=\\Varangle T D A=\\Varangle A E T=\\Varangle A C B+\\Varangle E A C\n$$\n\nthis means $\\Varangle T D B=\\Varangle E A C=\\Varangle D A B$ which means that $T D$ is tangent to the circumcircle of $\\triangle A B C$ at point $D$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_7db2aacb858e7f253ee0g-15.jpg?height=524&width=740&top_left_y=1619&top_left_x=649)\n\nUsing similar triangles $T A B$ and $T C A$ we get\n\n$$\n\\frac{A B}{A C}=\\frac{T A}{T C}\n$$\n\nUsing similar triangles $T B D$ and $T D C$ we get\n\n$$\n\\frac{B D}{C D}=\\frac{T D}{T C}\n$$\n\nUsing the fact that $T A=T D$ with (1) and (2) we get\n\n$$\n\\frac{A B}{A C}=\\frac{B D}{C D}\n$$\n\nNow since $\\Varangle D A B=\\Varangle C A E$ and $\\Varangle B D A=\\Varangle E C A$ we get that the triangles $D A B$ and $C A E$ are similar. Analogously, we get that triangles $C A D$ and $E A B$ are similar. These similarities give us\n\n$$\n\\frac{D B}{C E}=\\frac{A B}{A E} \\quad \\text { and } \\quad \\frac{C D}{E B}=\\frac{C A}{E A}\n$$\n\nwhich, when combined with (3) give us $B E=C E$ giving $E$ is the midpoint of side $B C$.\n\nUsing the fact that triangles $D A B$ and $C A E$ are similar with the fact that $E$ is the midpoint of $B C$ we get:\n\n$$\n\\frac{2 D S}{C A}=\\frac{D A}{C A}=\\frac{D B}{C E}=\\frac{D B}{\\frac{C B}{2}}=\\frac{2 D B}{C B}\n$$\n\nimplying that\n\n$$\n\\frac{D S}{D B}=\\frac{C A}{C B}\n$$\n\nSince $\\Varangle S D B=\\Varangle A D B=\\Varangle A C B$ we get from (4) that the triangles $S D B$ and $A C B$ are similar, giving us $\\Varangle B S D=\\Varangle B A C$. Analogously we get $\\triangle S D C$ and $\\triangle A B C$ are similar we get $\\Varangle C S D=\\Varangle C A B$. Combining the last two equalities we get\n\n$$\n2 \\Varangle B A C=\\Varangle B A C+\\Varangle C A B=\\Varangle C S D+\\Varangle B S D=\\Varangle C S B\n$$\n\nThis completes the proof.\n\n## Alternative solution (PSC).\n\nLemma 1. A point $P$ is such that $\\Varangle P X Y=\\Varangle P Y Z$ and $\\Varangle P Z Y=\\Varangle P Y X$. If $R$ is the midpoint of $X Z$ then $\\Varangle X Y P=\\Varangle Z Y R$.\n\nProof. We consider the case when $P$ is inside the triangle $X Y Z$ (the other case is treated in similar way). Let $Q$ be the conjugate of $P$ in $\\triangle X Y Z$ and let $Y Q$ intersects $X Z$ at $S$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_7db2aacb858e7f253ee0g-17.jpg?height=594&width=491&top_left_y=386&top_left_x=790)\n\nThen $\\Varangle Q X Z=\\Varangle Q Y X$ and $\\Varangle Q Z X=\\Varangle Q Y Z$ and therefore $\\triangle S X Y \\sim \\triangle S Q X$ and $\\triangle S Z Y \\sim$ $\\triangle S Q Z$. Thus $S X^{2}=S Q . S Y=S Z^{2}$ and we conclude that $S \\equiv R$. This completes the proof of the Lemma.\n\nFor $\\triangle D C A$ we have $\\Varangle C D E=\\Varangle E C A$ and $\\Varangle E A C=\\Varangle E C D$. By the Lemma 1 for $\\triangle D C A$ and point $E$ we have that $\\Varangle S C A=\\Varangle D C E$. Therefore\n\n$$\n\\Varangle D S C=\\Varangle S A C+\\Varangle S C A=\\Varangle S A C+\\Varangle D C E=\\Varangle S A C+\\Varangle B A D=\\Varangle B A C .\n$$\n\nBy analogy, Lemma 1 applied for $\\triangle B D A$ and point $E$ gives $\\Varangle B S D=\\Varangle B A C$. Thus, $\\Varangle B S C=$ $2 \\Varangle B A C$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nProblem G3.", "solution_match": "\nSolution."}}
-{"year": "2017", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C$ be a scalene triangle with circumcircle $\\Gamma$ and circumcenter $O$. Let $M$ be the midpoint of $B C$ and $D$ be a point on $\\Gamma$ such that $A D \\perp B C$. Let $T$ be a point such that $B D C T$ is a parallelogram and $Q$ a point on the same side of $B C$ as $A$ such that\n\n$$\n\\Varangle B Q M=\\Varangle B C A \\quad \\text { and } \\quad \\Varangle C Q M=\\Varangle C B A\n$$\n\nLet $A O$ intersect $\\Gamma$ again at $E$ and let the circumcircle of $E T Q$ intersect $\\Gamma$ at point $X \\neq E$. Prove that the points $A, M$, and $X$ are collinear.", "solution": "Let $X^{\\prime}$ be symmetric point to $Q$ in line $B C$. Now since $\\Varangle C B A=\\Varangle C Q M=$ $\\Varangle C X^{\\prime} M, \\Varangle B C A=\\Varangle B Q M=\\Varangle B X^{\\prime} M$, we have\n\n$$\n\\Varangle B X^{\\prime} C=\\Varangle B X^{\\prime} M+\\Varangle C X^{\\prime} M=\\Varangle C B A+\\Varangle B C A=180^{\\circ}-\\Varangle B A C\n$$\n\nwe have that $X^{\\prime} \\in \\Gamma$. Now since $\\Varangle A X^{\\prime} B=\\Varangle A C B=\\Varangle M X^{\\prime} B$ we have that $A, M, X^{\\prime}$ are collinear. Note that since\n\n$$\n\\Varangle D C B=\\Varangle D A B=90^{\\circ}-\\Varangle A B C=\\Varangle O A C=\\Varangle E A C\n$$\n\nwe get that $D B C E$ is an isosceles trapezoid.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_7db2aacb858e7f253ee0g-18.jpg?height=772&width=727&top_left_y=1625&top_left_x=669)\n\nSince $B D C T$ is a parallelogram we have $M T=M D$, with $M, D, T$ being collinear, $B D=C T$, and since $B D E C$ is an isosceles trapezoid we have $B D=C E$ and $M E=M D$. Since\n\n$$\n\\Varangle B T C=\\Varangle B D C=\\Varangle B E D, \\quad C E=B D=C T \\quad \\text { and } \\quad M E=M T\n$$\n\nwe have that $E$ and $T$ are symmetric with respect to the line $B C$. Now since $Q$ and $X^{\\prime}$ are symmetric with respect to the line $B C$ as well, this means that $Q X^{\\prime} E T$ is an isosceles trapezoid which means that $Q, X^{\\prime}, E, T$ are concyclic. Since $X^{\\prime} \\in \\Gamma$ this means that $X \\equiv X^{\\prime}$ and therefore $A, M, X$ are collinear.\n\nAlternative solution (PSC). Denote by $H$ the orthocenter of $\\triangle A B C$. We use the following well known properties:\n\n(i) Point $D$ is the symmetric point of $H$ with respect to $B C$. Indeed, if $H_{1}$ is the symmetric point of $H$ with respect to $B C$ then $\\Varangle B H_{1} C+\\Varangle B A C=180^{\\circ}$ and therefore $H_{1} \\equiv D$.\n\n(ii) The symmetric point of $H$ with respect to $M$ is the point $E$. Indeed, if $H_{2}$ is the symmetric point of $H$ with respect to $M$ then $B H_{2} C H$ is parallelogram, $\\Varangle B H_{2} C+\\Varangle B A C=180^{\\circ}$ and since $E B \\| C H$ we have $\\Varangle E B A=90^{\\circ}$.\n\nSince $D E T H$ is a parallelogram and $M H=M D$ we have that $D E T H$ is a rectangle. Therefore $M T=M E$ and $T E \\perp B C$ implying that $T$ and $E$ are symmetric with respect to $B C$. Denote by $Q^{\\prime}$ the symmetric point of $Q$ with respect to $B C$. Then $Q^{\\prime} E T Q$ is isosceles trapezoid, so $Q^{\\prime}$ is a point on the circumcircle of $\\triangle E T Q$. Moreover $\\Varangle B Q^{\\prime} C+\\Varangle B A C=180^{\\circ}$ and we conclude that $Q^{\\prime} \\in \\Gamma$. Therefore $Q^{\\prime} \\equiv X$.\n\nIt remains to observe that $\\Varangle C X M=\\Varangle C Q M=\\Varangle C B A$ and $\\Varangle C X A=\\Varangle C B A$ and we infer that $X, M$ and $A$ are collinear.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nProblem G4.", "solution_match": "\nSolution."}}
-{"year": "2017", "tier": "T3", "problem_label": "G5", "problem_type": "Geometry", "exam": "JBMO", "problem": "A point $P$ lies in the interior of the triangle $A B C$. The lines $A P, B P$, and $C P$ intersect $B C, C A$, and $A B$ at points $D, E$, and $F$, respectively. Prove that if two of the quadrilaterals $A B D E, B C E F, C A F D, A E P F, B F P D$, and $C D P E$ are concyclic, then all six are concyclic.", "solution": "We first prove the following lemma:\n\nLemma 1. Let $A B C D$ be a convex quadrilateral and let $A B \\cap C D=E$ and $B C \\cap D A=F$. Then the circumcircles of triangles $A B F, C D F, B C E$ and $D A E$ all pass through a common point $P$. This point lies on line $E F$ if and only if $A B C D$ in concyclic.\n\nProof. Let the circumcircles of $A B F$ and $B C F$ intersect at $P \\neq B$. We have\n\n$$\n\\begin{aligned}\n\\Varangle F P C & =\\Varangle F P B+\\Varangle B P C=\\Varangle B A D+\\Varangle B E C=\\Varangle E A D+\\Varangle A E D= \\\\\n& =180^{\\circ}-\\Varangle A D E=180^{\\circ}-\\Varangle F D C\n\\end{aligned}\n$$\n\nwhich gives us $F, P, C$ and $D$ are concyclic. Similarly we have\n\n$$\n\\begin{aligned}\n\\Varangle A P E & =\\Varangle A P B+\\Varangle B P E=\\Varangle A F B+\\Varangle B C D=\\Varangle D F C+\\Varangle F C D= \\\\\n& =180^{\\circ}-\\Varangle F D C=180^{\\circ}-\\Varangle A D E\n\\end{aligned}\n$$\n\nwhich gives us $E, P, A$ and $D$ are concyclic. Since $\\Varangle F P E=\\Varangle F P B+\\Varangle E P B=\\Varangle B A D+$ $\\Varangle B C D$ we get that $\\Varangle F P E=180^{\\circ}$ if and only if $\\Varangle B A D+\\Varangle B C D=180^{\\circ}$ which completes the lemma. We now divide the problem into cases:\n\nCase 1: $A E P F$ and $B F E C$ are concyclic. Here we get that\n\n$$\n180^{\\circ}=\\Varangle A E P+\\Varangle A F P=360^{\\circ}-\\Varangle C E B-\\Varangle B F C=360^{\\circ}-2 \\Varangle C E B\n$$\n\nand here we get that $\\Varangle C E B=\\Varangle C F B=90^{\\circ}$, from here it follows that $P$ is the ortocenter of $\\triangle A B C$ and that gives us $\\Varangle A D B=\\Varangle A D C=90^{\\circ}$. Now the quadrilaterals $C E P D$ and $B D P F$ are concyclic because\n\n$$\n\\Varangle C E P=\\Varangle C D P=\\Varangle P D B=\\Varangle P F B=90^{\\circ} .\n$$\n\nQuadrilaterals $A C D F$ and $A B D E$ are concyclic because\n\n$$\n\\Varangle A E B=\\Varangle A D B=\\Varangle A D C=\\Varangle A F C=90^{\\circ}\n$$\n\nCase 2: $A E P F$ and $C E P D$ are concyclic. Now by lemma 1 applied to the quadrilateral $A E P F$ we get that the circumcircles of $C E P, C A F, B P F$ and $B E A$ intersect at a point on $B C$. Since $D \\in B C$ and $C E P D$ is concyclic we get that $D$ is the desired point and it follows that $B D P F, B A E D, C A F D$ are all concylic and now we can finish same as Case 1 since $A E D B$ and $C E P D$ are concyclic.\n\nCase 3: $A E P F$ and $A E D B$ are concyclic. We apply lemma 1 as in Case 2 on the quadrilateral $A E P F$. From the lemma we get that $B D P F, C E P D$ and $C A F D$ are concylic and we finish off the same as in Case 1.\n\nCase 4: $A C D F$ and $A B D E$ are concyclic. We apply lemma 1 on the quadrilateral $A E P F$ and get that the circumcircles of $A C F, E C P, P F B$ and $B A E$ intersect at one point. Since this point is $D$ (because $A C D F$ and $A B D E$ are concyclic) we get that $A E P F, C E P D$ and $B F P D$ are concylic. We now finish off as in Case 1. These four cases prove the problem statement.\n\nRemark. A more natural approach is to solve each of the four cases by simple angle chasing.\n\n## Number Theory", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nProblem G5.", "solution_match": "\nSolution."}}
-{"year": "2017", "tier": "T3", "problem_label": "NT1", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Determine all sets of six consecutive positive integers such that the product of two of them, added to the the product of other two of them is equal to the product of the remaining two numbers.", "solution": "Exactly two of the six numbers are multiples of 3 and these two need to be multiplied together, otherwise two of the three terms of the equality are multiples of 3 but the third one is not.\n\nLet $n$ and $n+3$ denote these multiples of 3 . Two of the four remaining numbers give remainder 1 when divided by 3 , while the other two give remainder 2 , so the two other products are either\n\n$\\equiv 1 \\cdot 1=1(\\bmod 3)$ and $\\equiv 2 \\cdot 2 \\equiv 1(\\bmod 3)$, or they are both $\\equiv 1 \\cdot 2 \\equiv 2(\\bmod 3)$. In conclusion, the term $n(n+3)$ needs to be on the right hand side of the equality.\n\nLooking at parity, three of the numbers are odd, and three are even. One of $n$ and $n+3$ is odd, the other even, so exactly two of the other numbers are odd. As $n(n+3)$ is even, the two remaining odd numbers need to appear in different terms.\n\nWe distinguish the following cases:\n\nI. The numbers are $n-2, n-1, n, n+1, n+2, n+3$.\n\nThe product of the two numbers on the RHS needs to be larger than $n(n+3)$. The only possibility is $(n-2)(n-1)+n(n+3)=(n+1)(n+2)$ which leads to $n=3$. Indeed, $1 \\cdot 2+3 \\cdot 6=4 \\cdot 5$\n\nII. The numbers are $n-1, n, n+1, n+2, n+3, n+4$.\n\nAs $(n+4)(n-1)+n(n+3)=(n+1)(n+2)$ has no solutions, $n+4$ needs to be on the RHS, multiplied with a number having a different parity, so $n-1$ or $n+1$.\n\n$(n+2)(n-1)+n(n+3)=(n+1)(n+4)$ leads to $n=3$. Indeed, $2 \\cdot 5+3 \\cdot 6=4 \\cdot 7$. $(n+2)(n+1)+n(n+3)=(n-1)(n+4)$ has no solution.\n\nIII. The numbers are $n, n+1, n+2, n+3, n+4, n+5$.\n\nWe need to consider the following situations:\n\n$(n+1)(n+2)+n(n+3)=(n+4)(n+5)$ which leads to $n=6$; indeed $7 \\cdot 8+6 \\cdot 9=10 \\cdot 11$;\n\n$(n+2)(n+5)+n(n+3)=(n+1)(n+4)$ obviously without solutions, and\n\n$(n+1)(n+4)+n(n+3)=(n+2)(n+5)$ which leads to $n=2$ (not a multiple of 3 ).\n\nIn conclusion, the problem has three solutions:\n\n$$\n1 \\cdot 2+3 \\cdot 6=4 \\cdot 5, \\quad 2 \\cdot 5+3 \\cdot 6=4 \\cdot 7, \\quad \\text { and } \\quad 7 \\cdot 8+6 \\cdot 9=10 \\cdot 11\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nNT1.", "solution_match": "\nSolution."}}
-{"year": "2017", "tier": "T3", "problem_label": "NT3", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all pairs of positive integers $(x, y)$ such that $2^{x}+3^{y}$ is a perfect square.", "solution": "In order for the expression $2^{x}+3^{y}$ to be a perfect square, a positive integer $t$ such that $2^{x}+3^{y}=t^{2}$ should exist.\n\nCase 1. If $x$ is even, then there exists a positive integer $z$ such that $x=2 z$. Then\n\n$$\n\\left(t-2^{z}\\right)\\left(t+2^{z}\\right)=3^{y}\n$$\n\nSince $t+2^{z}-\\left(t-2^{z}\\right)=2^{z+1}$, which implies $g c d\\left(t-2^{z}, t+2^{z}\\right) \\mid 2^{z+1}$, it follows that $\\operatorname{gcd}\\left(t-2^{z}, t+\\right.$ $\\left.2^{z}\\right)=1$, hence $t-2^{z}=1$ and $t+2^{z}=3^{y}$, so we have $2^{z+1}+1=3^{y}$.\n\nFor $z=1$ we have $5=3^{y}$ which clearly have no solution. For $z \\geq 2$ we have (modulo 4) that $y$ is even. Let $y=2 k$. Then $2^{z+1}=\\left(3^{k}-1\\right)\\left(3^{k}+1\\right)$ which is possible only when $3^{k}-1=2$, i.e. $k=1, y=2$, which implies that $t=5$. So the pair $(4,2)$ is a solution to our problem. Case 2. If $y$ is even, then there exists a positive integer $w$ such that $y=2 w$, and\n\n$$\n\\left(t-3^{w}\\right)\\left(t+3^{w}\\right)=2^{x}\n$$\n\nSince $t+3^{w}-\\left(t-3^{w}\\right)=2 \\cdot 3^{w}$, we have $\\operatorname{gcd}\\left(t-2^{z}, t+2^{z}\\right) \\mid 2 \\cdot 3^{w}$, which means that $g c d(t-$ $\\left.3^{w}, t+3^{w}\\right)=2$. Hence $t-3^{w}=2$ and $t+3^{w}=2^{x-1}$. So we have\n\n$$\n2 \\cdot 3^{w}+2=2^{x-1} \\Rightarrow 3^{w}+1=2^{x-2}\n$$\n\nHere we see modulo 3 that $x-2$ is even. Let $x-2=2 m$, then $3^{w}=\\left(2^{m}-1\\right)\\left(2^{m}+1\\right)$, whence $m=1$ since $\\operatorname{gcd}\\left(2^{m}-1,2^{m}+1\\right)=1$. So we arrive again to the solution $(4,2)$.\n\nCase 3. Let $x$ and $y$ be odd. For $x \\geq 3$ we have $2^{x}+3^{y} \\equiv 3(\\bmod 4)$ while $t^{2} \\equiv 0,1(\\bmod 4)$, a contradiction. For $x=1$ we have $2+3^{y}=t^{2}$. For $y \\geq 2$ we have $2+3^{y} \\equiv 2(\\bmod 9)$ while $t^{2} \\equiv 0,1,4,7(\\bmod 9)$. For $y=1$ we have $5=2+3=t^{2}$ clearly this doesn't have solution. Note. The proposer's solution used Zsigmondy's theorem in the final steps of cases 1 and 2 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nNT3.", "solution_match": "\nSolution."}}
-{"year": "2017", "tier": "T3", "problem_label": "NT4", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Solve in nonnegative integers the equation $5^{t}+3^{x} 4^{y}=z^{2}$.", "solution": "If $x=0$ we have\n\n$$\nz^{2}-2^{2 y}=5^{t} \\Longleftrightarrow\\left(z+2^{y}\\right)\\left(z-2^{y}\\right)=5^{t}\n$$\n\nPutting $z+2^{y}=5^{a}$ and $z-2^{y}=5^{b}$ with $a+b=t$ we get $5^{a}-5^{b}=2^{y+1}$. This gives us $b=0$ and now we have $5^{t}-1=2^{y+1}$. If $y \\geq 2$ then consideration by modulo 8 gives $2 \\mid t$. Putting $t=2 s$ we get $\\left(5^{s}-1\\right)\\left(5^{s}+1\\right)=2^{y+1}$. This means $5^{s}-1=2^{c}$ and $5^{s}+1=2^{d}$ with $c+d=y+1$. Subtracting we get $2=2^{d}-2^{c}$. Then we have $c=1, d=2$, but the equation $5^{s}-1=2$ has no solutions over nonnegative integers. Therefore so $y \\geq 2$ in this case gives us no solutions. If $y=0$ we get again $5^{t}-1=2$ which again has no solutions in nonnegative integers. If $y=1$ we get $t=1$ and $z=3$ which gives us the solution $(t, x, y, z)=(1,0,1,3)$.\n\nNow if $x \\geq 1$ then by modulo 3 we have $2 \\mid t$. Putting $t=2 s$ we get\n\n$$\n3^{x} 4^{y}=z^{2}-5^{2 s} \\Longleftrightarrow 3^{x} 4^{y}=\\left(z+5^{s}\\right)\\left(z-5^{s}\\right)\n$$\n\nNow we have $z+5^{s}=3^{m} 2^{k}$ and $z-5^{s}=3^{n} 2^{l}$, with $k+l=2 y$ and $m+n=x \\geq 1$. Subtracting we get\n\n$$\n2.5^{s}=3^{m} 2^{k}-3^{n} 2^{l}\n$$\n\nHere we get that $\\min \\{m, n\\}=0$. We now have a couple of cases.\n\nCase 1. $k=l=0$. Now we have $n=0$ and we get the equation $2.5^{s}=3^{m}-1$. From modulo 4 we get that $m$ is odd. If $s \\geq 1$ we get modulo 5 that $4 \\mid m$, a contradiction. So $s=0$ and we get $m=1$. This gives us $t=0, x=1, y=0, z=2$.\n\nCase 2. $\\min \\{k, l\\}=1$. Now we deal with two subcases:\n\nCase 2a. $l>k=1$. We get $5^{s}=3^{m}-3^{n} 2^{l-1}$. Since $\\min \\{m, n\\}=0$, we get that $n=0$. Now the equation becomes $5^{s}=3^{m}-2^{l-1}$. Note that $l-1=2 y-2$ is even. By modulo 3 we get that $s$ is odd and this means $s \\geq 1$. Now by modulo 5 we get $3^{m} \\equiv 2^{2 y-2} \\equiv 1,-1(\\bmod 5)$. Here we get that $m$ is even as well, so we write $m=2 q$. Now we get $5^{s}=\\left(3^{q}-2^{y-1}\\right)\\left(3^{q}+2^{y-1}\\right)$.\n\nTherefore $3^{q}-2^{y-1}=5^{v}$ and $3^{q}+2^{y-1}=5^{u}$ with $u+v=s$. Then $2^{y}=5^{u}-5^{v}$, whence $v=0$ and we have $3^{q}-2^{y-1}=1$. Plugging in $y=1,2$ we get the solution $y=2, q=1$. This gives us $m=2, s=1, n=0, x=2, t=2$ and therefore $z=13$. Thus we have the solution $(t, x, y, z)=(2,2,2,13)$. If $y \\geq 3$ we get modulo 4 that $q, q=2 r$. Then $\\left(3^{r}-1\\right)\\left(3^{r}+1\\right)=2^{y-1}$. Putting $3^{r}-1=2^{e}$ and $3^{r}+1=2^{f}$ with $e+f=y-1$ and subtracting these two and dividing by 2 we get $2^{f-1}-2^{e-1}=1$, whence $e=1, f=2$. Therefore $r=1, q=2, y=4$. Now since $2^{4}=5^{u}-1$ does not have a solution, it follows that there are no more solutions in this case.\n\nCase $2 b . \\quad k>l=1$. We now get $5^{s}=3^{m} 2^{k-1}-3^{n}$. By modulo 4 (which we can use since $0<k-1=2 y-2)$ we get $3^{n} \\equiv-1(\\bmod 4)$ and therefore $n$ is odd. Now since $\\min \\{m, n\\}=0$ we get that $m=0,0+n=m+n=x \\geq 1$. The equation becomes $5^{s}=2^{2 y-2}-3^{x}$. By modulo 3 we see that $s$ is even. We now put $s=2 g$ and obtain $\\left(2^{y-1}-5^{g}\\right)\\left(5^{g}+2^{y-1}\\right)=3^{x}$. Putting $2^{y-1}-5^{g}=3^{h}, 2^{y-1}+5^{g}=3^{i}$, where $i+h=x$, and subtracting the equations we get $3^{i}-3^{h}=2^{y}$. This gives us $h=0$ and now we are solving the equation $3^{x}+1=2^{y}$.\n\nThe solution $x=0, y=1$ gives $1-5^{g}=1$ without solution. If $x \\geq 1$ then by modulo 3 we get that $y$ is even. Putting $y=2 y_{1}$ we obtain $3^{x}=\\left(2^{y_{1}}-1\\right)\\left(2^{y_{1}}+1\\right)$. Putting $2^{y_{1}}-1=3^{x_{1}}$ and $2^{y_{1}}+1=3^{x_{2}}$ and subtracting we get $3^{x_{2}}-3^{x_{1}}=2$. This equation gives us $x_{1}=0, x_{2}=1$. Then $y_{1}=1, x=1, y=2$ is the only solution to $3^{x}+1=2^{y}$ with $x \\geq 1$. Now from $2-5^{g}=1$ we get $g=0$. This gives us $t=0$. Now this gives us the solution $1+3.16=49$ and $(t, x, y, z)=(0,1,2,7)$.\n\nThis completes all the cases and thus the solutions are $(t, x, y, z)=(1,0,1,3),(0,1,0,2)$, $(2,2,2,13)$, and $(0,1,2,7)$.\n\nNote. The problem can be simplified by asking for solutions in positive integers (without significant loss in ideas).", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nNT4.", "solution_match": "\nSolution."}}
-{"year": "2017", "tier": "T3", "problem_label": "NT5", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all positive integers $n$ such that there exists a prime number $p$, such that\n\n$$\np^{n}-(p-1)^{n}\n$$\n\nis a power of 3 .\n\nNote. A power of 3 is a number of the form $3^{a}$ where $a$ is a positive integer.", "solution": "Suppose that the positive integer $n$ is such that\n\n$$\np^{n}-(p-1)^{n}=3^{a}\n$$\n\nfor some prime $p$ and positive integer $a$.\n\nIf $p=2$, then $2^{n}-1=3^{a}$ by $(1)$, whence $(-1)^{n}-1 \\equiv 0(\\bmod 3)$, so $n$ should be even. Setting $n=2 s$ we obtain $\\left(2^{s}-1\\right)\\left(2^{s}+1\\right)=3^{a}$. It follows that $2^{s}-1$ and $2^{s}+1$ are both powers of 3 , but since they are both odd, they are co-prime, and we have $2^{s}-1=1$, i.e. $s=1$ and $n=2$. If $p=3$, then (1) gives $3 \\mid 2^{n}$, which is impossible.\n\nLet $p \\geq 5$. Then it follows from (1) that we can not have $3 \\mid p-1$. This means that $2^{n}-1 \\equiv 0$ $(\\bmod 3)$, so $n$ should be even, and let $n=2 k$. Then\n\n$$\np^{2 k}-(p-1)^{2 k}=3^{a} \\Longleftrightarrow\\left(p^{k}-(p-1)^{k}\\right)\\left(p^{k}+(p-1)^{k}\\right)=3^{a}\n$$\n\nIf $d=\\left(p^{k}-(p-1)^{k}, p^{k}+(p-1)^{k}\\right)$, then $d \\mid 2 p^{k}$. However, both numbers are powers of 3 , so $d=1$ and $p^{k}-(p-1)^{k}=1, p^{k}+(p-1)^{k}=3^{a}$.\n\nIf $k=1$, then $n=2$ and we can take $p=5$. For $k \\geq 2$ we have $1=p^{k}-(p-1)^{k} \\geq p^{2}-(p-1)^{2}$ (this inequality is equivalent to $p^{2}\\left(p^{k-2}-1\\right) \\geq(p-1)^{2}\\left((p-1)^{k-2}-1\\right)$, which is obviously true). Then $1 \\geq p^{2}-(p-1)^{2}=2 p-1 \\geq 9$, which is absurd.\n\nIt follows that the only solution is $n=2$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nNT5.", "solution_match": "\nSolution."}}
+{"year": "2017", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $a, b, c$ be positive real numbers such that $a+b+c+a b+b c+c a+a b c=7$. Prove that\n\n$$\n\\sqrt{a^{2}+b^{2}+2}+\\sqrt{b^{2}+c^{2}+2}+\\sqrt{c^{2}+a^{2}+2} \\geq 6\n$$", "solution": "First we see that $x^{2}+y^{2}+1 \\geq x y+x+y$. Indeed, this is equivalent to\n\n$$\n(x-y)^{2}+(x-1)^{2}+(y-1)^{2} \\geq 0\n$$\n\nTherefore\n\n$$\n\\begin{aligned}\n& \\sqrt{a^{2}+b^{2}+2}+\\sqrt{b^{2}+c^{2}+2}+\\sqrt{c^{2}+a^{2}+2} \\\\\n\\geq & \\sqrt{a b+a+b+1}+\\sqrt{b c+b+c+1}+\\sqrt{c a+c+a+1} \\\\\n= & \\sqrt{(a+1)(b+1)}+\\sqrt{(b+1)(a+1)}+\\sqrt{(c+1)(a+1)}\n\\end{aligned}\n$$\n\nIt follows from the AM-GM inequality that\n\n$$\n\\begin{aligned}\n& \\sqrt{(a+1)(b+1)}+\\sqrt{(b+1)(a+1)}+\\sqrt{(c+1)(a+1)} \\\\\n\\geq & 3 \\sqrt[3]{\\sqrt{(a+1)(b+1)} \\cdot \\sqrt{(b+1)(a+1)} \\cdot \\sqrt{(c+1)(a+1)}} \\\\\n= & 3 \\sqrt[3]{(a+1)(b+1)(c+1)}\n\\end{aligned}\n$$\n\nOn the other hand, the given condition is equivalent to $(a+1)(b+1)(c+1)=8$ and we get the desired inequality.\n\nObviously, equality is attained if and only if $a=b=c=1$.\n\nRemark. The condition of positivity of $a, b, c$ is superfluous and the equality $\\cdots=7$ can be replaced by the inequality $\\cdots \\geq 7$. Indeed, the above proof and the triangle inequality imply that\n\n$$\n\\begin{aligned}\n\\sqrt{a^{2}+b^{2}+2}+\\sqrt{b^{2}+c^{2}+2}+\\sqrt{c^{2}+a^{2}+2} & \\geq 3 \\sqrt[3]{(|a|+1)(|b|+1)(|c|+1)} \\\\\n& \\geq 3 \\sqrt[3]{|a+1| \\cdot|b+1| \\cdot|c+1|} \\geq 6\n\\end{aligned}\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nA1.", "solution_match": "\nSolution."}}
+{"year": "2017", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $a$ and $b$ be positive real numbers such that $3 a^{2}+2 b^{2}=3 a+2 b$. Find the minimum value of\n\n$$\nA=\\sqrt{\\frac{a}{b(3 a+2)}}+\\sqrt{\\frac{b}{a(2 b+3)}}\n$$", "solution": "By the Cauchy-Schwarz inequality we have that\n\n$$\n5\\left(3 a^{2}+2 b^{2}\\right)=5\\left(a^{2}+a^{2}+a^{2}+b^{2}+b^{2}\\right) \\geq(3 a+2 b)^{2}\n$$\n\n(or use that the last inequality is equivalent to $(a-b)^{2} \\geq 0$ ).\n\nSo, with the help of the given condition we get that $3 a+2 b \\leq 5$. Now, by the AM-GM inequality we have that\n\n$$\nA \\geq 2 \\sqrt{\\sqrt{\\frac{a}{b(3 a+2)}} \\cdot \\sqrt{\\frac{b}{a(2 b+3)}}}=\\frac{2}{\\sqrt[4]{(3 a+2)(2 b+3)}}\n$$\n\nFinally, using again the AM-GM inequality, we get that\n\n$$\n(3 a+2)(2 b+3) \\leq\\left(\\frac{3 a+2 b+5}{2}\\right)^{2} \\leq 25\n$$\n\nso $A \\geq 2 / \\sqrt{5}$ and the equality holds if and only if $a=b=1$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nA2.", "solution_match": "\nSolution."}}
+{"year": "2017", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $a, b, c, d$ be real numbers such that $0 \\leq a \\leq b \\leq c \\leq d$. Prove the inequality\n\n$$\na b^{3}+b c^{3}+c d^{3}+d a^{3} \\geq a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}\n$$", "solution": "The inequality is equivalent to\n\n$$\n\\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\\right)^{2} \\geq\\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}\\right)^{2}\n$$\n\nBy the Cauchy-Schwarz inequality,\n\n$$\n\\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\\right)\\left(a^{3} b+b^{3} c+c^{3} d+d^{3} a\\right) \\geq\\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}\\right)^{2}\n$$\n\nHence it is sufficient to prove that\n\n$$\n\\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\\right)^{2} \\geq\\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\\right)\\left(a^{3} b+b^{3} c+c^{3} d+d^{3} a\\right)\n$$\n\ni.e. to prove $a b^{3}+b c^{3}+c d^{3}+d a^{3} \\geq a^{3} b+b^{3} c+c^{3} d+d^{3} a$.\n\nThis inequality can be written successively\n\n$$\na\\left(b^{3}-d^{3}\\right)+b\\left(c^{3}-a^{3}\\right)+c\\left(d^{3}-b^{3}\\right)+d\\left(a^{3}-c^{3}\\right) \\geq 0\n$$\n\nor\n\n$$\n(a-c)\\left(b^{3}-d^{3}\\right)-(b-d)\\left(a^{3}-c^{3}\\right) \\geq 0\n$$\n\nwhich comes down to\n\n$$\n(a-c)(b-d)\\left(b^{2}+b d+d^{2}-a^{2}-a c-c^{2}\\right) \\geq 0\n$$\n\nThe last inequality is true because $a-c \\leq 0, b-d \\leq 0$, and $\\left(b^{2}-a^{2}\\right)+(b d-a c)+\\left(d^{2}-c^{2}\\right) \\geq 0$ as a sum of three non-negative numbers.\n\nThe last inequality is satisfied with equality whence $a=b$ and $c=d$. Combining this with the equality cases in the Cauchy-Schwarz inequality we obtain the equality cases for the initial inequality: $a=b=c=d$.\n\nRemark. Instead of using the Cauchy-Schwarz inequality, once the inequality $a b^{3}+b c^{3}+c d^{3}+$ $d a^{3} \\geq a^{3} b+b^{3} c+c^{3} d+d^{3} a$ is established, we have $2\\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\\right) \\geq\\left(a b^{3}+b c^{3}+c d^{3}+\\right.$ $\\left.d a^{3}\\right)+\\left(a^{3} b+b^{3} c+c^{3} d+d^{3} a\\right)=\\left(a b^{3}+a^{3} b\\right)+\\left(b c^{3}+b^{3} c\\right)+\\left(c d^{3}+c^{3} d\\right)+\\left(d a^{3}+d^{3} a\\right) \\stackrel{A M-G M}{\\geq}$ $2 a^{2} b^{2}+2 b^{2} c^{2}+2 c^{2} d^{2}+2 d^{2} a^{2}$ which gives the conclusion.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nA3.", "solution_match": "\nSolution."}}
+{"year": "2017", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $x, y, z$ be three distinct positive integers. Prove that\n\n$$\n(x+y+z)(x y+y z+z x-2) \\geq 9 x y z\n$$\n\nWhen does the equality hold?", "solution": "Since $x, y, z$ are distinct positive integers, the required inequality is symmetric and WLOG we can suppose that $x \\geq y+1 \\geq z+2$. We consider 2 possible cases:\n\nCase 1. $y \\geq z+2$. Since $x \\geq y+1 \\geq z+3$ it follows that\n\n$$\n(x-y)^{2} \\geq 1, \\quad(y-z)^{2} \\geq 4, \\quad(x-z)^{2} \\geq 9\n$$\n\nwhich are equivalent to\n\n$$\nx^{2}+y^{2} \\geq 2 x y+1, \\quad y^{2}+z^{2} \\geq 2 y z+4, \\quad x^{2}+z^{2} \\geq 2 x z+9\n$$\n\nor otherwise\n\n$$\nz x^{2}+z y^{2} \\geq 2 x y z+z, \\quad x y^{2}+x z^{2} \\geq 2 x y z+4 x, \\quad y x^{2}+y z^{2} \\geq 2 x y z+9 y\n$$\n\nAdding up the last three inequalities we have\n\n$$\nx y(x+y)+y z(y+z)+z x(z+x) \\geq 6 x y z+4 x+9 y+z\n$$\n\nwhich implies that $(x+y+z)(x y+y z+z x-2) \\geq 9 x y z+2 x+7 y-z$.\n\nSince $x \\geq z+3$ it follows that $2 x+7 y-z \\geq 0$ and our inequality follows.\n\nCase 2. $y=z+1$. Since $x \\geq y+1=z+2$ it follows that $x \\geq z+2$, and replacing $y=z+1$ in the required inequality we have to prove\n\n$$\n(x+z+1+z)(x(z+1)+(z+1) z+z x-2) \\geq 9 x(z+1) z\n$$\n\nwhich is equivalent to\n\n$$\n(x+2 z+1)\\left(z^{2}+2 z x+z+x-2\\right)-9 x(z+1) z \\geq 0\n$$\n\nDoing easy algebraic manipulations, this is equivalent to prove\n\n$$\n(x-z-2)(x-z+1)(2 z+1) \\geq 0\n$$\n\nwhich is satisfied since $x \\geq z+2$.\n\nThe equality is achieved only in the Case 2 for $x=z+2$, so we have equality when $(x, y, z)=$ $(k+2, k+1, k)$ and all the permutations for any positive integer $k$.\n\n## Combinatorics", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nA4.", "solution_match": "\nSolution."}}
+{"year": "2017", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Consider a regular $2 n+1$-gon $P$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$, if the line segment $S E$ contains no other points that lie on the sides of $P$ except $S$. We want to color the sides of $P$ in 3 colors, such that every side is colored in exactly one color, and each color must be used at least once. Moreover, from every point in the plane external to $P$, at most 2 different colors on $P$ can be seen (ignore the vertices of $P$, we consider them colorless). Find the largest positive integer for which such a coloring is possible.", "solution": "Answer: $n=1$ is clearly a solution, we can just color each side of the equilateral triangle in a different color, and the conditions are satisfied. We prove there is no larger $n$ that fulfills the requirements.\n\nLemma 1. Given a regular $2 n+1$-gon in the plane, and a sequence of $n+1$ consecutive sides $s_{1}, s_{2}, \\ldots, s_{n+1}$ there is an external point $Q$ in the plane, such that the color of each $s_{i}$ can be seen from $Q$, for $i=1,2, \\ldots, n+1$.\n\nProof. It is obvious that for a semi-circle $S$, there is a point $R$ in the plane far enough on the perpendicular bisector of the diameter of $S$ such that almost the entire semi-circle can be seen from $R$.\n\nNow, it is clear that looking at the circumscribed circle around the $2 n+1$-gon, there is a semi-circle $S$ such that each $s_{i}$ either has both endpoints on it, or has an endpoint that is on the semi-circle, and is not on the semicircle's end. So, take $Q$ to be a point in the plane from which almost all of $S$ can be seen, clearly, the color of each $s_{i}$ can be seen from $Q$. $\\diamond$ Take $n \\geq 2$, denote the sides $a_{1}, a_{2}, \\ldots, a_{2 n+1}$ in that order, and suppose we have a coloring that satisfies the condition of the problem. Let's call the 3 colors red, green and blue. We must have 2 adjacent sides of different colors, say $a_{1}$ is red and $a_{2}$ is green. Then, by Lemma 1 :\n\n(i) We cannot have a blue side among $a_{1}, a_{2}, \\ldots, a_{n+1}$.\n(ii) We cannot have a blue side among $a_{2}, a_{1}, a_{2 n+1}, \\ldots, a_{n+3}$.\n\nWe are required to have at least one blue side, and according to 1 ) and 2), that can only be $a_{n+2}$, so $a_{n+2}$ is blue. Now, applying Lemma 1 on the sequence of sides $a_{2}, a_{3}, \\ldots, a_{n+2}$ we get that $a_{2}, a_{3}, \\ldots, a_{n+1}$ are all green. Applying Lemma 1 on the sequence of sides $a_{1}, a_{2 n+1}, a_{2 n}, \\ldots, a_{n+2}$ we get that $a_{2 n+1}, a_{2 n}, \\ldots, a_{n+3}$ are all red.\n\nTherefore $a_{n+1}, a_{n+2}$ and $a_{n+3}$ are all of different colors, and for $n \\geq 2$ they can all be seen from the same point according to Lemma 1 , so we have a contradiction.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nC1.", "solution_match": "\nSolution."}}
+{"year": "2017", "tier": "T3", "problem_label": "C2", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Consider a regular $2 n$-gon $P$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$, if the line segment $S E$ contains no other points that lie on the sides of $P$ except $S$. We want to color the sides of $P$ in 3 colors, such that every side is colored in exactly one color, and each color must be used at least once. Moreover, from every point in the plane external to $P$, at most 2 different colors on $P$ can be seen (ignore the vertices of $P$, we consider them colorless). Find the number of distinct such colorings of $P$ (two colorings are considered distinct if at least one side is colored differently).", "solution": "Answer: For $n=2$, the answer is 36 ; for $n=3$, the answer is 30 and for $n \\geq 4$, the answer is $6 n$.\n\nLemma 1. Given a regular $2 n$-gon in the plane and a sequence of $n$ consecutive sides $s_{1}, s_{2}, \\ldots, s_{n}$ there is an external point $Q$ in the plane, such that the color of each $s_{i}$ can be seen from $Q$, for $i=1,2, \\ldots, n$.\n\nProof. It is obvious that for a semi-circle $S$, there is a point $R$ in the plane far enough on the bisector of its diameter such that almost the entire semi-circle can be seen from $R$.\n\nNow, it is clear that looking at the circumscribed circle around the $2 n$-gon, there is a semi-circle $S$ such that each $s_{i}$ either has both endpoints on it, or has an endpoint that's on the semi-circle, and is not on the semi-circle's end. So, take $Q$ to be a point in the plane from which almost all of $S$ can be seen, clearly, the color of each $s_{i}$ can be seen from $Q$.\n\nLemma 2. Given a regular $2 n$-gon in the plane, and a sequence of $n+1$ consecutive sides $s_{1}, s_{2}, \\ldots, s_{n+1}$ there is no external point $Q$ in the plane, such that the color of each $s_{i}$ can be seen from $Q$, for $i=1,2, \\ldots, n+1$.\n\nProof. Since $s_{1}$ and $s_{n+1}$ are parallel opposite sides of the $2 n$-gon, they cannot be seen at the same time from an external point.\n\nFor $n=2$, we have a square, so all we have to do is make sure each color is used. Two sides will be of the same color, and we have to choose which are these 2 sides, and then assign colors\naccording to this choice, so the answer is $\\binom{4}{2} \\cdot 3 \\cdot 2=36$.\n\nFor $n=3$, we have a hexagon. Denote the sides as $a_{1}, a_{2}, \\ldots q_{6}$, in that order. There must be 2 consecutive sides of different colors, say $a_{1}$ is red, $a_{2}$ is blue. We must have a green side, and only $a_{4}$ and $a_{5}$ can be green. We have 3 possibilities:\n\n1) $a_{4}$ is green, $a_{5}$ is not. So, $a_{3}$ must be blue and $a_{5}$ must be blue (by elimination) and $a_{6}$ must be blue, so we get a valid coloring.\n2) Both $a_{4}$ and $a_{5}$ are green, thus $a_{6}$ must be red and $a_{5}$ must be blue, and we get the coloring rbbggr.\n3) $a_{5}$ is green, $a_{4}$ is not. Then $a_{6}$ must be red. Subsequently, $a_{4}$ must be red (we assume it is not green). It remains that $a_{3}$ must be red, and the coloring is rbrrgr.\n\nThus, we have 2 kinds of configurations:\n\ni) 2 opposite sides have 2 opposite colors and all other sides are of the third color. This can happen in 3.(3.2.1) $=18$ ways (first choosing the pair of opposite sides, then assigning colors), ii) 3 pairs of consecutive sides, each pair in one of the 3 colors. This can happen in $2.6=12$ ways (2 partitioning into pairs of consecutive sides, for each partitioning, 6 ways to assign the colors).\n\nThus, for $n=3$, the answer is $18+12=30$.\n\nFinally, let's address the case $n \\geq 4$. The important thing now is that any 4 consecutive sides can be seen from an external point, by Lemma 1.\n\nDenote the sides as $a_{1}, a_{2}, \\ldots, a_{2 n}$. Again, there must be 2 adjacent sides that are of different colors, say $a_{1}$ is blue and $a_{2}$ is red. We must have a green side, and by Lemma 1, that can only be $a_{n+1}$ or $a_{n+2}$. So, we have 2 cases:\n\nCase 1: $a_{n+1}$ is green, so $a_{n}$ must be red (cannot be green due to Lemma 1 applied to $a_{1}, a_{2}, \\ldots, a_{n}$, cannot be blue for the sake of $a_{2}, \\ldots, a_{n+1}$. If $a_{n+2}$ is red, so are $a_{n+3}, \\ldots, a_{2 n}$, and we get a valid coloring: $a_{1}$ is blue, $a_{n+1}$ is green, and all the others are red.\n\nIf $a_{n+2}$ is green:\n\na) $a_{n+3}$ cannot be green, because of $a_{2}, a_{1}, a_{2 n} \\ldots, a_{n+3}$.\nb) $a_{n+3}$ cannot be blue, because the 4 adjacent sides $a_{n}, \\ldots, a_{n+3}$ can be seen (this is the case that makes the separate treatment of $n \\geq 4$ necessary)\n\nc) $a_{n+3}$ cannot be red, because of $a_{1}, a_{2 n}, \\ldots, a_{n+2}$.\n\nSo, in the case that $a_{n+2}$ is also green, we cannot get a valid coloring.\n\nCase 2: $a_{n+2}$ is green is treated the same way as Case 1.\n\nThis means that the only valid configuration for $n \\geq 4$ is having 2 opposite sides colored in 2 different colors, and all other sides colored in the third color. This can be done in $n \\cdot 3 \\cdot 2=6 n$ ways.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nC2.", "solution_match": "\nSolution."}}
+{"year": "2017", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "We have two piles with 2000 and 2017 coins respectively. Ann and Bob take alternate turns making the following moves: The player whose turn is to move picks a pile with at least two coins, removes from that pile $t$ coins for some $2 \\leqslant t \\leqslant 4$, and adds to the other pile 1 coin. The players can choose a different $t$ at each turn, and the player who cannot make a move loses. If Ann plays first determine which player has a winning strategy.", "solution": "Denote the number of coins in the two piles by $X$ and $Y$. We say that the pair $(X, Y)$ is losing if the player who begins the game loses and that the pair $(X, Y)$ is winning otherwise. We shall prove that $(X, Y)$ is loosing if $X-Y \\equiv 0,1,7 \\bmod 8$, and winning if $X-Y \\equiv 2,3,4,5,6 \\bmod 8$.\n\nLemma 1. If we have a winning pair $(X, Y)$ then we can always play in such a way that the other player is then faced with a losing pair.\n\nProof of Lemma 1. Assume $X \\geq Y$ and write $X=Y+8 k+\\ell$ for some non-negative integer $k$ and some $\\ell \\in\\{2,3,4,5,6\\}$. If $\\ell=2,3,4$ then we remove two coins from the first pile and add one coin to the second pile. If $\\ell=5,6$ then we remove four coins from the first pile and add one coin to the second pile. In each case we then obtain loosing pair\n\nLemma 2. If we are faced with a losing distribution then either we cannot play, or, however we play, the other player is faced with a winning distribution.\n\nProof of Lemma 2. Without loss of generality we may assume that we remove $k$ coins from the first pile. The following table show the new difference for all possible values of $k$ and all possible differences $X-Y$. So however we move, the other player will be faced with a winning distribution.\n\n| $k \\backslash X-Y$ | 0 | 1 | 7 |\n| :---: | :---: | :---: | :---: |\n| 2 | 5 | 6 | 4 |\n| 3 | 4 | 5 | 3 |\n| 4 | 3 | 4 | 2 |\n\nSince initially the coin difference is 1 mod 8 , by Lemmas 1 and 2 Bob has a winning strategy: He can play so that he is always faced with a winning distribution while Ann is always faced with a losing distribution. So Bob cannot lose. On the other hand the game finishes after at most 4017 moves, so Ann has to lose.\n\n## Geometry", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nC3.", "solution_match": "\nSolution."}}
+{"year": "2017", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Given a parallelogram $A B C D$. The line perpendicular to $A C$ passing through $C$ and the line perpendicular to $B D$ passing through $A$ intersect at point $P$. The circle centered at point $P$ and radius $P C$ intersects the line $B C$ at point $X,(X \\neq C)$ and the line $D C$ at point $Y$, $(Y \\neq C)$. Prove that the line $A X$ passes through the point $Y$.", "solution": "Denote the feet of the perpendiculars from $P$ to the lines $B C$ and $D C$ by $M$ and $N$ respectively and let $O=A C \\cap B D$. Since the points $O, M$ and $N$ are midpoints of $C A, C X$ and $C Y$ respectively it suffices to prove that $M, N$ and $O$ are collinear. According to Menelaus's theorem for $\\triangle B C D$ and points $M, N$ and $O$ we have to prove that\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_7db2aacb858e7f253ee0g-11.jpg?height=575&width=616&top_left_y=905&top_left_x=1226)\n\n$$\n\\frac{B M}{M C} \\cdot \\frac{C N}{N D} \\cdot \\frac{D O}{O B}=1\n$$\n\nSince $D O=O B$ the above simplifies to $\\frac{B M}{C M}=\\frac{D N}{C N}$. It follows from $B M=B C+C M$ and $D N=D C-C N=A B-C N$ that the last equality is equivalent to:\n\n$$\n\\frac{B C}{C M}+2=\\frac{A B}{C N}\n$$\n\nDenote by $S$ the foot of the perpendicular from $B$ to $A C$. Since $\\Varangle B C S=\\Varangle C P M=\\varphi$ and $\\Varangle B A C=\\Varangle A C D=\\Varangle C P N=\\psi$ we conclude that $\\triangle C B S \\sim \\triangle P C M$ and $\\triangle A B S \\sim \\triangle P C N$. Therefore\n\n$$\n\\frac{C M}{B S}=\\frac{C P}{B C} \\text { and } \\frac{C N}{B S}=\\frac{C P}{A B}\n$$\n\nand thus,\n\n$$\nC M=\\frac{C P . B S}{B C} \\text { and } C N=\\frac{C P . B S}{A B}\n$$\n\nNow equality (1) becomes $A B^{2}-B C^{2}=2 C P . B S$. It follows from\n\n$$\nA B^{2}-B C^{2}=A S^{2}-C S^{2}=(A S-C S)(A S+C S)=2 O S . A C\n$$\n\nthat\n\n$$\nD C^{2}-B C^{2}=2 C P \\cdot B S \\Longleftrightarrow 2 O S \\cdot A C=2 C P . B S \\Longleftrightarrow O S \\cdot A C=C P \\cdot B S .\n$$\n\nSince $\\Varangle A C P=\\Varangle B S O=90^{\\circ}$ and $\\Varangle C A P=\\Varangle S B O$ we conclude that $\\triangle A C P \\sim \\triangle B S O$. This implies $O S . A C=C P . B S$, which completes the proof.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nG1.", "solution_match": "\nSolution."}}
+{"year": "2017", "tier": "T3", "problem_label": "G2", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C$ be an acute triangle such that $A B$ is the shortest side of the triangle. Let $D$ be the midpoint of the side $A B$ and $P$ be an interior point of the triangle such that\n\n$$\n\\Varangle C A P=\\Varangle C B P=\\Varangle A C B\n$$\n\nDenote by $M$ and $N$ the feet of the perpendiculars from $P$ to $B C$ and $A C$, respectively. Let $p$ be the line through $M$ parallel to $A C$ and $q$ be the line through $N$ parallel to $B C$. If $p$ and $q$ intersect at $K$ prove that $D$ is the circumcenter of triangle $M N K$.", "solution": "If $\\gamma=\\Varangle A C B$ then $\\Varangle C A P=\\Varangle C B P=\\Varangle A C B=\\gamma$. Let $E=K N \\cap A P$ and $F=K M \\cap B P$. We show that points $E$ and $F$ are midpoints of $A P$ and $B P$, respectively.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_7db2aacb858e7f253ee0g-13.jpg?height=545&width=1009&top_left_y=1137&top_left_x=539)\n\nIndeed, consider the triangle $A E N$. Since $K N \\| B C$, we have $\\Varangle E N A=\\Varangle B C A=\\gamma$. Moreover $\\Varangle E A N=\\gamma$ giving that triangle $A E N$ is isosceles, i.e. $A E=E N$. Next, consider the triangle $E N P$. Since $\\Varangle E N A=\\gamma$ we find that\n\n$$\n\\Varangle P N E=90^{\\circ}-\\Varangle E N A=90^{\\circ}-\\gamma\n$$\n\nNow $\\Varangle E P N=90^{\\circ}-\\gamma$ implies that the triangle $E N P$ is isosceles triangle, i.e. $E N=E P$.\n\nSince $A E=E N=E P$ point $E$ is the midpoint of $A P$ and analogously, $F$ is the midpoint of $B P$. Moreover, $D$ is also midpoint of $A B$ and we conclude that $D F P E$ is parallelogram.\n\nIt follows from $D E \\| A P$ and $K E \\| B C$ that $\\Varangle D E K=\\Varangle C B P=\\gamma$ and analogously $\\Varangle D F K=\\gamma$.\n\nWe conclude that $\\triangle E D N \\cong \\triangle F M D(E D=F P=F M, E N=E P=F D$ and $\\Varangle D E N=$ $\\left.\\Varangle M F D=180^{\\circ}-\\gamma\\right)$ and thus $N D=M D$. Therefore $D$ is a point on the perpendicular bisector of $M N$. Further,\n\n$$\n\\begin{aligned}\n\\Varangle F D E & =\\Varangle F P E=360^{\\circ}-\\Varangle B P M-\\Varangle M P N-\\Varangle N P A= \\\\\n& =360^{\\circ}-\\left(90^{\\circ}-\\gamma\\right)-\\left(180^{\\circ}-\\gamma\\right)-\\left(90^{\\circ}-\\gamma\\right)=3 \\gamma\n\\end{aligned}\n$$\n\nIt follows that\n\n$$\n\\begin{aligned}\n\\Varangle M D N & =\\Varangle F D E-\\Varangle F D M-\\Varangle E D N=\\Varangle F D E-\\Varangle E N D-\\Varangle E D N= \\\\\n& =\\Varangle F D E-(\\Varangle E N D+\\Varangle E D N)=3 \\gamma-\\gamma=2 \\gamma .\n\\end{aligned}\n$$\n\nFianlly, $K M C N$ is parallelogram, i.e. $\\Varangle M K N=\\Varangle M C N=\\gamma$. Therefore $D$ is a point on the perpendicular bisector of $M N$ and $\\Varangle M D N=2 \\Varangle M K N$, so $D$ is the circumcenter of $\\triangle M N K$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nG2.", "solution_match": "\nSolution."}}
+{"year": "2017", "tier": "T3", "problem_label": "G3", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Consider triangle $A B C$ such that $A B \\leq A C$. Point $D$ on the arc $B C$ of the circumcirle of $A B C$ not containing point $A$ and point $E$ on side $B C$ are such that\n\n$$\n\\Varangle B A D=\\Varangle C A E<\\frac{1}{2} \\Varangle B A C .\n$$\n\nLet $S$ be the midpoint of segment $A D$. If $\\Varangle A D E=\\Varangle A B C-\\Varangle A C B$ prove that\n\n$$\n\\Varangle B S C=2 \\Varangle B A C\n$$", "solution": "Let the tangent to the circumcircle of $\\triangle A B C$ at point $A$ intersect line $B C$ at $T$. Since $A B \\leq A C$ we get that $B$ lies between $T$ and $C$. Since $\\Varangle B A T=\\Varangle A C B$ and $\\Varangle A B T=\\Varangle 180^{\\circ}-\\Varangle A B C$ we get $\\Varangle E T A=\\Varangle B T A=\\Varangle A B C-\\Varangle A C B=\\Varangle A D E$ which gives that $A, E, D, T$ are concyclic. Since\n\n$$\n\\Varangle T D B+\\Varangle B C A=\\Varangle T D B+\\Varangle B D A=\\Varangle T D A=\\Varangle A E T=\\Varangle A C B+\\Varangle E A C\n$$\n\nthis means $\\Varangle T D B=\\Varangle E A C=\\Varangle D A B$ which means that $T D$ is tangent to the circumcircle of $\\triangle A B C$ at point $D$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_7db2aacb858e7f253ee0g-15.jpg?height=524&width=740&top_left_y=1619&top_left_x=649)\n\nUsing similar triangles $T A B$ and $T C A$ we get\n\n$$\n\\frac{A B}{A C}=\\frac{T A}{T C}\n$$\n\nUsing similar triangles $T B D$ and $T D C$ we get\n\n$$\n\\frac{B D}{C D}=\\frac{T D}{T C}\n$$\n\nUsing the fact that $T A=T D$ with (1) and (2) we get\n\n$$\n\\frac{A B}{A C}=\\frac{B D}{C D}\n$$\n\nNow since $\\Varangle D A B=\\Varangle C A E$ and $\\Varangle B D A=\\Varangle E C A$ we get that the triangles $D A B$ and $C A E$ are similar. Analogously, we get that triangles $C A D$ and $E A B$ are similar. These similarities give us\n\n$$\n\\frac{D B}{C E}=\\frac{A B}{A E} \\quad \\text { and } \\quad \\frac{C D}{E B}=\\frac{C A}{E A}\n$$\n\nwhich, when combined with (3) give us $B E=C E$ giving $E$ is the midpoint of side $B C$.\n\nUsing the fact that triangles $D A B$ and $C A E$ are similar with the fact that $E$ is the midpoint of $B C$ we get:\n\n$$\n\\frac{2 D S}{C A}=\\frac{D A}{C A}=\\frac{D B}{C E}=\\frac{D B}{\\frac{C B}{2}}=\\frac{2 D B}{C B}\n$$\n\nimplying that\n\n$$\n\\frac{D S}{D B}=\\frac{C A}{C B}\n$$\n\nSince $\\Varangle S D B=\\Varangle A D B=\\Varangle A C B$ we get from (4) that the triangles $S D B$ and $A C B$ are similar, giving us $\\Varangle B S D=\\Varangle B A C$. Analogously we get $\\triangle S D C$ and $\\triangle A B C$ are similar we get $\\Varangle C S D=\\Varangle C A B$. Combining the last two equalities we get\n\n$$\n2 \\Varangle B A C=\\Varangle B A C+\\Varangle C A B=\\Varangle C S D+\\Varangle B S D=\\Varangle C S B\n$$\n\nThis completes the proof.\n\n## Alternative solution (PSC).\n\nLemma 1. A point $P$ is such that $\\Varangle P X Y=\\Varangle P Y Z$ and $\\Varangle P Z Y=\\Varangle P Y X$. If $R$ is the midpoint of $X Z$ then $\\Varangle X Y P=\\Varangle Z Y R$.\n\nProof. We consider the case when $P$ is inside the triangle $X Y Z$ (the other case is treated in similar way). Let $Q$ be the conjugate of $P$ in $\\triangle X Y Z$ and let $Y Q$ intersects $X Z$ at $S$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_7db2aacb858e7f253ee0g-17.jpg?height=594&width=491&top_left_y=386&top_left_x=790)\n\nThen $\\Varangle Q X Z=\\Varangle Q Y X$ and $\\Varangle Q Z X=\\Varangle Q Y Z$ and therefore $\\triangle S X Y \\sim \\triangle S Q X$ and $\\triangle S Z Y \\sim$ $\\triangle S Q Z$. Thus $S X^{2}=S Q . S Y=S Z^{2}$ and we conclude that $S \\equiv R$. This completes the proof of the Lemma.\n\nFor $\\triangle D C A$ we have $\\Varangle C D E=\\Varangle E C A$ and $\\Varangle E A C=\\Varangle E C D$. By the Lemma 1 for $\\triangle D C A$ and point $E$ we have that $\\Varangle S C A=\\Varangle D C E$. Therefore\n\n$$\n\\Varangle D S C=\\Varangle S A C+\\Varangle S C A=\\Varangle S A C+\\Varangle D C E=\\Varangle S A C+\\Varangle B A D=\\Varangle B A C .\n$$\n\nBy analogy, Lemma 1 applied for $\\triangle B D A$ and point $E$ gives $\\Varangle B S D=\\Varangle B A C$. Thus, $\\Varangle B S C=$ $2 \\Varangle B A C$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nProblem G3.", "solution_match": "\nSolution."}}
+{"year": "2017", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C$ be a scalene triangle with circumcircle $\\Gamma$ and circumcenter $O$. Let $M$ be the midpoint of $B C$ and $D$ be a point on $\\Gamma$ such that $A D \\perp B C$. Let $T$ be a point such that $B D C T$ is a parallelogram and $Q$ a point on the same side of $B C$ as $A$ such that\n\n$$\n\\Varangle B Q M=\\Varangle B C A \\quad \\text { and } \\quad \\Varangle C Q M=\\Varangle C B A\n$$\n\nLet $A O$ intersect $\\Gamma$ again at $E$ and let the circumcircle of $E T Q$ intersect $\\Gamma$ at point $X \\neq E$. Prove that the points $A, M$, and $X$ are collinear.", "solution": "Let $X^{\\prime}$ be symmetric point to $Q$ in line $B C$. Now since $\\Varangle C B A=\\Varangle C Q M=$ $\\Varangle C X^{\\prime} M, \\Varangle B C A=\\Varangle B Q M=\\Varangle B X^{\\prime} M$, we have\n\n$$\n\\Varangle B X^{\\prime} C=\\Varangle B X^{\\prime} M+\\Varangle C X^{\\prime} M=\\Varangle C B A+\\Varangle B C A=180^{\\circ}-\\Varangle B A C\n$$\n\nwe have that $X^{\\prime} \\in \\Gamma$. Now since $\\Varangle A X^{\\prime} B=\\Varangle A C B=\\Varangle M X^{\\prime} B$ we have that $A, M, X^{\\prime}$ are collinear. Note that since\n\n$$\n\\Varangle D C B=\\Varangle D A B=90^{\\circ}-\\Varangle A B C=\\Varangle O A C=\\Varangle E A C\n$$\n\nwe get that $D B C E$ is an isosceles trapezoid.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_7db2aacb858e7f253ee0g-18.jpg?height=772&width=727&top_left_y=1625&top_left_x=669)\n\nSince $B D C T$ is a parallelogram we have $M T=M D$, with $M, D, T$ being collinear, $B D=C T$, and since $B D E C$ is an isosceles trapezoid we have $B D=C E$ and $M E=M D$. Since\n\n$$\n\\Varangle B T C=\\Varangle B D C=\\Varangle B E D, \\quad C E=B D=C T \\quad \\text { and } \\quad M E=M T\n$$\n\nwe have that $E$ and $T$ are symmetric with respect to the line $B C$. Now since $Q$ and $X^{\\prime}$ are symmetric with respect to the line $B C$ as well, this means that $Q X^{\\prime} E T$ is an isosceles trapezoid which means that $Q, X^{\\prime}, E, T$ are concyclic. Since $X^{\\prime} \\in \\Gamma$ this means that $X \\equiv X^{\\prime}$ and therefore $A, M, X$ are collinear.\n\nAlternative solution (PSC). Denote by $H$ the orthocenter of $\\triangle A B C$. We use the following well known properties:\n\n(i) Point $D$ is the symmetric point of $H$ with respect to $B C$. Indeed, if $H_{1}$ is the symmetric point of $H$ with respect to $B C$ then $\\Varangle B H_{1} C+\\Varangle B A C=180^{\\circ}$ and therefore $H_{1} \\equiv D$.\n\n(ii) The symmetric point of $H$ with respect to $M$ is the point $E$. Indeed, if $H_{2}$ is the symmetric point of $H$ with respect to $M$ then $B H_{2} C H$ is parallelogram, $\\Varangle B H_{2} C+\\Varangle B A C=180^{\\circ}$ and since $E B \\| C H$ we have $\\Varangle E B A=90^{\\circ}$.\n\nSince $D E T H$ is a parallelogram and $M H=M D$ we have that $D E T H$ is a rectangle. Therefore $M T=M E$ and $T E \\perp B C$ implying that $T$ and $E$ are symmetric with respect to $B C$. Denote by $Q^{\\prime}$ the symmetric point of $Q$ with respect to $B C$. Then $Q^{\\prime} E T Q$ is isosceles trapezoid, so $Q^{\\prime}$ is a point on the circumcircle of $\\triangle E T Q$. Moreover $\\Varangle B Q^{\\prime} C+\\Varangle B A C=180^{\\circ}$ and we conclude that $Q^{\\prime} \\in \\Gamma$. Therefore $Q^{\\prime} \\equiv X$.\n\nIt remains to observe that $\\Varangle C X M=\\Varangle C Q M=\\Varangle C B A$ and $\\Varangle C X A=\\Varangle C B A$ and we infer that $X, M$ and $A$ are collinear.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nProblem G4.", "solution_match": "\nSolution."}}
+{"year": "2017", "tier": "T3", "problem_label": "G5", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "A point $P$ lies in the interior of the triangle $A B C$. The lines $A P, B P$, and $C P$ intersect $B C, C A$, and $A B$ at points $D, E$, and $F$, respectively. Prove that if two of the quadrilaterals $A B D E, B C E F, C A F D, A E P F, B F P D$, and $C D P E$ are concyclic, then all six are concyclic.", "solution": "We first prove the following lemma:\n\nLemma 1. Let $A B C D$ be a convex quadrilateral and let $A B \\cap C D=E$ and $B C \\cap D A=F$. Then the circumcircles of triangles $A B F, C D F, B C E$ and $D A E$ all pass through a common point $P$. This point lies on line $E F$ if and only if $A B C D$ in concyclic.\n\nProof. Let the circumcircles of $A B F$ and $B C F$ intersect at $P \\neq B$. We have\n\n$$\n\\begin{aligned}\n\\Varangle F P C & =\\Varangle F P B+\\Varangle B P C=\\Varangle B A D+\\Varangle B E C=\\Varangle E A D+\\Varangle A E D= \\\\\n& =180^{\\circ}-\\Varangle A D E=180^{\\circ}-\\Varangle F D C\n\\end{aligned}\n$$\n\nwhich gives us $F, P, C$ and $D$ are concyclic. Similarly we have\n\n$$\n\\begin{aligned}\n\\Varangle A P E & =\\Varangle A P B+\\Varangle B P E=\\Varangle A F B+\\Varangle B C D=\\Varangle D F C+\\Varangle F C D= \\\\\n& =180^{\\circ}-\\Varangle F D C=180^{\\circ}-\\Varangle A D E\n\\end{aligned}\n$$\n\nwhich gives us $E, P, A$ and $D$ are concyclic. Since $\\Varangle F P E=\\Varangle F P B+\\Varangle E P B=\\Varangle B A D+$ $\\Varangle B C D$ we get that $\\Varangle F P E=180^{\\circ}$ if and only if $\\Varangle B A D+\\Varangle B C D=180^{\\circ}$ which completes the lemma. We now divide the problem into cases:\n\nCase 1: $A E P F$ and $B F E C$ are concyclic. Here we get that\n\n$$\n180^{\\circ}=\\Varangle A E P+\\Varangle A F P=360^{\\circ}-\\Varangle C E B-\\Varangle B F C=360^{\\circ}-2 \\Varangle C E B\n$$\n\nand here we get that $\\Varangle C E B=\\Varangle C F B=90^{\\circ}$, from here it follows that $P$ is the ortocenter of $\\triangle A B C$ and that gives us $\\Varangle A D B=\\Varangle A D C=90^{\\circ}$. Now the quadrilaterals $C E P D$ and $B D P F$ are concyclic because\n\n$$\n\\Varangle C E P=\\Varangle C D P=\\Varangle P D B=\\Varangle P F B=90^{\\circ} .\n$$\n\nQuadrilaterals $A C D F$ and $A B D E$ are concyclic because\n\n$$\n\\Varangle A E B=\\Varangle A D B=\\Varangle A D C=\\Varangle A F C=90^{\\circ}\n$$\n\nCase 2: $A E P F$ and $C E P D$ are concyclic. Now by lemma 1 applied to the quadrilateral $A E P F$ we get that the circumcircles of $C E P, C A F, B P F$ and $B E A$ intersect at a point on $B C$. Since $D \\in B C$ and $C E P D$ is concyclic we get that $D$ is the desired point and it follows that $B D P F, B A E D, C A F D$ are all concylic and now we can finish same as Case 1 since $A E D B$ and $C E P D$ are concyclic.\n\nCase 3: $A E P F$ and $A E D B$ are concyclic. We apply lemma 1 as in Case 2 on the quadrilateral $A E P F$. From the lemma we get that $B D P F, C E P D$ and $C A F D$ are concylic and we finish off the same as in Case 1.\n\nCase 4: $A C D F$ and $A B D E$ are concyclic. We apply lemma 1 on the quadrilateral $A E P F$ and get that the circumcircles of $A C F, E C P, P F B$ and $B A E$ intersect at one point. Since this point is $D$ (because $A C D F$ and $A B D E$ are concyclic) we get that $A E P F, C E P D$ and $B F P D$ are concylic. We now finish off as in Case 1. These four cases prove the problem statement.\n\nRemark. A more natural approach is to solve each of the four cases by simple angle chasing.\n\n## Number Theory", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nProblem G5.", "solution_match": "\nSolution."}}
+{"year": "2017", "tier": "T3", "problem_label": "NT1", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Determine all sets of six consecutive positive integers such that the product of two of them, added to the the product of other two of them is equal to the product of the remaining two numbers.", "solution": "Exactly two of the six numbers are multiples of 3 and these two need to be multiplied together, otherwise two of the three terms of the equality are multiples of 3 but the third one is not.\n\nLet $n$ and $n+3$ denote these multiples of 3 . Two of the four remaining numbers give remainder 1 when divided by 3 , while the other two give remainder 2 , so the two other products are either\n\n$\\equiv 1 \\cdot 1=1(\\bmod 3)$ and $\\equiv 2 \\cdot 2 \\equiv 1(\\bmod 3)$, or they are both $\\equiv 1 \\cdot 2 \\equiv 2(\\bmod 3)$. In conclusion, the term $n(n+3)$ needs to be on the right hand side of the equality.\n\nLooking at parity, three of the numbers are odd, and three are even. One of $n$ and $n+3$ is odd, the other even, so exactly two of the other numbers are odd. As $n(n+3)$ is even, the two remaining odd numbers need to appear in different terms.\n\nWe distinguish the following cases:\n\nI. The numbers are $n-2, n-1, n, n+1, n+2, n+3$.\n\nThe product of the two numbers on the RHS needs to be larger than $n(n+3)$. The only possibility is $(n-2)(n-1)+n(n+3)=(n+1)(n+2)$ which leads to $n=3$. Indeed, $1 \\cdot 2+3 \\cdot 6=4 \\cdot 5$\n\nII. The numbers are $n-1, n, n+1, n+2, n+3, n+4$.\n\nAs $(n+4)(n-1)+n(n+3)=(n+1)(n+2)$ has no solutions, $n+4$ needs to be on the RHS, multiplied with a number having a different parity, so $n-1$ or $n+1$.\n\n$(n+2)(n-1)+n(n+3)=(n+1)(n+4)$ leads to $n=3$. Indeed, $2 \\cdot 5+3 \\cdot 6=4 \\cdot 7$. $(n+2)(n+1)+n(n+3)=(n-1)(n+4)$ has no solution.\n\nIII. The numbers are $n, n+1, n+2, n+3, n+4, n+5$.\n\nWe need to consider the following situations:\n\n$(n+1)(n+2)+n(n+3)=(n+4)(n+5)$ which leads to $n=6$; indeed $7 \\cdot 8+6 \\cdot 9=10 \\cdot 11$;\n\n$(n+2)(n+5)+n(n+3)=(n+1)(n+4)$ obviously without solutions, and\n\n$(n+1)(n+4)+n(n+3)=(n+2)(n+5)$ which leads to $n=2$ (not a multiple of 3 ).\n\nIn conclusion, the problem has three solutions:\n\n$$\n1 \\cdot 2+3 \\cdot 6=4 \\cdot 5, \\quad 2 \\cdot 5+3 \\cdot 6=4 \\cdot 7, \\quad \\text { and } \\quad 7 \\cdot 8+6 \\cdot 9=10 \\cdot 11\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nNT1.", "solution_match": "\nSolution."}}
+{"year": "2017", "tier": "T3", "problem_label": "NT3", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all pairs of positive integers $(x, y)$ such that $2^{x}+3^{y}$ is a perfect square.", "solution": "In order for the expression $2^{x}+3^{y}$ to be a perfect square, a positive integer $t$ such that $2^{x}+3^{y}=t^{2}$ should exist.\n\nCase 1. If $x$ is even, then there exists a positive integer $z$ such that $x=2 z$. Then\n\n$$\n\\left(t-2^{z}\\right)\\left(t+2^{z}\\right)=3^{y}\n$$\n\nSince $t+2^{z}-\\left(t-2^{z}\\right)=2^{z+1}$, which implies $g c d\\left(t-2^{z}, t+2^{z}\\right) \\mid 2^{z+1}$, it follows that $\\operatorname{gcd}\\left(t-2^{z}, t+\\right.$ $\\left.2^{z}\\right)=1$, hence $t-2^{z}=1$ and $t+2^{z}=3^{y}$, so we have $2^{z+1}+1=3^{y}$.\n\nFor $z=1$ we have $5=3^{y}$ which clearly have no solution. For $z \\geq 2$ we have (modulo 4) that $y$ is even. Let $y=2 k$. Then $2^{z+1}=\\left(3^{k}-1\\right)\\left(3^{k}+1\\right)$ which is possible only when $3^{k}-1=2$, i.e. $k=1, y=2$, which implies that $t=5$. So the pair $(4,2)$ is a solution to our problem. Case 2. If $y$ is even, then there exists a positive integer $w$ such that $y=2 w$, and\n\n$$\n\\left(t-3^{w}\\right)\\left(t+3^{w}\\right)=2^{x}\n$$\n\nSince $t+3^{w}-\\left(t-3^{w}\\right)=2 \\cdot 3^{w}$, we have $\\operatorname{gcd}\\left(t-2^{z}, t+2^{z}\\right) \\mid 2 \\cdot 3^{w}$, which means that $g c d(t-$ $\\left.3^{w}, t+3^{w}\\right)=2$. Hence $t-3^{w}=2$ and $t+3^{w}=2^{x-1}$. So we have\n\n$$\n2 \\cdot 3^{w}+2=2^{x-1} \\Rightarrow 3^{w}+1=2^{x-2}\n$$\n\nHere we see modulo 3 that $x-2$ is even. Let $x-2=2 m$, then $3^{w}=\\left(2^{m}-1\\right)\\left(2^{m}+1\\right)$, whence $m=1$ since $\\operatorname{gcd}\\left(2^{m}-1,2^{m}+1\\right)=1$. So we arrive again to the solution $(4,2)$.\n\nCase 3. Let $x$ and $y$ be odd. For $x \\geq 3$ we have $2^{x}+3^{y} \\equiv 3(\\bmod 4)$ while $t^{2} \\equiv 0,1(\\bmod 4)$, a contradiction. For $x=1$ we have $2+3^{y}=t^{2}$. For $y \\geq 2$ we have $2+3^{y} \\equiv 2(\\bmod 9)$ while $t^{2} \\equiv 0,1,4,7(\\bmod 9)$. For $y=1$ we have $5=2+3=t^{2}$ clearly this doesn't have solution. Note. The proposer's solution used Zsigmondy's theorem in the final steps of cases 1 and 2 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nNT3.", "solution_match": "\nSolution."}}
+{"year": "2017", "tier": "T3", "problem_label": "NT4", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Solve in nonnegative integers the equation $5^{t}+3^{x} 4^{y}=z^{2}$.", "solution": "If $x=0$ we have\n\n$$\nz^{2}-2^{2 y}=5^{t} \\Longleftrightarrow\\left(z+2^{y}\\right)\\left(z-2^{y}\\right)=5^{t}\n$$\n\nPutting $z+2^{y}=5^{a}$ and $z-2^{y}=5^{b}$ with $a+b=t$ we get $5^{a}-5^{b}=2^{y+1}$. This gives us $b=0$ and now we have $5^{t}-1=2^{y+1}$. If $y \\geq 2$ then consideration by modulo 8 gives $2 \\mid t$. Putting $t=2 s$ we get $\\left(5^{s}-1\\right)\\left(5^{s}+1\\right)=2^{y+1}$. This means $5^{s}-1=2^{c}$ and $5^{s}+1=2^{d}$ with $c+d=y+1$. Subtracting we get $2=2^{d}-2^{c}$. Then we have $c=1, d=2$, but the equation $5^{s}-1=2$ has no solutions over nonnegative integers. Therefore so $y \\geq 2$ in this case gives us no solutions. If $y=0$ we get again $5^{t}-1=2$ which again has no solutions in nonnegative integers. If $y=1$ we get $t=1$ and $z=3$ which gives us the solution $(t, x, y, z)=(1,0,1,3)$.\n\nNow if $x \\geq 1$ then by modulo 3 we have $2 \\mid t$. Putting $t=2 s$ we get\n\n$$\n3^{x} 4^{y}=z^{2}-5^{2 s} \\Longleftrightarrow 3^{x} 4^{y}=\\left(z+5^{s}\\right)\\left(z-5^{s}\\right)\n$$\n\nNow we have $z+5^{s}=3^{m} 2^{k}$ and $z-5^{s}=3^{n} 2^{l}$, with $k+l=2 y$ and $m+n=x \\geq 1$. Subtracting we get\n\n$$\n2.5^{s}=3^{m} 2^{k}-3^{n} 2^{l}\n$$\n\nHere we get that $\\min \\{m, n\\}=0$. We now have a couple of cases.\n\nCase 1. $k=l=0$. Now we have $n=0$ and we get the equation $2.5^{s}=3^{m}-1$. From modulo 4 we get that $m$ is odd. If $s \\geq 1$ we get modulo 5 that $4 \\mid m$, a contradiction. So $s=0$ and we get $m=1$. This gives us $t=0, x=1, y=0, z=2$.\n\nCase 2. $\\min \\{k, l\\}=1$. Now we deal with two subcases:\n\nCase 2a. $l>k=1$. We get $5^{s}=3^{m}-3^{n} 2^{l-1}$. Since $\\min \\{m, n\\}=0$, we get that $n=0$. Now the equation becomes $5^{s}=3^{m}-2^{l-1}$. Note that $l-1=2 y-2$ is even. By modulo 3 we get that $s$ is odd and this means $s \\geq 1$. Now by modulo 5 we get $3^{m} \\equiv 2^{2 y-2} \\equiv 1,-1(\\bmod 5)$. Here we get that $m$ is even as well, so we write $m=2 q$. Now we get $5^{s}=\\left(3^{q}-2^{y-1}\\right)\\left(3^{q}+2^{y-1}\\right)$.\n\nTherefore $3^{q}-2^{y-1}=5^{v}$ and $3^{q}+2^{y-1}=5^{u}$ with $u+v=s$. Then $2^{y}=5^{u}-5^{v}$, whence $v=0$ and we have $3^{q}-2^{y-1}=1$. Plugging in $y=1,2$ we get the solution $y=2, q=1$. This gives us $m=2, s=1, n=0, x=2, t=2$ and therefore $z=13$. Thus we have the solution $(t, x, y, z)=(2,2,2,13)$. If $y \\geq 3$ we get modulo 4 that $q, q=2 r$. Then $\\left(3^{r}-1\\right)\\left(3^{r}+1\\right)=2^{y-1}$. Putting $3^{r}-1=2^{e}$ and $3^{r}+1=2^{f}$ with $e+f=y-1$ and subtracting these two and dividing by 2 we get $2^{f-1}-2^{e-1}=1$, whence $e=1, f=2$. Therefore $r=1, q=2, y=4$. Now since $2^{4}=5^{u}-1$ does not have a solution, it follows that there are no more solutions in this case.\n\nCase $2 b . \\quad k>l=1$. We now get $5^{s}=3^{m} 2^{k-1}-3^{n}$. By modulo 4 (which we can use since $0<k-1=2 y-2)$ we get $3^{n} \\equiv-1(\\bmod 4)$ and therefore $n$ is odd. Now since $\\min \\{m, n\\}=0$ we get that $m=0,0+n=m+n=x \\geq 1$. The equation becomes $5^{s}=2^{2 y-2}-3^{x}$. By modulo 3 we see that $s$ is even. We now put $s=2 g$ and obtain $\\left(2^{y-1}-5^{g}\\right)\\left(5^{g}+2^{y-1}\\right)=3^{x}$. Putting $2^{y-1}-5^{g}=3^{h}, 2^{y-1}+5^{g}=3^{i}$, where $i+h=x$, and subtracting the equations we get $3^{i}-3^{h}=2^{y}$. This gives us $h=0$ and now we are solving the equation $3^{x}+1=2^{y}$.\n\nThe solution $x=0, y=1$ gives $1-5^{g}=1$ without solution. If $x \\geq 1$ then by modulo 3 we get that $y$ is even. Putting $y=2 y_{1}$ we obtain $3^{x}=\\left(2^{y_{1}}-1\\right)\\left(2^{y_{1}}+1\\right)$. Putting $2^{y_{1}}-1=3^{x_{1}}$ and $2^{y_{1}}+1=3^{x_{2}}$ and subtracting we get $3^{x_{2}}-3^{x_{1}}=2$. This equation gives us $x_{1}=0, x_{2}=1$. Then $y_{1}=1, x=1, y=2$ is the only solution to $3^{x}+1=2^{y}$ with $x \\geq 1$. Now from $2-5^{g}=1$ we get $g=0$. This gives us $t=0$. Now this gives us the solution $1+3.16=49$ and $(t, x, y, z)=(0,1,2,7)$.\n\nThis completes all the cases and thus the solutions are $(t, x, y, z)=(1,0,1,3),(0,1,0,2)$, $(2,2,2,13)$, and $(0,1,2,7)$.\n\nNote. The problem can be simplified by asking for solutions in positive integers (without significant loss in ideas).", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nNT4.", "solution_match": "\nSolution."}}
+{"year": "2017", "tier": "T3", "problem_label": "NT5", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all positive integers $n$ such that there exists a prime number $p$, such that\n\n$$\np^{n}-(p-1)^{n}\n$$\n\nis a power of 3 .\n\nNote. A power of 3 is a number of the form $3^{a}$ where $a$ is a positive integer.", "solution": "Suppose that the positive integer $n$ is such that\n\n$$\np^{n}-(p-1)^{n}=3^{a}\n$$\n\nfor some prime $p$ and positive integer $a$.\n\nIf $p=2$, then $2^{n}-1=3^{a}$ by $(1)$, whence $(-1)^{n}-1 \\equiv 0(\\bmod 3)$, so $n$ should be even. Setting $n=2 s$ we obtain $\\left(2^{s}-1\\right)\\left(2^{s}+1\\right)=3^{a}$. It follows that $2^{s}-1$ and $2^{s}+1$ are both powers of 3 , but since they are both odd, they are co-prime, and we have $2^{s}-1=1$, i.e. $s=1$ and $n=2$. If $p=3$, then (1) gives $3 \\mid 2^{n}$, which is impossible.\n\nLet $p \\geq 5$. Then it follows from (1) that we can not have $3 \\mid p-1$. This means that $2^{n}-1 \\equiv 0$ $(\\bmod 3)$, so $n$ should be even, and let $n=2 k$. Then\n\n$$\np^{2 k}-(p-1)^{2 k}=3^{a} \\Longleftrightarrow\\left(p^{k}-(p-1)^{k}\\right)\\left(p^{k}+(p-1)^{k}\\right)=3^{a}\n$$\n\nIf $d=\\left(p^{k}-(p-1)^{k}, p^{k}+(p-1)^{k}\\right)$, then $d \\mid 2 p^{k}$. However, both numbers are powers of 3 , so $d=1$ and $p^{k}-(p-1)^{k}=1, p^{k}+(p-1)^{k}=3^{a}$.\n\nIf $k=1$, then $n=2$ and we can take $p=5$. For $k \\geq 2$ we have $1=p^{k}-(p-1)^{k} \\geq p^{2}-(p-1)^{2}$ (this inequality is equivalent to $p^{2}\\left(p^{k-2}-1\\right) \\geq(p-1)^{2}\\left((p-1)^{k-2}-1\\right)$, which is obviously true). Then $1 \\geq p^{2}-(p-1)^{2}=2 p-1 \\geq 9$, which is absurd.\n\nIt follows that the only solution is $n=2$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl", "problem_match": "\nNT5.", "solution_match": "\nSolution."}}

JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl

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-{"year": "2020", "tier": "T3", "problem_label": "A 1", "problem_type": "Algebra", "exam": "JBMO", "problem": "Find all triples $(a, b, c)$ of real numbers such that the following system holds:\n\n$$\n\\left\\{\\begin{array}{l}\na+b+c=\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} \\\\\na^{2}+b^{2}+c^{2}=\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}\n\\end{array}\\right.\n$$", "solution": "First of all if $(a, b, c)$ is a solution of the system then also $(-a,-b,-c)$ is a solution. Hence we can suppose that $a b c>0$. From the first condition we have\n\n$$\na+b+c=\\frac{a b+b c+c a}{a b c}\n$$\n\nNow, from the first condition and the second condition we get\n\n$$\n(a+b+c)^{2}-\\left(a^{2}+b^{2}+c^{2}\\right)=\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)^{2}-\\left(\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}\\right) .\n$$\n\nThe last one simplifies to\n\n$$\na b+b c+c a=\\frac{a+b+c}{a b c}\n$$\n\nFirst we show that $a+b+c$ and $a b+b c+c a$ are different from 0 . Suppose on contrary then from relation (1) or (2) we have $a+b+c=a b+b c+c a=0$. But then we would have\n\n$$\na^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(a b+b c+c a)=0\n$$\n\nwhich means that $a=b=c=0$. This is not possible since $a, b, c$ should be different from 0 . Now multiplying (1) and (2) we have\n\n$$\n(a+b+c)(a b+b c+c a)=\\frac{(a+b+c)(a b+b c+c a)}{(a b c)^{2}}\n$$\n\nSince $a+b+c$ and $a b+b c+c a$ are different from 0 , we get $(a b c)^{2}=1$ and using the fact that $a b c>0$ we obtain that $a b c=1$. So relations (1) and (2) transform to\n\n$$\na+b+c=a b+b c+c a\n$$\n\nTherefore,\n\n$$\n(a-1)(b-1)(c-1)=a b c-a b-b c-c a+a+b+c-1=0 .\n$$\n\nThis means that at least one of the numbers $a, b, c$ is equal to 1 . Suppose that $c=1$ then relations (1) and (2) transform to $a+b+1=a b+a+b \\Rightarrow a b=1$. Taking $a=t$ then we have $b=\\frac{1}{t}$. We can now verify that any triple $(a, b, c)=\\left(t, \\frac{1}{t}, 1\\right)$ satisfies both conditions. $t \\in \\mathbb{R} \\backslash\\{0\\}$. From the initial observation any triple $(a, b, c)=\\left(t, \\frac{1}{t},-1\\right)$ satisfies both conditions. $t \\in \\mathbb{R} \\backslash\\{0\\}$. So, all triples that satisfy both conditions are $(a, b, c)=\\left(t, \\frac{1}{t}, 1\\right),\\left(t, \\frac{1}{t},-1\\right)$ and all permutations for any $t \\in \\mathbb{R} \\backslash\\{0\\}$.\n\nComment by PSC. After finding that $a b c=1$ and\n\n$$\na+b+c=a b+b c+c a\n$$\n\nwe can avoid the trick considering $(a-1)(b-1)(c-1)$ as follows. By the Vieta's relations we have that $a, b, c$ are roots of the polynomial\n\n$$\nP(x)=x^{3}-s x^{2}+s x-1\n$$\n\nwhich has one root equal to 1 . Then, we can conclude as in the above solution.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nA 1.", "solution_match": "\nSolution."}}
-{"year": "2020", "tier": "T3", "problem_label": "A 2", "problem_type": "Algebra", "exam": "JBMO", "problem": "Consider the sequence $a_{1}, a_{2}, a_{3}, \\ldots$ defined by $a_{1}=9$ and\n\n$$\na_{n+1}=\\frac{(n+5) a_{n}+22}{n+3}\n$$\n\nfor $n \\geqslant 1$.\n\nFind all natural numbers $n$ for which $a_{n}$ is a perfect square of an integer.", "solution": "Define $b_{n}=a_{n}+11$. Then\n\n$$\n22=(n+3) a_{n+1}-(n+5) a_{n}=(n+3) b_{n+1}-11 n-33-(n+5) b_{n}+11 n+55\n$$\n\ngiving $(n+3) b_{n+1}=(n+5) b_{n}$. Then\n\n$b_{n+1}=\\frac{n+5}{n+3} b_{n}=\\frac{(n+5)(n+4)}{(n+3)(n+2)} b_{n-1}=\\frac{(n+5)(n+4)}{(n+2)(n+1)} b_{n-2}=\\cdots=\\frac{(n+5)(n+4)}{5 \\cdot 4} b_{1}=(n+5)(n+4)$.\n\nTherefore $b_{n}=(n+4)(n+3)=n^{2}+7 n+12$ and $a_{n}=n^{2}+7 n+1$.\n\nSince $(n+1)^{2}=n^{2}+2 n+1<a_{n}<n^{2}+8 n+16=(n+4)^{2}$, if $a_{n}$ is a perfect square, then $a_{n}=(n+2)^{2}$ or $a_{n}=(n+3)^{2}$.\n\nIf $a_{n}=(n+2)^{2}$, then $n^{2}+4 n+4=n^{2}+7 n+1$ giving $n=1$. If $a_{n}=(n+3)^{2}$, then $n^{2}+6 n+9=$ $n^{2}+7 n+1$ giving $n=8$.\n\nComment. We provide some other methods to find $a_{n}$.\n\nMethod 1: Define $b_{n}=\\frac{a_{n}+11}{n+3}$. Then $b_{1}=5$ and $a_{n}=(n+3) b_{n}-11$. So\n\n$$\na_{n+1}=(n+4) b_{n+1}-11=\\frac{(n+5) a_{n}+22}{n+3}=a_{n}+\\frac{2\\left(a_{n}+11\\right)}{n+3}=(n+3) b_{n}-11+2 b_{n}\n$$\n\ngiving $(n+4) b_{n+1}=(n+5) b_{n}$. Then\n\n$$\nb_{n+1}=\\frac{n+5}{n+4} b_{n}=\\frac{n+5}{n+3} b_{n-1}=\\cdots=\\frac{n+5}{5} b_{1}=n+5\n$$\n\nThen $b_{n}=n+4$, so $a_{n}=(n+3)(n+4)-11=n^{2}+7 n+1$.\n\nMethod 2: We have\n\n$$\n(n+3) a_{n+1}-(n+5) a_{n}=22\n$$\n\nand therefore\n\n$$\n\\frac{a_{n+1}}{(n+5)(n+4)}-\\frac{a_{n}}{(n+4)(n+3)}=\\frac{22}{(n+3)(n+4)(n+5)}=11\\left[\\frac{1}{n+3}-\\frac{2}{n+4}+\\frac{1}{n+5}\\right]\n$$\n\nNow define $b_{n}=\\frac{a_{n}}{(n+4)(n+3)}$ to get\n\n$$\nb_{n+1}=b_{n}+11\\left[\\frac{1}{n+3}-\\frac{2}{n+4}+\\frac{1}{n+5}\\right]\n$$\n\nwhich telescopically gives\n\n$$\n\\begin{aligned}\nb_{n+1} & =b_{1}+11\\left[\\left(\\frac{1}{4}-\\frac{2}{5}+\\frac{1}{6}\\right)+\\left(\\frac{1}{5}-\\frac{2}{6}+\\frac{1}{7}\\right)+\\cdots+\\left(\\frac{1}{n+3}-\\frac{2}{n+4}+\\frac{1}{n+5}\\right)\\right] \\\\\n& =b_{1}+11\\left(\\frac{1}{4}-\\frac{1}{5}-\\frac{1}{n+4}+\\frac{1}{n+5}\\right)=\\frac{9}{20}+\\frac{11}{20}-\\frac{11}{(n+4)(n+5)}\n\\end{aligned}\n$$\n\nWe get $b_{n+1}=(n+4)(n+5)-11$ from which it follows that $b_{n}=(n+3)(n+4)-11=n^{2}+7 n+1$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nA 2.", "solution_match": "\nSolution:"}}
-{"year": "2020", "tier": "T3", "problem_label": "A 3", "problem_type": "Algebra", "exam": "JBMO", "problem": "Find all triples of positive real numbers $(a, b, c)$ so that the expression\n\n$$\nM=\\frac{(a+b)(b+c)(a+b+c)}{a b c}\n$$\n\ngets its least value.", "solution": "The expression $M$ is homogeneous, therefore we can assume that $a b c=1$. We set $s=a+c$ and $p=a c$ and using $b=\\frac{1}{a c}$, we get\n\n$$\nM=\\left(a+\\frac{1}{a c}\\right)\\left(\\frac{1}{a c}+c\\right)\\left(a+\\frac{1}{a c}+c\\right)=\\left(a+p^{-1}\\right)\\left(c+p^{-1}\\right)\\left(s+p^{-1}\\right)\n$$\n\nExpanding the right-hand side we get\n\n$$\nM=p s+\\frac{s^{2}}{p}+1+\\frac{2 s}{p^{2}}+\\frac{1}{p^{3}}\n$$\n\nNow by $s \\geq 2 \\sqrt{p}$ and setting $x=p \\sqrt{p}>0$ we get\n\n$$\nM \\geq 2 x+5+\\frac{4}{x}+\\frac{1}{x^{2}}\n$$\n\nWe will now prove that\n\n$$\n2 x+5+\\frac{4}{x}+\\frac{1}{x^{2}} \\geq \\frac{11+5 \\sqrt{5}}{2} \\text {. }\n$$\n\nIndeed, the latter is equivalent to $4 x^{3}-(5 \\sqrt{5}+1) x^{2}+8 x+2 \\geq 0$, which can be rewritten as\n\n$$\n\\left(x-\\frac{1+\\sqrt{5}}{2}\\right)^{2}(4 x+3-\\sqrt{5}) \\geq 0\n$$\n\nwhich is true.\n\nRemark: Notice that the equality holds for $a=c=\\sqrt{p}=\\sqrt[3]{\\frac{1+\\sqrt{5}}{2}}$ and $b=\\frac{1}{a c}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nA 3.", "solution_match": "\nSolution."}}
-{"year": "2020", "tier": "T3", "problem_label": "C 1", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Alice and Bob play the following game: starting with the number 2 written on a blackboard, each player in turn changes the current number $n$ to a number $n+p$, where $p$ is a prime divisor of $n$. Alice goes first and the players alternate in turn. The game is lost by the one who is forced to write a number greater than $\\underbrace{2 \\ldots 2}_{2020}$. Assuming perfect play, who will win the game.", "solution": "We prove that Alice wins the game. For argument's sake, suppose that Bob can win by proper play regardless of what Alice does on each of her moves. Note that Alice can force the line $2 \\rightarrow \\mathbf{4} \\rightarrow 6 \\rightarrow \\mathbf{8} \\rightarrow 10 \\rightarrow \\mathbf{1 2}$ at the beginning stages of the game. (As each intermediate 'position' from which Bob has to play is a prime power.) Thus the player on turn when the number 12 is written on the blackboard must be in a 'winning position', i.e., can win the game with skillful play. However, Alice can place herself in that position through the following line that is once again forced for Bob: $2 \\rightarrow \\mathbf{4} \\rightarrow 6 \\rightarrow \\mathbf{9} \\rightarrow 12$. (This time she is in turn with 12 written on the blackboard.) The obtained contradiction proves our point.\n\nComment by the PSC. Notice that this is a game of two players which always ends in a finite number of moves with a winner. For games like this, a player whose turn is to make a move, may be in position to force a win for him. If not, then the other player is in position to force a win for him.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nC 1.", "solution_match": "\nSolution."}}
-{"year": "2020", "tier": "T3", "problem_label": "C 2", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Viktor and Natalia bought 2020 buckets of ice-cream and want to organize a degustation schedule with 2020 rounds such that:\n\n- In every round, each one of them tries 1 ice-cream, and those 2 ice-creams tried in a single round are different from each other.\n- At the end of the 2020 rounds, each one of them has tried each ice-cream exactly once.\n\nWe will call a degustation schedule fair if the number of ice-creams that were tried by Viktor before Natalia is equal to the number of ice creams tried by Natalia before Viktor.\n\nProve that the number of fair schedules is strictly larger than $2020!\\left(2^{1010}+(1010!)^{2}\\right)$.", "solution": "If we fix the order in which Natalia tries the ice-creams, we may consider 2 types of fair schedules:\n\n1) Her last 1010 ice-creams get assigned as Viktor's first 1010 ice-creams, and vice versa: Viktor's first 1010 ice-creams are assigned as Natalia's last 1010 ice-creams. This generates (1010!) $)^{2}$ distinct fair schedules by permuting the ice-creams within each group.\n2) We divide all ice-creams into disjoint groups of 4 , and in each group we swap the first 2 ice-creams with the last 2 , which gives us $\\left((2!)^{2}\\right)^{504}=2^{1010}$ distinct schedules.\n\nNow, to make the inequality strict, we consider 1 more schedule like 2 ), but with groups of 2 ice-creams instead of 4 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nC 2.", "solution_match": "\nSolution."}}
-{"year": "2020", "tier": "T3", "problem_label": "C 3", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Alice and Bob play the following game: Alice begins by picking a natural number $n \\geqslant 2$. Then, with Bob starting first, they alternately choose one number from the set $A=\\{1,2, \\ldots, n\\}$ according to the following condition: The number chosen at each step should be distinct from all the already chosen numbers, and should differ by 1 from an already chosen number. (At the very first step Bob can choose any number he wants.) The game ends when all numbers from the set $A$ are chosen.\n\nFor example, if Alice picks $n=4$, then a valid game would be for Bob to choose 2, then Alice to choose 3, then Bob to choose 1, and then Alice to choose 4 .\n\nAlice wins if the sum $S$ of all of the numbers that she has chosen is composite. Otherwise Bob wins. (In the above example $S=7$, so Bob wins.)\n\nDecide which player has a winning strategy.", "solution": "Alice has a winning strategy. She initially picks $n=8$. We will give a strategy so that she can end up with $S$ even, or $S=15$, or $S=21$, so she wins.\n\nCase 1: If Bob chooses 1, then the game ends with Alice choosing 2,4,6,8 so $S$ is even (larger than 2) and Alice wins.\n\nCase 2: If Bob chooses 2, then Alice chooses 3. Bob can now choose either 1 or 3 .\n\nCase 2A: If Bob chooses 1, then Alice's numbers are $3,4,6,8$. So $S=21$ and Alice wins.\n\nCase 2B: If Bob chooses 4, then Alice chooses 1 and ends with the numbers 1,3,6,8. So $S$ is even and alice wins.\n\nCase 3: If Bob chooses 3, then Alice chooses 2. Bob can now choose either 1 or 4 .\n\nCase 3A: If Bob chooses 1, then Alice's numbers are 2,4,6,8. So $S$ is even and Alice wins.\n\nCase 3B: If Bob chooses 4, then Alice chooses 5. Bob can now choose either 1 or 6 .\n\nCase 3Bi: If Bob chooses 1, then Alice's numbers are 2,5,6,8. So $S=21$ and Alice wins.\n\nCase 3Bii: If Bob chooses 6 , then Alice chooses 1. Then Alice's numbers are 2,5,1,8. So $S$ is even and Alice wins.\n\nCase 4: If Bob chooses 4, then Alice chooses 5. Bob can now choose either 3 or 6 .\n\nCase 4A: If Bob chooses 3, then Alice chooses 6. Bob can now choose either 2 or 7 .\n\nCase 4Ai: If Bob chooses 2, then Alice chooses 1 and ends up with 5,6,1,8. So $S$ is even and Alice wins.\n\nCase 4Aii: If Bob chooses 7, then Alice chooses 8 and ends up with $5,6,8,1$. So $S$ is even and Alice wins.\n\nCase 4B: If Bob chooses 6, then Alice chooses 7. Bob can now choose either 3 or 8 .\n\nCase 4Bi: If Bob chooses 3, then Alice chooses 2 and ends up with 5,7,2 and either 1 or 8 . So $S=15$ or $S=22$ and Alice wins.\n\nCase 4Bii: If Bob chooses 8, then Alice's numbers are $5,7,3,1$. So $S$ is even and Alice wins.\n\nCases 5-8: If Bob chooses $k \\in\\{5,6,7,8\\}$ then Alice follows the strategy in case $9-k$ but whenever she had to choose $\\ell$, she instead chooses $9-\\ell$. If at the end of that strategy she ended up with $S$, she will now end up with $S^{\\prime}=4 \\cdot 9-S=36-S$. Then $S^{\\prime}$ is even or $S^{\\prime}=15$ or $S^{\\prime}=21$ so again she wins.\n\n## GEOMETRY", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nC 3.", "solution_match": "\nSolution."}}
-{"year": "2020", "tier": "T3", "problem_label": "G 1", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $\\triangle A B C$ be an acute triangle. The line through $A$ perpendicular to $B C$ intersects $B C$ at $D$. Let $E$ be the midpoint of $A D$ and $\\omega$ the the circle with center $E$ and radius equal to $A E$. The line $B E$ intersects $\\omega$ at a point $X$ such that $X$ and $B$ are not on the same side of $A D$ and the line $C E$ intersects $\\omega$ at a point $Y$ such that $C$ and $Y$ are not on the same side of $A D$. If both of the intersection points of the circumcircles of $\\triangle B D X$ and $\\triangle C D Y$ lie on the line $A D$, prove that $A B=A C$.", "solution": "Alternative Solution. Let the circle $\\odot C D Y$ intersects the line $A D$ at another point $Z$. Then we have $\\angle C E D=\\angle Y E Z$. We also have $E D=E Y$ because $E$ is the center of the circle $\\omega$. Also note that\n\n$$\n\\angle D C E=\\angle D C Y=\\angle D Z Y=\\angle E Z Y\n$$\n\nWe conclude that $\\triangle C D E$ and $\\triangle Z Y E$ are congruent. From here we have that $E Z=E C$. Now denote by $Z^{\\prime}$ the other intersection point of $A D$ and $\\odot B D X$. In the same way we prove that $E Z^{\\prime}=E B$. By the assumption of the problem, we must have that $Z=Z^{\\prime}$. We now conclude that\n\n$$\nB E=C E=E Z=E Z^{\\prime}\n$$\n\nAlso, $\\angle B D E=90^{\\circ}=\\angle C D E$. Now we see that $\\triangle B D E$ and $\\triangle C D E$ are congruent (they share the side $E D)$, so $B D=C D$. But $D$ is both the midpoint of $B C$ and the foot of the altitude from $A$, which means that $A B=A C$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_8956532c2a2d8e9f5e66g-12.jpg?height=989&width=1326&top_left_y=991&top_left_x=367)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nG 1.", "solution_match": "\nSolution."}}
-{"year": "2020", "tier": "T3", "problem_label": "G 1", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $\\triangle A B C$ be an acute triangle. The line through $A$ perpendicular to $B C$ intersects $B C$ at $D$. Let $E$ be the midpoint of $A D$ and $\\omega$ the the circle with center $E$ and radius equal to $A E$. The line $B E$ intersects $\\omega$ at a point $X$ such that $X$ and $B$ are not on the same side of $A D$ and the line $C E$ intersects $\\omega$ at a point $Y$ such that $C$ and $Y$ are not on the same side of $A D$. If both of the intersection points of the circumcircles of $\\triangle B D X$ and $\\triangle C D Y$ lie on the line $A D$, prove that $A B=A C$.", "solution": "Alternative Solution. Let $\\alpha=\\angle B X D$. Denote by $T$ the second intersection point of the circumcircles of $\\triangle B D X$ and $\\triangle C D Y$, which is on $A D$. We have\n\n$$\nE D=E A=E X\n$$\n\nbecause $E$ is the center of $\\omega$. Now $E X=E D$ implies $\\angle E D X=\\alpha$. From here we have that $\\angle B E D=2 \\alpha$. Using that $B T X D$ is cyclic we obtain $\\angle B T D=\\angle B X D=\\alpha$. We also have that\n\n$$\n\\angle T B E=180^{\\circ}-\\angle B E T-\\angle E T B=\n$$\n\n$$\n=180^{\\circ}-\\left(180^{\\circ}-2 \\alpha\\right)-\\alpha=\\alpha=\\angle B T E\n$$\n\nThis gives us $B E=T E$. We similarly show that $C E=T E$, and so $B E=C E$. In the same way as in the second solution, this now gives us that $B D=C D$, so $D$ is also the midpoint of $B C$ and we must have $A B=A C$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_8956532c2a2d8e9f5e66g-13.jpg?height=938&width=1181&top_left_y=526&top_left_x=453)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nG 1.", "solution_match": "\nSolution."}}
-{"year": "2020", "tier": "T3", "problem_label": "G 1", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $\\triangle A B C$ be an acute triangle. The line through $A$ perpendicular to $B C$ intersects $B C$ at $D$. Let $E$ be the midpoint of $A D$ and $\\omega$ the the circle with center $E$ and radius equal to $A E$. The line $B E$ intersects $\\omega$ at a point $X$ such that $X$ and $B$ are not on the same side of $A D$ and the line $C E$ intersects $\\omega$ at a point $Y$ such that $C$ and $Y$ are not on the same side of $A D$. If both of the intersection points of the circumcircles of $\\triangle B D X$ and $\\triangle C D Y$ lie on the line $A D$, prove that $A B=A C$.", "solution": "Alternative Solution. We can solve the problem using only calculations. Note that the condition of the problem is that $E$ lies on the radical axis of the circumcircles of $\\triangle B D X$ and $\\triangle C D Y$. This gives us $E B \\cdot E X=E C \\cdot E Y$. However, $E X=E Y$ because $E$ is the center of $\\omega$ and this means that $B E=C E$. Now using Pythagoras' theorem we have the following:\n\n$$\nA B^{2}=A D^{2}+B D^{2}=A D^{2}+\\left(B E^{2}-D E^{2}\\right)=A D^{2}+\\left(C E^{2}-D E^{2}\\right)=A D^{2}+C D^{2}=A C^{2}\n$$\n\nFrom here we obtain $A B=A C$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nG 1.", "solution_match": "\nSolution."}}
-{"year": "2020", "tier": "T3", "problem_label": "G 2", "problem_type": "Geometry", "exam": "JBMO", "problem": "Problem: Let $\\triangle A B C$ be a right-angled triangle with $\\angle B A C=90^{\\circ}$, and let $E$ be the foot of the perpendicular from $A$ on $B C$. Let $Z \\neq A$ be a point on the line $A B$ with $A B=B Z$. Let (c), $\\left(c_{1}\\right)$ be the circmucircles of the triangles $\\triangle A E Z$ and $\\triangle B E Z$, respectively. Let ( $c_{2}$ ) be an arbitrary circle passing through the points $A$ and $E$. Suppose $\\left(c_{1}\\right)$ meets the line $C Z$ again at the point $F$, and meets $\\left(c_{2}\\right)$ again at the point $N$. If $P$ is the other point of intesection of $\\left(c_{2}\\right)$ with $A F$, prove that the points $N, B, P$ are collinear.", "solution": "Since the triangles $\\triangle A E B$ and $\\triangle C A B$ are similar, then\n\n$$\n\\frac{A B}{E B}=\\frac{C B}{A B}\n$$\n\nSince $A B=B Z$ we get\n\n$$\n\\frac{B Z}{E B}=\\frac{C B}{B Z}\n$$\n\nfrom which it follows that the triangles $\\triangle Z B E$ and $\\triangle C B Z$ are also similar. Since $F E B Z$ is cyclic,\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_8956532c2a2d8e9f5e66g-14.jpg?height=1087&width=1401&top_left_y=1012&top_left_x=248)\n\nthen $\\angle B E Z=\\angle B F Z$. So by the similarity of triangles $\\triangle Z B E$ and $\\triangle C B Z$ we get\n\n$$\n\\angle B F Z=\\angle B E Z=\\angle B Z C=\\angle B Z F\n$$\n\nand therefore the triangle $\\triangle B F Z$ is isosceles. Since $B F=B Z=A B$, then the triangle $\\triangle A F Z$ is right-angled with $\\angle A F Z=90^{\\circ}$.\n\nIt now follows that the points $A, E, F, C$ are concyclic. Since $A, P, E, N$ are also concyclic, then\n\n$$\n\\angle E N P=\\angle E A P=\\angle E A F=\\angle E C F=\\angle B C Z=\\angle B Z E,\n$$\n\nwhere in the last equality we used again the similarity of the triangles $\\triangle Z B E$ and $\\triangle C B Z$. Since $N, B, E, Z$ are concyclic, then $\\angle E N P=\\angle B Z E=\\angle E N B$, from which it follows that the points $N, B, P$ are collinear.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nG 2.", "solution_match": "\nSolution."}}
-{"year": "2020", "tier": "T3", "problem_label": "G 3", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $\\triangle A B C$ be a right-angled triangle with $\\angle B A C=90^{\\circ}$ and let $E$ be the foot of the perpendicular from $A$ on $B C$. Let $Z \\neq A$ be a point on the line $A B$ with $A B=B Z$. Let (c) be the circumcircle of the triangle $\\triangle A E Z$. Let $D$ be the second point of intersection of $(c)$ with $Z C$ and let $F$ be the antidiametric point of $D$ with respect to $(c)$. Let $P$ be the point of intersection of the lines $F E$ and $C Z$. If the tangent to $(c)$ at $Z$ meets $P A$ at $T$, prove that the points $T, E, B, Z$ are concyclic.", "solution": "We will first show that $P A$ is tangent to $(c)$ at $A$.\n\nSince $E, D, Z, A$ are concyclic, then $\\angle E D C=\\angle E A Z=\\angle E A B$. Since also the triangles $\\triangle A B C$ and $\\triangle E B A$ are similar, then $\\angle E A B=\\angle B C A$, therefore $\\angle E D C=\\angle B C A$.\n\nSince $\\angle F E D=90^{\\circ}$, then $\\angle P E D=90^{\\circ}$ and so\n\n$$\n\\angle E P D=90^{\\circ}-\\angle E D C=90^{\\circ}-\\angle B C A=\\angle E A C\n$$\n\nTherefore the points $E, A, C, P$ are concyclic. It follows that $\\angle C P A=90^{\\circ}$ and therefore the triangle $\\angle P A Z$ is right-angled. Since also $B$ is the midpoint of $A Z$, then $P B=A B=B Z$ and so $\\angle Z P B=$ $\\angle P Z B$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_8956532c2a2d8e9f5e66g-16.jpg?height=1215&width=1016&top_left_y=1008&top_left_x=533)\n\nFurthermore, $\\angle E P D=\\angle E A C=\\angle C B A=\\angle E B A$ from which it follows that the points $P, E, B, Z$ are also concyclic.\n\nNow observe that\n\n$$\n\\angle P A E=\\angle P C E=\\angle Z P B-\\angle P B E=\\angle P Z B-\\angle P Z E=\\angle E Z B\n$$\n\nTherefore $P A$ is tangent to $(c)$ at $A$ as claimed.\n\nIt now follows that $T A=T Z$. Therefore\n\n$$\n\\begin{aligned}\n\\angle P T Z & =180^{\\circ}-2(\\angle T A B)=180^{\\circ}-2(\\angle P A E+\\angle E A B)=180^{\\circ}-2(\\angle E C P+\\angle A C B) \\\\\n& =180^{\\circ}-2\\left(90^{\\circ}-\\angle P Z B\\right)=2(\\angle P Z B)=\\angle P Z B+\\angle B P Z=\\angle P B A .\n\\end{aligned}\n$$\n\nThus $T, P, B, Z$ are concyclic, and since $P, E, B, Z$ are also concyclic then $T, E, B, Z$ are concyclic as required.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nG 3.", "solution_match": "\nSolution."}}
-{"year": "2020", "tier": "T3", "problem_label": "NT 1", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Determine whether there is a natural number $n$ for which $8^{n}+47$ is prime.", "solution": "The number $m=8^{n}+47$ is never prime.\n\nIf $n$ is even, say $n=2 k$, then $m=64^{k}+47 \\equiv 1+2 \\equiv 0 \\bmod 3$. Since also $m>3$, then $m$ is not prime.\n\nIf $n \\equiv 1 \\bmod 4$, say $n=4 k+1$, then $m=8 \\cdot\\left(8^{k}\\right)^{4}+47 \\equiv 3+2 \\equiv 0 \\bmod 5$. Since also $m>3$, then $m$ is not prime.\n\nIf $n \\equiv 3 \\bmod 4$, say $n=4 k+3$, then $m=8\\left(64^{2 k+1}+1\\right) \\equiv 8\\left((-1)^{2 k+1}+1\\right) \\equiv 0 \\bmod 13$. Since also $m>13$, then $m$ is not prime.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nNT 1.", "solution_match": "\nSolution."}}
-{"year": "2020", "tier": "T3", "problem_label": "NT 2", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all positive integers $a, b, c$ and $p$, where $p$ is a prime number, such that\n\n$$\n73 p^{2}+6=9 a^{2}+17 b^{2}+17 c^{2}\n$$", "solution": "Since the equation is symmetric with respect to the numbers $b$ and $c$, we assume that $b \\geq c$.\n\nIf $p \\geq 3$, then $p$ is an odd number. We consider the equation modulo 8 . Since,\n\n$$\n73 p^{2}+6 \\equiv 79 \\equiv 7 \\quad(\\bmod 8)\n$$\n\nwe get that\n\n$$\na^{2}+b^{2}+c^{2} \\equiv 7 \\quad(\\bmod 8)\n$$\n\nThis cannot happen since for any integer $x$ we have that\n\n$$\nx^{2} \\equiv 0,1,4 \\quad(\\bmod 8)\n$$\n\nHence, $p$ must be an even prime number, which means that $p=2$. In this case, we obtain the equation\n\n$$\n9 a^{2}+17\\left(b^{2}+c^{2}\\right)=289\n$$\n\nIt follows that $b^{2}+c^{2} \\leq 17$. This is possible only for\n\n$$\n(b, c) \\in\\{(4,1),(3,2),(3,1),(2,2),(2,1),(1,1)\\}\n$$\n\nIt is easy to check that among these pairs only the $(4,1)$ gives an integer solution for $a$, namely $a=1$. Therefore, the given equation has only two solutions,\n\n$$\n(a, b, c, p) \\in\\{(1,1,4,2),(1,4,1,2)\\}\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nNT 2.", "solution_match": "\nSolution."}}
-{"year": "2020", "tier": "T3", "problem_label": "NT 3", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find the largest integer $k(k \\geq 2)$, for which there exists an integer $n(n \\geq k)$ such that from any collection of $n$ consecutive positive integers one can always choose $k$ numbers, which verify the following conditions:\n\n1. each chosen number is not divisible by 6 , by 7 and by 8 ;\n2. the positive difference of any two different chosen numbers is not divisible by at least one of the numbers 6,7 or 8 .", "solution": "An integer is divisible by 6,7 and 8 if and only if it is divisible by their Least Common Multiple, which equals $6 \\times 7 \\times 4=168$.\n\nLet $n$ be a positive integer and let $A$ be an arbitrary set of $n$ consecutive positive integers. Replace each number $a_{i}$ from $A$ with its remainder $r_{i}$ ( mod 168). The number $a_{i}$ is divisible by 6 ( 7 or 8 ) if and only if its remainder $r_{i}$ is divisible by 6 (respectively 7 or 8 ). The difference $\\left|a_{i}-a_{j}\\right|$ is divisible by 168 if and only if their remainders $r_{i}=r_{j}$.\n\nChoosing $k$ numbers from the initial set $A$, which verify the required conditions, is the same as choosing $k$ their remainders ( mod 168) such that:\n\n1. each chosen remainder is not divisible by 6,7 and 8 ;\n2. all chosen remainders are different.\n\nSuppose we have chosen $k$ numbers from $A$, which verify the conditions. Therefore, all remainders are different and $k \\leq 168$ (otherwise, there would be two equal remainders).\n\nDenote by $B=\\{0,1,2,3, \\ldots, 167\\}$ the set of all possible remainders ( $\\bmod 168)$ and by $B_{m}$ the subset of all elements of $B$, which are divisible by $m$. Compute the number of elements of the following subsets:\n\n$$\n\\begin{gathered}\n\\left|B_{6}\\right|=168: 6=28, \\quad\\left|B_{7}\\right|=168: 7=24, \\quad\\left|B_{8}\\right|=168: 8=21 \\\\\n\\left|B_{6} \\cap B_{7}\\right|=\\left|B_{42}\\right|=168: 42=4, \\quad\\left|B_{6} \\cap B_{8}\\right|=\\left|B_{24}\\right|=168: 24=7 \\\\\n\\left|B_{7} \\cap B_{8}\\right|=\\left|B_{56}\\right|=168: 56=3, \\quad\\left|B_{6} \\cap B_{7} \\cap B_{8}\\right|=\\left|B_{168}\\right|=1\n\\end{gathered}\n$$\n\nDenote by $D=B_{6} \\cup B_{7} \\cup B_{8}$, the subset of all elements of $B$, which are divisible by at least one of the numbers 6,7 or 8 . By the Inclusion-Exclusion principle we got\n\n$$\n\\begin{gathered}\n|D|=\\left|B_{6}\\right|+\\left|B_{7}\\right|+\\left|B_{8}\\right|-\\left(\\left|B_{6} \\cap B_{7}\\right|+\\left|B_{6} \\cap B_{8}\\right|+\\left|B_{7} \\cap B_{8}\\right|\\right)+\\left|B_{6} \\cap B_{7} \\cap B_{8}\\right|= \\\\\n28+24+21-(4+7+3)+1=60 .\n\\end{gathered}\n$$\n\nEach chosen remainder belongs to the subset $B \\backslash D$, since it is not divisible by 6,7 and 8 . Hence, $k \\leq|B \\backslash D|=168-60=108$.\n\nLet us show that the greatest possible value is $k=108$. Consider $n=168$. Given any collection $A$ of 168 consecutive positive integers, replace each number with its remainder ( $\\bmod 168$ ). Choose from these remainders 108 numbers, which constitute the set $B \\backslash D$. Finally, take 108 numbers from the initial set $A$, having exactly these remainders. These $k=108$ numbers verify the required conditions.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nNT 3.", "solution_match": "\nSolution."}}
-{"year": "2020", "tier": "T3", "problem_label": "NT 4", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all prime numbers $p$ such that\n\n$$\n(x+y)^{19}-x^{19}-y^{19}\n$$\n\nis a multiple of $p$ for any positive integers $x, y$.", "solution": "If $x=y=1$ then $p$ divides\n\n$$\n2^{19}-2=2\\left(2^{18}-1\\right)=2\\left(2^{9}-1\\right)\\left(2^{9}+1\\right)=2 \\cdot 511 \\cdot 513=2 \\cdot 3^{3} \\cdot 7 \\cdot 19 \\cdot 73\n$$\n\nIf $x=2, y=1$ then\n\n$$\np \\mid 3^{19}-2^{19}-1\n$$\n\nWe will show that $3^{19}-2^{19}-1$ is not a multiple of 73 . Indeed,\n\n$$\n3^{19} \\equiv 3^{3} \\cdot\\left(3^{4}\\right)^{4} \\equiv 3^{3} \\cdot 8^{4} \\equiv 3^{3} \\cdot(-9)^{2} \\equiv 27 \\cdot 81 \\equiv 27 \\cdot 8 \\equiv 70 \\quad(\\bmod 73)\n$$\n\nand\n\n$$\n2^{19} \\equiv 2 \\cdot 64^{3} \\equiv 2 \\cdot(-9)^{3} \\equiv-18 \\cdot 81 \\equiv-18 \\cdot 8 \\equiv-144 \\equiv 2 \\quad(\\bmod 73)\n$$\n\nThus $p$ can be only among $2,3,7,19$. We will prove all these work.\n\n- For $p=19$ this follows by Fermat's Theorem as\n\n$$\n(x+y)^{19} \\equiv x+y \\quad(\\bmod 19), \\quad x^{19} \\equiv x \\quad(\\bmod 19), \\quad y^{19} \\equiv y \\quad(\\bmod 19)\n$$\n\n- For $p=7$, we have that\n\n$$\na^{19} \\equiv a \\quad(\\bmod 7)\n$$\n\nfor every integer $a$. Indeed, if $7 \\mid a$, it is trivial, while if $7 \\nmid a$, then by Fermat's Theorem we have\n\n$$\n7\\left|a^{6}-1\\right| a^{18}-1\n$$\n\ntherefore\n\n$$\n7 \\mid a\\left(a^{18}-1\\right)\n$$\n\n- For $p=3$, we will prove that\n\n$$\nb^{19} \\equiv b \\quad(\\bmod 3)\n$$\n\nIndeed, if $3 \\mid b$, it is trivial, while if $3 \\nmid b$, then by Fermat's Theorem we have\n\n$$\n3\\left|b^{2}-1\\right| b^{18}-1\n$$\n\ntherefore\n\n$$\n3 \\mid b\\left(b^{18}-1\\right)\n$$\n\n- For $p=2$ it is true, since among $x+y, x$ and $y$ there are 0 or 2 odd numbers.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nNT 4.", "solution_match": "\nSolution."}}
-{"year": "2020", "tier": "T3", "problem_label": "NT 5", "problem_type": "Number Theory", "exam": "JBMO", "problem": "The positive integer $k$ and the set $A$ of different integers from 1 to $3 k$ inclusive are such that there are no distinct $a, b, c$ in $A$ satisfying $2 b=a+c$. The numbers from $A$ in the interval $[1, k]$ will be called small; those in $[k+1,2 k]$ - medium and those in $[2 k+1,3 k]$ - large. Is it always true that there are no positive integers $x$ and $d$ such that if $x, x+d$ and $x+2 d$ are divided by $3 k$ then the remainders belong to $A$ and those of $x$ and $x+d$ are different and are:\na) small?\nb) medium?\nc) large?\n\n(In this problem we assume that if a multiple of $3 k$ is divided by $3 k$ then the remainder is $3 k$ rather than 0. )", "solution": "Solution. A counterexample for a) is $k=3, A=\\{1,2,9\\}, x=2$ and $d=8$. A counterexample for c) is $k=3, A=\\{1,8,9\\}, x=8$ and $d=1$.\n\nWe will prove that b) is true.\n\nSuppose the contrary and let $x, d$ have the above properties. We can assume $0<d<3 k, 0<x \\leq 3 k$ (since for $d=3 k$ the remainders for $x$ and $x+d$ are equal). Hence $0<x+d<6 k$ and there are two cases:\n\n- If $x+d>3 k$, then since the remainder for $x+d$ is medium we have $4 k<x+d \\leq 5 k$. This means that the remainder of $x+d$ when it is divided by $3 k$ is\n\n$$\nx+d-3 k\n$$\n\nSince $x$ is medium we have $x \\leq 2 k$ so $d=(x+d)-x>2 k$. Therefore $6 k=4 k+2 k<(x+d)+d<$ $8 k$. This means that the remainder of $x+2 d$ when it is divided by $3 k$ is\n\n$$\nx+2 d-6 k \\text {. }\n$$\n\nThus the remainders $(x+2 d-6 k),(x+d-3 k)$ and $x$ are in $[1,3 k]$, they belong to $A$ and\n\n$$\n2(x+d-3 k)=(x+2 d-6 k)+x\n$$\n\na contradiction.\n\n- If $x+d \\leq 3 k$ then as $x+d$ is medium we have $k<x+d \\leq 2 k$. From the limitations on $x$, we have $x>k$ so $d=(x+d)-x<k$. Hence $0 \\leq x+2 d=(x+d)+d<3 k$. Thus the remainders $x, x+d$ and $x+2 d$ are in $A$ and\n\n$$\n2(x+d)=(x+2 d)+x\n$$\n\na contradiction.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nNT 5.", "solution_match": "## Solution."}}
-{"year": "2020", "tier": "T3", "problem_label": "NT 6", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Are there any positive integers $m$ and $n$ satisfying the equation\n\n$$\nm^{3}=9 n^{4}+170 n^{2}+289 ?\n$$", "solution": "We will prove that the answer is no. Note that\n\n$$\nm^{3}=9 n^{4}+170 n^{2}+289=\\left(9 n^{2}+17\\right)\\left(n^{2}+17\\right)\n$$\n\nIf $n$ is odd then $m$ is even, therefore $8 \\mid m^{3}$. However,\n\n$$\n9 n^{4}+170 n^{2}+289 \\equiv 9+170+289 \\equiv 4(\\bmod 8)\n$$\n\nwhich leads to a contradiction. If $n$ is a multiple of 17 then so is $m$ and hence 289 is a multiple of $17^{3}$, which is absurd. For $n$ even and not multiple of 17 , since\n\n$$\n\\operatorname{gcd}\\left(9 n^{2}+17, n^{2}+17\\right) \\mid 9\\left(n^{2}+17\\right)-\\left(9 n^{2}+17\\right)=2^{3} \\cdot 17\n$$\n\nthis gcd must be 1 . Therefore $n^{2}+17=a^{3}$ for an odd $a$, so\n\n$$\nn^{2}+25=(a+2)\\left(a^{2}-2 a+4\\right)\n$$\n\nFor $a \\equiv 1(\\bmod 4)$ we have $a+2 \\equiv 3(\\bmod 4)$, while for $a \\equiv 3(\\bmod 4)$ we have $a^{2}-2 a+4 \\equiv 3$ $(\\bmod 4)$. Thus $(a+2)\\left(a^{2}-2 a+4\\right)$ has a prime divisor of type $4 \\ell+3$. As it divides $n^{2}+25$, it has to divide $n$ and 5 , which is absurd.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nNT 6.", "solution_match": "\nSolution."}}
-{"year": "2020", "tier": "T3", "problem_label": "NT 7", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Prove that there doesn't exist any prime $p$ such that every power of $p$ is a palindrome (palindrome is a number that is read the same from the left as it is from the right; in particular, number that ends in one or more zeros cannot be a palindrome).", "solution": "Note that by criterion for divisibility by 11 and the definition of a palindrome we have that every palindrome that has even number of digits is divisible by 11 .\n\nSince $11^{5}=161051$ is not a palindrome and since 11 cannot divide $p^{k}$ for any prime other than 11 we are now left to prove that no prime whose all powers have odd number of digits exists.\n\nAssume the contrary. It means that the difference between the numbers of digits of $p^{m}$ and $p^{m+1}$ is even number. We will prove that for every natural $m$, the difference is the same even number.\n\nIf we assume not, that means that the difference for some $m_{1}$ has at least 2 digits more than the difference for some $m_{2}$. We will prove that this is impossible.\n\nLet $p^{m_{1}}=10^{t_{1}} \\cdot a_{1}, p^{m_{2}}=10^{t_{2}} \\cdot a_{2}$ and $p=10^{h} \\cdot z$, where\n\n$$\n1<a_{1}, a_{2}, z<10\n$$\n\nThis implies that\n\n$$\n1<a_{1} \\cdot z, a_{2} \\cdot z<100\n$$\n\nwhich further implies that multiplying these powers of $p$ by $p$ can increase their number of digits by either $h$ or $h+1$.\n\nThis is a contradiction. Call the difference between numbers of digits of consecutive powers $d$. Number $p$ clearly cannot be equal to $10^{d}$ for $d \\geq 1$ because 10 is divisible by two primes, but for $d=0$, we would have that 1 is a prime which is not true.\n\nCase 1. $p>10^{d}$. Let $p=10^{d} \\cdot a$, for some real number $a$ greater than 1. (1)\n\nFrom the definition of $d$ we also see that $a$ is smaller than 10. (2)\n\nFrom (1) we see that powering $a$ gives us arbitrarily large numbers and from (2) we conclude that there is some natural power of $a$, call it $b$, greater than 1 , such that\n\n$$\n10<a^{b}<100\n$$\n\nIt is clear that $p^{b}$ has exactly $(b-1) d+1$ digits more than $p$ has, which is an odd number, but sum of even numbers is even.\n\nCase 2. $p<10^{d}$. Let $p=\\frac{10^{d}}{a}$, for some real number $a$ greater than 1. (1) From the definition of $d$ we also see that $a$ is smaller than 10. (2)\n\nFrom (1) we see that powering $a$ gives us arbitrarily large numbers and from (2) we conclude that there is some natural power of $a$, call it $b$, greater than 1 , such that\n\n$$\n10<a^{b}<100\n$$\n\nIt is clear that $p^{b}$ has exactly $(b-1) d-1$ digits more than $p$ has, which is an odd number, but sum of even numbers is even.\n\nWe have now arrived at the desired contradiction for both cases and have thus finished the proof.\n\nAlternative solution. Note that the sequence $\\left\\{p^{n}\\right\\}$ is periodic $(\\bmod 10)$. Let the period be $d$. Also, let $p^{d}=g$.\n\nSince all powers of $p$ are palindromes, all powers of $g$ are as well. Since $\\left\\{g^{n}\\right\\}$ is constant (mod 10), the leftmost digit of each power of $g$ is equal to some $f$.\n\nWe will prove that the difference between numbers of digits of $g^{m}$ and $g^{m+1}$ is equal to some $r$ for every natural number $m$.\n\nThis is true due to size reasons. Namely, to add exactly $k$ digits, and yet to have the same leftmost digit, we need to multiply the number by at least $5 \\cdot 10^{k-1}$ (if $k=0$ then it's 1 ) and by at most $2 \\cdot 10^{k}$ (values depend on the leftmost digit, it can easily be seen that leftmost digit being 1 yields the extremal values). Notice that\n\n$$\n2 \\cdot 10^{k}<5 \\cdot 10^{k+1-1}\n$$\n\nSince this inequality has clearly shown that the interval of multipliers which add exactly $k$ digits and leave the leftmost digit the same is disjunct from the same kind of interval for $k+1$ digits, which implies that no number can belong to both intervals, we have successfully proven the claim.\n\nClearly, $g$ cannot be equal to $10^{r}$ for $r \\geq 1$ because a palindrome cannot be divisible by 10 , but for $r=0$ we again cannot have the equality because 1 is not a natural power of a prime.\n\nCase 1. $g>10^{r}$. Let $g=10^{r} \\cdot a$, where $a$ is a real number, $10>a>1$. (1) Here, $a$ is less than 10 because if it was not, multiplying by $g$ would add at least $r+1$ digits, which is impossible.\n\nFrom (1) we see that powering $a$ gives us arbitrarily large numbers and that there is some natural power of $a$, call it $b$, greater than 1 , such that\n\n$$\n10<a^{b}<100\n$$\n\nPick smallest such $b$.\n\nNow we easily see that the difference between numbers of digits of numbers $g^{b-1}$ and $g^{b}$ is exactly $r+1$.\n\nCase 2. $g<10^{r}$. Let $g=\\frac{10^{r}}{a}$, where $a$ is a real number, $10>a>1$. (2) Here, $a$ is less than 10 because if it was not, multiplying by $g$ would add at most $r-1$ digits, which is impossible.\n\nFrom (2) we see that powering $a$ gives us arbitrarily large numbers and that there is some natural power of $a$, call it $b$, greater than 1 , such that\n\n$$\n10<a^{b}<100\n$$\n\nPick smallest such $b$.\n\nNow we easily see that the difference between numbers of digits of numbers $g^{b-1}$ and $g^{b}$ is exactly $r-1$.\n\nWe have arrived at the desired contradiction for both cases and have thus finished the proof.\n\nComment. In both solutions, after introducing $a$, there are multiple ways to finish the problem. In particular, solution 1 and solution 2 could be finished in the same way, but distinct finishes were purposely offered. Third, maybe even the most intuitive finishing argument, could be using non-exact size arguments; namely, just the fact that powers of $a$ grow arbitrarily large is enough to reach the contradiction.\n\nAlternative problem. Find all positive integers $n$ such that every power of $n$ is a palindrome (palindrome is a number that is read the same from the left as it is from the right; in particular, number that ends in one or more zeros cannot be a palindrome).\n\nSuggested difficulty for this problem is hard. Noting why solution 2 doesn't work for $n=1$ (because\n$g$ now could be equal to 1 ) and saying that $n=1$ actually works are all necessary modifications to solution 2 to make it work for the alternative problem as well.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nNT 7.", "solution_match": "\nSolution."}}
-{"year": "2020", "tier": "T3", "problem_label": "NT 8", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all pairs $(p, q)$ of prime numbers such that\n\n$$\n1+\\frac{p^{q}-q^{p}}{p+q}\n$$\n\nis a prime number.", "solution": "It is clear that $p \\neq q$. We set\n\n$$\n1+\\frac{p^{q}-q^{p}}{p+q}=r\n$$\n\nand we have that\n\n$$\np^{q}-q^{p}=(r-1)(p+q)\n$$\n\nFrom Fermat's Little Theorem we have\n\n$$\np^{q}-q^{p} \\equiv-q \\quad(\\bmod p)\n$$\n\nSince we also have that\n\n$$\n(r-1)(p+q) \\equiv-r q-q \\quad(\\bmod p)\n$$\n\nfrom (1) we get that\n\n$$\nr q \\equiv 0 \\quad(\\bmod p) \\Rightarrow p \\mid q r\n$$\n\nhence $p \\mid r$, which means that $p=r$. Therefore, (1) takes the form\n\n$$\np^{q}-q^{p}=(p-1)(p+q)\n$$\n\nWe will prove that $p=2$. Indeed, if $p$ is odd, then from Fermat's Little Theorem we have\n\n$$\np^{q}-q^{p} \\equiv p \\quad(\\bmod q)\n$$\n\nand since\n\n$$\n(p-1)(p+q) \\equiv p(p-1) \\quad(\\bmod q)\n$$\n\nwe have\n\n$$\np(p-2) \\equiv 0 \\quad(\\bmod q) \\Rightarrow q|p(p-2) \\Rightarrow q| p-2 \\Rightarrow q \\leq p-2<p\n$$\n\nNow, from (2) we have\n\n$$\np^{q}-q^{p} \\equiv 0 \\quad(\\bmod p-1) \\Rightarrow 1-q^{p} \\equiv 0 \\quad(\\bmod p-1) \\Rightarrow q^{p} \\equiv 1 \\quad(\\bmod p-1)\n$$\n\nClearly $\\operatorname{gcd}(q, p-1)=1$ and if we set $k=\\operatorname{ord}_{p-1}(q)$, it is well-known that $k \\mid p$ and $k<p$, therefore $k=1$. It follows that\n\n$$\nq \\equiv 1 \\quad(\\bmod p-1) \\Rightarrow p-1 \\mid q-1 \\Rightarrow p-1 \\leq q-1 \\Rightarrow p \\leq q\n$$\n\na contradiction.\n\nTherefore, $p=2$ and (2) transforms to\n\n$$\n2^{q}=q^{2}+q+2\n$$\n\nWe can easily check by induction that for every positive integer $n \\geq 6$ we have $2^{n}>n^{2}+n+2$. This means that $q \\leq 5$ and the only solution is for $q=5$. Hence the only pair which satisfy the condition is $(p, q)=(2,5)$.\n\nComment by the PSC. From the problem condition, we get that $p^{q}$ should be bigger than $q^{p}$, which gives\n\n$$\nq \\ln p>p \\ln q \\Longleftrightarrow \\frac{\\ln p}{p}>\\frac{\\ln q}{q}\n$$\n\nThe function $\\frac{\\ln x}{x}$ is decreasing for $x>e$, thus if $p$ and $q$ are odd primes, we obtain $q>p$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nNT 8.", "solution_match": "\nSolution."}}
+{"year": "2020", "tier": "T3", "problem_label": "A 1", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Find all triples $(a, b, c)$ of real numbers such that the following system holds:\n\n$$\n\\left\\{\\begin{array}{l}\na+b+c=\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} \\\\\na^{2}+b^{2}+c^{2}=\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}\n\\end{array}\\right.\n$$", "solution": "First of all if $(a, b, c)$ is a solution of the system then also $(-a,-b,-c)$ is a solution. Hence we can suppose that $a b c>0$. From the first condition we have\n\n$$\na+b+c=\\frac{a b+b c+c a}{a b c}\n$$\n\nNow, from the first condition and the second condition we get\n\n$$\n(a+b+c)^{2}-\\left(a^{2}+b^{2}+c^{2}\\right)=\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)^{2}-\\left(\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}\\right) .\n$$\n\nThe last one simplifies to\n\n$$\na b+b c+c a=\\frac{a+b+c}{a b c}\n$$\n\nFirst we show that $a+b+c$ and $a b+b c+c a$ are different from 0 . Suppose on contrary then from relation (1) or (2) we have $a+b+c=a b+b c+c a=0$. But then we would have\n\n$$\na^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(a b+b c+c a)=0\n$$\n\nwhich means that $a=b=c=0$. This is not possible since $a, b, c$ should be different from 0 . Now multiplying (1) and (2) we have\n\n$$\n(a+b+c)(a b+b c+c a)=\\frac{(a+b+c)(a b+b c+c a)}{(a b c)^{2}}\n$$\n\nSince $a+b+c$ and $a b+b c+c a$ are different from 0 , we get $(a b c)^{2}=1$ and using the fact that $a b c>0$ we obtain that $a b c=1$. So relations (1) and (2) transform to\n\n$$\na+b+c=a b+b c+c a\n$$\n\nTherefore,\n\n$$\n(a-1)(b-1)(c-1)=a b c-a b-b c-c a+a+b+c-1=0 .\n$$\n\nThis means that at least one of the numbers $a, b, c$ is equal to 1 . Suppose that $c=1$ then relations (1) and (2) transform to $a+b+1=a b+a+b \\Rightarrow a b=1$. Taking $a=t$ then we have $b=\\frac{1}{t}$. We can now verify that any triple $(a, b, c)=\\left(t, \\frac{1}{t}, 1\\right)$ satisfies both conditions. $t \\in \\mathbb{R} \\backslash\\{0\\}$. From the initial observation any triple $(a, b, c)=\\left(t, \\frac{1}{t},-1\\right)$ satisfies both conditions. $t \\in \\mathbb{R} \\backslash\\{0\\}$. So, all triples that satisfy both conditions are $(a, b, c)=\\left(t, \\frac{1}{t}, 1\\right),\\left(t, \\frac{1}{t},-1\\right)$ and all permutations for any $t \\in \\mathbb{R} \\backslash\\{0\\}$.\n\nComment by PSC. After finding that $a b c=1$ and\n\n$$\na+b+c=a b+b c+c a\n$$\n\nwe can avoid the trick considering $(a-1)(b-1)(c-1)$ as follows. By the Vieta's relations we have that $a, b, c$ are roots of the polynomial\n\n$$\nP(x)=x^{3}-s x^{2}+s x-1\n$$\n\nwhich has one root equal to 1 . Then, we can conclude as in the above solution.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nA 1.", "solution_match": "\nSolution."}}
+{"year": "2020", "tier": "T3", "problem_label": "A 2", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Consider the sequence $a_{1}, a_{2}, a_{3}, \\ldots$ defined by $a_{1}=9$ and\n\n$$\na_{n+1}=\\frac{(n+5) a_{n}+22}{n+3}\n$$\n\nfor $n \\geqslant 1$.\n\nFind all natural numbers $n$ for which $a_{n}$ is a perfect square of an integer.", "solution": "Define $b_{n}=a_{n}+11$. Then\n\n$$\n22=(n+3) a_{n+1}-(n+5) a_{n}=(n+3) b_{n+1}-11 n-33-(n+5) b_{n}+11 n+55\n$$\n\ngiving $(n+3) b_{n+1}=(n+5) b_{n}$. Then\n\n$b_{n+1}=\\frac{n+5}{n+3} b_{n}=\\frac{(n+5)(n+4)}{(n+3)(n+2)} b_{n-1}=\\frac{(n+5)(n+4)}{(n+2)(n+1)} b_{n-2}=\\cdots=\\frac{(n+5)(n+4)}{5 \\cdot 4} b_{1}=(n+5)(n+4)$.\n\nTherefore $b_{n}=(n+4)(n+3)=n^{2}+7 n+12$ and $a_{n}=n^{2}+7 n+1$.\n\nSince $(n+1)^{2}=n^{2}+2 n+1<a_{n}<n^{2}+8 n+16=(n+4)^{2}$, if $a_{n}$ is a perfect square, then $a_{n}=(n+2)^{2}$ or $a_{n}=(n+3)^{2}$.\n\nIf $a_{n}=(n+2)^{2}$, then $n^{2}+4 n+4=n^{2}+7 n+1$ giving $n=1$. If $a_{n}=(n+3)^{2}$, then $n^{2}+6 n+9=$ $n^{2}+7 n+1$ giving $n=8$.\n\nComment. We provide some other methods to find $a_{n}$.\n\nMethod 1: Define $b_{n}=\\frac{a_{n}+11}{n+3}$. Then $b_{1}=5$ and $a_{n}=(n+3) b_{n}-11$. So\n\n$$\na_{n+1}=(n+4) b_{n+1}-11=\\frac{(n+5) a_{n}+22}{n+3}=a_{n}+\\frac{2\\left(a_{n}+11\\right)}{n+3}=(n+3) b_{n}-11+2 b_{n}\n$$\n\ngiving $(n+4) b_{n+1}=(n+5) b_{n}$. Then\n\n$$\nb_{n+1}=\\frac{n+5}{n+4} b_{n}=\\frac{n+5}{n+3} b_{n-1}=\\cdots=\\frac{n+5}{5} b_{1}=n+5\n$$\n\nThen $b_{n}=n+4$, so $a_{n}=(n+3)(n+4)-11=n^{2}+7 n+1$.\n\nMethod 2: We have\n\n$$\n(n+3) a_{n+1}-(n+5) a_{n}=22\n$$\n\nand therefore\n\n$$\n\\frac{a_{n+1}}{(n+5)(n+4)}-\\frac{a_{n}}{(n+4)(n+3)}=\\frac{22}{(n+3)(n+4)(n+5)}=11\\left[\\frac{1}{n+3}-\\frac{2}{n+4}+\\frac{1}{n+5}\\right]\n$$\n\nNow define $b_{n}=\\frac{a_{n}}{(n+4)(n+3)}$ to get\n\n$$\nb_{n+1}=b_{n}+11\\left[\\frac{1}{n+3}-\\frac{2}{n+4}+\\frac{1}{n+5}\\right]\n$$\n\nwhich telescopically gives\n\n$$\n\\begin{aligned}\nb_{n+1} & =b_{1}+11\\left[\\left(\\frac{1}{4}-\\frac{2}{5}+\\frac{1}{6}\\right)+\\left(\\frac{1}{5}-\\frac{2}{6}+\\frac{1}{7}\\right)+\\cdots+\\left(\\frac{1}{n+3}-\\frac{2}{n+4}+\\frac{1}{n+5}\\right)\\right] \\\\\n& =b_{1}+11\\left(\\frac{1}{4}-\\frac{1}{5}-\\frac{1}{n+4}+\\frac{1}{n+5}\\right)=\\frac{9}{20}+\\frac{11}{20}-\\frac{11}{(n+4)(n+5)}\n\\end{aligned}\n$$\n\nWe get $b_{n+1}=(n+4)(n+5)-11$ from which it follows that $b_{n}=(n+3)(n+4)-11=n^{2}+7 n+1$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nA 2.", "solution_match": "\nSolution:"}}
+{"year": "2020", "tier": "T3", "problem_label": "A 3", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Find all triples of positive real numbers $(a, b, c)$ so that the expression\n\n$$\nM=\\frac{(a+b)(b+c)(a+b+c)}{a b c}\n$$\n\ngets its least value.", "solution": "The expression $M$ is homogeneous, therefore we can assume that $a b c=1$. We set $s=a+c$ and $p=a c$ and using $b=\\frac{1}{a c}$, we get\n\n$$\nM=\\left(a+\\frac{1}{a c}\\right)\\left(\\frac{1}{a c}+c\\right)\\left(a+\\frac{1}{a c}+c\\right)=\\left(a+p^{-1}\\right)\\left(c+p^{-1}\\right)\\left(s+p^{-1}\\right)\n$$\n\nExpanding the right-hand side we get\n\n$$\nM=p s+\\frac{s^{2}}{p}+1+\\frac{2 s}{p^{2}}+\\frac{1}{p^{3}}\n$$\n\nNow by $s \\geq 2 \\sqrt{p}$ and setting $x=p \\sqrt{p}>0$ we get\n\n$$\nM \\geq 2 x+5+\\frac{4}{x}+\\frac{1}{x^{2}}\n$$\n\nWe will now prove that\n\n$$\n2 x+5+\\frac{4}{x}+\\frac{1}{x^{2}} \\geq \\frac{11+5 \\sqrt{5}}{2} \\text {. }\n$$\n\nIndeed, the latter is equivalent to $4 x^{3}-(5 \\sqrt{5}+1) x^{2}+8 x+2 \\geq 0$, which can be rewritten as\n\n$$\n\\left(x-\\frac{1+\\sqrt{5}}{2}\\right)^{2}(4 x+3-\\sqrt{5}) \\geq 0\n$$\n\nwhich is true.\n\nRemark: Notice that the equality holds for $a=c=\\sqrt{p}=\\sqrt[3]{\\frac{1+\\sqrt{5}}{2}}$ and $b=\\frac{1}{a c}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nA 3.", "solution_match": "\nSolution."}}
+{"year": "2020", "tier": "T3", "problem_label": "C 1", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Alice and Bob play the following game: starting with the number 2 written on a blackboard, each player in turn changes the current number $n$ to a number $n+p$, where $p$ is a prime divisor of $n$. Alice goes first and the players alternate in turn. The game is lost by the one who is forced to write a number greater than $\\underbrace{2 \\ldots 2}_{2020}$. Assuming perfect play, who will win the game.", "solution": "We prove that Alice wins the game. For argument's sake, suppose that Bob can win by proper play regardless of what Alice does on each of her moves. Note that Alice can force the line $2 \\rightarrow \\mathbf{4} \\rightarrow 6 \\rightarrow \\mathbf{8} \\rightarrow 10 \\rightarrow \\mathbf{1 2}$ at the beginning stages of the game. (As each intermediate 'position' from which Bob has to play is a prime power.) Thus the player on turn when the number 12 is written on the blackboard must be in a 'winning position', i.e., can win the game with skillful play. However, Alice can place herself in that position through the following line that is once again forced for Bob: $2 \\rightarrow \\mathbf{4} \\rightarrow 6 \\rightarrow \\mathbf{9} \\rightarrow 12$. (This time she is in turn with 12 written on the blackboard.) The obtained contradiction proves our point.\n\nComment by the PSC. Notice that this is a game of two players which always ends in a finite number of moves with a winner. For games like this, a player whose turn is to make a move, may be in position to force a win for him. If not, then the other player is in position to force a win for him.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nC 1.", "solution_match": "\nSolution."}}
+{"year": "2020", "tier": "T3", "problem_label": "C 2", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Viktor and Natalia bought 2020 buckets of ice-cream and want to organize a degustation schedule with 2020 rounds such that:\n\n- In every round, each one of them tries 1 ice-cream, and those 2 ice-creams tried in a single round are different from each other.\n- At the end of the 2020 rounds, each one of them has tried each ice-cream exactly once.\n\nWe will call a degustation schedule fair if the number of ice-creams that were tried by Viktor before Natalia is equal to the number of ice creams tried by Natalia before Viktor.\n\nProve that the number of fair schedules is strictly larger than $2020!\\left(2^{1010}+(1010!)^{2}\\right)$.", "solution": "If we fix the order in which Natalia tries the ice-creams, we may consider 2 types of fair schedules:\n\n1) Her last 1010 ice-creams get assigned as Viktor's first 1010 ice-creams, and vice versa: Viktor's first 1010 ice-creams are assigned as Natalia's last 1010 ice-creams. This generates (1010!) $)^{2}$ distinct fair schedules by permuting the ice-creams within each group.\n2) We divide all ice-creams into disjoint groups of 4 , and in each group we swap the first 2 ice-creams with the last 2 , which gives us $\\left((2!)^{2}\\right)^{504}=2^{1010}$ distinct schedules.\n\nNow, to make the inequality strict, we consider 1 more schedule like 2 ), but with groups of 2 ice-creams instead of 4 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nC 2.", "solution_match": "\nSolution."}}
+{"year": "2020", "tier": "T3", "problem_label": "C 3", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Alice and Bob play the following game: Alice begins by picking a natural number $n \\geqslant 2$. Then, with Bob starting first, they alternately choose one number from the set $A=\\{1,2, \\ldots, n\\}$ according to the following condition: The number chosen at each step should be distinct from all the already chosen numbers, and should differ by 1 from an already chosen number. (At the very first step Bob can choose any number he wants.) The game ends when all numbers from the set $A$ are chosen.\n\nFor example, if Alice picks $n=4$, then a valid game would be for Bob to choose 2, then Alice to choose 3, then Bob to choose 1, and then Alice to choose 4 .\n\nAlice wins if the sum $S$ of all of the numbers that she has chosen is composite. Otherwise Bob wins. (In the above example $S=7$, so Bob wins.)\n\nDecide which player has a winning strategy.", "solution": "Alice has a winning strategy. She initially picks $n=8$. We will give a strategy so that she can end up with $S$ even, or $S=15$, or $S=21$, so she wins.\n\nCase 1: If Bob chooses 1, then the game ends with Alice choosing 2,4,6,8 so $S$ is even (larger than 2) and Alice wins.\n\nCase 2: If Bob chooses 2, then Alice chooses 3. Bob can now choose either 1 or 3 .\n\nCase 2A: If Bob chooses 1, then Alice's numbers are $3,4,6,8$. So $S=21$ and Alice wins.\n\nCase 2B: If Bob chooses 4, then Alice chooses 1 and ends with the numbers 1,3,6,8. So $S$ is even and alice wins.\n\nCase 3: If Bob chooses 3, then Alice chooses 2. Bob can now choose either 1 or 4 .\n\nCase 3A: If Bob chooses 1, then Alice's numbers are 2,4,6,8. So $S$ is even and Alice wins.\n\nCase 3B: If Bob chooses 4, then Alice chooses 5. Bob can now choose either 1 or 6 .\n\nCase 3Bi: If Bob chooses 1, then Alice's numbers are 2,5,6,8. So $S=21$ and Alice wins.\n\nCase 3Bii: If Bob chooses 6 , then Alice chooses 1. Then Alice's numbers are 2,5,1,8. So $S$ is even and Alice wins.\n\nCase 4: If Bob chooses 4, then Alice chooses 5. Bob can now choose either 3 or 6 .\n\nCase 4A: If Bob chooses 3, then Alice chooses 6. Bob can now choose either 2 or 7 .\n\nCase 4Ai: If Bob chooses 2, then Alice chooses 1 and ends up with 5,6,1,8. So $S$ is even and Alice wins.\n\nCase 4Aii: If Bob chooses 7, then Alice chooses 8 and ends up with $5,6,8,1$. So $S$ is even and Alice wins.\n\nCase 4B: If Bob chooses 6, then Alice chooses 7. Bob can now choose either 3 or 8 .\n\nCase 4Bi: If Bob chooses 3, then Alice chooses 2 and ends up with 5,7,2 and either 1 or 8 . So $S=15$ or $S=22$ and Alice wins.\n\nCase 4Bii: If Bob chooses 8, then Alice's numbers are $5,7,3,1$. So $S$ is even and Alice wins.\n\nCases 5-8: If Bob chooses $k \\in\\{5,6,7,8\\}$ then Alice follows the strategy in case $9-k$ but whenever she had to choose $\\ell$, she instead chooses $9-\\ell$. If at the end of that strategy she ended up with $S$, she will now end up with $S^{\\prime}=4 \\cdot 9-S=36-S$. Then $S^{\\prime}$ is even or $S^{\\prime}=15$ or $S^{\\prime}=21$ so again she wins.\n\n## GEOMETRY", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nC 3.", "solution_match": "\nSolution."}}
+{"year": "2020", "tier": "T3", "problem_label": "G 1", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $\\triangle A B C$ be an acute triangle. The line through $A$ perpendicular to $B C$ intersects $B C$ at $D$. Let $E$ be the midpoint of $A D$ and $\\omega$ the the circle with center $E$ and radius equal to $A E$. The line $B E$ intersects $\\omega$ at a point $X$ such that $X$ and $B$ are not on the same side of $A D$ and the line $C E$ intersects $\\omega$ at a point $Y$ such that $C$ and $Y$ are not on the same side of $A D$. If both of the intersection points of the circumcircles of $\\triangle B D X$ and $\\triangle C D Y$ lie on the line $A D$, prove that $A B=A C$.", "solution": "Alternative Solution. Let the circle $\\odot C D Y$ intersects the line $A D$ at another point $Z$. Then we have $\\angle C E D=\\angle Y E Z$. We also have $E D=E Y$ because $E$ is the center of the circle $\\omega$. Also note that\n\n$$\n\\angle D C E=\\angle D C Y=\\angle D Z Y=\\angle E Z Y\n$$\n\nWe conclude that $\\triangle C D E$ and $\\triangle Z Y E$ are congruent. From here we have that $E Z=E C$. Now denote by $Z^{\\prime}$ the other intersection point of $A D$ and $\\odot B D X$. In the same way we prove that $E Z^{\\prime}=E B$. By the assumption of the problem, we must have that $Z=Z^{\\prime}$. We now conclude that\n\n$$\nB E=C E=E Z=E Z^{\\prime}\n$$\n\nAlso, $\\angle B D E=90^{\\circ}=\\angle C D E$. Now we see that $\\triangle B D E$ and $\\triangle C D E$ are congruent (they share the side $E D)$, so $B D=C D$. But $D$ is both the midpoint of $B C$ and the foot of the altitude from $A$, which means that $A B=A C$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_8956532c2a2d8e9f5e66g-12.jpg?height=989&width=1326&top_left_y=991&top_left_x=367)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nG 1.", "solution_match": "\nSolution."}}
+{"year": "2020", "tier": "T3", "problem_label": "G 1", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $\\triangle A B C$ be an acute triangle. The line through $A$ perpendicular to $B C$ intersects $B C$ at $D$. Let $E$ be the midpoint of $A D$ and $\\omega$ the the circle with center $E$ and radius equal to $A E$. The line $B E$ intersects $\\omega$ at a point $X$ such that $X$ and $B$ are not on the same side of $A D$ and the line $C E$ intersects $\\omega$ at a point $Y$ such that $C$ and $Y$ are not on the same side of $A D$. If both of the intersection points of the circumcircles of $\\triangle B D X$ and $\\triangle C D Y$ lie on the line $A D$, prove that $A B=A C$.", "solution": "Alternative Solution. Let $\\alpha=\\angle B X D$. Denote by $T$ the second intersection point of the circumcircles of $\\triangle B D X$ and $\\triangle C D Y$, which is on $A D$. We have\n\n$$\nE D=E A=E X\n$$\n\nbecause $E$ is the center of $\\omega$. Now $E X=E D$ implies $\\angle E D X=\\alpha$. From here we have that $\\angle B E D=2 \\alpha$. Using that $B T X D$ is cyclic we obtain $\\angle B T D=\\angle B X D=\\alpha$. We also have that\n\n$$\n\\angle T B E=180^{\\circ}-\\angle B E T-\\angle E T B=\n$$\n\n$$\n=180^{\\circ}-\\left(180^{\\circ}-2 \\alpha\\right)-\\alpha=\\alpha=\\angle B T E\n$$\n\nThis gives us $B E=T E$. We similarly show that $C E=T E$, and so $B E=C E$. In the same way as in the second solution, this now gives us that $B D=C D$, so $D$ is also the midpoint of $B C$ and we must have $A B=A C$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_8956532c2a2d8e9f5e66g-13.jpg?height=938&width=1181&top_left_y=526&top_left_x=453)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nG 1.", "solution_match": "\nSolution."}}
+{"year": "2020", "tier": "T3", "problem_label": "G 1", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $\\triangle A B C$ be an acute triangle. The line through $A$ perpendicular to $B C$ intersects $B C$ at $D$. Let $E$ be the midpoint of $A D$ and $\\omega$ the the circle with center $E$ and radius equal to $A E$. The line $B E$ intersects $\\omega$ at a point $X$ such that $X$ and $B$ are not on the same side of $A D$ and the line $C E$ intersects $\\omega$ at a point $Y$ such that $C$ and $Y$ are not on the same side of $A D$. If both of the intersection points of the circumcircles of $\\triangle B D X$ and $\\triangle C D Y$ lie on the line $A D$, prove that $A B=A C$.", "solution": "Alternative Solution. We can solve the problem using only calculations. Note that the condition of the problem is that $E$ lies on the radical axis of the circumcircles of $\\triangle B D X$ and $\\triangle C D Y$. This gives us $E B \\cdot E X=E C \\cdot E Y$. However, $E X=E Y$ because $E$ is the center of $\\omega$ and this means that $B E=C E$. Now using Pythagoras' theorem we have the following:\n\n$$\nA B^{2}=A D^{2}+B D^{2}=A D^{2}+\\left(B E^{2}-D E^{2}\\right)=A D^{2}+\\left(C E^{2}-D E^{2}\\right)=A D^{2}+C D^{2}=A C^{2}\n$$\n\nFrom here we obtain $A B=A C$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nG 1.", "solution_match": "\nSolution."}}
+{"year": "2020", "tier": "T3", "problem_label": "G 2", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Problem: Let $\\triangle A B C$ be a right-angled triangle with $\\angle B A C=90^{\\circ}$, and let $E$ be the foot of the perpendicular from $A$ on $B C$. Let $Z \\neq A$ be a point on the line $A B$ with $A B=B Z$. Let (c), $\\left(c_{1}\\right)$ be the circmucircles of the triangles $\\triangle A E Z$ and $\\triangle B E Z$, respectively. Let ( $c_{2}$ ) be an arbitrary circle passing through the points $A$ and $E$. Suppose $\\left(c_{1}\\right)$ meets the line $C Z$ again at the point $F$, and meets $\\left(c_{2}\\right)$ again at the point $N$. If $P$ is the other point of intesection of $\\left(c_{2}\\right)$ with $A F$, prove that the points $N, B, P$ are collinear.", "solution": "Since the triangles $\\triangle A E B$ and $\\triangle C A B$ are similar, then\n\n$$\n\\frac{A B}{E B}=\\frac{C B}{A B}\n$$\n\nSince $A B=B Z$ we get\n\n$$\n\\frac{B Z}{E B}=\\frac{C B}{B Z}\n$$\n\nfrom which it follows that the triangles $\\triangle Z B E$ and $\\triangle C B Z$ are also similar. Since $F E B Z$ is cyclic,\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_8956532c2a2d8e9f5e66g-14.jpg?height=1087&width=1401&top_left_y=1012&top_left_x=248)\n\nthen $\\angle B E Z=\\angle B F Z$. So by the similarity of triangles $\\triangle Z B E$ and $\\triangle C B Z$ we get\n\n$$\n\\angle B F Z=\\angle B E Z=\\angle B Z C=\\angle B Z F\n$$\n\nand therefore the triangle $\\triangle B F Z$ is isosceles. Since $B F=B Z=A B$, then the triangle $\\triangle A F Z$ is right-angled with $\\angle A F Z=90^{\\circ}$.\n\nIt now follows that the points $A, E, F, C$ are concyclic. Since $A, P, E, N$ are also concyclic, then\n\n$$\n\\angle E N P=\\angle E A P=\\angle E A F=\\angle E C F=\\angle B C Z=\\angle B Z E,\n$$\n\nwhere in the last equality we used again the similarity of the triangles $\\triangle Z B E$ and $\\triangle C B Z$. Since $N, B, E, Z$ are concyclic, then $\\angle E N P=\\angle B Z E=\\angle E N B$, from which it follows that the points $N, B, P$ are collinear.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nG 2.", "solution_match": "\nSolution."}}
+{"year": "2020", "tier": "T3", "problem_label": "G 3", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $\\triangle A B C$ be a right-angled triangle with $\\angle B A C=90^{\\circ}$ and let $E$ be the foot of the perpendicular from $A$ on $B C$. Let $Z \\neq A$ be a point on the line $A B$ with $A B=B Z$. Let (c) be the circumcircle of the triangle $\\triangle A E Z$. Let $D$ be the second point of intersection of $(c)$ with $Z C$ and let $F$ be the antidiametric point of $D$ with respect to $(c)$. Let $P$ be the point of intersection of the lines $F E$ and $C Z$. If the tangent to $(c)$ at $Z$ meets $P A$ at $T$, prove that the points $T, E, B, Z$ are concyclic.", "solution": "We will first show that $P A$ is tangent to $(c)$ at $A$.\n\nSince $E, D, Z, A$ are concyclic, then $\\angle E D C=\\angle E A Z=\\angle E A B$. Since also the triangles $\\triangle A B C$ and $\\triangle E B A$ are similar, then $\\angle E A B=\\angle B C A$, therefore $\\angle E D C=\\angle B C A$.\n\nSince $\\angle F E D=90^{\\circ}$, then $\\angle P E D=90^{\\circ}$ and so\n\n$$\n\\angle E P D=90^{\\circ}-\\angle E D C=90^{\\circ}-\\angle B C A=\\angle E A C\n$$\n\nTherefore the points $E, A, C, P$ are concyclic. It follows that $\\angle C P A=90^{\\circ}$ and therefore the triangle $\\angle P A Z$ is right-angled. Since also $B$ is the midpoint of $A Z$, then $P B=A B=B Z$ and so $\\angle Z P B=$ $\\angle P Z B$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_8956532c2a2d8e9f5e66g-16.jpg?height=1215&width=1016&top_left_y=1008&top_left_x=533)\n\nFurthermore, $\\angle E P D=\\angle E A C=\\angle C B A=\\angle E B A$ from which it follows that the points $P, E, B, Z$ are also concyclic.\n\nNow observe that\n\n$$\n\\angle P A E=\\angle P C E=\\angle Z P B-\\angle P B E=\\angle P Z B-\\angle P Z E=\\angle E Z B\n$$\n\nTherefore $P A$ is tangent to $(c)$ at $A$ as claimed.\n\nIt now follows that $T A=T Z$. Therefore\n\n$$\n\\begin{aligned}\n\\angle P T Z & =180^{\\circ}-2(\\angle T A B)=180^{\\circ}-2(\\angle P A E+\\angle E A B)=180^{\\circ}-2(\\angle E C P+\\angle A C B) \\\\\n& =180^{\\circ}-2\\left(90^{\\circ}-\\angle P Z B\\right)=2(\\angle P Z B)=\\angle P Z B+\\angle B P Z=\\angle P B A .\n\\end{aligned}\n$$\n\nThus $T, P, B, Z$ are concyclic, and since $P, E, B, Z$ are also concyclic then $T, E, B, Z$ are concyclic as required.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nG 3.", "solution_match": "\nSolution."}}
+{"year": "2020", "tier": "T3", "problem_label": "NT 1", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Determine whether there is a natural number $n$ for which $8^{n}+47$ is prime.", "solution": "The number $m=8^{n}+47$ is never prime.\n\nIf $n$ is even, say $n=2 k$, then $m=64^{k}+47 \\equiv 1+2 \\equiv 0 \\bmod 3$. Since also $m>3$, then $m$ is not prime.\n\nIf $n \\equiv 1 \\bmod 4$, say $n=4 k+1$, then $m=8 \\cdot\\left(8^{k}\\right)^{4}+47 \\equiv 3+2 \\equiv 0 \\bmod 5$. Since also $m>3$, then $m$ is not prime.\n\nIf $n \\equiv 3 \\bmod 4$, say $n=4 k+3$, then $m=8\\left(64^{2 k+1}+1\\right) \\equiv 8\\left((-1)^{2 k+1}+1\\right) \\equiv 0 \\bmod 13$. Since also $m>13$, then $m$ is not prime.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nNT 1.", "solution_match": "\nSolution."}}
+{"year": "2020", "tier": "T3", "problem_label": "NT 2", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all positive integers $a, b, c$ and $p$, where $p$ is a prime number, such that\n\n$$\n73 p^{2}+6=9 a^{2}+17 b^{2}+17 c^{2}\n$$", "solution": "Since the equation is symmetric with respect to the numbers $b$ and $c$, we assume that $b \\geq c$.\n\nIf $p \\geq 3$, then $p$ is an odd number. We consider the equation modulo 8 . Since,\n\n$$\n73 p^{2}+6 \\equiv 79 \\equiv 7 \\quad(\\bmod 8)\n$$\n\nwe get that\n\n$$\na^{2}+b^{2}+c^{2} \\equiv 7 \\quad(\\bmod 8)\n$$\n\nThis cannot happen since for any integer $x$ we have that\n\n$$\nx^{2} \\equiv 0,1,4 \\quad(\\bmod 8)\n$$\n\nHence, $p$ must be an even prime number, which means that $p=2$. In this case, we obtain the equation\n\n$$\n9 a^{2}+17\\left(b^{2}+c^{2}\\right)=289\n$$\n\nIt follows that $b^{2}+c^{2} \\leq 17$. This is possible only for\n\n$$\n(b, c) \\in\\{(4,1),(3,2),(3,1),(2,2),(2,1),(1,1)\\}\n$$\n\nIt is easy to check that among these pairs only the $(4,1)$ gives an integer solution for $a$, namely $a=1$. Therefore, the given equation has only two solutions,\n\n$$\n(a, b, c, p) \\in\\{(1,1,4,2),(1,4,1,2)\\}\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nNT 2.", "solution_match": "\nSolution."}}
+{"year": "2020", "tier": "T3", "problem_label": "NT 3", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find the largest integer $k(k \\geq 2)$, for which there exists an integer $n(n \\geq k)$ such that from any collection of $n$ consecutive positive integers one can always choose $k$ numbers, which verify the following conditions:\n\n1. each chosen number is not divisible by 6 , by 7 and by 8 ;\n2. the positive difference of any two different chosen numbers is not divisible by at least one of the numbers 6,7 or 8 .", "solution": "An integer is divisible by 6,7 and 8 if and only if it is divisible by their Least Common Multiple, which equals $6 \\times 7 \\times 4=168$.\n\nLet $n$ be a positive integer and let $A$ be an arbitrary set of $n$ consecutive positive integers. Replace each number $a_{i}$ from $A$ with its remainder $r_{i}$ ( mod 168). The number $a_{i}$ is divisible by 6 ( 7 or 8 ) if and only if its remainder $r_{i}$ is divisible by 6 (respectively 7 or 8 ). The difference $\\left|a_{i}-a_{j}\\right|$ is divisible by 168 if and only if their remainders $r_{i}=r_{j}$.\n\nChoosing $k$ numbers from the initial set $A$, which verify the required conditions, is the same as choosing $k$ their remainders ( mod 168) such that:\n\n1. each chosen remainder is not divisible by 6,7 and 8 ;\n2. all chosen remainders are different.\n\nSuppose we have chosen $k$ numbers from $A$, which verify the conditions. Therefore, all remainders are different and $k \\leq 168$ (otherwise, there would be two equal remainders).\n\nDenote by $B=\\{0,1,2,3, \\ldots, 167\\}$ the set of all possible remainders ( $\\bmod 168)$ and by $B_{m}$ the subset of all elements of $B$, which are divisible by $m$. Compute the number of elements of the following subsets:\n\n$$\n\\begin{gathered}\n\\left|B_{6}\\right|=168: 6=28, \\quad\\left|B_{7}\\right|=168: 7=24, \\quad\\left|B_{8}\\right|=168: 8=21 \\\\\n\\left|B_{6} \\cap B_{7}\\right|=\\left|B_{42}\\right|=168: 42=4, \\quad\\left|B_{6} \\cap B_{8}\\right|=\\left|B_{24}\\right|=168: 24=7 \\\\\n\\left|B_{7} \\cap B_{8}\\right|=\\left|B_{56}\\right|=168: 56=3, \\quad\\left|B_{6} \\cap B_{7} \\cap B_{8}\\right|=\\left|B_{168}\\right|=1\n\\end{gathered}\n$$\n\nDenote by $D=B_{6} \\cup B_{7} \\cup B_{8}$, the subset of all elements of $B$, which are divisible by at least one of the numbers 6,7 or 8 . By the Inclusion-Exclusion principle we got\n\n$$\n\\begin{gathered}\n|D|=\\left|B_{6}\\right|+\\left|B_{7}\\right|+\\left|B_{8}\\right|-\\left(\\left|B_{6} \\cap B_{7}\\right|+\\left|B_{6} \\cap B_{8}\\right|+\\left|B_{7} \\cap B_{8}\\right|\\right)+\\left|B_{6} \\cap B_{7} \\cap B_{8}\\right|= \\\\\n28+24+21-(4+7+3)+1=60 .\n\\end{gathered}\n$$\n\nEach chosen remainder belongs to the subset $B \\backslash D$, since it is not divisible by 6,7 and 8 . Hence, $k \\leq|B \\backslash D|=168-60=108$.\n\nLet us show that the greatest possible value is $k=108$. Consider $n=168$. Given any collection $A$ of 168 consecutive positive integers, replace each number with its remainder ( $\\bmod 168$ ). Choose from these remainders 108 numbers, which constitute the set $B \\backslash D$. Finally, take 108 numbers from the initial set $A$, having exactly these remainders. These $k=108$ numbers verify the required conditions.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nNT 3.", "solution_match": "\nSolution."}}
+{"year": "2020", "tier": "T3", "problem_label": "NT 4", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all prime numbers $p$ such that\n\n$$\n(x+y)^{19}-x^{19}-y^{19}\n$$\n\nis a multiple of $p$ for any positive integers $x, y$.", "solution": "If $x=y=1$ then $p$ divides\n\n$$\n2^{19}-2=2\\left(2^{18}-1\\right)=2\\left(2^{9}-1\\right)\\left(2^{9}+1\\right)=2 \\cdot 511 \\cdot 513=2 \\cdot 3^{3} \\cdot 7 \\cdot 19 \\cdot 73\n$$\n\nIf $x=2, y=1$ then\n\n$$\np \\mid 3^{19}-2^{19}-1\n$$\n\nWe will show that $3^{19}-2^{19}-1$ is not a multiple of 73 . Indeed,\n\n$$\n3^{19} \\equiv 3^{3} \\cdot\\left(3^{4}\\right)^{4} \\equiv 3^{3} \\cdot 8^{4} \\equiv 3^{3} \\cdot(-9)^{2} \\equiv 27 \\cdot 81 \\equiv 27 \\cdot 8 \\equiv 70 \\quad(\\bmod 73)\n$$\n\nand\n\n$$\n2^{19} \\equiv 2 \\cdot 64^{3} \\equiv 2 \\cdot(-9)^{3} \\equiv-18 \\cdot 81 \\equiv-18 \\cdot 8 \\equiv-144 \\equiv 2 \\quad(\\bmod 73)\n$$\n\nThus $p$ can be only among $2,3,7,19$. We will prove all these work.\n\n- For $p=19$ this follows by Fermat's Theorem as\n\n$$\n(x+y)^{19} \\equiv x+y \\quad(\\bmod 19), \\quad x^{19} \\equiv x \\quad(\\bmod 19), \\quad y^{19} \\equiv y \\quad(\\bmod 19)\n$$\n\n- For $p=7$, we have that\n\n$$\na^{19} \\equiv a \\quad(\\bmod 7)\n$$\n\nfor every integer $a$. Indeed, if $7 \\mid a$, it is trivial, while if $7 \\nmid a$, then by Fermat's Theorem we have\n\n$$\n7\\left|a^{6}-1\\right| a^{18}-1\n$$\n\ntherefore\n\n$$\n7 \\mid a\\left(a^{18}-1\\right)\n$$\n\n- For $p=3$, we will prove that\n\n$$\nb^{19} \\equiv b \\quad(\\bmod 3)\n$$\n\nIndeed, if $3 \\mid b$, it is trivial, while if $3 \\nmid b$, then by Fermat's Theorem we have\n\n$$\n3\\left|b^{2}-1\\right| b^{18}-1\n$$\n\ntherefore\n\n$$\n3 \\mid b\\left(b^{18}-1\\right)\n$$\n\n- For $p=2$ it is true, since among $x+y, x$ and $y$ there are 0 or 2 odd numbers.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nNT 4.", "solution_match": "\nSolution."}}
+{"year": "2020", "tier": "T3", "problem_label": "NT 5", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "The positive integer $k$ and the set $A$ of different integers from 1 to $3 k$ inclusive are such that there are no distinct $a, b, c$ in $A$ satisfying $2 b=a+c$. The numbers from $A$ in the interval $[1, k]$ will be called small; those in $[k+1,2 k]$ - medium and those in $[2 k+1,3 k]$ - large. Is it always true that there are no positive integers $x$ and $d$ such that if $x, x+d$ and $x+2 d$ are divided by $3 k$ then the remainders belong to $A$ and those of $x$ and $x+d$ are different and are:\na) small?\nb) medium?\nc) large?\n\n(In this problem we assume that if a multiple of $3 k$ is divided by $3 k$ then the remainder is $3 k$ rather than 0. )", "solution": "Solution. A counterexample for a) is $k=3, A=\\{1,2,9\\}, x=2$ and $d=8$. A counterexample for c) is $k=3, A=\\{1,8,9\\}, x=8$ and $d=1$.\n\nWe will prove that b) is true.\n\nSuppose the contrary and let $x, d$ have the above properties. We can assume $0<d<3 k, 0<x \\leq 3 k$ (since for $d=3 k$ the remainders for $x$ and $x+d$ are equal). Hence $0<x+d<6 k$ and there are two cases:\n\n- If $x+d>3 k$, then since the remainder for $x+d$ is medium we have $4 k<x+d \\leq 5 k$. This means that the remainder of $x+d$ when it is divided by $3 k$ is\n\n$$\nx+d-3 k\n$$\n\nSince $x$ is medium we have $x \\leq 2 k$ so $d=(x+d)-x>2 k$. Therefore $6 k=4 k+2 k<(x+d)+d<$ $8 k$. This means that the remainder of $x+2 d$ when it is divided by $3 k$ is\n\n$$\nx+2 d-6 k \\text {. }\n$$\n\nThus the remainders $(x+2 d-6 k),(x+d-3 k)$ and $x$ are in $[1,3 k]$, they belong to $A$ and\n\n$$\n2(x+d-3 k)=(x+2 d-6 k)+x\n$$\n\na contradiction.\n\n- If $x+d \\leq 3 k$ then as $x+d$ is medium we have $k<x+d \\leq 2 k$. From the limitations on $x$, we have $x>k$ so $d=(x+d)-x<k$. Hence $0 \\leq x+2 d=(x+d)+d<3 k$. Thus the remainders $x, x+d$ and $x+2 d$ are in $A$ and\n\n$$\n2(x+d)=(x+2 d)+x\n$$\n\na contradiction.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nNT 5.", "solution_match": "## Solution."}}
+{"year": "2020", "tier": "T3", "problem_label": "NT 6", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Are there any positive integers $m$ and $n$ satisfying the equation\n\n$$\nm^{3}=9 n^{4}+170 n^{2}+289 ?\n$$", "solution": "We will prove that the answer is no. Note that\n\n$$\nm^{3}=9 n^{4}+170 n^{2}+289=\\left(9 n^{2}+17\\right)\\left(n^{2}+17\\right)\n$$\n\nIf $n$ is odd then $m$ is even, therefore $8 \\mid m^{3}$. However,\n\n$$\n9 n^{4}+170 n^{2}+289 \\equiv 9+170+289 \\equiv 4(\\bmod 8)\n$$\n\nwhich leads to a contradiction. If $n$ is a multiple of 17 then so is $m$ and hence 289 is a multiple of $17^{3}$, which is absurd. For $n$ even and not multiple of 17 , since\n\n$$\n\\operatorname{gcd}\\left(9 n^{2}+17, n^{2}+17\\right) \\mid 9\\left(n^{2}+17\\right)-\\left(9 n^{2}+17\\right)=2^{3} \\cdot 17\n$$\n\nthis gcd must be 1 . Therefore $n^{2}+17=a^{3}$ for an odd $a$, so\n\n$$\nn^{2}+25=(a+2)\\left(a^{2}-2 a+4\\right)\n$$\n\nFor $a \\equiv 1(\\bmod 4)$ we have $a+2 \\equiv 3(\\bmod 4)$, while for $a \\equiv 3(\\bmod 4)$ we have $a^{2}-2 a+4 \\equiv 3$ $(\\bmod 4)$. Thus $(a+2)\\left(a^{2}-2 a+4\\right)$ has a prime divisor of type $4 \\ell+3$. As it divides $n^{2}+25$, it has to divide $n$ and 5 , which is absurd.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nNT 6.", "solution_match": "\nSolution."}}
+{"year": "2020", "tier": "T3", "problem_label": "NT 7", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Prove that there doesn't exist any prime $p$ such that every power of $p$ is a palindrome (palindrome is a number that is read the same from the left as it is from the right; in particular, number that ends in one or more zeros cannot be a palindrome).", "solution": "Note that by criterion for divisibility by 11 and the definition of a palindrome we have that every palindrome that has even number of digits is divisible by 11 .\n\nSince $11^{5}=161051$ is not a palindrome and since 11 cannot divide $p^{k}$ for any prime other than 11 we are now left to prove that no prime whose all powers have odd number of digits exists.\n\nAssume the contrary. It means that the difference between the numbers of digits of $p^{m}$ and $p^{m+1}$ is even number. We will prove that for every natural $m$, the difference is the same even number.\n\nIf we assume not, that means that the difference for some $m_{1}$ has at least 2 digits more than the difference for some $m_{2}$. We will prove that this is impossible.\n\nLet $p^{m_{1}}=10^{t_{1}} \\cdot a_{1}, p^{m_{2}}=10^{t_{2}} \\cdot a_{2}$ and $p=10^{h} \\cdot z$, where\n\n$$\n1<a_{1}, a_{2}, z<10\n$$\n\nThis implies that\n\n$$\n1<a_{1} \\cdot z, a_{2} \\cdot z<100\n$$\n\nwhich further implies that multiplying these powers of $p$ by $p$ can increase their number of digits by either $h$ or $h+1$.\n\nThis is a contradiction. Call the difference between numbers of digits of consecutive powers $d$. Number $p$ clearly cannot be equal to $10^{d}$ for $d \\geq 1$ because 10 is divisible by two primes, but for $d=0$, we would have that 1 is a prime which is not true.\n\nCase 1. $p>10^{d}$. Let $p=10^{d} \\cdot a$, for some real number $a$ greater than 1. (1)\n\nFrom the definition of $d$ we also see that $a$ is smaller than 10. (2)\n\nFrom (1) we see that powering $a$ gives us arbitrarily large numbers and from (2) we conclude that there is some natural power of $a$, call it $b$, greater than 1 , such that\n\n$$\n10<a^{b}<100\n$$\n\nIt is clear that $p^{b}$ has exactly $(b-1) d+1$ digits more than $p$ has, which is an odd number, but sum of even numbers is even.\n\nCase 2. $p<10^{d}$. Let $p=\\frac{10^{d}}{a}$, for some real number $a$ greater than 1. (1) From the definition of $d$ we also see that $a$ is smaller than 10. (2)\n\nFrom (1) we see that powering $a$ gives us arbitrarily large numbers and from (2) we conclude that there is some natural power of $a$, call it $b$, greater than 1 , such that\n\n$$\n10<a^{b}<100\n$$\n\nIt is clear that $p^{b}$ has exactly $(b-1) d-1$ digits more than $p$ has, which is an odd number, but sum of even numbers is even.\n\nWe have now arrived at the desired contradiction for both cases and have thus finished the proof.\n\nAlternative solution. Note that the sequence $\\left\\{p^{n}\\right\\}$ is periodic $(\\bmod 10)$. Let the period be $d$. Also, let $p^{d}=g$.\n\nSince all powers of $p$ are palindromes, all powers of $g$ are as well. Since $\\left\\{g^{n}\\right\\}$ is constant (mod 10), the leftmost digit of each power of $g$ is equal to some $f$.\n\nWe will prove that the difference between numbers of digits of $g^{m}$ and $g^{m+1}$ is equal to some $r$ for every natural number $m$.\n\nThis is true due to size reasons. Namely, to add exactly $k$ digits, and yet to have the same leftmost digit, we need to multiply the number by at least $5 \\cdot 10^{k-1}$ (if $k=0$ then it's 1 ) and by at most $2 \\cdot 10^{k}$ (values depend on the leftmost digit, it can easily be seen that leftmost digit being 1 yields the extremal values). Notice that\n\n$$\n2 \\cdot 10^{k}<5 \\cdot 10^{k+1-1}\n$$\n\nSince this inequality has clearly shown that the interval of multipliers which add exactly $k$ digits and leave the leftmost digit the same is disjunct from the same kind of interval for $k+1$ digits, which implies that no number can belong to both intervals, we have successfully proven the claim.\n\nClearly, $g$ cannot be equal to $10^{r}$ for $r \\geq 1$ because a palindrome cannot be divisible by 10 , but for $r=0$ we again cannot have the equality because 1 is not a natural power of a prime.\n\nCase 1. $g>10^{r}$. Let $g=10^{r} \\cdot a$, where $a$ is a real number, $10>a>1$. (1) Here, $a$ is less than 10 because if it was not, multiplying by $g$ would add at least $r+1$ digits, which is impossible.\n\nFrom (1) we see that powering $a$ gives us arbitrarily large numbers and that there is some natural power of $a$, call it $b$, greater than 1 , such that\n\n$$\n10<a^{b}<100\n$$\n\nPick smallest such $b$.\n\nNow we easily see that the difference between numbers of digits of numbers $g^{b-1}$ and $g^{b}$ is exactly $r+1$.\n\nCase 2. $g<10^{r}$. Let $g=\\frac{10^{r}}{a}$, where $a$ is a real number, $10>a>1$. (2) Here, $a$ is less than 10 because if it was not, multiplying by $g$ would add at most $r-1$ digits, which is impossible.\n\nFrom (2) we see that powering $a$ gives us arbitrarily large numbers and that there is some natural power of $a$, call it $b$, greater than 1 , such that\n\n$$\n10<a^{b}<100\n$$\n\nPick smallest such $b$.\n\nNow we easily see that the difference between numbers of digits of numbers $g^{b-1}$ and $g^{b}$ is exactly $r-1$.\n\nWe have arrived at the desired contradiction for both cases and have thus finished the proof.\n\nComment. In both solutions, after introducing $a$, there are multiple ways to finish the problem. In particular, solution 1 and solution 2 could be finished in the same way, but distinct finishes were purposely offered. Third, maybe even the most intuitive finishing argument, could be using non-exact size arguments; namely, just the fact that powers of $a$ grow arbitrarily large is enough to reach the contradiction.\n\nAlternative problem. Find all positive integers $n$ such that every power of $n$ is a palindrome (palindrome is a number that is read the same from the left as it is from the right; in particular, number that ends in one or more zeros cannot be a palindrome).\n\nSuggested difficulty for this problem is hard. Noting why solution 2 doesn't work for $n=1$ (because\n$g$ now could be equal to 1 ) and saying that $n=1$ actually works are all necessary modifications to solution 2 to make it work for the alternative problem as well.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nNT 7.", "solution_match": "\nSolution."}}
+{"year": "2020", "tier": "T3", "problem_label": "NT 8", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all pairs $(p, q)$ of prime numbers such that\n\n$$\n1+\\frac{p^{q}-q^{p}}{p+q}\n$$\n\nis a prime number.", "solution": "It is clear that $p \\neq q$. We set\n\n$$\n1+\\frac{p^{q}-q^{p}}{p+q}=r\n$$\n\nand we have that\n\n$$\np^{q}-q^{p}=(r-1)(p+q)\n$$\n\nFrom Fermat's Little Theorem we have\n\n$$\np^{q}-q^{p} \\equiv-q \\quad(\\bmod p)\n$$\n\nSince we also have that\n\n$$\n(r-1)(p+q) \\equiv-r q-q \\quad(\\bmod p)\n$$\n\nfrom (1) we get that\n\n$$\nr q \\equiv 0 \\quad(\\bmod p) \\Rightarrow p \\mid q r\n$$\n\nhence $p \\mid r$, which means that $p=r$. Therefore, (1) takes the form\n\n$$\np^{q}-q^{p}=(p-1)(p+q)\n$$\n\nWe will prove that $p=2$. Indeed, if $p$ is odd, then from Fermat's Little Theorem we have\n\n$$\np^{q}-q^{p} \\equiv p \\quad(\\bmod q)\n$$\n\nand since\n\n$$\n(p-1)(p+q) \\equiv p(p-1) \\quad(\\bmod q)\n$$\n\nwe have\n\n$$\np(p-2) \\equiv 0 \\quad(\\bmod q) \\Rightarrow q|p(p-2) \\Rightarrow q| p-2 \\Rightarrow q \\leq p-2<p\n$$\n\nNow, from (2) we have\n\n$$\np^{q}-q^{p} \\equiv 0 \\quad(\\bmod p-1) \\Rightarrow 1-q^{p} \\equiv 0 \\quad(\\bmod p-1) \\Rightarrow q^{p} \\equiv 1 \\quad(\\bmod p-1)\n$$\n\nClearly $\\operatorname{gcd}(q, p-1)=1$ and if we set $k=\\operatorname{ord}_{p-1}(q)$, it is well-known that $k \\mid p$ and $k<p$, therefore $k=1$. It follows that\n\n$$\nq \\equiv 1 \\quad(\\bmod p-1) \\Rightarrow p-1 \\mid q-1 \\Rightarrow p-1 \\leq q-1 \\Rightarrow p \\leq q\n$$\n\na contradiction.\n\nTherefore, $p=2$ and (2) transforms to\n\n$$\n2^{q}=q^{2}+q+2\n$$\n\nWe can easily check by induction that for every positive integer $n \\geq 6$ we have $2^{n}>n^{2}+n+2$. This means that $q \\leq 5$ and the only solution is for $q=5$. Hence the only pair which satisfy the condition is $(p, q)=(2,5)$.\n\nComment by the PSC. From the problem condition, we get that $p^{q}$ should be bigger than $q^{p}$, which gives\n\n$$\nq \\ln p>p \\ln q \\Longleftrightarrow \\frac{\\ln p}{p}>\\frac{\\ln q}{q}\n$$\n\nThe function $\\frac{\\ln x}{x}$ is decreasing for $x>e$, thus if $p$ and $q$ are odd primes, we obtain $q>p$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl", "problem_match": "\nNT 8.", "solution_match": "\nSolution."}}

JBMO/segmented/en-shortlist/en-nt20111.jsonl

@@ -1,5 +1,5 @@
-{"year": "2011", "tier": "T3", "problem_label": "NT1", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Solve in positive integers the equation $1005^{x}+2011^{y}=1006^{z}$.", "solution": "We have $1006^{z}>2011^{y}>2011$, hence $z \\geq 2$. Then $1005^{x}+2011^{y} \\equiv 0(\\bmod 4)$.\n\nBut $1005^{x} \\equiv 1(\\bmod 4)$, so $2011^{y} \\equiv-1(\\bmod 4) \\Rightarrow y$ is odd, i.e. $2011^{y} \\equiv-1(\\bmod 1006)$.\n\nSince $1005^{x}+2011^{y} \\equiv 0(\\bmod 1006)$, we get $1005^{x} \\equiv 1(\\bmod 1006) \\Rightarrow x$ is even.\n\nNow $1005^{x} \\equiv 1(\\bmod 8)$ and $2011^{y} \\equiv 3(\\bmod 8)$, hence $1006^{z} \\equiv 4(\\bmod 8) \\Rightarrow z=2$.\n\nIt follows that $y<2 \\Rightarrow y=1$ and $x=2$. The solution is $(x, y, z)=(2,1,2)$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-nt20111.jsonl", "problem_match": "\nNT1 ", "solution_match": "## Solution"}}
-{"year": "2011", "tier": "T3", "problem_label": "NT2", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all prime numbers $p$ such that there exist positive integers $x, y$ that satisfy the relation $x\\left(y^{2}-p\\right)+y\\left(x^{2}-p\\right)=5 p$.", "solution": "The given equation is equivalent to $(x+y)(x y-p)=5 p$. Obviously $x+y \\geq 2$.\n\nWe will consider the following three cases:\n\nCase 1: $x+y=5$ and $x y=2 p$. The equation $x^{2}-5 x+2 p=0$ has at least a solution, so we must have $0 \\leq \\Delta=25-8 p$ which implies $p \\in\\{2,3\\}$. For $p=2$ we obtain the solutions $(1,4)$ and $(4,1)$ and for $p=3$ we obtain the solutions $(2,3)$ and $(3,2)$.\n\nCase 2: $x+y=p$ and $x y=p+5$. We have $x y-x-y=5$ or $(x-1)(y-1)=6$ which implies $(x, y) \\in\\{(2,7) ;(3,4) ;(4,3) ;(7,2)\\}$. Since $p$ is prime, we get $p=7$.\n\nCase 3: $x+y=5 p$ and $x y=p+1$. We have $(x-1)(y-1)=x y-x-y+1=2-4 p<0$. But this is impossible since $x, y \\geq 1$.\n\nFinally, the equation has solutions in positive integers only for $p \\in\\{2,3,7\\}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-nt20111.jsonl", "problem_match": "\nNT2 ", "solution_match": "## Solution"}}
-{"year": "2011", "tier": "T3", "problem_label": "NT3", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all positive integers $n$ such that the equation $y^{2}+x y+3 x=n\\left(x^{2}+x y+3 y\\right)$ has at least a solution $(x, y)$ in positive integers.", "solution": "Clearly for $n=1$, each pair $(x, y)$ with $x=y$ is a solution. Now, suppose that $n>1$ which implies $x \\neq y$. We have $0<n-1=\\frac{y^{2}+x y+3 x}{x^{2}+x y+3 y}-1=\\frac{(x+y-3)(y-x)}{x^{2}+x y+3 y}$.\n\nSince $x+y \\geq 3$, we conclude that $x+y>3$ and $y>x$. Take $d=\\operatorname{gcd}(x+y-$ $\\left.3 ; x^{2}+x y+3 y\\right)$. Then $d$ divides $x^{2}+x y+3 y-x(x+y-3)=3(x+y)$. Then $d$ also divides $3(x+y)-3(x+y-3)=9$, hence $d \\in\\{1,3,9\\}$. As $n-1=\\frac{\\frac{x+y-3}{d}(y-x)}{\\frac{x^{2}+x y+3 y}{d}}$ and $\\operatorname{gcd}\\left(\\frac{x+y-3}{d} ; \\frac{x^{2}+x y+3 y}{d}\\right)=1$, it follows that $\\frac{x^{2}+x y+3 y}{d}$ divides $y-x$, which leads to $x^{2}+x y+3 y \\leq d y-d x \\Leftrightarrow x^{2}+d x \\leq(d-3-x) y$. It is necessary that\n$d-3-x>0 \\Rightarrow d>3$, therefore $d=9$ and $x<6$. Take $x+y-3=9 k, k \\in \\mathbb{N}^{*}$ since $d \\mid x+y-3$ and we get $y=9 k+3-x$. Hence $n-1=\\frac{k(9 k+3-2 x)}{k(x+3)+1}$. Because $k$ and $k(x+3)+1$ are relatively prime, the number $t=\\frac{9 k+3-2 x}{k(x+3)+1}$ must be integer for some positive integers $x<6$. It remains to consider these values of $x$ :\n\n1) For $x=1$, then $t=\\frac{9 k+1}{4 k+1}$ and since $1<t<3$, we get $t=2, k=1, y=11$, so $n=3$.\n2) For $x=2$, then $t=\\frac{9 k-1}{5 k+1}$ and since $1<t<2$, there are no solutions in this case.\n3) For $x=3$, then $t=\\frac{9 k-3}{6 k+1}$ and since $1 \\neq t<2$, there are no solutions in this case.\n4) For $x=4$, then $t=\\frac{9 k-5}{7 k+1}<2$,i.e. $t=1$ which leads to $k=3, y=26$, so $n=4$.\n5) For $x=5$, then $t=\\frac{9 k-7}{8 k+1}<2$,i.e. $t=1$ which leads to $k=8, y=70$, so $n=9$.\n\nFinally, the answer is $n \\in\\{1,3,4,9\\}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-nt20111.jsonl", "problem_match": "\nNT3 ", "solution_match": "## Solution"}}
-{"year": "2011", "tier": "T3", "problem_label": "NT4", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all prime positive integers $p, q$ such that $2 p^{3}-q^{2}=2(p+q)^{2}$.", "solution": "The given equation can be rewritten as $2 p^{2}(p-1)=q(3 q+4 p)$.\n\nHence $p\\left|3 q^{2}+4 p q \\Rightarrow p\\right| 3 q^{2} \\Rightarrow p \\mid 3 q$ (since $p$ is a prime number) $\\Rightarrow p \\mid 3$ or $p \\mid q$. If $p \\mid q$, then $p=q$. The equation becomes $2 p^{3}-9 p^{2}=0$ which has no prime solution. If $p \\mid 3$, then $p=3$. The equation becomes $q^{2}+4 q-12=0 \\Leftrightarrow(q-2)(q+6)=0$.\n\nSince $q>0$, we get $q=2$, so we have the solution $(p, q)=(3,2)$.\n\n## Solution 2\n\nSince $2 p^{3}$ and $2\\left(p+q^{2}\\right)$ are even, $q^{2}$ is also even, thus $q=2$ because it is a prime number.\n\nThe equation becomes $p^{3}-p^{2}-4 p-6=0 \\Leftrightarrow\\left(p^{2}-4\\right)(p-1)=10$.\n\nIf $p \\geq 4$, then $\\left(p^{2}-4\\right)(p-1) \\geq 12 \\cdot 3>10$, so $p \\leq 3$. A direct verification gives $p=3$.\n\nFinally, the unique solution is $(p, q)=(3,2)$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-nt20111.jsonl", "problem_match": "\nNT4 ", "solution_match": "## Solution 1"}}
-{"year": "2011", "tier": "T3", "problem_label": "NT5", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find the least positive integer such that the sum of its digits is 2011 and the product of its digits is a power of 6 .", "solution": "Denote this number by $N$. Then $N$ can not contain the digits $0,5,7$ and its digits must be written in increasing order. Suppose that $N$ has $x_{1}$ ones, $x_{2}$ twos, $x_{3}$ threes, $x_{4}$ fours, $x_{6}$ sixes, $x_{8}$ eights and $x_{9}$ nines, then $x_{1}+2 x_{2}+3 x_{3}+4 x_{4}+6 x_{6}+8 x_{8}+9 x_{9}=2011$. (1) The product of digits of the number $N$ is a power of 6 when we have the relation $x_{2}+2 x_{4}+x_{6}+3 x_{8}=x_{3}+x_{6}+2 x_{9}$, hence $x_{2}-x_{3}+2 x_{4}+3 x_{8}-2 x_{9}=0$.\n\nDenote by $S$ the number of digits of $N\\left(S=x_{1}+x_{2}+\\ldots+x_{9}\\right)$. In order to make the coefficients of $x_{8}$ and $x_{9}$ equal, we multiply relation (1) by 5 , then we add relation (2). We get $43 x_{9}+43 x_{8}+30 x_{6}+22 x_{4}+14 x_{3}+11 x_{2}+5 x_{1}=10055 \\Leftrightarrow 43 S=10055+13 x_{6}+$ $21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1}$. Then $10055+13 x_{6}+21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1}$ is a multiple of 43 not less than 10055. The least such number is 10062 , but the relation $10062=$ $10055+13 x_{6}+21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1}$ means that among $x_{1}, x_{2}, \\ldots, x_{6}$ there is at least one positive, so $10062=10055+13 x_{6}+21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1} \\geq 10055+13=10068$ which is obviously false. The next multiple of 43 is 10105 and from relation $10105=$ $10055+13 x_{6}+21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1}$ we get $50=13 x_{6}+21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1}$. By writing as $13 x_{6}+21\\left(x_{4}-1\\right)+29\\left(x_{3}-1\\right)+32 x_{2}+38 x_{1}=0$, we can easily see that the only possibility is $x_{1}=x_{2}=x_{6}=0$ and $x_{3}=x_{4}=1$. Then $S=235, x_{8}=93, x_{9}=140$. Since $S$ is strictly minimal, we conclude that $N=34 \\underbrace{88 \\ldots 8}_{93} \\underbrace{99 \\ldots 9}_{140}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-nt20111.jsonl", "problem_match": "\nNT5 ", "solution_match": "## Solution"}}
+{"year": "2011", "tier": "T3", "problem_label": "NT1", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Solve in positive integers the equation $1005^{x}+2011^{y}=1006^{z}$.", "solution": "We have $1006^{z}>2011^{y}>2011$, hence $z \\geq 2$. Then $1005^{x}+2011^{y} \\equiv 0(\\bmod 4)$.\n\nBut $1005^{x} \\equiv 1(\\bmod 4)$, so $2011^{y} \\equiv-1(\\bmod 4) \\Rightarrow y$ is odd, i.e. $2011^{y} \\equiv-1(\\bmod 1006)$.\n\nSince $1005^{x}+2011^{y} \\equiv 0(\\bmod 1006)$, we get $1005^{x} \\equiv 1(\\bmod 1006) \\Rightarrow x$ is even.\n\nNow $1005^{x} \\equiv 1(\\bmod 8)$ and $2011^{y} \\equiv 3(\\bmod 8)$, hence $1006^{z} \\equiv 4(\\bmod 8) \\Rightarrow z=2$.\n\nIt follows that $y<2 \\Rightarrow y=1$ and $x=2$. The solution is $(x, y, z)=(2,1,2)$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-nt20111.jsonl", "problem_match": "\nNT1 ", "solution_match": "## Solution"}}
+{"year": "2011", "tier": "T3", "problem_label": "NT2", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all prime numbers $p$ such that there exist positive integers $x, y$ that satisfy the relation $x\\left(y^{2}-p\\right)+y\\left(x^{2}-p\\right)=5 p$.", "solution": "The given equation is equivalent to $(x+y)(x y-p)=5 p$. Obviously $x+y \\geq 2$.\n\nWe will consider the following three cases:\n\nCase 1: $x+y=5$ and $x y=2 p$. The equation $x^{2}-5 x+2 p=0$ has at least a solution, so we must have $0 \\leq \\Delta=25-8 p$ which implies $p \\in\\{2,3\\}$. For $p=2$ we obtain the solutions $(1,4)$ and $(4,1)$ and for $p=3$ we obtain the solutions $(2,3)$ and $(3,2)$.\n\nCase 2: $x+y=p$ and $x y=p+5$. We have $x y-x-y=5$ or $(x-1)(y-1)=6$ which implies $(x, y) \\in\\{(2,7) ;(3,4) ;(4,3) ;(7,2)\\}$. Since $p$ is prime, we get $p=7$.\n\nCase 3: $x+y=5 p$ and $x y=p+1$. We have $(x-1)(y-1)=x y-x-y+1=2-4 p<0$. But this is impossible since $x, y \\geq 1$.\n\nFinally, the equation has solutions in positive integers only for $p \\in\\{2,3,7\\}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-nt20111.jsonl", "problem_match": "\nNT2 ", "solution_match": "## Solution"}}
+{"year": "2011", "tier": "T3", "problem_label": "NT3", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all positive integers $n$ such that the equation $y^{2}+x y+3 x=n\\left(x^{2}+x y+3 y\\right)$ has at least a solution $(x, y)$ in positive integers.", "solution": "Clearly for $n=1$, each pair $(x, y)$ with $x=y$ is a solution. Now, suppose that $n>1$ which implies $x \\neq y$. We have $0<n-1=\\frac{y^{2}+x y+3 x}{x^{2}+x y+3 y}-1=\\frac{(x+y-3)(y-x)}{x^{2}+x y+3 y}$.\n\nSince $x+y \\geq 3$, we conclude that $x+y>3$ and $y>x$. Take $d=\\operatorname{gcd}(x+y-$ $\\left.3 ; x^{2}+x y+3 y\\right)$. Then $d$ divides $x^{2}+x y+3 y-x(x+y-3)=3(x+y)$. Then $d$ also divides $3(x+y)-3(x+y-3)=9$, hence $d \\in\\{1,3,9\\}$. As $n-1=\\frac{\\frac{x+y-3}{d}(y-x)}{\\frac{x^{2}+x y+3 y}{d}}$ and $\\operatorname{gcd}\\left(\\frac{x+y-3}{d} ; \\frac{x^{2}+x y+3 y}{d}\\right)=1$, it follows that $\\frac{x^{2}+x y+3 y}{d}$ divides $y-x$, which leads to $x^{2}+x y+3 y \\leq d y-d x \\Leftrightarrow x^{2}+d x \\leq(d-3-x) y$. It is necessary that\n$d-3-x>0 \\Rightarrow d>3$, therefore $d=9$ and $x<6$. Take $x+y-3=9 k, k \\in \\mathbb{N}^{*}$ since $d \\mid x+y-3$ and we get $y=9 k+3-x$. Hence $n-1=\\frac{k(9 k+3-2 x)}{k(x+3)+1}$. Because $k$ and $k(x+3)+1$ are relatively prime, the number $t=\\frac{9 k+3-2 x}{k(x+3)+1}$ must be integer for some positive integers $x<6$. It remains to consider these values of $x$ :\n\n1) For $x=1$, then $t=\\frac{9 k+1}{4 k+1}$ and since $1<t<3$, we get $t=2, k=1, y=11$, so $n=3$.\n2) For $x=2$, then $t=\\frac{9 k-1}{5 k+1}$ and since $1<t<2$, there are no solutions in this case.\n3) For $x=3$, then $t=\\frac{9 k-3}{6 k+1}$ and since $1 \\neq t<2$, there are no solutions in this case.\n4) For $x=4$, then $t=\\frac{9 k-5}{7 k+1}<2$,i.e. $t=1$ which leads to $k=3, y=26$, so $n=4$.\n5) For $x=5$, then $t=\\frac{9 k-7}{8 k+1}<2$,i.e. $t=1$ which leads to $k=8, y=70$, so $n=9$.\n\nFinally, the answer is $n \\in\\{1,3,4,9\\}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-nt20111.jsonl", "problem_match": "\nNT3 ", "solution_match": "## Solution"}}
+{"year": "2011", "tier": "T3", "problem_label": "NT4", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all prime positive integers $p, q$ such that $2 p^{3}-q^{2}=2(p+q)^{2}$.", "solution": "The given equation can be rewritten as $2 p^{2}(p-1)=q(3 q+4 p)$.\n\nHence $p\\left|3 q^{2}+4 p q \\Rightarrow p\\right| 3 q^{2} \\Rightarrow p \\mid 3 q$ (since $p$ is a prime number) $\\Rightarrow p \\mid 3$ or $p \\mid q$. If $p \\mid q$, then $p=q$. The equation becomes $2 p^{3}-9 p^{2}=0$ which has no prime solution. If $p \\mid 3$, then $p=3$. The equation becomes $q^{2}+4 q-12=0 \\Leftrightarrow(q-2)(q+6)=0$.\n\nSince $q>0$, we get $q=2$, so we have the solution $(p, q)=(3,2)$.\n\n## Solution 2\n\nSince $2 p^{3}$ and $2\\left(p+q^{2}\\right)$ are even, $q^{2}$ is also even, thus $q=2$ because it is a prime number.\n\nThe equation becomes $p^{3}-p^{2}-4 p-6=0 \\Leftrightarrow\\left(p^{2}-4\\right)(p-1)=10$.\n\nIf $p \\geq 4$, then $\\left(p^{2}-4\\right)(p-1) \\geq 12 \\cdot 3>10$, so $p \\leq 3$. A direct verification gives $p=3$.\n\nFinally, the unique solution is $(p, q)=(3,2)$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-nt20111.jsonl", "problem_match": "\nNT4 ", "solution_match": "## Solution 1"}}
+{"year": "2011", "tier": "T3", "problem_label": "NT5", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find the least positive integer such that the sum of its digits is 2011 and the product of its digits is a power of 6 .", "solution": "Denote this number by $N$. Then $N$ can not contain the digits $0,5,7$ and its digits must be written in increasing order. Suppose that $N$ has $x_{1}$ ones, $x_{2}$ twos, $x_{3}$ threes, $x_{4}$ fours, $x_{6}$ sixes, $x_{8}$ eights and $x_{9}$ nines, then $x_{1}+2 x_{2}+3 x_{3}+4 x_{4}+6 x_{6}+8 x_{8}+9 x_{9}=2011$. (1) The product of digits of the number $N$ is a power of 6 when we have the relation $x_{2}+2 x_{4}+x_{6}+3 x_{8}=x_{3}+x_{6}+2 x_{9}$, hence $x_{2}-x_{3}+2 x_{4}+3 x_{8}-2 x_{9}=0$.\n\nDenote by $S$ the number of digits of $N\\left(S=x_{1}+x_{2}+\\ldots+x_{9}\\right)$. In order to make the coefficients of $x_{8}$ and $x_{9}$ equal, we multiply relation (1) by 5 , then we add relation (2). We get $43 x_{9}+43 x_{8}+30 x_{6}+22 x_{4}+14 x_{3}+11 x_{2}+5 x_{1}=10055 \\Leftrightarrow 43 S=10055+13 x_{6}+$ $21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1}$. Then $10055+13 x_{6}+21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1}$ is a multiple of 43 not less than 10055. The least such number is 10062 , but the relation $10062=$ $10055+13 x_{6}+21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1}$ means that among $x_{1}, x_{2}, \\ldots, x_{6}$ there is at least one positive, so $10062=10055+13 x_{6}+21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1} \\geq 10055+13=10068$ which is obviously false. The next multiple of 43 is 10105 and from relation $10105=$ $10055+13 x_{6}+21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1}$ we get $50=13 x_{6}+21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1}$. By writing as $13 x_{6}+21\\left(x_{4}-1\\right)+29\\left(x_{3}-1\\right)+32 x_{2}+38 x_{1}=0$, we can easily see that the only possibility is $x_{1}=x_{2}=x_{6}=0$ and $x_{3}=x_{4}=1$. Then $S=235, x_{8}=93, x_{9}=140$. Since $S$ is strictly minimal, we conclude that $N=34 \\underbrace{88 \\ldots 8}_{93} \\underbrace{99 \\ldots 9}_{140}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-nt20111.jsonl", "problem_match": "\nNT5 ", "solution_match": "## Solution"}}

JBMO/segmented/en-shortlist/en-shl-2012.jsonl

@@ -1,18 +1,18 @@
-{"year": "2012", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $a, b, c$ be positive real numbers such that abc $=1$. Show that\n\n$$\n\\frac{1}{a^{3}+b c}+\\frac{1}{b^{3}+c a}+\\frac{1}{c^{3}+a b} \\leq \\frac{(a b+b c+c a)^{2}}{6}\n$$\n\nso", "solution": "By the AM-GM inequality we have $a^{3}+b c \\geq 2 \\sqrt{a^{3} b c}=2 \\sqrt{a^{2}(a b c)}=2 a$ and\n\n$$\n\\frac{1}{a^{3}+b c} \\leq \\frac{1}{2 a}\n$$\n\nSimilarly; $\\frac{1}{b^{3}+c a} \\leq \\frac{1}{2 b} \\cdot \\frac{1}{c^{3}+a b} \\leq \\frac{1}{2 c}$ and then\n\n$$\n\\frac{1}{a^{3}+b c}+\\frac{1}{b^{3}+c a}+\\frac{1}{c^{3}+a b} \\leq \\frac{1}{2 a}+\\frac{1}{2 b}+\\frac{1}{2 c}=\\frac{1}{2} \\frac{a b+b c+c a}{a b c} \\leq \\frac{(a b+b c+c a)^{2}}{6}\n$$\n\nTherefore it is enongil to prove $\\frac{(a h+b c+c a)^{2}}{6} \\leq \\frac{(a b+b c+c a)^{2}}{6}$. This mequalits is trivially shomn to be equivalent to $3 \\leq a b+b c+c a$ which is true because of the AM-GM inequalit: $3=\\sqrt[3]{(a b c)^{2}} \\leq a b+b c+c a$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nA2.", "solution_match": "\nSolution."}}
-{"year": "2012", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $a . b$ c ce positue real numbers such that $a+b+c=a^{2}+b^{2}+c^{2}$. Shou that\n\n$$\n\\frac{a^{2}}{a^{2}+a b}+\\frac{b^{2}}{b^{2}+b c}+\\frac{c^{2}}{c^{2}+c a} \\geq \\frac{a+b+c}{2}\n$$", "solution": "By the Cauchy-Schwarz inequality it is\n\n$$\n\\begin{aligned}\n& \\left(\\frac{a^{2}}{a^{2}+a b}+\\frac{b^{2}}{b^{2}+b c}+\\frac{c^{2}}{c^{2}+c a}\\right)\\left(\\left(a^{2}+a b\\right)+\\left(b^{2}+b c\\right)+\\left(c^{2}+c a\\right)\\right) \\geq(a+b+c)^{2} \\\\\n\\Rightarrow & \\frac{a^{2}}{a^{2}+a b}+\\frac{b^{2}}{b^{2}+b c}+\\frac{c^{2}}{c^{2}+c a} \\geq \\frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}+a b+b c+c a}\n\\end{aligned}\n$$\n\nSo in is enough to prove $\\frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}+a b+b c+c a} \\geq \\frac{a+b+c}{2}$, that is to prove\n\n$$\n2(a+b+c) \\geq a^{2}+b^{2}+c^{2}+a b+b c+c a\n$$\n\nSubstituting $a^{2}+b^{2}+c^{2}$ for $a+b+c$ into the left hand side we wish equivalently to prove\n\n$$\na^{2}+b^{2}+c^{2} \\geq a b+b c+c a\n$$\n\nBut the $a^{2}+b^{2} \\geq 2 a b, b^{2}+c^{2} \\geq 2 b c, c^{2}+a^{2} \\geq 2 c a$ which by addition imply the desired inequality.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nA3.", "solution_match": "\nSolution."}}
-{"year": "2012", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "exam": "JBMO", "problem": "Solve the following equation for $x, y, z \\in \\mathbb{N}$\n\n$$\n\\left(1+\\frac{x}{y+z}\\right)^{2}+\\left(1+\\frac{y}{z+x}\\right)^{2}+\\left(1+\\frac{z}{x+y}\\right)^{2}=\\frac{27}{4}\n$$", "solution": "Call $a=1+\\frac{x}{y+z}, b=1+\\frac{y}{z+x}, c=1+\\frac{z}{x+y}$ to get\n\n$$\na^{2}+b^{2}+c^{2}=\\frac{27}{4}\n$$\n\nSince it is also true that\n\n$$\n\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=2\n$$\n\nthe quadratic-harmonic means inequality implies\n\n$$\n\\frac{3}{2}=\\sqrt{\\frac{a^{2}+b^{2}+c^{2}}{3}} \\geq \\frac{3}{\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}}=\\frac{3}{2}\n$$\n\nSo the inequality in the middle holds as an equality, and this happens whenever $a=b=c$, from which $1+\\frac{x}{y+z}=1+\\frac{y}{z+x}=1+\\frac{z}{x+y}$.\n\nBut $1+\\frac{x}{y+z}=1+\\frac{y}{z+x} \\Leftrightarrow x^{2}+x z=y^{2}+y z \\Leftrightarrow(x-y)(x+y)=z(y-\\dot{x})$ and the two sides of this equality will be of different sign, unless $x=y$ in which case both sides become 0 . So $x=y$, and similarly $y=z$, thus $x=y=z$.\n\nIndeed, any triad of equal natural numbers $x=y=z$ is a solution for the given equation, and so these are all its solutions.\n\nSolution 2. The given equation is equivalent to\n\n$$\n\\frac{27}{4}=(x+y+z)^{2}\\left(\\frac{1}{(y+z)^{2}}+\\frac{1}{(z+x)^{2}}+\\frac{1}{(x+y)^{2}}\\right)\n$$\n\nNow observe that by the well known inequality $a^{2}+b^{2}+c^{2} \\geq a b+b c+c a$, with $\\frac{1}{y+z}, \\frac{1}{z+x}$, $\\frac{1}{x+y}$ in place of $a, b, c$; we get:\n\n$$\n\\begin{aligned}\n\\frac{27}{4} & =(x+y+z)^{2}\\left(\\frac{1}{(y+z)^{2}}+\\frac{1}{(z+x)^{2}}+\\frac{1}{(x+y)^{2}}\\right) \\\\\n& \\geq(x+y+z)^{2}\\left(\\frac{1}{(y+z)(z+x)}+\\frac{1}{(z+x)(x+y)}+\\frac{1}{(x+y)(y+z)}\\right)=\\frac{2(x+y+z)^{3}}{(x+y)(y+z)(z+x)} \\\\\n& =\\frac{(2(x+y+z))^{3}}{4(x+y)(y+z)(z+x)}=\\frac{((x+y)+(y+z)+(z+x)))^{3}}{4(x+y)(y+z)(z+x)} \\stackrel{\\text { AM-GM }}{\\geq \\frac{(3 \\sqrt[3]{(x+y)(y+z)(z+x)})^{3}}{4(x+y)(y+z)(z+x)} .} \\\\\n& =\\frac{27}{4}\n\\end{aligned}\n$$\n\nThis means all inequalities in the above calculations are equalities, and this holds exactly whenever $x+y=y+z=z+x$, that is $x=y=z$. By the statement's demand we need to have $a, b, c$ integers. And conversely, any triad of equal natural numbers $x=y=z$ is indeed a solution for the given equation, and so these are all its solutions.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nA4.", "solution_match": "\nSolution 1."}}
-{"year": "2012", "tier": "T3", "problem_label": "A5", "problem_type": "Algebra", "exam": "JBMO", "problem": "Find the largest positive integer $n$ for which the inequality\n\n$$\n\\frac{a+b+c}{a b c+1}+\\sqrt[n]{a b c} \\leq \\frac{5}{2}\n$$\n\nholds for all $a, b, c \\in[0,1]$. Here $\\sqrt[1]{a b c}=a b c$.", "solution": "Let $n_{\\max }$ be the sought largest value of $n$, and let $E_{a, b, c}(n)=\\frac{a+b+c}{a b c+1}+\\sqrt[n]{a b c}$. Then $E_{a, b, c}(m)-E_{a, b, c}(n)=\\sqrt[m]{a b c}-\\sqrt[n]{a b c}$ and since $a . b c \\leq 1$ we clearly have $E_{a, b, c}(m) \\geq$ $E_{a, b, c}(n)$ for $m \\geq n$. So if $E_{a, b, c}(n) \\geq \\frac{5}{2}$ for some choice of $a, b, c \\in[0,1]$, it must be $n_{\\max } \\leq n$. We use this remark to determine the upper bound $n_{\\max } \\leq 3$ by plugging some particular values of $a, b, c$ into the given inequality as follow's:\n\n$$\n\\text { For }(a, b, c)=(1,1, c), c \\in[0,1] \\text {, inequality (1) implies } \\frac{c+2}{c+1}+\\sqrt[n]{c} \\leq \\frac{5}{2} \\Leftrightarrow \\frac{1}{c+1}+\\sqrt[n]{c} \\leq\n$$\n\n$\\frac{3}{2}$. Obviously, every $x \\in[0 ; 1]$ is written as $\\sqrt[n]{c}$ for some $c \\in[0 ; 1]$. So the last inequality is equivalent to:\n\n$$\n\\begin{aligned}\n& \\frac{1}{x^{n}+1}+x \\leq \\frac{3}{2} \\Leftrightarrow 2+2 x^{n+1}+2 x \\leq 3 x^{n}+3 \\Leftrightarrow 3 x^{n}+1 \\geq 2 x^{n+1}+2 x \\\\\n\\Leftrightarrow & 2 x^{n}(1-x)+(1-x)+(x-1)\\left(x^{n-1}+\\cdots+x\\right) \\geq 0 \\\\\n\\Leftrightarrow & (1-x)\\left[2 x^{n}+1-\\left(x^{n-1}+x^{n-2}+\\ldots+x\\right)\\right] \\geq 0, \\forall x \\in[0,1]\n\\end{aligned}\n$$\n\nFor $n=4$, the left hand side of the above becomes $(1-x)\\left(2 x^{4}+1-x^{3}-x^{2}-x\\right)=$ $(1-x)(x-1)\\left(2 x^{3}+x^{2}-1\\right)=-(1-x)^{2}\\left(2 x^{3}+x^{2}-1\\right)$ which for $x=0.9$ is negative. Thus. $n_{\\max } \\leq 3$ as claimed.\n\nNow, we shall prove that for $n=3$ inequality (1) holds for all $a, b, c \\in[0,1]$, and this would mean $n_{\\max }=3$. We shall use the following Lemma:\n\nLemma. For all $a, b, c \\in[0 ; 1]: a+b+c \\leq a b c+2$.\nProof of the Lemma: The required result comes by adding the following two inequalities side by side\n\n$$\n\\begin{aligned}\n& 0 \\leq(a-1)(b-1) \\Leftrightarrow a+b \\leq a b+1 \\Leftrightarrow a+b-a b \\leq 1 \\\\\n& 0 \\leq(a b-1)(c-1) \\Leftrightarrow a b+c \\leq a b c+1\n\\end{aligned}\n$$\n\nBecause of the Lemma, our inequality (1) for $n=3$ wrill be proved if the following weaker inequality is proved for all $a, b, c \\in[0,1]$ :\n\n$$\n\\frac{a b c+2}{a b c+1}+\\sqrt[3]{a b c} \\leq \\frac{5}{2} \\Leftrightarrow \\frac{1}{a b c+1}+\\sqrt[3]{a b c} \\leq \\frac{3}{2}\n$$\n\nDenoting $\\sqrt[3]{a b c}=y \\in[0 ; 1]$, this inequality becomes:\n\n$$\n\\begin{aligned}\n& \\frac{1}{y^{3}+1}+y \\leq \\frac{3}{2} \\Leftrightarrow 2+2 y^{4}+2 y \\leq 3 y^{3}+3 \\Leftrightarrow-2 y^{4}+3 y^{3}-2 y+1 \\geq 0 \\\\\n\\Leftrightarrow & 2 y^{3}(1-y)+(y-1) y(y+1)+(1-y) \\geq 0 \\Leftrightarrow(1-y)\\left(2 y^{3}+1-y^{2}-y\\right) \\geq 0\n\\end{aligned}\n$$\n\nThe last inequality is obvious because $1-y \\geq 0$ and $2 y^{3}+1-y^{2}-y=y^{3}+(y-1)^{2}(y+1) \\geq 0$.\n\n## Geometry\n\n2", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nA5.", "solution_match": "\nSolution."}}
-{"year": "2012", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C$ be an equilateral triangle, and $P$ a point on the circumcircle of the triangle $A B C$ and distinct from $A, B$ and $C$. If the lines through $P$ and parallel to $B C, C A, A B$ intersect the lines $C A, A B, B C$ at $M, N$ and $Q$ respectively, prove that $M, N$ and $Q$ are collinear.", "solution": "Solution 1. Let lines $K H, A B$ intersect at $M$ (Figure 5a). From the quadrelateral $K M A C$ we have\n\n$\\angle K M A=360^{\\circ}-\\angle A-\\angle A C K-\\angle C K M=360^{\\circ}-\\angle A-90^{\\circ}-\\left(180^{\\circ}-2 \\angle K C H\\right)=$ $90-\\angle A+2 \\angle K C H=90-\\angle A+2\\left(90^{\\circ}-\\angle A C B\\right)=270^{\\circ}-\\angle A-2 \\angle A C B=270-\\angle A-$ $\\angle A C B-\\angle A B C=270^{\\circ}-180^{\\circ}=90^{\\circ}$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_61d3145f18c90bf9f370g-06.jpg?height=519&width=574&top_left_y=320&top_left_x=455)\n\n(a)\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_61d3145f18c90bf9f370g-06.jpg?height=634&width=571&top_left_y=317&top_left_x=1093)\n\n(b)\n\nFigure 2: Exercise G2.\n\nso $K H \\perp A B$ as wanted.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nG1.", "solution_match": "\nSolution."}}
-{"year": "2012", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C$ be an equilateral triangle, and $P$ a point on the circumcircle of the triangle $A B C$ and distinct from $A, B$ and $C$. If the lines through $P$ and parallel to $B C, C A, A B$ intersect the lines $C A, A B, B C$ at $M, N$ and $Q$ respectively, prove that $M, N$ and $Q$ are collinear.", "solution": "Solution 2. Let $D$ be a point on $c$ such that $A D<A C$, and let $E, Z$ be the second points of intersection of lines $A D$ and $B D$ with $c$ respectively. Let also $N$ be the second point of intersection of line $B E$ with the circle $c$. Figure $5 \\mathrm{~b}$ shows $Z$ between $B, D$. The argument below can be trivially modified to apply in case $D$ is in the segment $B, Z$ as well. It is\n\n$$\nA B^{2}=A C^{2}=A D \\cdot A E \\Rightarrow \\frac{A B}{A E}=\\frac{A D}{A B}\n$$\n\nThis relation arid the fact that $\\angle B A E=\\angle B A D$ implies that the triangles $A B E, A D B$ are similar. Thus $\\angle A B E=\\angle A D B$. Also from the cyclic quadrilateral we get $\\angle A D B=\\angle Z N E$. Therefore $\\angle A B E=\\angle Z N E$, so $A B \\| N Z$.\n\nCall $P$ the intersection point of $B C, N Z$. Since $A C$ is tangent to $c$ it is\n\n$$\n\\angle C E H=\\angle B C A\n$$\n\nand then\n\n$$\n\\begin{aligned}\n& \\angle Z N H+\\angle C N Z=\\angle H N C=\\angle C E H \\stackrel{(3)}{=} \\angle B C A=\\angle A B C=\\angle Z P C=\\angle B C N+\\angle C N Z \\\\\n\\Rightarrow \\quad & \\angle Z N H=\\angle B C N \\\\\n\\Rightarrow \\quad & \\angle Z N H=\\angle H Z N\n\\end{aligned}\n$$\n\nTherefore $H$ is the midpoint of the arc $N Z$, so $K H \\perp N Z$ and as $A B \\| N Z$ we finally get $K H \\perp A B$ as wanted.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_61d3145f18c90bf9f370g-07.jpg?height=494&width=1167&top_left_y=290&top_left_x=501)\n\nFigure 3: Exercise G3.\n\n2-3 G3. Let $A B$ and $C D$ be chords in a circle of center $O$ with $A, B, C, D$ distinct, and let the lines $A B$ and $C D$ meet at a right angle at point $E$. Let also $M$ and $N$ be the midpoints of $A C$ and $B D$ respectively. If $M N \\perp O E$, prove that $A D \\| B C$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nG1.", "solution_match": "\nSolution."}}
-{"year": "2012", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C$ be an equilateral triangle, and $P$ a point on the circumcircle of the triangle $A B C$ and distinct from $A, B$ and $C$. If the lines through $P$ and parallel to $B C, C A, A B$ intersect the lines $C A, A B, B C$ at $M, N$ and $Q$ respectively, prove that $M, N$ and $Q$ are collinear.", "solution": "Solution. $E$ can be inside, or outside the circle (Figure 3) but the proof below holds in both cases; notice that $E$ cannot be on the circle as $A, B, C, D$ are distinct. Let lines $A C$ and $N E$ meet at point $P$. Then $E N=D N=B N$ (median in a right triangle), so $\\angle P E C=\\angle N E D=\\angle N D E=\\angle B D C=\\angle B A C=\\angle E A P$. Now $A B \\perp C D$ so $E N \\perp A C$. But $O M \\perp A C$ so $O M \\| E N$. Similarly $O N \\| E M$ so $N E M O$ is a parallelogram (possibly degenerated). As $M N \\perp O E$, this parallelogram is a rhombus. Then the chords $A C$ and $B D$, being equidistant from $O$, are equal. Hence their minor arcs are equal, which means that either $A D \\| B C$ or $A B \\| C D$; the latter contradicts the fact that $A B$ and $C D$ meet at $E$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nG1.", "solution_match": "\nSolution."}}
-{"year": "2012", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C$ be an acute-angled triangle with circumcircle $\\Gamma$, and let $O, H$ be the triangle's circumcenter and orthocenter respectively. Let also $A^{\\prime}$ be the point where the angle bisector of angle $B A C$ meets $\\Gamma$. If $A^{\\prime} H=A H$, find the measure of angle $B A C$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_61d3145f18c90bf9f370g-07.jpg?height=519&width=1042&top_left_y=1756&top_left_x=541)\n\nFigure 4: Exercise G4.", "solution": "Solution. Let $P$ be the symmetric of $A$ with respect to $M$ (Figure 5). Then $A M=M P$ and $t \\perp A P$, hence the triangle $A P N$ is isosceles with $\\mathrm{AP}$ as its base, so $\\angle N A P=\\angle N P A$. We have $\\angle B A P=\\angle B A M=\\angle B M N$ and $\\angle B A N=\\angle B N M$. Thus\n\n$$\n180^{\\circ}-\\angle N B M=\\angle B N M+B M N=\\angle B A N+\\angle B A P=\\angle N A P=\\angle N P A\n$$\n\nso the quadrangle $M B N P$ is cyclic (since the points $B$ and $P$ lie on different sides of $M N$ ). Hence $\\angle A P B=\\angle M P B=\\angle M N B$ and the triangles $A P B$ and $M N B$ are congruent $(M N=2 A M=A M+M P=A P)$. From that we get $A B=M B$, i.e. the triangle $A M B$ is\nisosceles, and since $t$ is tangent to $k_{1}$ and perpendicular to $A M$, the center of $k_{1}$ is on $A M$, hence $A M B$ is a right-angled triangle. From the last two statements we infer $\\angle A M B=45^{\\circ}$, and so $\\angle N M B=90^{\\circ}-\\angle A M B=45^{\\circ}$.\n\n4. G6. Let $O_{1}$ be a point in the exterior of the circle $c(O, R)$ and let $\\dot{O}_{1} N, O_{1} D$ be the tangent segments from $O_{1}$ to the circle. On the segment $O_{1} N$ consider the point $B$ such that $B N=R$. Let the line from $B$ parallel to $O N$ intersect the segment $O_{1} D$ at $C$. If $A$ is a point on the segment $O_{1} D$ other than $C$ so that $B C=B A=a$, and if $\\dot{C}^{\\prime}(K, r)$ is the incircle of the triangle $O_{1} A B$; find the area of $A B C$ in terms of a, $R, r$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nG4.", "solution_match": "\nSolution."}}
-{"year": "2012", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C$ be an acute-angled triangle with circumcircle $\\Gamma$, and let $O, H$ be the triangle's circumcenter and orthocenter respectively. Let also $A^{\\prime}$ be the point where the angle bisector of angle $B A C$ meets $\\Gamma$. If $A^{\\prime} H=A H$, find the measure of angle $B A C$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_61d3145f18c90bf9f370g-07.jpg?height=519&width=1042&top_left_y=1756&top_left_x=541)\n\nFigure 4: Exercise G4.", "solution": "Solution. Obviously; the segment $B C$ is taingent to the circle $c$. Let $M$ be the point of tangency (Figure 6). Call $Q, M$ the tangency points of $B A, B C$ with $c^{\\prime}$ and $c$ respectively, and call $H$ the midpoint of segment $A C$. It is well known that\n\n$$\nA Q=\\frac{1}{2}\\left(A O_{1}+A B-B O_{1}\\right) \\text { and } C M=\\frac{1}{2}\\left(B O_{1}+B C-C O_{1}\\right)\n$$\n\nand so\n\n$$\nA Q+C M=\\frac{1}{2}(2 B C-A C)=a-\\frac{1}{2} A C=a-H C\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_61d3145f18c90bf9f370g-09.jpg?height=509&width=1019&top_left_y=1247&top_left_x=564)\n\nFigure 6: Exercise G6.\n\nThe triangles $K A Q$ and $O C M$ are similar and this implies\n\n$$\n\\begin{aligned}\n\\frac{K Q}{O M} & =\\frac{A Q}{C M} \\Leftrightarrow \\frac{K Q}{A Q}=\\frac{O M}{C M} \\Leftrightarrow \\frac{r}{A Q}=\\frac{R}{C M}=\\frac{R+r}{A Q+C M}=\\frac{R+r}{a-H C} \\Leftrightarrow \\\\\n\\frac{r}{A Q} & =\\frac{R+r}{a-H A}\n\\end{aligned}\n$$\n\nIf $A Z$ is the bisector segment of triangle $B A H$ it holds\n\n$$\n\\angle A Z H=90^{\\circ}=\\frac{1}{2} \\angle B A C \\text { and } \\angle K A Q=\\frac{1}{2}\\left(180^{\\circ}-\\angle B A C\\right)=90^{\\circ}-\\frac{1}{2} \\angle B A C\n$$\n\nTherefore, from the similar triangles $K Q A$ and $A H Z$ we get\n\n$$\n\\frac{-A H}{Z H}=\\frac{r}{A Q} \\stackrel{(6)}{=} \\frac{R+r}{a-H A}\n$$\n\nAlso, from the bisector-theorem in triangle $A B H$ it holds\n\n$$\nZ H=\\frac{A H \\cdot B H}{a+A H}\n$$\n\nand from (7) it follows\n\n$$\n\\frac{R+r}{a-H A}=\\frac{A H}{\\frac{A H \\cdot B H}{a+A H}} \\Rightarrow R+r=\\frac{a^{2}-A H^{2}}{B H}=\\frac{B H^{2}}{B H}=B H\n$$\n\nSo\n\n$$\nH A^{2}=a^{2}-(R+r)^{2} \\Leftrightarrow H A=\\sqrt{a^{2}-(R+r)^{2}}\n$$\n\nand finally the area of triangle $A B C$ in terms of $a, R, r$ is:\n\n$$\n(A B C)=A H \\cdot B H=(R+r) \\sqrt{a^{2}-(R+r)^{2}}\n$$\n\n4 GZ, Let $M N P Q$ be a square of side length 1 , and $A, B, C, D$ points on the sides $M N$, $N P, P Q$, and $Q M$ respectively such that $A C \\cdot B D=\\frac{5}{4}$. Can the set $\\{A B, B C, C D, D A\\}$ be partitioned into two subsets $S_{1}$ and $S_{2}$ of two elements each, so that.each one of the sums of the elements of $S_{1}$ and $S_{2}$ are positive integers?", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nG4.", "solution_match": "\nSolution."}}
-{"year": "2012", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C$ be an acute-angled triangle with circumcircle $\\Gamma$, and let $O, H$ be the triangle's circumcenter and orthocenter respectively. Let also $A^{\\prime}$ be the point where the angle bisector of angle $B A C$ meets $\\Gamma$. If $A^{\\prime} H=A H$, find the measure of angle $B A C$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_61d3145f18c90bf9f370g-07.jpg?height=519&width=1042&top_left_y=1756&top_left_x=541)\n\nFigure 4: Exercise G4.", "solution": "Solution. The answer is negative.\n\nSuppose such a partitioning was possible (Figure 7). Then $A B+B C+C D+D A \\in \\mathbb{N}$. But $(A B+B C)+(C D+D A)>A C+A C \\geq 2$, hence $A B+B C+C D+D A>2$.\n\nOn the other hand, $A B+B C+C D+D A<(A N+N B)+(B P+P C)+(C Q+Q D)+$ $(D M+M A)=4$, hence $A B+B C+C D+D A=3$.\n\nObviously one of the aums of the elerients of $S_{1}$ and $S_{2}$ must be 1 and the other 2 . Without any loss of generality, we may assume that the sum of the elements of $S_{1}$ is 1 and the sum of the elements of $S_{2}$ is 2 . As $A B+B C>A C \\geq 1$ we find that $S_{1} \\neq\\{A B, B C\\}$. Similarly, $S_{1}$ cannot contain two adjacent sides of the quadrilateral $A B C D$. Therefore, without any loss of generality, we may assume that $S_{1}=\\{A D, B C\\}$ and $S_{2}=\\{A B, C D\\}$. Then $A D+B C=1$ and $A B+C D=2$.\n\nWe have $A D \\cdot B C \\leq \\frac{1}{4} \\cdot(A D+C B)^{2}=\\frac{1}{4}$ and $A B \\cdot C D \\leq \\frac{1}{4} \\cdot(A B+C D)^{2}=1$.\n\nAccording to Ptolemy's inequality, we have\n\n$$\n\\frac{5}{4}=A C \\cdot B D \\leq A B \\cdot C D+A D \\cdot B C=\\frac{1}{4}+1=\\frac{5}{4}\n$$\n\nhence we have equality all around, which means the quadrilateral $A B C D$ is cyclic, $A D=$ $B C=\\frac{1}{2}$ and $A B=C D=1$, hence $A B C D$ is a rectangle of dimensions 1 and $\\frac{1}{2}$.\n\nThere are many different ways of proving that this configuration is not possible. For example: - Suppose $A B C D$ is a rectangle with $A D=\\frac{1}{2}, A B=1$. Then we have $A C=B D=\\frac{\\sqrt{5}}{2}$ and $\\triangle A N B \\equiv \\triangle C Q D$ (Angle-Side-Angle). Denoting $A M=x, M D=y$ we have $A N=$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_61d3145f18c90bf9f370g-11.jpg?height=448&width=1127&top_left_y=318&top_left_x=561)\n\nFigure 7: Exercise G7.\n\n$1-x, B N=1-y$ and the following conditions need to be fulfilled for some $x, y \\in[0 ; 1]$ (Pythagorean Theorem in triangles $A M D, A N B, B B^{\\prime} C$, where $B^{\\prime}$ is the projection of $B$ on $M Q)$ :\n\n$$\nx^{2}+y^{2}=\\frac{1}{4}, \\quad(1-x)^{2}+(1-y)^{2}=1 \\text { and } 1+(2 y-1)^{2}=\\frac{5}{4}\n$$\n\nBut $1+(2 y-1)^{2}=\\frac{5}{4}$ implies $y \\in\\left\\{\\frac{1}{4}, \\frac{3}{4}\\right\\}$. If $y=\\frac{3}{4}$, then $x^{2}+y^{2}=\\frac{1}{4}$ cannot hold. If on the other hand $y=\\frac{1}{4}$, then $(1-x)^{2}+(1-y)^{2}=1$ implies $x=0$, but then $(1-x)^{2}+(1-y)^{2}=1$ cannot hold. Therefore such a configuration is not possible.\n\n## Combinatorics", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nG4.", "solution_match": "\nSolution."}}
-{"year": "2012", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Along a round table are arranged 11 cards with the names (all distinct) of the 11 members of the $16^{\\text {th }}$ JBMO Problem Selection Committee. The distances between each two consecutive cards are equal. Assume that in the first meeting of the Committee none of its 11 members sits in front of the card with his name. Is it possible to rotate the table by some angle so that at the end at least two members of sit in front of the card with their names?", "solution": "Yes it is: Rotating the table by the angles $\\frac{360^{\\circ}}{11}, 2 \\cdot \\frac{360^{\\circ}}{11}, 3 \\cdot \\frac{360^{\\circ}}{11}, \\ldots, 10 \\cdot \\frac{360^{\\circ}}{11}$, we obtain 10 new positions of the table. By the assumption, it is obvious that every one of the 11 members of the Committee will be seated in front of the card with his name in exactly one of these 10 positions. Then by the Pigeonhole Principle there should exist one among these 10 positions in which at least two of the $11(>10)$ members of the Committee will be placed in their positions, as claimed.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nC1.", "solution_match": "\nSolution."}}
-{"year": "2012", "tier": "T3", "problem_label": "C2", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "$n$ nails nailed on a board are connected by two via a string. Each string is colored in one of $n$ given colors. For any three colors there exist three nails conne.cted by two with strings in these three colors. Can $n$ be: (a) 6, (b) 7?", "solution": "(a) The answer is no:\n\nSuppose it is possible. Consider some color, say blue. Each blue string is the side of 4 triangles formed with vertices on the given points. As there exist $\\binom{5}{2}=\\frac{5 \\cdot 4}{2}=10$ pairs of colors other than blue, and for any such pair of colors together with the blue color there exists a triangle with strings in these colors, we conclude that there exist at least 3 blue strings (otherwise the number of triangles with a blue string as a side would be at most $2 \\cdot 4=8$, a contradiction). The same is true for any color, so altogether there exist at least $6 \\cdot 3=18$ strings, while we have just $\\binom{6}{2}=\\frac{6 \\cdot 5}{2}=15$ of them.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_61d3145f18c90bf9f370g-12.jpg?height=519&width=488&top_left_y=1850&top_left_x=835)\n\nFigure 8: Exercise C2.\n\n(b) The answer is yes (Figure 8):\n\nPut the nails at the vertices of a regular 7-gon (Figure 8) and color each one of its sides in a different color. Now color each diagonal in the color of the unique side parallel to it. It can be checked directly that each triple of colors appears in some triangle (because of symmetry, it is enough to check only the triples containing the first color).\n\nRemark. The argument in (a) can be applied to any even $n$. The argument. in (b) can be applied to any odd $n=2 k+1$ as follows: first number the nails as $0,1,2 \\ldots, 2 k$ and similarly number the colors as $0,1,2 \\ldots, 2 k$. Then connect nail $x$ with nail $y$ by a string of color $x+y(\\bmod n)$. For each triple of colors $(p, q, r)$ there are vertices $x, y, z$ connected by these three colors. Indeed, we need to solve (modn.) the system\n\n$$\n(*)(x+y \\equiv p, x+z \\equiv q, y+z \\equiv r)\n$$\n\nAdding all three, we get $2(x+y+z) \\equiv p+q+r$ and multiplying by $k+1$ we get $x+y+z \\equiv$ $(k+1)(p+q+r)$. We can now find $x, y, z$ from the identities $(*)$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nC2.", "solution_match": "\nSolution."}}
-{"year": "2012", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "In a circle of diameter 1 consider 65 points no three of which are collinear. Prove that there exist 3 among these points which form a triangle with area less then or equal to $\\frac{1}{72}$.", "solution": "Solution. It is $s=(a+b)^{3}-63 a b(a+b)$ which gives the same residue module 7 as $(a+b)^{3}$. But the residues modulo 7 of perfect cubes can only be 0,1 or 6 . So the residue of $s$ modulo 7 is 0,1 or 6 . Now for $a=6, b=-1$ we get $s=2015 \\geq 2012$ and this is the least possible value of $s$ because the numbers 2012, 2013, 2014 give 3,4 and 5 as residues $\\bmod 7$ which are distinct from $0,1,6$, and so 2012, 2013, 2014 cannot be $s$ for any choice of $a, b$. $\\square$\n\n2-3 NT2. Do there exist prime numbers $p$ and $q$ such that $p^{2}\\left(p^{3}-1\\right)=q(q+1)$ ?", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nC3.", "solution_match": "\nSolution."}}
-{"year": "2012", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "In a circle of diameter 1 consider 65 points no three of which are collinear. Prove that there exist 3 among these points which form a triangle with area less then or equal to $\\frac{1}{72}$.", "solution": "Solution. Write the given equation in the form\n\n$$\np^{2}(p-1)\\left(p^{2}+p+1\\right)=q(q+1)\n$$\n\nFirst observe that it must not be $p=q$, since in this case the left hand side of (9) is greater than its right hand side. Hence, since $p$ and $q$ are distinct prines, (9) immediately yields $p^{2} \\mid q+1$, that is\n\n$$\nq=a p^{2}-1\n$$\n\nfor some $a \\in \\mathrm{N}$. Since $p$ and $q$ are both primes, by (9) we get the following cases:\n\nCase 1: $q \\mid p-1$, that is\n\n$$\np=b q+1\n$$\n\nfor some $b \\in N$. Substituting (11) into (10), and using the fact that $a \\geq 1$ and $b \\geq 1$, we obtain\n\n$$\nq=a(b q+1)^{2}-1 \\geq(q+1)^{2}-1=q^{2}+2 q\n$$\n\na contradiction.\n\nCase 2: $q \\mid p^{2}+p+1$, that is\n\n$$\np^{2}+p+1=b q\n$$\n\nfor some $b \\in$ N. Substituting (10) into (12), we get\n\n$$\np^{2}+p+1=b\\left(a p^{2}-1\\right)\n$$\n\nIf $a \\geq 2$, then from (13) it follows that\n\n$$\np^{2}+p+1 \\geq 2 p^{2}-1\n$$\n\nor equivalently, $p+1 \\geq(p-1)(p+1)$, that is, $(p+1)(2-p) \\geq 0$. This implies that $p=2$, and so $q \\mid 2^{2}+2+1=7$. Hence, $q=7$, but the pair $p=2$ and $q=7$ does not satisfy the equation ( 9 ).\n\nHence, it must be $a=1$. Then if $b \\geq 3$, (13) implies\n\n$$\np^{2}+p+1 \\geq 3\\left(p^{2}-1\\right)\n$$\n\nor equivalently, $4 \\geq p(2 p-1)$, which is obviously impossible.\n\nThus, it must be $a=1$ and $b \\in\\{1,2\\}$. For $a=b=1$, (13) implies that $p=2$, which by (12) again yields $q=7$, which is impossible. Finally, for $a=1$ and $b=2$, (13) gives $p(p-1)=3$, which is clearly not satisfied for any prime $p$.\n\nHence, there do not exist prime numbers $p$ and $q$ which satisfy given equation.\n\n## 3 NT3. Decipher the equality\n\n$$\n(\\overline{V E R}-\\overline{I A}):(\\overline{G R E}+\\overline{E C E})=G^{R^{E}}\n$$\n\nassuming that the number $\\overline{\\text { GREECE }}$ has a maximum value. It is supposed that each letter corresponds to a unique digit from 0 to 9 and different letters correspond to different digits, and also that all letters $G, E, V$ and $I$ are different from 0 . Also, the notation $\\overline{a_{n} \\ldots a_{1} a_{0}}$ stands for the number $a_{n} \\cdot 10^{n}+\\cdots+10^{1} \\cdot a_{1}+a_{0}$ :", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nC3.", "solution_match": "\nSolution."}}
-{"year": "2012", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "In a circle of diameter 1 consider 65 points no three of which are collinear. Prove that there exist 3 among these points which form a triangle with area less then or equal to $\\frac{1}{72}$.", "solution": "Solution. Denote\n\n$$\nx=\\overline{V E R}-\\overline{I A}, y=\\overline{G R E}+\\overline{E C E}, z=G^{R^{E}}\n$$\n\nThen obviously, we have\n\n$$\n\\begin{aligned}\n& (201+131 \\text { or } 231+101) \\leq y \\leq(879+969 \\text { or } 869+979 \\text { or. } 769+989) \\\\\n\\Rightarrow \\quad & 332 \\leq y \\leq 1848 \\Rightarrow 102-98 \\leq x \\leq 987-10 \\Rightarrow 4 \\leq x \\leq 977,\n\\end{aligned}\n$$\n\nhence it follows that\n\n$$\n\\frac{4}{1848} \\leq \\frac{x}{y}=z \\leq \\frac{977}{332} \\Rightarrow 1 \\leq z \\leq 2\n$$\n\nThis shows that $z=G^{R^{E}} \\in\\{1,2\\}$. Hence, if $R \\geq 1$, then $R^{E} \\geq 1$, which implies that $2 \\geq G^{R^{E}} \\geq G$. Thus, if $R \\geq 1$, then it must be $G \\leq 2$. In view of this and the assumption of the problem that the number $\\overline{G R E E C E}$ has a maximum value, we will consider the case when $R=0$ hoping to get a solution with $G>2$. Then $G^{R^{E}}=G^{0}=1$ for all digits $G$ and $E$ with $1 \\leq G, E \\leq 9$, and therefore, the above equality becomes\n\n$$\n\\overline{V E R}-\\overline{I A}=\\overline{G R E}+\\overline{E C E}\n$$\n\nwhich substituting $R=0$, can be written as\n\n$$\n\\overline{V E O}=\\overline{G O E}+\\overline{E C E}+\\overline{I A}\n$$\n\nNow we consider the following cases:\n\nCase 1: $G=9$. Then $V \\leq 8$, so $\\overline{V E 0} \\leq 900$, while the right hand side of (1) is greater than 900 . This is impossible, and no solution exists in this case.\n\nCase 2: $G=8$. Then (14) becomes\n\n$$\n\\overline{V E 0}=\\overline{80 E}+\\overline{E C E}+\\overline{I A}\n$$\n\nhence it immediately follows that $V=9$. For $V=9$, (15) becomes\n\n$$\n\\overline{9 E 0}=\\overline{80 E}+\\overline{E C E}+\\overline{I A}\n$$\n\nNotice that for $E \\geq 2$, the right hand side of (16) is greater than 1000 , while the left hand side of (16) is less than 1000 . Therefore, it must be $E \\leq 1$, that is, $E=1$ in view of the fact that $R=0$. Substituting $E=1$ into (16), we get\n\n$$\n\\overline{910}=\\overline{801}+\\overline{1 C 1}+\\overline{I A}\n$$\n\nhence it follows that\n\n$$\n109=\\overline{1 C 1}+\\overline{I A}\n$$\n\nBut the right hand side of (18) is greater than 121. This shows that $G=8$ does not lead to any solution.\n\nCase 3: $G=7$. Then (14) becomes\n\n$$\n\\overline{V E O}=\\overline{70 E}+\\overline{E C E}+\\overline{I A}\n$$\n\nThus it must be $V \\geq 8$.\n\nSubcase 3(a): $V=8$. Then (19) gives\n\n$$\n\\overline{\\delta E 0}=\\overline{70 E}+\\overline{E C E}+\\overline{I A}\n$$\n\nhence we immediately obtain $E=1$ (since the right hand side of (6) must be lass than 900). For $E=1,(20)$ reduces to\n\n$$\n109=\\overline{1 C 1}+\\overline{I A}\n$$\n\nwhich is impossible since $\\overline{1 C 1} \\geq 121$.\n\nSubcase 3(b): $V=9$. Then (19) gives\n\n$$\n\\overline{9 E 0}=\\overline{70 E}+\\overline{E C E}+\\overline{I A}\n$$\n\nhence we immediately obtain $E \\leq 2$ (since the right hand side of (7) must be less than 1000). For $E=2$, (22) reduces to\n\n$$\n218=\\overline{2 C 2}+\\overline{I A}\n$$\n\nwhich is impossible since $\\overline{2 C 2} \\geq 232$. Finally, for $E=1$, (22) reduces to\n\n$$\n209=\\overline{1 C 1}+\\overline{I A}\n$$\n\nSince it is required that the number $\\overline{G R E E C E}$ has a maximum value, taking $C=8$ into (24) we find that\n\n$$\n28=\\overline{I A}\n$$\n\nwhich yields $8=A=C$. This is impossible since must be $A \\neq C$. Since $C \\neq G=7$, then taking $C=6$ into (24) we obtain\n\n$$\n48=\\overline{I A}\n$$\n\nhence we have $I=4$ and $A=8$. Previously, we have obtain $G=7, R=0, V=9, E=1$ and $C=6$. For these values, we obtain that $\\overline{G R E E C E}=701161$ is the desired maximum value.\n\n4 NT4. Determine all triples $(m, n, p)$ satisfying\n\n$$\nn^{2 p}=m^{2}+n^{2}+p+1\n$$\n\nwhere $m$ and $n$ are integers and $p$ is a prime number.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nC3.", "solution_match": "\nSolution."}}
-{"year": "2012", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "In a circle of diameter 1 consider 65 points no three of which are collinear. Prove that there exist 3 among these points which form a triangle with area less then or equal to $\\frac{1}{72}$.", "solution": "Solution. By Fermat's theorem $n^{2 p} \\equiv n^{2}(\\bmod p)$, therefore $m^{2}+n^{2}+p+1 \\equiv n^{2}(\\bmod p) \\Rightarrow$ $m^{2} \\equiv-1(\\bmod p)$.\n\nCase 1: $p=4 k+3$. We have $\\left(m^{2}\\right)^{2 k+1} \\equiv(-1)^{2 k+1}(\\bmod p)$. Therefore,\n\n$$\nm^{p-1} \\equiv-1(\\bmod p)\n$$\n\nand $p$ does not divide $m$. On the other hand, by Fermat's theorem\n\n$$\nm^{p-1} \\equiv 1(\\bmod p)\n$$\n\n(28) and (29) yield $p=2$. Thus, $p \\neq 4 k+3$.\n\nCase 2: $p=4 k+1$. Let us consider (27) in mod4. $n^{2}=0$ or 1 in mod4. In both cases $n^{2 p}=n^{2}(\\bmod 4)$. From $(27)$ we get $n^{2} \\equiv m^{2}+n^{2}+1+1(\\bmod 4)$. Therefore, $m^{2} \\equiv-2(\\bmod 4)$, and again there is no solution.\n\nCase 3: $p=2$. The given equation is written as\n\n$$\nn^{4}-n^{2}-3=m^{2}\n$$\n\nLet $l=n^{2}$. Readily, we do not get any solution for $l=0$, 1 . If $l=4$, then there are four solutions: $(3,2,2),(-3,2,2),(3,-2,2),(-3,-2,2)$. There is no solution for $l>4$, since in this case\n\n$$\n(l-1)^{2}=l^{2}-2 l+1<m^{2}=l^{2}-l-3<l^{2}\n$$\n\nThus, (27) has four solutions: $(3,2,2),(-3,2,2),(3,-2,2),(-3,-2,2)$ and we are done.\n\n4", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nC3.", "solution_match": "\nSolution."}}
-{"year": "2012", "tier": "T3", "problem_label": "NT5", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all the positive integers $x, y, z, t$ such that $2^{x} \\cdot 3^{y}+5^{z}=7^{t}$.", "solution": "Solution. First we prove the following Lemma\n\nLemma: If $x, y, z$ real numbers such that $x+y+z=0$, then $2\\left(x^{4}+y^{4}+z^{4}\\right)=\\left(x^{2}+y^{2}+z^{2}\\right)^{2}$. Proof of the Lemma:\n\n$$\n\\begin{aligned}\nx^{4}+y^{4}+z^{4} & =x^{2} x^{2}+y^{2} y^{2}+z^{2} z^{2}=x^{2}(y+z)^{2}+y^{2}(z+x)^{2}+z^{2}(x+y)^{2} \\\\\n& =2\\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\\right)+2 x y z(x+y+z) \\\\\n& =\\left(x^{2}+y^{2}+z^{2}\\right)^{2}-x^{4}-y^{4}-z^{4}\n\\end{aligned}\n$$\n\nand the claim follows.\n\nNow back to our problem notice that $(a-2 b+c)+(b-2 c+a)+(c-2 a+b)=0$, thus according to the lemma it holds\n\n$$\n\\begin{aligned}\nA & =2(a-2 b+c)^{4}+2(b-2 c+a)^{4}+2(c-2 a+b)^{4} \\\\\n& =\\left[(a-2 b+c)^{2}+(b-2 c+a)^{2}+(c-2 a+b)^{2}\\right]=\\left[6\\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\\right)\\right]^{2}\n\\end{aligned}\n$$\n\nSince $a^{2}+b^{2}+c^{2} \\geq a b+b c+c a$ we have that\n\n$$\n\\sqrt{A}+1=6\\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\\right)+1\n$$\n\nIn addition, it is easy to check that\n\n$$\nB=d(d+1)(d+2)(d+3)=\\left(d^{2}+3 d+1\\right)^{2}\n$$\n\nLet us set $6\\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\\right)+1=m, d^{2}+3 d+1=n$. We need to prove that the number $(\\sqrt{A}+1)^{2}+B=m^{2}+n^{2}$ is not a perfect square.\n\nSince both $m, n$ are odd integers, both $m^{2}, n^{2}$ are integers of the form $4 k+1$, so the number $m^{2}+n^{2}$ is an integer of the form $4 k+2$. But it is well known that all perfect squares are of the form $4 k$ or $4 k+1$, and we are done.\n\n$4 \\quad \\mathrm{NT7}$. Find all natural numbers a,b, c for which $1997^{a}+15^{b}=2012^{c}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nNT5.", "solution_match": "\nSolution."}}
-{"year": "2012", "tier": "T3", "problem_label": "NT5", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all the positive integers $x, y, z, t$ such that $2^{x} \\cdot 3^{y}+5^{z}=7^{t}$.", "solution": "Solution. $1997^{a}+15^{b}=2012^{c} \\Rightarrow 1+(-1)^{b} \\equiv 0(\\bmod 4)$, so $b$ is an odd number.\n\n$1997^{a}+15^{b}=2012^{c} \\Rightarrow 1+0 \\equiv 2^{c}(\\bmod 3)$, so $c$ is even, say $c=2 c_{1}$.\n\nWe intend to consider the given equation modulo 8 and for this reason we discern two cases:\n\n(1): $c=1$. Clearly then $a=b=1$ and $a=b=c=1$ is a solution. This is actually the only solution of the given equation since in the remaining case where $c>1$ it will be shown that there exist no solution.\n\n(2): $c>1$. Then $2012^{c}=(4 \\cdot 503)^{c}$ is a multiple of 8 and\n\n$1997^{a}+15^{b}=2012^{c} \\Rightarrow 5^{a}+(-1)^{b} \\equiv 5^{a}+(-1) \\equiv 0(\\bmod 8)$, so $a$ is even, say $a=2 a_{1}$. Hence\n\n$$\n3^{\\&} \\cdot 5^{d 1}=15^{b}=2012^{c}-1997^{a}=\\left(2012^{c_{1}}-1997^{a_{1}}\\right) \\cdot\\left(2012^{c_{1}}+1997^{a_{1}}\\right)\n$$\n\nObserve that $2012^{c_{1}}-1997^{a_{1}}, 2012^{c_{1}}+199^{\\prime} 7^{a_{1}}$ are both greater than 1 and prime to each other as $\\operatorname{gcd}\\left(2012^{c_{1}}-1997^{a_{1}}, 2012^{c_{1}}+1997^{a_{1}}\\right)=\\operatorname{gcd}\\left(2012^{c_{1}}-1997^{a_{1}}, 2 \\cdot 1997^{a_{1}}\\right)=1$. So there exist two cases:\n\n$$\n\\text { Case 1: } \\begin{aligned}\n& 2012^{c_{1}}-1997^{a_{1}}=5^{b} \\\\\n& 2012^{c_{1}}+1997^{a_{1}}=3^{b}\n\\end{aligned}, \\quad \\text { Case 2: } \\begin{aligned}\n& 2012^{c_{1}}-1997^{a_{1}}=3^{b} \\\\\n& 2012^{c_{1}}+1997^{a_{1}}=5^{b}\n\\end{aligned}\n$$\n\nCase 1:\n\n$$\n\\begin{gathered}\n2012^{c_{1}}-1997^{a_{1}}=5^{b} \\Rightarrow 2^{c_{1}}-2^{a_{1}} \\equiv 0(\\bmod 5) \\Rightarrow c_{1} \\equiv a_{1}(\\bmod 5) \\\\\n2012^{c_{1}}+1997^{a_{1}}=3^{b} \\Rightarrow 2^{c_{1}}+2^{a_{1}} \\equiv 0(\\bmod 5) \\Rightarrow c_{1} \\equiv a_{1}+1(\\bmod 5)\n\\end{gathered}\n$$\n\na contradiction.\n\nCase 2:\n\n$$\n\\begin{aligned}\n& 2012^{c_{1}}-1997^{a_{1}}=3^{b} \\\\\n& 2012^{c_{1}}+1997^{a_{1}}=5^{b}\n\\end{aligned}\n$$\n\nSince $b$ is an odd number we get $2012^{c_{1}}+1997^{a_{1}} \\equiv 5^{b}(\\bmod 3) \\Rightarrow 2^{c_{1}}+2^{a_{1}} \\equiv 5^{b} \\equiv 2(\\bmod 3)$ so $a_{1}, c_{1}$ are even numbers, say $a_{1}=2 a_{2}, c_{1}=2 c_{2}$. Then\n\n$$\n\\left(2012^{c_{2}}-1997^{a_{2}}\\right) \\cdot\\left(2012^{c_{2}}+1997^{a_{2}}\\right)=3^{b}\n$$\n\nBut $\\operatorname{gcd}\\left(2012^{c_{2}}-1997^{a_{2}}, 2012^{c_{2}}+1997^{a_{2}}\\right)=1$ and the above implies $2012^{c_{2}}-1997^{a_{2}}=1$. But then mod4, we get $0-1 \\equiv(\\bmod 4)$, a contradiction.\n\nTherefore there exists no solution for $c>1$.\n\nHence $a=b=c=1$ is the only solution.\n\n$$\n\\therefore\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nNT5.", "solution_match": "\nSolution."}}
+{"year": "2012", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $a, b, c$ be positive real numbers such that abc $=1$. Show that\n\n$$\n\\frac{1}{a^{3}+b c}+\\frac{1}{b^{3}+c a}+\\frac{1}{c^{3}+a b} \\leq \\frac{(a b+b c+c a)^{2}}{6}\n$$\n\nso", "solution": "By the AM-GM inequality we have $a^{3}+b c \\geq 2 \\sqrt{a^{3} b c}=2 \\sqrt{a^{2}(a b c)}=2 a$ and\n\n$$\n\\frac{1}{a^{3}+b c} \\leq \\frac{1}{2 a}\n$$\n\nSimilarly; $\\frac{1}{b^{3}+c a} \\leq \\frac{1}{2 b} \\cdot \\frac{1}{c^{3}+a b} \\leq \\frac{1}{2 c}$ and then\n\n$$\n\\frac{1}{a^{3}+b c}+\\frac{1}{b^{3}+c a}+\\frac{1}{c^{3}+a b} \\leq \\frac{1}{2 a}+\\frac{1}{2 b}+\\frac{1}{2 c}=\\frac{1}{2} \\frac{a b+b c+c a}{a b c} \\leq \\frac{(a b+b c+c a)^{2}}{6}\n$$\n\nTherefore it is enongil to prove $\\frac{(a h+b c+c a)^{2}}{6} \\leq \\frac{(a b+b c+c a)^{2}}{6}$. This mequalits is trivially shomn to be equivalent to $3 \\leq a b+b c+c a$ which is true because of the AM-GM inequalit: $3=\\sqrt[3]{(a b c)^{2}} \\leq a b+b c+c a$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nA2.", "solution_match": "\nSolution."}}
+{"year": "2012", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $a . b$ c ce positue real numbers such that $a+b+c=a^{2}+b^{2}+c^{2}$. Shou that\n\n$$\n\\frac{a^{2}}{a^{2}+a b}+\\frac{b^{2}}{b^{2}+b c}+\\frac{c^{2}}{c^{2}+c a} \\geq \\frac{a+b+c}{2}\n$$", "solution": "By the Cauchy-Schwarz inequality it is\n\n$$\n\\begin{aligned}\n& \\left(\\frac{a^{2}}{a^{2}+a b}+\\frac{b^{2}}{b^{2}+b c}+\\frac{c^{2}}{c^{2}+c a}\\right)\\left(\\left(a^{2}+a b\\right)+\\left(b^{2}+b c\\right)+\\left(c^{2}+c a\\right)\\right) \\geq(a+b+c)^{2} \\\\\n\\Rightarrow & \\frac{a^{2}}{a^{2}+a b}+\\frac{b^{2}}{b^{2}+b c}+\\frac{c^{2}}{c^{2}+c a} \\geq \\frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}+a b+b c+c a}\n\\end{aligned}\n$$\n\nSo in is enough to prove $\\frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}+a b+b c+c a} \\geq \\frac{a+b+c}{2}$, that is to prove\n\n$$\n2(a+b+c) \\geq a^{2}+b^{2}+c^{2}+a b+b c+c a\n$$\n\nSubstituting $a^{2}+b^{2}+c^{2}$ for $a+b+c$ into the left hand side we wish equivalently to prove\n\n$$\na^{2}+b^{2}+c^{2} \\geq a b+b c+c a\n$$\n\nBut the $a^{2}+b^{2} \\geq 2 a b, b^{2}+c^{2} \\geq 2 b c, c^{2}+a^{2} \\geq 2 c a$ which by addition imply the desired inequality.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nA3.", "solution_match": "\nSolution."}}
+{"year": "2012", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Solve the following equation for $x, y, z \\in \\mathbb{N}$\n\n$$\n\\left(1+\\frac{x}{y+z}\\right)^{2}+\\left(1+\\frac{y}{z+x}\\right)^{2}+\\left(1+\\frac{z}{x+y}\\right)^{2}=\\frac{27}{4}\n$$", "solution": "Call $a=1+\\frac{x}{y+z}, b=1+\\frac{y}{z+x}, c=1+\\frac{z}{x+y}$ to get\n\n$$\na^{2}+b^{2}+c^{2}=\\frac{27}{4}\n$$\n\nSince it is also true that\n\n$$\n\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=2\n$$\n\nthe quadratic-harmonic means inequality implies\n\n$$\n\\frac{3}{2}=\\sqrt{\\frac{a^{2}+b^{2}+c^{2}}{3}} \\geq \\frac{3}{\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}}=\\frac{3}{2}\n$$\n\nSo the inequality in the middle holds as an equality, and this happens whenever $a=b=c$, from which $1+\\frac{x}{y+z}=1+\\frac{y}{z+x}=1+\\frac{z}{x+y}$.\n\nBut $1+\\frac{x}{y+z}=1+\\frac{y}{z+x} \\Leftrightarrow x^{2}+x z=y^{2}+y z \\Leftrightarrow(x-y)(x+y)=z(y-\\dot{x})$ and the two sides of this equality will be of different sign, unless $x=y$ in which case both sides become 0 . So $x=y$, and similarly $y=z$, thus $x=y=z$.\n\nIndeed, any triad of equal natural numbers $x=y=z$ is a solution for the given equation, and so these are all its solutions.\n\nSolution 2. The given equation is equivalent to\n\n$$\n\\frac{27}{4}=(x+y+z)^{2}\\left(\\frac{1}{(y+z)^{2}}+\\frac{1}{(z+x)^{2}}+\\frac{1}{(x+y)^{2}}\\right)\n$$\n\nNow observe that by the well known inequality $a^{2}+b^{2}+c^{2} \\geq a b+b c+c a$, with $\\frac{1}{y+z}, \\frac{1}{z+x}$, $\\frac{1}{x+y}$ in place of $a, b, c$; we get:\n\n$$\n\\begin{aligned}\n\\frac{27}{4} & =(x+y+z)^{2}\\left(\\frac{1}{(y+z)^{2}}+\\frac{1}{(z+x)^{2}}+\\frac{1}{(x+y)^{2}}\\right) \\\\\n& \\geq(x+y+z)^{2}\\left(\\frac{1}{(y+z)(z+x)}+\\frac{1}{(z+x)(x+y)}+\\frac{1}{(x+y)(y+z)}\\right)=\\frac{2(x+y+z)^{3}}{(x+y)(y+z)(z+x)} \\\\\n& =\\frac{(2(x+y+z))^{3}}{4(x+y)(y+z)(z+x)}=\\frac{((x+y)+(y+z)+(z+x)))^{3}}{4(x+y)(y+z)(z+x)} \\stackrel{\\text { AM-GM }}{\\geq \\frac{(3 \\sqrt[3]{(x+y)(y+z)(z+x)})^{3}}{4(x+y)(y+z)(z+x)} .} \\\\\n& =\\frac{27}{4}\n\\end{aligned}\n$$\n\nThis means all inequalities in the above calculations are equalities, and this holds exactly whenever $x+y=y+z=z+x$, that is $x=y=z$. By the statement's demand we need to have $a, b, c$ integers. And conversely, any triad of equal natural numbers $x=y=z$ is indeed a solution for the given equation, and so these are all its solutions.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nA4.", "solution_match": "\nSolution 1."}}
+{"year": "2012", "tier": "T3", "problem_label": "A5", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Find the largest positive integer $n$ for which the inequality\n\n$$\n\\frac{a+b+c}{a b c+1}+\\sqrt[n]{a b c} \\leq \\frac{5}{2}\n$$\n\nholds for all $a, b, c \\in[0,1]$. Here $\\sqrt[1]{a b c}=a b c$.", "solution": "Let $n_{\\max }$ be the sought largest value of $n$, and let $E_{a, b, c}(n)=\\frac{a+b+c}{a b c+1}+\\sqrt[n]{a b c}$. Then $E_{a, b, c}(m)-E_{a, b, c}(n)=\\sqrt[m]{a b c}-\\sqrt[n]{a b c}$ and since $a . b c \\leq 1$ we clearly have $E_{a, b, c}(m) \\geq$ $E_{a, b, c}(n)$ for $m \\geq n$. So if $E_{a, b, c}(n) \\geq \\frac{5}{2}$ for some choice of $a, b, c \\in[0,1]$, it must be $n_{\\max } \\leq n$. We use this remark to determine the upper bound $n_{\\max } \\leq 3$ by plugging some particular values of $a, b, c$ into the given inequality as follow's:\n\n$$\n\\text { For }(a, b, c)=(1,1, c), c \\in[0,1] \\text {, inequality (1) implies } \\frac{c+2}{c+1}+\\sqrt[n]{c} \\leq \\frac{5}{2} \\Leftrightarrow \\frac{1}{c+1}+\\sqrt[n]{c} \\leq\n$$\n\n$\\frac{3}{2}$. Obviously, every $x \\in[0 ; 1]$ is written as $\\sqrt[n]{c}$ for some $c \\in[0 ; 1]$. So the last inequality is equivalent to:\n\n$$\n\\begin{aligned}\n& \\frac{1}{x^{n}+1}+x \\leq \\frac{3}{2} \\Leftrightarrow 2+2 x^{n+1}+2 x \\leq 3 x^{n}+3 \\Leftrightarrow 3 x^{n}+1 \\geq 2 x^{n+1}+2 x \\\\\n\\Leftrightarrow & 2 x^{n}(1-x)+(1-x)+(x-1)\\left(x^{n-1}+\\cdots+x\\right) \\geq 0 \\\\\n\\Leftrightarrow & (1-x)\\left[2 x^{n}+1-\\left(x^{n-1}+x^{n-2}+\\ldots+x\\right)\\right] \\geq 0, \\forall x \\in[0,1]\n\\end{aligned}\n$$\n\nFor $n=4$, the left hand side of the above becomes $(1-x)\\left(2 x^{4}+1-x^{3}-x^{2}-x\\right)=$ $(1-x)(x-1)\\left(2 x^{3}+x^{2}-1\\right)=-(1-x)^{2}\\left(2 x^{3}+x^{2}-1\\right)$ which for $x=0.9$ is negative. Thus. $n_{\\max } \\leq 3$ as claimed.\n\nNow, we shall prove that for $n=3$ inequality (1) holds for all $a, b, c \\in[0,1]$, and this would mean $n_{\\max }=3$. We shall use the following Lemma:\n\nLemma. For all $a, b, c \\in[0 ; 1]: a+b+c \\leq a b c+2$.\nProof of the Lemma: The required result comes by adding the following two inequalities side by side\n\n$$\n\\begin{aligned}\n& 0 \\leq(a-1)(b-1) \\Leftrightarrow a+b \\leq a b+1 \\Leftrightarrow a+b-a b \\leq 1 \\\\\n& 0 \\leq(a b-1)(c-1) \\Leftrightarrow a b+c \\leq a b c+1\n\\end{aligned}\n$$\n\nBecause of the Lemma, our inequality (1) for $n=3$ wrill be proved if the following weaker inequality is proved for all $a, b, c \\in[0,1]$ :\n\n$$\n\\frac{a b c+2}{a b c+1}+\\sqrt[3]{a b c} \\leq \\frac{5}{2} \\Leftrightarrow \\frac{1}{a b c+1}+\\sqrt[3]{a b c} \\leq \\frac{3}{2}\n$$\n\nDenoting $\\sqrt[3]{a b c}=y \\in[0 ; 1]$, this inequality becomes:\n\n$$\n\\begin{aligned}\n& \\frac{1}{y^{3}+1}+y \\leq \\frac{3}{2} \\Leftrightarrow 2+2 y^{4}+2 y \\leq 3 y^{3}+3 \\Leftrightarrow-2 y^{4}+3 y^{3}-2 y+1 \\geq 0 \\\\\n\\Leftrightarrow & 2 y^{3}(1-y)+(y-1) y(y+1)+(1-y) \\geq 0 \\Leftrightarrow(1-y)\\left(2 y^{3}+1-y^{2}-y\\right) \\geq 0\n\\end{aligned}\n$$\n\nThe last inequality is obvious because $1-y \\geq 0$ and $2 y^{3}+1-y^{2}-y=y^{3}+(y-1)^{2}(y+1) \\geq 0$.\n\n## Geometry\n\n2", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nA5.", "solution_match": "\nSolution."}}
+{"year": "2012", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C$ be an equilateral triangle, and $P$ a point on the circumcircle of the triangle $A B C$ and distinct from $A, B$ and $C$. If the lines through $P$ and parallel to $B C, C A, A B$ intersect the lines $C A, A B, B C$ at $M, N$ and $Q$ respectively, prove that $M, N$ and $Q$ are collinear.", "solution": "Solution 1. Let lines $K H, A B$ intersect at $M$ (Figure 5a). From the quadrelateral $K M A C$ we have\n\n$\\angle K M A=360^{\\circ}-\\angle A-\\angle A C K-\\angle C K M=360^{\\circ}-\\angle A-90^{\\circ}-\\left(180^{\\circ}-2 \\angle K C H\\right)=$ $90-\\angle A+2 \\angle K C H=90-\\angle A+2\\left(90^{\\circ}-\\angle A C B\\right)=270^{\\circ}-\\angle A-2 \\angle A C B=270-\\angle A-$ $\\angle A C B-\\angle A B C=270^{\\circ}-180^{\\circ}=90^{\\circ}$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_61d3145f18c90bf9f370g-06.jpg?height=519&width=574&top_left_y=320&top_left_x=455)\n\n(a)\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_61d3145f18c90bf9f370g-06.jpg?height=634&width=571&top_left_y=317&top_left_x=1093)\n\n(b)\n\nFigure 2: Exercise G2.\n\nso $K H \\perp A B$ as wanted.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nG1.", "solution_match": "\nSolution."}}
+{"year": "2012", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C$ be an equilateral triangle, and $P$ a point on the circumcircle of the triangle $A B C$ and distinct from $A, B$ and $C$. If the lines through $P$ and parallel to $B C, C A, A B$ intersect the lines $C A, A B, B C$ at $M, N$ and $Q$ respectively, prove that $M, N$ and $Q$ are collinear.", "solution": "Solution 2. Let $D$ be a point on $c$ such that $A D<A C$, and let $E, Z$ be the second points of intersection of lines $A D$ and $B D$ with $c$ respectively. Let also $N$ be the second point of intersection of line $B E$ with the circle $c$. Figure $5 \\mathrm{~b}$ shows $Z$ between $B, D$. The argument below can be trivially modified to apply in case $D$ is in the segment $B, Z$ as well. It is\n\n$$\nA B^{2}=A C^{2}=A D \\cdot A E \\Rightarrow \\frac{A B}{A E}=\\frac{A D}{A B}\n$$\n\nThis relation arid the fact that $\\angle B A E=\\angle B A D$ implies that the triangles $A B E, A D B$ are similar. Thus $\\angle A B E=\\angle A D B$. Also from the cyclic quadrilateral we get $\\angle A D B=\\angle Z N E$. Therefore $\\angle A B E=\\angle Z N E$, so $A B \\| N Z$.\n\nCall $P$ the intersection point of $B C, N Z$. Since $A C$ is tangent to $c$ it is\n\n$$\n\\angle C E H=\\angle B C A\n$$\n\nand then\n\n$$\n\\begin{aligned}\n& \\angle Z N H+\\angle C N Z=\\angle H N C=\\angle C E H \\stackrel{(3)}{=} \\angle B C A=\\angle A B C=\\angle Z P C=\\angle B C N+\\angle C N Z \\\\\n\\Rightarrow \\quad & \\angle Z N H=\\angle B C N \\\\\n\\Rightarrow \\quad & \\angle Z N H=\\angle H Z N\n\\end{aligned}\n$$\n\nTherefore $H$ is the midpoint of the arc $N Z$, so $K H \\perp N Z$ and as $A B \\| N Z$ we finally get $K H \\perp A B$ as wanted.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_61d3145f18c90bf9f370g-07.jpg?height=494&width=1167&top_left_y=290&top_left_x=501)\n\nFigure 3: Exercise G3.\n\n2-3 G3. Let $A B$ and $C D$ be chords in a circle of center $O$ with $A, B, C, D$ distinct, and let the lines $A B$ and $C D$ meet at a right angle at point $E$. Let also $M$ and $N$ be the midpoints of $A C$ and $B D$ respectively. If $M N \\perp O E$, prove that $A D \\| B C$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nG1.", "solution_match": "\nSolution."}}
+{"year": "2012", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C$ be an equilateral triangle, and $P$ a point on the circumcircle of the triangle $A B C$ and distinct from $A, B$ and $C$. If the lines through $P$ and parallel to $B C, C A, A B$ intersect the lines $C A, A B, B C$ at $M, N$ and $Q$ respectively, prove that $M, N$ and $Q$ are collinear.", "solution": "Solution. $E$ can be inside, or outside the circle (Figure 3) but the proof below holds in both cases; notice that $E$ cannot be on the circle as $A, B, C, D$ are distinct. Let lines $A C$ and $N E$ meet at point $P$. Then $E N=D N=B N$ (median in a right triangle), so $\\angle P E C=\\angle N E D=\\angle N D E=\\angle B D C=\\angle B A C=\\angle E A P$. Now $A B \\perp C D$ so $E N \\perp A C$. But $O M \\perp A C$ so $O M \\| E N$. Similarly $O N \\| E M$ so $N E M O$ is a parallelogram (possibly degenerated). As $M N \\perp O E$, this parallelogram is a rhombus. Then the chords $A C$ and $B D$, being equidistant from $O$, are equal. Hence their minor arcs are equal, which means that either $A D \\| B C$ or $A B \\| C D$; the latter contradicts the fact that $A B$ and $C D$ meet at $E$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nG1.", "solution_match": "\nSolution."}}
+{"year": "2012", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C$ be an acute-angled triangle with circumcircle $\\Gamma$, and let $O, H$ be the triangle's circumcenter and orthocenter respectively. Let also $A^{\\prime}$ be the point where the angle bisector of angle $B A C$ meets $\\Gamma$. If $A^{\\prime} H=A H$, find the measure of angle $B A C$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_61d3145f18c90bf9f370g-07.jpg?height=519&width=1042&top_left_y=1756&top_left_x=541)\n\nFigure 4: Exercise G4.", "solution": "Solution. Let $P$ be the symmetric of $A$ with respect to $M$ (Figure 5). Then $A M=M P$ and $t \\perp A P$, hence the triangle $A P N$ is isosceles with $\\mathrm{AP}$ as its base, so $\\angle N A P=\\angle N P A$. We have $\\angle B A P=\\angle B A M=\\angle B M N$ and $\\angle B A N=\\angle B N M$. Thus\n\n$$\n180^{\\circ}-\\angle N B M=\\angle B N M+B M N=\\angle B A N+\\angle B A P=\\angle N A P=\\angle N P A\n$$\n\nso the quadrangle $M B N P$ is cyclic (since the points $B$ and $P$ lie on different sides of $M N$ ). Hence $\\angle A P B=\\angle M P B=\\angle M N B$ and the triangles $A P B$ and $M N B$ are congruent $(M N=2 A M=A M+M P=A P)$. From that we get $A B=M B$, i.e. the triangle $A M B$ is\nisosceles, and since $t$ is tangent to $k_{1}$ and perpendicular to $A M$, the center of $k_{1}$ is on $A M$, hence $A M B$ is a right-angled triangle. From the last two statements we infer $\\angle A M B=45^{\\circ}$, and so $\\angle N M B=90^{\\circ}-\\angle A M B=45^{\\circ}$.\n\n4. G6. Let $O_{1}$ be a point in the exterior of the circle $c(O, R)$ and let $\\dot{O}_{1} N, O_{1} D$ be the tangent segments from $O_{1}$ to the circle. On the segment $O_{1} N$ consider the point $B$ such that $B N=R$. Let the line from $B$ parallel to $O N$ intersect the segment $O_{1} D$ at $C$. If $A$ is a point on the segment $O_{1} D$ other than $C$ so that $B C=B A=a$, and if $\\dot{C}^{\\prime}(K, r)$ is the incircle of the triangle $O_{1} A B$; find the area of $A B C$ in terms of a, $R, r$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nG4.", "solution_match": "\nSolution."}}
+{"year": "2012", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C$ be an acute-angled triangle with circumcircle $\\Gamma$, and let $O, H$ be the triangle's circumcenter and orthocenter respectively. Let also $A^{\\prime}$ be the point where the angle bisector of angle $B A C$ meets $\\Gamma$. If $A^{\\prime} H=A H$, find the measure of angle $B A C$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_61d3145f18c90bf9f370g-07.jpg?height=519&width=1042&top_left_y=1756&top_left_x=541)\n\nFigure 4: Exercise G4.", "solution": "Solution. Obviously; the segment $B C$ is taingent to the circle $c$. Let $M$ be the point of tangency (Figure 6). Call $Q, M$ the tangency points of $B A, B C$ with $c^{\\prime}$ and $c$ respectively, and call $H$ the midpoint of segment $A C$. It is well known that\n\n$$\nA Q=\\frac{1}{2}\\left(A O_{1}+A B-B O_{1}\\right) \\text { and } C M=\\frac{1}{2}\\left(B O_{1}+B C-C O_{1}\\right)\n$$\n\nand so\n\n$$\nA Q+C M=\\frac{1}{2}(2 B C-A C)=a-\\frac{1}{2} A C=a-H C\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_61d3145f18c90bf9f370g-09.jpg?height=509&width=1019&top_left_y=1247&top_left_x=564)\n\nFigure 6: Exercise G6.\n\nThe triangles $K A Q$ and $O C M$ are similar and this implies\n\n$$\n\\begin{aligned}\n\\frac{K Q}{O M} & =\\frac{A Q}{C M} \\Leftrightarrow \\frac{K Q}{A Q}=\\frac{O M}{C M} \\Leftrightarrow \\frac{r}{A Q}=\\frac{R}{C M}=\\frac{R+r}{A Q+C M}=\\frac{R+r}{a-H C} \\Leftrightarrow \\\\\n\\frac{r}{A Q} & =\\frac{R+r}{a-H A}\n\\end{aligned}\n$$\n\nIf $A Z$ is the bisector segment of triangle $B A H$ it holds\n\n$$\n\\angle A Z H=90^{\\circ}=\\frac{1}{2} \\angle B A C \\text { and } \\angle K A Q=\\frac{1}{2}\\left(180^{\\circ}-\\angle B A C\\right)=90^{\\circ}-\\frac{1}{2} \\angle B A C\n$$\n\nTherefore, from the similar triangles $K Q A$ and $A H Z$ we get\n\n$$\n\\frac{-A H}{Z H}=\\frac{r}{A Q} \\stackrel{(6)}{=} \\frac{R+r}{a-H A}\n$$\n\nAlso, from the bisector-theorem in triangle $A B H$ it holds\n\n$$\nZ H=\\frac{A H \\cdot B H}{a+A H}\n$$\n\nand from (7) it follows\n\n$$\n\\frac{R+r}{a-H A}=\\frac{A H}{\\frac{A H \\cdot B H}{a+A H}} \\Rightarrow R+r=\\frac{a^{2}-A H^{2}}{B H}=\\frac{B H^{2}}{B H}=B H\n$$\n\nSo\n\n$$\nH A^{2}=a^{2}-(R+r)^{2} \\Leftrightarrow H A=\\sqrt{a^{2}-(R+r)^{2}}\n$$\n\nand finally the area of triangle $A B C$ in terms of $a, R, r$ is:\n\n$$\n(A B C)=A H \\cdot B H=(R+r) \\sqrt{a^{2}-(R+r)^{2}}\n$$\n\n4 GZ, Let $M N P Q$ be a square of side length 1 , and $A, B, C, D$ points on the sides $M N$, $N P, P Q$, and $Q M$ respectively such that $A C \\cdot B D=\\frac{5}{4}$. Can the set $\\{A B, B C, C D, D A\\}$ be partitioned into two subsets $S_{1}$ and $S_{2}$ of two elements each, so that.each one of the sums of the elements of $S_{1}$ and $S_{2}$ are positive integers?", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nG4.", "solution_match": "\nSolution."}}
+{"year": "2012", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C$ be an acute-angled triangle with circumcircle $\\Gamma$, and let $O, H$ be the triangle's circumcenter and orthocenter respectively. Let also $A^{\\prime}$ be the point where the angle bisector of angle $B A C$ meets $\\Gamma$. If $A^{\\prime} H=A H$, find the measure of angle $B A C$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_61d3145f18c90bf9f370g-07.jpg?height=519&width=1042&top_left_y=1756&top_left_x=541)\n\nFigure 4: Exercise G4.", "solution": "Solution. The answer is negative.\n\nSuppose such a partitioning was possible (Figure 7). Then $A B+B C+C D+D A \\in \\mathbb{N}$. But $(A B+B C)+(C D+D A)>A C+A C \\geq 2$, hence $A B+B C+C D+D A>2$.\n\nOn the other hand, $A B+B C+C D+D A<(A N+N B)+(B P+P C)+(C Q+Q D)+$ $(D M+M A)=4$, hence $A B+B C+C D+D A=3$.\n\nObviously one of the aums of the elerients of $S_{1}$ and $S_{2}$ must be 1 and the other 2 . Without any loss of generality, we may assume that the sum of the elements of $S_{1}$ is 1 and the sum of the elements of $S_{2}$ is 2 . As $A B+B C>A C \\geq 1$ we find that $S_{1} \\neq\\{A B, B C\\}$. Similarly, $S_{1}$ cannot contain two adjacent sides of the quadrilateral $A B C D$. Therefore, without any loss of generality, we may assume that $S_{1}=\\{A D, B C\\}$ and $S_{2}=\\{A B, C D\\}$. Then $A D+B C=1$ and $A B+C D=2$.\n\nWe have $A D \\cdot B C \\leq \\frac{1}{4} \\cdot(A D+C B)^{2}=\\frac{1}{4}$ and $A B \\cdot C D \\leq \\frac{1}{4} \\cdot(A B+C D)^{2}=1$.\n\nAccording to Ptolemy's inequality, we have\n\n$$\n\\frac{5}{4}=A C \\cdot B D \\leq A B \\cdot C D+A D \\cdot B C=\\frac{1}{4}+1=\\frac{5}{4}\n$$\n\nhence we have equality all around, which means the quadrilateral $A B C D$ is cyclic, $A D=$ $B C=\\frac{1}{2}$ and $A B=C D=1$, hence $A B C D$ is a rectangle of dimensions 1 and $\\frac{1}{2}$.\n\nThere are many different ways of proving that this configuration is not possible. For example: - Suppose $A B C D$ is a rectangle with $A D=\\frac{1}{2}, A B=1$. Then we have $A C=B D=\\frac{\\sqrt{5}}{2}$ and $\\triangle A N B \\equiv \\triangle C Q D$ (Angle-Side-Angle). Denoting $A M=x, M D=y$ we have $A N=$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_61d3145f18c90bf9f370g-11.jpg?height=448&width=1127&top_left_y=318&top_left_x=561)\n\nFigure 7: Exercise G7.\n\n$1-x, B N=1-y$ and the following conditions need to be fulfilled for some $x, y \\in[0 ; 1]$ (Pythagorean Theorem in triangles $A M D, A N B, B B^{\\prime} C$, where $B^{\\prime}$ is the projection of $B$ on $M Q)$ :\n\n$$\nx^{2}+y^{2}=\\frac{1}{4}, \\quad(1-x)^{2}+(1-y)^{2}=1 \\text { and } 1+(2 y-1)^{2}=\\frac{5}{4}\n$$\n\nBut $1+(2 y-1)^{2}=\\frac{5}{4}$ implies $y \\in\\left\\{\\frac{1}{4}, \\frac{3}{4}\\right\\}$. If $y=\\frac{3}{4}$, then $x^{2}+y^{2}=\\frac{1}{4}$ cannot hold. If on the other hand $y=\\frac{1}{4}$, then $(1-x)^{2}+(1-y)^{2}=1$ implies $x=0$, but then $(1-x)^{2}+(1-y)^{2}=1$ cannot hold. Therefore such a configuration is not possible.\n\n## Combinatorics", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nG4.", "solution_match": "\nSolution."}}
+{"year": "2012", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Along a round table are arranged 11 cards with the names (all distinct) of the 11 members of the $16^{\\text {th }}$ JBMO Problem Selection Committee. The distances between each two consecutive cards are equal. Assume that in the first meeting of the Committee none of its 11 members sits in front of the card with his name. Is it possible to rotate the table by some angle so that at the end at least two members of sit in front of the card with their names?", "solution": "Yes it is: Rotating the table by the angles $\\frac{360^{\\circ}}{11}, 2 \\cdot \\frac{360^{\\circ}}{11}, 3 \\cdot \\frac{360^{\\circ}}{11}, \\ldots, 10 \\cdot \\frac{360^{\\circ}}{11}$, we obtain 10 new positions of the table. By the assumption, it is obvious that every one of the 11 members of the Committee will be seated in front of the card with his name in exactly one of these 10 positions. Then by the Pigeonhole Principle there should exist one among these 10 positions in which at least two of the $11(>10)$ members of the Committee will be placed in their positions, as claimed.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nC1.", "solution_match": "\nSolution."}}
+{"year": "2012", "tier": "T3", "problem_label": "C2", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "$n$ nails nailed on a board are connected by two via a string. Each string is colored in one of $n$ given colors. For any three colors there exist three nails conne.cted by two with strings in these three colors. Can $n$ be: (a) 6, (b) 7?", "solution": "(a) The answer is no:\n\nSuppose it is possible. Consider some color, say blue. Each blue string is the side of 4 triangles formed with vertices on the given points. As there exist $\\binom{5}{2}=\\frac{5 \\cdot 4}{2}=10$ pairs of colors other than blue, and for any such pair of colors together with the blue color there exists a triangle with strings in these colors, we conclude that there exist at least 3 blue strings (otherwise the number of triangles with a blue string as a side would be at most $2 \\cdot 4=8$, a contradiction). The same is true for any color, so altogether there exist at least $6 \\cdot 3=18$ strings, while we have just $\\binom{6}{2}=\\frac{6 \\cdot 5}{2}=15$ of them.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_61d3145f18c90bf9f370g-12.jpg?height=519&width=488&top_left_y=1850&top_left_x=835)\n\nFigure 8: Exercise C2.\n\n(b) The answer is yes (Figure 8):\n\nPut the nails at the vertices of a regular 7-gon (Figure 8) and color each one of its sides in a different color. Now color each diagonal in the color of the unique side parallel to it. It can be checked directly that each triple of colors appears in some triangle (because of symmetry, it is enough to check only the triples containing the first color).\n\nRemark. The argument in (a) can be applied to any even $n$. The argument. in (b) can be applied to any odd $n=2 k+1$ as follows: first number the nails as $0,1,2 \\ldots, 2 k$ and similarly number the colors as $0,1,2 \\ldots, 2 k$. Then connect nail $x$ with nail $y$ by a string of color $x+y(\\bmod n)$. For each triple of colors $(p, q, r)$ there are vertices $x, y, z$ connected by these three colors. Indeed, we need to solve (modn.) the system\n\n$$\n(*)(x+y \\equiv p, x+z \\equiv q, y+z \\equiv r)\n$$\n\nAdding all three, we get $2(x+y+z) \\equiv p+q+r$ and multiplying by $k+1$ we get $x+y+z \\equiv$ $(k+1)(p+q+r)$. We can now find $x, y, z$ from the identities $(*)$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nC2.", "solution_match": "\nSolution."}}
+{"year": "2012", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "In a circle of diameter 1 consider 65 points no three of which are collinear. Prove that there exist 3 among these points which form a triangle with area less then or equal to $\\frac{1}{72}$.", "solution": "Solution. It is $s=(a+b)^{3}-63 a b(a+b)$ which gives the same residue module 7 as $(a+b)^{3}$. But the residues modulo 7 of perfect cubes can only be 0,1 or 6 . So the residue of $s$ modulo 7 is 0,1 or 6 . Now for $a=6, b=-1$ we get $s=2015 \\geq 2012$ and this is the least possible value of $s$ because the numbers 2012, 2013, 2014 give 3,4 and 5 as residues $\\bmod 7$ which are distinct from $0,1,6$, and so 2012, 2013, 2014 cannot be $s$ for any choice of $a, b$. $\\square$\n\n2-3 NT2. Do there exist prime numbers $p$ and $q$ such that $p^{2}\\left(p^{3}-1\\right)=q(q+1)$ ?", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nC3.", "solution_match": "\nSolution."}}
+{"year": "2012", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "In a circle of diameter 1 consider 65 points no three of which are collinear. Prove that there exist 3 among these points which form a triangle with area less then or equal to $\\frac{1}{72}$.", "solution": "Solution. Write the given equation in the form\n\n$$\np^{2}(p-1)\\left(p^{2}+p+1\\right)=q(q+1)\n$$\n\nFirst observe that it must not be $p=q$, since in this case the left hand side of (9) is greater than its right hand side. Hence, since $p$ and $q$ are distinct prines, (9) immediately yields $p^{2} \\mid q+1$, that is\n\n$$\nq=a p^{2}-1\n$$\n\nfor some $a \\in \\mathrm{N}$. Since $p$ and $q$ are both primes, by (9) we get the following cases:\n\nCase 1: $q \\mid p-1$, that is\n\n$$\np=b q+1\n$$\n\nfor some $b \\in N$. Substituting (11) into (10), and using the fact that $a \\geq 1$ and $b \\geq 1$, we obtain\n\n$$\nq=a(b q+1)^{2}-1 \\geq(q+1)^{2}-1=q^{2}+2 q\n$$\n\na contradiction.\n\nCase 2: $q \\mid p^{2}+p+1$, that is\n\n$$\np^{2}+p+1=b q\n$$\n\nfor some $b \\in$ N. Substituting (10) into (12), we get\n\n$$\np^{2}+p+1=b\\left(a p^{2}-1\\right)\n$$\n\nIf $a \\geq 2$, then from (13) it follows that\n\n$$\np^{2}+p+1 \\geq 2 p^{2}-1\n$$\n\nor equivalently, $p+1 \\geq(p-1)(p+1)$, that is, $(p+1)(2-p) \\geq 0$. This implies that $p=2$, and so $q \\mid 2^{2}+2+1=7$. Hence, $q=7$, but the pair $p=2$ and $q=7$ does not satisfy the equation ( 9 ).\n\nHence, it must be $a=1$. Then if $b \\geq 3$, (13) implies\n\n$$\np^{2}+p+1 \\geq 3\\left(p^{2}-1\\right)\n$$\n\nor equivalently, $4 \\geq p(2 p-1)$, which is obviously impossible.\n\nThus, it must be $a=1$ and $b \\in\\{1,2\\}$. For $a=b=1$, (13) implies that $p=2$, which by (12) again yields $q=7$, which is impossible. Finally, for $a=1$ and $b=2$, (13) gives $p(p-1)=3$, which is clearly not satisfied for any prime $p$.\n\nHence, there do not exist prime numbers $p$ and $q$ which satisfy given equation.\n\n## 3 NT3. Decipher the equality\n\n$$\n(\\overline{V E R}-\\overline{I A}):(\\overline{G R E}+\\overline{E C E})=G^{R^{E}}\n$$\n\nassuming that the number $\\overline{\\text { GREECE }}$ has a maximum value. It is supposed that each letter corresponds to a unique digit from 0 to 9 and different letters correspond to different digits, and also that all letters $G, E, V$ and $I$ are different from 0 . Also, the notation $\\overline{a_{n} \\ldots a_{1} a_{0}}$ stands for the number $a_{n} \\cdot 10^{n}+\\cdots+10^{1} \\cdot a_{1}+a_{0}$ :", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nC3.", "solution_match": "\nSolution."}}
+{"year": "2012", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "In a circle of diameter 1 consider 65 points no three of which are collinear. Prove that there exist 3 among these points which form a triangle with area less then or equal to $\\frac{1}{72}$.", "solution": "Solution. Denote\n\n$$\nx=\\overline{V E R}-\\overline{I A}, y=\\overline{G R E}+\\overline{E C E}, z=G^{R^{E}}\n$$\n\nThen obviously, we have\n\n$$\n\\begin{aligned}\n& (201+131 \\text { or } 231+101) \\leq y \\leq(879+969 \\text { or } 869+979 \\text { or. } 769+989) \\\\\n\\Rightarrow \\quad & 332 \\leq y \\leq 1848 \\Rightarrow 102-98 \\leq x \\leq 987-10 \\Rightarrow 4 \\leq x \\leq 977,\n\\end{aligned}\n$$\n\nhence it follows that\n\n$$\n\\frac{4}{1848} \\leq \\frac{x}{y}=z \\leq \\frac{977}{332} \\Rightarrow 1 \\leq z \\leq 2\n$$\n\nThis shows that $z=G^{R^{E}} \\in\\{1,2\\}$. Hence, if $R \\geq 1$, then $R^{E} \\geq 1$, which implies that $2 \\geq G^{R^{E}} \\geq G$. Thus, if $R \\geq 1$, then it must be $G \\leq 2$. In view of this and the assumption of the problem that the number $\\overline{G R E E C E}$ has a maximum value, we will consider the case when $R=0$ hoping to get a solution with $G>2$. Then $G^{R^{E}}=G^{0}=1$ for all digits $G$ and $E$ with $1 \\leq G, E \\leq 9$, and therefore, the above equality becomes\n\n$$\n\\overline{V E R}-\\overline{I A}=\\overline{G R E}+\\overline{E C E}\n$$\n\nwhich substituting $R=0$, can be written as\n\n$$\n\\overline{V E O}=\\overline{G O E}+\\overline{E C E}+\\overline{I A}\n$$\n\nNow we consider the following cases:\n\nCase 1: $G=9$. Then $V \\leq 8$, so $\\overline{V E 0} \\leq 900$, while the right hand side of (1) is greater than 900 . This is impossible, and no solution exists in this case.\n\nCase 2: $G=8$. Then (14) becomes\n\n$$\n\\overline{V E 0}=\\overline{80 E}+\\overline{E C E}+\\overline{I A}\n$$\n\nhence it immediately follows that $V=9$. For $V=9$, (15) becomes\n\n$$\n\\overline{9 E 0}=\\overline{80 E}+\\overline{E C E}+\\overline{I A}\n$$\n\nNotice that for $E \\geq 2$, the right hand side of (16) is greater than 1000 , while the left hand side of (16) is less than 1000 . Therefore, it must be $E \\leq 1$, that is, $E=1$ in view of the fact that $R=0$. Substituting $E=1$ into (16), we get\n\n$$\n\\overline{910}=\\overline{801}+\\overline{1 C 1}+\\overline{I A}\n$$\n\nhence it follows that\n\n$$\n109=\\overline{1 C 1}+\\overline{I A}\n$$\n\nBut the right hand side of (18) is greater than 121. This shows that $G=8$ does not lead to any solution.\n\nCase 3: $G=7$. Then (14) becomes\n\n$$\n\\overline{V E O}=\\overline{70 E}+\\overline{E C E}+\\overline{I A}\n$$\n\nThus it must be $V \\geq 8$.\n\nSubcase 3(a): $V=8$. Then (19) gives\n\n$$\n\\overline{\\delta E 0}=\\overline{70 E}+\\overline{E C E}+\\overline{I A}\n$$\n\nhence we immediately obtain $E=1$ (since the right hand side of (6) must be lass than 900). For $E=1,(20)$ reduces to\n\n$$\n109=\\overline{1 C 1}+\\overline{I A}\n$$\n\nwhich is impossible since $\\overline{1 C 1} \\geq 121$.\n\nSubcase 3(b): $V=9$. Then (19) gives\n\n$$\n\\overline{9 E 0}=\\overline{70 E}+\\overline{E C E}+\\overline{I A}\n$$\n\nhence we immediately obtain $E \\leq 2$ (since the right hand side of (7) must be less than 1000). For $E=2$, (22) reduces to\n\n$$\n218=\\overline{2 C 2}+\\overline{I A}\n$$\n\nwhich is impossible since $\\overline{2 C 2} \\geq 232$. Finally, for $E=1$, (22) reduces to\n\n$$\n209=\\overline{1 C 1}+\\overline{I A}\n$$\n\nSince it is required that the number $\\overline{G R E E C E}$ has a maximum value, taking $C=8$ into (24) we find that\n\n$$\n28=\\overline{I A}\n$$\n\nwhich yields $8=A=C$. This is impossible since must be $A \\neq C$. Since $C \\neq G=7$, then taking $C=6$ into (24) we obtain\n\n$$\n48=\\overline{I A}\n$$\n\nhence we have $I=4$ and $A=8$. Previously, we have obtain $G=7, R=0, V=9, E=1$ and $C=6$. For these values, we obtain that $\\overline{G R E E C E}=701161$ is the desired maximum value.\n\n4 NT4. Determine all triples $(m, n, p)$ satisfying\n\n$$\nn^{2 p}=m^{2}+n^{2}+p+1\n$$\n\nwhere $m$ and $n$ are integers and $p$ is a prime number.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nC3.", "solution_match": "\nSolution."}}
+{"year": "2012", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "In a circle of diameter 1 consider 65 points no three of which are collinear. Prove that there exist 3 among these points which form a triangle with area less then or equal to $\\frac{1}{72}$.", "solution": "Solution. By Fermat's theorem $n^{2 p} \\equiv n^{2}(\\bmod p)$, therefore $m^{2}+n^{2}+p+1 \\equiv n^{2}(\\bmod p) \\Rightarrow$ $m^{2} \\equiv-1(\\bmod p)$.\n\nCase 1: $p=4 k+3$. We have $\\left(m^{2}\\right)^{2 k+1} \\equiv(-1)^{2 k+1}(\\bmod p)$. Therefore,\n\n$$\nm^{p-1} \\equiv-1(\\bmod p)\n$$\n\nand $p$ does not divide $m$. On the other hand, by Fermat's theorem\n\n$$\nm^{p-1} \\equiv 1(\\bmod p)\n$$\n\n(28) and (29) yield $p=2$. Thus, $p \\neq 4 k+3$.\n\nCase 2: $p=4 k+1$. Let us consider (27) in mod4. $n^{2}=0$ or 1 in mod4. In both cases $n^{2 p}=n^{2}(\\bmod 4)$. From $(27)$ we get $n^{2} \\equiv m^{2}+n^{2}+1+1(\\bmod 4)$. Therefore, $m^{2} \\equiv-2(\\bmod 4)$, and again there is no solution.\n\nCase 3: $p=2$. The given equation is written as\n\n$$\nn^{4}-n^{2}-3=m^{2}\n$$\n\nLet $l=n^{2}$. Readily, we do not get any solution for $l=0$, 1 . If $l=4$, then there are four solutions: $(3,2,2),(-3,2,2),(3,-2,2),(-3,-2,2)$. There is no solution for $l>4$, since in this case\n\n$$\n(l-1)^{2}=l^{2}-2 l+1<m^{2}=l^{2}-l-3<l^{2}\n$$\n\nThus, (27) has four solutions: $(3,2,2),(-3,2,2),(3,-2,2),(-3,-2,2)$ and we are done.\n\n4", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nC3.", "solution_match": "\nSolution."}}
+{"year": "2012", "tier": "T3", "problem_label": "NT5", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all the positive integers $x, y, z, t$ such that $2^{x} \\cdot 3^{y}+5^{z}=7^{t}$.", "solution": "Solution. First we prove the following Lemma\n\nLemma: If $x, y, z$ real numbers such that $x+y+z=0$, then $2\\left(x^{4}+y^{4}+z^{4}\\right)=\\left(x^{2}+y^{2}+z^{2}\\right)^{2}$. Proof of the Lemma:\n\n$$\n\\begin{aligned}\nx^{4}+y^{4}+z^{4} & =x^{2} x^{2}+y^{2} y^{2}+z^{2} z^{2}=x^{2}(y+z)^{2}+y^{2}(z+x)^{2}+z^{2}(x+y)^{2} \\\\\n& =2\\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\\right)+2 x y z(x+y+z) \\\\\n& =\\left(x^{2}+y^{2}+z^{2}\\right)^{2}-x^{4}-y^{4}-z^{4}\n\\end{aligned}\n$$\n\nand the claim follows.\n\nNow back to our problem notice that $(a-2 b+c)+(b-2 c+a)+(c-2 a+b)=0$, thus according to the lemma it holds\n\n$$\n\\begin{aligned}\nA & =2(a-2 b+c)^{4}+2(b-2 c+a)^{4}+2(c-2 a+b)^{4} \\\\\n& =\\left[(a-2 b+c)^{2}+(b-2 c+a)^{2}+(c-2 a+b)^{2}\\right]=\\left[6\\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\\right)\\right]^{2}\n\\end{aligned}\n$$\n\nSince $a^{2}+b^{2}+c^{2} \\geq a b+b c+c a$ we have that\n\n$$\n\\sqrt{A}+1=6\\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\\right)+1\n$$\n\nIn addition, it is easy to check that\n\n$$\nB=d(d+1)(d+2)(d+3)=\\left(d^{2}+3 d+1\\right)^{2}\n$$\n\nLet us set $6\\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\\right)+1=m, d^{2}+3 d+1=n$. We need to prove that the number $(\\sqrt{A}+1)^{2}+B=m^{2}+n^{2}$ is not a perfect square.\n\nSince both $m, n$ are odd integers, both $m^{2}, n^{2}$ are integers of the form $4 k+1$, so the number $m^{2}+n^{2}$ is an integer of the form $4 k+2$. But it is well known that all perfect squares are of the form $4 k$ or $4 k+1$, and we are done.\n\n$4 \\quad \\mathrm{NT7}$. Find all natural numbers a,b, c for which $1997^{a}+15^{b}=2012^{c}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nNT5.", "solution_match": "\nSolution."}}
+{"year": "2012", "tier": "T3", "problem_label": "NT5", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all the positive integers $x, y, z, t$ such that $2^{x} \\cdot 3^{y}+5^{z}=7^{t}$.", "solution": "Solution. $1997^{a}+15^{b}=2012^{c} \\Rightarrow 1+(-1)^{b} \\equiv 0(\\bmod 4)$, so $b$ is an odd number.\n\n$1997^{a}+15^{b}=2012^{c} \\Rightarrow 1+0 \\equiv 2^{c}(\\bmod 3)$, so $c$ is even, say $c=2 c_{1}$.\n\nWe intend to consider the given equation modulo 8 and for this reason we discern two cases:\n\n(1): $c=1$. Clearly then $a=b=1$ and $a=b=c=1$ is a solution. This is actually the only solution of the given equation since in the remaining case where $c>1$ it will be shown that there exist no solution.\n\n(2): $c>1$. Then $2012^{c}=(4 \\cdot 503)^{c}$ is a multiple of 8 and\n\n$1997^{a}+15^{b}=2012^{c} \\Rightarrow 5^{a}+(-1)^{b} \\equiv 5^{a}+(-1) \\equiv 0(\\bmod 8)$, so $a$ is even, say $a=2 a_{1}$. Hence\n\n$$\n3^{\\&} \\cdot 5^{d 1}=15^{b}=2012^{c}-1997^{a}=\\left(2012^{c_{1}}-1997^{a_{1}}\\right) \\cdot\\left(2012^{c_{1}}+1997^{a_{1}}\\right)\n$$\n\nObserve that $2012^{c_{1}}-1997^{a_{1}}, 2012^{c_{1}}+199^{\\prime} 7^{a_{1}}$ are both greater than 1 and prime to each other as $\\operatorname{gcd}\\left(2012^{c_{1}}-1997^{a_{1}}, 2012^{c_{1}}+1997^{a_{1}}\\right)=\\operatorname{gcd}\\left(2012^{c_{1}}-1997^{a_{1}}, 2 \\cdot 1997^{a_{1}}\\right)=1$. So there exist two cases:\n\n$$\n\\text { Case 1: } \\begin{aligned}\n& 2012^{c_{1}}-1997^{a_{1}}=5^{b} \\\\\n& 2012^{c_{1}}+1997^{a_{1}}=3^{b}\n\\end{aligned}, \\quad \\text { Case 2: } \\begin{aligned}\n& 2012^{c_{1}}-1997^{a_{1}}=3^{b} \\\\\n& 2012^{c_{1}}+1997^{a_{1}}=5^{b}\n\\end{aligned}\n$$\n\nCase 1:\n\n$$\n\\begin{gathered}\n2012^{c_{1}}-1997^{a_{1}}=5^{b} \\Rightarrow 2^{c_{1}}-2^{a_{1}} \\equiv 0(\\bmod 5) \\Rightarrow c_{1} \\equiv a_{1}(\\bmod 5) \\\\\n2012^{c_{1}}+1997^{a_{1}}=3^{b} \\Rightarrow 2^{c_{1}}+2^{a_{1}} \\equiv 0(\\bmod 5) \\Rightarrow c_{1} \\equiv a_{1}+1(\\bmod 5)\n\\end{gathered}\n$$\n\na contradiction.\n\nCase 2:\n\n$$\n\\begin{aligned}\n& 2012^{c_{1}}-1997^{a_{1}}=3^{b} \\\\\n& 2012^{c_{1}}+1997^{a_{1}}=5^{b}\n\\end{aligned}\n$$\n\nSince $b$ is an odd number we get $2012^{c_{1}}+1997^{a_{1}} \\equiv 5^{b}(\\bmod 3) \\Rightarrow 2^{c_{1}}+2^{a_{1}} \\equiv 5^{b} \\equiv 2(\\bmod 3)$ so $a_{1}, c_{1}$ are even numbers, say $a_{1}=2 a_{2}, c_{1}=2 c_{2}$. Then\n\n$$\n\\left(2012^{c_{2}}-1997^{a_{2}}\\right) \\cdot\\left(2012^{c_{2}}+1997^{a_{2}}\\right)=3^{b}\n$$\n\nBut $\\operatorname{gcd}\\left(2012^{c_{2}}-1997^{a_{2}}, 2012^{c_{2}}+1997^{a_{2}}\\right)=1$ and the above implies $2012^{c_{2}}-1997^{a_{2}}=1$. But then mod4, we get $0-1 \\equiv(\\bmod 4)$, a contradiction.\n\nTherefore there exists no solution for $c>1$.\n\nHence $a=b=c=1$ is the only solution.\n\n$$\n\\therefore\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl", "problem_match": "\nNT5.", "solution_match": "\nSolution."}}

JBMO/segmented/en-shortlist/en-shl-2015.jsonl

@@ -1,17 +1,17 @@
-{"year": "2015", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO", "problem": "MLD\n\nLet $x, y, z$ be real numbers, satisfying the relations\n\n$$\n\\left\\{\\begin{array}{l}\nx \\geq 20 \\\\\ny \\geq 40 \\\\\nz \\geq 1675 \\\\\nx+y+z=2015\n\\end{array}\\right.\n$$\n\nFind the greatest value of the product $P=x \\cdot y \\cdot z$.", "solution": "By virtue of $z \\geq 1675$ we have\n\n$$\ny+z<2015 \\Leftrightarrow y<2015-z \\leq 2015-1675<1675\n$$\n\nIt follows that $(1675-y) \\cdot(1675-z) \\leq 0 \\Leftrightarrow y \\cdot z \\leq 1675 \\cdot(y+z-1675)$.\n\nBy using the inequality $u \\cdot v \\leq\\left(\\frac{u+v}{2}\\right)^{2}$ for all real numbers $u, v$ we obtain\n\n$$\n\\begin{gathered}\nP=x \\cdot y \\cdot z \\leq 1675 \\cdot x \\cdot(y+z-1675) \\leq 1675 \\cdot\\left(\\frac{x+y+z-1675}{2}\\right)^{2}= \\\\\n1675 \\cdot\\left(\\frac{2015-1675}{2}\\right)^{2}=1675 \\cdot 170^{2}=48407500\n\\end{gathered}\n$$\n\n$$\n\\text { We have } P=x \\cdot y \\cdot z=48407500 \\Leftrightarrow\\left\\{\\begin{array} { l } \n{ x + y + z = 2 0 1 5 , } \\\\\n{ z = 1 6 7 5 , } \\\\\n{ x = y + z - 1 6 7 5 }\n\\end{array} \\Leftrightarrow \\left\\{\\begin{array}{l}\nx=170 \\\\\ny=170 \\\\\nz=1675\n\\end{array}\\right.\\right.\n$$\n\nSo, the greatest value of the product is $P=x \\cdot y \\cdot z=48407500$.\n\n## Solution 2:\n\nLet $S=\\{(x, y, z) \\mid x \\geq 20, y \\geq 40, z \\geq 1675, x+y+z=2015\\}$ and $\\Pi=\\{|x \\cdot y \\cdot z|(x, y, z) \\in S\\}$ We have to find the biggest element of $\\Pi$. By using the given inequalities we obtain:\n\n$$\n\\left\\{\\begin{array}{l}\n20 \\leq x \\leq 300 \\\\\n40 \\leq y \\leq 320 \\\\\n1675 \\leq z \\leq 1955 \\\\\ny<1000<z\n\\end{array}\\right.\n$$\n\nLet $z=1675+d$. Since $x \\leq 300$ so $(1675+d) \\cdot x=1675 x+d x \\leq 1675 x+1675 d=1675 \\cdot(x+d)$ That means that if $(x, y, 1675+d) \\in S$ then $(x+d, y, 1675) \\in S$, and $x \\cdot y \\cdot(1675+d) \\leq(x+d) \\cdot y \\cdot 1675$. Therefore $z=1675$ must be for the greatest product.\n\nFurthermore, $x \\cdot y \\leq\\left(\\frac{x+y}{2}\\right)^{2}=\\left(\\frac{2015-1675}{2}\\right)^{2}=\\left(\\frac{340}{2}\\right)^{2}=170^{2}$. Since $(170,170,1675) \\in S$ that means that the biggest element of $\\Pi$ is $170 \\cdot 170 \\cdot 1675=48407500$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "## A1 ", "solution_match": "## Solution 1:"}}
-{"year": "2015", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO", "problem": "ALB\n\n3) If $x^{3}-3 \\sqrt{3} x^{2}+9 x-3 \\sqrt{3}-64=0$, find the value of $x^{6}-8 x^{5}+13 x^{4}-5 x^{3}+49 x^{2}-137 x+2015$.", "solution": "$x^{3}-3 \\sqrt{3} x^{2}+9 x-3 \\sqrt{3}-64=0 \\Leftrightarrow(x-\\sqrt{3})^{3}=64 \\Leftrightarrow(x-\\sqrt{3})=4 \\Leftrightarrow x-4=\\sqrt{3} \\Leftrightarrow x^{2}-8 x+16=3 \\Leftrightarrow$ $x^{2}-8 x+13=0$\n\n$x^{6}-8 x^{5}+13 x^{4}-5 x^{3}+49 x^{2}-137 x+2015=\\left(x^{2}-8 x+13\\right)\\left(x^{4}-5 x+9\\right)+1898=0+1898=1898$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "## A2 ", "solution_match": "\nSolution"}}
-{"year": "2015", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO", "problem": "MNE\n\nLet $a, b, c$ be positive real numbers. Prove that\n\n$$\n\\frac{a}{b}+\\sqrt{\\frac{b}{c}}+\\sqrt[3]{\\frac{c}{a}}>2\n$$", "solution": "Starting from the double expression on the left-hand side of given inequality, and applying twice the Arithmetic-Geometric mean inequality, we find that\n\n$$\n\\begin{aligned}\n2 \\frac{a}{b}+2 \\sqrt{\\frac{b}{c}}+2 \\sqrt[3]{\\frac{c}{a}} & =\\frac{a}{b}+\\left(\\frac{a}{b}+\\sqrt{\\frac{b}{c}}+\\sqrt{\\frac{b}{c}}\\right)+2 \\sqrt[3]{\\frac{c}{a}} \\\\\n& \\geq \\frac{a}{b}+3 \\sqrt[3]{\\frac{a}{b}} \\sqrt{\\frac{b}{c}} \\sqrt{\\frac{b}{c}}+2 \\sqrt[3]{\\frac{c}{a}} \\\\\n& =\\frac{a}{b}+3 \\sqrt[3]{\\frac{a}{c}}+2 \\sqrt[3]{\\frac{c}{a}} \\\\\n& \\left.=\\frac{a}{b}+\\sqrt[3]{\\frac{a}{c}}+2 \\sqrt[3]{\\frac{a}{c}}+\\sqrt[3]{\\frac{c}{a}}\\right) \\\\\n& \\geq \\frac{a}{b}+\\sqrt[3]{\\frac{a}{c}}+2 \\cdot 2 \\sqrt{\\sqrt[3]{\\frac{a}{c}} \\sqrt[3]{\\frac{c}{a}}} \\\\\n& =\\frac{a}{b}+\\sqrt[3]{\\frac{a}{c}}+4 \\\\\n& >4\n\\end{aligned}\n$$\n\nwhich yields the given inequality.\n\n(A4) GRE\n\nLet $a, b, c$ be positive real numbers such that $a+b+c=3$. Find the minimum value of\n\n$$\nA=\\frac{2-a^{3}}{a}+\\frac{2-b^{3}}{b}+\\frac{2-c^{3}}{c}\n$$\n\n## Solution:\n\nWe rewrite $A$ as follows:\n\n$$\n\\begin{aligned}\n& A=\\frac{2-a^{3}}{a}+\\frac{2-b^{3}}{b}+\\frac{2-c^{3}}{c}=2\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)-a^{2}-b^{2}-c^{2}= \\\\\n& 2\\left(\\frac{a b+b c+c a}{a b c}\\right)-\\left(a^{2}+b^{2}+c^{2}\\right)=2\\left(\\frac{a b+b c+c a}{a b c}\\right)-\\left((a+b+c)^{2}-2(a b+b c+c a)\\right)= \\\\\n& 2\\left(\\frac{a b+b c+c a}{a b c}\\right)-(9-2(a b+b c+c a))=2\\left(\\frac{a b+b c+c a}{a b c}\\right)+2(a b+b c+c a)-9= \\\\\n& 2(a b+b c+c a)\\left(\\frac{1}{a b c}+1\\right)-9\n\\end{aligned}\n$$\n\nRecall now the well-known inequality $(x+y+z)^{2} \\geq 3(x y+y z+z x)$ and set $x=a b, y=b c, z=c a$, to obtain $(a b+b c+c a)^{2} \\geq 3 a b c(a+b+c)=9 a b c$ where we have used $a+b+c=3$. By taking the square roots on both sides of the last one we obtain:\n\n$a b+b c+c a \\geq 3 \\sqrt{a b c}$. (1)\n\nAlso by using AM-GM inequality we get that\n\n$$\n\\frac{1}{a b c}+1 \\geq 2 \\sqrt{\\frac{1}{a b c}}\n$$\n\nMultiplication of (1) and (2) gives:\n\n$(a b+b c+c a)\\left(\\frac{1}{a b c}+1\\right) \\geq 3 \\sqrt{a b c} \\cdot 2 \\sqrt{\\frac{1}{a b c}}=6$.\n\nSo $A \\geq 2 \\cdot 6-9=3$ and the equality holds if and only if $a=b=c=1$, so the minimum value is 3.\n\nRemark: Note that if $f(x)=\\frac{2-x^{3}}{x}, x \\in(0,3)$ then $f^{\\prime \\prime}(x)=\\frac{4}{x^{3}}-2$, so the function is convex on $x \\in(0, \\sqrt[3]{2})$ and concave on $x \\in(\\sqrt[3]{2}, 3)$. This means that we cannot apply Jensen's inequality.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "## A3 ", "solution_match": "## Solution:"}}
-{"year": "2015", "tier": "T3", "problem_label": "A5", "problem_type": "Algebra", "exam": "JBMO", "problem": "MKCD\n\nLet $x, y, z$ be positive real numbers that satisfy the equality $x^{2}+y^{2}+z^{2}=3$. Prove that\n\n$$\n\\frac{x^{2}+y z}{x^{2}+y z+1}+\\frac{y^{2}+z x}{y^{2}+z x+1}+\\frac{z^{2}+x y}{z^{2}+x y+1} \\leq 2\n$$", "solution": "We have\n\n$$\n\\begin{aligned}\n& \\frac{x^{2}+y z}{x^{2}+y z+1}+\\frac{y^{2}+z x}{y^{2}+z x+1}+\\frac{z^{2}+x y}{z^{2}+x y+1} \\leq 2 \\Leftrightarrow \\\\\n& \\frac{x^{2}+y z+1}{x^{2}+y z+1}+\\frac{y^{2}+z x+1}{y^{2}+z x+1}+\\frac{z^{2}+x y+1}{z^{2}+x y+1} \\leq 2+\\frac{1}{x^{2}+y z+1}+\\frac{1}{y^{2}+z x+1}+\\frac{1}{z^{2}+x y+1} \\Leftrightarrow \\\\\n& 3 \\leq 2+\\frac{1}{x^{2}+y z+1}+\\frac{1}{y^{2}+z x+1}+\\frac{1}{z^{2}+x y+1} \\Leftrightarrow \\\\\n& 1 \\leq \\frac{1}{x^{2}+y z+1}+\\frac{1}{y^{2}+z x+1}+\\frac{1}{z^{2}+x y+1} \\\\\n& \\frac{1}{x^{2}+y z+1}+\\frac{1}{y^{2}+z x+1}+\\frac{1}{z^{2}+x y+1} \\geq \\frac{9}{x^{2}+y z+1+y^{2}+z x+1+z^{2}+x y+1}= \\\\\n& \\frac{9}{x^{2}+y^{2}+z^{2}+x y+y z+z x+3} \\geq \\frac{9}{2 x^{2}+y^{2}+z^{2}+3}=1\n\\end{aligned}\n$$\n\nThe first inequality: AM-GM inequality (also can be achieved with Cauchy-BunjakowskiSchwarz inequality). The second inequality: $x y+y z+z x £ x^{2}+y^{2}+z^{2}$ (there are more ways to prove it, AM-GM, full squares etc.)\n\n## GEOMETRY", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "## A5 ", "solution_match": "## Solution:"}}
-{"year": "2015", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO", "problem": "MNE\n\nAround the triangle $A B C$ the circle is circumscribed, and at the vertex $C$ tangent $t$ to this circle is drawn. The line $p$ which is parallel to this tangent intersects the lines $B C$ and $A C$ at the points $D$ and $E$, respectively. Prove that the points $A, B, D, E$ belong to the same circle.", "solution": "Let $O$ be the center of a circumscribed circle $k$ of the triangle $A B C$, and let $F$ and $G$ be the points of intersection of the line $C O$ with the line $p$ and the circle $k$, respectively (see Figure). From $p \\| t$ it follows that $p \\perp C O$. Furthermore, $\\angle A B C=\\angle A G C$, because these angles are peripheral over the same chord. The quadrilateral $A G F E$ has two right angles at the vertices $A$ and $F$, and hence, $\\angle A E D+\\angle A B D=\\angle A E F+\\angle A G F=180^{\\circ}$. Hence, the quadrilateral $A B D E$ is cyclic, as asserted.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-06.jpg?height=406&width=483&top_left_y=2006&top_left_x=843)\n\nFigure", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "## G1 ", "solution_match": "## Solution:"}}
-{"year": "2015", "tier": "T3", "problem_label": "G2", "problem_type": "Geometry", "exam": "JBMO", "problem": "MLD\n\nThe point $P$ is outside of the circle $\\Omega$. Two tangent lines, passing from the point $P$, touch the circle $\\Omega$ at the points $A$ and $B$. The median $A M, M \\in(B P)$, intersects the circle $\\Omega$ at the point $C$ and the line $P C$ intersects again the circle $\\Omega$ at the point $D$. Prove that the lines $A D$ and $B P$ are parallel.", "solution": "Since $\\angle B A C=\\angle B A M=\\angle M B C$, we have $\\triangle M A B \\cong \\triangle M B C$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-07.jpg?height=498&width=745&top_left_y=735&top_left_x=625)\n\nWe obtain $\\frac{M A}{M B}=\\frac{M B}{M C}=\\frac{A B}{B C}$. The equality $\\quad M B=M P$ implies $\\frac{M A}{M P}=\\frac{M P}{M C}$ and $\\angle P M C \\equiv \\angle P M A$ gives the relation $\\triangle P M A \\cong \\triangle C M P$. It follows that $\\angle B P D \\equiv \\angle M P C \\equiv \\angle M A P \\equiv \\angle C A P \\equiv \\angle C D A \\equiv \\angle P D A$. So, the lines $A D$ and $B P$ are parallel.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "## G2 ", "solution_match": "## Solution:"}}
-{"year": "2015", "tier": "T3", "problem_label": "G3", "problem_type": "Geometry", "exam": "JBMO", "problem": "GRE\n\nLet $c \\equiv c(O, K)$ be a circle with center $O$ and radius $R$ and $A, B$ be two points on it, not belonging to the same diameter. The bisector of the angle $A \\hat{B} O$ intersects the circle $c$ at point $C$, the circumcircle of the triangle $A O B$, say ${ }^{c_{1}}$ at point $K$ and the circumcircle of the triangle $A O C$, say ${ }^{c_{2}}$, at point $L$. Prove that the point $K$ is the circumcenter of the triangle $A O C$ and the point $L$ is the incenter of the triangle $A O B$.", "solution": "The segments $O B, O C$ are equal, as radii of the circle ${ }^{c}$. Hence $O B C$ is an isosceles triangle and\n\n$$\n\\hat{B}_{1}=\\hat{C}_{1}=\\hat{x}\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-07.jpg?height=414&width=545&top_left_y=2097&top_left_x=714)\n\nThe chord $B C$ is the bisector of the angle $O \\hat{B} A$, and hence\n\n$$\n\\hat{B}_{1}=\\hat{B}_{2}=\\hat{x}\n$$\n\nThe angles $\\hat{B}_{2}$ and $\\hat{O}_{1}$ are inscribed to the same arc $O K$ of the circle ${ }^{c_{1}}$ and hence\n\n$$\n\\hat{B}_{2}=\\hat{\\mathrm{O}}_{1}=\\hat{x}\n$$\n\nThe segments $K O, K C$ are equal, as radii of the circle ${ }^{c_{2}}$. Hence the triangle $K O C$ is isosceles and so\n\n$$\n\\hat{O}_{2}=\\hat{C}_{1}=\\hat{x}\n$$\n\nFrom equalities $(1),(2),(3)$ we conclude that\n\n$$\n\\hat{O}_{1}=\\hat{O}_{2}=\\hat{x}\n$$\n\nand so $O K$ is the bisector, and hence perpendicular bisector of the isosceles triangle $O A C$. The point $K$ is the middle of the arc $O K$ (since $B K$ bisects the angle $O \\hat{B} A$ ). Hence the perpendicular bisector of the chord $A O$ of the circle ${ }^{c_{1}}$ is passing through point $K$. It means that $K$ is the circumcenter of the triangle $O A C$.\n\nFrom equalities (1),(2),(3) we conclude that $\\hat{B}_{2}=\\hat{C}_{1}=\\hat{x}$ and so $A B / / O C \\Rightarrow O \\hat{A} B=A \\hat{O} C$, that is $\\hat{A}_{1}+\\hat{A}_{2}=\\hat{O}_{1}+\\hat{O}_{2}$ and since $\\hat{O}_{1}=\\hat{O}_{2}=\\hat{x}$, we conclude that\n\n$$\n\\hat{A}_{1}+\\hat{A}_{2}=2 \\hat{O}_{1}=2 \\hat{x}\n$$\n\nThe angles $\\hat{A}_{I}$ and $\\hat{C}_{I}$ are inscribed into the circle ${ }^{c_{2}}$ and correspond to the same arc $O L$. Hence\n\n$$\n\\hat{A}_{I}=\\hat{C}_{1}=\\hat{x}\n$$\n\nFrom (5) and (6) we have $\\hat{A}_{1}=\\hat{A}_{2}$, i.e. $A L$ is the bisector of the angle $B \\hat{A} O$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "## G3 ", "solution_match": "## Solution:"}}
-{"year": "2015", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO", "problem": "CYP\n\nLet $\\triangle A B C$ be an acute triangle. The lines $\\left(\\varepsilon_{1}\\right),\\left(\\varepsilon_{2}\\right)$ are perpendicular to $A B$ at the points $A$, $B$, respectively. The perpendicular lines from the midpoint $M$ of $A B$ to the sides of the triangle $A C_{;} B C$ intersect the lines $\\left(\\varepsilon_{1}\\right),\\left(\\xi_{2}\\right)$ at the points $E, F$, respectively. If $I$ is the intersection point of $E F, M C$, prove that\n\n$$\n\\angle A I B=\\angle E M F=\\angle C A B+\\angle C B A\n$$", "solution": "Let $H, G$ be the points of intersection of $M E, M F$, with $A C, B C$ respectively. From the similarity of triangles $\\triangle M H A$ and $\\triangle M A E$ we get\n\n$$\n\\frac{M H}{M A}=\\frac{M A}{M E}\n$$\n\nthus, $M A^{2}=M H \\cdot M E$\n\nSimilarly, from the similarity of triangles $\\triangle M B G$ and $\\triangle M F B$ we get\n\n$$\n\\frac{M B}{M F}=\\frac{M G}{M B}\n$$\n\nthus, $M B^{2}=M F \\cdot M G$\n\nSince $M A=M B$, from (1), (2), we have that the points $E, H, G, F$ are concyclic.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-09.jpg?height=710&width=1025&top_left_y=298&top_left_x=298)\n\nTherefore, we get that $\\angle F E H=\\angle F E M=\\angle H G M$. Also, the quadrilateral $C H M G$ is cyclic, so $\\angle C M H=\\angle H G C$. We have\n\n$$\n\\angle F E H+\\angle C M H=\\angle H G M+\\angle H G C=90^{\\circ}\n$$\n\nThus $C M \\perp E F$. Now, from the cyclic quadrilaterals $F I M B$ and EIMA, we get that\n\n$\\angle I F M=\\angle I B M$ and $\\angle I E M=\\angle I A M$. Therefore, the triangles $\\triangle E M F$ and $\\triangle A I B$ are similar, so $\\angle A J B=\\angle E M F$. Finally\n\n$$\n\\angle A I B=\\angle A I M+\\angle M I B=\\angle A E M+\\angle M F B=\\angle C A B+\\angle C B A\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "\nG4", "solution_match": "## Solution:"}}
-{"year": "2015", "tier": "T3", "problem_label": "G5", "problem_type": "Geometry", "exam": "JBMO", "problem": "ROU\n\nLet $A B C$ be an acute triangle with $A B \\neq A C$. The incircle $\\omega$ of the triangle touches the sides $B C, C A$ and $A B$ at $D, E$ and $F$, respectively. The perpendicular line erected at $C$ onto $B C$ meets $E F$ at $M$, and similarly, the perpendicular line erected at $B$ onto $B C$ meets $E F$ at $N$. The line $D M$ meets $\\omega$ again in $P$, and the line $D N$ meets $\\omega$ again at $Q$. Prove that $D P=D Q$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-09.jpg?height=583&width=737&top_left_y=1595&top_left_x=607)", "solution": "## Proof 1.1.\n\nLet $\\{T\\}=E F \\cap B C$. Applying Menelaus' theorem to the triangle $A B C$ and the transversal line $E-F-T$ we obtain $\\frac{T B}{T C} \\cdot \\frac{E C}{E A} \\cdot \\frac{F A}{F B}=1$, i.e. $\\frac{T B}{T C} \\cdot \\frac{s-c}{s-a} \\cdot \\frac{s-a}{s-b}=1$, or $\\frac{T B}{T C}=\\frac{s-b}{s-c}$, where the notations are the usual ones.\n\nThis means that triangles $T B N$ and $T C M$ are similar, therefore $\\frac{T B}{T C}=\\frac{B N}{C M}$. From the above it follows $\\frac{B N}{C M}=\\frac{s-b}{s-c}, \\frac{B D}{C D}=\\frac{s-b}{s-c}$, and $\\angle D B N=\\angle D C M=90^{\\circ}$, which means that triangles $B D N$ and $C D M$ are similar, hence angles $B D N$ and $C D M$ are equal. This leads to the arcs $D Q$ and $D P$ being equal, and finally to $D P=D Q$.\n\n## Proof 1.2.\n\nLet $S$ be the meeting point of the altitude from $A$ with the line $E F$. Lines $B N, A S, C M$ are parallel, therefore triangles $B N F$ and $A S F$ are similar, as are triangles $A S E$ and $C M E$. We obtain $\\frac{B N}{A S}=\\frac{B F}{F A}$ and $\\frac{A S}{C M}=\\frac{A E}{E C}$.\n\nMultiplying the two relations, we obtain $\\frac{B N}{C M}=\\frac{B F}{F A} \\cdot \\frac{A E}{E C}=\\frac{B F}{E C}=\\frac{B D}{D C}$ (we have used that $A E=A F, B F=B D$ and $C E=C D)$.\n\nIt follows that the right triangles $B D N$ and $C D M$ are similar (SAS), which leads to the same ending as in the first proof.\n\n## NUMBER THEORY", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "\nG5 ", "solution_match": "## Solution:"}}
-{"year": "2015", "tier": "T3", "problem_label": "NT1", "problem_type": "Number Theory", "exam": "JBMO", "problem": "SAU\n\nWhat is the greatest number of integers that can be selected from a set of 2015 consecutive numbers so that no sum of any two selected numbers is divisible by their difference?", "solution": "We take any two chosen numbers. If their difference is 1 , it is clear that their sum is divisible by their difference. If their difference is 2 , they will be of the same parity, and their sum is divisible by their difference. Therefore, the difference between any chosen numbers will be at least 3 . In other words, we can choose at most one number of any three consecutive numbers. This implies that we can choose at most 672 numbers.\n\nNow, we will show that we can choose 672 such numbers from any 2015 consecutive numbers. Suppose that these numbers are $a, a+1, \\ldots, a+2014$. If $a$ is divisible by 3 , we can choose $a+1, a+4, \\ldots, a+2014$. If $a$ is not divisible by 3 , we can choose $a, a+3, \\ldots, a+$ 2013.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "\nNT1 ", "solution_match": "## Solution:"}}
-{"year": "2015", "tier": "T3", "problem_label": "NT2", "problem_type": "Number Theory", "exam": "JBMO", "problem": "BUL\n\nA positive integer is called a repunit, if it is written only by ones. The repunit with $n$ digits will be denoted by $\\underbrace{11 \\ldots 1}_{n}$. Prove that:\n\na) the repunit $\\underbrace{11 \\ldots 1}_{n}$ is divisible by 37 if and only if $n$ is divisible by 3 ;\n\nb) there exists a positive integer $k$ such that the repunit $\\underbrace{11 \\ldots 1}_{n}$ is divisible by 41 if and only if $n$ is divisible by $k$.", "solution": "a) Let $n=3 m+r$, where $m$ and $r$ are non-negative integers and $r<3$.\n\nDenote by $\\underbrace{00 \\ldots 0}_{p}$ a recording with $p$ zeroes and $\\underbrace{a b c a b c \\ldots a b c}_{p x a b c c}$ recording with $p$ times $a b c$. We\n\nhave: $\\quad \\underbrace{11 \\ldots 1}_{n}=\\underbrace{11 \\ldots 1}_{3 m+r}=\\underbrace{11 \\ldots 1}_{3 m} \\cdot \\underbrace{00 \\ldots 0}_{r}+\\underbrace{11 \\ldots 1}_{\\tau}=111 \\cdot \\underbrace{100100 \\ldots 100100 \\ldots 0}_{(m-1) \\times 100}+\\underbrace{11 \\ldots 1}_{r}$.\n\nSince $111=37.3$, the numbers $\\underbrace{11 \\ldots 1}_{n}$ and $\\underbrace{11 \\ldots 1}_{r}$ are equal modulo 37 . On the other hand the numbers 1 and 11 are not divisible by 37 . We conclude that $\\underbrace{11 \\ldots 1}_{n}$ is divisible by 37 if only if $r=0$, i.e. if and only if $n$ is divisible by 3 .\n\nb) Using the idea from a), we look for a repunit, which is divisible by 41 . Obviously, 1 and 11 are not divisible by 41 , while the residues of 111 and 1111 are 29 and 4 , respectively. We have $11111=41 \\cdot 271$. Since 11111 is a repunit with 5 digits, it follows in the same way as in a) that $\\underbrace{1}_{11 \\ldots 1}$ is divisible by 41 if and only if $n$ is divisible by 5 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "\nNT2 ", "solution_match": "## Solution:"}}
-{"year": "2015", "tier": "T3", "problem_label": "NT3", "problem_type": "Number Theory", "exam": "JBMO", "problem": "ALB\n\na) Show that the product of all differences of possible couples of six given positive integers is divisible by 960 (original from Albania).\n\nb) Show that the product of all differences of possible couples of six given positive integers. is divisible by 34560 (modified by problem selecting committee).", "solution": "a) Since we have six numbers then at least two of them have a same residue when divided by 3 , so at least one of the differences in our product is divisible by 3 .\n\nSince we have six numbers then at least two of them have a same residuc when divided by 5 , so at least one of the differences in our product is divisible by 5 .\n\nWe may have:\n\na) six numbers with the same parity\n\nb) five numbers with the same parity\n\nc) four numbers with the same parity\n\nd) three numbers with the same parity\n\nThere are $C_{6}^{2}=15$ different pairs, so there are 15 different differences in this product.\n\na) The six numbers have the same parity; then each difference is divisible by 2 , therefore our product is divisible by $2^{15}$.\nb) If we have five numbers with the same parity, then the couples that have their difference odd are formed by taking one number from these five numbers, and the second will be the sixth one. Then $C_{5}^{1}=15=5$ differences are odd. So $15-5=10$ differences are even, so product is divisible by $2^{10}$.\n\nc) If we have four numbers with the same parity, then the couples that have their difference odd are formed by taking one number from these four numbers, and the second will be from the two others numbers. Then $2 \\cdot C_{4}^{1}=8$ differences are odd. So $15-8=7$ differences are even, so our product is divisible by $2^{7}$.\n\nd) If we have three numbers with the same parity, then the couples that have their difference odd are formed by taking one from each triple. Then $C_{3}^{1} \\cdot C_{3}^{1}=9$ differences are odd, therefore $15-9=6$ differences are even, so our product is divisible by $2^{6}$.\n\nThus, our production is divisible by $2^{6} \\cdot 3 \\cdot 5=960$.\n\nb) Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ be these numbers. Since we have six numbers then at least two of them when divided by 5 have the same residue, so at least one of these differences in our product is divisible by 5 .\n\nSince we have six numbers, and we have three possible residues at the division by 3 , then at least three of them replies the residue of previous numbers, so at least three of these differences in our product are divisible by 3 .\n\nSince we have six numbers, and we have two possible residues at the division by 2 , then at least four of them replies the residue of previous numbers, and two of them replies replied residues, so at least six of these differences in our product are divisible by 2 .\n\nSince we have six numbers, and we have four possible residues at the division by 4 , then at least two of them replies the residue of previous numbers, so at least two of these differences in our product are divisible by 4 . That means that two of these differences are divisible by 4 and moreover four of them are divisible by 2 .\n\nThus, our production is divisible by $2^{4} \\cdot 4^{2} \\cdot 3^{3} \\cdot 5=34560$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "## NT3 ", "solution_match": "## Solution:"}}
-{"year": "2015", "tier": "T3", "problem_label": "NT4", "problem_type": "Number Theory", "exam": "JBMO", "problem": "MLD\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-12.jpg?height=136&width=181&top_left_y=1553&top_left_x=907)\n\nFind all prime numbers $a, b, c$ and integers $k$ which satisfy the equation $a^{2}+b^{2}+16 \\cdot c^{2}=9 \\cdot k^{2}+1$.", "solution": "The relation $9 \\cdot k^{2}+1 \\equiv 1(\\bmod 3)$ implies\n\n$$\na^{2}+b^{2}+16 \\cdot c^{2} \\equiv 1(\\bmod 3) \\Leftrightarrow a^{2}+b^{2}+c^{2} \\equiv 1(\\bmod 3)\n$$\n\nSince $a^{2} \\equiv 0,1(\\bmod 3), b^{2} \\equiv 0,1(\\bmod 3), c^{2} \\equiv 0,1(\\bmod 3)$, we have:\n\n| $a^{2}$ | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| $b^{2}$ | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 |\n| $c^{2}$ | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 |\n| $a^{2}+b^{2}+c^{2}$ | 0 | 1 | 1 | 2 | 1 | 2 | 2 | 0 |\n\nFrom the previous table it follows that two of three prime numbers $a, b, c$ are equal to 3 .\n\nCase 1. $a=b=3$. We have\n\n$$\na^{2}+b^{2}+16 \\cdot c^{2}=9 \\cdot k^{2}+1 \\Leftrightarrow 9 \\cdot k^{2}-16 \\cdot c^{2}=17 \\Leftrightarrow(3 k-4 c) \\cdot(3 k+4 c)=17\n$$\n\nIf $\\left\\{\\begin{array}{l}3 k-4 c=1, \\\\ 3 k+4 c=17,\\end{array}\\right.$ then $\\left\\{\\begin{array}{l}c=2, \\\\ k=3,\\end{array}\\right.$ and $(a, b, c, k)=(3,3,2,3)$.\n\nIf $\\left\\{\\begin{array}{l}3 k-4 c=-17, \\\\ 3 k+4 c=-1,\\end{array}\\right.$ then $\\left\\{\\begin{array}{l}c=2, \\\\ k=-3,\\end{array}\\right.$ and $(a, b, c, k)=(3,3,2,-3)$.\n\nCase 2. $c=3$. If $\\left(3, b_{0}, c, k\\right)$ is a solution of the given equation, then $\\left(b_{0}, 3, c, k\\right)$ is a solution, too.\n\nLet $a=3$. We have\n\n$$\na^{2}+b^{2}+16 \\cdot c^{2}=9 \\cdot k^{2}+1 \\Leftrightarrow 9 \\cdot k^{2}-b^{2}=152 \\Leftrightarrow(3 k-b) \\cdot(3 k+b)=152 .\n$$\n\nBoth factors shall have the same parity and we obtain only 4 cases:\n\nIf $\\left\\{\\begin{array}{l}3 k-b=2, \\\\ 3 k+b=76,\\end{array}\\right.$ then $\\left\\{\\begin{array}{l}b=37, \\\\ k=13,\\end{array}\\right.$ and $(a, b, c, k)=(3,37,3,13)$.\n\nIf $\\left\\{\\begin{array}{l}3 k-b=4, \\\\ 3 k+b=38,\\end{array}\\right.$ then $\\left\\{\\begin{array}{l}b=17, \\\\ k=7,\\end{array}\\right.$ and $(a, b, c, k)=(3,17,3,7)$.\n\nIf $\\left\\{\\begin{array}{l}3 k-b=-76, \\\\ 3 k+b=-2,\\end{array}\\right.$, .\n\nIf $\\left\\{\\begin{array}{l}3 k-b=-38, \\\\ 3 k+b=-4,\\end{array}\\right.$ then $\\left\\{\\begin{array}{l}b=17, \\\\ k=-7,\\end{array}\\right.$ and $(a, b, c, k)=(3,17,3,-7)$.\n\nIn addition, $(a, b, c, k) \\in\\{(37,3,3,13),(17,3,3,7),(37,3,3,-13),(17,3,3,-7)\\}$.\n\nSo, the given equation has 10 solutions:\n\n$S=\\left\\{\\begin{array}{l}(37,3,3,13),(17,3,3,7),(37,3,3,-13),(17,3,3,-7),(3,37,3,13),(3,17,3,7),(3,37,3,-13), \\\\ (3,17,3,-7),(3,3,2,3),(3,3,2,-3)\\end{array}\\right\\}$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "\nNT4 ", "solution_match": "## Solution:"}}
-{"year": "2015", "tier": "T3", "problem_label": "NT5", "problem_type": "Number Theory", "exam": "JBMO", "problem": "MNE\n\nDoes there exist positive integers $a, b$ and a prime $p$ such that\n\n$$\na^{3}-b^{3}=4 p^{2} ?\n$$", "solution": "The given equality may be written as\n\n$$\n(a-b)\\left(a^{2}+a b+b^{2}\\right)=4 p^{2}\n$$\n\nSince $a-b<a^{2}+a b+b^{2}$, it follows from (1) that\n\n(2) $a-b<2 p$.\n\nNow consider two cases: 1. $p=2$, and 2. $p$ is an odd prime.\n\nCase 1: $p=2$. Then (1) becomes\n\n(3) $(a-b)\\left(a^{2}+a b+b^{2}\\right)=16$ :\n\nIn view of (2) and (3), it must be $a-b=1$ or $a-b=2$. If $a-b=1$, then substituting $a=b+1$\n\nin (3) we obtain\n\n$$\nb(b+1)=5\n$$\n\nwhich is impossible since $b(b+1)$ is an even integer.\n\nIf $a-b=2$, then substituting $a=b+2$ in (3) we get\n\n$$\n3 b(b+2)=4\n$$\n\nwhich is obviously impossible.\n\nCase 2: $\\mathrm{p}$ is an odd prime. Then (1) yields $a-b \\mid 4 p^{2}$. This together with the facts that\n\n$p$ is a prime and that by (2) $a-b<2 p$, yields $a-b \\in\\{1,2,4, p\\}$.\n\nIf $a-b=1$, then substituting $a=b+1$ in (1) we obtain\n\n$$\n3 b(b+1)+1=4 p^{2}\n$$\n\nwhich is impossible since $3 b(b+1)+1$ is an odd integer.\n\nIf $a-b=2$, then substituting $a=b+2$ in (1) we obtain\n\n$$\n3 b^{2}+6 b+4=2 p^{2}\n$$\n\nwhence it follows that\n\n$$\n2\\left(p^{2}-2\\right)=3\\left(b^{2}+2 b\\right) \\equiv 0(\\bmod 3)\n$$\n\nand hence\n\n$$\np^{2} \\equiv 2(\\bmod 3)\n$$\n\nSince $p^{2} \\equiv 1(\\bmod 3)$ for each odd prime $p>3$ and $3^{2} \\equiv 0(\\bmod 3)$, it follows that the congruence (5) is not satisfied for any odd prime $p$.\n\nIf $a-b=4$, then substituting $a=b+4$ in (1) we obtain\n\n$$\n3 b^{2}+12 b+16=p^{2}\n$$\n\nwhence it follows that $b$ is an odd integer such that\n\n$$\n3 b^{2} \\equiv p^{2}(\\bmod 4)\n$$\n\nwhence since $p^{2} \\equiv 1(\\bmod 4)$ for each odd prime $p$, we have\n\n$$\n3 b^{2} \\equiv 1(\\bmod 4)\n$$\n\nHowever, since $b^{2} \\equiv 1(\\bmod 4)$ for each odd integer $b$, it follows that the congruence (6) is not satisfied for any odd integer $b$.\n\nIf $a-b=p$, then substituting $a=b+p$ in (1) we obtain\n\n$$\np\\left(3 b^{2}+3 b p+p^{2}-4 p\\right)=0\n$$\n\ni.e.,\n\n(7)\n\n$$\n3 b^{2}+3 b p+p^{2}-4 p=0\n$$\n\nIf $p \\geq 5$, then $p^{2}-4 p>0$, and thus (7) cannot be satisfied for any positive integer $b$. If $p=3$, then $(7)$ becomes\n\n$$\n3\\left(b^{2}+3 b-1\\right)=0\n$$\n\nwhich is obviously not satisfied for any positive integer $b$.\n\nHence, there does not exist positive integers $a, b$ and a prime $p$ such that $a^{3}-b^{3}=4 p^{2}$.\n\n## COMBINATORICS", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "## NT5 ", "solution_match": "## Solution:"}}
-{"year": "2015", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "BUL\n\nA board $n \\times n(n \\geq 3)$ is divided into $n^{2}$ unit squares. Integers from 0 to $n$ included are written down: one integer in each unit square, in such a way that the sums of integers in each $2 \\times 2$ square of the board are different. Find all $n$ for which such boards exist.", "solution": "The number of the $2 \\times 2$ squares in a board $n \\times n$ is equal to $(n-1)^{2}$. All possible sums of the numbers in such squares are $0,1, \\ldots, 4 n$. A necessary condition for the existence of a board with the required property is $4 n+1 \\geq(n-1)^{2}$ and consequently $n(n-6) \\leq 0$. Thus $n \\leq 6$. The examples show the existence of boards $n \\times n$ for all $3 \\leq n \\leq 6$.\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-15.jpg?height=378&width=890&top_left_y=392&top_left_x=304)\n\n| 6 | 6 | 6 | 6 | 5 | 5 |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| 6 | 6 | 5 | 5 | 5 | 5 |\n| 1 | 2 | 3 | 4 | 4 | 5 |\n| 3 | 5 | 0 | 5 | 0 | 5 |\n| 1 | 0 | 2 | 1 | 0 | 0 |\n| 1 | 0 | 1 | 0 | 0 | 0 |\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-15.jpg?height=98&width=228&top_left_y=810&top_left_x=271)\n\n2015 points are given in a plane such that from any five points we can choose two points with distance less than 1 unit. Prove that 504 of the given points lie on a unit disc.\n\n## Solution:\n\nStart from an arbitrary point $A$ and draw a unit disc with center $A$. If all other points belong to this dise then we are done. Otherwise, take any point $B$ outside of the disc. Draw a unit disc with center $B$. If two drawn discs cover all 2015 points, by PHP, one of the discs contains at least 1008 points.\n\nSuppose that there is a point $C$ outside of the two drawn discs. Draw a unit disc with center $C$, If three drawn discs cover all 2015 points, by PHP, one of the discs contains at least 672 points.\n\nFinally, if there is a point $D$ outside of the three drawn discs, draw a unit disc with center $D$. By the given condition, any other point belongs to one of the four drawn discs. By PHP, one of the discs contains at least 504 points, concluding the solution.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "## C1 ", "solution_match": "## Solution:"}}
-{"year": "2015", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "ALB\n\nPositive integers are put into the following table\n\n| 1 | 3 | 6 | 10 | 15 | 21 | 28 | 36 |  |  |\n| ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |\n| 2 | 5 | 9 | 14 | 20 | 27 | 35 | 44 |  |  |\n| 4 | 8 | 13 | 19 | 26 | 34 | 43 | 53 |  |  |\n| 7 | 12 | 18 | 25 | 33 | 42 |  |  |  |  |\n| 11 | 17 | 24 | 32 | 41 |  |  |  |  |  |\n| 16 | 23 |  |  |  |  |  |  |  |  |\n| $\\ldots$ |  |  |  |  |  |  |  |  |  |\n| $\\ldots$ |  |  |  |  |  |  |  |  |  |\n\nFind the number of the line and column where the number 2015 stays.", "solution": "We shall observe straights lines as on the next picture. We can call these lines diagonals.\n\n| 1 | $\\sqrt{3}$ | 6 | 10 | 15 | 21 | 28 | 36 |  |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 2 | 5 | 9 | 14 | 20 | 27 | 35 | 44 |  |\n| 4 | 8 | 13 | 19 | 26 | 34 | 43 | 53 |  |\n|  | 12 | 18 | 25 | 33 | 42 |  |  |  |\n| 11 | 17 | 24 | 32 | 41 |  |  |  |  |\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-16.jpg?height=152&width=799&top_left_y=301&top_left_x=327)\n\nOn the first diagonal is number 1 .\n\nOn the second diagonal are two numbers: 2 and 3 .\n\nOn the 3rd diagonal are three numbers: 4,5 and 6\n\n.\n\nOn the $n$-th diagonal are $n$ numbers. These numbers are greater then $\\frac{(n-1) n}{2}$ and not greater than $\\frac{n(n+1)}{2}$ (see the next sentence!).\n\nOn the first $n$ diagonals are $1+2+3+\\ldots+n=\\frac{n(n+1)}{2}$ numbers.\n\nIf $m$ is in the $k$-th row $l$-th column and on the $n$-th diagonal, then it is $m=\\frac{(n-1) n}{2}+l$ and $n+1=k+l$. So, $m=\\frac{(k+l-2)(k+l-1)}{2}+l$.\n\nWe have to find such numbers $n, k$ and $l$ for which:\n\n$$\n\\begin{gathered}\n\\frac{(n-1) n}{2}<2015 \\leq \\frac{n(n+1)}{2} \\\\\nn+1=k+l \\\\\n2015=\\frac{(k+l-2)(k+l-1)}{2}+l\n\\end{gathered}\n$$\n\n(1), (2), (3) $\\Rightarrow n^{2}-n<4030 \\leq n^{2}+n \\Rightarrow n=63, k+l=64,2015=\\frac{(64-2)(64-1)}{2}+l \\Rightarrow$ $t=2015-31 \\cdot 63=62, k=64-62=2$\n\nTherefore 2015 is located in the second row and 62 -th column.\n\n## Solution 2:\n\nFirstly, we can see that the first elements of the columns are triangular numbers. If $a_{i}$ is the first element of the line $i$, we have $a_{i}=\\frac{i(i-1)}{2}$.\n\nThe second elements of the first row obtained by adding the first element 2 .\n\nThe second elements of the second row is obtained by adding the first element 3 .\n\nAnd so on, then the second element on the $n$-th row is obtained by adding the first element $n+1$.\n\nThen the third element of the $n$-th row is obtained by adding $n+2$, and the $k$-th element of it is obtained by adding $k$.\n\nSince the first element of the n-th row is $\\frac{(n-1) n}{2}+1$, the second one is\n\n$\\frac{(n-1) n}{2}+1+(n+1)=\\frac{n(n+1)}{2}+2$.\n\nThe third one $\\frac{n(n+1)}{2}+1+(n+2)=\\frac{(n+1)(n+2)}{2}+3$ so the $\\mathrm{k}$-th one should be $\\frac{(n+k-2)(n+k-1)}{2}+k$\n\n$$\n\\frac{(n+k-2)(n+k-1)}{2}+k=2015 \\Leftrightarrow n^{2}+n(2 k-3)+k^{2}-k-4028=0\n$$\n\nTo have a positive integer solution $(2 k-3)^{2}-4\\left(k^{2}-k-4028\\right)=16121-8 k$ must be a perfect square.\n\nFrom $16121-8 k=x^{2}$, it is noticed that the maximum of $x$ is 126 (since $\\mathrm{k}>0$ ).\n\nSimultaneously can be seen that $\\mathrm{x}$ is odd, so $x \\leq 125$.\n\n$16121-8 k=x^{2} \\Leftrightarrow 125^{2}+496-8 k=x^{2}$\n\nSo $496-8 k=0$, form that $k=62$.\n\nFrom that we can find $n=2$.\n\nSo 2015 is located on the second row and 62-th column.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "## C3 ", "solution_match": "## Solution 1:"}}
-{"year": "2015", "tier": "T3", "problem_label": "C4", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "GRE\n\nLet $n \\geq 1$ be a positive integer. A square of side length $n$ is divided by lines parallel to each side into $n^{2}$ squares of side length 1 . Find the number of parallelograms which have vertices among the vertices of the $n^{2}$ squares of side length 1 , with both sides smaller or equal to 2 , and which have the area equal to 2 .", "solution": "We can divide all these parallelograms into 7 classes (types I-VII), according to Figure.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-17.jpg?height=632&width=1277&top_left_y=1207&top_left_x=291)\n\nType 1: There are $n$ ways to choose the strip for the horizontal (shorter) side of the parallelogram, and $(n-1)$ ways to choose the strip (of the width 2 ) for the vertical (longer) side. So there are $n(n-1)$ parallelograms of the type I.\n\nType II: There are $(n-1)$ ways to choose the strip (of the width 2 ) for the horizontal (longer) side, and $n$ ways to choose the strip for the vertical (shorter) side. So the number of the parallelogram of this type is also $n(n-1)$.\n\nType III: Each parallelogram of this type is a square inscribed in a unique square $2 \\times 2$ of our grid. The number of such squares is $(n-1)^{2}$. So there are $(n-1)^{2}$ parallelograms of type III. For each of the types IV, V, VI, VII, the strip of the width 1 in which the parallelogram is located can be chosen in $n$ ways and for each such choice there are $n-2$ parallelograms located in the chosen strip.\n\nSumming we obtain that the total number of parallelograms is:\n\n$$\n2 n(n-1)+(n-1)^{2}+4 n(n-2)=7 n^{2}-12 n+1\n$$\n\n(C5) CYP\n\nWe have a $5 \\times 5$ chessboard and a supply of $\\mathrm{L}$-shaped triominoes, i.e. $2 \\times 2$ squares with one comer missing. Two players $A$ and $B$ play the following game: A positive integer $k \\leq 25$ is chosen. Starting with $A$, the players take alternating turns marking squares of the chessboard until they mark a total of $k$ squares. (In each turn a player has to mark exactly one new square.)\n\nAt the end of the process, player $A$ wins if he can cover without overlapping all but at most 2 unmarked squares with $\\mathrm{L}$-shaped triominoes, otherwise player $\\boldsymbol{B}$ wins. It is not permitted any\nmarked squares to be covered.\n\nFind the smallest $\\boldsymbol{k}$, if it exists, such that player $\\boldsymbol{B}$ has a winning strategy.\n\n## Solution:\n\nWe will show that player $A$ wins if $k=1,2$ or 3 , but player $B$ wins if $k=4$. Thus the smallest $k$ for which $B$ has a winning strategy exists and is equal to 4 .\n\nIf $k=1$, player $A$ marks the upper left corner of the square and then fills it as follows.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-18.jpg?height=575&width=572&top_left_y=1114&top_left_x=798)\n\nIf $k=2$, player $A$ marks the upper left comer of the square. Whatever square player $B$ marks, then player $\\boldsymbol{A}$ can fill in the square in exactly the same pattern as above except that he doesn't put the triomino which covers the marked square of $B$. Player $A$ wins because he has left only two unmarked squares uncovered.\n\nFor $k=3$, player $\\boldsymbol{A}$ wins by following the same strategy. When he has to mark a square for the second time, he marks any yet unmarked square of the triomino that covers the marked square of $B$.\n\nLet us now show that for $k=4$ player $B$ has a wimning strategy. Since there will be 21 unmarked squares, player $\\boldsymbol{A}$ will need to cover all of them with seven L-shaped triominoes. We can assume that in his first move, player $\\boldsymbol{A}$ does not mark any square in the bottom two rows of the chessboard (otherwise just rotate the chessboard). In his first move player $\\boldsymbol{B}$ marks the square labeled 1 in the following figure.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-19.jpg?height=550&width=564&top_left_y=300&top_left_x=737)\n\nIf player $\\boldsymbol{A}$ in his next move does not mark any of the squares labeled 2,3 and 4 then player $\\boldsymbol{B}$ marks the square labeled 3. Player $\\boldsymbol{B}$ wins as the square labeled 2 is left unmarked but cannot be covered with an L-shaped triomino.\n\nIf player $\\boldsymbol{A}$ in his next move marks the square labeled 2, then player $\\boldsymbol{B}$ marks the square labeled 5 . Player $B$ wins as the square labeled 3 is left unmarked but cannot be covered with an L-shaped triomino.\n\nFinally, if player $\\boldsymbol{A}$ in his next move marks one of the squares labeled 3 or 4 , player $\\boldsymbol{B}$ marks the other of these two squares. Player $\\boldsymbol{B}$ wins as the square labeled 2 is left unmarked but cannot be covered with an L-shaped triomino.\n\nSince we have covered all possible cases, player $B$ wins when $k=4$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "\nC4", "solution_match": "\nSolution:"}}
+{"year": "2015", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "MLD\n\nLet $x, y, z$ be real numbers, satisfying the relations\n\n$$\n\\left\\{\\begin{array}{l}\nx \\geq 20 \\\\\ny \\geq 40 \\\\\nz \\geq 1675 \\\\\nx+y+z=2015\n\\end{array}\\right.\n$$\n\nFind the greatest value of the product $P=x \\cdot y \\cdot z$.", "solution": "By virtue of $z \\geq 1675$ we have\n\n$$\ny+z<2015 \\Leftrightarrow y<2015-z \\leq 2015-1675<1675\n$$\n\nIt follows that $(1675-y) \\cdot(1675-z) \\leq 0 \\Leftrightarrow y \\cdot z \\leq 1675 \\cdot(y+z-1675)$.\n\nBy using the inequality $u \\cdot v \\leq\\left(\\frac{u+v}{2}\\right)^{2}$ for all real numbers $u, v$ we obtain\n\n$$\n\\begin{gathered}\nP=x \\cdot y \\cdot z \\leq 1675 \\cdot x \\cdot(y+z-1675) \\leq 1675 \\cdot\\left(\\frac{x+y+z-1675}{2}\\right)^{2}= \\\\\n1675 \\cdot\\left(\\frac{2015-1675}{2}\\right)^{2}=1675 \\cdot 170^{2}=48407500\n\\end{gathered}\n$$\n\n$$\n\\text { We have } P=x \\cdot y \\cdot z=48407500 \\Leftrightarrow\\left\\{\\begin{array} { l } \n{ x + y + z = 2 0 1 5 , } \\\\\n{ z = 1 6 7 5 , } \\\\\n{ x = y + z - 1 6 7 5 }\n\\end{array} \\Leftrightarrow \\left\\{\\begin{array}{l}\nx=170 \\\\\ny=170 \\\\\nz=1675\n\\end{array}\\right.\\right.\n$$\n\nSo, the greatest value of the product is $P=x \\cdot y \\cdot z=48407500$.\n\n## Solution 2:\n\nLet $S=\\{(x, y, z) \\mid x \\geq 20, y \\geq 40, z \\geq 1675, x+y+z=2015\\}$ and $\\Pi=\\{|x \\cdot y \\cdot z|(x, y, z) \\in S\\}$ We have to find the biggest element of $\\Pi$. By using the given inequalities we obtain:\n\n$$\n\\left\\{\\begin{array}{l}\n20 \\leq x \\leq 300 \\\\\n40 \\leq y \\leq 320 \\\\\n1675 \\leq z \\leq 1955 \\\\\ny<1000<z\n\\end{array}\\right.\n$$\n\nLet $z=1675+d$. Since $x \\leq 300$ so $(1675+d) \\cdot x=1675 x+d x \\leq 1675 x+1675 d=1675 \\cdot(x+d)$ That means that if $(x, y, 1675+d) \\in S$ then $(x+d, y, 1675) \\in S$, and $x \\cdot y \\cdot(1675+d) \\leq(x+d) \\cdot y \\cdot 1675$. Therefore $z=1675$ must be for the greatest product.\n\nFurthermore, $x \\cdot y \\leq\\left(\\frac{x+y}{2}\\right)^{2}=\\left(\\frac{2015-1675}{2}\\right)^{2}=\\left(\\frac{340}{2}\\right)^{2}=170^{2}$. Since $(170,170,1675) \\in S$ that means that the biggest element of $\\Pi$ is $170 \\cdot 170 \\cdot 1675=48407500$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "## A1 ", "solution_match": "## Solution 1:"}}
+{"year": "2015", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "ALB\n\n3) If $x^{3}-3 \\sqrt{3} x^{2}+9 x-3 \\sqrt{3}-64=0$, find the value of $x^{6}-8 x^{5}+13 x^{4}-5 x^{3}+49 x^{2}-137 x+2015$.", "solution": "$x^{3}-3 \\sqrt{3} x^{2}+9 x-3 \\sqrt{3}-64=0 \\Leftrightarrow(x-\\sqrt{3})^{3}=64 \\Leftrightarrow(x-\\sqrt{3})=4 \\Leftrightarrow x-4=\\sqrt{3} \\Leftrightarrow x^{2}-8 x+16=3 \\Leftrightarrow$ $x^{2}-8 x+13=0$\n\n$x^{6}-8 x^{5}+13 x^{4}-5 x^{3}+49 x^{2}-137 x+2015=\\left(x^{2}-8 x+13\\right)\\left(x^{4}-5 x+9\\right)+1898=0+1898=1898$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "## A2 ", "solution_match": "\nSolution"}}
+{"year": "2015", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "MNE\n\nLet $a, b, c$ be positive real numbers. Prove that\n\n$$\n\\frac{a}{b}+\\sqrt{\\frac{b}{c}}+\\sqrt[3]{\\frac{c}{a}}>2\n$$", "solution": "Starting from the double expression on the left-hand side of given inequality, and applying twice the Arithmetic-Geometric mean inequality, we find that\n\n$$\n\\begin{aligned}\n2 \\frac{a}{b}+2 \\sqrt{\\frac{b}{c}}+2 \\sqrt[3]{\\frac{c}{a}} & =\\frac{a}{b}+\\left(\\frac{a}{b}+\\sqrt{\\frac{b}{c}}+\\sqrt{\\frac{b}{c}}\\right)+2 \\sqrt[3]{\\frac{c}{a}} \\\\\n& \\geq \\frac{a}{b}+3 \\sqrt[3]{\\frac{a}{b}} \\sqrt{\\frac{b}{c}} \\sqrt{\\frac{b}{c}}+2 \\sqrt[3]{\\frac{c}{a}} \\\\\n& =\\frac{a}{b}+3 \\sqrt[3]{\\frac{a}{c}}+2 \\sqrt[3]{\\frac{c}{a}} \\\\\n& \\left.=\\frac{a}{b}+\\sqrt[3]{\\frac{a}{c}}+2 \\sqrt[3]{\\frac{a}{c}}+\\sqrt[3]{\\frac{c}{a}}\\right) \\\\\n& \\geq \\frac{a}{b}+\\sqrt[3]{\\frac{a}{c}}+2 \\cdot 2 \\sqrt{\\sqrt[3]{\\frac{a}{c}} \\sqrt[3]{\\frac{c}{a}}} \\\\\n& =\\frac{a}{b}+\\sqrt[3]{\\frac{a}{c}}+4 \\\\\n& >4\n\\end{aligned}\n$$\n\nwhich yields the given inequality.\n\n(A4) GRE\n\nLet $a, b, c$ be positive real numbers such that $a+b+c=3$. Find the minimum value of\n\n$$\nA=\\frac{2-a^{3}}{a}+\\frac{2-b^{3}}{b}+\\frac{2-c^{3}}{c}\n$$\n\n## Solution:\n\nWe rewrite $A$ as follows:\n\n$$\n\\begin{aligned}\n& A=\\frac{2-a^{3}}{a}+\\frac{2-b^{3}}{b}+\\frac{2-c^{3}}{c}=2\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)-a^{2}-b^{2}-c^{2}= \\\\\n& 2\\left(\\frac{a b+b c+c a}{a b c}\\right)-\\left(a^{2}+b^{2}+c^{2}\\right)=2\\left(\\frac{a b+b c+c a}{a b c}\\right)-\\left((a+b+c)^{2}-2(a b+b c+c a)\\right)= \\\\\n& 2\\left(\\frac{a b+b c+c a}{a b c}\\right)-(9-2(a b+b c+c a))=2\\left(\\frac{a b+b c+c a}{a b c}\\right)+2(a b+b c+c a)-9= \\\\\n& 2(a b+b c+c a)\\left(\\frac{1}{a b c}+1\\right)-9\n\\end{aligned}\n$$\n\nRecall now the well-known inequality $(x+y+z)^{2} \\geq 3(x y+y z+z x)$ and set $x=a b, y=b c, z=c a$, to obtain $(a b+b c+c a)^{2} \\geq 3 a b c(a+b+c)=9 a b c$ where we have used $a+b+c=3$. By taking the square roots on both sides of the last one we obtain:\n\n$a b+b c+c a \\geq 3 \\sqrt{a b c}$. (1)\n\nAlso by using AM-GM inequality we get that\n\n$$\n\\frac{1}{a b c}+1 \\geq 2 \\sqrt{\\frac{1}{a b c}}\n$$\n\nMultiplication of (1) and (2) gives:\n\n$(a b+b c+c a)\\left(\\frac{1}{a b c}+1\\right) \\geq 3 \\sqrt{a b c} \\cdot 2 \\sqrt{\\frac{1}{a b c}}=6$.\n\nSo $A \\geq 2 \\cdot 6-9=3$ and the equality holds if and only if $a=b=c=1$, so the minimum value is 3.\n\nRemark: Note that if $f(x)=\\frac{2-x^{3}}{x}, x \\in(0,3)$ then $f^{\\prime \\prime}(x)=\\frac{4}{x^{3}}-2$, so the function is convex on $x \\in(0, \\sqrt[3]{2})$ and concave on $x \\in(\\sqrt[3]{2}, 3)$. This means that we cannot apply Jensen's inequality.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "## A3 ", "solution_match": "## Solution:"}}
+{"year": "2015", "tier": "T3", "problem_label": "A5", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "MKCD\n\nLet $x, y, z$ be positive real numbers that satisfy the equality $x^{2}+y^{2}+z^{2}=3$. Prove that\n\n$$\n\\frac{x^{2}+y z}{x^{2}+y z+1}+\\frac{y^{2}+z x}{y^{2}+z x+1}+\\frac{z^{2}+x y}{z^{2}+x y+1} \\leq 2\n$$", "solution": "We have\n\n$$\n\\begin{aligned}\n& \\frac{x^{2}+y z}{x^{2}+y z+1}+\\frac{y^{2}+z x}{y^{2}+z x+1}+\\frac{z^{2}+x y}{z^{2}+x y+1} \\leq 2 \\Leftrightarrow \\\\\n& \\frac{x^{2}+y z+1}{x^{2}+y z+1}+\\frac{y^{2}+z x+1}{y^{2}+z x+1}+\\frac{z^{2}+x y+1}{z^{2}+x y+1} \\leq 2+\\frac{1}{x^{2}+y z+1}+\\frac{1}{y^{2}+z x+1}+\\frac{1}{z^{2}+x y+1} \\Leftrightarrow \\\\\n& 3 \\leq 2+\\frac{1}{x^{2}+y z+1}+\\frac{1}{y^{2}+z x+1}+\\frac{1}{z^{2}+x y+1} \\Leftrightarrow \\\\\n& 1 \\leq \\frac{1}{x^{2}+y z+1}+\\frac{1}{y^{2}+z x+1}+\\frac{1}{z^{2}+x y+1} \\\\\n& \\frac{1}{x^{2}+y z+1}+\\frac{1}{y^{2}+z x+1}+\\frac{1}{z^{2}+x y+1} \\geq \\frac{9}{x^{2}+y z+1+y^{2}+z x+1+z^{2}+x y+1}= \\\\\n& \\frac{9}{x^{2}+y^{2}+z^{2}+x y+y z+z x+3} \\geq \\frac{9}{2 x^{2}+y^{2}+z^{2}+3}=1\n\\end{aligned}\n$$\n\nThe first inequality: AM-GM inequality (also can be achieved with Cauchy-BunjakowskiSchwarz inequality). The second inequality: $x y+y z+z x £ x^{2}+y^{2}+z^{2}$ (there are more ways to prove it, AM-GM, full squares etc.)\n\n## GEOMETRY", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "## A5 ", "solution_match": "## Solution:"}}
+{"year": "2015", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "MNE\n\nAround the triangle $A B C$ the circle is circumscribed, and at the vertex $C$ tangent $t$ to this circle is drawn. The line $p$ which is parallel to this tangent intersects the lines $B C$ and $A C$ at the points $D$ and $E$, respectively. Prove that the points $A, B, D, E$ belong to the same circle.", "solution": "Let $O$ be the center of a circumscribed circle $k$ of the triangle $A B C$, and let $F$ and $G$ be the points of intersection of the line $C O$ with the line $p$ and the circle $k$, respectively (see Figure). From $p \\| t$ it follows that $p \\perp C O$. Furthermore, $\\angle A B C=\\angle A G C$, because these angles are peripheral over the same chord. The quadrilateral $A G F E$ has two right angles at the vertices $A$ and $F$, and hence, $\\angle A E D+\\angle A B D=\\angle A E F+\\angle A G F=180^{\\circ}$. Hence, the quadrilateral $A B D E$ is cyclic, as asserted.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-06.jpg?height=406&width=483&top_left_y=2006&top_left_x=843)\n\nFigure", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "## G1 ", "solution_match": "## Solution:"}}
+{"year": "2015", "tier": "T3", "problem_label": "G2", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "MLD\n\nThe point $P$ is outside of the circle $\\Omega$. Two tangent lines, passing from the point $P$, touch the circle $\\Omega$ at the points $A$ and $B$. The median $A M, M \\in(B P)$, intersects the circle $\\Omega$ at the point $C$ and the line $P C$ intersects again the circle $\\Omega$ at the point $D$. Prove that the lines $A D$ and $B P$ are parallel.", "solution": "Since $\\angle B A C=\\angle B A M=\\angle M B C$, we have $\\triangle M A B \\cong \\triangle M B C$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-07.jpg?height=498&width=745&top_left_y=735&top_left_x=625)\n\nWe obtain $\\frac{M A}{M B}=\\frac{M B}{M C}=\\frac{A B}{B C}$. The equality $\\quad M B=M P$ implies $\\frac{M A}{M P}=\\frac{M P}{M C}$ and $\\angle P M C \\equiv \\angle P M A$ gives the relation $\\triangle P M A \\cong \\triangle C M P$. It follows that $\\angle B P D \\equiv \\angle M P C \\equiv \\angle M A P \\equiv \\angle C A P \\equiv \\angle C D A \\equiv \\angle P D A$. So, the lines $A D$ and $B P$ are parallel.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "## G2 ", "solution_match": "## Solution:"}}
+{"year": "2015", "tier": "T3", "problem_label": "G3", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "GRE\n\nLet $c \\equiv c(O, K)$ be a circle with center $O$ and radius $R$ and $A, B$ be two points on it, not belonging to the same diameter. The bisector of the angle $A \\hat{B} O$ intersects the circle $c$ at point $C$, the circumcircle of the triangle $A O B$, say ${ }^{c_{1}}$ at point $K$ and the circumcircle of the triangle $A O C$, say ${ }^{c_{2}}$, at point $L$. Prove that the point $K$ is the circumcenter of the triangle $A O C$ and the point $L$ is the incenter of the triangle $A O B$.", "solution": "The segments $O B, O C$ are equal, as radii of the circle ${ }^{c}$. Hence $O B C$ is an isosceles triangle and\n\n$$\n\\hat{B}_{1}=\\hat{C}_{1}=\\hat{x}\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-07.jpg?height=414&width=545&top_left_y=2097&top_left_x=714)\n\nThe chord $B C$ is the bisector of the angle $O \\hat{B} A$, and hence\n\n$$\n\\hat{B}_{1}=\\hat{B}_{2}=\\hat{x}\n$$\n\nThe angles $\\hat{B}_{2}$ and $\\hat{O}_{1}$ are inscribed to the same arc $O K$ of the circle ${ }^{c_{1}}$ and hence\n\n$$\n\\hat{B}_{2}=\\hat{\\mathrm{O}}_{1}=\\hat{x}\n$$\n\nThe segments $K O, K C$ are equal, as radii of the circle ${ }^{c_{2}}$. Hence the triangle $K O C$ is isosceles and so\n\n$$\n\\hat{O}_{2}=\\hat{C}_{1}=\\hat{x}\n$$\n\nFrom equalities $(1),(2),(3)$ we conclude that\n\n$$\n\\hat{O}_{1}=\\hat{O}_{2}=\\hat{x}\n$$\n\nand so $O K$ is the bisector, and hence perpendicular bisector of the isosceles triangle $O A C$. The point $K$ is the middle of the arc $O K$ (since $B K$ bisects the angle $O \\hat{B} A$ ). Hence the perpendicular bisector of the chord $A O$ of the circle ${ }^{c_{1}}$ is passing through point $K$. It means that $K$ is the circumcenter of the triangle $O A C$.\n\nFrom equalities (1),(2),(3) we conclude that $\\hat{B}_{2}=\\hat{C}_{1}=\\hat{x}$ and so $A B / / O C \\Rightarrow O \\hat{A} B=A \\hat{O} C$, that is $\\hat{A}_{1}+\\hat{A}_{2}=\\hat{O}_{1}+\\hat{O}_{2}$ and since $\\hat{O}_{1}=\\hat{O}_{2}=\\hat{x}$, we conclude that\n\n$$\n\\hat{A}_{1}+\\hat{A}_{2}=2 \\hat{O}_{1}=2 \\hat{x}\n$$\n\nThe angles $\\hat{A}_{I}$ and $\\hat{C}_{I}$ are inscribed into the circle ${ }^{c_{2}}$ and correspond to the same arc $O L$. Hence\n\n$$\n\\hat{A}_{I}=\\hat{C}_{1}=\\hat{x}\n$$\n\nFrom (5) and (6) we have $\\hat{A}_{1}=\\hat{A}_{2}$, i.e. $A L$ is the bisector of the angle $B \\hat{A} O$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "## G3 ", "solution_match": "## Solution:"}}
+{"year": "2015", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "CYP\n\nLet $\\triangle A B C$ be an acute triangle. The lines $\\left(\\varepsilon_{1}\\right),\\left(\\varepsilon_{2}\\right)$ are perpendicular to $A B$ at the points $A$, $B$, respectively. The perpendicular lines from the midpoint $M$ of $A B$ to the sides of the triangle $A C_{;} B C$ intersect the lines $\\left(\\varepsilon_{1}\\right),\\left(\\xi_{2}\\right)$ at the points $E, F$, respectively. If $I$ is the intersection point of $E F, M C$, prove that\n\n$$\n\\angle A I B=\\angle E M F=\\angle C A B+\\angle C B A\n$$", "solution": "Let $H, G$ be the points of intersection of $M E, M F$, with $A C, B C$ respectively. From the similarity of triangles $\\triangle M H A$ and $\\triangle M A E$ we get\n\n$$\n\\frac{M H}{M A}=\\frac{M A}{M E}\n$$\n\nthus, $M A^{2}=M H \\cdot M E$\n\nSimilarly, from the similarity of triangles $\\triangle M B G$ and $\\triangle M F B$ we get\n\n$$\n\\frac{M B}{M F}=\\frac{M G}{M B}\n$$\n\nthus, $M B^{2}=M F \\cdot M G$\n\nSince $M A=M B$, from (1), (2), we have that the points $E, H, G, F$ are concyclic.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-09.jpg?height=710&width=1025&top_left_y=298&top_left_x=298)\n\nTherefore, we get that $\\angle F E H=\\angle F E M=\\angle H G M$. Also, the quadrilateral $C H M G$ is cyclic, so $\\angle C M H=\\angle H G C$. We have\n\n$$\n\\angle F E H+\\angle C M H=\\angle H G M+\\angle H G C=90^{\\circ}\n$$\n\nThus $C M \\perp E F$. Now, from the cyclic quadrilaterals $F I M B$ and EIMA, we get that\n\n$\\angle I F M=\\angle I B M$ and $\\angle I E M=\\angle I A M$. Therefore, the triangles $\\triangle E M F$ and $\\triangle A I B$ are similar, so $\\angle A J B=\\angle E M F$. Finally\n\n$$\n\\angle A I B=\\angle A I M+\\angle M I B=\\angle A E M+\\angle M F B=\\angle C A B+\\angle C B A\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "\nG4", "solution_match": "## Solution:"}}
+{"year": "2015", "tier": "T3", "problem_label": "G5", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "ROU\n\nLet $A B C$ be an acute triangle with $A B \\neq A C$. The incircle $\\omega$ of the triangle touches the sides $B C, C A$ and $A B$ at $D, E$ and $F$, respectively. The perpendicular line erected at $C$ onto $B C$ meets $E F$ at $M$, and similarly, the perpendicular line erected at $B$ onto $B C$ meets $E F$ at $N$. The line $D M$ meets $\\omega$ again in $P$, and the line $D N$ meets $\\omega$ again at $Q$. Prove that $D P=D Q$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-09.jpg?height=583&width=737&top_left_y=1595&top_left_x=607)", "solution": "## Proof 1.1.\n\nLet $\\{T\\}=E F \\cap B C$. Applying Menelaus' theorem to the triangle $A B C$ and the transversal line $E-F-T$ we obtain $\\frac{T B}{T C} \\cdot \\frac{E C}{E A} \\cdot \\frac{F A}{F B}=1$, i.e. $\\frac{T B}{T C} \\cdot \\frac{s-c}{s-a} \\cdot \\frac{s-a}{s-b}=1$, or $\\frac{T B}{T C}=\\frac{s-b}{s-c}$, where the notations are the usual ones.\n\nThis means that triangles $T B N$ and $T C M$ are similar, therefore $\\frac{T B}{T C}=\\frac{B N}{C M}$. From the above it follows $\\frac{B N}{C M}=\\frac{s-b}{s-c}, \\frac{B D}{C D}=\\frac{s-b}{s-c}$, and $\\angle D B N=\\angle D C M=90^{\\circ}$, which means that triangles $B D N$ and $C D M$ are similar, hence angles $B D N$ and $C D M$ are equal. This leads to the arcs $D Q$ and $D P$ being equal, and finally to $D P=D Q$.\n\n## Proof 1.2.\n\nLet $S$ be the meeting point of the altitude from $A$ with the line $E F$. Lines $B N, A S, C M$ are parallel, therefore triangles $B N F$ and $A S F$ are similar, as are triangles $A S E$ and $C M E$. We obtain $\\frac{B N}{A S}=\\frac{B F}{F A}$ and $\\frac{A S}{C M}=\\frac{A E}{E C}$.\n\nMultiplying the two relations, we obtain $\\frac{B N}{C M}=\\frac{B F}{F A} \\cdot \\frac{A E}{E C}=\\frac{B F}{E C}=\\frac{B D}{D C}$ (we have used that $A E=A F, B F=B D$ and $C E=C D)$.\n\nIt follows that the right triangles $B D N$ and $C D M$ are similar (SAS), which leads to the same ending as in the first proof.\n\n## NUMBER THEORY", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "\nG5 ", "solution_match": "## Solution:"}}
+{"year": "2015", "tier": "T3", "problem_label": "NT1", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "SAU\n\nWhat is the greatest number of integers that can be selected from a set of 2015 consecutive numbers so that no sum of any two selected numbers is divisible by their difference?", "solution": "We take any two chosen numbers. If their difference is 1 , it is clear that their sum is divisible by their difference. If their difference is 2 , they will be of the same parity, and their sum is divisible by their difference. Therefore, the difference between any chosen numbers will be at least 3 . In other words, we can choose at most one number of any three consecutive numbers. This implies that we can choose at most 672 numbers.\n\nNow, we will show that we can choose 672 such numbers from any 2015 consecutive numbers. Suppose that these numbers are $a, a+1, \\ldots, a+2014$. If $a$ is divisible by 3 , we can choose $a+1, a+4, \\ldots, a+2014$. If $a$ is not divisible by 3 , we can choose $a, a+3, \\ldots, a+$ 2013.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "\nNT1 ", "solution_match": "## Solution:"}}
+{"year": "2015", "tier": "T3", "problem_label": "NT2", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "BUL\n\nA positive integer is called a repunit, if it is written only by ones. The repunit with $n$ digits will be denoted by $\\underbrace{11 \\ldots 1}_{n}$. Prove that:\n\na) the repunit $\\underbrace{11 \\ldots 1}_{n}$ is divisible by 37 if and only if $n$ is divisible by 3 ;\n\nb) there exists a positive integer $k$ such that the repunit $\\underbrace{11 \\ldots 1}_{n}$ is divisible by 41 if and only if $n$ is divisible by $k$.", "solution": "a) Let $n=3 m+r$, where $m$ and $r$ are non-negative integers and $r<3$.\n\nDenote by $\\underbrace{00 \\ldots 0}_{p}$ a recording with $p$ zeroes and $\\underbrace{a b c a b c \\ldots a b c}_{p x a b c c}$ recording with $p$ times $a b c$. We\n\nhave: $\\quad \\underbrace{11 \\ldots 1}_{n}=\\underbrace{11 \\ldots 1}_{3 m+r}=\\underbrace{11 \\ldots 1}_{3 m} \\cdot \\underbrace{00 \\ldots 0}_{r}+\\underbrace{11 \\ldots 1}_{\\tau}=111 \\cdot \\underbrace{100100 \\ldots 100100 \\ldots 0}_{(m-1) \\times 100}+\\underbrace{11 \\ldots 1}_{r}$.\n\nSince $111=37.3$, the numbers $\\underbrace{11 \\ldots 1}_{n}$ and $\\underbrace{11 \\ldots 1}_{r}$ are equal modulo 37 . On the other hand the numbers 1 and 11 are not divisible by 37 . We conclude that $\\underbrace{11 \\ldots 1}_{n}$ is divisible by 37 if only if $r=0$, i.e. if and only if $n$ is divisible by 3 .\n\nb) Using the idea from a), we look for a repunit, which is divisible by 41 . Obviously, 1 and 11 are not divisible by 41 , while the residues of 111 and 1111 are 29 and 4 , respectively. We have $11111=41 \\cdot 271$. Since 11111 is a repunit with 5 digits, it follows in the same way as in a) that $\\underbrace{1}_{11 \\ldots 1}$ is divisible by 41 if and only if $n$ is divisible by 5 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "\nNT2 ", "solution_match": "## Solution:"}}
+{"year": "2015", "tier": "T3", "problem_label": "NT3", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "ALB\n\na) Show that the product of all differences of possible couples of six given positive integers is divisible by 960 (original from Albania).\n\nb) Show that the product of all differences of possible couples of six given positive integers. is divisible by 34560 (modified by problem selecting committee).", "solution": "a) Since we have six numbers then at least two of them have a same residue when divided by 3 , so at least one of the differences in our product is divisible by 3 .\n\nSince we have six numbers then at least two of them have a same residuc when divided by 5 , so at least one of the differences in our product is divisible by 5 .\n\nWe may have:\n\na) six numbers with the same parity\n\nb) five numbers with the same parity\n\nc) four numbers with the same parity\n\nd) three numbers with the same parity\n\nThere are $C_{6}^{2}=15$ different pairs, so there are 15 different differences in this product.\n\na) The six numbers have the same parity; then each difference is divisible by 2 , therefore our product is divisible by $2^{15}$.\nb) If we have five numbers with the same parity, then the couples that have their difference odd are formed by taking one number from these five numbers, and the second will be the sixth one. Then $C_{5}^{1}=15=5$ differences are odd. So $15-5=10$ differences are even, so product is divisible by $2^{10}$.\n\nc) If we have four numbers with the same parity, then the couples that have their difference odd are formed by taking one number from these four numbers, and the second will be from the two others numbers. Then $2 \\cdot C_{4}^{1}=8$ differences are odd. So $15-8=7$ differences are even, so our product is divisible by $2^{7}$.\n\nd) If we have three numbers with the same parity, then the couples that have their difference odd are formed by taking one from each triple. Then $C_{3}^{1} \\cdot C_{3}^{1}=9$ differences are odd, therefore $15-9=6$ differences are even, so our product is divisible by $2^{6}$.\n\nThus, our production is divisible by $2^{6} \\cdot 3 \\cdot 5=960$.\n\nb) Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ be these numbers. Since we have six numbers then at least two of them when divided by 5 have the same residue, so at least one of these differences in our product is divisible by 5 .\n\nSince we have six numbers, and we have three possible residues at the division by 3 , then at least three of them replies the residue of previous numbers, so at least three of these differences in our product are divisible by 3 .\n\nSince we have six numbers, and we have two possible residues at the division by 2 , then at least four of them replies the residue of previous numbers, and two of them replies replied residues, so at least six of these differences in our product are divisible by 2 .\n\nSince we have six numbers, and we have four possible residues at the division by 4 , then at least two of them replies the residue of previous numbers, so at least two of these differences in our product are divisible by 4 . That means that two of these differences are divisible by 4 and moreover four of them are divisible by 2 .\n\nThus, our production is divisible by $2^{4} \\cdot 4^{2} \\cdot 3^{3} \\cdot 5=34560$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "## NT3 ", "solution_match": "## Solution:"}}
+{"year": "2015", "tier": "T3", "problem_label": "NT4", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "MLD\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-12.jpg?height=136&width=181&top_left_y=1553&top_left_x=907)\n\nFind all prime numbers $a, b, c$ and integers $k$ which satisfy the equation $a^{2}+b^{2}+16 \\cdot c^{2}=9 \\cdot k^{2}+1$.", "solution": "The relation $9 \\cdot k^{2}+1 \\equiv 1(\\bmod 3)$ implies\n\n$$\na^{2}+b^{2}+16 \\cdot c^{2} \\equiv 1(\\bmod 3) \\Leftrightarrow a^{2}+b^{2}+c^{2} \\equiv 1(\\bmod 3)\n$$\n\nSince $a^{2} \\equiv 0,1(\\bmod 3), b^{2} \\equiv 0,1(\\bmod 3), c^{2} \\equiv 0,1(\\bmod 3)$, we have:\n\n| $a^{2}$ | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| $b^{2}$ | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 |\n| $c^{2}$ | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 |\n| $a^{2}+b^{2}+c^{2}$ | 0 | 1 | 1 | 2 | 1 | 2 | 2 | 0 |\n\nFrom the previous table it follows that two of three prime numbers $a, b, c$ are equal to 3 .\n\nCase 1. $a=b=3$. We have\n\n$$\na^{2}+b^{2}+16 \\cdot c^{2}=9 \\cdot k^{2}+1 \\Leftrightarrow 9 \\cdot k^{2}-16 \\cdot c^{2}=17 \\Leftrightarrow(3 k-4 c) \\cdot(3 k+4 c)=17\n$$\n\nIf $\\left\\{\\begin{array}{l}3 k-4 c=1, \\\\ 3 k+4 c=17,\\end{array}\\right.$ then $\\left\\{\\begin{array}{l}c=2, \\\\ k=3,\\end{array}\\right.$ and $(a, b, c, k)=(3,3,2,3)$.\n\nIf $\\left\\{\\begin{array}{l}3 k-4 c=-17, \\\\ 3 k+4 c=-1,\\end{array}\\right.$ then $\\left\\{\\begin{array}{l}c=2, \\\\ k=-3,\\end{array}\\right.$ and $(a, b, c, k)=(3,3,2,-3)$.\n\nCase 2. $c=3$. If $\\left(3, b_{0}, c, k\\right)$ is a solution of the given equation, then $\\left(b_{0}, 3, c, k\\right)$ is a solution, too.\n\nLet $a=3$. We have\n\n$$\na^{2}+b^{2}+16 \\cdot c^{2}=9 \\cdot k^{2}+1 \\Leftrightarrow 9 \\cdot k^{2}-b^{2}=152 \\Leftrightarrow(3 k-b) \\cdot(3 k+b)=152 .\n$$\n\nBoth factors shall have the same parity and we obtain only 4 cases:\n\nIf $\\left\\{\\begin{array}{l}3 k-b=2, \\\\ 3 k+b=76,\\end{array}\\right.$ then $\\left\\{\\begin{array}{l}b=37, \\\\ k=13,\\end{array}\\right.$ and $(a, b, c, k)=(3,37,3,13)$.\n\nIf $\\left\\{\\begin{array}{l}3 k-b=4, \\\\ 3 k+b=38,\\end{array}\\right.$ then $\\left\\{\\begin{array}{l}b=17, \\\\ k=7,\\end{array}\\right.$ and $(a, b, c, k)=(3,17,3,7)$.\n\nIf $\\left\\{\\begin{array}{l}3 k-b=-76, \\\\ 3 k+b=-2,\\end{array}\\right.$, .\n\nIf $\\left\\{\\begin{array}{l}3 k-b=-38, \\\\ 3 k+b=-4,\\end{array}\\right.$ then $\\left\\{\\begin{array}{l}b=17, \\\\ k=-7,\\end{array}\\right.$ and $(a, b, c, k)=(3,17,3,-7)$.\n\nIn addition, $(a, b, c, k) \\in\\{(37,3,3,13),(17,3,3,7),(37,3,3,-13),(17,3,3,-7)\\}$.\n\nSo, the given equation has 10 solutions:\n\n$S=\\left\\{\\begin{array}{l}(37,3,3,13),(17,3,3,7),(37,3,3,-13),(17,3,3,-7),(3,37,3,13),(3,17,3,7),(3,37,3,-13), \\\\ (3,17,3,-7),(3,3,2,3),(3,3,2,-3)\\end{array}\\right\\}$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "\nNT4 ", "solution_match": "## Solution:"}}
+{"year": "2015", "tier": "T3", "problem_label": "NT5", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "MNE\n\nDoes there exist positive integers $a, b$ and a prime $p$ such that\n\n$$\na^{3}-b^{3}=4 p^{2} ?\n$$", "solution": "The given equality may be written as\n\n$$\n(a-b)\\left(a^{2}+a b+b^{2}\\right)=4 p^{2}\n$$\n\nSince $a-b<a^{2}+a b+b^{2}$, it follows from (1) that\n\n(2) $a-b<2 p$.\n\nNow consider two cases: 1. $p=2$, and 2. $p$ is an odd prime.\n\nCase 1: $p=2$. Then (1) becomes\n\n(3) $(a-b)\\left(a^{2}+a b+b^{2}\\right)=16$ :\n\nIn view of (2) and (3), it must be $a-b=1$ or $a-b=2$. If $a-b=1$, then substituting $a=b+1$\n\nin (3) we obtain\n\n$$\nb(b+1)=5\n$$\n\nwhich is impossible since $b(b+1)$ is an even integer.\n\nIf $a-b=2$, then substituting $a=b+2$ in (3) we get\n\n$$\n3 b(b+2)=4\n$$\n\nwhich is obviously impossible.\n\nCase 2: $\\mathrm{p}$ is an odd prime. Then (1) yields $a-b \\mid 4 p^{2}$. This together with the facts that\n\n$p$ is a prime and that by (2) $a-b<2 p$, yields $a-b \\in\\{1,2,4, p\\}$.\n\nIf $a-b=1$, then substituting $a=b+1$ in (1) we obtain\n\n$$\n3 b(b+1)+1=4 p^{2}\n$$\n\nwhich is impossible since $3 b(b+1)+1$ is an odd integer.\n\nIf $a-b=2$, then substituting $a=b+2$ in (1) we obtain\n\n$$\n3 b^{2}+6 b+4=2 p^{2}\n$$\n\nwhence it follows that\n\n$$\n2\\left(p^{2}-2\\right)=3\\left(b^{2}+2 b\\right) \\equiv 0(\\bmod 3)\n$$\n\nand hence\n\n$$\np^{2} \\equiv 2(\\bmod 3)\n$$\n\nSince $p^{2} \\equiv 1(\\bmod 3)$ for each odd prime $p>3$ and $3^{2} \\equiv 0(\\bmod 3)$, it follows that the congruence (5) is not satisfied for any odd prime $p$.\n\nIf $a-b=4$, then substituting $a=b+4$ in (1) we obtain\n\n$$\n3 b^{2}+12 b+16=p^{2}\n$$\n\nwhence it follows that $b$ is an odd integer such that\n\n$$\n3 b^{2} \\equiv p^{2}(\\bmod 4)\n$$\n\nwhence since $p^{2} \\equiv 1(\\bmod 4)$ for each odd prime $p$, we have\n\n$$\n3 b^{2} \\equiv 1(\\bmod 4)\n$$\n\nHowever, since $b^{2} \\equiv 1(\\bmod 4)$ for each odd integer $b$, it follows that the congruence (6) is not satisfied for any odd integer $b$.\n\nIf $a-b=p$, then substituting $a=b+p$ in (1) we obtain\n\n$$\np\\left(3 b^{2}+3 b p+p^{2}-4 p\\right)=0\n$$\n\ni.e.,\n\n(7)\n\n$$\n3 b^{2}+3 b p+p^{2}-4 p=0\n$$\n\nIf $p \\geq 5$, then $p^{2}-4 p>0$, and thus (7) cannot be satisfied for any positive integer $b$. If $p=3$, then $(7)$ becomes\n\n$$\n3\\left(b^{2}+3 b-1\\right)=0\n$$\n\nwhich is obviously not satisfied for any positive integer $b$.\n\nHence, there does not exist positive integers $a, b$ and a prime $p$ such that $a^{3}-b^{3}=4 p^{2}$.\n\n## COMBINATORICS", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "## NT5 ", "solution_match": "## Solution:"}}
+{"year": "2015", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "BUL\n\nA board $n \\times n(n \\geq 3)$ is divided into $n^{2}$ unit squares. Integers from 0 to $n$ included are written down: one integer in each unit square, in such a way that the sums of integers in each $2 \\times 2$ square of the board are different. Find all $n$ for which such boards exist.", "solution": "The number of the $2 \\times 2$ squares in a board $n \\times n$ is equal to $(n-1)^{2}$. All possible sums of the numbers in such squares are $0,1, \\ldots, 4 n$. A necessary condition for the existence of a board with the required property is $4 n+1 \\geq(n-1)^{2}$ and consequently $n(n-6) \\leq 0$. Thus $n \\leq 6$. The examples show the existence of boards $n \\times n$ for all $3 \\leq n \\leq 6$.\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-15.jpg?height=378&width=890&top_left_y=392&top_left_x=304)\n\n| 6 | 6 | 6 | 6 | 5 | 5 |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| 6 | 6 | 5 | 5 | 5 | 5 |\n| 1 | 2 | 3 | 4 | 4 | 5 |\n| 3 | 5 | 0 | 5 | 0 | 5 |\n| 1 | 0 | 2 | 1 | 0 | 0 |\n| 1 | 0 | 1 | 0 | 0 | 0 |\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-15.jpg?height=98&width=228&top_left_y=810&top_left_x=271)\n\n2015 points are given in a plane such that from any five points we can choose two points with distance less than 1 unit. Prove that 504 of the given points lie on a unit disc.\n\n## Solution:\n\nStart from an arbitrary point $A$ and draw a unit disc with center $A$. If all other points belong to this dise then we are done. Otherwise, take any point $B$ outside of the disc. Draw a unit disc with center $B$. If two drawn discs cover all 2015 points, by PHP, one of the discs contains at least 1008 points.\n\nSuppose that there is a point $C$ outside of the two drawn discs. Draw a unit disc with center $C$, If three drawn discs cover all 2015 points, by PHP, one of the discs contains at least 672 points.\n\nFinally, if there is a point $D$ outside of the three drawn discs, draw a unit disc with center $D$. By the given condition, any other point belongs to one of the four drawn discs. By PHP, one of the discs contains at least 504 points, concluding the solution.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "## C1 ", "solution_match": "## Solution:"}}
+{"year": "2015", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "ALB\n\nPositive integers are put into the following table\n\n| 1 | 3 | 6 | 10 | 15 | 21 | 28 | 36 |  |  |\n| ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |\n| 2 | 5 | 9 | 14 | 20 | 27 | 35 | 44 |  |  |\n| 4 | 8 | 13 | 19 | 26 | 34 | 43 | 53 |  |  |\n| 7 | 12 | 18 | 25 | 33 | 42 |  |  |  |  |\n| 11 | 17 | 24 | 32 | 41 |  |  |  |  |  |\n| 16 | 23 |  |  |  |  |  |  |  |  |\n| $\\ldots$ |  |  |  |  |  |  |  |  |  |\n| $\\ldots$ |  |  |  |  |  |  |  |  |  |\n\nFind the number of the line and column where the number 2015 stays.", "solution": "We shall observe straights lines as on the next picture. We can call these lines diagonals.\n\n| 1 | $\\sqrt{3}$ | 6 | 10 | 15 | 21 | 28 | 36 |  |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 2 | 5 | 9 | 14 | 20 | 27 | 35 | 44 |  |\n| 4 | 8 | 13 | 19 | 26 | 34 | 43 | 53 |  |\n|  | 12 | 18 | 25 | 33 | 42 |  |  |  |\n| 11 | 17 | 24 | 32 | 41 |  |  |  |  |\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-16.jpg?height=152&width=799&top_left_y=301&top_left_x=327)\n\nOn the first diagonal is number 1 .\n\nOn the second diagonal are two numbers: 2 and 3 .\n\nOn the 3rd diagonal are three numbers: 4,5 and 6\n\n.\n\nOn the $n$-th diagonal are $n$ numbers. These numbers are greater then $\\frac{(n-1) n}{2}$ and not greater than $\\frac{n(n+1)}{2}$ (see the next sentence!).\n\nOn the first $n$ diagonals are $1+2+3+\\ldots+n=\\frac{n(n+1)}{2}$ numbers.\n\nIf $m$ is in the $k$-th row $l$-th column and on the $n$-th diagonal, then it is $m=\\frac{(n-1) n}{2}+l$ and $n+1=k+l$. So, $m=\\frac{(k+l-2)(k+l-1)}{2}+l$.\n\nWe have to find such numbers $n, k$ and $l$ for which:\n\n$$\n\\begin{gathered}\n\\frac{(n-1) n}{2}<2015 \\leq \\frac{n(n+1)}{2} \\\\\nn+1=k+l \\\\\n2015=\\frac{(k+l-2)(k+l-1)}{2}+l\n\\end{gathered}\n$$\n\n(1), (2), (3) $\\Rightarrow n^{2}-n<4030 \\leq n^{2}+n \\Rightarrow n=63, k+l=64,2015=\\frac{(64-2)(64-1)}{2}+l \\Rightarrow$ $t=2015-31 \\cdot 63=62, k=64-62=2$\n\nTherefore 2015 is located in the second row and 62 -th column.\n\n## Solution 2:\n\nFirstly, we can see that the first elements of the columns are triangular numbers. If $a_{i}$ is the first element of the line $i$, we have $a_{i}=\\frac{i(i-1)}{2}$.\n\nThe second elements of the first row obtained by adding the first element 2 .\n\nThe second elements of the second row is obtained by adding the first element 3 .\n\nAnd so on, then the second element on the $n$-th row is obtained by adding the first element $n+1$.\n\nThen the third element of the $n$-th row is obtained by adding $n+2$, and the $k$-th element of it is obtained by adding $k$.\n\nSince the first element of the n-th row is $\\frac{(n-1) n}{2}+1$, the second one is\n\n$\\frac{(n-1) n}{2}+1+(n+1)=\\frac{n(n+1)}{2}+2$.\n\nThe third one $\\frac{n(n+1)}{2}+1+(n+2)=\\frac{(n+1)(n+2)}{2}+3$ so the $\\mathrm{k}$-th one should be $\\frac{(n+k-2)(n+k-1)}{2}+k$\n\n$$\n\\frac{(n+k-2)(n+k-1)}{2}+k=2015 \\Leftrightarrow n^{2}+n(2 k-3)+k^{2}-k-4028=0\n$$\n\nTo have a positive integer solution $(2 k-3)^{2}-4\\left(k^{2}-k-4028\\right)=16121-8 k$ must be a perfect square.\n\nFrom $16121-8 k=x^{2}$, it is noticed that the maximum of $x$ is 126 (since $\\mathrm{k}>0$ ).\n\nSimultaneously can be seen that $\\mathrm{x}$ is odd, so $x \\leq 125$.\n\n$16121-8 k=x^{2} \\Leftrightarrow 125^{2}+496-8 k=x^{2}$\n\nSo $496-8 k=0$, form that $k=62$.\n\nFrom that we can find $n=2$.\n\nSo 2015 is located on the second row and 62-th column.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "## C3 ", "solution_match": "## Solution 1:"}}
+{"year": "2015", "tier": "T3", "problem_label": "C4", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "GRE\n\nLet $n \\geq 1$ be a positive integer. A square of side length $n$ is divided by lines parallel to each side into $n^{2}$ squares of side length 1 . Find the number of parallelograms which have vertices among the vertices of the $n^{2}$ squares of side length 1 , with both sides smaller or equal to 2 , and which have the area equal to 2 .", "solution": "We can divide all these parallelograms into 7 classes (types I-VII), according to Figure.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-17.jpg?height=632&width=1277&top_left_y=1207&top_left_x=291)\n\nType 1: There are $n$ ways to choose the strip for the horizontal (shorter) side of the parallelogram, and $(n-1)$ ways to choose the strip (of the width 2 ) for the vertical (longer) side. So there are $n(n-1)$ parallelograms of the type I.\n\nType II: There are $(n-1)$ ways to choose the strip (of the width 2 ) for the horizontal (longer) side, and $n$ ways to choose the strip for the vertical (shorter) side. So the number of the parallelogram of this type is also $n(n-1)$.\n\nType III: Each parallelogram of this type is a square inscribed in a unique square $2 \\times 2$ of our grid. The number of such squares is $(n-1)^{2}$. So there are $(n-1)^{2}$ parallelograms of type III. For each of the types IV, V, VI, VII, the strip of the width 1 in which the parallelogram is located can be chosen in $n$ ways and for each such choice there are $n-2$ parallelograms located in the chosen strip.\n\nSumming we obtain that the total number of parallelograms is:\n\n$$\n2 n(n-1)+(n-1)^{2}+4 n(n-2)=7 n^{2}-12 n+1\n$$\n\n(C5) CYP\n\nWe have a $5 \\times 5$ chessboard and a supply of $\\mathrm{L}$-shaped triominoes, i.e. $2 \\times 2$ squares with one comer missing. Two players $A$ and $B$ play the following game: A positive integer $k \\leq 25$ is chosen. Starting with $A$, the players take alternating turns marking squares of the chessboard until they mark a total of $k$ squares. (In each turn a player has to mark exactly one new square.)\n\nAt the end of the process, player $A$ wins if he can cover without overlapping all but at most 2 unmarked squares with $\\mathrm{L}$-shaped triominoes, otherwise player $\\boldsymbol{B}$ wins. It is not permitted any\nmarked squares to be covered.\n\nFind the smallest $\\boldsymbol{k}$, if it exists, such that player $\\boldsymbol{B}$ has a winning strategy.\n\n## Solution:\n\nWe will show that player $A$ wins if $k=1,2$ or 3 , but player $B$ wins if $k=4$. Thus the smallest $k$ for which $B$ has a winning strategy exists and is equal to 4 .\n\nIf $k=1$, player $A$ marks the upper left corner of the square and then fills it as follows.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-18.jpg?height=575&width=572&top_left_y=1114&top_left_x=798)\n\nIf $k=2$, player $A$ marks the upper left comer of the square. Whatever square player $B$ marks, then player $\\boldsymbol{A}$ can fill in the square in exactly the same pattern as above except that he doesn't put the triomino which covers the marked square of $B$. Player $A$ wins because he has left only two unmarked squares uncovered.\n\nFor $k=3$, player $\\boldsymbol{A}$ wins by following the same strategy. When he has to mark a square for the second time, he marks any yet unmarked square of the triomino that covers the marked square of $B$.\n\nLet us now show that for $k=4$ player $B$ has a wimning strategy. Since there will be 21 unmarked squares, player $\\boldsymbol{A}$ will need to cover all of them with seven L-shaped triominoes. We can assume that in his first move, player $\\boldsymbol{A}$ does not mark any square in the bottom two rows of the chessboard (otherwise just rotate the chessboard). In his first move player $\\boldsymbol{B}$ marks the square labeled 1 in the following figure.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-19.jpg?height=550&width=564&top_left_y=300&top_left_x=737)\n\nIf player $\\boldsymbol{A}$ in his next move does not mark any of the squares labeled 2,3 and 4 then player $\\boldsymbol{B}$ marks the square labeled 3. Player $\\boldsymbol{B}$ wins as the square labeled 2 is left unmarked but cannot be covered with an L-shaped triomino.\n\nIf player $\\boldsymbol{A}$ in his next move marks the square labeled 2, then player $\\boldsymbol{B}$ marks the square labeled 5 . Player $B$ wins as the square labeled 3 is left unmarked but cannot be covered with an L-shaped triomino.\n\nFinally, if player $\\boldsymbol{A}$ in his next move marks one of the squares labeled 3 or 4 , player $\\boldsymbol{B}$ marks the other of these two squares. Player $\\boldsymbol{B}$ wins as the square labeled 2 is left unmarked but cannot be covered with an L-shaped triomino.\n\nSince we have covered all possible cases, player $B$ wins when $k=4$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl", "problem_match": "\nC4", "solution_match": "\nSolution:"}}

JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl

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-{"year": "2016", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $a, b, c$ be positive real numbers such that $a b c=8$. Prove that\n\n$$\n\\frac{a b+4}{a+2}+\\frac{b c+4}{b+2}+\\frac{c a+4}{c+2} \\geq 6\n$$", "solution": "We have $a b+4=\\frac{8}{c}+4=\\frac{4(c+2)}{c}$ and similarly $b c+4=\\frac{4(a+2)}{a}$ and $c a+4=\\frac{4(b+2)}{b}$. It follows that\n\n$$\n(a b+4)(b c+4)(c a+4)=\\frac{64}{a b c}(a+2)(b+2)(c+2)=8(a+2)(b+2)(c+2)\n$$\n\nso that\n\n$$\n\\frac{(a b+4)(b c+4)(c a+4)}{(a+2)(b+2)(c+2)}=8\n$$\n\nApplying AM-GM, we conclude:\n\n$$\n\\frac{a b+4}{a+2}+\\frac{b c+4}{b+2}+\\frac{c a+4}{c+2} \\geq 3 \\cdot \\sqrt[3]{\\frac{(a b+4)(b c+4)(c a+4)}{(a+2)(b+2)(c+2)}}=6\n$$\n\nAlternatively, we can write LHS as\n\n$$\n\\frac{b c(a b+4)}{2(b c+4)}+\\frac{a c(b c+4)}{2(a c+4)}+\\frac{a b(c a+4)}{2(a b+4)}\n$$\n\nand then apply AM-GM.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nA1.", "solution_match": "\nSolution."}}
-{"year": "2016", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO", "problem": "Given positive real numbers $a, b, c$, prove that\n\n$$\n\\frac{8}{(a+b)^{2}+4 a b c}+\\frac{8}{(a+c)^{2}+4 a b c}+\\frac{8}{(b+c)^{2}+4 a b c}+a^{2}+b^{2}+c^{2} \\geq \\frac{8}{a+3}+\\frac{8}{b+3}+\\frac{8}{c+3}\n$$", "solution": "Since $2 a b \\leq a^{2}+b^{2}$, it follows that $(a+b)^{2} \\leq 2\\left(a^{2}+b^{2}\\right)$ and $4 a b c \\leq 2 c\\left(a^{2}+b^{2}\\right)$, for any positive reals $a, b, c$. Adding these inequalities, we find\n\n$$\n(a+b)^{2}+4 a b c \\leq 2\\left(a^{2}+b^{2}\\right)(c+1)\n$$\n\nso that\n\n$$\n\\frac{8}{(a+b)^{2}+4 a b c} \\geq \\frac{4}{\\left(a^{2}+b^{2}\\right)(c+1)}\n$$\n\nUsing the AM-GM inequality, we have\n\n$$\n\\frac{4}{\\left(a^{2}+b^{2}\\right)(c+1)}+\\frac{a^{2}+b^{2}}{2} \\geq 2 \\sqrt{\\frac{2}{c+1}}=\\frac{4}{\\sqrt{2(c+1)}}\n$$\n\nrespectively\n\n$$\n\\frac{c+3}{8}=\\frac{(c+1)+2}{8} \\geq \\frac{\\sqrt{2(c+1)}}{4}\n$$\n\nWe conclude that\n\n$$\n\\frac{4}{\\left(a^{2}+b^{2}\\right)(c+1)}+\\frac{a^{2}+b^{2}}{2} \\geq \\frac{8}{c+3}\n$$\n\nand finally\n\n$\\frac{8}{(a+b)^{2}+4 a b c}+\\frac{8}{(a+c)^{2}+4 a b c}+\\frac{8}{(b+c)^{2}+4 a b c}+a^{2}+b^{2}+c^{2} \\geq \\frac{8}{a+3}+\\frac{8}{b+3}+\\frac{8}{c+3}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nA2.", "solution_match": "\nSolution."}}
-{"year": "2016", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO", "problem": "Determine the number of pairs of integers $(m, n)$ such that\n\n$$\n\\sqrt{n+\\sqrt{2016}}+\\sqrt{m-\\sqrt{2016}} \\in \\mathbb{Q}\n$$", "solution": "Let $r=\\sqrt{n+\\sqrt{2016}}+\\sqrt{m-\\sqrt{2016}}$. Then\n\n$$\nn+m+2 \\sqrt{n+\\sqrt{2016}} \\cdot \\sqrt{m-\\sqrt{2016}}=r^{2}\n$$\n\nand\n\n$$\n(m-n) \\sqrt{2106}=\\frac{1}{4}\\left(r^{2}-m-n\\right)^{2}-m n+2016 \\in \\mathbb{Q}\n$$\n\nSince $\\sqrt{2016} \\notin \\mathbb{Q}$, it follows that $m=n$. Then\n\n$$\n\\sqrt{n^{2}-2016}=\\frac{1}{2}\\left(r^{2}-2 n\\right) \\in \\mathbb{Q}\n$$\n\nHence, there is some nonnegative integer $p$ such that $n^{2}-2016=p^{2}$ and (1) becomes $2 n+2 p=r^{2}$.\n\nIt follows that $2(n+p)=r^{2}$ is the square of a rational and also an integer, hence a perfect square. On the other hand, $2016=(n-p)(n+p)$ and $n+p$ is a divisor of 2016, larger than $\\sqrt{2016}$. Since $n+p$ is even, so is also $n-p$, and $r^{2}=2(n+p)$ is a divisor of $2016=2^{5} \\cdot 3^{2} \\cdot 7$, larger than $2 \\sqrt{2016}>88$. The only possibility is $r^{2}=2^{4} \\cdot 3^{2}=12^{2}$. Hence, $n+p=72$ and $n-p=28$, and we conclude that $n=m=50$. Thus, there is only one such pair.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nA3.", "solution_match": "\nSolution."}}
-{"year": "2016", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "exam": "JBMO", "problem": "If $x, y, z$ are non-negative real numbers such that $x^{2}+y^{2}+z^{2}=x+y+z$, then show that:\n\n$$\n\\frac{x+1}{\\sqrt{x^{5}+x+1}}+\\frac{y+1}{\\sqrt{y^{5}+y+1}}+\\frac{z+1}{\\sqrt{z^{5}+z+1}} \\geq 3\n$$\n\nWhen does the equality hold?", "solution": "First we factor $x^{5}+x+1$ as follows:\n\n$$\n\\begin{aligned}\nx^{5}+x+1 & =x^{5}-x^{2}+x^{2}+x+1=x^{2}\\left(x^{3}-1\\right)+x^{2}+x+1=x^{2}(x-1)\\left(x^{2}+x+1\\right)+x^{2}+x+1 \\\\\n& =\\left(x^{2}+x+1\\right)\\left(x^{2}(x-1)+1\\right)=\\left(x^{2}+x+1\\right)\\left(x^{3}-x^{2}+1\\right)\n\\end{aligned}\n$$\n\nUsing the $A M-G M$ inequality, we have\n\n$$\n\\sqrt{x^{5}+x+1}=\\sqrt{\\left(x^{2}+x+1\\right)\\left(x^{3}-x^{2}+1\\right)} \\leq \\frac{x^{2}+x+1+x^{3}-x^{2}+1}{2}=\\frac{x^{3}+x+2}{2}\n$$\n\nand since\n\n$x^{3}+x+2=x^{3}+1+x+1=(x+1)\\left(x^{2}-x+1\\right)+x+1=(x+1)\\left(x^{2}-x+1+1\\right)=(x+1)\\left(x^{2}-x+2\\right)$,\n\nthen\n\n$$\n\\sqrt{x^{5}+x+1} \\leq \\frac{(x+1)\\left(x^{2}-x+2\\right)}{2}\n$$\n\nUsing $x^{2}-x+2=\\left(x-\\frac{1}{2}\\right)^{2}+\\frac{7}{4}>0$, we obtain $\\frac{x+1}{\\sqrt{x^{5}+x+1}} \\geq \\frac{2}{x^{2}-x+2}$ Applying the CauchySchwarz inequality and the given condition, we get\n\n$$\n\\sum_{c y c} \\frac{x+1}{\\sqrt{x^{5}+x+1}} \\geq \\sum_{c y c} \\frac{2}{x^{2}-x+2} \\geq \\frac{18}{\\sum_{c y c}\\left(x^{2}-x+2\\right)}=\\frac{18}{6}=3\n$$\n\nwhich is the desired result.\n\nFor the equality both conditions: $x^{2}-x+2=y^{2}-y+2=z^{2}-z+2$ (equality in CBS) and $x^{3}-x^{2}+1=x^{2}+x+1$ (equality in AM-GM) have to be satisfied.\n\nBy using the given condition it follows that $x^{2}-x+2+y^{2}-y+2+z^{2}-z+2=6$, hence $3\\left(x^{2}-x+2\\right)=6$, implying $x=0$ or $x=1$. Of these, only $x=0$ satisfies the second condition. We conclude that equality can only hold for $x=y=z=0$.\n\nIt is an immediate check that indeed for these values equality holds.\n\n## Alternative solution\n\nLet us present an heuristic argument to reach the key inequality $\\frac{x+1}{\\sqrt{x^{5}+x+1}} \\geq \\frac{2}{x^{2}-x+2}$.\n\nIn order to exploit the condition $x^{2}+y^{2}+z^{2}=x+y+z$ when applying CBS in Engel form, we are looking for $\\alpha, \\beta, \\gamma>0$ such that\n\n$$\n\\frac{x+1}{\\sqrt{x^{5}+x+1}} \\geq \\frac{\\gamma}{\\alpha\\left(x^{2}-x\\right)+\\beta}\n$$\n\nAfter squaring and cancelling the denominators, we get\n\n$$\n(x+1)^{2}\\left(\\alpha\\left(x^{2}-x\\right)+\\beta\\right)^{2} \\geq \\gamma^{2}\\left(x^{5}+x+1\\right)\n$$\n\nfor all $x \\geq 0$, and, after some manipulations, we reach to $f(x) \\geq 0$ for all $x \\geq 0$, where $f(x)=\\alpha^{2} x^{6}-\\gamma^{2} x^{5}+\\left(2 \\alpha \\beta-2 \\alpha^{2}\\right) x^{4}+2 \\alpha \\beta x^{3}+(\\alpha-\\beta)^{2} x^{2}+\\left(2 \\beta^{2}-2 \\alpha \\beta-\\gamma^{2}\\right) x+\\beta^{2}-\\gamma^{2}$.\n\nAs we are expecting the equality to hold for $x=0$, we naturally impose the condition that $f$ has 0 as a double root. This implies $\\beta^{2}-\\gamma^{2}=0$ and $2 \\beta^{2}-2 \\alpha \\beta-\\gamma^{2}=0$, that is, $\\beta=\\gamma$ and $\\gamma=2 \\alpha$.\n\nThus the inequality $f(x) \\geq 0$ becomes\n\n$$\n\\alpha^{2} x^{6}-4 \\alpha^{2} x^{5}+2 \\alpha^{2} x^{4}+4 \\alpha^{2} x^{3}+\\alpha^{2} x^{2} \\geq 0, \\forall x \\geq 0\n$$\n\nthat is,\n\n$$\n\\alpha^{2} x^{2}\\left(x^{2}-2 x-1\\right)^{2} \\geq 0 \\forall x \\geq 0\n$$\n\nwhich is obviously true.\n\nTherefore, the inequality $\\frac{x+1}{\\sqrt{x^{5}+x+1}} \\geq \\frac{2}{x^{2}-x+2}$ holds for all $x \\geq 0$ and now we can continue as in the first solution.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nA4.", "solution_match": "\nSolution."}}
-{"year": "2016", "tier": "T3", "problem_label": "A5", "problem_type": "Algebra", "exam": "JBMO", "problem": "Let $x, y, z$ be positive real numbers such that $x+y+z=\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}$.\n\na) Prove the inequality\n\n$$\nx+y+z \\geq \\sqrt{\\frac{x y+1}{2}}+\\sqrt{\\frac{y z+1}{2}}+\\sqrt{\\frac{z x+1}{2}}\n$$\n\nb) (Added by the problem selecting committee) When does the equality hold?", "solution": "a) We rewrite the inequality as\n\n$$\n(\\sqrt{x y+1}+\\sqrt{y z+1}+\\sqrt{z x+1})^{2} \\leq 2 \\cdot(x+y+z)^{2}\n$$\n\nand note that, from CBS,\n\n$$\n\\text { LHS } \\leq\\left(\\frac{x y+1}{x}+\\frac{y z+1}{y}+\\frac{z x+1}{z}\\right)(x+y+z)\n$$\n\nBut\n\n$$\n\\frac{x y+1}{x}+\\frac{y z+1}{y}+\\frac{z x+1}{z}=x+y+z+\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=2(x+y+z)\n$$\n\nwhich proves (1).\n\nb) The equality occurs when we have equality in CBS, i.e. when\n\n$$\n\\frac{x y+1}{x^{2}}=\\frac{y z+1}{y^{2}}=\\frac{z x+1}{z^{2}}\\left(=\\frac{x y+y z+z x+3}{x^{2}+y^{2}+z^{2}}\\right)\n$$\n\nSince we can also write\n\n$$\n(\\sqrt{x y+1}+\\sqrt{y z+1}+\\sqrt{z x+1})^{2} \\leq\\left(\\frac{x y+1}{y}+\\frac{y z+1}{z}+\\frac{z x+1}{x}\\right)(y+z+x)=2(x+y+z)^{2}\n$$\n\nthe equality implies also\n\n$$\n\\frac{x y+1}{y^{2}}=\\frac{y z+1}{z^{2}}=\\frac{z x+1}{x^{2}}\\left(=\\frac{x y+y z+z x+3}{x^{2}+y^{2}+z^{2}}\\right)\n$$\n\nBut then $x=y=z$, and since $x+y+z=\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}$, we conclude that $x=\\frac{1}{x}=1=y=z$.\n\nAlternative solution to $b$ ): The equality condition\n\n$$\n\\frac{x y+1}{x^{2}}=\\frac{y z+1}{y^{2}}=\\frac{z x+1}{z^{2}}\n$$\n\ncan be rewritten as\n\n$$\n\\frac{y+\\frac{1}{x}}{x}=\\frac{z+\\frac{1}{y}}{y}=\\frac{x+\\frac{1}{z}}{z}=\\frac{x+y+z+\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}}{x+y+z}=2\n$$\n\nand thus we obtain the system:\n\n$$\n\\left\\{\\begin{array}{l}\ny=2 x-\\frac{1}{x} \\\\\nz=2 y-\\frac{1}{y} \\\\\nx=2 z-\\frac{1}{z}\n\\end{array}\\right.\n$$\n\nWe show that $x=y=z$. Indeed, if for example $x>y$, then $2 x-\\frac{1}{x}>2 y-\\frac{1}{y}$, that is, $y>z$ and $z=2 y-\\frac{1}{y}>2 z-\\frac{1}{z}=x$, and we obtain the contradiction $x>y>z>x$. Similarly, if $x<y$, we obtain $x<y<z<x$.\n\nHence, the numbers are equal, and as above we get $x=y=z=1$.\n\n## Chapter 2\n\n## Combinatorics", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nA5.", "solution_match": "## Solution."}}
-{"year": "2016", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Let $S_{n}$ be the sum of reciprocal values of non-zero digits of all positive integers up to (and including) $n$. For instance, $S_{13}=\\frac{1}{1}+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}+\\frac{1}{6}+\\frac{1}{7}+\\frac{1}{8}+\\frac{1}{9}+\\frac{1}{1}+\\frac{1}{1}+\\frac{1}{1}+\\frac{1}{1}+\\frac{1}{2}+\\frac{1}{1}+\\frac{1}{3}$. Find the least positive integer $k$ making the number $k!\\cdot S_{2016}$ an integer.", "solution": "We will first calculate $S_{999}$, then $S_{1999}-S_{999}$, and then $S_{2016}-S_{1999}$.\n\nWriting the integers from 1 to 999 as 001 to 999, adding eventually also 000 (since 0 digits actually do not matter), each digit appears exactly 100 times in each position(as unit, ten, or hundred). Hence\n\n$$\nS_{999}=300 \\cdot\\left(\\frac{1}{1}+\\frac{1}{2}+\\cdots+\\frac{1}{9}\\right)\n$$\n\nFor the numbers in the interval $1000 \\rightarrow 1999$, compared to $0 \\rightarrow 999$, there are precisely 1000 more digits 1 . We get\n\n$$\nS_{1999}-S_{999}=1000+S_{999} \\Longrightarrow S_{1999}=1000+600 \\cdot\\left(\\frac{1}{1}+\\frac{1}{2}+\\cdots+\\frac{1}{9}\\right)\n$$\n\nFinally, in the interval $2000 \\rightarrow 2016$, the digit 1 appears twice as unit and seven times as a ten, the digit 2 twice as a unit and 17 times as a thousand, the digits $3,4,5$, and 6 each appear exactly twice as units, and the digits 7,8 , and 9 each appear exactly once as a unit. Hence\n\n$$\nS_{2016}-S_{1999}=9 \\cdot 1+19 \\cdot \\frac{1}{2}+2 \\cdot\\left(\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}+\\frac{1}{6}\\right)+1 \\cdot\\left(\\frac{1}{7}+\\frac{1}{8}+\\frac{1}{9}\\right)\n$$\n\nIn the end, we get\n\n$$\n\\begin{aligned}\nS_{2016} & =1609 \\cdot 1+619 \\cdot \\frac{1}{2}+602 \\cdot\\left(\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}+\\frac{1}{6}\\right)+601 \\cdot\\left(\\frac{1}{7}+\\frac{1}{8}+\\frac{1}{9}\\right) \\\\\n& =m+\\frac{1}{2}+\\frac{2}{3}+\\frac{2}{4}+\\frac{2}{5}+\\frac{2}{6}+\\frac{6}{7}+\\frac{1}{8}+\\frac{7}{9}=n+\\frac{p}{2^{3} \\cdot 3^{2} \\cdot 5 \\cdot 7}\n\\end{aligned}\n$$\n\nwhere $m, n$, and $p$ are positive integers, $p$ coprime to $2^{3} \\cdot 3^{2} \\cdot 5 \\cdot 7$. Then $k!\\cdot S_{2016}$ is an integer precisely when $k$ ! is a multiple of $2^{3} \\cdot 3^{2} \\cdot 5 \\cdot 7$. Since $7 \\mid k!$, it follows that $k \\geq 7$. Also, $7!=2^{4} \\cdot 3^{2} \\cdot 5 \\cdot 7$, implying that the least $k$ satisfying $k!\\cdot S_{2016} \\in \\mathbb{Z}$ is $k=7$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nC1.", "solution_match": "## Solution."}}
-{"year": "2016", "tier": "T3", "problem_label": "C2", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "The natural numbers from 1 to 50 are written down on the blackboard. At least how many of them should be deleted, in order that the sum of any two of the remaining numbers is not a prime?", "solution": "Notice that if the odd, respectively even, numbers are all deleted, then the sum of any two remaining numbers is even and exceeds 2 , so it is certainly not a prime. We prove that 25 is the minimal number of deleted numbers. To this end, we group the positive integers from 1 to 50 in 25 pairs, such that the sum of the numbers within each pair is a prime:\n\n$$\n\\begin{aligned}\n& (1,2),(3,4),(5,6),(7,10),(8,9),(11,12),(13,16),(14,15),(17,20) \\\\\n& (18,19),(21,22),(23,24),(25,28),(26,27),(29,30),(31,36),(32,35) \\\\\n& (33,34),(37,42),(38,41),(39,40),(43,46),(44,45),(47,50),(48,49)\n\\end{aligned}\n$$\n\nSince at least one number from each pair has to be deleted, the minimal number is 25 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nC2.", "solution_match": "\nSolution."}}
-{"year": "2016", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "Consider any four pairwise distinct real numbers and write one of these numbers in each cell of a $5 \\times 5$ array so that each number occurs exactly once in every $2 \\times 2$ subarray. The sum over all entries of the array is called the total sum of that array. Determine the maximum number of distinct total sums that may be obtained in this way.", "solution": "We will prove that the maximum number of total sums is 60 .\n\nThe proof is based on the following claim.\n\nClaim. Either each row contains exactly two of the numbers, or each column contains exactly two of the numbers.\n\nProof of the Claim. Indeed, let $R$ be a row containing at least three of the numbers. Then, in row $R$ we can find three of the numbers in consecutive positions, let $x, y, z$ be the numbers in consecutive positions(where $\\{x, y, s, z\\}=\\{a, b, c, d\\}$ ). Due to our hypothesis that in every $2 \\times 2$ subarray each number is used exactly once, in the row above $\\mathrm{R}$ (if there is such a row), precisely above the numbers $x, y, z$ will be the numbers $z, t, x$ in this order. And above them will be the numbers $x, y, z$ in this order. The same happens in the rows below $R$ (see at the following figure).\n\n$$\n\\left(\\begin{array}{lllll}\n\\bullet & x & y & z & \\bullet \\\\\n\\bullet & z & t & x & \\bullet \\\\\n\\bullet & x & y & z & \\bullet \\\\\n\\bullet & z & t & x & \\bullet \\\\\n\\bullet & x & y & z & \\bullet\n\\end{array}\\right)\n$$\n\nCompleting all the array, it easily follows that each column contains exactly two of the numbers and our claim has been proven.\n\nRotating the matrix (if it is necessary), we may assume that each row contains exactly two of the numbers. If we forget the first row and column from the array, we obtain a $4 \\times 4$ array, that can be divided into four $2 \\times 2$ subarrays, containing thus each number exactly four times, with a total sum of $4(a+b+c+d)$. It suffices to find how many different ways are there to put the numbers in the first row $R_{1}$ and the first column $C_{1}$.\n\nDenoting by $a_{1}, b_{1}, c_{1}, d_{1}$ the number of appearances of $a, b, c$, and respectively $d$ in $R_{1}$ and $C_{1}$, the total sum of the numbers in the entire $5 \\times 5$ array will be\n\n$$\nS=4(a+b+c+d)+a_{1} \\cdot a+b_{1} \\cdot b+c_{1} \\cdot c+d_{1} \\cdot d\n$$\n\nIf the first, the third and the fifth row contain the numbers $x, y$, with $x$ denoting the number at the entry $(1,1)$, then the second and the fourth row will contain only the numbers $z, t$, with $z$ denoting the number at the entry $(2,1)$. Then $x_{1}+y_{1}=7$ and $x_{1} \\geqslant 3$, $y_{1} \\geqslant 2, z_{1}+t_{1}=2$, and $z_{1} \\geqslant t_{1}$. Then $\\left\\{x_{1}, y_{1}\\right\\}=\\{5,2\\}$ or $\\left\\{x_{1}, y_{1}\\right\\}=\\{4,3\\}$, respectively $\\left\\{z_{1}, t_{1}\\right\\}=\\{2,0\\}$ or $\\left\\{z_{1}, t_{1}\\right\\}=\\{1,1\\}$. Then $\\left(a_{1}, b_{1}, c_{1}, d_{1}\\right)$ is obtained by permuting one of the following quadruples:\n\n$$\n(5,2,2,0),(5,2,1,1),(4,3,2,0),(4,3,1,1)\n$$\n\nThere are a total of $\\frac{4!}{2!}=12$ permutations of $(5,2,2,0)$, also 12 permutations of $(5,2,1,1)$, 24 permutations of $(4,3,2,0)$ and finally, there are 12 permutations of $(4,3,1,1)$. Hence, there are at most 60 different possible total sums.\n\nWe can obtain indeed each of these 60 combinations: take three rows ababa alternating\nwith two rows $c d c d c$ to get $(5,2,2,0)$; take three rows ababa alternating with one row $c d c d c$ and a row $(d c d c d)$ to get $(5,2,1,1)$; take three rows $a b a b c$ alternating with two rows $c d c d a$ to get $(4,3,2,0)$; take three rows abcda alternating with two rows $c d a b c$ to get $(4,3,1,1)$. By choosing for example $a=10^{3}, b=10^{2}, c=10, d=1$, we can make all these sums different. Hence, 60 is indeed the maximum possible number of different sums.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nC3.", "solution_match": "\nSolution."}}
-{"year": "2016", "tier": "T3", "problem_label": "C4", "problem_type": "Combinatorics", "exam": "JBMO", "problem": "A splitting of a planar polygon is a finite set of triangles whose interiors are pairwise disjoint, and whose union is the polygon in question. Given an integer $n \\geq 3$, determine the largest integer $m$ such that no planar $n$-gon splits into less than $m$ triangles.", "solution": "The required maximum is $\\lceil n / 3\\rceil$, the least integer greater than or equal to $n / 3$. To describe a planar $n$-gon splitting into this many triangles, write $n=3 m-r$, where $m$ is a positive integer and $r=0,1,2$, and consider $m$ coplanar equilateral triangles $A_{3 i} A_{3 i+1} A_{3 i+2}$, $i=0, \\ldots, m-1$, where the $A_{i}$ are pairwise distinct, the $A$ 's of rank congruent to 0 or 2 modulo 3 are all collinear, $A_{2}, A_{3}, A_{5}, \\ldots, A_{3 m-3}, A_{3 m-1}, A_{0}$, in order, and the line $A_{0} A_{2}$ separates $A_{1}$ from the remaining $A$ 's of rank congruent to 1 modulo 3 . The polygon $A_{0} A_{1} A_{2} A_{3} A_{4} A_{5} \\ldots A_{3 m-3} A_{3 m-2} A_{3 m-1}$ settles the case $r=0$; removal of $A_{3}$ from the list settles the case $r=1$; and removal of $A_{3}$ and $A_{3 m-1}$ settles the case $r=2$.\n\nNext, we prove that no planar $n$-gon splits into less than $n / 3$ triangles. Alternatively, but equivalently, if a planar polygon splits into $t$ triangles, then its boundary has (combinatorial) length at most $3 t$. Proceed by induction on $t$. The base case $t=1$ is clear, so let $t>1$.\n\nThe vertices of the triangles in the splitting may subdivide the boundary of the polygon, making it into a possibly combinatorially longer simple loop $\\Omega$. Clearly, it is sufficient to prove that the length of $\\Omega$ does not exceed $3 t$.\n\nTo this end, consider a triangle in the splitting whose boundary $\\omega$ meets $\\Omega$ along at least one of its edges. Trace $\\Omega$ counterclockwise and let $\\alpha_{1}, \\ldots, \\alpha_{k}$, in order, be the connected components of $\\Omega-\\omega$. Each $\\alpha_{i}$ is a path along $\\Omega$ with distinct end points, whose terminal point is joined to the starting point of $\\alpha_{i+1}$ by a (possibly constant) path $\\beta_{i}$ along $\\omega$. Trace $\\omega$ clockwise from the terminal point of $\\alpha_{i}$ to its starting point to obtain a path $\\alpha_{i}^{\\prime}$ of positive length, and notice that $\\alpha_{i}+\\alpha_{i}^{\\prime}$ is the boundary of a polygon split into $t_{i}<t$ triangles. By the induction hypothesis, the length of $\\alpha_{i}+\\alpha_{i}^{\\prime}$ does not exceed $3 t_{i}$, and since $\\alpha_{i}^{\\prime}$ has positive length, the length of $\\alpha_{i}$ is at most $3 t_{i}-1$. Consequently, the length of $\\Omega-\\omega$ does not exceed $\\sum_{i=1}^{k}\\left(3 t_{i}-1\\right)=3 t-3-k$.\n\nFinally, we prove that the total length of the $\\beta_{i}$ does not exceed $k+3$. Begin by noticing that no $\\beta_{i}$ has length greater than 4 , at most one has length greater than 2 , and at most three have length 2 . If some $\\beta_{i}$ has length 4 , then the remaining $k-1$ are all of length at most 1 , so the total length of the $\\beta$ 's does not exceed $4+(k-1)=k+3$. Otherwise, either some $\\beta_{i}$ has length 3 , in which case at most one other has length 2 and the remaining $k-2$ all have length at most 1 , or the $\\beta_{i}$ all have length less than 3 , in which case there are at most three of length 2 and the remaining $k-3$ all have length at most 1 ; in the former case, the total length of the $\\beta$ 's does not exceed $3+2+(k-2)=k+3$, and in the latter, the total length of the $\\beta$ 's does not exceed $3 \\cdot 2+(k-3)=k+3$. The conclusion follows.\n\n## Chapter 3\n\n## Geometry", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nC4.", "solution_match": "\nSolution."}}
-{"year": "2016", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C$ be an acute angled triangle, let $O$ be its circumcentre, and let $D, E, F$ be points on the sides $B C, A C, A B$, respectively. The circle $\\left(c_{1}\\right)$ of radius $F A$, centred at $F$, crosses the segment $(O A)$ at $A^{\\prime}$ and the circumcircle (c) of the triangle $A B C$ again at $K$. Similarly, the circle $\\left(c_{2}\\right)$ of radius $D B$, centred at $D$, crosses the segment $(O B)$ at $B^{\\prime}$ and the circle (c) again at $L$. Finally, the circle $\\left(c_{3}\\right)$ of radius $E C$, centred at $E$, crosses the segment $(O C)$ at $C^{\\prime}$ and the circle (c) again at $M$. Prove that the quadrilaterals $B K F A^{\\prime}$, $C L D B^{\\prime}$ and $A M E C^{\\prime}$ are all cyclic, and their circumcircles share a common point.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_76c244c1de22b5795a02g-16.jpg?height=523&width=539&top_left_y=1429&top_left_x=770)", "solution": "We will prove that the quadrilateral $B K F A^{\\prime}$ is cyclic and its circumcircle passes through the center $O$ of the circle (c).\n\nThe triangle $A F K$ is isosceles, so $m(\\widehat{K F B})=2 m(\\widehat{K A B})=m(\\widehat{K O B})$. It follows that the quadrilateral $B K F O$ is cyclic.\n\nThe triangles $O F K$ and $O F A$ are congruent (S.S.S.), hence $m(\\widehat{O K F})=m(\\widehat{O A F})$. The triangle $F A A^{\\prime}$ is isosceles, so $m\\left(\\widehat{F A^{\\prime} A}\\right)=m(\\widehat{O A F})$. Therefore $m\\left(\\widehat{F A^{\\prime} A}\\right)=m(\\widehat{O K F})$, so the quadrilateral $O K F A^{\\prime}$ is cyclic.\n\n(1) and (2) prove the initial claim.\n\nAlong the same lines, we can prove that the points $C, D, L, B^{\\prime}, O$ and $A, M, E, C^{\\prime}, O$ are concylic, respectively, so their circumcircles also pass through $O$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nG1.", "solution_match": "\nSolution."}}
-{"year": "2016", "tier": "T3", "problem_label": "G2", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C$ be a triangle with $m(\\widehat{B A C})=60^{\\circ}$. Let $D$ and $E$ be the feet of the perpendiculars from $A$ to the external angle bisectors of $\\widehat{A B C}$ and $\\widehat{A C B}$, respectively. Let $O$ be the circumcenter of the triangle $A B C$. Prove that the circumcircles of the triangles $\\triangle A D E$ and $\\triangle B O C$ are tangent to each other.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_76c244c1de22b5795a02g-17.jpg?height=480&width=443&top_left_y=591&top_left_x=818)", "solution": "Let $X$ be the intersection of the lines $B D$ and $C E$.\n\nWe will prove that $X$ lies on the circumcircles of both triangles $\\triangle A D E$ and $\\triangle B O C$ and then we will prove that the centers of these circles and the point $X$ are collinear, which is enough for proving that the circles are tangent to each other.\n\nIn this proof we will use the notation $(M N P)$ to denote the circumcircle of the triangle $\\triangle M N P$.\n\nObviously, the quadrilateral $A D X E$ is cyclic, and the circle $(D A E)$ has $[A X]$ as diameter. (1)\n\nLet $I$ be the incenter of triangle $A B C$. So, the point $I$ lies on the segment $[A X]$ (2), and the quadrilateral $X B I C$ is cyclic because $I C \\perp X C$ and $I B \\perp X B$. So, the circle (BIC) has $[I X]$ as diameter.\n\nFinally, $m(\\widehat{B I C})=90^{\\circ}+\\frac{1}{2} m(\\widehat{B A C})=120^{\\circ}$ and $m(\\widehat{B O C})=2 m(\\widehat{B A C})=120^{\\circ}$.\n\nSo, the quadrilateral $B O I C$ is cyclic and the circle $(B O C)$ has $[I X]$ as diameter. (3)\n\n(1), (2), (3) imply the conclusion.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nG2.", "solution_match": "\nSolution."}}
-{"year": "2016", "tier": "T3", "problem_label": "G3", "problem_type": "Geometry", "exam": "JBMO", "problem": "A trapezoid $A B C D(A B \\| C D, A B>C D)$ is circumscribed. The incircle of triangle $A B C$ touches the lines $A B$ and $A C$ at $M$ and $N$, respectively. Prove that the incenter of the trapezoid lies on the line $M N$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_76c244c1de22b5795a02g-18.jpg?height=428&width=646&top_left_y=540&top_left_x=725)", "solution": "Let $I$ be the incenter of triangle $A B C$ and $R$ be the common point of the lines $B I$ and $M N$. Since\n\n$$\nm(\\widehat{A N M})=90^{\\circ}-\\frac{1}{2} m(\\widehat{M A N}) \\quad \\text { and } \\quad m(\\widehat{B I C})=90^{\\circ}+\\frac{1}{2} m(\\widehat{M A N})\n$$\n\nthe quadrilateral $I R N C$ is cyclic.\n\nIt follows that $m(\\widehat{B R C})=90^{\\circ}$ and therefore\n\n$$\nm(\\widehat{B C R})=90^{\\circ}-m(\\widehat{C B R})=90^{\\circ}-\\frac{1}{2}\\left(180^{\\circ}-m(\\widehat{B C D})\\right)=\\frac{1}{2} m(\\widehat{B C D})\n$$\n\nSo, $(C R$ is the angle bisector of $\\widehat{D C B}$ and $R$ is the incenter of the trapezoid.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nG3.", "solution_match": "\nSolution."}}
-{"year": "2016", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C$ be an acute angled triangle whose shortest side is $[B C]$. Consider a variable point $P$ on the side $[B C]$, and let $D$ and $E$ be points on $(A B]$ and $(A C]$, respectively, such that $B D=B P$ and $C P=C E$. Prove that, as $P$ traces $[B C]$, the circumcircle of the triangle $A D E$ passes through a fixed point.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_76c244c1de22b5795a02g-19.jpg?height=554&width=445&top_left_y=591&top_left_x=820)", "solution": "We claim that the fixed point is the center of the incircle of $A B C$.\n\nLet $I$ be the center of the incircle of $A B C$. Since $B D=B P$ and $[B I$ is the bisector of $\\widehat{D B P}$, the line $B I$ is the perpendicular bisector of $[D P]$. This yields $D I=P I$. Analogously we get $E I=P I$. So, the point $I$ is the circumcenter of the triangle $D E P$.\n\nThis means $m(\\widehat{D I E})=2 m(\\widehat{D P E})$.\n\nOn the other hand\n\n$$\n\\begin{aligned}\nm(\\widehat{D P E}) & =180^{\\circ}-m(\\widehat{D P B})-m(\\widehat{E P C}) \\\\\n& =180^{\\circ}-\\left(90^{\\circ}-\\frac{1}{2} m(\\widehat{D B P})\\right)-\\left(90^{\\circ}-\\frac{1}{2} m(\\widehat{E C P})\\right) \\\\\n& =90^{\\circ}-\\frac{1}{2} m(\\widehat{B A C})\n\\end{aligned}\n$$\n\nSo, $m(\\widehat{D I E})=2 m(\\widehat{D P E})=180^{\\circ}-m(\\widehat{D A E})$, which means that the points $A, D, E$ and $I$ are cocyclic.\n\nRemark. The fact that the incentre $I$ of the triangle $A B C$ is the required fixed point could be guessed by considering the two extremal positions of $P$. Thus, if $P=B$, then $D=D_{B}=B$ as well, and $C E=C E_{B}=B C$, so $m(\\angle A E B)=m(\\angle C)+m(\\angle E B C)=$ $m(\\angle C)+\\frac{180^{\\circ}-m(\\angle C)}{2}=90^{\\circ}+\\frac{m(\\angle C)}{2}=m(\\angle A I B)$. Hence the points $A, E=E_{B}, I, D_{B}=B$ are cocyclic. Similarly, letting $P=C$, the points $A, D=D_{C}, I, E_{C}=C$ are cocyclic. Consequently, the circles $A D_{B} E_{B}$ and $A D_{C} E_{C}$ meet again at $I$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nG4.", "solution_match": "\nSolution."}}
-{"year": "2016", "tier": "T3", "problem_label": "G5", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $A B C$ be an acute angled triangle with orthocenter $H$ and circumcenter $O$. Assume the circumcenter $X$ of $B H C$ lies on the circumcircle of $A B C$. Reflect $O$ across $X$ to obtain $O^{\\prime}$, and let the lines $X H$ and $O^{\\prime} A$ meet at $K$. Let $L, M$ and $N$ be the midpoints of $[X B],[X C]$ and $[B C]$, respectively. Prove that the points $K, L, M$ and $N$ are cocyclic.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_76c244c1de22b5795a02g-20.jpg?height=440&width=651&top_left_y=588&top_left_x=720)", "solution": "The circumcircles of $A B C$ and $B H C$ have the same radius. So, $X B=$ $X C=X H=X O=r$ (where $r$ is the radius of the circle $A B C$ ) and $O^{\\prime}$ lies on $C(X, r)$. We conclude that $O X$ is the perpendicular bisector for $[B C]$. So, $B O X$ and $C O X$ are equilateral triangles.\n\nIt is known that $A H=2 O N=r$. So, $A H O^{\\prime} X$ is parallelogram, and $X K=K H=r / 2$. Finally, $X L=X K=X N=X M=r / 2$. So, $K, L, M$ and $N$ lie on the circle $c(X, r / 2)$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nG5.", "solution_match": "\nSolution."}}
-{"year": "2016", "tier": "T3", "problem_label": "G6", "problem_type": "Geometry", "exam": "JBMO", "problem": "Given an acute triangle $A B C$, erect triangles $A B D$ and $A C E$ externally, so that $m(\\widehat{A D B})=m(\\widehat{A E C})=90^{\\circ}$ and $\\widehat{B A D} \\equiv \\widehat{C A E}$. Let $A_{1} \\in B C, B_{1} \\in A C$ and $C_{1} \\in A B$ be the feet of the altitudes of the triangle $A B C$, and let $K$ and $L$ be the midpoints of $\\left[B C_{1}\\right]$ and $\\left[C B_{1}\\right]$, respectively. Prove that the circumcenters of the triangles $A K L, A_{1} B_{1} C_{1}$ and $D E A_{1}$ are collinear.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_76c244c1de22b5795a02g-21.jpg?height=537&width=944&top_left_y=648&top_left_x=567)", "solution": "Let $M, P$ and $Q$ be the midpoints of $[B C],[C A]$ and $[A B]$, respectively.\n\nThe circumcircle of the triangle $A_{1} B_{1} C_{1}$ is the Euler's circle. So, the point $M$ lies on this circle.\n\nIt is enough to prove now that $\\left[A_{1} M\\right]$ is a common chord of the three circles $\\left(A_{1} B_{1} C_{1}\\right)$, $(A K L)$ and $\\left(D E A_{1}\\right)$.\n\nThe segments $[M K]$ and $[M L]$ are midlines for the triangles $B C C_{1}$ and $B C B_{1}$ respectively, hence $M K \\| C C_{1} \\perp A B$ and $M L \\| B B_{1} \\perp A C$. So, the circle $(A K L)$ has diameter $[A M]$ and therefore passes through $M$.\n\nFinally, we prove that the quadrilateral $D A_{1} M E$ is cyclic.\n\nFrom the cyclic quadrilaterals $A D B A_{1}$ and $A E C A_{1}, \\widehat{A A_{1} D} \\equiv \\widehat{A B D}$ and $\\widehat{A A_{1} E} \\equiv \\widehat{A C E} \\equiv$ $\\widehat{A B D}$, so $m\\left(\\widehat{D A_{1} E}\\right)=2 m(\\widehat{A B D})=180^{\\circ}-2 m(\\widehat{D A B})$.\n\nWe notice now that $D Q=A B / 2=M P, Q M=A C / 2=P E$ and\n\n$$\n\\begin{aligned}\n& m(\\widehat{D Q M})=m(\\widehat{D Q B})+m(\\widehat{B Q M})=2 m(\\widehat{D A B})+m(\\widehat{B A C}) \\\\\n& m(\\widehat{E P M})=m(\\widehat{E P C})+m(\\widehat{C P M})=2 m(\\widehat{E A C})+m(\\widehat{C A B})\n\\end{aligned}\n$$\n\nso $\\triangle M P E \\equiv \\triangle D Q M$ (S.A.S.). This leads to $m(\\widehat{D M E})=m(\\widehat{D M Q})+m(Q M P)+$ $m(P M E)=m(\\widehat{D M Q})+m(B Q M)+m(Q D M)=180^{\\circ}-m(\\widehat{D Q B})=180^{\\circ}-2 m(\\widehat{D A B})$. Since $m\\left(\\widehat{D A_{1} E}\\right)=m(\\widehat{D M E})$, the quadrilateral $D A_{1} M E$ is cyclic.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nG6.", "solution_match": "\nSolution."}}
-{"year": "2016", "tier": "T3", "problem_label": "G7", "problem_type": "Geometry", "exam": "JBMO", "problem": "Let $[A B]$ be a chord of a circle $(c)$ centered at $O$, and let $K$ be a point on the segment $(A B)$ such that $A K<B K$. Two circles through $K$, internally tangent to (c) at $A$ and $B$, respectively, meet again at $L$. Let $P$ be one of the points of intersection of the line $K L$ and the circle (c), and let the lines $A B$ and $L O$ meet at $M$. Prove that the line $M P$ is tangent to the circle $(c)$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_76c244c1de22b5795a02g-22.jpg?height=497&width=606&top_left_y=657&top_left_x=742)", "solution": "Let $\\left(c_{1}\\right)$ and $\\left(c_{2}\\right)$ be circles through $K$, internally tangent to (c) at $A$ and $B$, respectively, and meeting again at $L$, and let the common tangent to $\\left(c_{1}\\right)$ and $(c)$ meet the common tangent to $\\left(c_{2}\\right)$ and $(c)$ at $Q$. Then the point $Q$ is the radical center of the circles $\\left(c_{1}\\right),\\left(c_{2}\\right)$ and $(c)$, and the line $K L$ passes through $Q$.\n\nWe have $m(\\widehat{Q L B})=m(\\widehat{Q B K})=m(\\widehat{Q B A})=\\frac{1}{2} m(\\overparen{B A})=m(\\widehat{Q O B})$. So, the quadrilateral $O B Q L$ is cyclic. We conclude that $m(\\widehat{Q L O})=90^{\\circ}$ and the points $O, B, Q, A$ and $L$ are cocyclic on a circle $(k)$.\n\nIn the sequel, we will denote $\\mathcal{P}_{\\omega}(X)$ the power of the point $X$ with respect of the circle $\\omega$. The first continuation.\n\nFrom $M O^{2}-O P^{2}=\\mathcal{P}_{c}(M)=M A \\cdot M B=\\mathcal{P}_{k}(M)=M L \\cdot M O=(M O-O L) \\cdot M O=$ $M O^{2}-O L \\cdot M O$ follows that $O P^{2}=O L \\cdot O M$. Since $P L \\perp O M$, this shows that the triangle $M P O$ is right at point $P$. Thus, the line $M P$ is tangent to the circle (c).\n\nThe second continuation.\n\nLet $R \\in(c)$ be so that $B R \\perp M O$. The triangle $L B R$ is isosceles with $L B=L R$, so $\\widehat{O L R} \\equiv \\widehat{O L B} \\equiv \\widehat{O Q B} \\equiv \\widehat{O Q A} \\equiv \\widehat{M L A}$. We conclude that the points $A, L$ and $R$ are collinear.\n\nNow $m(\\widehat{A M R})+m(\\widehat{A O R})=m(\\widehat{A M R})+2 m(\\widehat{A B R})=m(\\widehat{A M R})+m(\\widehat{A B R})+m(\\widehat{M R B})=$ $180^{\\circ}$, since the triangle $M B R$ is isosceles. So, the quadrilateral $M A O R$ is cyclic.\n\nThis yields $L M \\cdot L O=-\\mathcal{P}_{(\\text {MAOR })}(L)=L A \\cdot L R=-\\mathcal{P}_{c}(L)=L P^{2}$, which as above, shows that $O P \\perp P M$.\n\nThe third continuation.\n\n$\\widehat{K L A} \\equiv \\widehat{K A Q} \\equiv \\widehat{K L B}$ and $m(\\widehat{M L K})=90^{\\circ}$ show that $[L K$ and $[L M$ are the internal and external bisectors af the angle $\\widehat{A L B}$, so $(M, K)$ and $(A, B)$ are harmonic conjugates. So, $L K$ is the polar line of $M$ in the circle $(c)$.\n\n## Chapter 4\n\n## Number Theory", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nG7.", "solution_match": "\nSolution."}}
-{"year": "2016", "tier": "T3", "problem_label": "N1", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Determine the largest positive integer $n$ that divides $p^{6}-1$ for all primes $p>7$.", "solution": "Note that\n\n$$\np^{6}-1=(p-1)(p+1)\\left(p^{2}-p+1\\right)\\left(p^{2}+p+1\\right)\n$$\n\nFor $p=11$ we have\n\n$$\np^{6}-1=1771560=2^{3} \\cdot 3^{2} \\cdot 5 \\cdot 7 \\cdot 19 \\cdot 37\n$$\n\nFor $p=13$ we have\n\n$$\np^{6}-1=2^{3} \\cdot 3^{2} \\cdot 7 \\cdot 61 \\cdot 157\n$$\n\nFrom the last two calculations we find evidence to try showing that $p^{6}-1$ is divisible by $2^{3} \\cdot 3^{2} \\cdot 7=504$ and this would be the largest positive integer that divides $p^{6}-1$ for all primes greater than 7 .\n\nBy Fermat's theorem, $7 \\mid p^{6}-1$.\n\nNext, since $p$ is odd, $8 \\mid p^{2}-1=(p-1)(p+1)$, hence $8 \\mid p^{6}-1$.\n\nIt remains to show that $9 \\mid p^{6}-1$.\n\nAny prime number $p, p>3$ is 1 or -1 modulo 3 .\n\nIn the first case both $p-1$ and $p^{2}+p+1$ are divisible by 3 , and in the second case, both $p+1$ and $p^{2}-p+1$ are divisible by 3 .\n\nConsequently, the required number is indeed 504\n\n## Alternative solution\n\nLet $q$ be a (positive) prime factor of $n$. Then $q \\leq 7$, as $q \\nmid q^{6}-1$. Also, $q$ is not 5 , as the last digit of $13^{6}-1$ is 8 .\n\nHence, the prime factors of $n$ are among 2,3 , and 7 .\n\nNext, from $11^{6}-1=2^{3} \\cdot 3^{2} \\cdot 5 \\cdot 7 \\cdot 19 \\cdot 37$ it follows that the largest integer $n$ such that $n \\mid p^{6}-1$ for all primes $p$ greater than 7 is at most $2^{3} \\cdot 3^{2} \\cdot 7$, and it remains to prove that 504 divides $p^{6}-1$ for all primes grater than 7 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nN1.", "solution_match": "\nSolution."}}
-{"year": "2016", "tier": "T3", "problem_label": "N2", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find the maximum number of natural numbers $x_{1}, x_{2}, \\ldots, x_{m}$ satisfying the conditions:\n\na) No $x_{i}-x_{j}, 1 \\leq i<j \\leq m$ is divisible by 11 ; and\n\nb) The sum $x_{2} x_{3} \\ldots x_{m}+x_{1} x_{3} \\ldots x_{m}+\\cdots+x_{1} x_{2} \\ldots x_{m-1}$ is divisible by 11 .", "solution": "The required maximum is 10 .\n\nAccording to a), the numbers $x_{i}, 1 \\leq i \\leq m$, are all different $(\\bmod 11)$ (1)\n\nHence, the number of natural numbers satisfying the conditions is at most 11.\n\nIf $x_{j} \\equiv 0(\\bmod 11)$ for some $j$, then\n\n$$\nx_{2} x_{3} \\ldots x_{m}+x_{1} x_{3} \\ldots x_{m}+\\cdots+x_{1} x_{2} \\ldots x_{m-1} \\equiv x_{1} \\ldots x_{j-1} x_{j+1} \\ldots x_{m} \\quad(\\bmod 11)\n$$\n\nwhich would lead to $x_{i} \\equiv 0(\\bmod 11)$ for some $i \\neq j$, contradicting (1).\n\nWe now prove that 10 is indeed the required maximum.\n\nConsider $x_{i}=i$, for all $i \\in\\{1,2, \\ldots, 10\\}$. The products $2 \\cdot 3 \\cdots \\cdot 10,1 \\cdot 3 \\cdots \\cdots 10, \\ldots$, $1 \\cdot 2 \\cdots \\cdot 9$ are all different $(\\bmod 11)$, and so\n\n$$\n2 \\cdot 3 \\cdots \\cdots 10+1 \\cdot 3 \\cdots \\cdots 10+\\cdots+1 \\cdot 2 \\cdots \\cdot 9 \\equiv 1+2+\\cdots+10 \\quad(\\bmod 11)\n$$\n\nand condition b) is satisfied, since $1+2+\\cdots+10=55=5 \\cdot 11$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nN2.", "solution_match": "\nSolution."}}
-{"year": "2016", "tier": "T3", "problem_label": "N3", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all positive integers $n$ such that the number $A_{n}=\\frac{2^{4 n+2}+1}{65}$ is\n\na) an integer;\n\nb) a prime.", "solution": "a) Note that $65=5 \\cdot 13$.\n\nObviously, $5=2^{2}+1$ is a divisor of $\\left(2^{2}\\right)^{2 n+1}+1=2^{4 n+2}+1$ for any positive integer $n$. Since $2^{12} \\equiv 1(\\bmod 13)$, if $n \\equiv r(\\bmod 3)$, then $2^{4 n+2}+1 \\equiv 2^{4 r+2}+1(\\bmod 13)$. Now, $2^{4 \\cdot 0+2}+1=5,2^{4 \\cdot 1+2}+1=65$, and $2^{4 \\cdot 2+2}+1=1025=13 \\cdot 78+11$. Hence 13 is a divisor of $2^{4 n+2}+1$ precisely when $n \\equiv 1(\\bmod 3)$. Hence, $A_{n}$ is an integer iff $n \\equiv 1(\\bmod 3)$.\n\nb) Applying the identity $4 x^{4}+1=\\left(2 x^{2}-2 x+1\\right)\\left(2 x^{2}+2 x+1\\right)$, we have $2^{4 n+2}+1=$ $\\left(2^{2 n+1}-2^{n+1}+1\\right)\\left(2^{2 n+1}+2^{n+1}+1\\right)$. For $n=1, A_{1}=1$, which is not a prime. According to a), if $n \\neq 1$, then $n \\geq 4$. But then $2^{2 n+1}+2^{n+1}+1>2^{2 n+1}-2^{n+1}+1>65$, and $A_{n}$ has at least two factors. We conclude that $A_{n}$ can never be a prime.\n\nAlternative Solution to b): Knowing that $n=3 k+1$ in order for $A_{n}$ to be an integer, $2^{4 n+2}+1=2^{12 k+6}+1=\\left(2^{4 k+2}\\right)^{3}+1=\\left(2^{4 k+2}+1\\right)\\left(2^{8 k+4}-2^{4 k+2}+1\\right) \\quad(*)$. As in the previous solution, if $k=0$, then $A_{1}=1$, if $k=1$, then $A_{4}=2^{12}-2^{6}+1=4033=37 \\cdot 109$, and for $k \\geq 2$ both factors in $(*)$ are larger than 65 , so $A_{3 k+1}$ is not a prime.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nN3.", "solution_match": "\nSolution."}}
-{"year": "2016", "tier": "T3", "problem_label": "N4", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Find all triples of integers $(a, b, c)$ such that the number\n\n$$\nN=\\frac{(a-b)(b-c)(c-a)}{2}+2\n$$\n\nis a power of 2016 .", "solution": "Let $z$ be a positive integer such that\n\n$$\n(a-b)(b-c)(c-a)+4=2 \\cdot 2016^{z}\n$$\n\nWe set $a-b=-x, b-c=-y$ and we rewrite the equation as\n\n$$\nx y(x+y)+4=2 \\cdot 2016^{z}\n$$\n\nNote that the right hand side is divisible by 7 , so we have that\n\n$$\nx y(x+y)+4 \\equiv 0 \\quad(\\bmod 7)\n$$\n\nor\n\n$$\n3 x y(x+y) \\equiv 2 \\quad(\\bmod 7)\n$$\n\nor\n\n$$\n(x+y)^{3}-x^{3}-y^{3} \\equiv 2 \\quad(\\bmod 7)\n$$\n\nNote that, by Fermat's Little Theorem, we have that for any integer $k$ the cubic residues are $k^{3} \\equiv-1,0,1 \\bmod 7$. It follows that in (4.1) some of $(x+y)^{3}, x^{3}$ and $y^{3}$ should be divisible by 7 , but in this case, $x y(x+y)$ is divisible by 7 and this is a contradiction. So, the only possibility is to have $z=0$ and consequently, $x y(x+y)+4=2$, or, equivalently, $x y(x+y)=-2$. The only solution of the latter is $(x, y)=(-1,-1)$, so the required triples are $(a, b, c)=(k+2, k+1, k), k \\in \\mathbb{Z}$, and all their cyclic permutations.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nN4.", "solution_match": "\nSolution."}}
-{"year": "2016", "tier": "T3", "problem_label": "N5", "problem_type": "Number Theory", "exam": "JBMO", "problem": "Determine all four-digit numbers $\\overline{a b c d}$ such that\n\n$$\n(a+b)(a+c)(a+d)(b+c)(b+d)(c+d)=\\overline{a b c d}\n$$", "solution": "Depending on the parity of $a, b, c, d$, at least two of the factors $(a+b),(a+c)$, $(a+d),(b+c),(b+d),(c+d)$ are even, so that $4 \\mid \\overline{a b c d}$.\n\nWe claim that $3 \\mid \\overline{a b c d}$.\n\nAssume $a+b+c+d \\equiv 2(\\bmod 3)$. Then $x+y \\equiv 1(\\bmod 3)$, for all distinct $x, y \\in\\{a, b, c, d\\}$. But then the left hand side in the above equality is congruent to $1(\\bmod 3)$ and the right hand side congruent to $2(\\bmod 3)$, contradiction.\n\nAssume $a+b+c+d \\equiv 1(\\bmod 3)$. Then $x+y \\equiv 2(\\bmod 3)$, for all distinct $x, y \\in\\{a, b, c, d\\}$, and $x \\equiv 1(\\bmod 3)$, for all $x, y \\in\\{a, b, c, d\\}$. Hence, $a, b, c, d \\in\\{1,4,7\\}$, and since $4 \\mid \\overline{a b c d}$, we have $c=d=4$. Therefore, $8 \\mid \\overline{a b 44}$, and since at least one more factor is even, it follows that $16 \\overline{a b 44}$. Then $b \\neq 4$, and the only possibilities are $b=1$, implying $a=4$, which is impossible because 4144 is not divisible by $5=1+4$, or $b=7$, implying $11 \\mid \\overline{a 744}$, hence $a=7$, which is also impossible because 7744 is not divisible by $14=7+7$.\n\nWe conclude that $3 \\mid \\overline{a b c d}$, hence also $3 \\mid a+b+c+d$. Then at least one factor $x+y$ of $(a+b),(a+c),(a+d),(b+c),(b+d),(c+d)$ is a multiple of 3 , implying that also $3 \\mid a+b+c+d-x-y$, so $9 \\mid \\overline{a b c d}$. Then $9 \\mid a+b+c+d$, and $a+b+c+d \\in\\{9,18,27,36\\}$. Using the inequality $x y \\geq x+y-1$, valid for all $x, y \\in \\mathbb{N}^{*}$, if $a+b+c+d \\in\\{27,36\\}$, then\n\n$$\n\\overline{a b c d}=(a+b)(a+c)(a+d)(b+c)(b+d)(c+d) \\geq 26^{3}>10^{4}\n$$\n\nwhich is impossible.\n\nUsing the inequality $x y \\geq 2(x+y)-4$ for all $x, y \\geq 2$, if $a+b+c+d=18$ and all two-digit sums are greater than 1 , then $\\overline{a b c d} \\geq 32^{3}>10^{4}$. Hence, if $a+b+c+d=18$, some two-digit sum must be 1 , hence the complementary sum will be 17 , and the digits are $\\{a, b, c, d\\}=\\{0,1,8,9\\}$. But then $\\overline{a b c d}=1 \\cdot 17 \\cdot 8 \\cdot 9^{2} \\cdot 10>10^{4}$.\n\nWe conclude that $a+b+c+d=9$. Then among $a, b, c, d$ there are either three odd or three even numbers, and $8 \\mid \\overline{a b c d}$.\n\nIf three of the digits are odd, then $d$ is even and since $c$ is odd, divisibility by 8 implies that $d \\in\\{2,6\\}$. If $d=6$, then $a=b=c=1$. But 1116 is not divisible by 7 , so this is not a solution. If $d=2$, then $a, b, c$ are either $1,1,5$ or $1,3,3$ in some order. In the first case $2 \\cdot 6^{2} \\cdot 3^{2} \\cdot 7=4536 \\neq \\overline{a b c d}$. The second case cannot hold because the resulting number is not a multiple of 5 .\n\nHence, there has to be one odd and three even digits. At least one of the two-digits sums of even digits is a multiple of 4 , and since there cannot be two zero digits, we have either $x+y=4$ and $z+t=5$, or $x+y=8$ and $z+t=1$ for some ordering $x, y, z, t$ of $a, b, c, d$. In the first case we have $d=0$ and the digits are $0,1,4,4$, or $0,2,3,4$, or $0,2,2,5$. None of these is a solution because $1 \\cdot 4^{2} \\cdot 5^{2} \\cdot 8=3200,2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdot 7=5040$ and $2^{2} \\cdot 5 \\cdot 4 \\cdot 7^{2}=3920$. In the second case two of the digits are 0 and 1 , and the other two have to be either 4 and 4 , or 2 and 6 . We already know that the first possibility fails. For the second, we get\n\n$$\n(0+1) \\cdot(0+2) \\cdot(0+6) \\cdot(1+2) \\cdot(1+6) \\cdot(2+6)=2016\n$$\n\nand $\\overline{a b c d}=2016$ is the only solution.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nN5.", "solution_match": "\nSolution."}}
+{"year": "2016", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $a, b, c$ be positive real numbers such that $a b c=8$. Prove that\n\n$$\n\\frac{a b+4}{a+2}+\\frac{b c+4}{b+2}+\\frac{c a+4}{c+2} \\geq 6\n$$", "solution": "We have $a b+4=\\frac{8}{c}+4=\\frac{4(c+2)}{c}$ and similarly $b c+4=\\frac{4(a+2)}{a}$ and $c a+4=\\frac{4(b+2)}{b}$. It follows that\n\n$$\n(a b+4)(b c+4)(c a+4)=\\frac{64}{a b c}(a+2)(b+2)(c+2)=8(a+2)(b+2)(c+2)\n$$\n\nso that\n\n$$\n\\frac{(a b+4)(b c+4)(c a+4)}{(a+2)(b+2)(c+2)}=8\n$$\n\nApplying AM-GM, we conclude:\n\n$$\n\\frac{a b+4}{a+2}+\\frac{b c+4}{b+2}+\\frac{c a+4}{c+2} \\geq 3 \\cdot \\sqrt[3]{\\frac{(a b+4)(b c+4)(c a+4)}{(a+2)(b+2)(c+2)}}=6\n$$\n\nAlternatively, we can write LHS as\n\n$$\n\\frac{b c(a b+4)}{2(b c+4)}+\\frac{a c(b c+4)}{2(a c+4)}+\\frac{a b(c a+4)}{2(a b+4)}\n$$\n\nand then apply AM-GM.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nA1.", "solution_match": "\nSolution."}}
+{"year": "2016", "tier": "T3", "problem_label": "A2", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Given positive real numbers $a, b, c$, prove that\n\n$$\n\\frac{8}{(a+b)^{2}+4 a b c}+\\frac{8}{(a+c)^{2}+4 a b c}+\\frac{8}{(b+c)^{2}+4 a b c}+a^{2}+b^{2}+c^{2} \\geq \\frac{8}{a+3}+\\frac{8}{b+3}+\\frac{8}{c+3}\n$$", "solution": "Since $2 a b \\leq a^{2}+b^{2}$, it follows that $(a+b)^{2} \\leq 2\\left(a^{2}+b^{2}\\right)$ and $4 a b c \\leq 2 c\\left(a^{2}+b^{2}\\right)$, for any positive reals $a, b, c$. Adding these inequalities, we find\n\n$$\n(a+b)^{2}+4 a b c \\leq 2\\left(a^{2}+b^{2}\\right)(c+1)\n$$\n\nso that\n\n$$\n\\frac{8}{(a+b)^{2}+4 a b c} \\geq \\frac{4}{\\left(a^{2}+b^{2}\\right)(c+1)}\n$$\n\nUsing the AM-GM inequality, we have\n\n$$\n\\frac{4}{\\left(a^{2}+b^{2}\\right)(c+1)}+\\frac{a^{2}+b^{2}}{2} \\geq 2 \\sqrt{\\frac{2}{c+1}}=\\frac{4}{\\sqrt{2(c+1)}}\n$$\n\nrespectively\n\n$$\n\\frac{c+3}{8}=\\frac{(c+1)+2}{8} \\geq \\frac{\\sqrt{2(c+1)}}{4}\n$$\n\nWe conclude that\n\n$$\n\\frac{4}{\\left(a^{2}+b^{2}\\right)(c+1)}+\\frac{a^{2}+b^{2}}{2} \\geq \\frac{8}{c+3}\n$$\n\nand finally\n\n$\\frac{8}{(a+b)^{2}+4 a b c}+\\frac{8}{(a+c)^{2}+4 a b c}+\\frac{8}{(b+c)^{2}+4 a b c}+a^{2}+b^{2}+c^{2} \\geq \\frac{8}{a+3}+\\frac{8}{b+3}+\\frac{8}{c+3}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nA2.", "solution_match": "\nSolution."}}
+{"year": "2016", "tier": "T3", "problem_label": "A3", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Determine the number of pairs of integers $(m, n)$ such that\n\n$$\n\\sqrt{n+\\sqrt{2016}}+\\sqrt{m-\\sqrt{2016}} \\in \\mathbb{Q}\n$$", "solution": "Let $r=\\sqrt{n+\\sqrt{2016}}+\\sqrt{m-\\sqrt{2016}}$. Then\n\n$$\nn+m+2 \\sqrt{n+\\sqrt{2016}} \\cdot \\sqrt{m-\\sqrt{2016}}=r^{2}\n$$\n\nand\n\n$$\n(m-n) \\sqrt{2106}=\\frac{1}{4}\\left(r^{2}-m-n\\right)^{2}-m n+2016 \\in \\mathbb{Q}\n$$\n\nSince $\\sqrt{2016} \\notin \\mathbb{Q}$, it follows that $m=n$. Then\n\n$$\n\\sqrt{n^{2}-2016}=\\frac{1}{2}\\left(r^{2}-2 n\\right) \\in \\mathbb{Q}\n$$\n\nHence, there is some nonnegative integer $p$ such that $n^{2}-2016=p^{2}$ and (1) becomes $2 n+2 p=r^{2}$.\n\nIt follows that $2(n+p)=r^{2}$ is the square of a rational and also an integer, hence a perfect square. On the other hand, $2016=(n-p)(n+p)$ and $n+p$ is a divisor of 2016, larger than $\\sqrt{2016}$. Since $n+p$ is even, so is also $n-p$, and $r^{2}=2(n+p)$ is a divisor of $2016=2^{5} \\cdot 3^{2} \\cdot 7$, larger than $2 \\sqrt{2016}>88$. The only possibility is $r^{2}=2^{4} \\cdot 3^{2}=12^{2}$. Hence, $n+p=72$ and $n-p=28$, and we conclude that $n=m=50$. Thus, there is only one such pair.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nA3.", "solution_match": "\nSolution."}}
+{"year": "2016", "tier": "T3", "problem_label": "A4", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "If $x, y, z$ are non-negative real numbers such that $x^{2}+y^{2}+z^{2}=x+y+z$, then show that:\n\n$$\n\\frac{x+1}{\\sqrt{x^{5}+x+1}}+\\frac{y+1}{\\sqrt{y^{5}+y+1}}+\\frac{z+1}{\\sqrt{z^{5}+z+1}} \\geq 3\n$$\n\nWhen does the equality hold?", "solution": "First we factor $x^{5}+x+1$ as follows:\n\n$$\n\\begin{aligned}\nx^{5}+x+1 & =x^{5}-x^{2}+x^{2}+x+1=x^{2}\\left(x^{3}-1\\right)+x^{2}+x+1=x^{2}(x-1)\\left(x^{2}+x+1\\right)+x^{2}+x+1 \\\\\n& =\\left(x^{2}+x+1\\right)\\left(x^{2}(x-1)+1\\right)=\\left(x^{2}+x+1\\right)\\left(x^{3}-x^{2}+1\\right)\n\\end{aligned}\n$$\n\nUsing the $A M-G M$ inequality, we have\n\n$$\n\\sqrt{x^{5}+x+1}=\\sqrt{\\left(x^{2}+x+1\\right)\\left(x^{3}-x^{2}+1\\right)} \\leq \\frac{x^{2}+x+1+x^{3}-x^{2}+1}{2}=\\frac{x^{3}+x+2}{2}\n$$\n\nand since\n\n$x^{3}+x+2=x^{3}+1+x+1=(x+1)\\left(x^{2}-x+1\\right)+x+1=(x+1)\\left(x^{2}-x+1+1\\right)=(x+1)\\left(x^{2}-x+2\\right)$,\n\nthen\n\n$$\n\\sqrt{x^{5}+x+1} \\leq \\frac{(x+1)\\left(x^{2}-x+2\\right)}{2}\n$$\n\nUsing $x^{2}-x+2=\\left(x-\\frac{1}{2}\\right)^{2}+\\frac{7}{4}>0$, we obtain $\\frac{x+1}{\\sqrt{x^{5}+x+1}} \\geq \\frac{2}{x^{2}-x+2}$ Applying the CauchySchwarz inequality and the given condition, we get\n\n$$\n\\sum_{c y c} \\frac{x+1}{\\sqrt{x^{5}+x+1}} \\geq \\sum_{c y c} \\frac{2}{x^{2}-x+2} \\geq \\frac{18}{\\sum_{c y c}\\left(x^{2}-x+2\\right)}=\\frac{18}{6}=3\n$$\n\nwhich is the desired result.\n\nFor the equality both conditions: $x^{2}-x+2=y^{2}-y+2=z^{2}-z+2$ (equality in CBS) and $x^{3}-x^{2}+1=x^{2}+x+1$ (equality in AM-GM) have to be satisfied.\n\nBy using the given condition it follows that $x^{2}-x+2+y^{2}-y+2+z^{2}-z+2=6$, hence $3\\left(x^{2}-x+2\\right)=6$, implying $x=0$ or $x=1$. Of these, only $x=0$ satisfies the second condition. We conclude that equality can only hold for $x=y=z=0$.\n\nIt is an immediate check that indeed for these values equality holds.\n\n## Alternative solution\n\nLet us present an heuristic argument to reach the key inequality $\\frac{x+1}{\\sqrt{x^{5}+x+1}} \\geq \\frac{2}{x^{2}-x+2}$.\n\nIn order to exploit the condition $x^{2}+y^{2}+z^{2}=x+y+z$ when applying CBS in Engel form, we are looking for $\\alpha, \\beta, \\gamma>0$ such that\n\n$$\n\\frac{x+1}{\\sqrt{x^{5}+x+1}} \\geq \\frac{\\gamma}{\\alpha\\left(x^{2}-x\\right)+\\beta}\n$$\n\nAfter squaring and cancelling the denominators, we get\n\n$$\n(x+1)^{2}\\left(\\alpha\\left(x^{2}-x\\right)+\\beta\\right)^{2} \\geq \\gamma^{2}\\left(x^{5}+x+1\\right)\n$$\n\nfor all $x \\geq 0$, and, after some manipulations, we reach to $f(x) \\geq 0$ for all $x \\geq 0$, where $f(x)=\\alpha^{2} x^{6}-\\gamma^{2} x^{5}+\\left(2 \\alpha \\beta-2 \\alpha^{2}\\right) x^{4}+2 \\alpha \\beta x^{3}+(\\alpha-\\beta)^{2} x^{2}+\\left(2 \\beta^{2}-2 \\alpha \\beta-\\gamma^{2}\\right) x+\\beta^{2}-\\gamma^{2}$.\n\nAs we are expecting the equality to hold for $x=0$, we naturally impose the condition that $f$ has 0 as a double root. This implies $\\beta^{2}-\\gamma^{2}=0$ and $2 \\beta^{2}-2 \\alpha \\beta-\\gamma^{2}=0$, that is, $\\beta=\\gamma$ and $\\gamma=2 \\alpha$.\n\nThus the inequality $f(x) \\geq 0$ becomes\n\n$$\n\\alpha^{2} x^{6}-4 \\alpha^{2} x^{5}+2 \\alpha^{2} x^{4}+4 \\alpha^{2} x^{3}+\\alpha^{2} x^{2} \\geq 0, \\forall x \\geq 0\n$$\n\nthat is,\n\n$$\n\\alpha^{2} x^{2}\\left(x^{2}-2 x-1\\right)^{2} \\geq 0 \\forall x \\geq 0\n$$\n\nwhich is obviously true.\n\nTherefore, the inequality $\\frac{x+1}{\\sqrt{x^{5}+x+1}} \\geq \\frac{2}{x^{2}-x+2}$ holds for all $x \\geq 0$ and now we can continue as in the first solution.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nA4.", "solution_match": "\nSolution."}}
+{"year": "2016", "tier": "T3", "problem_label": "A5", "problem_type": "Algebra", "exam": "JBMO-SL", "problem": "Let $x, y, z$ be positive real numbers such that $x+y+z=\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}$.\n\na) Prove the inequality\n\n$$\nx+y+z \\geq \\sqrt{\\frac{x y+1}{2}}+\\sqrt{\\frac{y z+1}{2}}+\\sqrt{\\frac{z x+1}{2}}\n$$\n\nb) (Added by the problem selecting committee) When does the equality hold?", "solution": "a) We rewrite the inequality as\n\n$$\n(\\sqrt{x y+1}+\\sqrt{y z+1}+\\sqrt{z x+1})^{2} \\leq 2 \\cdot(x+y+z)^{2}\n$$\n\nand note that, from CBS,\n\n$$\n\\text { LHS } \\leq\\left(\\frac{x y+1}{x}+\\frac{y z+1}{y}+\\frac{z x+1}{z}\\right)(x+y+z)\n$$\n\nBut\n\n$$\n\\frac{x y+1}{x}+\\frac{y z+1}{y}+\\frac{z x+1}{z}=x+y+z+\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=2(x+y+z)\n$$\n\nwhich proves (1).\n\nb) The equality occurs when we have equality in CBS, i.e. when\n\n$$\n\\frac{x y+1}{x^{2}}=\\frac{y z+1}{y^{2}}=\\frac{z x+1}{z^{2}}\\left(=\\frac{x y+y z+z x+3}{x^{2}+y^{2}+z^{2}}\\right)\n$$\n\nSince we can also write\n\n$$\n(\\sqrt{x y+1}+\\sqrt{y z+1}+\\sqrt{z x+1})^{2} \\leq\\left(\\frac{x y+1}{y}+\\frac{y z+1}{z}+\\frac{z x+1}{x}\\right)(y+z+x)=2(x+y+z)^{2}\n$$\n\nthe equality implies also\n\n$$\n\\frac{x y+1}{y^{2}}=\\frac{y z+1}{z^{2}}=\\frac{z x+1}{x^{2}}\\left(=\\frac{x y+y z+z x+3}{x^{2}+y^{2}+z^{2}}\\right)\n$$\n\nBut then $x=y=z$, and since $x+y+z=\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}$, we conclude that $x=\\frac{1}{x}=1=y=z$.\n\nAlternative solution to $b$ ): The equality condition\n\n$$\n\\frac{x y+1}{x^{2}}=\\frac{y z+1}{y^{2}}=\\frac{z x+1}{z^{2}}\n$$\n\ncan be rewritten as\n\n$$\n\\frac{y+\\frac{1}{x}}{x}=\\frac{z+\\frac{1}{y}}{y}=\\frac{x+\\frac{1}{z}}{z}=\\frac{x+y+z+\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}}{x+y+z}=2\n$$\n\nand thus we obtain the system:\n\n$$\n\\left\\{\\begin{array}{l}\ny=2 x-\\frac{1}{x} \\\\\nz=2 y-\\frac{1}{y} \\\\\nx=2 z-\\frac{1}{z}\n\\end{array}\\right.\n$$\n\nWe show that $x=y=z$. Indeed, if for example $x>y$, then $2 x-\\frac{1}{x}>2 y-\\frac{1}{y}$, that is, $y>z$ and $z=2 y-\\frac{1}{y}>2 z-\\frac{1}{z}=x$, and we obtain the contradiction $x>y>z>x$. Similarly, if $x<y$, we obtain $x<y<z<x$.\n\nHence, the numbers are equal, and as above we get $x=y=z=1$.\n\n## Chapter 2\n\n## Combinatorics", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nA5.", "solution_match": "## Solution."}}
+{"year": "2016", "tier": "T3", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Let $S_{n}$ be the sum of reciprocal values of non-zero digits of all positive integers up to (and including) $n$. For instance, $S_{13}=\\frac{1}{1}+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}+\\frac{1}{6}+\\frac{1}{7}+\\frac{1}{8}+\\frac{1}{9}+\\frac{1}{1}+\\frac{1}{1}+\\frac{1}{1}+\\frac{1}{1}+\\frac{1}{2}+\\frac{1}{1}+\\frac{1}{3}$. Find the least positive integer $k$ making the number $k!\\cdot S_{2016}$ an integer.", "solution": "We will first calculate $S_{999}$, then $S_{1999}-S_{999}$, and then $S_{2016}-S_{1999}$.\n\nWriting the integers from 1 to 999 as 001 to 999, adding eventually also 000 (since 0 digits actually do not matter), each digit appears exactly 100 times in each position(as unit, ten, or hundred). Hence\n\n$$\nS_{999}=300 \\cdot\\left(\\frac{1}{1}+\\frac{1}{2}+\\cdots+\\frac{1}{9}\\right)\n$$\n\nFor the numbers in the interval $1000 \\rightarrow 1999$, compared to $0 \\rightarrow 999$, there are precisely 1000 more digits 1 . We get\n\n$$\nS_{1999}-S_{999}=1000+S_{999} \\Longrightarrow S_{1999}=1000+600 \\cdot\\left(\\frac{1}{1}+\\frac{1}{2}+\\cdots+\\frac{1}{9}\\right)\n$$\n\nFinally, in the interval $2000 \\rightarrow 2016$, the digit 1 appears twice as unit and seven times as a ten, the digit 2 twice as a unit and 17 times as a thousand, the digits $3,4,5$, and 6 each appear exactly twice as units, and the digits 7,8 , and 9 each appear exactly once as a unit. Hence\n\n$$\nS_{2016}-S_{1999}=9 \\cdot 1+19 \\cdot \\frac{1}{2}+2 \\cdot\\left(\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}+\\frac{1}{6}\\right)+1 \\cdot\\left(\\frac{1}{7}+\\frac{1}{8}+\\frac{1}{9}\\right)\n$$\n\nIn the end, we get\n\n$$\n\\begin{aligned}\nS_{2016} & =1609 \\cdot 1+619 \\cdot \\frac{1}{2}+602 \\cdot\\left(\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}+\\frac{1}{6}\\right)+601 \\cdot\\left(\\frac{1}{7}+\\frac{1}{8}+\\frac{1}{9}\\right) \\\\\n& =m+\\frac{1}{2}+\\frac{2}{3}+\\frac{2}{4}+\\frac{2}{5}+\\frac{2}{6}+\\frac{6}{7}+\\frac{1}{8}+\\frac{7}{9}=n+\\frac{p}{2^{3} \\cdot 3^{2} \\cdot 5 \\cdot 7}\n\\end{aligned}\n$$\n\nwhere $m, n$, and $p$ are positive integers, $p$ coprime to $2^{3} \\cdot 3^{2} \\cdot 5 \\cdot 7$. Then $k!\\cdot S_{2016}$ is an integer precisely when $k$ ! is a multiple of $2^{3} \\cdot 3^{2} \\cdot 5 \\cdot 7$. Since $7 \\mid k!$, it follows that $k \\geq 7$. Also, $7!=2^{4} \\cdot 3^{2} \\cdot 5 \\cdot 7$, implying that the least $k$ satisfying $k!\\cdot S_{2016} \\in \\mathbb{Z}$ is $k=7$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nC1.", "solution_match": "## Solution."}}
+{"year": "2016", "tier": "T3", "problem_label": "C2", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "The natural numbers from 1 to 50 are written down on the blackboard. At least how many of them should be deleted, in order that the sum of any two of the remaining numbers is not a prime?", "solution": "Notice that if the odd, respectively even, numbers are all deleted, then the sum of any two remaining numbers is even and exceeds 2 , so it is certainly not a prime. We prove that 25 is the minimal number of deleted numbers. To this end, we group the positive integers from 1 to 50 in 25 pairs, such that the sum of the numbers within each pair is a prime:\n\n$$\n\\begin{aligned}\n& (1,2),(3,4),(5,6),(7,10),(8,9),(11,12),(13,16),(14,15),(17,20) \\\\\n& (18,19),(21,22),(23,24),(25,28),(26,27),(29,30),(31,36),(32,35) \\\\\n& (33,34),(37,42),(38,41),(39,40),(43,46),(44,45),(47,50),(48,49)\n\\end{aligned}\n$$\n\nSince at least one number from each pair has to be deleted, the minimal number is 25 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nC2.", "solution_match": "\nSolution."}}
+{"year": "2016", "tier": "T3", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "Consider any four pairwise distinct real numbers and write one of these numbers in each cell of a $5 \\times 5$ array so that each number occurs exactly once in every $2 \\times 2$ subarray. The sum over all entries of the array is called the total sum of that array. Determine the maximum number of distinct total sums that may be obtained in this way.", "solution": "We will prove that the maximum number of total sums is 60 .\n\nThe proof is based on the following claim.\n\nClaim. Either each row contains exactly two of the numbers, or each column contains exactly two of the numbers.\n\nProof of the Claim. Indeed, let $R$ be a row containing at least three of the numbers. Then, in row $R$ we can find three of the numbers in consecutive positions, let $x, y, z$ be the numbers in consecutive positions(where $\\{x, y, s, z\\}=\\{a, b, c, d\\}$ ). Due to our hypothesis that in every $2 \\times 2$ subarray each number is used exactly once, in the row above $\\mathrm{R}$ (if there is such a row), precisely above the numbers $x, y, z$ will be the numbers $z, t, x$ in this order. And above them will be the numbers $x, y, z$ in this order. The same happens in the rows below $R$ (see at the following figure).\n\n$$\n\\left(\\begin{array}{lllll}\n\\bullet & x & y & z & \\bullet \\\\\n\\bullet & z & t & x & \\bullet \\\\\n\\bullet & x & y & z & \\bullet \\\\\n\\bullet & z & t & x & \\bullet \\\\\n\\bullet & x & y & z & \\bullet\n\\end{array}\\right)\n$$\n\nCompleting all the array, it easily follows that each column contains exactly two of the numbers and our claim has been proven.\n\nRotating the matrix (if it is necessary), we may assume that each row contains exactly two of the numbers. If we forget the first row and column from the array, we obtain a $4 \\times 4$ array, that can be divided into four $2 \\times 2$ subarrays, containing thus each number exactly four times, with a total sum of $4(a+b+c+d)$. It suffices to find how many different ways are there to put the numbers in the first row $R_{1}$ and the first column $C_{1}$.\n\nDenoting by $a_{1}, b_{1}, c_{1}, d_{1}$ the number of appearances of $a, b, c$, and respectively $d$ in $R_{1}$ and $C_{1}$, the total sum of the numbers in the entire $5 \\times 5$ array will be\n\n$$\nS=4(a+b+c+d)+a_{1} \\cdot a+b_{1} \\cdot b+c_{1} \\cdot c+d_{1} \\cdot d\n$$\n\nIf the first, the third and the fifth row contain the numbers $x, y$, with $x$ denoting the number at the entry $(1,1)$, then the second and the fourth row will contain only the numbers $z, t$, with $z$ denoting the number at the entry $(2,1)$. Then $x_{1}+y_{1}=7$ and $x_{1} \\geqslant 3$, $y_{1} \\geqslant 2, z_{1}+t_{1}=2$, and $z_{1} \\geqslant t_{1}$. Then $\\left\\{x_{1}, y_{1}\\right\\}=\\{5,2\\}$ or $\\left\\{x_{1}, y_{1}\\right\\}=\\{4,3\\}$, respectively $\\left\\{z_{1}, t_{1}\\right\\}=\\{2,0\\}$ or $\\left\\{z_{1}, t_{1}\\right\\}=\\{1,1\\}$. Then $\\left(a_{1}, b_{1}, c_{1}, d_{1}\\right)$ is obtained by permuting one of the following quadruples:\n\n$$\n(5,2,2,0),(5,2,1,1),(4,3,2,0),(4,3,1,1)\n$$\n\nThere are a total of $\\frac{4!}{2!}=12$ permutations of $(5,2,2,0)$, also 12 permutations of $(5,2,1,1)$, 24 permutations of $(4,3,2,0)$ and finally, there are 12 permutations of $(4,3,1,1)$. Hence, there are at most 60 different possible total sums.\n\nWe can obtain indeed each of these 60 combinations: take three rows ababa alternating\nwith two rows $c d c d c$ to get $(5,2,2,0)$; take three rows ababa alternating with one row $c d c d c$ and a row $(d c d c d)$ to get $(5,2,1,1)$; take three rows $a b a b c$ alternating with two rows $c d c d a$ to get $(4,3,2,0)$; take three rows abcda alternating with two rows $c d a b c$ to get $(4,3,1,1)$. By choosing for example $a=10^{3}, b=10^{2}, c=10, d=1$, we can make all these sums different. Hence, 60 is indeed the maximum possible number of different sums.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nC3.", "solution_match": "\nSolution."}}
+{"year": "2016", "tier": "T3", "problem_label": "C4", "problem_type": "Combinatorics", "exam": "JBMO-SL", "problem": "A splitting of a planar polygon is a finite set of triangles whose interiors are pairwise disjoint, and whose union is the polygon in question. Given an integer $n \\geq 3$, determine the largest integer $m$ such that no planar $n$-gon splits into less than $m$ triangles.", "solution": "The required maximum is $\\lceil n / 3\\rceil$, the least integer greater than or equal to $n / 3$. To describe a planar $n$-gon splitting into this many triangles, write $n=3 m-r$, where $m$ is a positive integer and $r=0,1,2$, and consider $m$ coplanar equilateral triangles $A_{3 i} A_{3 i+1} A_{3 i+2}$, $i=0, \\ldots, m-1$, where the $A_{i}$ are pairwise distinct, the $A$ 's of rank congruent to 0 or 2 modulo 3 are all collinear, $A_{2}, A_{3}, A_{5}, \\ldots, A_{3 m-3}, A_{3 m-1}, A_{0}$, in order, and the line $A_{0} A_{2}$ separates $A_{1}$ from the remaining $A$ 's of rank congruent to 1 modulo 3 . The polygon $A_{0} A_{1} A_{2} A_{3} A_{4} A_{5} \\ldots A_{3 m-3} A_{3 m-2} A_{3 m-1}$ settles the case $r=0$; removal of $A_{3}$ from the list settles the case $r=1$; and removal of $A_{3}$ and $A_{3 m-1}$ settles the case $r=2$.\n\nNext, we prove that no planar $n$-gon splits into less than $n / 3$ triangles. Alternatively, but equivalently, if a planar polygon splits into $t$ triangles, then its boundary has (combinatorial) length at most $3 t$. Proceed by induction on $t$. The base case $t=1$ is clear, so let $t>1$.\n\nThe vertices of the triangles in the splitting may subdivide the boundary of the polygon, making it into a possibly combinatorially longer simple loop $\\Omega$. Clearly, it is sufficient to prove that the length of $\\Omega$ does not exceed $3 t$.\n\nTo this end, consider a triangle in the splitting whose boundary $\\omega$ meets $\\Omega$ along at least one of its edges. Trace $\\Omega$ counterclockwise and let $\\alpha_{1}, \\ldots, \\alpha_{k}$, in order, be the connected components of $\\Omega-\\omega$. Each $\\alpha_{i}$ is a path along $\\Omega$ with distinct end points, whose terminal point is joined to the starting point of $\\alpha_{i+1}$ by a (possibly constant) path $\\beta_{i}$ along $\\omega$. Trace $\\omega$ clockwise from the terminal point of $\\alpha_{i}$ to its starting point to obtain a path $\\alpha_{i}^{\\prime}$ of positive length, and notice that $\\alpha_{i}+\\alpha_{i}^{\\prime}$ is the boundary of a polygon split into $t_{i}<t$ triangles. By the induction hypothesis, the length of $\\alpha_{i}+\\alpha_{i}^{\\prime}$ does not exceed $3 t_{i}$, and since $\\alpha_{i}^{\\prime}$ has positive length, the length of $\\alpha_{i}$ is at most $3 t_{i}-1$. Consequently, the length of $\\Omega-\\omega$ does not exceed $\\sum_{i=1}^{k}\\left(3 t_{i}-1\\right)=3 t-3-k$.\n\nFinally, we prove that the total length of the $\\beta_{i}$ does not exceed $k+3$. Begin by noticing that no $\\beta_{i}$ has length greater than 4 , at most one has length greater than 2 , and at most three have length 2 . If some $\\beta_{i}$ has length 4 , then the remaining $k-1$ are all of length at most 1 , so the total length of the $\\beta$ 's does not exceed $4+(k-1)=k+3$. Otherwise, either some $\\beta_{i}$ has length 3 , in which case at most one other has length 2 and the remaining $k-2$ all have length at most 1 , or the $\\beta_{i}$ all have length less than 3 , in which case there are at most three of length 2 and the remaining $k-3$ all have length at most 1 ; in the former case, the total length of the $\\beta$ 's does not exceed $3+2+(k-2)=k+3$, and in the latter, the total length of the $\\beta$ 's does not exceed $3 \\cdot 2+(k-3)=k+3$. The conclusion follows.\n\n## Chapter 3\n\n## Geometry", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nC4.", "solution_match": "\nSolution."}}
+{"year": "2016", "tier": "T3", "problem_label": "G1", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C$ be an acute angled triangle, let $O$ be its circumcentre, and let $D, E, F$ be points on the sides $B C, A C, A B$, respectively. The circle $\\left(c_{1}\\right)$ of radius $F A$, centred at $F$, crosses the segment $(O A)$ at $A^{\\prime}$ and the circumcircle (c) of the triangle $A B C$ again at $K$. Similarly, the circle $\\left(c_{2}\\right)$ of radius $D B$, centred at $D$, crosses the segment $(O B)$ at $B^{\\prime}$ and the circle (c) again at $L$. Finally, the circle $\\left(c_{3}\\right)$ of radius $E C$, centred at $E$, crosses the segment $(O C)$ at $C^{\\prime}$ and the circle (c) again at $M$. Prove that the quadrilaterals $B K F A^{\\prime}$, $C L D B^{\\prime}$ and $A M E C^{\\prime}$ are all cyclic, and their circumcircles share a common point.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_76c244c1de22b5795a02g-16.jpg?height=523&width=539&top_left_y=1429&top_left_x=770)", "solution": "We will prove that the quadrilateral $B K F A^{\\prime}$ is cyclic and its circumcircle passes through the center $O$ of the circle (c).\n\nThe triangle $A F K$ is isosceles, so $m(\\widehat{K F B})=2 m(\\widehat{K A B})=m(\\widehat{K O B})$. It follows that the quadrilateral $B K F O$ is cyclic.\n\nThe triangles $O F K$ and $O F A$ are congruent (S.S.S.), hence $m(\\widehat{O K F})=m(\\widehat{O A F})$. The triangle $F A A^{\\prime}$ is isosceles, so $m\\left(\\widehat{F A^{\\prime} A}\\right)=m(\\widehat{O A F})$. Therefore $m\\left(\\widehat{F A^{\\prime} A}\\right)=m(\\widehat{O K F})$, so the quadrilateral $O K F A^{\\prime}$ is cyclic.\n\n(1) and (2) prove the initial claim.\n\nAlong the same lines, we can prove that the points $C, D, L, B^{\\prime}, O$ and $A, M, E, C^{\\prime}, O$ are concylic, respectively, so their circumcircles also pass through $O$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nG1.", "solution_match": "\nSolution."}}
+{"year": "2016", "tier": "T3", "problem_label": "G2", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C$ be a triangle with $m(\\widehat{B A C})=60^{\\circ}$. Let $D$ and $E$ be the feet of the perpendiculars from $A$ to the external angle bisectors of $\\widehat{A B C}$ and $\\widehat{A C B}$, respectively. Let $O$ be the circumcenter of the triangle $A B C$. Prove that the circumcircles of the triangles $\\triangle A D E$ and $\\triangle B O C$ are tangent to each other.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_76c244c1de22b5795a02g-17.jpg?height=480&width=443&top_left_y=591&top_left_x=818)", "solution": "Let $X$ be the intersection of the lines $B D$ and $C E$.\n\nWe will prove that $X$ lies on the circumcircles of both triangles $\\triangle A D E$ and $\\triangle B O C$ and then we will prove that the centers of these circles and the point $X$ are collinear, which is enough for proving that the circles are tangent to each other.\n\nIn this proof we will use the notation $(M N P)$ to denote the circumcircle of the triangle $\\triangle M N P$.\n\nObviously, the quadrilateral $A D X E$ is cyclic, and the circle $(D A E)$ has $[A X]$ as diameter. (1)\n\nLet $I$ be the incenter of triangle $A B C$. So, the point $I$ lies on the segment $[A X]$ (2), and the quadrilateral $X B I C$ is cyclic because $I C \\perp X C$ and $I B \\perp X B$. So, the circle (BIC) has $[I X]$ as diameter.\n\nFinally, $m(\\widehat{B I C})=90^{\\circ}+\\frac{1}{2} m(\\widehat{B A C})=120^{\\circ}$ and $m(\\widehat{B O C})=2 m(\\widehat{B A C})=120^{\\circ}$.\n\nSo, the quadrilateral $B O I C$ is cyclic and the circle $(B O C)$ has $[I X]$ as diameter. (3)\n\n(1), (2), (3) imply the conclusion.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nG2.", "solution_match": "\nSolution."}}
+{"year": "2016", "tier": "T3", "problem_label": "G3", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "A trapezoid $A B C D(A B \\| C D, A B>C D)$ is circumscribed. The incircle of triangle $A B C$ touches the lines $A B$ and $A C$ at $M$ and $N$, respectively. Prove that the incenter of the trapezoid lies on the line $M N$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_76c244c1de22b5795a02g-18.jpg?height=428&width=646&top_left_y=540&top_left_x=725)", "solution": "Let $I$ be the incenter of triangle $A B C$ and $R$ be the common point of the lines $B I$ and $M N$. Since\n\n$$\nm(\\widehat{A N M})=90^{\\circ}-\\frac{1}{2} m(\\widehat{M A N}) \\quad \\text { and } \\quad m(\\widehat{B I C})=90^{\\circ}+\\frac{1}{2} m(\\widehat{M A N})\n$$\n\nthe quadrilateral $I R N C$ is cyclic.\n\nIt follows that $m(\\widehat{B R C})=90^{\\circ}$ and therefore\n\n$$\nm(\\widehat{B C R})=90^{\\circ}-m(\\widehat{C B R})=90^{\\circ}-\\frac{1}{2}\\left(180^{\\circ}-m(\\widehat{B C D})\\right)=\\frac{1}{2} m(\\widehat{B C D})\n$$\n\nSo, $(C R$ is the angle bisector of $\\widehat{D C B}$ and $R$ is the incenter of the trapezoid.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nG3.", "solution_match": "\nSolution."}}
+{"year": "2016", "tier": "T3", "problem_label": "G4", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C$ be an acute angled triangle whose shortest side is $[B C]$. Consider a variable point $P$ on the side $[B C]$, and let $D$ and $E$ be points on $(A B]$ and $(A C]$, respectively, such that $B D=B P$ and $C P=C E$. Prove that, as $P$ traces $[B C]$, the circumcircle of the triangle $A D E$ passes through a fixed point.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_76c244c1de22b5795a02g-19.jpg?height=554&width=445&top_left_y=591&top_left_x=820)", "solution": "We claim that the fixed point is the center of the incircle of $A B C$.\n\nLet $I$ be the center of the incircle of $A B C$. Since $B D=B P$ and $[B I$ is the bisector of $\\widehat{D B P}$, the line $B I$ is the perpendicular bisector of $[D P]$. This yields $D I=P I$. Analogously we get $E I=P I$. So, the point $I$ is the circumcenter of the triangle $D E P$.\n\nThis means $m(\\widehat{D I E})=2 m(\\widehat{D P E})$.\n\nOn the other hand\n\n$$\n\\begin{aligned}\nm(\\widehat{D P E}) & =180^{\\circ}-m(\\widehat{D P B})-m(\\widehat{E P C}) \\\\\n& =180^{\\circ}-\\left(90^{\\circ}-\\frac{1}{2} m(\\widehat{D B P})\\right)-\\left(90^{\\circ}-\\frac{1}{2} m(\\widehat{E C P})\\right) \\\\\n& =90^{\\circ}-\\frac{1}{2} m(\\widehat{B A C})\n\\end{aligned}\n$$\n\nSo, $m(\\widehat{D I E})=2 m(\\widehat{D P E})=180^{\\circ}-m(\\widehat{D A E})$, which means that the points $A, D, E$ and $I$ are cocyclic.\n\nRemark. The fact that the incentre $I$ of the triangle $A B C$ is the required fixed point could be guessed by considering the two extremal positions of $P$. Thus, if $P=B$, then $D=D_{B}=B$ as well, and $C E=C E_{B}=B C$, so $m(\\angle A E B)=m(\\angle C)+m(\\angle E B C)=$ $m(\\angle C)+\\frac{180^{\\circ}-m(\\angle C)}{2}=90^{\\circ}+\\frac{m(\\angle C)}{2}=m(\\angle A I B)$. Hence the points $A, E=E_{B}, I, D_{B}=B$ are cocyclic. Similarly, letting $P=C$, the points $A, D=D_{C}, I, E_{C}=C$ are cocyclic. Consequently, the circles $A D_{B} E_{B}$ and $A D_{C} E_{C}$ meet again at $I$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nG4.", "solution_match": "\nSolution."}}
+{"year": "2016", "tier": "T3", "problem_label": "G5", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $A B C$ be an acute angled triangle with orthocenter $H$ and circumcenter $O$. Assume the circumcenter $X$ of $B H C$ lies on the circumcircle of $A B C$. Reflect $O$ across $X$ to obtain $O^{\\prime}$, and let the lines $X H$ and $O^{\\prime} A$ meet at $K$. Let $L, M$ and $N$ be the midpoints of $[X B],[X C]$ and $[B C]$, respectively. Prove that the points $K, L, M$ and $N$ are cocyclic.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_76c244c1de22b5795a02g-20.jpg?height=440&width=651&top_left_y=588&top_left_x=720)", "solution": "The circumcircles of $A B C$ and $B H C$ have the same radius. So, $X B=$ $X C=X H=X O=r$ (where $r$ is the radius of the circle $A B C$ ) and $O^{\\prime}$ lies on $C(X, r)$. We conclude that $O X$ is the perpendicular bisector for $[B C]$. So, $B O X$ and $C O X$ are equilateral triangles.\n\nIt is known that $A H=2 O N=r$. So, $A H O^{\\prime} X$ is parallelogram, and $X K=K H=r / 2$. Finally, $X L=X K=X N=X M=r / 2$. So, $K, L, M$ and $N$ lie on the circle $c(X, r / 2)$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nG5.", "solution_match": "\nSolution."}}
+{"year": "2016", "tier": "T3", "problem_label": "G6", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Given an acute triangle $A B C$, erect triangles $A B D$ and $A C E$ externally, so that $m(\\widehat{A D B})=m(\\widehat{A E C})=90^{\\circ}$ and $\\widehat{B A D} \\equiv \\widehat{C A E}$. Let $A_{1} \\in B C, B_{1} \\in A C$ and $C_{1} \\in A B$ be the feet of the altitudes of the triangle $A B C$, and let $K$ and $L$ be the midpoints of $\\left[B C_{1}\\right]$ and $\\left[C B_{1}\\right]$, respectively. Prove that the circumcenters of the triangles $A K L, A_{1} B_{1} C_{1}$ and $D E A_{1}$ are collinear.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_76c244c1de22b5795a02g-21.jpg?height=537&width=944&top_left_y=648&top_left_x=567)", "solution": "Let $M, P$ and $Q$ be the midpoints of $[B C],[C A]$ and $[A B]$, respectively.\n\nThe circumcircle of the triangle $A_{1} B_{1} C_{1}$ is the Euler's circle. So, the point $M$ lies on this circle.\n\nIt is enough to prove now that $\\left[A_{1} M\\right]$ is a common chord of the three circles $\\left(A_{1} B_{1} C_{1}\\right)$, $(A K L)$ and $\\left(D E A_{1}\\right)$.\n\nThe segments $[M K]$ and $[M L]$ are midlines for the triangles $B C C_{1}$ and $B C B_{1}$ respectively, hence $M K \\| C C_{1} \\perp A B$ and $M L \\| B B_{1} \\perp A C$. So, the circle $(A K L)$ has diameter $[A M]$ and therefore passes through $M$.\n\nFinally, we prove that the quadrilateral $D A_{1} M E$ is cyclic.\n\nFrom the cyclic quadrilaterals $A D B A_{1}$ and $A E C A_{1}, \\widehat{A A_{1} D} \\equiv \\widehat{A B D}$ and $\\widehat{A A_{1} E} \\equiv \\widehat{A C E} \\equiv$ $\\widehat{A B D}$, so $m\\left(\\widehat{D A_{1} E}\\right)=2 m(\\widehat{A B D})=180^{\\circ}-2 m(\\widehat{D A B})$.\n\nWe notice now that $D Q=A B / 2=M P, Q M=A C / 2=P E$ and\n\n$$\n\\begin{aligned}\n& m(\\widehat{D Q M})=m(\\widehat{D Q B})+m(\\widehat{B Q M})=2 m(\\widehat{D A B})+m(\\widehat{B A C}) \\\\\n& m(\\widehat{E P M})=m(\\widehat{E P C})+m(\\widehat{C P M})=2 m(\\widehat{E A C})+m(\\widehat{C A B})\n\\end{aligned}\n$$\n\nso $\\triangle M P E \\equiv \\triangle D Q M$ (S.A.S.). This leads to $m(\\widehat{D M E})=m(\\widehat{D M Q})+m(Q M P)+$ $m(P M E)=m(\\widehat{D M Q})+m(B Q M)+m(Q D M)=180^{\\circ}-m(\\widehat{D Q B})=180^{\\circ}-2 m(\\widehat{D A B})$. Since $m\\left(\\widehat{D A_{1} E}\\right)=m(\\widehat{D M E})$, the quadrilateral $D A_{1} M E$ is cyclic.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nG6.", "solution_match": "\nSolution."}}
+{"year": "2016", "tier": "T3", "problem_label": "G7", "problem_type": "Geometry", "exam": "JBMO-SL", "problem": "Let $[A B]$ be a chord of a circle $(c)$ centered at $O$, and let $K$ be a point on the segment $(A B)$ such that $A K<B K$. Two circles through $K$, internally tangent to (c) at $A$ and $B$, respectively, meet again at $L$. Let $P$ be one of the points of intersection of the line $K L$ and the circle (c), and let the lines $A B$ and $L O$ meet at $M$. Prove that the line $M P$ is tangent to the circle $(c)$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_76c244c1de22b5795a02g-22.jpg?height=497&width=606&top_left_y=657&top_left_x=742)", "solution": "Let $\\left(c_{1}\\right)$ and $\\left(c_{2}\\right)$ be circles through $K$, internally tangent to (c) at $A$ and $B$, respectively, and meeting again at $L$, and let the common tangent to $\\left(c_{1}\\right)$ and $(c)$ meet the common tangent to $\\left(c_{2}\\right)$ and $(c)$ at $Q$. Then the point $Q$ is the radical center of the circles $\\left(c_{1}\\right),\\left(c_{2}\\right)$ and $(c)$, and the line $K L$ passes through $Q$.\n\nWe have $m(\\widehat{Q L B})=m(\\widehat{Q B K})=m(\\widehat{Q B A})=\\frac{1}{2} m(\\overparen{B A})=m(\\widehat{Q O B})$. So, the quadrilateral $O B Q L$ is cyclic. We conclude that $m(\\widehat{Q L O})=90^{\\circ}$ and the points $O, B, Q, A$ and $L$ are cocyclic on a circle $(k)$.\n\nIn the sequel, we will denote $\\mathcal{P}_{\\omega}(X)$ the power of the point $X$ with respect of the circle $\\omega$. The first continuation.\n\nFrom $M O^{2}-O P^{2}=\\mathcal{P}_{c}(M)=M A \\cdot M B=\\mathcal{P}_{k}(M)=M L \\cdot M O=(M O-O L) \\cdot M O=$ $M O^{2}-O L \\cdot M O$ follows that $O P^{2}=O L \\cdot O M$. Since $P L \\perp O M$, this shows that the triangle $M P O$ is right at point $P$. Thus, the line $M P$ is tangent to the circle (c).\n\nThe second continuation.\n\nLet $R \\in(c)$ be so that $B R \\perp M O$. The triangle $L B R$ is isosceles with $L B=L R$, so $\\widehat{O L R} \\equiv \\widehat{O L B} \\equiv \\widehat{O Q B} \\equiv \\widehat{O Q A} \\equiv \\widehat{M L A}$. We conclude that the points $A, L$ and $R$ are collinear.\n\nNow $m(\\widehat{A M R})+m(\\widehat{A O R})=m(\\widehat{A M R})+2 m(\\widehat{A B R})=m(\\widehat{A M R})+m(\\widehat{A B R})+m(\\widehat{M R B})=$ $180^{\\circ}$, since the triangle $M B R$ is isosceles. So, the quadrilateral $M A O R$ is cyclic.\n\nThis yields $L M \\cdot L O=-\\mathcal{P}_{(\\text {MAOR })}(L)=L A \\cdot L R=-\\mathcal{P}_{c}(L)=L P^{2}$, which as above, shows that $O P \\perp P M$.\n\nThe third continuation.\n\n$\\widehat{K L A} \\equiv \\widehat{K A Q} \\equiv \\widehat{K L B}$ and $m(\\widehat{M L K})=90^{\\circ}$ show that $[L K$ and $[L M$ are the internal and external bisectors af the angle $\\widehat{A L B}$, so $(M, K)$ and $(A, B)$ are harmonic conjugates. So, $L K$ is the polar line of $M$ in the circle $(c)$.\n\n## Chapter 4\n\n## Number Theory", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nG7.", "solution_match": "\nSolution."}}
+{"year": "2016", "tier": "T3", "problem_label": "N1", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Determine the largest positive integer $n$ that divides $p^{6}-1$ for all primes $p>7$.", "solution": "Note that\n\n$$\np^{6}-1=(p-1)(p+1)\\left(p^{2}-p+1\\right)\\left(p^{2}+p+1\\right)\n$$\n\nFor $p=11$ we have\n\n$$\np^{6}-1=1771560=2^{3} \\cdot 3^{2} \\cdot 5 \\cdot 7 \\cdot 19 \\cdot 37\n$$\n\nFor $p=13$ we have\n\n$$\np^{6}-1=2^{3} \\cdot 3^{2} \\cdot 7 \\cdot 61 \\cdot 157\n$$\n\nFrom the last two calculations we find evidence to try showing that $p^{6}-1$ is divisible by $2^{3} \\cdot 3^{2} \\cdot 7=504$ and this would be the largest positive integer that divides $p^{6}-1$ for all primes greater than 7 .\n\nBy Fermat's theorem, $7 \\mid p^{6}-1$.\n\nNext, since $p$ is odd, $8 \\mid p^{2}-1=(p-1)(p+1)$, hence $8 \\mid p^{6}-1$.\n\nIt remains to show that $9 \\mid p^{6}-1$.\n\nAny prime number $p, p>3$ is 1 or -1 modulo 3 .\n\nIn the first case both $p-1$ and $p^{2}+p+1$ are divisible by 3 , and in the second case, both $p+1$ and $p^{2}-p+1$ are divisible by 3 .\n\nConsequently, the required number is indeed 504\n\n## Alternative solution\n\nLet $q$ be a (positive) prime factor of $n$. Then $q \\leq 7$, as $q \\nmid q^{6}-1$. Also, $q$ is not 5 , as the last digit of $13^{6}-1$ is 8 .\n\nHence, the prime factors of $n$ are among 2,3 , and 7 .\n\nNext, from $11^{6}-1=2^{3} \\cdot 3^{2} \\cdot 5 \\cdot 7 \\cdot 19 \\cdot 37$ it follows that the largest integer $n$ such that $n \\mid p^{6}-1$ for all primes $p$ greater than 7 is at most $2^{3} \\cdot 3^{2} \\cdot 7$, and it remains to prove that 504 divides $p^{6}-1$ for all primes grater than 7 .", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nN1.", "solution_match": "\nSolution."}}
+{"year": "2016", "tier": "T3", "problem_label": "N2", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find the maximum number of natural numbers $x_{1}, x_{2}, \\ldots, x_{m}$ satisfying the conditions:\n\na) No $x_{i}-x_{j}, 1 \\leq i<j \\leq m$ is divisible by 11 ; and\n\nb) The sum $x_{2} x_{3} \\ldots x_{m}+x_{1} x_{3} \\ldots x_{m}+\\cdots+x_{1} x_{2} \\ldots x_{m-1}$ is divisible by 11 .", "solution": "The required maximum is 10 .\n\nAccording to a), the numbers $x_{i}, 1 \\leq i \\leq m$, are all different $(\\bmod 11)$ (1)\n\nHence, the number of natural numbers satisfying the conditions is at most 11.\n\nIf $x_{j} \\equiv 0(\\bmod 11)$ for some $j$, then\n\n$$\nx_{2} x_{3} \\ldots x_{m}+x_{1} x_{3} \\ldots x_{m}+\\cdots+x_{1} x_{2} \\ldots x_{m-1} \\equiv x_{1} \\ldots x_{j-1} x_{j+1} \\ldots x_{m} \\quad(\\bmod 11)\n$$\n\nwhich would lead to $x_{i} \\equiv 0(\\bmod 11)$ for some $i \\neq j$, contradicting (1).\n\nWe now prove that 10 is indeed the required maximum.\n\nConsider $x_{i}=i$, for all $i \\in\\{1,2, \\ldots, 10\\}$. The products $2 \\cdot 3 \\cdots \\cdot 10,1 \\cdot 3 \\cdots \\cdots 10, \\ldots$, $1 \\cdot 2 \\cdots \\cdot 9$ are all different $(\\bmod 11)$, and so\n\n$$\n2 \\cdot 3 \\cdots \\cdots 10+1 \\cdot 3 \\cdots \\cdots 10+\\cdots+1 \\cdot 2 \\cdots \\cdot 9 \\equiv 1+2+\\cdots+10 \\quad(\\bmod 11)\n$$\n\nand condition b) is satisfied, since $1+2+\\cdots+10=55=5 \\cdot 11$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nN2.", "solution_match": "\nSolution."}}
+{"year": "2016", "tier": "T3", "problem_label": "N3", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all positive integers $n$ such that the number $A_{n}=\\frac{2^{4 n+2}+1}{65}$ is\n\na) an integer;\n\nb) a prime.", "solution": "a) Note that $65=5 \\cdot 13$.\n\nObviously, $5=2^{2}+1$ is a divisor of $\\left(2^{2}\\right)^{2 n+1}+1=2^{4 n+2}+1$ for any positive integer $n$. Since $2^{12} \\equiv 1(\\bmod 13)$, if $n \\equiv r(\\bmod 3)$, then $2^{4 n+2}+1 \\equiv 2^{4 r+2}+1(\\bmod 13)$. Now, $2^{4 \\cdot 0+2}+1=5,2^{4 \\cdot 1+2}+1=65$, and $2^{4 \\cdot 2+2}+1=1025=13 \\cdot 78+11$. Hence 13 is a divisor of $2^{4 n+2}+1$ precisely when $n \\equiv 1(\\bmod 3)$. Hence, $A_{n}$ is an integer iff $n \\equiv 1(\\bmod 3)$.\n\nb) Applying the identity $4 x^{4}+1=\\left(2 x^{2}-2 x+1\\right)\\left(2 x^{2}+2 x+1\\right)$, we have $2^{4 n+2}+1=$ $\\left(2^{2 n+1}-2^{n+1}+1\\right)\\left(2^{2 n+1}+2^{n+1}+1\\right)$. For $n=1, A_{1}=1$, which is not a prime. According to a), if $n \\neq 1$, then $n \\geq 4$. But then $2^{2 n+1}+2^{n+1}+1>2^{2 n+1}-2^{n+1}+1>65$, and $A_{n}$ has at least two factors. We conclude that $A_{n}$ can never be a prime.\n\nAlternative Solution to b): Knowing that $n=3 k+1$ in order for $A_{n}$ to be an integer, $2^{4 n+2}+1=2^{12 k+6}+1=\\left(2^{4 k+2}\\right)^{3}+1=\\left(2^{4 k+2}+1\\right)\\left(2^{8 k+4}-2^{4 k+2}+1\\right) \\quad(*)$. As in the previous solution, if $k=0$, then $A_{1}=1$, if $k=1$, then $A_{4}=2^{12}-2^{6}+1=4033=37 \\cdot 109$, and for $k \\geq 2$ both factors in $(*)$ are larger than 65 , so $A_{3 k+1}$ is not a prime.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nN3.", "solution_match": "\nSolution."}}
+{"year": "2016", "tier": "T3", "problem_label": "N4", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Find all triples of integers $(a, b, c)$ such that the number\n\n$$\nN=\\frac{(a-b)(b-c)(c-a)}{2}+2\n$$\n\nis a power of 2016 .", "solution": "Let $z$ be a positive integer such that\n\n$$\n(a-b)(b-c)(c-a)+4=2 \\cdot 2016^{z}\n$$\n\nWe set $a-b=-x, b-c=-y$ and we rewrite the equation as\n\n$$\nx y(x+y)+4=2 \\cdot 2016^{z}\n$$\n\nNote that the right hand side is divisible by 7 , so we have that\n\n$$\nx y(x+y)+4 \\equiv 0 \\quad(\\bmod 7)\n$$\n\nor\n\n$$\n3 x y(x+y) \\equiv 2 \\quad(\\bmod 7)\n$$\n\nor\n\n$$\n(x+y)^{3}-x^{3}-y^{3} \\equiv 2 \\quad(\\bmod 7)\n$$\n\nNote that, by Fermat's Little Theorem, we have that for any integer $k$ the cubic residues are $k^{3} \\equiv-1,0,1 \\bmod 7$. It follows that in (4.1) some of $(x+y)^{3}, x^{3}$ and $y^{3}$ should be divisible by 7 , but in this case, $x y(x+y)$ is divisible by 7 and this is a contradiction. So, the only possibility is to have $z=0$ and consequently, $x y(x+y)+4=2$, or, equivalently, $x y(x+y)=-2$. The only solution of the latter is $(x, y)=(-1,-1)$, so the required triples are $(a, b, c)=(k+2, k+1, k), k \\in \\mathbb{Z}$, and all their cyclic permutations.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nN4.", "solution_match": "\nSolution."}}
+{"year": "2016", "tier": "T3", "problem_label": "N5", "problem_type": "Number Theory", "exam": "JBMO-SL", "problem": "Determine all four-digit numbers $\\overline{a b c d}$ such that\n\n$$\n(a+b)(a+c)(a+d)(b+c)(b+d)(c+d)=\\overline{a b c d}\n$$", "solution": "Depending on the parity of $a, b, c, d$, at least two of the factors $(a+b),(a+c)$, $(a+d),(b+c),(b+d),(c+d)$ are even, so that $4 \\mid \\overline{a b c d}$.\n\nWe claim that $3 \\mid \\overline{a b c d}$.\n\nAssume $a+b+c+d \\equiv 2(\\bmod 3)$. Then $x+y \\equiv 1(\\bmod 3)$, for all distinct $x, y \\in\\{a, b, c, d\\}$. But then the left hand side in the above equality is congruent to $1(\\bmod 3)$ and the right hand side congruent to $2(\\bmod 3)$, contradiction.\n\nAssume $a+b+c+d \\equiv 1(\\bmod 3)$. Then $x+y \\equiv 2(\\bmod 3)$, for all distinct $x, y \\in\\{a, b, c, d\\}$, and $x \\equiv 1(\\bmod 3)$, for all $x, y \\in\\{a, b, c, d\\}$. Hence, $a, b, c, d \\in\\{1,4,7\\}$, and since $4 \\mid \\overline{a b c d}$, we have $c=d=4$. Therefore, $8 \\mid \\overline{a b 44}$, and since at least one more factor is even, it follows that $16 \\overline{a b 44}$. Then $b \\neq 4$, and the only possibilities are $b=1$, implying $a=4$, which is impossible because 4144 is not divisible by $5=1+4$, or $b=7$, implying $11 \\mid \\overline{a 744}$, hence $a=7$, which is also impossible because 7744 is not divisible by $14=7+7$.\n\nWe conclude that $3 \\mid \\overline{a b c d}$, hence also $3 \\mid a+b+c+d$. Then at least one factor $x+y$ of $(a+b),(a+c),(a+d),(b+c),(b+d),(c+d)$ is a multiple of 3 , implying that also $3 \\mid a+b+c+d-x-y$, so $9 \\mid \\overline{a b c d}$. Then $9 \\mid a+b+c+d$, and $a+b+c+d \\in\\{9,18,27,36\\}$. Using the inequality $x y \\geq x+y-1$, valid for all $x, y \\in \\mathbb{N}^{*}$, if $a+b+c+d \\in\\{27,36\\}$, then\n\n$$\n\\overline{a b c d}=(a+b)(a+c)(a+d)(b+c)(b+d)(c+d) \\geq 26^{3}>10^{4}\n$$\n\nwhich is impossible.\n\nUsing the inequality $x y \\geq 2(x+y)-4$ for all $x, y \\geq 2$, if $a+b+c+d=18$ and all two-digit sums are greater than 1 , then $\\overline{a b c d} \\geq 32^{3}>10^{4}$. Hence, if $a+b+c+d=18$, some two-digit sum must be 1 , hence the complementary sum will be 17 , and the digits are $\\{a, b, c, d\\}=\\{0,1,8,9\\}$. But then $\\overline{a b c d}=1 \\cdot 17 \\cdot 8 \\cdot 9^{2} \\cdot 10>10^{4}$.\n\nWe conclude that $a+b+c+d=9$. Then among $a, b, c, d$ there are either three odd or three even numbers, and $8 \\mid \\overline{a b c d}$.\n\nIf three of the digits are odd, then $d$ is even and since $c$ is odd, divisibility by 8 implies that $d \\in\\{2,6\\}$. If $d=6$, then $a=b=c=1$. But 1116 is not divisible by 7 , so this is not a solution. If $d=2$, then $a, b, c$ are either $1,1,5$ or $1,3,3$ in some order. In the first case $2 \\cdot 6^{2} \\cdot 3^{2} \\cdot 7=4536 \\neq \\overline{a b c d}$. The second case cannot hold because the resulting number is not a multiple of 5 .\n\nHence, there has to be one odd and three even digits. At least one of the two-digits sums of even digits is a multiple of 4 , and since there cannot be two zero digits, we have either $x+y=4$ and $z+t=5$, or $x+y=8$ and $z+t=1$ for some ordering $x, y, z, t$ of $a, b, c, d$. In the first case we have $d=0$ and the digits are $0,1,4,4$, or $0,2,3,4$, or $0,2,2,5$. None of these is a solution because $1 \\cdot 4^{2} \\cdot 5^{2} \\cdot 8=3200,2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdot 7=5040$ and $2^{2} \\cdot 5 \\cdot 4 \\cdot 7^{2}=3920$. In the second case two of the digits are 0 and 1 , and the other two have to be either 4 and 4 , or 2 and 6 . We already know that the first possibility fails. For the second, we get\n\n$$\n(0+1) \\cdot(0+2) \\cdot(0+6) \\cdot(1+2) \\cdot(1+6) \\cdot(2+6)=2016\n$$\n\nand $\\overline{a b c d}=2016$ is the only solution.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-shortlist_jbmo_2016_v7-1.jsonl", "problem_match": "\nN5.", "solution_match": "\nSolution."}}