diff --git "a/JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl" "b/JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl" --- "a/JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl" +++ "b/JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl" @@ -1,19 +1,19 @@ -{"year": "2003", "tier": "T3", "problem_label": "ALG 1", "problem_type": "Algebra", "problem": "A number $A$ is written with $2 n$ digits, each of whish is 4 , and a number $B$ is written with $n$ digits, each of which is 8 . Prove that for each $n, A+2 B+4$ is a total square.", "solution": "$$\n\\begin{aligned}\nA & =\\underbrace{44 \\ldots 44}_{2 n}=\\underbrace{44 \\ldots 4}_{n} \\underbrace{44 \\ldots 4}_{n}=\\underbrace{44 \\ldots 4}_{n} \\underbrace{400 \\ldots 0}_{n}-\\underbrace{44 \\ldots 4}_{n}+\\underbrace{88 \\ldots 8}_{n}=\\underbrace{44 \\ldots 4}_{n} \\cdot\\left(10^{n}-1\\right)+B \\\\\n& =4 \\cdot \\underbrace{11 \\ldots 1}_{n} \\cdot \\underbrace{99 \\ldots 9}_{n}+B=2^{2} \\cdot \\underbrace{11 \\ldots 1}_{n} \\cdot 3^{2} \\cdot \\underbrace{11 \\ldots 1}_{n}+B=\\underbrace{66}_{n} \\ldots 6 \\\\\n& =[\\frac{36}{4} \\cdot \\underbrace{88 \\ldots 8}_{n}+B=[3 \\cdot \\underbrace{22 \\ldots 2}_{n}]^{2}+B=\\left(\\frac{3}{4} B\\right)^{2}+B .\n\\end{aligned}\n$$\n\nSo,\n\n$$\n\\begin{aligned}\nA+2 B+4 & =\\left(\\frac{3}{4} B\\right)^{2}+B+2 B+4=\\left(\\frac{3}{4} B\\right)^{2}+2 \\cdot \\frac{3}{4} B \\cdot 2+2^{2}=\\left(\\frac{3}{4} B+2\\right)^{2}=(\\frac{3}{4} \\cdot \\underbrace{88 \\ldots 8}_{n}+2)^{2} \\\\\n& =(3 \\cdot \\underbrace{22 \\ldots 2}_{n}+2)^{2}=\\underbrace{66 \\ldots 68^{2}}_{n-1}\n\\end{aligned}\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_tag": "\nALG 1.", "solution_tag": "## Solution."}} -{"year": "2003", "tier": "T3", "problem_label": "ALG 2", "problem_type": "Algebra", "problem": "Let $a, b, c$ be lengths of triangle sides, $p=\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}$ and $q=\\frac{a}{c}+\\frac{c}{b}+\\frac{b}{a}$.\n\nProve that $|p-q|<1$.", "solution": "One has\n\n$$\n\\begin{aligned}\na b c|p-q| & =a b c\\left|\\frac{c-b}{a}+\\frac{a-c}{b}+\\frac{b-a}{c}\\right| \\\\\n& =\\left|b c^{2}-b^{2} c+a^{2} c-a c^{2}+a b^{2}-a^{2} b\\right|= \\\\\n& =\\left|a b c-a c^{2}-a^{2} b+a^{2} c-b^{2} c+b c^{2}+a b^{2}-a b c\\right|= \\\\\n& =\\left|(b-c)\\left(a c-a^{2}-b c+a b\\right)\\right|= \\\\\n& =|(b-c)(c-a)(a-b)| .\n\\end{aligned}\n$$\n\nSince $|b-c|0$. The contradiction shows that $P=-\\frac{5}{2}$.\n\nComment: By the Cauchy Schwarz inequality $|t| \\leq \\sqrt{3}$, so the smallest value of $P$ is attained at $t=\\sqrt{3}$ and equals $1-2 \\sqrt{3} \\approx-2.46$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_tag": "\nALG 3:", "solution_tag": "\nSolution:"}} -{"year": "2003", "tier": "T3", "problem_label": "ALG 4", "problem_type": "Algebra", "problem": "Let $a, b, c$ be rational numbers such that\n\n$$\n\\frac{1}{a+b c}+\\frac{1}{b+a c}=\\frac{1}{a+b}\n$$\n\nProve that $\\sqrt{\\frac{c-3}{c+1}}$ is also a rational number", "solution": "By cancelling the denominators\n\n$$\n(a+b)^{2}(1+c)=a b+c\\left(a^{2}+b^{2}\\right)+a b c^{2}\n$$\n\nand\n\n$$\na b(c-1)^{2}=(a+b)^{2}\n$$\n\nIf $c=-1$, we obtrin the contradiction\n\n$$\n\\frac{1}{a-b}+\\frac{1}{b-a}=\\frac{1}{a+b}\n$$\n\nFurtherrdore,\n\n$$\n\\begin{aligned}\n(c-3)(c+1) & =(c-1)^{2}-4=\\frac{(a+b)^{2}}{a b}-4 \\\\\n& =\\frac{(a-b)^{2}}{a b}=\\left(\\frac{(a-b)(c-1)}{a+b}\\right)^{2}\n\\end{aligned}\n$$\n\nThus\n\n$$\n\\sqrt{\\frac{c-3}{c+1}}=\\frac{\\sqrt{(c-3)(c+1)}}{c+1}=\\frac{|a-b||c-1|}{(c+1)|a+b|} \\in \\mathrm{Q}\n$$\n\nas needed.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_tag": "## ALG 4.", "solution_tag": "\nSolution."}} -{"year": "2003", "tier": "T3", "problem_label": "ALG 5", "problem_type": "Algebra", "problem": "Let $A B C$ be a scalene triangle with $B C=a, A C=b$ and $A B=c$, where $a_{r} b, c$ are positive integers. Prove that\n\n$$\n\\left|a b^{2}+b c^{2}+c a^{2}-a^{2} b-b^{2} c-c^{2} a\\right| \\geq 2\n$$", "solution": "Denote $E=a b^{2}+b c^{2}+c a^{2}-a^{2} b-b^{2} c-c^{2} a$. We have\n\n$$\n\\begin{aligned}\nE= & \\left(a b c-c^{2} a\\right)+\\left(c a^{2}-a^{2} b\\right)+\\left(b c^{2}-b^{2} c\\right)+\\left(a b^{2}-a b c\\right)= \\\\\n& (b-c)\\left(a c-a^{2}-b c+a b\\right)=(b-c)\\left(a a^{2}-b\\right)(c-a)\n\\end{aligned}\n$$\n\nSo, $|E|=|a-b| \\cdot|b-c| \\cdot|c-a|$. By hypothesis each factor from $|E|$ is a positive integer. We shall prove that at least one factor from $|E|$ is greater than 1. Suppose that $|a-b|=|b-c|=|c-a|=1$. It follows that the numbers $a-b, b-c, c-a$ are odd. So, the number $0=(a-b)+(b-c) \\div(c-a)$ is olso odd, a contradiction. Hence, $|E| \\geq 1 \\cdot 1 \\cdot 2=2$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_tag": "\nALG 5.", "solution_tag": "\nSolution."}} -{"year": "2003", "tier": "T3", "problem_label": "ALG 6", "problem_type": "Algebra", "problem": "Let $a, b, c$ be positive numbers such that $a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=3$. Prove that\n\n$$\na+b+c \\geq a b c+2\n$$", "solution": "We can consider the case $a \\geq b \\geq c$ which implies $c \\leq 1$. The given inequality writes\n\n$$\na+b-2 \\geq(a b-1) c \\geq(a b-1) c^{2}=(a b-1) \\frac{3-a^{2} b^{2}}{a^{2}+b^{2}}\n$$\n\nPut $x=\\sqrt{a b}$. From the inequality $3 a^{2} b^{2} \\geq a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=3$ we infer $x \\geq 1$ and from $a^{2} b^{2}0$, by $A M-G M$ inequality we have\n\n$$\n\\frac{A}{B}+\\frac{B}{C}+\\frac{C}{A} \\geq 3 \\sqrt[3]{\\frac{A}{B} \\cdot \\frac{B}{C} \\cdot \\frac{C}{A}}\n$$\n\nand\n\n$$\n\\frac{B}{A}+\\frac{C}{B}+\\frac{A}{C} \\geq 3\n$$\n\nand we are done.\n\nAlternative solution for inequality (1).\n\nBy the Cauchy-Schwarz inequality,\n\n$$\n\\frac{a}{2 c+b}+\\frac{b}{2 a+c}+\\frac{c}{2 b+a}=\\frac{a^{2}}{2 a c+a b}+\\frac{b^{2}}{2 a b+c b}+\\frac{c^{2}}{2 b c+a c} \\geq \\frac{(a+b+c)^{2}}{3(a b+b c+c a)} \\geq 1\n$$\n\nThe last inequality reduces immediately to the obvious $a^{2}+b^{2}+c^{2} \\geq a b+b c+c a$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_tag": "\nALG 7 ", "solution_tag": "\nSolution."}} -{"year": "2003", "tier": "T3", "problem_label": "ALG 8", "problem_type": "Algebra", "problem": "Prove that there exist two sets $A=\\{x, y, z\\}$ and $B=\\{m, n, p\\}$ of positive integers greater than 2003 such that the sets have no common elements and the equalities $x+y+z=m+n+p$ and $x^{2}+y^{2}+z^{2}=m^{2}+n^{2}+p^{2}$ hold.", "solution": "Let $A B C$ be a triangle with $B C=a, A C=b, A B=c$ and $ak+3=c\n$$\n\na triangle with such length sides there exist. After the simple calculations we have\n\n$$\n\\begin{gathered}\nA=\\left\\{3(k+1)^{2}-2,3(k+2)^{2}+4,3(k+3)^{2}-2\\right\\} \\\\\nB=\\left\\{3(k+1)^{2}, 3(k+2)^{2}, 3(k+3)^{2}\\right\\}\n\\end{gathered}\n$$\n\nIt easy to prove that\n\n$$\n\\begin{gathered}\nx+y+z=m+n+p=3\\left[(k+1)^{2}+(k+2)^{2}+(k+3)^{2}\\right] \\\\\nx^{2}+y^{2}+z^{2}=m^{2}+n^{2}+p^{2}=9\\left[(k+1)^{4}+(k+2)^{4}+(k+3)^{4}\\right]\n\\end{gathered}\n$$\n\n$>$ From the inequality $3(k+1)^{2}-2>2003$ we obtain $k \\geq 25$. For $k=25$ we have an example of two sets\n\n$$\nA=\\{2026,2191,2350\\}, \\quad B=\\{2028,2187,2352\\}\n$$\n\nwith desired properties.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_tag": "\nALG 8.", "solution_tag": "\nSolution."}} -{"year": "2003", "tier": "T3", "problem_label": "COM 1", "problem_type": "Combinatorics", "problem": "In a group of 60 students: 40 speak English; 30 speak French; 8 speak all the three languages; the number of students that speak English and French but not German is equal to \"the sum of the number of students that speak English and German but not French plus the number of students that speak French and German but not English; and the number of students that speak at least 2 of those fanguages is 28 . How many students speak:\na) German;\nb) only English;\nc) only German?", "solution": "We use the following notation.\n\n$E=\\#$ students that speak English, $F=\\#$ students that speak French,\n\n$G=\\#$ students that speak German; $m=$ \\# students that speak all the three languages,\n\n$x=\\#$ students that speak English and French but not German,\n\n$y=\\#$ students that speak German and French but not English,\n\n$z=\\#$ students that speak English and German but not French.\n\nThe conditions $x+y=z$ and $x+y+z+8=28$, imply that $z=x+y=10$, i.e. 10 students speak German and French, but not English. Then: $G+E-y-8+F-x-8-10=60$, implies that $G+70-$ $36=60$. Hence: a) $\\mathrm{G}=36$; b) only English speak $40-10-8=22$ students; c) the information given is not enough to find the number of students that speak only German. This ${ }^{n}$ number can be any one from 8 to 18 .\n\nComment: There are some mistakes in the solution. The corrections are as follows:\n\n1. The given condition is $x=y+z($ not $x+y=z)$; thus $x=y+z=10$.\n2. From $G+70-36=60$ one gets $G=26$ (not $G=36$ ).\n3. One gets \"only German speakers\" as $G-y-z-8=8$.\n4. \"Only English speakers\" are $E-x-z-8=22-z$, so this number can not be determined.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_tag": "\nCOM 1.", "solution_tag": "\nSolution:"}} -{"year": "2003", "tier": "T3", "problem_label": "COM 2", "problem_type": "Combinatorics", "problem": "Natural numbers 1,2,3, .., 2003 are written in an arbitrary sequence $a_{1}, a_{2}, a_{3}, \\ldots a_{2003}$. Let $b_{1}=1 a_{1}, b_{2}=2 a_{2}, b_{3}=3 a_{3}, \\ldots, b_{2003}=2003 a_{2003}$, and $B$ be the maximum of the numbers $b_{1}, b_{2}, b_{3}, \\ldots, b_{2003}$.\n\na) If $a_{1}=2003, a_{2}=2002, a_{3}=2001, \\ldots, a_{2002}=2, a_{2003}=1$, find the value of $B$.\n\nb) Prove that $B \\geq 1002^{2}$.", "solution": "a) Using the inequality between the arithmetical and geometrical mean, we obtain that $b_{n}=n(2004-n) \\leq\\left(\\frac{n+(2004-n)}{2}\\right)^{2}=1002^{2}$ for $n=1,2,3, \\ldots, 2003$. The equality holds if and only if $n=2004-n$, i.e. $n=1002$. Therefore, $B=b_{1002}=1002 \\times(2004-1002)=1002^{2}$. b) Let $a_{1}, a_{2}, a_{3}, \\ldots a_{2003}$ be an arbitrary order of the numbers $1,2,3, \\ldots, 2003$. First, we will show that numbers $1002,1003,1004, \\ldots, 2003$ cannot occulpy the places numbered $1,2,3$, $\\ldots, 1001$ only. Indeed, we have $(2003-1002)+1=1002$ numbers and 1002 places. This means that at least one of the numbers $1002,1003,1004, \\ldots, 2003$, say $a_{m}$, lies on a place which number $m$ is greater than 1001 . Therefore, $B \\geq m a \\geq 1002 \\times 1002=1002^{2}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_tag": "\nCOM 2 ", "solution_tag": "\nSolution:"}} -{"year": "2003", "tier": "T3", "problem_label": "COM 3", "problem_type": "Combinatorics", "problem": "Prove that amongst any 29 natural numbers there are 15 such that sum of them is divisible by 15 .", "solution": "Amongst any 5 natural numbers there are 3 such that sum of them is divisible by 3 . Amongst any 29 natural numbers we can choose 9 groups with 3 numbers such that sum of numbers in every group is divisible by 3. In that way we get 9 natural numbers such that all of them are divisiblc by 3. It is easy to see that amongst any 9 natural numbers there are 5 such that sum of them is divisible by 5 . Since we have 9 numbers, all of them are divisible by 3 , there are 5 such that sum of them is divisible by 15 .\n\n## $\\operatorname{COM} 4$.\n\n$n$ points are given in a plane, not three of them colinear. One observes that no matter how we label the points from 1 to $n$, the broken line joining the points $1,2,3, \\ldots, n$ (in this order) do not intersect itself.\n\nFind the maximal value of $n$.\n\nSolution. Notice that $n=4$ satisfies the condition. Indeed, for a\n\nconcave quadrilateral, this can be checked immediately.\n\nThen, observe that for $n \\geq 5$ one can choose four points $A, B, C, D$ such that $A B C D$ is a convex quadrilateral. The diagonals $A C$ and $B D$ intersect at a point, hence labeling $A, B, C, D$ with $1,2,3,4$ we reach a contradiction.\n\nThus, it is sufficient to proove that from five points we can select four that are vertices of a convex quadrilateral. Consider the convex hull of the five points set. If this is not a triangle we are done. If it is a triangle, then draw the line through the two points inside the triangle. This line meet exactly two sides of the triangle. Let $A$ be the common vertex of these sides. Then the four remaining points solve the claim.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_tag": "\nCOM 3.", "solution_tag": "\nSolution:"}} -{"year": "2003", "tier": "T3", "problem_label": "COM 5", "problem_type": "Combinatorics", "problem": "If $m$ is a number from the set $\\{1,2,3,4\\}$ and each point of the plane is painted in red or blue, prove that in the plane there exists at least an equilateral triangle with the vertices of the same colour and with length side $m$.", "solution": "Suppose that in the plane there no exists an equilateral triangle with the vertices of the same colour and length side $m=1,2,3,4$.\n\nFirst assertion: we shall prove that in the plane there no exists a segment with the length 2 such that the ends and the midpint of this segment have the same colour. Suppose that the segment $X Y$ with length 2 have the midpoint $T$ such that the points $X, Y, T$ have the same colour (for example, red). We construct the equilateral triangle. $X Y Z$. Hence, the point $Z$ is blue. Let $U$ and $V$ be the midpoints of the segments $X Z$ and $Y Z$ respectively. So, the points $U$ and $V$ are blue. We obtain a contradiction, because the equilateral triangle $U V Z$ have three blue vertices.\n\nSecond assertion: in the same way we prove that in the plane there no exists a segment with the length 4 such that the ends and the midpoint of this segment have the same colour.\n\nConsider the equilateral triangle $A B C$ with length side 4 and divide it into 16 equilateral triangles with length sides 1. L $0: D$ be the midpoint of the segment $A B$. The vertices $A, B, C$ don't have the same colour. WLOG we suppose that $A$ and $B$ are red and $C$ is blue. So, the point $D$ is blue too. We shall investigate the following cases:\n\na) The midpoints $E$ and $F$ of the sides $A C$ and, respectively, $B C$ are red. From the first assertion it follows that the midpoints $M$ and $N$ of the segments $A E$ and, respectively, $B F$ are blue. Hence, the equilateral triangle $M N C$ have three blue vertices, a contradiction.\n\nb) Let $E$ is red and $F$ is blue. The second one position of $E$ and $F$ is simmetrical. If $P, K, L$ are the midpoints of the segments $C F, A D, B D$ respectively, then by first assertion $P$ is red, $M$ is blue and $N$ is red. This imply that $K$ and $L$ are blue. So, the segment $K L$ with length 2 has the blue ends and blue midpoint, a contradiction.\n\nc) If $E$ and $F$ are blue, then the equilateral triangle $E F C$ has three blue vertices, a contradiction.\n\nHence, in the plane there exists at least an equilateral triangle with the vertices of the same colour and with length side $m$, where $m \\in\\{1,2,3,4\\}$.\n\nComment: The formuation of the problem suggests that one has to find 4 triangles, one for each $m$ from the set $\\{1,2,3,4\\}$ whereas the solution is for one $m$. A better formulation is:\n\nEach point of the plane is painted in red or blue. Prove that in the plane there exists at least an equilateral triangle with the vertices of the same colour and with length side $m$, where $m$ is some number from the set $\\{1,2,3,4\\}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_tag": "\nCOM 5.", "solution_tag": "\nSolution."}} -{"year": "2003", "tier": "T3", "problem_label": "GEO 1", "problem_type": "Geometry", "problem": "Is there a convex quadrilateral, whose diagonals divide it into four triangles, such that their areas are four distinct prime integers.", "solution": "No. Let the areas of those triangles be the prime numbers $p, q, r$ and $t$. But for the areas of the triangles we have $\\mathrm{pq}=\\mathrm{rt}$, where the triangles with areas $\\mathrm{p}$ and $\\mathrm{q}$ have only a common vertex. This is not possible for distinct primes.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_tag": "\nGEO 1.", "solution_tag": "\nSolution."}} -{"year": "2003", "tier": "T3", "problem_label": "GEO 2", "problem_type": "Geometry", "problem": "Is there a triangle whose area is $12 \\mathrm{~cm}^{2}$ and whose perimeter is $12 \\mathrm{~cm}$.", "solution": "No. Let $\\mathrm{r}$ be the radius of the inscribed circle. Then $12=6 \\mathrm{r}$, i.e. $\\mathrm{r}=2 \\mathrm{~cm}$. But the area of the inscribed circle is $4 \\pi>12$, and it is known that the area of any triangle is bigger than the area of its inscribed circle.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_tag": "\nGEO 2.", "solution_tag": "\nSolution."}} -{"year": "2003", "tier": "T3", "problem_label": "GEO 3", "problem_type": "Geometry", "problem": "Let $G$ be the centroid of the the triangle $A B C$. Reflect point $A$ across $C$ at $A^{\\prime}$. Prove that $G, B, C, A^{\\prime}$ are on the same circle if and only if $G A$ is perpendicular to $G C$.", "solution": "Observe first that $G A \\perp G C$ if and only if $5 A C^{2}=A B^{2}+B C^{2}$. Indeed,\n\n$$\nG A \\perp G C \\Leftrightarrow \\frac{4}{9} m_{a}^{2}+\\frac{4}{9} m_{c}^{2}=b^{2} \\Leftrightarrow 5 b^{2}=a^{2}+c^{2}\n$$\n\nMoreover,\n\n$$\nG B^{2}=\\frac{4}{9} m_{b}^{2}=\\frac{2 a^{2}+2 c^{2}-b^{2}}{9}=\\frac{9 b^{2}}{9}=b^{2}\n$$\n\nhence $G B=A C=C A^{\\prime}$ (1). Let $C^{\\prime}$ be the intersection point of the lines $G C$ and $A B$. Then $C C^{\\prime}$ is the middle line of the triangle $A B A^{\\prime}$, hence $G C \\| B A^{\\prime}$. Consequently, $G C A^{\\prime} B$ is a trapezoid. From (1) we find that $G C A^{\\prime} B$ is isosceles, thus cyclic, as needed.\n\nConversely, since $G C A^{\\prime} B$ is a cyclic trapezoid, then it is also isosceles. Thus $C A^{\\prime}=$ $G B$, which leads to (1).\n\nComment: An alternate proof is as follows:\n\nLet $M$ be the midpoint of $A C$. Then the triangles $M C G$ and $M A^{\\prime} B$ are similar. So $G C$ is parallel to $A^{\\prime} B$.\n\n$G A \\perp G C$ if and only if $G M=M C$. By the above similarity, this happen if and only if $A^{\\prime} C=G B$; if and only if the trapezoid is cyclic.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_tag": "## GEO 3.", "solution_tag": "\nSolution."}} -{"year": "2003", "tier": "T3", "problem_label": "GEO 5", "problem_type": "Geometry", "problem": "Let three congruent circles intersect in one point $M$ and $A_{1}, A_{2}$ and $A_{3}$ be the other intersection points for those circles. Prove that $M$ is a.orthocenter for a triangle $A_{1} A_{2} A_{3}$.", "solution": "The quadrilaterals $\\mathrm{O}_{3} M O_{2} A_{1}, \\mathrm{O}_{3} M O_{1} A_{2}$ and $O_{1} M O_{2} A_{3}$ are rombes. Therefore, $O_{2} A_{1} \\| M O_{3}$ and $M O_{3} \\| O_{1} A_{2}$, which imply $O_{2} A_{1} \\| O_{1} A_{2}$. Because $O_{2} A_{1}=O_{3}{ }^{*} M=O_{1} A_{2}$ the quadrilateral $O_{2} A_{1} A_{2} O_{1}$ is parallelogram and then $A_{1} A_{2} \\| O_{1} O_{2}$ and $A_{1} A_{2}=O_{1} O_{2}$. Similary, $A_{2} A_{3} \\| O_{2} O_{3}$ and $A_{2} A_{3}=O_{2} O_{3} ; A_{3} A_{1} \\| O_{3} O_{1}$ and $A_{3} A_{1}=O_{3} O_{1}$. The triangles $A_{1} A_{2} A_{3}$ and $\\mathrm{O}_{1} \\mathrm{O}_{2} \\mathrm{O}_{3}$ are congruent.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f45e5cc675871c830e68g-16.jpg?height=453&width=399&top_left_y=772&top_left_x=776)\n\nSince $A_{3} M \\perp O_{1} O_{2}$ and $O_{1} O_{2} \\| A_{1} A_{2}$ we infere $A_{3} M \\perp A_{1} A_{2}$. Similary, $A_{2} M \\perp A_{1} A_{3}$ and $A_{1} M \\perp A_{2} A_{3}$. Thus, $M$ is the orthocenter for the triangle $A_{1} A_{2} A_{3}$.\n\n## GEO.6.\n\nConsider an isosceles triangle $A B C$ with $A B=A C$. A semicircle of diameter $E F$, lying on the side $B C$, is tangent to the lines $A B$ and $A C$ at $M$ and $N$, respectively. The line $A E$ intersects again the semicircle at point $P$.\n\nProve that the line PF passes through the midpoint of the chord $M N$.\n\nSolution. Let $O$ be the center of the semicircle and let $R$ be the midpoint of $M N$. It is obvious that $M N$ is perpendicular to $A O$ at point $R$. Since $\\angle A N O$ is right, then from the leg theorem we have $A N^{2}=A R \\cdot A O$. From the powen of a point theorem,\n\n$$\nA P \\cdot A E=A N^{2}=A M^{2}=A R \\cdot A O\n$$\n\nUsing the same theorem we infer that points $P, R, O$ and $E$ are concyclic, hence $\\angle R P E$ is right. As $\\angle F P E$ is also a right angle, the conclusion follows.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_tag": "\nGEO 5.", "solution_tag": "\nSolution:"}} -{"year": "2003", "tier": "T3", "problem_label": "GEO 7", "problem_type": "Geometry", "problem": "Through a interior point of a triangle, three lines parallel to the sides of the triangle are constructed. In that way the triangle is divided on six figures, areas equal $a, b, c, \\alpha, \\beta, \\gamma$ (see the picture).\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f45e5cc675871c830e68g-18.jpg?height=300&width=417&top_left_y=432&top_left_x=730)\n\nProve that\n\n$$\n\\frac{a}{\\alpha}+\\frac{b}{\\beta}+\\frac{c}{\\gamma} \\geqslant \\frac{3}{2}\n$$", "solution": "We will prove the inequality in two steps. First one is the following\n\nLemma: Let $A B C$ be a triangle, $E$ arbitrary point on the side $A C$. Parallel lines to $A B$ and $B C$, drown through $E$ meet sides $B C$ and $A B$ in points $F$. and $D$ respectively. Then: $P_{B D E F}=2 \\sqrt{P_{A D E} \\cdot P_{E F C}}$ ( $P_{X}$ is area for the figure $X)$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f45e5cc675871c830e68g-18.jpg?height=291&width=423&top_left_y=1304&top_left_x=747)\n\nThe triangles $A D E$ and $E F C$ are similar. Then:\n\n$$\n\\frac{P_{B D E F}}{2 P_{A D E}}=\\frac{P_{B D E}}{P_{A D E}}=\\frac{B D}{A D}=\\frac{E F}{A D}=\\frac{\\sqrt{P_{E F C}}}{\\sqrt{P_{A D E}}}\n$$\n\nHence, $P_{B D E F}=2 \\sqrt{P_{A D E} \\cdot P_{E F C}}$.\n\nUsing this lemma one has $\\alpha=2 \\sqrt{b c}, \\beta=2 \\sqrt{a c}, \\gamma=2 \\sqrt{a b}$. The GML-AM mean inequality provides\n\n$$\n\\frac{a}{\\alpha}+\\frac{b}{\\beta}+\\frac{c}{\\gamma} \\geqslant 3 \\sqrt[3]{\\frac{a b c}{\\alpha \\beta \\gamma}}=3 \\sqrt[3]{\\frac{a b c}{2^{3} \\sqrt{a^{2} b^{2} c^{2}}}}=\\frac{3}{2}\n$$\n\nBULGARIA\n\n| Leader: | Chavdar Lozanov |\n| :--- | :--- |\n| Deputy Leader: | Ivan Tonov |\n| Contestants: | Asparuh Vladislavov Hriston |\n| | Tzvetelina Kirilova Tzeneva |\n| | Vladislav Vladilenon Petkov |\n| | Alexander Sotirov Bikov |\n| | Deyan Stanislavov Simeonov |\n| | Anton Sotirov Bikov |\n\n## CYPRUS\n\n| Leader: | Efthyvoulos Liasides |\n| :--- | :--- |\n| Deputy Leader: | Andreas Savvides |\n| Contestants: | Marina Kouyiali |\n| | Yiannis loannides |\n| | Anastasia Solea |\n| | Nansia Drakou |\n| | Michalis Rossides |\n| | Domna Fanidou |\n| Observer: | Myrianthi Savvidou |\n\nFORMER YUGOSLAV\n\nREPUBLIC of MACEDONIA\n\n| Leader: | Slavica Grkovska |\n| :--- | :--- |\n| Deputy Leader: | Misko Mitkovski |\n| Contestants: | Aleksandar lliovski |\n| | Viktor Simjanovski |\n| | Maja Tasevska |\n| | Tanja Velkova |\n| | Matej Dobrevski |\n| | Oliver Metodijev |\n\n## GREECE\n\nLeader: Anargyros Felouris\n\nDeputy Leader: Ageliki Vlachou\n\nContestants: Theodosios Douvropoulos\n\nMarina lliopoulou\n\nFaethontas Karagiannopoulos\n\nStefanos Kasselakis\n\nFragiskos Koufogiannis\n\nEfrosyni Sarla\n\nROMANIA\n\n| Leader: | Dan Branzei |\n| :--- | :--- |\n| Deputy Leader: | Dinu Serbanescu |\n| Contestants: | Dragos Michnea |\n| | Adrian Zahariuc |\n| | Cristian Talau |\n| | Beniämin Bogosel |\n| | Sebastian Dumitrescu |\n| | Lucian Turea |\n\n## TURKEY\n\nLeader:\n\nHalil Ibrahim Karakaş\n\n\\&Deputy Leader: Duru Türkoğlu\n\nContestants: Sait Tunç\n\nAnmet Kabakulak\n\nTürkü Çobanoğlu\n\nBurak Sağlam\n\nIbrahim Çimentepe\n\nHale Nur Kazaçeşme\n\n## YUGOSLAVIA\n\n(SERBIA and MONTENEGRO)\n\n| Leader: | Branislav Popovic |\n| :--- | :--- |\n| Deputy Leader: | Marija Stanic |\n| Contestants: | Radojevic Mladen |\n| | Jevremovic Marko |\n| | Djoric Milos |\n| | Lukic Dragan |\n| | Andric Jelena |\n| | Pajovic Jelena |\n\n## TURKEY-B\n\n## Leader:\n\nDeputy Leader: Contestants:\nAhmet Karahan\nDeniz Ahçihoca ..... Havva Yeşildağl|\nÇağıl Şentip\nBuse Uslu\nAli Yilmaz\nDemirhan Çetereisi\nYakup Yildirim\n\n## REPUBLIC of MOLDOVA\n\n| Leader: | Ion Goian |\n| :--- | :--- |\n| Deputy Leader: | Ana Costas |\n| Contestants: | lurie Boreico |\n| | Andrei Frimu |\n| | Mihaela Rusu |\n| | Vladimir Vanovschi |\n| | Da Vier: |\n| | Alexandru Zamorzaev |\n\n1.Prove that $7^{n}-1$ is not divisible by $6^{n}-1$ for any positive integer $n$.\n\n2. 2003 denars were divided in several bags and the bags were placed in several pockets. The number of bags is greater than the number of denars in each pocket. Is it true that the number of pockets is greater than the number of denars in one of the bags?\n3. In the triangle $\\mathrm{ABC}, R$ and $r$ are the radii of the circumcircle and the incircle, respectively; $a$ is the longest side and $h$ is the shortest altitude. Prove that $R / r>a / h$.\n4. Prove that for all positive numbers $x, y, z$ such that $x+y+z=1$ the following inequality holds\n\n$$\n\\frac{x^{2}}{1+y}+\\frac{y^{2}}{1+z}+\\frac{z^{2}}{1+x} \\leq 1\n$$\n\n5.Is it possible to cover a $2003 \\times 2003$ board with $1 \\times 2$ dominoes placed horizontally and $1 \\times 3$ threeminoes placed vertically?\n\n## THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA Chişinău, March 9-12, 2003\n\n7.1 Let $m>n$ be pozitive integers. For every positive integers $k$ we define the number $a_{k}=(\\sqrt{5}+2)^{k}+$ $(\\sqrt{5}-2)^{k}$. Show that $a_{m+n}+a_{m-n}=a_{m} \\cdot a_{n}$.\n\nT. Fild all five digits numbers $\\overline{a b c d e}$, written in decimal system, if it is known that $\\overline{a b} c d e-\\overline{e b c d a}=69993$, $\\overline{b c d}-\\overline{d c b}=792, \\overline{b c}-\\overline{c b}=72$.\n\n7.3 In the triangle $A B C$ with semiperemeter $p$ the points $M, N$ and $P$ lie on the sides $(B C),(C A)$ and - (AB) respectively. Show that $pb \\geq$ 10. Prove that this equation has two irrational solutions. (The number $m$ is triangular, if $m=n(n-1) / 2$ for certain positive integer $n \\geq 1$ ).\n\n9.3 The distinct points $M$ and $N$ lie on the hypotenuse ( $A C)$ of the right isosceles triangle $A B C$ so that $M \\in(A N)$ and $M N^{2}=A M^{2}+C N^{2}$. Prove that $m(\\angle M B N)=45^{\\circ}$.\n\n9.4 Find all the functions $f: N^{*} \\rightarrow N^{*}$ which verify the relation $f(2 x+3 y)=2 f(x)+3 f(y)+4$ for every positive integers $x, y \\geq 1$.\n\n9.5 The numbers $a_{1}, a_{2}, \\ldots, a_{n}$ are the first $n$ positive integers with the property that the number $8 a_{k}+1$ is a perfect square for every $k=1,2, \\ldots, n$. Find the sum $S_{n}=a_{1}+a_{2}+\\ldots+a_{n}$.\n\n9.6 Find all real solutions of the equation $x^{4}+7 x^{3}+6 x^{2}+5 \\sqrt{2003} x-2003=0$.\n\n9.7 The side lengths of the triangle $A B C$ satisfy the relations $a>b \\geq 2 c$. Prove that the altitudes of the triangle $A B C$ can not be the sides of any triangle.\n\n9.8 The base of a pyramid is a convex polygon with 9 sides. All the lateral edges of the pyramid and all the liagunads ui the base are coloured in a random way in red or blue. Pröve that there exist at least three vertices of the pyramid which belong to a triangle with the sides coloured in the same colour.\n\n10.1 Find all prime numbers $a, b$ and $c$ for which the equality $(a-2)!+2 b!=22 c-1$ holds.\n\n10.2 Solve the system $x+y+z+t=6, \\sqrt{1-x^{2}}+\\sqrt{4-y^{2}}+\\sqrt{9-z^{2}}+\\sqrt{16-t^{2}}=8$.\n\n10.3 In the scalen triangle $A B C$ the points $A_{1}$ and $B_{1}$ are the bissectrices feets, drawing from the vertices $A$ and $B$ respectively. The straight line $A_{1} B_{1}$ intersect the line $A B$ at the point $D$. Prove that one of the angles $\\angle A C D$ or $\\angle B C D$ is obtuze and $m(\\angle A C D)+m(\\angle B C D)=180^{\\circ}$.\n\n10.4 Let $a>1$ be not integer number and $a \\neq \\sqrt[2]{q}$ for every positive integers $p \\geq 2$ and $q \\geq 1$, $k=\\left[\\log _{a} n\\right] \\geq 1$, where $[x]$ is the integral part of the real number $x$. Prove that for every positive integer $n \\geq 1$ the equality\n\n$$\n\\left[\\log _{a} 2\\right]+\\left[\\log _{a} 3\\right]+\\ldots+\\left[\\log _{a} n\\right]+[a]+\\left[a^{2}\\right]+\\ldots+\\left[a^{k}\\right]=n k\n$$\n\nholds.\n\n10.5 The rational numbers $p, q, r$ satisfy the relation $p q+p r+q r=1$. Prove that the number $\\left(1+p^{2}\\right)\\left(1+q^{3}\\right)\\left(1+r^{2}\\right)$ is a square of any rational number.\n\n10.6 Let $n \\geq 1$ be a positive integer. For every $k=1,2, \\ldots, n$ the functions $f_{k}: R \\rightarrow R, f_{k}(x)=$ $a_{k} x^{2}+b_{k} x+c_{k}$ with $a_{k} \\neq 0$ are given. Find the greatest possible number of parts of the rectangular plane $x O y$ which can be obtained by the intersection of the graphs of the functions $f_{k}(k=1,2, \\ldots, n)$.\n\n10.7 The circle with the center $O$ is tangent to the sides $[A B],[B C],[C D]$ and $[D A]$ of the convex quadrilateral $A B C D$ at the points $M, N, \\mathcal{K}$ and $L$ respectively. The straight lines $M N$ and $A C$ are parallel and the straight line $M K$ intersect the line $L N$ at the point $P$. Prove that the points $A, M, P, O$ and $L$ are concyclic.\n\n10.8 Find all integers $n$ for which the number $\\log _{2 n-1}\\left(n^{2}+2\\right)$ is rational.\n\n11.1 Let $a, b, c, d \\geq 1$ be arbitrary positive numbers. Prove that the equations system $a x-y z=$ $c, \\quad b x-y t=-d$. has at least a solution $(x, y, z, t)$ in positive integers.\n\n11.2 The sequences $\\left(a_{n}\\right)_{n \\geq 0}$ and $\\left(b_{n}\\right)_{n \\geq 0}$ satisfy the conditions $(1+\\sqrt{3})^{2 n+1}=a_{n}+b_{n} \\sqrt{3}$ and $a_{n}, b_{n} \\in Z$. Find the recurrent relation for each of the sequences $\\left(a_{n}\\right)$ and $\\left(b_{n}\\right)$.\n\n11.3 The triangle $A B C$ is rightangled in $A, A C=b, A B=c$ and $B C=a$. The halfstraight line ( $A z$ is perpendicular to the plane $(A B C), M \\in(A z$ so that $\\alpha, \\beta, \\gamma$ are the mesures of the angles, formed by the edges $M B, M C$ and the plane ( $M B C$ ) with the plane ( $A B C$ ) respectively. In the set of the triangular pyramids MABC on consider the pyramids with the volumes $V_{1}$ and $V_{2}$ which satisfy the relations $\\alpha+\\beta+\\gamma=\\pi$ and $\\alpha+\\beta+\\gamma=\\pi / 2$ respectively. Prove the equality $\\left(V_{1} / V_{2}\\right)^{2}=(a+b+c)(1 / a+1 / b+1 / c)$.\n\n11.4 Find all the functions $f:[0 ;+\\infty) \\rightarrow[0 ;+\\infty)$ which satisfy the conditions: : $f(x f(y)) \\cdot f(y)=$ $f(x+y)$ for every $x, y \\in[0 ;+\\infty) ; f(2)=0 ; f(x) \\neq 0$ for every $x \\in[0 ; 2)$.\n\n11.5 Let $02 R \\sin \\alpha$.\n\n12.4 The real numbers $\\alpha, \\beta, \\gamma$ satisfy the relations $\\sin \\alpha+\\sin \\beta+\\sin \\gamma=0$ and $\\cos \\alpha+\\cos \\beta+\\cos \\gamma=0$. Find all positive integers $n \\geq 0$ for-which $\\sin (n \\alpha+\\pi / 4)+\\sin (n \\beta+\\pi / 4)+\\sin (n \\gamma+\\pi / 4)=0$.\n\n12.5 For every positive integer $n \\geq 1$ we define the polynomial $P(X)=X^{2 n}-X^{2 n-1}+\\ldots-X+1$, Find the remainder of the division of the polynomial $P\\left(X^{2 n+1}\\right)$ by the polynomial $P(X)$.\n\n12.6 Fie $n \\in N$. Find all the primitives of the function\n\n$$\nf: R \\rightarrow R, \\quad f(x)=\\frac{x^{3}-9 x^{2}+29 x-33}{\\left(x^{2}-6 x+10\\right)^{n}}\n$$\n\n12.7 In a rectangular system $x O y$ the graph of the function $f: R \\rightarrow R, f(x)=x^{2}$ is drawn. The ordered triple $B, A, C$ has distinct points on the parabola, the point $D \\in(B C)$ such that the straight line $A D$ is parallel to the axis $O y$ and the triangles $B A D$ and $C A D$ have the areas $s_{1}$ and $s_{2}$ respectively. Find the length of the segment $[A D]$.\n\n12.8 Let $\\left(F_{n}\\right)_{n \\in N^{*}}$ be the Fibonacci sequence so that: $F_{1}=1, F_{2}=1, F_{n+1}=F_{n}+F_{n-1}$ for every positive integer $n \\geq 2$. Shown that $F_{n}<3^{n / 2}$ and calculate the limit $\\lim _{n \\rightarrow \\infty}\\left(F_{1} / 2+F_{2} / 2^{2}+\\ldots+F_{n} / 2^{n}\\right)$.\n\n## The first selection test for IMO 2003 and BMO 2003, March 12, 2003\n\nB1. Each side of the arbitrary triangle is divided into 2002 congruent segments. After that each interior division point of the side is joined with opposite vertex. Prove that the number of obtained regions of the triangle is divisible by 6 .\n\nB2. The positive real numbers $x, y$ and $z$ satisfy the relation $x+y+z \\geq 1$. Prove the inequality\n\n$$\n\\frac{x \\sqrt{x}}{y+z}+\\frac{y \\sqrt{y}}{x+z}+\\frac{z \\sqrt{z}}{x+y} \\geq \\frac{\\sqrt{3}}{2}\n$$\n\nB3. The quadrilateral $A B C D$ is inscribed in the circle with center $O$, the points $M$ and $N$ are the middle points of the diagonals $[A C]$ and $[B D]$ respectively and $P$ is the intersection point of the diagonals. It is known that the points $O, M, N$ si $P$ are distinct. Prove that the points $O, M, B$ and $D$ are concyclic if and only if the points $O, N, A$ and $C$ are concyclic.\n\nB4. Prove that the equation $1 / a+1 / b+1 / c+1 /(a b c) \\doteq 12 /(a+b+c)$ has many solutions $(a, b, c)$ in strictly positive integers.\n\n## The second selection test for IMO 2003, March 22, 2003\n\nB5. Let $n \\geq 1$ be positive integer. Find all polynomials of degree $2 n$ with real coefficients\n\n$$\nP(X)=X^{2 n}+(2 n-10) X^{2 n-1}+a_{2} X^{2 n-2}+\\ldots+a_{2 n-2} X^{2}+(2 n-10) X+1\n$$\n\n-if it is known that they have positive real roots.\n\nB6. The triangle $A B C$ has the semiperimeter $p$, the circumradius $R$, the inradius $r$ and $l_{a,}, l_{b}, l_{c}$ are the lengths of internal bissecticies, drawing from the vertices $A, B$ and $C$ respectively. Prove the inequality $l_{a} l_{b}+l_{b} l_{c}+l_{c} l_{a} \\leq p \\sqrt{3 r^{2}+12 R r}$.\n\nB7. The points $M$ and $N$ are the tangent points of the sides $[A B]$ and $[A C]$ of the triangle $A B C$ to the incircle with the center $I$. The internal bissectrices, drawn from the vertices $B$ and $C$, intersect the straight line $M N$ at points $P$ and $Q$ respectively. If $F$ is the intersection point of the swtraight lines $C P$ and $B Q$, then prove that the straight lines $F I$ and $B C$ are perpendicular.\n\nB8. Let $n \\geq 4$ be the positive integer. On the checkmate table with dimensions $n \\times n$ we put the coins. One consider the diagonal of the table each diagonal with at least two unit squares. What is the smallest number of coins put on the table so that on the each horizontal, each vertical and each diagonal there exists att least one coin. Prove the answer.\n\n## The third selection test for IMO 2003, March 23, 2003\n\nB9. Let $n \\geq 1$ be positive integer. A permutation $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ of the numbers $(1,2, \\ldots, n)$ is called quadratique if among the numbers $a_{1}, a_{1}+a_{2}, \\ldots, a_{1}+a_{2}+\\ldots+a_{n}$ there exist at least a perfect square. Find the greatest number $n$, which is less than 2003 , such that every permutation of the numbers $(1,2, \\ldots, n)$ will be quadratique.\n\nB10. The real numbers $a_{1}, a_{2}, \\ldots, a_{2003}$ satisfy simultaneousiy the relations: $a_{i} \\geq 0$ for all $i=$ $1,2, \\ldots, 2003 ; \\quad a_{1}+a_{2}+\\ldots+a_{2003}=2 ; \\quad a_{1} a_{2}+a_{2} a_{3}+\\ldots+a_{2003} a_{1}=1$. Find the smallest value of the sum $a_{1}^{2}+a_{2}^{2}+\\ldots+a_{2003}^{2}$.\n\nB11. The arbitrary point $M$ on the plane of the triangle $A B C$ does not belong on the straight lines $A B, B C$ and $A C$. If $S_{1}, S_{2}$ and $S_{3}$ are the areas of the triangles $A M B, B M C$ and $A M C$ respectively, find the geometrical locus of the points $M$ which satisfy the relation $\\left(M A^{2}+M B^{2}+M C^{2}\\right)^{2}=16\\left(S_{1}^{2}+S_{2}^{2}+S_{3}^{2}\\right)$.\n\n812. Let $n \\geq 1$ be a positive integer. A square table of dimensions $n \\times n$ is full arbitrarly completed $\\because$ the numb so, shat every number appear exactly conce the table. from cack fine one select the smallest number and the greatest of them is denote by $x$. From each column one select the greatest number and the smallest of them is denote by $y$. The table is called equilibrated if $x=y$. How match equilibrated tables there exist?\n\n## The first selection test for JBMO 2003, April 12, 2003\n\nJB1. Let $n \\geq 2003$ be a positive integer such that the number $1+2003 n$ is a perfect square. Prove that the number $n+1$ is equal to the sum of 2003 positive perfect squares.\n\nJB2. The positive real numbers $a, b, c$ satisfy the relation $a^{2}+b^{2}+c^{2}=3 a b c$. Prove the inequality\n\n$$\n\\frac{a}{b^{2} c^{2}}+\\frac{b}{c^{2} a^{2}}+\\frac{c}{a^{2} b^{2}} \\geq \\frac{9}{a+b+c}\n$$\n\nJB3. The quadrilateral $A B C D$ with perpendicular diagonals is inscribed in the circle with center $O$, the points $M$ and $N$ are the middle points of the sides $[B C]$ and $[C D]$ respectively. Find the value of the ratio of areas of the figures $O M C N$ and $A B C D$.\n\nJB4. Let $m$ and $n$ be the arbitrary digits of the decimal system and $a, b, c$ be the positive distinct integers of the form $2^{m} \\cdot 5^{n}$. Find the number of the equations $a x^{2}-2 b x+c=0$, if it is known that each equation has a single real solution.\n\n## The second selection test for JMBO 2003, April 13, 2003\n\nJB5. Prove that each positive integer is equal to a difference of two positive integers with the same number of the prime divisors.\n\nJB6. The real numbers $x$ and $y$ satisfy the equalities\n\n$$\n\\sqrt{3 x}\\left(1+\\frac{1}{x+y}\\right)=2, \\quad \\sqrt{7 y}\\left(1-\\frac{1}{x+y}\\right)=4 \\sqrt{2}\n$$\n\nFind the numerical value of the ratio $y / x$.\n\n$J B 7$. The triangle $A B C$ is isosceles with $A B=B C$. The point $F$ on the side $[B C]$ and the point $D$ on the side $[A C]$ are the feets of the internal bissectrix drawn from $A$ and altitude drawn from $B$ respectively so that $A F=2 B D$. Find the measure of the angle $A B C$.\n\nJB8. In the rectangular coordinate system every point with integer coordinates is called laticeal point. Let $P_{n}(n, n+5)$ be a laticeal point and denote by $f(n)$ the number of laticeal points on the open segment $\\left(O P_{n}\\right)$, where the point $O(0,0)$ is the coordinates system origine. Calculate the number $f(1)+f(2)+$ $f(3)+\\ldots+f(2002)+f(2003)$.\n\n7 th Junior Balkan Mathematical O-lympiad\n\n$20-25$ Jun e, 20.03 I $\\mathrm{m}$ i r $\\quad$. $\\quad$ u rke y\n\n## English Version\n\n1. Let $n$ be a positive integer. A number $A$ consists of $2 n$ digits, each of which is 4 ; and a number $B$ consists of $n$ digits, each of which is 8 . Prove that $A+2 B+4$ is a perfect square.\n\n\\&\n\n2. Suppose there are $n$ points in a plane no three of which are collinear with the following property:\n\nIf we label these points as $A_{1}, A_{2}, \\ldots, A_{n}$ in any way whatsoever, the broken line $A_{1} A_{2} \\ldots A_{n}$ does not intersect itself.\n\nFind the maximal value that $n$ can have.\n\n3. Let $k$ be the circumcircle of the triangle $A B C$. Consider the arcs $\\overparen{A B}, \\widehat{B C}, \\widetilde{C A}$ such that $C \\notin \\widetilde{A B}, A \\notin \\widetilde{B C}, B \\notin \\widetilde{C A}$. Let $D, E$ and $F$ be the midpoints of the arcs $\\widehat{B C}, \\overparen{C A}, \\overparen{A B}$, respectively. Let $G$ and $H$ be the points of intersection of $D E$ with $C B$ and $C A$; let $I$ and $J$ be the points of intersection of $D F$ with $B C$ and $B A$, respectively. Denote the midpoints of $G H$ and $I J$ by $M$ and $N$, respectively.\n\na) Find the angles of the triangle $D M N$ in terms of the angles of the triangle $A B C$.\n\nb) If $O$ is the circumcentre of the triangle $D M N$ and $P$ is the intersection point of $A D$ and $E F$, prove that $O, P, M$ and $N$ lie on the same circle.\n\n4. Let $x, y, z$ be real numbers greater than -1 . Prove that\n\n$$\n\\frac{1+x^{2}}{1+y+z^{2}}+\\frac{1+y^{2}}{1+z+x^{2}}+\\frac{1+z^{2}}{1+x+y^{2}} \\geq 2\n$$\n\n## Romanian Version", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_tag": "\nGEO 7.", "solution_tag": "\nSolution:"}} -{"year": "2003", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "problem": "1. Fie $n$ un număr natural nenul.. Un număr $A$ conține $2 n$ cifre, fiecare fiind 4 ; și un număr $B$ conţine $n$ cifre, fiecare fiind 8 . Demonstratị că $A+2 B+4$ este un pătrat perfect.\n\n$$\n\\begin{aligned}\n& \\text { Macedrea } \\\\\n& \\text { Shucar Crevok. }\n\\end{aligned}\n$$\n\n2. Fie $n$ puncte în plan, oricare trei necoliniare, cu proprietátea:\n\noricum am numerota aceste puncte $A_{1}, A_{2}, \\ldots, A_{n}$, linia frântă $A_{1} A_{2} \\ldots A_{n}$ nu se autointersectează.\n\nGăsitị valoarea maximă a lui $n$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f45e5cc675871c830e68g-26.jpg?height=115&width=625&top_left_y=1112&top_left_x=1228)\n\n3. Fie $k$ cercul circumscris triunghiului $A B C$. Fie arcele $\\overparen{A B}, \\overparen{B C}, \\overparen{C A}$ astfel încât $C \\notin \\widehat{A B}, A \\notin \\widehat{B C}, B \\notin \\widehat{C A}$ siV․ $E$ mijloacele acestor arce. Fie $G, H$ punctele de intersectie ale lui $D E$ cu $C B, C A$; fie $I, J$ punctele de intersecție ale lui $D F$ cu $B C, B A$. Notăm mijloacele lui $G H, I J$ cu $M$, respectiv $N$.\n\na) Găsiți unghiurile triunghiului $D M N$ în funcție de unghiurile triunghiului $A B C$.\n\nb) Dacă $O$ este circumcentrul triunghiului $D M N$ şi $P$ este intersecția lui $A D \\mathrm{cu}$ $E F$, arătați că $O, P, M$ şi $N$ aparțin unui acelaşi cerc.\n\n4. Fie $x, y, z$ numere reale mai mari decât -1 . Demonstrați că:\n\n$$\n\\frac{1+x^{2}}{1+y+z^{2}}+\\frac{1+y^{2}}{1+z+x^{2}}+\\frac{1+z^{2}}{1+x+y^{2}} \\geq 2\n$$\n\nRomalala - Pmecityol.\n\nTimp de lucru: 4 ore și jumătate.\n\nFiecare problemăeste notată czu 10 مuncte\n\n## Question 1\n\nI. To do a special case $n \\geq 2$.\n\nII. To assert that $A+2 B+4=(\\underbrace{6 \\ldots 68}_{n-1})^{2}$.\n\nIII. To observe that $A=4 \\times \\frac{10^{2 n}-1}{9}$ and $B=8 \\times \\frac{10^{n}-1}{9}$.\n\nIV. To observe that $A=3^{2} \\times(\\underbrace{2 \\ldots 2}_{n})^{2}+4 \\times(\\underbrace{2 \\ldots 2}_{n})$ or $A=\\left(\\frac{3 B}{4}\\right)^{2}+B$.\n\n$$\n\\begin{aligned}\n& \\mathbf{I} \\rightarrow 1 \\text { point } \\\\\n& \\mathbf{I}+\\mathrm{II} \\rightarrow 2 \\text { points } \\\\\n& \\mathbf{I I I} \\rightarrow 4 \\text { points or } \\quad \\mathbf{I V} \\rightarrow 5 \\text { points }\n\\end{aligned}\n$$\n\n## Question 2\n\nI. To claim $n=4$ with example for $n=4$.\n\nII. To show impossibility of the case when the set of points includes 4 points that form a convex quadrilateral.\n\nIII. To show that every set of $n \\geq 5$ points contains 4 points forming a convex quadrilateral.\n\n$$\n\\begin{aligned}\n& \\text { I } \\rightarrow 2 \\text { points } \\\\\n& \\text { II } \\rightarrow 1 \\text { point } \\\\\n& \\text { III } \\rightarrow 4 \\text { points } \\\\\n& \\text { II }+ \\text { III } \\rightarrow 7 \\text { points }\n\\end{aligned}\n$$\n\n## Question 3\n\n## Part a\n\nI. Computing the angles of the triangle $D E F$.\n\nII. Observing that the lines $C F \\perp D E$ and that $B E \\perp D F$.\n\nI $\\rightarrow$ 1\"point\n\n$\\mathrm{I}+\\mathrm{II} \\rightarrow 3$ points\n\nOnly Part a $\\rightarrow 6$ points\n\n## Part b\n\nIII. Completing the figure by drawing $E F$.\n\nPart a + III $\\rightarrow 7$ points\n\nOnly Part $\\mathbf{b} \\rightarrow 6$ points\n\n## Question 4\n\nI. To observe that $y \\leq \\frac{y^{2}+1}{2}$.\n\nII. To observe that $1+y+z^{2}>0$ and to obtain $\\frac{1+x^{2}}{1+y+z^{2}} \\geq \\frac{1+x^{2}}{1+z^{2}+\\frac{1+y^{2}}{2}}$.\n\nIII. To reduce to $\\frac{C+4 B-2 A}{A}+\\frac{A+4 C-2 B}{B}+\\frac{B+4 A-2 C}{C} \\geq 9$.\n\n$$\n\\begin{aligned}\n& I \\rightarrow 1 \\text { point } \\\\\n& X \\rightarrow I Y \\rightarrow 3 \\text { points } \\\\\n& I+I I+I I \\rightarrow 5 \\text { points }\n\\end{aligned}\n$$\n\n| SCORES | | | | |\n| :---: | :---: | :---: | :---: | :---: |\n| $=1$ | ROM-6 | Adrian Zahariuc | 40 | First Prize |\n| 2 | ROM-3 $=$ | Dragos Michnea | 40 | First Prize |\n| 3 | MOL-6 | Alexandru Zamorzaev | 39 | First Prize |\n| 4 | MOL-1 | lurie Boreico | 38 | First Prize |\n| 5 | ROM-5 | Lucian Turea. | 38 | First Prize |\n| 6 | ROM-4 | Cristian Talau | 37 | First Prize |\n| 7 | BUL-4 | Vladislav Vladilenon Petkov | 33 | Second Prize |\n| 8 | HEL-1 | Theodosios Douvropaulos | 32 | Second Prize |\n| 9 | BUL-1 | Alexander Sotirov Bikov | 31 | Second Prize |\n| 10 | BUL-2 | Anton Sotirov Bikov | 31 | Second Prize |\n| 11 | TUR-4 | Hale Nur Kazaçeşme | 31 | Second Prize |\n| 12 | TUR-6 | Sait Tunç | 31 | Second Prize |\n| 13 | BUL-5 | Deyan Stanislavov Simeonov | 30 | Second Prize |\n| 14 | HEL-3 | Faethontas Karagiannopoulos | 30 | Second Prize |\n| 15 | MCD-5 | Maja Tasevska | 29 | Second Prize |\n| 16 | ROM-2 | Sebastian Dumitrescu | 29 | Second Prize |\n| 17 | $B U L-6$ | Tzvetelina Kirilova Tzeneva | 29 | Second Prize |\n| 18 | $B U L-3$ | Asparuh Vladislavov Hriston | 28 | Second Prize |\n| 19 | TUR-5 | Burak Sağlam | 24 | Third Prize |\n| 20 | TUR-1 | Ibrahim Çimentepe | 23 | Third Prize |\n| 21 | YUG-4 | Jevremovic Marko | 22 | Third Prize |\n| 22 | YUG-1 | Lukic Dragan | 22 | Third Prize |\n| 23 | ROM-1 | Beniamin Bogosel | 21 | Third Prize |\n| 24 | YUG-5 | Djoric.Milos | 21 | Third Prize |\n\n\n| 25 | MOL-4 | Vladimir Vanovschi | 21 | Third Prize |\n| :---: | :---: | :---: | :---: | :---: |\n| 26 | YUG-2 | Andric Jelena | 19 | Third Prize |\n| 27 | YUG-6 | Radojevic Mladen | 19 | Third Prize |\n| 28 | MCD-4 | Viktor Simjanovski | 17 | Third Prize |\n| 29 | HEL-6 | Efrosyni Sarla | 16 | Third Prize |\n| 30 | TUR-2 | Türkü Çobanoğlu | 13 | Third Prize |\n| 31 | YUG-3 | Pajovic Jelena | 12 | Third Prize |\n| 32 | MCD-2 | Aleksandar lliovski | 11 | Third Prize |\n| 33 | MCD-6 | Tanja Velkova | 11 | Third Prize |\n| 34 | MOL-2 | Andrei Frimu | 10 | Honorary Mention |\n| 35 | MOL-5 | Dan Vieru | 10 | Honorary Mention |\n| 36 | MCD-3 | Oliver Metodijev | 10 | Honorary Mention |\n| 37 | $H E L-4$ | Stefanos Kasselakis | 9 | |\n| 38 | HEL-5 | Fragiskos Koufogiannis | 8 | - |\n| 39 | MCD-1 | Matej Dobrevski | 8 | |\n| 40 | HEL-2 | Marina lliopoulou | 4 | |\n| 41 | MOL-3 | Mihaela Rusu | 4 | |\n| 42 | CYP-1 | Narısia Drakou | 4 | |\n| 43 | CYP-6 | Anastasia Solea | 3 | |\n| 44 | TUR-3 | Ahmet Kabakulak | 2 | |\n| 45 | CYP-4 | Marina Kouyiali | 2 | |\n| 46 | CYP-5 | Michalis Rossides | 2 | |\n| 47 | CYP-2 | Domna Fanidou | 1 | |\n| 48 | CYP-3 | Yiannis loannides | 0 | |\n\nEen de a 7-a Olingriodá Batcanicà de Mabematicepentre juniosi s-a elesfancuat in pesidada 20-25 iunce in Tercia in stationea Kusadasi ( ecirca $90 \\mathrm{~km} \\mathrm{la}$ sud de Izmir, pe malul mäci Egee). Gchecpa Ronaincer' a fost condusà de. Prof di. Dan Brainzei, ascsbat de. Prof. grad I Dinu Yerfinescu. In clasamentul nesficial pe nationi Romincia scupà primal loc usmatà de Bulgoria, Jurcia, Republica Molabra, Serbia, Macedonia, Mecia, Sipta. Eomponentic echipei Romainiei cu. oblimat ismattoorele punctaje ic medolic:\n\nDragos Michnen (Satu Mare) - 40p-Awr.\n\nAdricin Zahariuc (Bacciu) - $40 p-$ Aur\n\nLucian Turen (Bucuresti) - $38 p$ - Awr\n\nGeistian Taliu (Slacova) - 37p-Aar\n\nSebastian Sumitresce (Bucuegti) - 29p-Aegint\n\nBeniamin Bogosel (Hlad) - 21p-Aegint.\n\nMentionaim ca promic doi trax sealicat punctajid total.\n\nTnainte de a se deplasn im Turcia, exhipn Romannier a fost garduità thei zile la Bucuresti. bैxclesio on sosp de antenument, in accastà periondin, juniouric au participat le al 5-lea si al 6-lea test test de selectie pentru OIM. Pestatio junioscibs la aceste leste a fost excelenta-\n\n## Olimpiada Naţională de Matematică\n\nAl cincilea test de selectie pentru OIM - 19 iunie 2003\n\n## Subiectul 1\n\nUn parlament are $n$ deputati. Aceştia fac parte din 10 partide şi din 10 comisi parlamentare. Fiecare deputat face parte dintr-un singur partid si dintr-o singură comisie.\n\nDeterminati valoarea minimă a lui $n$ pentru care indiferent de componenţa numerică a partidelor şi indiferent de repartizarea în comisii, să existe o numerotare cu toate numerele $1,2, \\ldots, 10$ atât a partidelor cât şi a comisiilor, astfel încât cel puţin 11 deputaţi să facă parte dintr-un partid si o comisie cu număr identic.\n\n## Subiectul 2\n\nSe dă un romb $A B C D$ cu latura 1. Pe laturile $B C$ şi $C D$ există punctele $M$, respectiv $N$, astfel încât $M C+C N+N M=2$ si $\\angle M A N=\\frac{1}{2} \\angle B A D$.\n\nSă se afle unghiurile rombului.\n\n## Subiectul 3\n\nÎntr-un plan înzestrat cu un sistem de coordonate $X O Y$ se numeste punct laticial un punct $A(x, y)$ in care ambele coordonate sunt numere întregi. Un punct laticial $A$ se numeşte invizibil dacă pe segmentul deschis $O A$ există cel puţin un punct laticial.\n\nSă. se arate că pentru orice număr natural $n, n>0$, există un pătrat de latură $n$ în care toate punctele laticiale interioare, de pe laturi sau din vârfuri, sunt invizibile.\n\nTimp de lucru: 4 ore\n\n## Olimpiada Naţională de Matematică 2003\n\nAl şaselea test de selecţie pentru OIM - 20 iunie 2003\n\n## Problema 1.\n\nFie $A B C D E F$ un hexagon convex. Notăm cu $A^{\\prime}, B^{\\prime}, C^{\\prime}, D^{\\prime}, E^{\\prime}, F^{\\prime}$ mijloacele laturilor $A B, B C, C D, D E, E F, F A$ respectiv. Se cunosc arile triunghiurilor $A B C^{\\prime}, B C D^{\\prime}, C D E^{\\prime}$, $D E F^{\\prime}, E F A^{\\prime}, F A B^{\\prime}$.\n\nSă se afle aria hexagonului $A B C D E F$.\n\n## Problema 2.\n\nO permutare $\\sigma:\\{1,2, \\ldots, n\\} \\rightarrow\\{1,2, \\ldots, n\\}$ se numeşte strânsă dacă pentru orice $k=1,2, \\ldots, n-1$ avem\n\n$$\n|\\sigma(k)-\\sigma(k+1)| \\leq 2\n$$\n\nSă se găsească cel mai mic număr natural $n$ pentru care Єxistă cel puţin 2003 permutări strânse.\n\n## Problema 3.\n\nPentru orice număr natural $n$ notăm cu $C(n)$ suma cifrelor sale în baza 10. Arătaţi că oricare ar fi numărul natural $k$ există un număr natural $m$ astfel încât ecuaţia $x+C(x)=m$ are cel puţin $k$ soluţii.\n\nTimp de lucru 4 ore\n\n## Proposed Problem \\#72\n\n$==$ Valentin Vornicu $==$\n\nJune 20, 2003\n\nProblem: A permutation $\\sigma:\\{1,2, \\ldots, n\\} \\rightarrow\\{1,2, \\ldots, n\\}$ is called straight if and only if for each integer $k, 1 \\leq k \\leq n-1$ the following inequality is fulfilled\n\n$$\n|\\sigma(k)-\\sigma(k+1)| \\leq 2\n$$\n\nFind the smallest positive integer $n$ for which there exist at least 2003 straight permutations.", "solution": "The main trick is to look where $n$ is positioned. Tn that idea let us denote by $x_{n}$ the number of all the straight permutations and by $a_{n}$ the number of straight permutations having $n$ on the first or on the last position, i.e. $\\sigma(1)=n$ or $\\sigma(n)=n$. Also let us denote by $b_{n}$ the difference $x_{n}-a_{n}$ and by $a_{n}^{\\prime}$ the number of permutations having $n$ on the first position, and by $a_{n}^{\\prime \\prime}$ the number of permutations having $n$ on the last position. From symmetry we have that $2 a_{n}^{\\prime}=2 a_{n}^{\\prime \\prime}=a_{n}^{\\prime}+a_{n}^{\\prime \\prime}=a_{n}$, for all $n$-s. Therefore finding a recurrence relationship for $\\left\\{a_{n}\\right\\}_{n}$ is equivalent with finding one for $\\left\\{a_{n}^{\\prime}\\right\\}_{n}$.\n\nOne can simply compute: $a_{2}^{\\prime}=1, a_{3}^{\\prime}=2, a_{4}^{\\prime}=4$. Suppose that $n \\geq 5$. We have two possibilities for the second position: if $\\sigma(2)=n-1$ then we must complete the remaining positions with $3,4, \\ldots, n$ thus the number of ways in which we can do that is $a_{n-1}^{\\prime}$ (because the permutation $\\sigma^{\\prime}:\\{1,2, \\ldots, n-1\\} \\rightarrow$ $\\{1,2, \\ldots, n-1\\}, \\sigma^{\\prime}(k)=\\sigma(k+1)$, for all $k, 1 \\leq k \\leq n-1$, is also a straight permutation $)$.\n\nIf on the second position we have $n-2, \\sigma(2)=n-2$, then $n-1$ can only be in the last position of the permutation or on the third position, i.e. $\\sigma(3)=n-1$ or $\\sigma(n)=n-1$. If $\\sigma(n)=n-1$, then we caw only have $\\sigma(n-1)=n-3$ thus $\\sigma(3)=n-4$ and so on, thus there is only one permutation of this kind. On the other hand, if $\\sigma(3)=n-1$ then it follows that $\\sigma(4)=n-3$ and now we can complete the permutation in $n_{n-3}^{\\prime}$ ways (hecause the permutation $\\sigma^{\\prime}:\\{1,2, \\ldots, n-3\\} \\rightarrow\\{1,2, \\ldots, n-3\\}, \\sigma^{\\prime}(k)=\\sigma(k+3)$, for all $k$, $1 \\leq k \\leq n-3$, is also a straight permutation).\n\nSumming all up we get the recurrence:\n\n$$\na_{n}^{\\prime}=a_{n-1}^{\\prime}+1+a_{n-3}^{\\prime} \\Rightarrow a_{n}=a_{n-1}+a_{n-3}+2, \\forall n \\geq 5\n$$\n\nThe recurrence relationship for $\\left\\{b_{n}\\right\\}$ can be obtained by observing that for each straight permutation $\\tau:\\{1,2, \\ldots, n+1\\} \\rightarrow\\{1,2, \\ldots, n+1\\}$ for which $2 \\leq \\tau^{-1}(n+1) \\leq n$ we can obtain a straight permutation $\\sigma:\\{1,2, \\ldots, n\\} \\rightarrow\\{1,2, \\ldots, n\\}$ by removing $n+1$. Indeed $n+1$ is \"surrounded\" by $n$ and $n-1$, so by removing it, $n$ and $n-1$ become neighbors, and thus the newly formed permutation is indeed straight. Now, if $\\tau^{-1}(n) \\in\\{1, n+1\\}$ then the newly formed permutation $\\sigma$ was counted as one of the $a_{n}$-s, minus the two special cases in which $n$ and $n-1$ are on the first and last positions. If $\\tau^{-1}(n) \\notin\\{1, n+1\\}$ then certainly $\\sigma$ was counted with the $b_{n}$-s. Also, from any straight permutation of $n$ elements, not having $n$ and $n-1$ in the first and last position, thus $n$ certainly being neighbor with $n-1$, we can make a straight $n+1$-element permutation by inserting $n+1$ between $n$ and $n-1$.\n\nTherefore we have obtained the following relationship:\n\n$$\nb_{n+1}=a_{n}-2+b_{n}=x_{n}-2, \\forall n \\geq 4\n$$\n\nFrom (1) and (2) we get that\n\n$$\nx_{n}=x_{n-1}+a_{n-1}+a_{n-3}, \\forall n \\geq 5\n$$\n\nIt is obvious that $\\left\\{x_{n}\\right\\}_{n}$ is a \"fast\" increasing sequence, so we will compute the first terms using the relationships obtained above, which will prove that the number that we are looking for is $n=16$ :\n\n$$\n\\begin{aligned}\n& a_{2}=2 \\quad x_{2}=3 \\\\\n& a_{9}=62 \\quad x_{9}=164 \\\\\n& a_{3}=4 \\quad x_{3}=6 \\\\\n& a_{4}=8 \\quad x_{4}=12 \\\\\n& a_{5}=12 \\quad x_{5}=22 \\\\\n& a_{6}=18 \\quad x_{6}=38 \\\\\n& a_{7}=28 \\quad x_{7}=64 \\\\\n& a_{10}=92 \\quad x_{10}=254 \\\\\n& a_{8}=42 \\quad x_{8}=i 04 \\\\\n& a_{11}=136 \\quad x_{11}=388 \\\\\n& a_{12}=200 \\quad x_{12}=586 \\\\\n& a_{13}=294 \\quad x_{13}=878 \\\\\n& a_{14}=432 \\quad x_{14}=1308 \\\\\n& a_{15}=034 \\quad x_{15}=1940\n\\end{aligned}\n$$\n\n## ENUNȚURULE PROBLEMELOR DIN ATENTTIA JURIULUT LA CEA DE A 7-A JBMO (KUSADASI, TURCIA, 20-25 IUNIE 2003)\n\nA.1. Un număr A este scris cu 2 n cifre, fiecare dintre acestea fiind 4 ; un număr B este scris cu $n$ cifre, fiecare dintre acestea fiind 8 . Demonstrați că, pentru orice $n, A+2 B+4$ este pătrat perfect. A.2. Fie a, b, c lungimile laturilor unui triunghi, $p=\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}, q=\\frac{a}{c}+\\frac{c}{b}+\\frac{b}{a}$. Demonstrați că $|p-q|<1$.\n\nA.3. Fie $\\mathrm{a}, \\mathrm{b}, \\mathrm{c}$ numere reale astfel încât $a^{2}+b^{2}+c^{2}=1$. Demonstrați că $P=a b+b c+c a-2(a+b+c) \\geq-5 / 2$. Există valori pentru $\\mathrm{a}, \\mathrm{b}, \\mathrm{c}$ încât $\\mathrm{P}=-5 / 2$ ?\n\nA.4. Fie $\\mathrm{a}, \\mathrm{b}$, c numere raționale astfel încât $\\frac{1}{a+b c}+\\frac{1}{b+a c}=\\frac{1}{a+b}$. Demonstrați că $\\sqrt{\\frac{c-3}{c+1}}$ este de asemenea număr rațional.\n\nA.5. Fie $A B C$ triunghi neisoscel cu lungimile $a, b, c$ ale laturilor numere naturale. Dempnstraţi că $\\left|a b^{2}\\right|+\\left|\\tilde{b c} c^{2}\\right|+\\left|c a^{2}-a^{2} b-b^{2} c-c^{2} a\\right| \\geq 2$.\n\nA.6. Fie $\\mathrm{a}, \\mathrm{b}$, c c numere pozitive astfel ca $a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=3$. Demonstrați că $a+b+c \\geq a b c+2$.\n\nA.6'. Fie $\\mathrm{a}, \\mathrm{b}, \\mathrm{c}$ numere pozitive astfel ca $a b+b c+c a=3$. Demonstrați că $a+b+c \\geq a b c+2$.\n\nA.7. Fie $\\mathrm{x}, \\mathrm{y}$, $\\mathrm{z}$ numere mai mari câ -1. Demönstraţi că $\\frac{1+x^{2}}{1+y+z^{2}}+\\frac{1+y^{2}}{1+z+x^{2}}+\\frac{1+z^{2}}{1+x+y^{2}} \\geq 2$. A.8. Demonstraţi că există mulțimi disjuncte $A=\\{x, y, z\\}$ șì $B=\\{m, n, p\\}$ de numere naturale mai mari ca 2003 astfel ca $x+y+z=m+n+p$ si $x^{2}+y^{2}+z^{2}=m^{2}+n^{2}+p^{2}$.\n\nC.1. Într-un grup de 60 studenți: 40 vorbesc engleza, 30 vorbesc franceza, 8 vorbesc toate cele trei limbi. Numărul celor ce vorbesc doar engleza şi franceza este egal cu suma celor care vorbesc doar germana şi franceza cu a celor ce vorbesc doar engleza şi germana. Numărul celor ce vorbesc cel puțin două dintre aceste limbi este 28. Cât de mulți studenți vorbesc: a) germana; b) numai engleza; c) numai germana.\n\nC.2. Numerele $1,2,3, \\ldots, 2003$ sunt scrise într-un şir $a_{1}, a_{2}, a_{3}, \\ldots, a_{2003}$. Fie $b_{1}=1 \\exists a_{1}, b_{2}=2 \\exists a_{2}$, $b_{3}=3 \\exists a_{3}, \\ldots, b_{2003}=2003 \\exists a_{2003}$ şi B maximul numerelor $b_{1}, b_{2}, b_{3}, \\ldots, b_{2003}$.\n\na) Dacă $a_{1}=2003, a_{2}=2002, a_{3}=2001, \\ldots, a_{2003}=1$, găsiți valoarea lui $B$.\n\nb) Demonstrați că $B \\geq 1002^{2}$.\n\nC.3. Demonstrați că îtr-o mulțime de 29 numere naturale există 15 a căror sumă este divizibilă cu 15 . C.4. Fie n puncte în plan, oricare trei necoliniare, cu proprietatea că oricum le-am numerota $A_{1}$, $A_{2}, \\ldots, A_{n}$, linia frântă $A_{1} A_{2} \\ldots A_{n}$ nu se autointersectează. Găsiți valoarea maximă a lui $n$.\n\nC.5. Fie mulțmea $M=\\{1,2,3,4\\}$. Fiecare punct al planului este colorat în roşu sau albastru.\n\nDemonstrați că există cel puțin un triunghi echilateral cu latura $m \\in M$ cu vârfurile de aceeaşí culoare.\n\nG.1. Există un patrulater convex pe care diagonalele să-1 împartă în patru triunghiuri cu ariile numere prime distincte?\n\nG.2. Există un triunghi cu aria $12 \\mathrm{~cm}^{2}$ şi perimetrul 12 ?\n\nG.3. Fie $\\mathrm{G}$ centrul de greutate al triunghiului $\\mathrm{ABC}$ şi $\\mathrm{A}$ ' simetricul lui $\\mathrm{A}$ faţă de $\\mathrm{C}$. Demonstrați că punctele $\\mathrm{G}, \\mathrm{B}, \\mathrm{C}, \\mathrm{A}$ ' sunt conciclice dacă și numai dacă $\\mathrm{GA} \\zeta \\mathrm{GC}$.\n\nG.4. Fie $k$ cercul circumscris triunghiului $A B C$. Fie arcele $A B, B C, C A$ astfel incât\n$C \\notin A B, A \\notin B C, B \\notin \\mathcal{C} A$ şi $F, D, E$ mijloacele acestor arce. Fie $G, H$ punctele de intersectie ale lui $D E$ cu $C B, C A$; fie $I, J$ punctele de intersectie ale lui $D F$ cu $B C, B A$. Notăm mijloacele lui $G H, I J$ cu $M$, respectiv $N$.\n\na) Găsiți unghiurile triunghiului $D M N$ în funcție de unghiurile triunghiului $A B C$.\n\nb) Dacă $O$ este circumcentrul triunghiului $D M N$ şi $P$ este intersectia lui $A D \\operatorname{cu} E F$, arătați că $O, P, M$ şi $N$ aparțin unui acelaşi cerc.\n\nG.5. Trei cercuri egale au în comun un punct $\\mathrm{M}$ şi se intersectează câte două în puncte $\\mathrm{A}, \\mathrm{B}, \\mathrm{C}$. Demonstrați că $M$ este ortocentrul triunghiului $A B C$. ${ }^{1)}$\n\nG.6. Fie $\\mathrm{ABC}$ un triunghi isoscel $\\mathrm{cu} \\mathrm{AB}=\\mathrm{AC}$. Un semicerc de diametru $\\mathrm{EF}$ situat pe baza $\\mathrm{BC}$ este tangent laturilor $\\mathrm{AB}, \\mathrm{AC}$ în $\\mathrm{M}, \\mathrm{N}$. $\\mathrm{AE}$ retaie semicercul în $\\mathrm{P}$. Demonstraţi că dreapta $\\mathrm{PF}$ trece prin mijlocul corzii MN.\n\nG.7. Paralelele la laturile unui triunghi duse printr-un punct interior împart interiorul triunghiului în şase părți cu ariile notate ca în figură.\n\nDemonstrați că $\\frac{a}{\\alpha}+\\frac{b}{\\beta}+\\frac{c}{\\gamma} \\geq \\frac{3}{2}$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f45e5cc675871c830e68g-36.jpg?height=320&width=450&top_left_y=978&top_left_x=1381)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_tag": "\nA1 ", "solution_tag": "\nSolution:"}} +{"year": "2003", "tier": "T3", "problem_label": "ALG 1", "problem_type": "Algebra", "problem": "A number $A$ is written with $2 n$ digits, each of whish is 4 , and a number $B$ is written with $n$ digits, each of which is 8 . Prove that for each $n, A+2 B+4$ is a total square.", "solution": "$$\n\\begin{aligned}\nA & =\\underbrace{44 \\ldots 44}_{2 n}=\\underbrace{44 \\ldots 4}_{n} \\underbrace{44 \\ldots 4}_{n}=\\underbrace{44 \\ldots 4}_{n} \\underbrace{400 \\ldots 0}_{n}-\\underbrace{44 \\ldots 4}_{n}+\\underbrace{88 \\ldots 8}_{n}=\\underbrace{44 \\ldots 4}_{n} \\cdot\\left(10^{n}-1\\right)+B \\\\\n& =4 \\cdot \\underbrace{11 \\ldots 1}_{n} \\cdot \\underbrace{99 \\ldots 9}_{n}+B=2^{2} \\cdot \\underbrace{11 \\ldots 1}_{n} \\cdot 3^{2} \\cdot \\underbrace{11 \\ldots 1}_{n}+B=\\underbrace{66}_{n} \\ldots 6 \\\\\n& =[\\frac{36}{4} \\cdot \\underbrace{88 \\ldots 8}_{n}+B=[3 \\cdot \\underbrace{22 \\ldots 2}_{n}]^{2}+B=\\left(\\frac{3}{4} B\\right)^{2}+B .\n\\end{aligned}\n$$\n\nSo,\n\n$$\n\\begin{aligned}\nA+2 B+4 & =\\left(\\frac{3}{4} B\\right)^{2}+B+2 B+4=\\left(\\frac{3}{4} B\\right)^{2}+2 \\cdot \\frac{3}{4} B \\cdot 2+2^{2}=\\left(\\frac{3}{4} B+2\\right)^{2}=(\\frac{3}{4} \\cdot \\underbrace{88 \\ldots 8}_{n}+2)^{2} \\\\\n& =(3 \\cdot \\underbrace{22 \\ldots 2}_{n}+2)^{2}=\\underbrace{66 \\ldots 68^{2}}_{n-1}\n\\end{aligned}\n$$", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_match": "\nALG 1.", "solution_match": "## Solution."}} +{"year": "2003", "tier": "T3", "problem_label": "ALG 2", "problem_type": "Algebra", "problem": "Let $a, b, c$ be lengths of triangle sides, $p=\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}$ and $q=\\frac{a}{c}+\\frac{c}{b}+\\frac{b}{a}$.\n\nProve that $|p-q|<1$.", "solution": "One has\n\n$$\n\\begin{aligned}\na b c|p-q| & =a b c\\left|\\frac{c-b}{a}+\\frac{a-c}{b}+\\frac{b-a}{c}\\right| \\\\\n& =\\left|b c^{2}-b^{2} c+a^{2} c-a c^{2}+a b^{2}-a^{2} b\\right|= \\\\\n& =\\left|a b c-a c^{2}-a^{2} b+a^{2} c-b^{2} c+b c^{2}+a b^{2}-a b c\\right|= \\\\\n& =\\left|(b-c)\\left(a c-a^{2}-b c+a b\\right)\\right|= \\\\\n& =|(b-c)(c-a)(a-b)| .\n\\end{aligned}\n$$\n\nSince $|b-c|0$. The contradiction shows that $P=-\\frac{5}{2}$.\n\nComment: By the Cauchy Schwarz inequality $|t| \\leq \\sqrt{3}$, so the smallest value of $P$ is attained at $t=\\sqrt{3}$ and equals $1-2 \\sqrt{3} \\approx-2.46$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_match": "\nALG 3:", "solution_match": "\nSolution:"}} +{"year": "2003", "tier": "T3", "problem_label": "ALG 4", "problem_type": "Algebra", "problem": "Let $a, b, c$ be rational numbers such that\n\n$$\n\\frac{1}{a+b c}+\\frac{1}{b+a c}=\\frac{1}{a+b}\n$$\n\nProve that $\\sqrt{\\frac{c-3}{c+1}}$ is also a rational number", "solution": "By cancelling the denominators\n\n$$\n(a+b)^{2}(1+c)=a b+c\\left(a^{2}+b^{2}\\right)+a b c^{2}\n$$\n\nand\n\n$$\na b(c-1)^{2}=(a+b)^{2}\n$$\n\nIf $c=-1$, we obtrin the contradiction\n\n$$\n\\frac{1}{a-b}+\\frac{1}{b-a}=\\frac{1}{a+b}\n$$\n\nFurtherrdore,\n\n$$\n\\begin{aligned}\n(c-3)(c+1) & =(c-1)^{2}-4=\\frac{(a+b)^{2}}{a b}-4 \\\\\n& =\\frac{(a-b)^{2}}{a b}=\\left(\\frac{(a-b)(c-1)}{a+b}\\right)^{2}\n\\end{aligned}\n$$\n\nThus\n\n$$\n\\sqrt{\\frac{c-3}{c+1}}=\\frac{\\sqrt{(c-3)(c+1)}}{c+1}=\\frac{|a-b||c-1|}{(c+1)|a+b|} \\in \\mathrm{Q}\n$$\n\nas needed.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_match": "## ALG 4.", "solution_match": "\nSolution."}} +{"year": "2003", "tier": "T3", "problem_label": "ALG 5", "problem_type": "Algebra", "problem": "Let $A B C$ be a scalene triangle with $B C=a, A C=b$ and $A B=c$, where $a_{r} b, c$ are positive integers. Prove that\n\n$$\n\\left|a b^{2}+b c^{2}+c a^{2}-a^{2} b-b^{2} c-c^{2} a\\right| \\geq 2\n$$", "solution": "Denote $E=a b^{2}+b c^{2}+c a^{2}-a^{2} b-b^{2} c-c^{2} a$. We have\n\n$$\n\\begin{aligned}\nE= & \\left(a b c-c^{2} a\\right)+\\left(c a^{2}-a^{2} b\\right)+\\left(b c^{2}-b^{2} c\\right)+\\left(a b^{2}-a b c\\right)= \\\\\n& (b-c)\\left(a c-a^{2}-b c+a b\\right)=(b-c)\\left(a a^{2}-b\\right)(c-a)\n\\end{aligned}\n$$\n\nSo, $|E|=|a-b| \\cdot|b-c| \\cdot|c-a|$. By hypothesis each factor from $|E|$ is a positive integer. We shall prove that at least one factor from $|E|$ is greater than 1. Suppose that $|a-b|=|b-c|=|c-a|=1$. It follows that the numbers $a-b, b-c, c-a$ are odd. So, the number $0=(a-b)+(b-c) \\div(c-a)$ is olso odd, a contradiction. Hence, $|E| \\geq 1 \\cdot 1 \\cdot 2=2$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_match": "\nALG 5.", "solution_match": "\nSolution."}} +{"year": "2003", "tier": "T3", "problem_label": "ALG 6", "problem_type": "Algebra", "problem": "Let $a, b, c$ be positive numbers such that $a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=3$. Prove that\n\n$$\na+b+c \\geq a b c+2\n$$", "solution": "We can consider the case $a \\geq b \\geq c$ which implies $c \\leq 1$. The given inequality writes\n\n$$\na+b-2 \\geq(a b-1) c \\geq(a b-1) c^{2}=(a b-1) \\frac{3-a^{2} b^{2}}{a^{2}+b^{2}}\n$$\n\nPut $x=\\sqrt{a b}$. From the inequality $3 a^{2} b^{2} \\geq a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=3$ we infer $x \\geq 1$ and from $a^{2} b^{2}0$, by $A M-G M$ inequality we have\n\n$$\n\\frac{A}{B}+\\frac{B}{C}+\\frac{C}{A} \\geq 3 \\sqrt[3]{\\frac{A}{B} \\cdot \\frac{B}{C} \\cdot \\frac{C}{A}}\n$$\n\nand\n\n$$\n\\frac{B}{A}+\\frac{C}{B}+\\frac{A}{C} \\geq 3\n$$\n\nand we are done.\n\nAlternative solution for inequality (1).\n\nBy the Cauchy-Schwarz inequality,\n\n$$\n\\frac{a}{2 c+b}+\\frac{b}{2 a+c}+\\frac{c}{2 b+a}=\\frac{a^{2}}{2 a c+a b}+\\frac{b^{2}}{2 a b+c b}+\\frac{c^{2}}{2 b c+a c} \\geq \\frac{(a+b+c)^{2}}{3(a b+b c+c a)} \\geq 1\n$$\n\nThe last inequality reduces immediately to the obvious $a^{2}+b^{2}+c^{2} \\geq a b+b c+c a$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_match": "\nALG 7 ", "solution_match": "\nSolution."}} +{"year": "2003", "tier": "T3", "problem_label": "ALG 8", "problem_type": "Algebra", "problem": "Prove that there exist two sets $A=\\{x, y, z\\}$ and $B=\\{m, n, p\\}$ of positive integers greater than 2003 such that the sets have no common elements and the equalities $x+y+z=m+n+p$ and $x^{2}+y^{2}+z^{2}=m^{2}+n^{2}+p^{2}$ hold.", "solution": "Let $A B C$ be a triangle with $B C=a, A C=b, A B=c$ and $ak+3=c\n$$\n\na triangle with such length sides there exist. After the simple calculations we have\n\n$$\n\\begin{gathered}\nA=\\left\\{3(k+1)^{2}-2,3(k+2)^{2}+4,3(k+3)^{2}-2\\right\\} \\\\\nB=\\left\\{3(k+1)^{2}, 3(k+2)^{2}, 3(k+3)^{2}\\right\\}\n\\end{gathered}\n$$\n\nIt easy to prove that\n\n$$\n\\begin{gathered}\nx+y+z=m+n+p=3\\left[(k+1)^{2}+(k+2)^{2}+(k+3)^{2}\\right] \\\\\nx^{2}+y^{2}+z^{2}=m^{2}+n^{2}+p^{2}=9\\left[(k+1)^{4}+(k+2)^{4}+(k+3)^{4}\\right]\n\\end{gathered}\n$$\n\n$>$ From the inequality $3(k+1)^{2}-2>2003$ we obtain $k \\geq 25$. For $k=25$ we have an example of two sets\n\n$$\nA=\\{2026,2191,2350\\}, \\quad B=\\{2028,2187,2352\\}\n$$\n\nwith desired properties.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_match": "\nALG 8.", "solution_match": "\nSolution."}} +{"year": "2003", "tier": "T3", "problem_label": "COM 1", "problem_type": "Combinatorics", "problem": "In a group of 60 students: 40 speak English; 30 speak French; 8 speak all the three languages; the number of students that speak English and French but not German is equal to \"the sum of the number of students that speak English and German but not French plus the number of students that speak French and German but not English; and the number of students that speak at least 2 of those fanguages is 28 . How many students speak:\na) German;\nb) only English;\nc) only German?", "solution": "We use the following notation.\n\n$E=\\#$ students that speak English, $F=\\#$ students that speak French,\n\n$G=\\#$ students that speak German; $m=$ \\# students that speak all the three languages,\n\n$x=\\#$ students that speak English and French but not German,\n\n$y=\\#$ students that speak German and French but not English,\n\n$z=\\#$ students that speak English and German but not French.\n\nThe conditions $x+y=z$ and $x+y+z+8=28$, imply that $z=x+y=10$, i.e. 10 students speak German and French, but not English. Then: $G+E-y-8+F-x-8-10=60$, implies that $G+70-$ $36=60$. Hence: a) $\\mathrm{G}=36$; b) only English speak $40-10-8=22$ students; c) the information given is not enough to find the number of students that speak only German. This ${ }^{n}$ number can be any one from 8 to 18 .\n\nComment: There are some mistakes in the solution. The corrections are as follows:\n\n1. The given condition is $x=y+z($ not $x+y=z)$; thus $x=y+z=10$.\n2. From $G+70-36=60$ one gets $G=26$ (not $G=36$ ).\n3. One gets \"only German speakers\" as $G-y-z-8=8$.\n4. \"Only English speakers\" are $E-x-z-8=22-z$, so this number can not be determined.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_match": "\nCOM 1.", "solution_match": "\nSolution:"}} +{"year": "2003", "tier": "T3", "problem_label": "COM 2", "problem_type": "Combinatorics", "problem": "Natural numbers 1,2,3, .., 2003 are written in an arbitrary sequence $a_{1}, a_{2}, a_{3}, \\ldots a_{2003}$. Let $b_{1}=1 a_{1}, b_{2}=2 a_{2}, b_{3}=3 a_{3}, \\ldots, b_{2003}=2003 a_{2003}$, and $B$ be the maximum of the numbers $b_{1}, b_{2}, b_{3}, \\ldots, b_{2003}$.\n\na) If $a_{1}=2003, a_{2}=2002, a_{3}=2001, \\ldots, a_{2002}=2, a_{2003}=1$, find the value of $B$.\n\nb) Prove that $B \\geq 1002^{2}$.", "solution": "a) Using the inequality between the arithmetical and geometrical mean, we obtain that $b_{n}=n(2004-n) \\leq\\left(\\frac{n+(2004-n)}{2}\\right)^{2}=1002^{2}$ for $n=1,2,3, \\ldots, 2003$. The equality holds if and only if $n=2004-n$, i.e. $n=1002$. Therefore, $B=b_{1002}=1002 \\times(2004-1002)=1002^{2}$. b) Let $a_{1}, a_{2}, a_{3}, \\ldots a_{2003}$ be an arbitrary order of the numbers $1,2,3, \\ldots, 2003$. First, we will show that numbers $1002,1003,1004, \\ldots, 2003$ cannot occulpy the places numbered $1,2,3$, $\\ldots, 1001$ only. Indeed, we have $(2003-1002)+1=1002$ numbers and 1002 places. This means that at least one of the numbers $1002,1003,1004, \\ldots, 2003$, say $a_{m}$, lies on a place which number $m$ is greater than 1001 . Therefore, $B \\geq m a \\geq 1002 \\times 1002=1002^{2}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_match": "\nCOM 2 ", "solution_match": "\nSolution:"}} +{"year": "2003", "tier": "T3", "problem_label": "COM 3", "problem_type": "Combinatorics", "problem": "Prove that amongst any 29 natural numbers there are 15 such that sum of them is divisible by 15 .", "solution": "Amongst any 5 natural numbers there are 3 such that sum of them is divisible by 3 . Amongst any 29 natural numbers we can choose 9 groups with 3 numbers such that sum of numbers in every group is divisible by 3. In that way we get 9 natural numbers such that all of them are divisiblc by 3. It is easy to see that amongst any 9 natural numbers there are 5 such that sum of them is divisible by 5 . Since we have 9 numbers, all of them are divisible by 3 , there are 5 such that sum of them is divisible by 15 .\n\n## $\\operatorname{COM} 4$.\n\n$n$ points are given in a plane, not three of them colinear. One observes that no matter how we label the points from 1 to $n$, the broken line joining the points $1,2,3, \\ldots, n$ (in this order) do not intersect itself.\n\nFind the maximal value of $n$.\n\nSolution. Notice that $n=4$ satisfies the condition. Indeed, for a\n\nconcave quadrilateral, this can be checked immediately.\n\nThen, observe that for $n \\geq 5$ one can choose four points $A, B, C, D$ such that $A B C D$ is a convex quadrilateral. The diagonals $A C$ and $B D$ intersect at a point, hence labeling $A, B, C, D$ with $1,2,3,4$ we reach a contradiction.\n\nThus, it is sufficient to proove that from five points we can select four that are vertices of a convex quadrilateral. Consider the convex hull of the five points set. If this is not a triangle we are done. If it is a triangle, then draw the line through the two points inside the triangle. This line meet exactly two sides of the triangle. Let $A$ be the common vertex of these sides. Then the four remaining points solve the claim.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_match": "\nCOM 3.", "solution_match": "\nSolution:"}} +{"year": "2003", "tier": "T3", "problem_label": "COM 5", "problem_type": "Combinatorics", "problem": "If $m$ is a number from the set $\\{1,2,3,4\\}$ and each point of the plane is painted in red or blue, prove that in the plane there exists at least an equilateral triangle with the vertices of the same colour and with length side $m$.", "solution": "Suppose that in the plane there no exists an equilateral triangle with the vertices of the same colour and length side $m=1,2,3,4$.\n\nFirst assertion: we shall prove that in the plane there no exists a segment with the length 2 such that the ends and the midpint of this segment have the same colour. Suppose that the segment $X Y$ with length 2 have the midpoint $T$ such that the points $X, Y, T$ have the same colour (for example, red). We construct the equilateral triangle. $X Y Z$. Hence, the point $Z$ is blue. Let $U$ and $V$ be the midpoints of the segments $X Z$ and $Y Z$ respectively. So, the points $U$ and $V$ are blue. We obtain a contradiction, because the equilateral triangle $U V Z$ have three blue vertices.\n\nSecond assertion: in the same way we prove that in the plane there no exists a segment with the length 4 such that the ends and the midpoint of this segment have the same colour.\n\nConsider the equilateral triangle $A B C$ with length side 4 and divide it into 16 equilateral triangles with length sides 1. L $0: D$ be the midpoint of the segment $A B$. The vertices $A, B, C$ don't have the same colour. WLOG we suppose that $A$ and $B$ are red and $C$ is blue. So, the point $D$ is blue too. We shall investigate the following cases:\n\na) The midpoints $E$ and $F$ of the sides $A C$ and, respectively, $B C$ are red. From the first assertion it follows that the midpoints $M$ and $N$ of the segments $A E$ and, respectively, $B F$ are blue. Hence, the equilateral triangle $M N C$ have three blue vertices, a contradiction.\n\nb) Let $E$ is red and $F$ is blue. The second one position of $E$ and $F$ is simmetrical. If $P, K, L$ are the midpoints of the segments $C F, A D, B D$ respectively, then by first assertion $P$ is red, $M$ is blue and $N$ is red. This imply that $K$ and $L$ are blue. So, the segment $K L$ with length 2 has the blue ends and blue midpoint, a contradiction.\n\nc) If $E$ and $F$ are blue, then the equilateral triangle $E F C$ has three blue vertices, a contradiction.\n\nHence, in the plane there exists at least an equilateral triangle with the vertices of the same colour and with length side $m$, where $m \\in\\{1,2,3,4\\}$.\n\nComment: The formuation of the problem suggests that one has to find 4 triangles, one for each $m$ from the set $\\{1,2,3,4\\}$ whereas the solution is for one $m$. A better formulation is:\n\nEach point of the plane is painted in red or blue. Prove that in the plane there exists at least an equilateral triangle with the vertices of the same colour and with length side $m$, where $m$ is some number from the set $\\{1,2,3,4\\}$.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_match": "\nCOM 5.", "solution_match": "\nSolution."}} +{"year": "2003", "tier": "T3", "problem_label": "GEO 1", "problem_type": "Geometry", "problem": "Is there a convex quadrilateral, whose diagonals divide it into four triangles, such that their areas are four distinct prime integers.", "solution": "No. Let the areas of those triangles be the prime numbers $p, q, r$ and $t$. But for the areas of the triangles we have $\\mathrm{pq}=\\mathrm{rt}$, where the triangles with areas $\\mathrm{p}$ and $\\mathrm{q}$ have only a common vertex. This is not possible for distinct primes.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_match": "\nGEO 1.", "solution_match": "\nSolution."}} +{"year": "2003", "tier": "T3", "problem_label": "GEO 2", "problem_type": "Geometry", "problem": "Is there a triangle whose area is $12 \\mathrm{~cm}^{2}$ and whose perimeter is $12 \\mathrm{~cm}$.", "solution": "No. Let $\\mathrm{r}$ be the radius of the inscribed circle. Then $12=6 \\mathrm{r}$, i.e. $\\mathrm{r}=2 \\mathrm{~cm}$. But the area of the inscribed circle is $4 \\pi>12$, and it is known that the area of any triangle is bigger than the area of its inscribed circle.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_match": "\nGEO 2.", "solution_match": "\nSolution."}} +{"year": "2003", "tier": "T3", "problem_label": "GEO 3", "problem_type": "Geometry", "problem": "Let $G$ be the centroid of the the triangle $A B C$. Reflect point $A$ across $C$ at $A^{\\prime}$. Prove that $G, B, C, A^{\\prime}$ are on the same circle if and only if $G A$ is perpendicular to $G C$.", "solution": "Observe first that $G A \\perp G C$ if and only if $5 A C^{2}=A B^{2}+B C^{2}$. Indeed,\n\n$$\nG A \\perp G C \\Leftrightarrow \\frac{4}{9} m_{a}^{2}+\\frac{4}{9} m_{c}^{2}=b^{2} \\Leftrightarrow 5 b^{2}=a^{2}+c^{2}\n$$\n\nMoreover,\n\n$$\nG B^{2}=\\frac{4}{9} m_{b}^{2}=\\frac{2 a^{2}+2 c^{2}-b^{2}}{9}=\\frac{9 b^{2}}{9}=b^{2}\n$$\n\nhence $G B=A C=C A^{\\prime}$ (1). Let $C^{\\prime}$ be the intersection point of the lines $G C$ and $A B$. Then $C C^{\\prime}$ is the middle line of the triangle $A B A^{\\prime}$, hence $G C \\| B A^{\\prime}$. Consequently, $G C A^{\\prime} B$ is a trapezoid. From (1) we find that $G C A^{\\prime} B$ is isosceles, thus cyclic, as needed.\n\nConversely, since $G C A^{\\prime} B$ is a cyclic trapezoid, then it is also isosceles. Thus $C A^{\\prime}=$ $G B$, which leads to (1).\n\nComment: An alternate proof is as follows:\n\nLet $M$ be the midpoint of $A C$. Then the triangles $M C G$ and $M A^{\\prime} B$ are similar. So $G C$ is parallel to $A^{\\prime} B$.\n\n$G A \\perp G C$ if and only if $G M=M C$. By the above similarity, this happen if and only if $A^{\\prime} C=G B$; if and only if the trapezoid is cyclic.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_match": "## GEO 3.", "solution_match": "\nSolution."}} +{"year": "2003", "tier": "T3", "problem_label": "GEO 5", "problem_type": "Geometry", "problem": "Let three congruent circles intersect in one point $M$ and $A_{1}, A_{2}$ and $A_{3}$ be the other intersection points for those circles. Prove that $M$ is a.orthocenter for a triangle $A_{1} A_{2} A_{3}$.", "solution": "The quadrilaterals $\\mathrm{O}_{3} M O_{2} A_{1}, \\mathrm{O}_{3} M O_{1} A_{2}$ and $O_{1} M O_{2} A_{3}$ are rombes. Therefore, $O_{2} A_{1} \\| M O_{3}$ and $M O_{3} \\| O_{1} A_{2}$, which imply $O_{2} A_{1} \\| O_{1} A_{2}$. Because $O_{2} A_{1}=O_{3}{ }^{*} M=O_{1} A_{2}$ the quadrilateral $O_{2} A_{1} A_{2} O_{1}$ is parallelogram and then $A_{1} A_{2} \\| O_{1} O_{2}$ and $A_{1} A_{2}=O_{1} O_{2}$. Similary, $A_{2} A_{3} \\| O_{2} O_{3}$ and $A_{2} A_{3}=O_{2} O_{3} ; A_{3} A_{1} \\| O_{3} O_{1}$ and $A_{3} A_{1}=O_{3} O_{1}$. The triangles $A_{1} A_{2} A_{3}$ and $\\mathrm{O}_{1} \\mathrm{O}_{2} \\mathrm{O}_{3}$ are congruent.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f45e5cc675871c830e68g-16.jpg?height=453&width=399&top_left_y=772&top_left_x=776)\n\nSince $A_{3} M \\perp O_{1} O_{2}$ and $O_{1} O_{2} \\| A_{1} A_{2}$ we infere $A_{3} M \\perp A_{1} A_{2}$. Similary, $A_{2} M \\perp A_{1} A_{3}$ and $A_{1} M \\perp A_{2} A_{3}$. Thus, $M$ is the orthocenter for the triangle $A_{1} A_{2} A_{3}$.\n\n## GEO.6.\n\nConsider an isosceles triangle $A B C$ with $A B=A C$. A semicircle of diameter $E F$, lying on the side $B C$, is tangent to the lines $A B$ and $A C$ at $M$ and $N$, respectively. The line $A E$ intersects again the semicircle at point $P$.\n\nProve that the line PF passes through the midpoint of the chord $M N$.\n\nSolution. Let $O$ be the center of the semicircle and let $R$ be the midpoint of $M N$. It is obvious that $M N$ is perpendicular to $A O$ at point $R$. Since $\\angle A N O$ is right, then from the leg theorem we have $A N^{2}=A R \\cdot A O$. From the powen of a point theorem,\n\n$$\nA P \\cdot A E=A N^{2}=A M^{2}=A R \\cdot A O\n$$\n\nUsing the same theorem we infer that points $P, R, O$ and $E$ are concyclic, hence $\\angle R P E$ is right. As $\\angle F P E$ is also a right angle, the conclusion follows.", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_match": "\nGEO 5.", "solution_match": "\nSolution:"}} +{"year": "2003", "tier": "T3", "problem_label": "GEO 7", "problem_type": "Geometry", "problem": "Through a interior point of a triangle, three lines parallel to the sides of the triangle are constructed. In that way the triangle is divided on six figures, areas equal $a, b, c, \\alpha, \\beta, \\gamma$ (see the picture).\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f45e5cc675871c830e68g-18.jpg?height=300&width=417&top_left_y=432&top_left_x=730)\n\nProve that\n\n$$\n\\frac{a}{\\alpha}+\\frac{b}{\\beta}+\\frac{c}{\\gamma} \\geqslant \\frac{3}{2}\n$$", "solution": "We will prove the inequality in two steps. First one is the following\n\nLemma: Let $A B C$ be a triangle, $E$ arbitrary point on the side $A C$. Parallel lines to $A B$ and $B C$, drown through $E$ meet sides $B C$ and $A B$ in points $F$. and $D$ respectively. Then: $P_{B D E F}=2 \\sqrt{P_{A D E} \\cdot P_{E F C}}$ ( $P_{X}$ is area for the figure $X)$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f45e5cc675871c830e68g-18.jpg?height=291&width=423&top_left_y=1304&top_left_x=747)\n\nThe triangles $A D E$ and $E F C$ are similar. Then:\n\n$$\n\\frac{P_{B D E F}}{2 P_{A D E}}=\\frac{P_{B D E}}{P_{A D E}}=\\frac{B D}{A D}=\\frac{E F}{A D}=\\frac{\\sqrt{P_{E F C}}}{\\sqrt{P_{A D E}}}\n$$\n\nHence, $P_{B D E F}=2 \\sqrt{P_{A D E} \\cdot P_{E F C}}$.\n\nUsing this lemma one has $\\alpha=2 \\sqrt{b c}, \\beta=2 \\sqrt{a c}, \\gamma=2 \\sqrt{a b}$. The GML-AM mean inequality provides\n\n$$\n\\frac{a}{\\alpha}+\\frac{b}{\\beta}+\\frac{c}{\\gamma} \\geqslant 3 \\sqrt[3]{\\frac{a b c}{\\alpha \\beta \\gamma}}=3 \\sqrt[3]{\\frac{a b c}{2^{3} \\sqrt{a^{2} b^{2} c^{2}}}}=\\frac{3}{2}\n$$\n\nBULGARIA\n\n| Leader: | Chavdar Lozanov |\n| :--- | :--- |\n| Deputy Leader: | Ivan Tonov |\n| Contestants: | Asparuh Vladislavov Hriston |\n| | Tzvetelina Kirilova Tzeneva |\n| | Vladislav Vladilenon Petkov |\n| | Alexander Sotirov Bikov |\n| | Deyan Stanislavov Simeonov |\n| | Anton Sotirov Bikov |\n\n## CYPRUS\n\n| Leader: | Efthyvoulos Liasides |\n| :--- | :--- |\n| Deputy Leader: | Andreas Savvides |\n| Contestants: | Marina Kouyiali |\n| | Yiannis loannides |\n| | Anastasia Solea |\n| | Nansia Drakou |\n| | Michalis Rossides |\n| | Domna Fanidou |\n| Observer: | Myrianthi Savvidou |\n\nFORMER YUGOSLAV\n\nREPUBLIC of MACEDONIA\n\n| Leader: | Slavica Grkovska |\n| :--- | :--- |\n| Deputy Leader: | Misko Mitkovski |\n| Contestants: | Aleksandar lliovski |\n| | Viktor Simjanovski |\n| | Maja Tasevska |\n| | Tanja Velkova |\n| | Matej Dobrevski |\n| | Oliver Metodijev |\n\n## GREECE\n\nLeader: Anargyros Felouris\n\nDeputy Leader: Ageliki Vlachou\n\nContestants: Theodosios Douvropoulos\n\nMarina lliopoulou\n\nFaethontas Karagiannopoulos\n\nStefanos Kasselakis\n\nFragiskos Koufogiannis\n\nEfrosyni Sarla\n\nROMANIA\n\n| Leader: | Dan Branzei |\n| :--- | :--- |\n| Deputy Leader: | Dinu Serbanescu |\n| Contestants: | Dragos Michnea |\n| | Adrian Zahariuc |\n| | Cristian Talau |\n| | Beniämin Bogosel |\n| | Sebastian Dumitrescu |\n| | Lucian Turea |\n\n## TURKEY\n\nLeader:\n\nHalil Ibrahim Karakaş\n\n\\&Deputy Leader: Duru Türkoğlu\n\nContestants: Sait Tunç\n\nAnmet Kabakulak\n\nTürkü Çobanoğlu\n\nBurak Sağlam\n\nIbrahim Çimentepe\n\nHale Nur Kazaçeşme\n\n## YUGOSLAVIA\n\n(SERBIA and MONTENEGRO)\n\n| Leader: | Branislav Popovic |\n| :--- | :--- |\n| Deputy Leader: | Marija Stanic |\n| Contestants: | Radojevic Mladen |\n| | Jevremovic Marko |\n| | Djoric Milos |\n| | Lukic Dragan |\n| | Andric Jelena |\n| | Pajovic Jelena |\n\n## TURKEY-B\n\n## Leader:\n\nDeputy Leader: Contestants:\nAhmet Karahan\nDeniz Ahçihoca ..... Havva Yeşildağl|\nÇağıl Şentip\nBuse Uslu\nAli Yilmaz\nDemirhan Çetereisi\nYakup Yildirim\n\n## REPUBLIC of MOLDOVA\n\n| Leader: | Ion Goian |\n| :--- | :--- |\n| Deputy Leader: | Ana Costas |\n| Contestants: | lurie Boreico |\n| | Andrei Frimu |\n| | Mihaela Rusu |\n| | Vladimir Vanovschi |\n| | Da Vier: |\n| | Alexandru Zamorzaev |\n\n1.Prove that $7^{n}-1$ is not divisible by $6^{n}-1$ for any positive integer $n$.\n\n2. 2003 denars were divided in several bags and the bags were placed in several pockets. The number of bags is greater than the number of denars in each pocket. Is it true that the number of pockets is greater than the number of denars in one of the bags?\n3. In the triangle $\\mathrm{ABC}, R$ and $r$ are the radii of the circumcircle and the incircle, respectively; $a$ is the longest side and $h$ is the shortest altitude. Prove that $R / r>a / h$.\n4. Prove that for all positive numbers $x, y, z$ such that $x+y+z=1$ the following inequality holds\n\n$$\n\\frac{x^{2}}{1+y}+\\frac{y^{2}}{1+z}+\\frac{z^{2}}{1+x} \\leq 1\n$$\n\n5.Is it possible to cover a $2003 \\times 2003$ board with $1 \\times 2$ dominoes placed horizontally and $1 \\times 3$ threeminoes placed vertically?\n\n## THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA Chişinău, March 9-12, 2003\n\n7.1 Let $m>n$ be pozitive integers. For every positive integers $k$ we define the number $a_{k}=(\\sqrt{5}+2)^{k}+$ $(\\sqrt{5}-2)^{k}$. Show that $a_{m+n}+a_{m-n}=a_{m} \\cdot a_{n}$.\n\nT. Fild all five digits numbers $\\overline{a b c d e}$, written in decimal system, if it is known that $\\overline{a b} c d e-\\overline{e b c d a}=69993$, $\\overline{b c d}-\\overline{d c b}=792, \\overline{b c}-\\overline{c b}=72$.\n\n7.3 In the triangle $A B C$ with semiperemeter $p$ the points $M, N$ and $P$ lie on the sides $(B C),(C A)$ and - (AB) respectively. Show that $pb \\geq$ 10. Prove that this equation has two irrational solutions. (The number $m$ is triangular, if $m=n(n-1) / 2$ for certain positive integer $n \\geq 1$ ).\n\n9.3 The distinct points $M$ and $N$ lie on the hypotenuse ( $A C)$ of the right isosceles triangle $A B C$ so that $M \\in(A N)$ and $M N^{2}=A M^{2}+C N^{2}$. Prove that $m(\\angle M B N)=45^{\\circ}$.\n\n9.4 Find all the functions $f: N^{*} \\rightarrow N^{*}$ which verify the relation $f(2 x+3 y)=2 f(x)+3 f(y)+4$ for every positive integers $x, y \\geq 1$.\n\n9.5 The numbers $a_{1}, a_{2}, \\ldots, a_{n}$ are the first $n$ positive integers with the property that the number $8 a_{k}+1$ is a perfect square for every $k=1,2, \\ldots, n$. Find the sum $S_{n}=a_{1}+a_{2}+\\ldots+a_{n}$.\n\n9.6 Find all real solutions of the equation $x^{4}+7 x^{3}+6 x^{2}+5 \\sqrt{2003} x-2003=0$.\n\n9.7 The side lengths of the triangle $A B C$ satisfy the relations $a>b \\geq 2 c$. Prove that the altitudes of the triangle $A B C$ can not be the sides of any triangle.\n\n9.8 The base of a pyramid is a convex polygon with 9 sides. All the lateral edges of the pyramid and all the liagunads ui the base are coloured in a random way in red or blue. Pröve that there exist at least three vertices of the pyramid which belong to a triangle with the sides coloured in the same colour.\n\n10.1 Find all prime numbers $a, b$ and $c$ for which the equality $(a-2)!+2 b!=22 c-1$ holds.\n\n10.2 Solve the system $x+y+z+t=6, \\sqrt{1-x^{2}}+\\sqrt{4-y^{2}}+\\sqrt{9-z^{2}}+\\sqrt{16-t^{2}}=8$.\n\n10.3 In the scalen triangle $A B C$ the points $A_{1}$ and $B_{1}$ are the bissectrices feets, drawing from the vertices $A$ and $B$ respectively. The straight line $A_{1} B_{1}$ intersect the line $A B$ at the point $D$. Prove that one of the angles $\\angle A C D$ or $\\angle B C D$ is obtuze and $m(\\angle A C D)+m(\\angle B C D)=180^{\\circ}$.\n\n10.4 Let $a>1$ be not integer number and $a \\neq \\sqrt[2]{q}$ for every positive integers $p \\geq 2$ and $q \\geq 1$, $k=\\left[\\log _{a} n\\right] \\geq 1$, where $[x]$ is the integral part of the real number $x$. Prove that for every positive integer $n \\geq 1$ the equality\n\n$$\n\\left[\\log _{a} 2\\right]+\\left[\\log _{a} 3\\right]+\\ldots+\\left[\\log _{a} n\\right]+[a]+\\left[a^{2}\\right]+\\ldots+\\left[a^{k}\\right]=n k\n$$\n\nholds.\n\n10.5 The rational numbers $p, q, r$ satisfy the relation $p q+p r+q r=1$. Prove that the number $\\left(1+p^{2}\\right)\\left(1+q^{3}\\right)\\left(1+r^{2}\\right)$ is a square of any rational number.\n\n10.6 Let $n \\geq 1$ be a positive integer. For every $k=1,2, \\ldots, n$ the functions $f_{k}: R \\rightarrow R, f_{k}(x)=$ $a_{k} x^{2}+b_{k} x+c_{k}$ with $a_{k} \\neq 0$ are given. Find the greatest possible number of parts of the rectangular plane $x O y$ which can be obtained by the intersection of the graphs of the functions $f_{k}(k=1,2, \\ldots, n)$.\n\n10.7 The circle with the center $O$ is tangent to the sides $[A B],[B C],[C D]$ and $[D A]$ of the convex quadrilateral $A B C D$ at the points $M, N, \\mathcal{K}$ and $L$ respectively. The straight lines $M N$ and $A C$ are parallel and the straight line $M K$ intersect the line $L N$ at the point $P$. Prove that the points $A, M, P, O$ and $L$ are concyclic.\n\n10.8 Find all integers $n$ for which the number $\\log _{2 n-1}\\left(n^{2}+2\\right)$ is rational.\n\n11.1 Let $a, b, c, d \\geq 1$ be arbitrary positive numbers. Prove that the equations system $a x-y z=$ $c, \\quad b x-y t=-d$. has at least a solution $(x, y, z, t)$ in positive integers.\n\n11.2 The sequences $\\left(a_{n}\\right)_{n \\geq 0}$ and $\\left(b_{n}\\right)_{n \\geq 0}$ satisfy the conditions $(1+\\sqrt{3})^{2 n+1}=a_{n}+b_{n} \\sqrt{3}$ and $a_{n}, b_{n} \\in Z$. Find the recurrent relation for each of the sequences $\\left(a_{n}\\right)$ and $\\left(b_{n}\\right)$.\n\n11.3 The triangle $A B C$ is rightangled in $A, A C=b, A B=c$ and $B C=a$. The halfstraight line ( $A z$ is perpendicular to the plane $(A B C), M \\in(A z$ so that $\\alpha, \\beta, \\gamma$ are the mesures of the angles, formed by the edges $M B, M C$ and the plane ( $M B C$ ) with the plane ( $A B C$ ) respectively. In the set of the triangular pyramids MABC on consider the pyramids with the volumes $V_{1}$ and $V_{2}$ which satisfy the relations $\\alpha+\\beta+\\gamma=\\pi$ and $\\alpha+\\beta+\\gamma=\\pi / 2$ respectively. Prove the equality $\\left(V_{1} / V_{2}\\right)^{2}=(a+b+c)(1 / a+1 / b+1 / c)$.\n\n11.4 Find all the functions $f:[0 ;+\\infty) \\rightarrow[0 ;+\\infty)$ which satisfy the conditions: : $f(x f(y)) \\cdot f(y)=$ $f(x+y)$ for every $x, y \\in[0 ;+\\infty) ; f(2)=0 ; f(x) \\neq 0$ for every $x \\in[0 ; 2)$.\n\n11.5 Let $02 R \\sin \\alpha$.\n\n12.4 The real numbers $\\alpha, \\beta, \\gamma$ satisfy the relations $\\sin \\alpha+\\sin \\beta+\\sin \\gamma=0$ and $\\cos \\alpha+\\cos \\beta+\\cos \\gamma=0$. Find all positive integers $n \\geq 0$ for-which $\\sin (n \\alpha+\\pi / 4)+\\sin (n \\beta+\\pi / 4)+\\sin (n \\gamma+\\pi / 4)=0$.\n\n12.5 For every positive integer $n \\geq 1$ we define the polynomial $P(X)=X^{2 n}-X^{2 n-1}+\\ldots-X+1$, Find the remainder of the division of the polynomial $P\\left(X^{2 n+1}\\right)$ by the polynomial $P(X)$.\n\n12.6 Fie $n \\in N$. Find all the primitives of the function\n\n$$\nf: R \\rightarrow R, \\quad f(x)=\\frac{x^{3}-9 x^{2}+29 x-33}{\\left(x^{2}-6 x+10\\right)^{n}}\n$$\n\n12.7 In a rectangular system $x O y$ the graph of the function $f: R \\rightarrow R, f(x)=x^{2}$ is drawn. The ordered triple $B, A, C$ has distinct points on the parabola, the point $D \\in(B C)$ such that the straight line $A D$ is parallel to the axis $O y$ and the triangles $B A D$ and $C A D$ have the areas $s_{1}$ and $s_{2}$ respectively. Find the length of the segment $[A D]$.\n\n12.8 Let $\\left(F_{n}\\right)_{n \\in N^{*}}$ be the Fibonacci sequence so that: $F_{1}=1, F_{2}=1, F_{n+1}=F_{n}+F_{n-1}$ for every positive integer $n \\geq 2$. Shown that $F_{n}<3^{n / 2}$ and calculate the limit $\\lim _{n \\rightarrow \\infty}\\left(F_{1} / 2+F_{2} / 2^{2}+\\ldots+F_{n} / 2^{n}\\right)$.\n\n## The first selection test for IMO 2003 and BMO 2003, March 12, 2003\n\nB1. Each side of the arbitrary triangle is divided into 2002 congruent segments. After that each interior division point of the side is joined with opposite vertex. Prove that the number of obtained regions of the triangle is divisible by 6 .\n\nB2. The positive real numbers $x, y$ and $z$ satisfy the relation $x+y+z \\geq 1$. Prove the inequality\n\n$$\n\\frac{x \\sqrt{x}}{y+z}+\\frac{y \\sqrt{y}}{x+z}+\\frac{z \\sqrt{z}}{x+y} \\geq \\frac{\\sqrt{3}}{2}\n$$\n\nB3. The quadrilateral $A B C D$ is inscribed in the circle with center $O$, the points $M$ and $N$ are the middle points of the diagonals $[A C]$ and $[B D]$ respectively and $P$ is the intersection point of the diagonals. It is known that the points $O, M, N$ si $P$ are distinct. Prove that the points $O, M, B$ and $D$ are concyclic if and only if the points $O, N, A$ and $C$ are concyclic.\n\nB4. Prove that the equation $1 / a+1 / b+1 / c+1 /(a b c) \\doteq 12 /(a+b+c)$ has many solutions $(a, b, c)$ in strictly positive integers.\n\n## The second selection test for IMO 2003, March 22, 2003\n\nB5. Let $n \\geq 1$ be positive integer. Find all polynomials of degree $2 n$ with real coefficients\n\n$$\nP(X)=X^{2 n}+(2 n-10) X^{2 n-1}+a_{2} X^{2 n-2}+\\ldots+a_{2 n-2} X^{2}+(2 n-10) X+1\n$$\n\n-if it is known that they have positive real roots.\n\nB6. The triangle $A B C$ has the semiperimeter $p$, the circumradius $R$, the inradius $r$ and $l_{a,}, l_{b}, l_{c}$ are the lengths of internal bissecticies, drawing from the vertices $A, B$ and $C$ respectively. Prove the inequality $l_{a} l_{b}+l_{b} l_{c}+l_{c} l_{a} \\leq p \\sqrt{3 r^{2}+12 R r}$.\n\nB7. The points $M$ and $N$ are the tangent points of the sides $[A B]$ and $[A C]$ of the triangle $A B C$ to the incircle with the center $I$. The internal bissectrices, drawn from the vertices $B$ and $C$, intersect the straight line $M N$ at points $P$ and $Q$ respectively. If $F$ is the intersection point of the swtraight lines $C P$ and $B Q$, then prove that the straight lines $F I$ and $B C$ are perpendicular.\n\nB8. Let $n \\geq 4$ be the positive integer. On the checkmate table with dimensions $n \\times n$ we put the coins. One consider the diagonal of the table each diagonal with at least two unit squares. What is the smallest number of coins put on the table so that on the each horizontal, each vertical and each diagonal there exists att least one coin. Prove the answer.\n\n## The third selection test for IMO 2003, March 23, 2003\n\nB9. Let $n \\geq 1$ be positive integer. A permutation $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ of the numbers $(1,2, \\ldots, n)$ is called quadratique if among the numbers $a_{1}, a_{1}+a_{2}, \\ldots, a_{1}+a_{2}+\\ldots+a_{n}$ there exist at least a perfect square. Find the greatest number $n$, which is less than 2003 , such that every permutation of the numbers $(1,2, \\ldots, n)$ will be quadratique.\n\nB10. The real numbers $a_{1}, a_{2}, \\ldots, a_{2003}$ satisfy simultaneousiy the relations: $a_{i} \\geq 0$ for all $i=$ $1,2, \\ldots, 2003 ; \\quad a_{1}+a_{2}+\\ldots+a_{2003}=2 ; \\quad a_{1} a_{2}+a_{2} a_{3}+\\ldots+a_{2003} a_{1}=1$. Find the smallest value of the sum $a_{1}^{2}+a_{2}^{2}+\\ldots+a_{2003}^{2}$.\n\nB11. The arbitrary point $M$ on the plane of the triangle $A B C$ does not belong on the straight lines $A B, B C$ and $A C$. If $S_{1}, S_{2}$ and $S_{3}$ are the areas of the triangles $A M B, B M C$ and $A M C$ respectively, find the geometrical locus of the points $M$ which satisfy the relation $\\left(M A^{2}+M B^{2}+M C^{2}\\right)^{2}=16\\left(S_{1}^{2}+S_{2}^{2}+S_{3}^{2}\\right)$.\n\n812. Let $n \\geq 1$ be a positive integer. A square table of dimensions $n \\times n$ is full arbitrarly completed $\\because$ the numb so, shat every number appear exactly conce the table. from cack fine one select the smallest number and the greatest of them is denote by $x$. From each column one select the greatest number and the smallest of them is denote by $y$. The table is called equilibrated if $x=y$. How match equilibrated tables there exist?\n\n## The first selection test for JBMO 2003, April 12, 2003\n\nJB1. Let $n \\geq 2003$ be a positive integer such that the number $1+2003 n$ is a perfect square. Prove that the number $n+1$ is equal to the sum of 2003 positive perfect squares.\n\nJB2. The positive real numbers $a, b, c$ satisfy the relation $a^{2}+b^{2}+c^{2}=3 a b c$. Prove the inequality\n\n$$\n\\frac{a}{b^{2} c^{2}}+\\frac{b}{c^{2} a^{2}}+\\frac{c}{a^{2} b^{2}} \\geq \\frac{9}{a+b+c}\n$$\n\nJB3. The quadrilateral $A B C D$ with perpendicular diagonals is inscribed in the circle with center $O$, the points $M$ and $N$ are the middle points of the sides $[B C]$ and $[C D]$ respectively. Find the value of the ratio of areas of the figures $O M C N$ and $A B C D$.\n\nJB4. Let $m$ and $n$ be the arbitrary digits of the decimal system and $a, b, c$ be the positive distinct integers of the form $2^{m} \\cdot 5^{n}$. Find the number of the equations $a x^{2}-2 b x+c=0$, if it is known that each equation has a single real solution.\n\n## The second selection test for JMBO 2003, April 13, 2003\n\nJB5. Prove that each positive integer is equal to a difference of two positive integers with the same number of the prime divisors.\n\nJB6. The real numbers $x$ and $y$ satisfy the equalities\n\n$$\n\\sqrt{3 x}\\left(1+\\frac{1}{x+y}\\right)=2, \\quad \\sqrt{7 y}\\left(1-\\frac{1}{x+y}\\right)=4 \\sqrt{2}\n$$\n\nFind the numerical value of the ratio $y / x$.\n\n$J B 7$. The triangle $A B C$ is isosceles with $A B=B C$. The point $F$ on the side $[B C]$ and the point $D$ on the side $[A C]$ are the feets of the internal bissectrix drawn from $A$ and altitude drawn from $B$ respectively so that $A F=2 B D$. Find the measure of the angle $A B C$.\n\nJB8. In the rectangular coordinate system every point with integer coordinates is called laticeal point. Let $P_{n}(n, n+5)$ be a laticeal point and denote by $f(n)$ the number of laticeal points on the open segment $\\left(O P_{n}\\right)$, where the point $O(0,0)$ is the coordinates system origine. Calculate the number $f(1)+f(2)+$ $f(3)+\\ldots+f(2002)+f(2003)$.\n\n7 th Junior Balkan Mathematical O-lympiad\n\n$20-25$ Jun e, 20.03 I $\\mathrm{m}$ i r $\\quad$. $\\quad$ u rke y\n\n## English Version\n\n1. Let $n$ be a positive integer. A number $A$ consists of $2 n$ digits, each of which is 4 ; and a number $B$ consists of $n$ digits, each of which is 8 . Prove that $A+2 B+4$ is a perfect square.\n\n\\&\n\n2. Suppose there are $n$ points in a plane no three of which are collinear with the following property:\n\nIf we label these points as $A_{1}, A_{2}, \\ldots, A_{n}$ in any way whatsoever, the broken line $A_{1} A_{2} \\ldots A_{n}$ does not intersect itself.\n\nFind the maximal value that $n$ can have.\n\n3. Let $k$ be the circumcircle of the triangle $A B C$. Consider the arcs $\\overparen{A B}, \\widehat{B C}, \\widetilde{C A}$ such that $C \\notin \\widetilde{A B}, A \\notin \\widetilde{B C}, B \\notin \\widetilde{C A}$. Let $D, E$ and $F$ be the midpoints of the arcs $\\widehat{B C}, \\overparen{C A}, \\overparen{A B}$, respectively. Let $G$ and $H$ be the points of intersection of $D E$ with $C B$ and $C A$; let $I$ and $J$ be the points of intersection of $D F$ with $B C$ and $B A$, respectively. Denote the midpoints of $G H$ and $I J$ by $M$ and $N$, respectively.\n\na) Find the angles of the triangle $D M N$ in terms of the angles of the triangle $A B C$.\n\nb) If $O$ is the circumcentre of the triangle $D M N$ and $P$ is the intersection point of $A D$ and $E F$, prove that $O, P, M$ and $N$ lie on the same circle.\n\n4. Let $x, y, z$ be real numbers greater than -1 . Prove that\n\n$$\n\\frac{1+x^{2}}{1+y+z^{2}}+\\frac{1+y^{2}}{1+z+x^{2}}+\\frac{1+z^{2}}{1+x+y^{2}} \\geq 2\n$$\n\n## Romanian Version", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_match": "\nGEO 7.", "solution_match": "\nSolution:"}} +{"year": "2003", "tier": "T3", "problem_label": "A1", "problem_type": "Algebra", "problem": "1. Fie $n$ un număr natural nenul.. Un număr $A$ conține $2 n$ cifre, fiecare fiind 4 ; și un număr $B$ conţine $n$ cifre, fiecare fiind 8 . Demonstratị că $A+2 B+4$ este un pătrat perfect.\n\n$$\n\\begin{aligned}\n& \\text { Macedrea } \\\\\n& \\text { Shucar Crevok. }\n\\end{aligned}\n$$\n\n2. Fie $n$ puncte în plan, oricare trei necoliniare, cu proprietátea:\n\noricum am numerota aceste puncte $A_{1}, A_{2}, \\ldots, A_{n}$, linia frântă $A_{1} A_{2} \\ldots A_{n}$ nu se autointersectează.\n\nGăsitị valoarea maximă a lui $n$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f45e5cc675871c830e68g-26.jpg?height=115&width=625&top_left_y=1112&top_left_x=1228)\n\n3. Fie $k$ cercul circumscris triunghiului $A B C$. Fie arcele $\\overparen{A B}, \\overparen{B C}, \\overparen{C A}$ astfel încât $C \\notin \\widehat{A B}, A \\notin \\widehat{B C}, B \\notin \\widehat{C A}$ siV․ $E$ mijloacele acestor arce. Fie $G, H$ punctele de intersectie ale lui $D E$ cu $C B, C A$; fie $I, J$ punctele de intersecție ale lui $D F$ cu $B C, B A$. Notăm mijloacele lui $G H, I J$ cu $M$, respectiv $N$.\n\na) Găsiți unghiurile triunghiului $D M N$ în funcție de unghiurile triunghiului $A B C$.\n\nb) Dacă $O$ este circumcentrul triunghiului $D M N$ şi $P$ este intersecția lui $A D \\mathrm{cu}$ $E F$, arătați că $O, P, M$ şi $N$ aparțin unui acelaşi cerc.\n\n4. Fie $x, y, z$ numere reale mai mari decât -1 . Demonstrați că:\n\n$$\n\\frac{1+x^{2}}{1+y+z^{2}}+\\frac{1+y^{2}}{1+z+x^{2}}+\\frac{1+z^{2}}{1+x+y^{2}} \\geq 2\n$$\n\nRomalala - Pmecityol.\n\nTimp de lucru: 4 ore și jumătate.\n\nFiecare problemăeste notată czu 10 مuncte\n\n## Question 1\n\nI. To do a special case $n \\geq 2$.\n\nII. To assert that $A+2 B+4=(\\underbrace{6 \\ldots 68}_{n-1})^{2}$.\n\nIII. To observe that $A=4 \\times \\frac{10^{2 n}-1}{9}$ and $B=8 \\times \\frac{10^{n}-1}{9}$.\n\nIV. To observe that $A=3^{2} \\times(\\underbrace{2 \\ldots 2}_{n})^{2}+4 \\times(\\underbrace{2 \\ldots 2}_{n})$ or $A=\\left(\\frac{3 B}{4}\\right)^{2}+B$.\n\n$$\n\\begin{aligned}\n& \\mathbf{I} \\rightarrow 1 \\text { point } \\\\\n& \\mathbf{I}+\\mathrm{II} \\rightarrow 2 \\text { points } \\\\\n& \\mathbf{I I I} \\rightarrow 4 \\text { points or } \\quad \\mathbf{I V} \\rightarrow 5 \\text { points }\n\\end{aligned}\n$$\n\n## Question 2\n\nI. To claim $n=4$ with example for $n=4$.\n\nII. To show impossibility of the case when the set of points includes 4 points that form a convex quadrilateral.\n\nIII. To show that every set of $n \\geq 5$ points contains 4 points forming a convex quadrilateral.\n\n$$\n\\begin{aligned}\n& \\text { I } \\rightarrow 2 \\text { points } \\\\\n& \\text { II } \\rightarrow 1 \\text { point } \\\\\n& \\text { III } \\rightarrow 4 \\text { points } \\\\\n& \\text { II }+ \\text { III } \\rightarrow 7 \\text { points }\n\\end{aligned}\n$$\n\n## Question 3\n\n## Part a\n\nI. Computing the angles of the triangle $D E F$.\n\nII. Observing that the lines $C F \\perp D E$ and that $B E \\perp D F$.\n\nI $\\rightarrow$ 1\"point\n\n$\\mathrm{I}+\\mathrm{II} \\rightarrow 3$ points\n\nOnly Part a $\\rightarrow 6$ points\n\n## Part b\n\nIII. Completing the figure by drawing $E F$.\n\nPart a + III $\\rightarrow 7$ points\n\nOnly Part $\\mathbf{b} \\rightarrow 6$ points\n\n## Question 4\n\nI. To observe that $y \\leq \\frac{y^{2}+1}{2}$.\n\nII. To observe that $1+y+z^{2}>0$ and to obtain $\\frac{1+x^{2}}{1+y+z^{2}} \\geq \\frac{1+x^{2}}{1+z^{2}+\\frac{1+y^{2}}{2}}$.\n\nIII. To reduce to $\\frac{C+4 B-2 A}{A}+\\frac{A+4 C-2 B}{B}+\\frac{B+4 A-2 C}{C} \\geq 9$.\n\n$$\n\\begin{aligned}\n& I \\rightarrow 1 \\text { point } \\\\\n& X \\rightarrow I Y \\rightarrow 3 \\text { points } \\\\\n& I+I I+I I \\rightarrow 5 \\text { points }\n\\end{aligned}\n$$\n\n| SCORES | | | | |\n| :---: | :---: | :---: | :---: | :---: |\n| $=1$ | ROM-6 | Adrian Zahariuc | 40 | First Prize |\n| 2 | ROM-3 $=$ | Dragos Michnea | 40 | First Prize |\n| 3 | MOL-6 | Alexandru Zamorzaev | 39 | First Prize |\n| 4 | MOL-1 | lurie Boreico | 38 | First Prize |\n| 5 | ROM-5 | Lucian Turea. | 38 | First Prize |\n| 6 | ROM-4 | Cristian Talau | 37 | First Prize |\n| 7 | BUL-4 | Vladislav Vladilenon Petkov | 33 | Second Prize |\n| 8 | HEL-1 | Theodosios Douvropaulos | 32 | Second Prize |\n| 9 | BUL-1 | Alexander Sotirov Bikov | 31 | Second Prize |\n| 10 | BUL-2 | Anton Sotirov Bikov | 31 | Second Prize |\n| 11 | TUR-4 | Hale Nur Kazaçeşme | 31 | Second Prize |\n| 12 | TUR-6 | Sait Tunç | 31 | Second Prize |\n| 13 | BUL-5 | Deyan Stanislavov Simeonov | 30 | Second Prize |\n| 14 | HEL-3 | Faethontas Karagiannopoulos | 30 | Second Prize |\n| 15 | MCD-5 | Maja Tasevska | 29 | Second Prize |\n| 16 | ROM-2 | Sebastian Dumitrescu | 29 | Second Prize |\n| 17 | $B U L-6$ | Tzvetelina Kirilova Tzeneva | 29 | Second Prize |\n| 18 | $B U L-3$ | Asparuh Vladislavov Hriston | 28 | Second Prize |\n| 19 | TUR-5 | Burak Sağlam | 24 | Third Prize |\n| 20 | TUR-1 | Ibrahim Çimentepe | 23 | Third Prize |\n| 21 | YUG-4 | Jevremovic Marko | 22 | Third Prize |\n| 22 | YUG-1 | Lukic Dragan | 22 | Third Prize |\n| 23 | ROM-1 | Beniamin Bogosel | 21 | Third Prize |\n| 24 | YUG-5 | Djoric.Milos | 21 | Third Prize |\n\n\n| 25 | MOL-4 | Vladimir Vanovschi | 21 | Third Prize |\n| :---: | :---: | :---: | :---: | :---: |\n| 26 | YUG-2 | Andric Jelena | 19 | Third Prize |\n| 27 | YUG-6 | Radojevic Mladen | 19 | Third Prize |\n| 28 | MCD-4 | Viktor Simjanovski | 17 | Third Prize |\n| 29 | HEL-6 | Efrosyni Sarla | 16 | Third Prize |\n| 30 | TUR-2 | Türkü Çobanoğlu | 13 | Third Prize |\n| 31 | YUG-3 | Pajovic Jelena | 12 | Third Prize |\n| 32 | MCD-2 | Aleksandar lliovski | 11 | Third Prize |\n| 33 | MCD-6 | Tanja Velkova | 11 | Third Prize |\n| 34 | MOL-2 | Andrei Frimu | 10 | Honorary Mention |\n| 35 | MOL-5 | Dan Vieru | 10 | Honorary Mention |\n| 36 | MCD-3 | Oliver Metodijev | 10 | Honorary Mention |\n| 37 | $H E L-4$ | Stefanos Kasselakis | 9 | |\n| 38 | HEL-5 | Fragiskos Koufogiannis | 8 | - |\n| 39 | MCD-1 | Matej Dobrevski | 8 | |\n| 40 | HEL-2 | Marina lliopoulou | 4 | |\n| 41 | MOL-3 | Mihaela Rusu | 4 | |\n| 42 | CYP-1 | Narısia Drakou | 4 | |\n| 43 | CYP-6 | Anastasia Solea | 3 | |\n| 44 | TUR-3 | Ahmet Kabakulak | 2 | |\n| 45 | CYP-4 | Marina Kouyiali | 2 | |\n| 46 | CYP-5 | Michalis Rossides | 2 | |\n| 47 | CYP-2 | Domna Fanidou | 1 | |\n| 48 | CYP-3 | Yiannis loannides | 0 | |\n\nEen de a 7-a Olingriodá Batcanicà de Mabematicepentre juniosi s-a elesfancuat in pesidada 20-25 iunce in Tercia in stationea Kusadasi ( ecirca $90 \\mathrm{~km} \\mathrm{la}$ sud de Izmir, pe malul mäci Egee). Gchecpa Ronaincer' a fost condusà de. Prof di. Dan Brainzei, ascsbat de. Prof. grad I Dinu Yerfinescu. In clasamentul nesficial pe nationi Romincia scupà primal loc usmatà de Bulgoria, Jurcia, Republica Molabra, Serbia, Macedonia, Mecia, Sipta. Eomponentic echipei Romainiei cu. oblimat ismattoorele punctaje ic medolic:\n\nDragos Michnen (Satu Mare) - 40p-Awr.\n\nAdricin Zahariuc (Bacciu) - $40 p-$ Aur\n\nLucian Turen (Bucuresti) - $38 p$ - Awr\n\nGeistian Taliu (Slacova) - 37p-Aar\n\nSebastian Sumitresce (Bucuegti) - 29p-Aegint\n\nBeniamin Bogosel (Hlad) - 21p-Aegint.\n\nMentionaim ca promic doi trax sealicat punctajid total.\n\nTnainte de a se deplasn im Turcia, exhipn Romannier a fost garduità thei zile la Bucuresti. bैxclesio on sosp de antenument, in accastà periondin, juniouric au participat le al 5-lea si al 6-lea test test de selectie pentru OIM. Pestatio junioscibs la aceste leste a fost excelenta-\n\n## Olimpiada Naţională de Matematică\n\nAl cincilea test de selectie pentru OIM - 19 iunie 2003\n\n## Subiectul 1\n\nUn parlament are $n$ deputati. Aceştia fac parte din 10 partide şi din 10 comisi parlamentare. Fiecare deputat face parte dintr-un singur partid si dintr-o singură comisie.\n\nDeterminati valoarea minimă a lui $n$ pentru care indiferent de componenţa numerică a partidelor şi indiferent de repartizarea în comisii, să existe o numerotare cu toate numerele $1,2, \\ldots, 10$ atât a partidelor cât şi a comisiilor, astfel încât cel puţin 11 deputaţi să facă parte dintr-un partid si o comisie cu număr identic.\n\n## Subiectul 2\n\nSe dă un romb $A B C D$ cu latura 1. Pe laturile $B C$ şi $C D$ există punctele $M$, respectiv $N$, astfel încât $M C+C N+N M=2$ si $\\angle M A N=\\frac{1}{2} \\angle B A D$.\n\nSă se afle unghiurile rombului.\n\n## Subiectul 3\n\nÎntr-un plan înzestrat cu un sistem de coordonate $X O Y$ se numeste punct laticial un punct $A(x, y)$ in care ambele coordonate sunt numere întregi. Un punct laticial $A$ se numeşte invizibil dacă pe segmentul deschis $O A$ există cel puţin un punct laticial.\n\nSă. se arate că pentru orice număr natural $n, n>0$, există un pătrat de latură $n$ în care toate punctele laticiale interioare, de pe laturi sau din vârfuri, sunt invizibile.\n\nTimp de lucru: 4 ore\n\n## Olimpiada Naţională de Matematică 2003\n\nAl şaselea test de selecţie pentru OIM - 20 iunie 2003\n\n## Problema 1.\n\nFie $A B C D E F$ un hexagon convex. Notăm cu $A^{\\prime}, B^{\\prime}, C^{\\prime}, D^{\\prime}, E^{\\prime}, F^{\\prime}$ mijloacele laturilor $A B, B C, C D, D E, E F, F A$ respectiv. Se cunosc arile triunghiurilor $A B C^{\\prime}, B C D^{\\prime}, C D E^{\\prime}$, $D E F^{\\prime}, E F A^{\\prime}, F A B^{\\prime}$.\n\nSă se afle aria hexagonului $A B C D E F$.\n\n## Problema 2.\n\nO permutare $\\sigma:\\{1,2, \\ldots, n\\} \\rightarrow\\{1,2, \\ldots, n\\}$ se numeşte strânsă dacă pentru orice $k=1,2, \\ldots, n-1$ avem\n\n$$\n|\\sigma(k)-\\sigma(k+1)| \\leq 2\n$$\n\nSă se găsească cel mai mic număr natural $n$ pentru care Єxistă cel puţin 2003 permutări strânse.\n\n## Problema 3.\n\nPentru orice număr natural $n$ notăm cu $C(n)$ suma cifrelor sale în baza 10. Arătaţi că oricare ar fi numărul natural $k$ există un număr natural $m$ astfel încât ecuaţia $x+C(x)=m$ are cel puţin $k$ soluţii.\n\nTimp de lucru 4 ore\n\n## Proposed Problem \\#72\n\n$==$ Valentin Vornicu $==$\n\nJune 20, 2003\n\nProblem: A permutation $\\sigma:\\{1,2, \\ldots, n\\} \\rightarrow\\{1,2, \\ldots, n\\}$ is called straight if and only if for each integer $k, 1 \\leq k \\leq n-1$ the following inequality is fulfilled\n\n$$\n|\\sigma(k)-\\sigma(k+1)| \\leq 2\n$$\n\nFind the smallest positive integer $n$ for which there exist at least 2003 straight permutations.", "solution": "The main trick is to look where $n$ is positioned. Tn that idea let us denote by $x_{n}$ the number of all the straight permutations and by $a_{n}$ the number of straight permutations having $n$ on the first or on the last position, i.e. $\\sigma(1)=n$ or $\\sigma(n)=n$. Also let us denote by $b_{n}$ the difference $x_{n}-a_{n}$ and by $a_{n}^{\\prime}$ the number of permutations having $n$ on the first position, and by $a_{n}^{\\prime \\prime}$ the number of permutations having $n$ on the last position. From symmetry we have that $2 a_{n}^{\\prime}=2 a_{n}^{\\prime \\prime}=a_{n}^{\\prime}+a_{n}^{\\prime \\prime}=a_{n}$, for all $n$-s. Therefore finding a recurrence relationship for $\\left\\{a_{n}\\right\\}_{n}$ is equivalent with finding one for $\\left\\{a_{n}^{\\prime}\\right\\}_{n}$.\n\nOne can simply compute: $a_{2}^{\\prime}=1, a_{3}^{\\prime}=2, a_{4}^{\\prime}=4$. Suppose that $n \\geq 5$. We have two possibilities for the second position: if $\\sigma(2)=n-1$ then we must complete the remaining positions with $3,4, \\ldots, n$ thus the number of ways in which we can do that is $a_{n-1}^{\\prime}$ (because the permutation $\\sigma^{\\prime}:\\{1,2, \\ldots, n-1\\} \\rightarrow$ $\\{1,2, \\ldots, n-1\\}, \\sigma^{\\prime}(k)=\\sigma(k+1)$, for all $k, 1 \\leq k \\leq n-1$, is also a straight permutation $)$.\n\nIf on the second position we have $n-2, \\sigma(2)=n-2$, then $n-1$ can only be in the last position of the permutation or on the third position, i.e. $\\sigma(3)=n-1$ or $\\sigma(n)=n-1$. If $\\sigma(n)=n-1$, then we caw only have $\\sigma(n-1)=n-3$ thus $\\sigma(3)=n-4$ and so on, thus there is only one permutation of this kind. On the other hand, if $\\sigma(3)=n-1$ then it follows that $\\sigma(4)=n-3$ and now we can complete the permutation in $n_{n-3}^{\\prime}$ ways (hecause the permutation $\\sigma^{\\prime}:\\{1,2, \\ldots, n-3\\} \\rightarrow\\{1,2, \\ldots, n-3\\}, \\sigma^{\\prime}(k)=\\sigma(k+3)$, for all $k$, $1 \\leq k \\leq n-3$, is also a straight permutation).\n\nSumming all up we get the recurrence:\n\n$$\na_{n}^{\\prime}=a_{n-1}^{\\prime}+1+a_{n-3}^{\\prime} \\Rightarrow a_{n}=a_{n-1}+a_{n-3}+2, \\forall n \\geq 5\n$$\n\nThe recurrence relationship for $\\left\\{b_{n}\\right\\}$ can be obtained by observing that for each straight permutation $\\tau:\\{1,2, \\ldots, n+1\\} \\rightarrow\\{1,2, \\ldots, n+1\\}$ for which $2 \\leq \\tau^{-1}(n+1) \\leq n$ we can obtain a straight permutation $\\sigma:\\{1,2, \\ldots, n\\} \\rightarrow\\{1,2, \\ldots, n\\}$ by removing $n+1$. Indeed $n+1$ is \"surrounded\" by $n$ and $n-1$, so by removing it, $n$ and $n-1$ become neighbors, and thus the newly formed permutation is indeed straight. Now, if $\\tau^{-1}(n) \\in\\{1, n+1\\}$ then the newly formed permutation $\\sigma$ was counted as one of the $a_{n}$-s, minus the two special cases in which $n$ and $n-1$ are on the first and last positions. If $\\tau^{-1}(n) \\notin\\{1, n+1\\}$ then certainly $\\sigma$ was counted with the $b_{n}$-s. Also, from any straight permutation of $n$ elements, not having $n$ and $n-1$ in the first and last position, thus $n$ certainly being neighbor with $n-1$, we can make a straight $n+1$-element permutation by inserting $n+1$ between $n$ and $n-1$.\n\nTherefore we have obtained the following relationship:\n\n$$\nb_{n+1}=a_{n}-2+b_{n}=x_{n}-2, \\forall n \\geq 4\n$$\n\nFrom (1) and (2) we get that\n\n$$\nx_{n}=x_{n-1}+a_{n-1}+a_{n-3}, \\forall n \\geq 5\n$$\n\nIt is obvious that $\\left\\{x_{n}\\right\\}_{n}$ is a \"fast\" increasing sequence, so we will compute the first terms using the relationships obtained above, which will prove that the number that we are looking for is $n=16$ :\n\n$$\n\\begin{aligned}\n& a_{2}=2 \\quad x_{2}=3 \\\\\n& a_{9}=62 \\quad x_{9}=164 \\\\\n& a_{3}=4 \\quad x_{3}=6 \\\\\n& a_{4}=8 \\quad x_{4}=12 \\\\\n& a_{5}=12 \\quad x_{5}=22 \\\\\n& a_{6}=18 \\quad x_{6}=38 \\\\\n& a_{7}=28 \\quad x_{7}=64 \\\\\n& a_{10}=92 \\quad x_{10}=254 \\\\\n& a_{8}=42 \\quad x_{8}=i 04 \\\\\n& a_{11}=136 \\quad x_{11}=388 \\\\\n& a_{12}=200 \\quad x_{12}=586 \\\\\n& a_{13}=294 \\quad x_{13}=878 \\\\\n& a_{14}=432 \\quad x_{14}=1308 \\\\\n& a_{15}=034 \\quad x_{15}=1940\n\\end{aligned}\n$$\n\n## ENUNȚURULE PROBLEMELOR DIN ATENTTIA JURIULUT LA CEA DE A 7-A JBMO (KUSADASI, TURCIA, 20-25 IUNIE 2003)\n\nA.1. Un număr A este scris cu 2 n cifre, fiecare dintre acestea fiind 4 ; un număr B este scris cu $n$ cifre, fiecare dintre acestea fiind 8 . Demonstrați că, pentru orice $n, A+2 B+4$ este pătrat perfect. A.2. Fie a, b, c lungimile laturilor unui triunghi, $p=\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}, q=\\frac{a}{c}+\\frac{c}{b}+\\frac{b}{a}$. Demonstrați că $|p-q|<1$.\n\nA.3. Fie $\\mathrm{a}, \\mathrm{b}, \\mathrm{c}$ numere reale astfel încât $a^{2}+b^{2}+c^{2}=1$. Demonstrați că $P=a b+b c+c a-2(a+b+c) \\geq-5 / 2$. Există valori pentru $\\mathrm{a}, \\mathrm{b}, \\mathrm{c}$ încât $\\mathrm{P}=-5 / 2$ ?\n\nA.4. Fie $\\mathrm{a}, \\mathrm{b}$, c numere raționale astfel încât $\\frac{1}{a+b c}+\\frac{1}{b+a c}=\\frac{1}{a+b}$. Demonstrați că $\\sqrt{\\frac{c-3}{c+1}}$ este de asemenea număr rațional.\n\nA.5. Fie $A B C$ triunghi neisoscel cu lungimile $a, b, c$ ale laturilor numere naturale. Dempnstraţi că $\\left|a b^{2}\\right|+\\left|\\tilde{b c} c^{2}\\right|+\\left|c a^{2}-a^{2} b-b^{2} c-c^{2} a\\right| \\geq 2$.\n\nA.6. Fie $\\mathrm{a}, \\mathrm{b}$, c c numere pozitive astfel ca $a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=3$. Demonstrați că $a+b+c \\geq a b c+2$.\n\nA.6'. Fie $\\mathrm{a}, \\mathrm{b}, \\mathrm{c}$ numere pozitive astfel ca $a b+b c+c a=3$. Demonstrați că $a+b+c \\geq a b c+2$.\n\nA.7. Fie $\\mathrm{x}, \\mathrm{y}$, $\\mathrm{z}$ numere mai mari câ -1. Demönstraţi că $\\frac{1+x^{2}}{1+y+z^{2}}+\\frac{1+y^{2}}{1+z+x^{2}}+\\frac{1+z^{2}}{1+x+y^{2}} \\geq 2$. A.8. Demonstraţi că există mulțimi disjuncte $A=\\{x, y, z\\}$ șì $B=\\{m, n, p\\}$ de numere naturale mai mari ca 2003 astfel ca $x+y+z=m+n+p$ si $x^{2}+y^{2}+z^{2}=m^{2}+n^{2}+p^{2}$.\n\nC.1. Într-un grup de 60 studenți: 40 vorbesc engleza, 30 vorbesc franceza, 8 vorbesc toate cele trei limbi. Numărul celor ce vorbesc doar engleza şi franceza este egal cu suma celor care vorbesc doar germana şi franceza cu a celor ce vorbesc doar engleza şi germana. Numărul celor ce vorbesc cel puțin două dintre aceste limbi este 28. Cât de mulți studenți vorbesc: a) germana; b) numai engleza; c) numai germana.\n\nC.2. Numerele $1,2,3, \\ldots, 2003$ sunt scrise într-un şir $a_{1}, a_{2}, a_{3}, \\ldots, a_{2003}$. Fie $b_{1}=1 \\exists a_{1}, b_{2}=2 \\exists a_{2}$, $b_{3}=3 \\exists a_{3}, \\ldots, b_{2003}=2003 \\exists a_{2003}$ şi B maximul numerelor $b_{1}, b_{2}, b_{3}, \\ldots, b_{2003}$.\n\na) Dacă $a_{1}=2003, a_{2}=2002, a_{3}=2001, \\ldots, a_{2003}=1$, găsiți valoarea lui $B$.\n\nb) Demonstrați că $B \\geq 1002^{2}$.\n\nC.3. Demonstrați că îtr-o mulțime de 29 numere naturale există 15 a căror sumă este divizibilă cu 15 . C.4. Fie n puncte în plan, oricare trei necoliniare, cu proprietatea că oricum le-am numerota $A_{1}$, $A_{2}, \\ldots, A_{n}$, linia frântă $A_{1} A_{2} \\ldots A_{n}$ nu se autointersectează. Găsiți valoarea maximă a lui $n$.\n\nC.5. Fie mulțmea $M=\\{1,2,3,4\\}$. Fiecare punct al planului este colorat în roşu sau albastru.\n\nDemonstrați că există cel puțin un triunghi echilateral cu latura $m \\in M$ cu vârfurile de aceeaşí culoare.\n\nG.1. Există un patrulater convex pe care diagonalele să-1 împartă în patru triunghiuri cu ariile numere prime distincte?\n\nG.2. Există un triunghi cu aria $12 \\mathrm{~cm}^{2}$ şi perimetrul 12 ?\n\nG.3. Fie $\\mathrm{G}$ centrul de greutate al triunghiului $\\mathrm{ABC}$ şi $\\mathrm{A}$ ' simetricul lui $\\mathrm{A}$ faţă de $\\mathrm{C}$. Demonstrați că punctele $\\mathrm{G}, \\mathrm{B}, \\mathrm{C}, \\mathrm{A}$ ' sunt conciclice dacă și numai dacă $\\mathrm{GA} \\zeta \\mathrm{GC}$.\n\nG.4. Fie $k$ cercul circumscris triunghiului $A B C$. Fie arcele $A B, B C, C A$ astfel incât\n$C \\notin A B, A \\notin B C, B \\notin \\mathcal{C} A$ şi $F, D, E$ mijloacele acestor arce. Fie $G, H$ punctele de intersectie ale lui $D E$ cu $C B, C A$; fie $I, J$ punctele de intersectie ale lui $D F$ cu $B C, B A$. Notăm mijloacele lui $G H, I J$ cu $M$, respectiv $N$.\n\na) Găsiți unghiurile triunghiului $D M N$ în funcție de unghiurile triunghiului $A B C$.\n\nb) Dacă $O$ este circumcentrul triunghiului $D M N$ şi $P$ este intersectia lui $A D \\operatorname{cu} E F$, arătați că $O, P, M$ şi $N$ aparțin unui acelaşi cerc.\n\nG.5. Trei cercuri egale au în comun un punct $\\mathrm{M}$ şi se intersectează câte două în puncte $\\mathrm{A}, \\mathrm{B}, \\mathrm{C}$. Demonstrați că $M$ este ortocentrul triunghiului $A B C$. ${ }^{1)}$\n\nG.6. Fie $\\mathrm{ABC}$ un triunghi isoscel $\\mathrm{cu} \\mathrm{AB}=\\mathrm{AC}$. Un semicerc de diametru $\\mathrm{EF}$ situat pe baza $\\mathrm{BC}$ este tangent laturilor $\\mathrm{AB}, \\mathrm{AC}$ în $\\mathrm{M}, \\mathrm{N}$. $\\mathrm{AE}$ retaie semicercul în $\\mathrm{P}$. Demonstraţi că dreapta $\\mathrm{PF}$ trece prin mijlocul corzii MN.\n\nG.7. Paralelele la laturile unui triunghi duse printr-un punct interior împart interiorul triunghiului în şase părți cu ariile notate ca în figură.\n\nDemonstrați că $\\frac{a}{\\alpha}+\\frac{b}{\\beta}+\\frac{c}{\\gamma} \\geq \\frac{3}{2}$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_f45e5cc675871c830e68g-36.jpg?height=320&width=450&top_left_y=978&top_left_x=1381)", "metadata": {"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl", "problem_match": "\nA1 ", "solution_match": "\nSolution:"}}