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8. (ROM 2) $)^{\mathrm{IMO} 3}$ In a plane two different points $O$ and $A$ are given. For each point $X \neq O$ of the plane denote by $\alpha(X)$ the angle $A O X$ measured in radians $(0 \leq \alpha(X)<2 \pi)$ and by $C(X)$ the circle with center $O$ and radius $O X+\frac{\alpha(X)}{O X}$. Suppose each point of the plane is colored by one of a finite number of colors. Show that there exists a point $X$ with $\alpha(X)>0$ such that its color appears somewhere on the circle $C(X)$.
|
8. Suppose that the statement of the problem is false. Consider two arbitrary circles $R=(O, r)$ and $S=(O, s)$ with $0<r<s<1$. The point $X \in R$ with $\alpha(X)=r(s-r)<2 \pi$ satisfies that $C(X)=S$. It follows that the color of the point $X$ does not appear on $S$. Consequently, the set of colors that appear on $R$ is not the same as the set of colors that appear on $S$. Hence any two distinct circles with center at $O$ and radii less than 1 have distinct sets of colors. This is a contradiction, since there are infinitely many such circles but only finitely many possible sets of colors.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
8. (ROM 2) $)^{\mathrm{IMO} 3}$ In a plane two different points $O$ and $A$ are given. For each point $X \neq O$ of the plane denote by $\alpha(X)$ the angle $A O X$ measured in radians $(0 \leq \alpha(X)<2 \pi)$ and by $C(X)$ the circle with center $O$ and radius $O X+\frac{\alpha(X)}{O X}$. Suppose each point of the plane is colored by one of a finite number of colors. Show that there exists a point $X$ with $\alpha(X)>0$ such that its color appears somewhere on the circle $C(X)$.
|
8. Suppose that the statement of the problem is false. Consider two arbitrary circles $R=(O, r)$ and $S=(O, s)$ with $0<r<s<1$. The point $X \in R$ with $\alpha(X)=r(s-r)<2 \pi$ satisfies that $C(X)=S$. It follows that the color of the point $X$ does not appear on $S$. Consequently, the set of colors that appear on $R$ is not the same as the set of colors that appear on $S$. Hence any two distinct circles with center at $O$ and radii less than 1 have distinct sets of colors. This is a contradiction, since there are infinitely many such circles but only finitely many possible sets of colors.
|
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569fb5d7-2c57-5044-847b-5c4fee20f7b1
| 23,942
|
9. (POL 2) Let $a, b, c$ be positive numbers with $\sqrt{a}+\sqrt{b}+\sqrt{c}=\frac{\sqrt{3}}{2}$. Prove that the system of equations $$ \begin{aligned} & \sqrt{y-a}+\sqrt{z-a}=1, \\ & \sqrt{z-b}+\sqrt{x-b}=1 \\ & \sqrt{x-c}+\sqrt{y-c}=1 \end{aligned} $$ has exactly one solution $(x, y, z)$ in real numbers.
|
9. Let us show first that the system has at most one solution. Suppose that $(x, y, z)$ and $\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ are two distinct solutions and that w.l.o.g. $x<x^{\prime}$. Then the second and third equation imply that $y>y^{\prime}$ and $z>z^{\prime}$, but then $\sqrt{y-a}+\sqrt{z-a}>\sqrt{y^{\prime}-a}+\sqrt{z^{\prime}-a}$, which is a contradiction. We shall now prove the existence of at least one solution. Let $P$ be an arbitrary point in the plane and $K, L, M$ points such that $P K=\sqrt{a}$, $P L=\sqrt{b}, P M=\sqrt{c}$, and $\angle K P L=\angle L P M=\angle M P K=120^{\circ}$. The lines through $K, L, M$ perpendicular respectively to $P K, P L, P M$ form an equilateral triangle $A B C$, where $K \in B C, L \in A C$, and $M \in A B$. Since its area equals $A B^{2} \sqrt{3} / 4=S_{\triangle B P C}+S_{\triangle A P C}+$ $S_{\triangle A P B}=A B(\sqrt{a}+\sqrt{b}+\sqrt{c}) / 2$, it follows that $A B=1$. Therefore $x=P A^{2}, y=P B^{2}$, and $z=P C^{2}$ is a solution of the system (indeed, $\sqrt{y-a}+\sqrt{z-a}=\sqrt{P B^{2}-P K^{2}}+\sqrt{P C^{2}-P K^{2}}=B K+C K=1$, etc.).
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
9. (POL 2) Let $a, b, c$ be positive numbers with $\sqrt{a}+\sqrt{b}+\sqrt{c}=\frac{\sqrt{3}}{2}$. Prove that the system of equations $$ \begin{aligned} & \sqrt{y-a}+\sqrt{z-a}=1, \\ & \sqrt{z-b}+\sqrt{x-b}=1 \\ & \sqrt{x-c}+\sqrt{y-c}=1 \end{aligned} $$ has exactly one solution $(x, y, z)$ in real numbers.
|
9. Let us show first that the system has at most one solution. Suppose that $(x, y, z)$ and $\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ are two distinct solutions and that w.l.o.g. $x<x^{\prime}$. Then the second and third equation imply that $y>y^{\prime}$ and $z>z^{\prime}$, but then $\sqrt{y-a}+\sqrt{z-a}>\sqrt{y^{\prime}-a}+\sqrt{z^{\prime}-a}$, which is a contradiction. We shall now prove the existence of at least one solution. Let $P$ be an arbitrary point in the plane and $K, L, M$ points such that $P K=\sqrt{a}$, $P L=\sqrt{b}, P M=\sqrt{c}$, and $\angle K P L=\angle L P M=\angle M P K=120^{\circ}$. The lines through $K, L, M$ perpendicular respectively to $P K, P L, P M$ form an equilateral triangle $A B C$, where $K \in B C, L \in A C$, and $M \in A B$. Since its area equals $A B^{2} \sqrt{3} / 4=S_{\triangle B P C}+S_{\triangle A P C}+$ $S_{\triangle A P B}=A B(\sqrt{a}+\sqrt{b}+\sqrt{c}) / 2$, it follows that $A B=1$. Therefore $x=P A^{2}, y=P B^{2}$, and $z=P C^{2}$ is a solution of the system (indeed, $\sqrt{y-a}+\sqrt{z-a}=\sqrt{P B^{2}-P K^{2}}+\sqrt{P C^{2}-P K^{2}}=B K+C K=1$, etc.).
|
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02349cac-0860-5627-b8ce-63c302557959
| 23,946
|
1. (MON 1) ${ }^{\mathrm{IMO} 4}$ Given a set $M$ of 1985 positive integers, none of which has a prime divisor larger than 26 , prove that the set has four distinct elements whose geometric mean is an integer.
|
1. Since there are 9 primes ( $p_{1}=2<p_{2}=3<\cdots<p_{9}=23$ ) less than 26 , each number $x_{j} \in M$ is of the form $\prod_{i=1}^{9} p_{i}^{a_{i j}}$, where $0 \leq a_{i j}$. Now, $x_{j} x_{k}$ is a square if $a_{i j}+a_{i k} \equiv 0(\bmod 2)$ for $i=1, \ldots, 9$. Since the number of distinct ninetuples modulo 2 is $2^{9}$, any subset of $M$ with at least 513 elements contains two elements with square product. Starting from $M$ and eliminating such pairs, one obtains $(1985-513) / 2=736>513$ distinct two-element subsets of $M$ each having a square as the product of elements. Reasoning as above, we find at least one (in fact many) pair of such squares whose product is a fourth power.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
1. (MON 1) ${ }^{\mathrm{IMO} 4}$ Given a set $M$ of 1985 positive integers, none of which has a prime divisor larger than 26 , prove that the set has four distinct elements whose geometric mean is an integer.
|
1. Since there are 9 primes ( $p_{1}=2<p_{2}=3<\cdots<p_{9}=23$ ) less than 26 , each number $x_{j} \in M$ is of the form $\prod_{i=1}^{9} p_{i}^{a_{i j}}$, where $0 \leq a_{i j}$. Now, $x_{j} x_{k}$ is a square if $a_{i j}+a_{i k} \equiv 0(\bmod 2)$ for $i=1, \ldots, 9$. Since the number of distinct ninetuples modulo 2 is $2^{9}$, any subset of $M$ with at least 513 elements contains two elements with square product. Starting from $M$ and eliminating such pairs, one obtains $(1985-513) / 2=736>513$ distinct two-element subsets of $M$ each having a square as the product of elements. Reasoning as above, we find at least one (in fact many) pair of such squares whose product is a fourth power.
|
{
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94400678-c130-5c1a-9e6f-0d9547ab74a7
| 23,948
|
10. 2b.(VIE 1) Prove that for every point $M$ on the surface of a regular tetrahedron there exists a point $M^{\prime}$ such that there are at least three different curves on the surface joining $M$ to $M^{\prime}$ with the smallest possible length among all curves on the surface joining $M$ to $M^{\prime}$.
|
10. If $M$ is at a vertex of the regular tetrahedron $A B C D(A B=1)$, then one can take $M^{\prime}$ at the center of the opposite face of the tetrahedron. Let $M$ be on the face $(A B C)$ of the tetrahedron, excluding the vertices. Consider a continuous mapping $f$ of $\mathbb{C}$ onto the surface $S$ of $A B C D$ that maps $m+n e^{2 \pi / 3}$ for $m, n \in$ $\mathbb{Z}$ onto $A, B, C, D$ if $(m, n) \equiv$ $(1,1),(1,0),(0,1),(0,0)(\bmod 2)$ re-  spectively, and maps each unit equilateral triangle with vertices of the form $m+n e^{2 \pi / 3}$ isometrically onto the corresponding face of $A B C D$. The point $M$ then has one preimage $M_{j}, j=1,2, \ldots, 6$, in each of the six preimages of $\triangle A B C$ having two vertices on the unit circle. The $M_{j}$ 's form a convex centrally symmetric (possibly degenerate) hexagon. Of the triangles formed by two adjacent sides of this hexagon consider the one, say $M_{1} M_{2} M_{3}$, with the smallest radius of circumcircle and denote by $\widehat{M^{\prime}}$ its circumcenter. Then we can choose $M^{\prime}=f\left(\widehat{M^{\prime}}\right)$. Indeed, the images of the segments $M_{1} \widehat{M^{\prime}}, M_{2} \widehat{M^{\prime}}, M_{3} \widehat{M^{\prime}}$ are three different shortest paths on $S$ from $M$ to $M^{\prime}$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
10. 2b.(VIE 1) Prove that for every point $M$ on the surface of a regular tetrahedron there exists a point $M^{\prime}$ such that there are at least three different curves on the surface joining $M$ to $M^{\prime}$ with the smallest possible length among all curves on the surface joining $M$ to $M^{\prime}$.
|
10. If $M$ is at a vertex of the regular tetrahedron $A B C D(A B=1)$, then one can take $M^{\prime}$ at the center of the opposite face of the tetrahedron. Let $M$ be on the face $(A B C)$ of the tetrahedron, excluding the vertices. Consider a continuous mapping $f$ of $\mathbb{C}$ onto the surface $S$ of $A B C D$ that maps $m+n e^{2 \pi / 3}$ for $m, n \in$ $\mathbb{Z}$ onto $A, B, C, D$ if $(m, n) \equiv$ $(1,1),(1,0),(0,1),(0,0)(\bmod 2)$ re-  spectively, and maps each unit equilateral triangle with vertices of the form $m+n e^{2 \pi / 3}$ isometrically onto the corresponding face of $A B C D$. The point $M$ then has one preimage $M_{j}, j=1,2, \ldots, 6$, in each of the six preimages of $\triangle A B C$ having two vertices on the unit circle. The $M_{j}$ 's form a convex centrally symmetric (possibly degenerate) hexagon. Of the triangles formed by two adjacent sides of this hexagon consider the one, say $M_{1} M_{2} M_{3}$, with the smallest radius of circumcircle and denote by $\widehat{M^{\prime}}$ its circumcenter. Then we can choose $M^{\prime}=f\left(\widehat{M^{\prime}}\right)$. Indeed, the images of the segments $M_{1} \widehat{M^{\prime}}, M_{2} \widehat{M^{\prime}}, M_{3} \widehat{M^{\prime}}$ are three different shortest paths on $S$ from $M$ to $M^{\prime}$.
|
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738dd687-8f02-5f79-9938-6c59ebee07de
| 23,952
|
12. 3b.(GBR 4) A sequence of polynomials $P_{m}(x, y, z), m=0,1,2, \ldots$, in $x, y$, and $z$ is defined by $P_{0}(x, y, z)=1$ and by $$ P_{m}(x, y, z)=(x+z)(y+z) P_{m-1}(x, y, z+1)-z^{2} P_{m-1}(x, y, z) $$ for $m>0$. Prove that each $P_{m}(x, y, z)$ is symmetric, in other words, is unaltered by any permutation of $x, y, z$.
|
12. We shall prove by induction on $m$ that $P_{m}(x, y, z)$ is symmetric and that $$ (x+y) P_{m}(x, z, y+1)-(x+z) P_{m}(x, y, z+1)=(y-z) P_{m}(x, y, z) $$ holds for all $x, y, z$. This is trivial for $m=0$. Assume now that it holds for $m=n-1$. Since obviously $P_{n}(x, y, z)=P_{n}(y, x, z)$, the symmetry of $P_{n}$ will follow if we prove that $P_{n}(x, y, z)=P_{n}(x, z, y)$. Using (1) we have $P_{n}(x, z, y)-$ $P_{n}(x, y, z)=(y+z)\left[(x+y) P_{n-1}(x, z, y+1)-(x+z) P_{n-1}(x, y, z+1)\right]-\left(y^{2}-\right.$ $\left.z^{2}\right) P_{n-1}(x, y, z)=(y+z)(y-z) P_{n-1}(x, y, z)-\left(y^{2}-z^{2}\right) P_{n-1}(x, y, z)=0$. It remains to prove (1) for $m=n$. Using the already established symmetry we have $$ \begin{aligned} & (x+y) P_{n}(x, z, y+1)-(x+z) P_{n}(x, y, z+1) \\ & =(x+y) P_{n}(y+1, z, x)-(x+z) P_{n}(z+1, y, x) \\ & =(x+y)\left[(y+x+1)(z+x) P_{n-1}(y+1, z, x+1)-x^{2} P_{n-1}(y+1, z, x)\right] \\ & \quad-(x+z)\left[(z+x+1)(y+x) P_{n-1}(z+1, y, x+1)-x^{2} P_{n-1}(z+1, y, x)\right] \\ & =(x+y)(x+z)(y-z) P_{n-1}(x+1, y, z)-x^{2}(y-z) P_{n-1}(x, y, z) \\ & =(y-z) P_{n}(z, y, x)=(y-z) P_{n}(x, y, z), \end{aligned} $$ as claimed.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
12. 3b.(GBR 4) A sequence of polynomials $P_{m}(x, y, z), m=0,1,2, \ldots$, in $x, y$, and $z$ is defined by $P_{0}(x, y, z)=1$ and by $$ P_{m}(x, y, z)=(x+z)(y+z) P_{m-1}(x, y, z+1)-z^{2} P_{m-1}(x, y, z) $$ for $m>0$. Prove that each $P_{m}(x, y, z)$ is symmetric, in other words, is unaltered by any permutation of $x, y, z$.
|
12. We shall prove by induction on $m$ that $P_{m}(x, y, z)$ is symmetric and that $$ (x+y) P_{m}(x, z, y+1)-(x+z) P_{m}(x, y, z+1)=(y-z) P_{m}(x, y, z) $$ holds for all $x, y, z$. This is trivial for $m=0$. Assume now that it holds for $m=n-1$. Since obviously $P_{n}(x, y, z)=P_{n}(y, x, z)$, the symmetry of $P_{n}$ will follow if we prove that $P_{n}(x, y, z)=P_{n}(x, z, y)$. Using (1) we have $P_{n}(x, z, y)-$ $P_{n}(x, y, z)=(y+z)\left[(x+y) P_{n-1}(x, z, y+1)-(x+z) P_{n-1}(x, y, z+1)\right]-\left(y^{2}-\right.$ $\left.z^{2}\right) P_{n-1}(x, y, z)=(y+z)(y-z) P_{n-1}(x, y, z)-\left(y^{2}-z^{2}\right) P_{n-1}(x, y, z)=0$. It remains to prove (1) for $m=n$. Using the already established symmetry we have $$ \begin{aligned} & (x+y) P_{n}(x, z, y+1)-(x+z) P_{n}(x, y, z+1) \\ & =(x+y) P_{n}(y+1, z, x)-(x+z) P_{n}(z+1, y, x) \\ & =(x+y)\left[(y+x+1)(z+x) P_{n-1}(y+1, z, x+1)-x^{2} P_{n-1}(y+1, z, x)\right] \\ & \quad-(x+z)\left[(z+x+1)(y+x) P_{n-1}(z+1, y, x+1)-x^{2} P_{n-1}(z+1, y, x)\right] \\ & =(x+y)(x+z)(y-z) P_{n-1}(x+1, y, z)-x^{2}(y-z) P_{n-1}(x, y, z) \\ & =(y-z) P_{n}(z, y, x)=(y-z) P_{n}(x, y, z), \end{aligned} $$ as claimed.
|
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0cb4a3ac-7db7-5c08-a05a-78b9709cab20
| 23,955
|
13. 4a.(BUL 1) Let $m$ boxes be given, with some balls in each box. Let $n<m$ be a given integer. The following operation is performed: choose $n$ of the boxes and put 1 ball in each of them. Prove: (a) If $m$ and $n$ are relatively prime, then it is possible, by performing the operation a finite number of times, to arrive at the situation that all the boxes contain an equal number of balls. (b) If $m$ and $n$ are not relatively prime, there exist initial distributions of balls in the boxes such that an equal distribution is not possible to achieve.
|
13. If $m$ and $n$ are relatively prime, there exist positive integers $p, q$ such that $p m=q n+1$. Thus by putting $m$ balls in some boxes $p$ times we can achieve that one box receives $q+1$ balls while all others receive $q$ balls. Repeating this process sufficiently many times, we can obtain an equal distribution of the balls. Now assume $\operatorname{gcd}(m, n)>1$. If initially there is only one ball in the boxes, then after $k$ operations the number of balls will be $1+k m$, which is never divisible by $n$. Hence the task cannot be done.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
13. 4a.(BUL 1) Let $m$ boxes be given, with some balls in each box. Let $n<m$ be a given integer. The following operation is performed: choose $n$ of the boxes and put 1 ball in each of them. Prove: (a) If $m$ and $n$ are relatively prime, then it is possible, by performing the operation a finite number of times, to arrive at the situation that all the boxes contain an equal number of balls. (b) If $m$ and $n$ are not relatively prime, there exist initial distributions of balls in the boxes such that an equal distribution is not possible to achieve.
|
13. If $m$ and $n$ are relatively prime, there exist positive integers $p, q$ such that $p m=q n+1$. Thus by putting $m$ balls in some boxes $p$ times we can achieve that one box receives $q+1$ balls while all others receive $q$ balls. Repeating this process sufficiently many times, we can obtain an equal distribution of the balls. Now assume $\operatorname{gcd}(m, n)>1$. If initially there is only one ball in the boxes, then after $k$ operations the number of balls will be $1+k m$, which is never divisible by $n$. Hence the task cannot be done.
|
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21aaf2e5-a302-5baf-b683-c925b7ad862b
| 23,957
|
14. 4b.(IRE 4) A set of 1985 points is distributed around the circumference of a circle and each of the points is marked with 1 or -1 . A point is called "good" if the partial sums that can be formed by starting at that point and proceeding around the circle for any distance in either direction are all strictly positive. Show that if the number of points marked with -1 is less than 662 , there must be at least one good point.
|
14. It suffices to prove the existence of a good point in the case of exactly 661 -1 's. We prove by induction on $k$ that in any arrangement with $3 k+2$ points $k$ of which are -1 's a good point exists. For $k=1$ this is clear by inspection. Assume that the assertion holds for all arrangements of $3 n+2$ points and consider an arrangement of $3(n+1)+2$ points. Now there exists a sequence of consecutive -1 's surrounded by two +1 's. There is a point $P$ which is good for the arrangement obtained by removing the two +1 's bordering the sequence of -1 's and one of these -1 's. Since $P$ is out of this sequence, clearly the removal either leaves a partial sum as it was or diminishes it by 1 , so $P$ is good for the original arrangement. Second solution. Denote the number on an arbitrary point by $a_{1}$, and the numbers on successive points going in the positive direction by $a_{2}, a_{3}, \ldots$ (in particular, $a_{k+1985}=a_{k}$ ). We define the partial sums $s_{0}=0, s_{n}=$ $a_{1}+a_{2}+\cdots+a_{n}$ for all positive integers $n$; then $s_{k+1985}=s_{k}+s_{1985}$ and $s_{1985} \geq 663$. Since $s_{1985 m} \geq 663 m$ and $3 \cdot 663 m>1985(m+2)+1$ for large $m$, not all values $0,1,2, \ldots 663 m$ can appear thrice among the $1985(m+2)+1$ sums $s_{-1985}, s_{-1984}, \ldots, s_{1985(m+1)}$ (and none of them appears out of this set). Thus there is an integral value $s>0$ that appears at most twice as a partial sum, say $s_{k}=s_{l}=s, k<l$. Then either $a_{k}$ or $a_{l}$ is a good point. Actually, $s_{i}>s$ must hold for all $i>l$, and $s_{i}<s$ for all $i<k$ (otherwise, the sum $s$ would appear more than twice). Also, for the same reason there cannot exist indices $p, q$ between $k$ and $l$ such that $s_{p}>s$ and $s_{q}<s$; i.e., for $k<p<l, s_{p}$ 's are either all greater than or equal to $s$, or smaller than or equal to $s$. In the former case $a_{k}$ is good, while in the latter $a_{l}$ is good.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
14. 4b.(IRE 4) A set of 1985 points is distributed around the circumference of a circle and each of the points is marked with 1 or -1 . A point is called "good" if the partial sums that can be formed by starting at that point and proceeding around the circle for any distance in either direction are all strictly positive. Show that if the number of points marked with -1 is less than 662 , there must be at least one good point.
|
14. It suffices to prove the existence of a good point in the case of exactly 661 -1 's. We prove by induction on $k$ that in any arrangement with $3 k+2$ points $k$ of which are -1 's a good point exists. For $k=1$ this is clear by inspection. Assume that the assertion holds for all arrangements of $3 n+2$ points and consider an arrangement of $3(n+1)+2$ points. Now there exists a sequence of consecutive -1 's surrounded by two +1 's. There is a point $P$ which is good for the arrangement obtained by removing the two +1 's bordering the sequence of -1 's and one of these -1 's. Since $P$ is out of this sequence, clearly the removal either leaves a partial sum as it was or diminishes it by 1 , so $P$ is good for the original arrangement. Second solution. Denote the number on an arbitrary point by $a_{1}$, and the numbers on successive points going in the positive direction by $a_{2}, a_{3}, \ldots$ (in particular, $a_{k+1985}=a_{k}$ ). We define the partial sums $s_{0}=0, s_{n}=$ $a_{1}+a_{2}+\cdots+a_{n}$ for all positive integers $n$; then $s_{k+1985}=s_{k}+s_{1985}$ and $s_{1985} \geq 663$. Since $s_{1985 m} \geq 663 m$ and $3 \cdot 663 m>1985(m+2)+1$ for large $m$, not all values $0,1,2, \ldots 663 m$ can appear thrice among the $1985(m+2)+1$ sums $s_{-1985}, s_{-1984}, \ldots, s_{1985(m+1)}$ (and none of them appears out of this set). Thus there is an integral value $s>0$ that appears at most twice as a partial sum, say $s_{k}=s_{l}=s, k<l$. Then either $a_{k}$ or $a_{l}$ is a good point. Actually, $s_{i}>s$ must hold for all $i>l$, and $s_{i}<s$ for all $i<k$ (otherwise, the sum $s$ would appear more than twice). Also, for the same reason there cannot exist indices $p, q$ between $k$ and $l$ such that $s_{p}>s$ and $s_{q}<s$; i.e., for $k<p<l, s_{p}$ 's are either all greater than or equal to $s$, or smaller than or equal to $s$. In the former case $a_{k}$ is good, while in the latter $a_{l}$ is good.
|
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61272903-6ce9-55ed-bc2e-1e84feb2915b
| 23,960
|
15. 5a.(FRA 3) Let $K$ and $K^{\prime}$ be two squares in the same plane, their sides of equal length. Is it possible to decompose $K$ into a finite number of triangles $T_{1}, T_{2}, \ldots, T_{p}$ with mutually disjoint interiors and find translations $t_{1}, t_{2}, \ldots, t_{p}$ such that $$ K^{\prime}=\bigcup_{i=1}^{p} t_{i}\left(T_{i}\right) ? $$
|
15. There is no loss of generality if we assume $K=A B C D, K^{\prime}=$ $A B^{\prime} C^{\prime} D^{\prime}$, and that $K^{\prime}$ is obtained from $K$ bya clockwise rotation around $A$ by $\phi, 0 \leq \phi \leq \pi / 4$. Let $C^{\prime} D^{\prime}, B^{\prime} C^{\prime}$, and the parallel to $A B$ through $D^{\prime}$ meet the line $B C$ at $E$, $F$, and $G$ respectively. Let us now choose points $E^{\prime} \in A B^{\prime}, G^{\prime} \in A B$, $C^{\prime \prime} \in A D^{\prime}$, and $E^{\prime \prime} \in A D$ such that  the triangles $A E^{\prime} G^{\prime}$ and $A C^{\prime \prime} E^{\prime \prime}$ are translates of the triangles $D^{\prime} E G$ and $F C^{\prime} E$ respectively. Since $A E^{\prime}=D^{\prime} E$ and $A C^{\prime \prime}=F C^{\prime}$, we have $C^{\prime \prime} E^{\prime \prime}=C^{\prime} E=B^{\prime} E^{\prime}$ and $C^{\prime \prime} D^{\prime}=B^{\prime} F$, which imply that $\triangle E^{\prime \prime} C^{\prime \prime} D^{\prime}$ is a translate of $\triangle E^{\prime} B^{\prime} F$, and consequently $E^{\prime \prime} D^{\prime}=E^{\prime} F$ and $E^{\prime \prime} D^{\prime} \| E^{\prime} F$. It follows that there exist points $H \in C D, H^{\prime} \in B F$, and $D^{\prime \prime} \in E^{\prime} G^{\prime}$ such that $E^{\prime \prime} D^{\prime} H D$ is a translate of $E^{\prime} F H^{\prime} D^{\prime \prime}$. The remaining parts of $K$ and $K^{\prime}$ are the rectangles $D^{\prime} G C H$ and $D^{\prime \prime} H^{\prime} B G^{\prime}$ of equal area. We shall now show that two rectangles with parallel sides and equal areas can be decomposed into translation invariant parts. Let the sides of the rectangles $X Y Z T$ and $X^{\prime} Y^{\prime} Z^{\prime} T^{\prime}\left(X Y \| X^{\prime} Y^{\prime}\right)$ satisfy $X^{\prime} Y^{\prime}<X Y$, $Y^{\prime} Z^{\prime}>Y Z$, and $X^{\prime} Y^{\prime} \cdot Y^{\prime} Z^{\prime}=X Y \cdot Y Z$. Suppose that $2 X^{\prime} Y^{\prime}>X Y$ (otherwise, we may cut off congruent rectangles from both the original ones until we reduce them to the case of $2 X^{\prime} Y^{\prime}>X Y$ ). Let $U \in X Y$ and $V \in Z T$ be points such that $Y U=T V=X^{\prime} Y^{\prime}$ and $W \in X V$ be a point such that $U W \| X T$. Then translating $\triangle X U W$ to a triangle $V Z R$ and $\triangle X V T$ to a triangle $W R S$ results in a rectangle $U Y R S$ congruent to $X^{\prime} Y^{\prime} Z^{\prime} T^{\prime}$. Thus we have partitioned $K$ and $K^{\prime}$ into translation-invariant parts. Although not all the parts are triangles, we may simply triangulate them.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
15. 5a.(FRA 3) Let $K$ and $K^{\prime}$ be two squares in the same plane, their sides of equal length. Is it possible to decompose $K$ into a finite number of triangles $T_{1}, T_{2}, \ldots, T_{p}$ with mutually disjoint interiors and find translations $t_{1}, t_{2}, \ldots, t_{p}$ such that $$ K^{\prime}=\bigcup_{i=1}^{p} t_{i}\left(T_{i}\right) ? $$
|
15. There is no loss of generality if we assume $K=A B C D, K^{\prime}=$ $A B^{\prime} C^{\prime} D^{\prime}$, and that $K^{\prime}$ is obtained from $K$ bya clockwise rotation around $A$ by $\phi, 0 \leq \phi \leq \pi / 4$. Let $C^{\prime} D^{\prime}, B^{\prime} C^{\prime}$, and the parallel to $A B$ through $D^{\prime}$ meet the line $B C$ at $E$, $F$, and $G$ respectively. Let us now choose points $E^{\prime} \in A B^{\prime}, G^{\prime} \in A B$, $C^{\prime \prime} \in A D^{\prime}$, and $E^{\prime \prime} \in A D$ such that  the triangles $A E^{\prime} G^{\prime}$ and $A C^{\prime \prime} E^{\prime \prime}$ are translates of the triangles $D^{\prime} E G$ and $F C^{\prime} E$ respectively. Since $A E^{\prime}=D^{\prime} E$ and $A C^{\prime \prime}=F C^{\prime}$, we have $C^{\prime \prime} E^{\prime \prime}=C^{\prime} E=B^{\prime} E^{\prime}$ and $C^{\prime \prime} D^{\prime}=B^{\prime} F$, which imply that $\triangle E^{\prime \prime} C^{\prime \prime} D^{\prime}$ is a translate of $\triangle E^{\prime} B^{\prime} F$, and consequently $E^{\prime \prime} D^{\prime}=E^{\prime} F$ and $E^{\prime \prime} D^{\prime} \| E^{\prime} F$. It follows that there exist points $H \in C D, H^{\prime} \in B F$, and $D^{\prime \prime} \in E^{\prime} G^{\prime}$ such that $E^{\prime \prime} D^{\prime} H D$ is a translate of $E^{\prime} F H^{\prime} D^{\prime \prime}$. The remaining parts of $K$ and $K^{\prime}$ are the rectangles $D^{\prime} G C H$ and $D^{\prime \prime} H^{\prime} B G^{\prime}$ of equal area. We shall now show that two rectangles with parallel sides and equal areas can be decomposed into translation invariant parts. Let the sides of the rectangles $X Y Z T$ and $X^{\prime} Y^{\prime} Z^{\prime} T^{\prime}\left(X Y \| X^{\prime} Y^{\prime}\right)$ satisfy $X^{\prime} Y^{\prime}<X Y$, $Y^{\prime} Z^{\prime}>Y Z$, and $X^{\prime} Y^{\prime} \cdot Y^{\prime} Z^{\prime}=X Y \cdot Y Z$. Suppose that $2 X^{\prime} Y^{\prime}>X Y$ (otherwise, we may cut off congruent rectangles from both the original ones until we reduce them to the case of $2 X^{\prime} Y^{\prime}>X Y$ ). Let $U \in X Y$ and $V \in Z T$ be points such that $Y U=T V=X^{\prime} Y^{\prime}$ and $W \in X V$ be a point such that $U W \| X T$. Then translating $\triangle X U W$ to a triangle $V Z R$ and $\triangle X V T$ to a triangle $W R S$ results in a rectangle $U Y R S$ congruent to $X^{\prime} Y^{\prime} Z^{\prime} T^{\prime}$. Thus we have partitioned $K$ and $K^{\prime}$ into translation-invariant parts. Although not all the parts are triangles, we may simply triangulate them.
|
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|
2fac151b-8a73-5a45-aafe-af04bb8af831
| 23,962
|
17. 6a. (SWE 3) ${ }^{\mathrm{IMO} O}$ The sequence $f_{1}, f_{2}, \ldots, f_{n}, \ldots$ of functions is defined for $x>0$ recursively by $$ f_{1}(x)=x, \quad f_{n+1}(x)=f_{n}(x)\left(f_{n}(x)+\frac{1}{n}\right) . $$ Prove that there exists one and only one positive number $a$ such that $0<f_{n}(a)<f_{n+1}(a)<1$ for all integers $n \geq 1$.
|
17. The statement of the problem is equivalent to the statement that there is one and only one $a$ such that $1-1 / n<f_{n}(a)<1$ for all $n$. We note that each $f_{n}$ is a polynomial with positive coefficients, and therefore increasing and convex in $\mathbb{R}^{+}$. Define $x_{n}$ and $y_{n}$ by $f_{n}\left(x_{n}\right)=1-1 / n$ and $f_{n}\left(y_{n}\right)=1$. Since $$ f_{n+1}\left(x_{n}\right)=\left(1-\frac{1}{n}\right)^{2}+\left(1-\frac{1}{n}\right) \frac{1}{n}=1-\frac{1}{n} $$ and $f_{n+1}\left(y_{n}\right)=1+1 / n$, it follows that $x_{n}<x_{n+1}<y_{n+1}<y_{n}$. Moreover, the convexity of $f_{n}$ together with the fact that $f_{n}(x)>x$ for all $x>0$ implies that $y_{n}-x_{n}<f_{n}\left(y_{n}\right)-f_{n}\left(x_{n}\right)=1 / n$. Therefore the sequences have a common limit $a$, which is the only number lying between $x_{n}$ and $y_{n}$ for all $n$. By the definition of $x_{n}$ and $y_{n}$, the statement immediately follows.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
17. 6a. (SWE 3) ${ }^{\mathrm{IMO} O}$ The sequence $f_{1}, f_{2}, \ldots, f_{n}, \ldots$ of functions is defined for $x>0$ recursively by $$ f_{1}(x)=x, \quad f_{n+1}(x)=f_{n}(x)\left(f_{n}(x)+\frac{1}{n}\right) . $$ Prove that there exists one and only one positive number $a$ such that $0<f_{n}(a)<f_{n+1}(a)<1$ for all integers $n \geq 1$.
|
17. The statement of the problem is equivalent to the statement that there is one and only one $a$ such that $1-1 / n<f_{n}(a)<1$ for all $n$. We note that each $f_{n}$ is a polynomial with positive coefficients, and therefore increasing and convex in $\mathbb{R}^{+}$. Define $x_{n}$ and $y_{n}$ by $f_{n}\left(x_{n}\right)=1-1 / n$ and $f_{n}\left(y_{n}\right)=1$. Since $$ f_{n+1}\left(x_{n}\right)=\left(1-\frac{1}{n}\right)^{2}+\left(1-\frac{1}{n}\right) \frac{1}{n}=1-\frac{1}{n} $$ and $f_{n+1}\left(y_{n}\right)=1+1 / n$, it follows that $x_{n}<x_{n+1}<y_{n+1}<y_{n}$. Moreover, the convexity of $f_{n}$ together with the fact that $f_{n}(x)>x$ for all $x>0$ implies that $y_{n}-x_{n}<f_{n}\left(y_{n}\right)-f_{n}\left(x_{n}\right)=1 / n$. Therefore the sequences have a common limit $a$, which is the only number lying between $x_{n}$ and $y_{n}$ for all $n$. By the definition of $x_{n}$ and $y_{n}$, the statement immediately follows.
|
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|
6de3f8c7-ab23-5459-883d-f2fc3ccf7663
| 23,967
|
18. 6b.(CAN 5) Let $x_{1}, x_{2}, \ldots, x_{n}$ be positive numbers. Prove that $$ \frac{x_{1}^{2}}{x_{1}^{2}+x_{2} x_{3}}+\frac{x_{2}^{2}}{x_{2}^{2}+x_{3} x_{4}}+\cdots+\frac{x_{n-1}^{2}}{x_{n-1}^{2}+x_{n} x_{1}}+\frac{x_{n}^{2}}{x_{n}^{2}+x_{1} x_{2}} \leq n-1 . $$ Supplementary Problems
|
18. Set $y_{i}=\frac{x_{i}^{2}}{x_{i+1} x_{i+2}}$, where $x_{n+i}=x_{i}$. Then $\prod_{i=1}^{n} y_{i}=1$ and the inequality to be proved becomes $\sum_{i=1}^{n} \frac{y_{i}}{1+y_{i}} \leq n-1$, or equivalently $$ \sum_{i=1}^{n} \frac{1}{1+y_{i}} \geq 1 $$ We prove this inequality by induction on $n$. Since $\frac{1}{1+y}+\frac{1}{1+y^{-1}}=1$, the inequality is true for $n=2$. Assume that it is true for $n-1$, and let there be given $y_{1}, \ldots, y_{n}>0$ with $\prod_{i=1}^{n} y_{i}=1$. Then $\frac{1}{1+y_{n-1}}+\frac{1}{1+y_{n}}>\frac{1}{1+y_{n-1} y_{n}}$, which is equivalent to $1+y_{n} y_{n-1}(1+$ $\left.y_{n}+y_{n-1}\right)>0$. Hence by the inductive hypothesis $$ \sum_{i=1}^{n} \frac{1}{1+y_{i}} \geq \sum_{i=1}^{n-2} \frac{1}{1+y_{i}}+\frac{1}{1+y_{n-1} y_{n}} \geq 1 $$ Remark. The constant $n-1$ is best possible (take for example $x_{i}=a^{i}$ with $a$ arbitrarily large).
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
18. 6b.(CAN 5) Let $x_{1}, x_{2}, \ldots, x_{n}$ be positive numbers. Prove that $$ \frac{x_{1}^{2}}{x_{1}^{2}+x_{2} x_{3}}+\frac{x_{2}^{2}}{x_{2}^{2}+x_{3} x_{4}}+\cdots+\frac{x_{n-1}^{2}}{x_{n-1}^{2}+x_{n} x_{1}}+\frac{x_{n}^{2}}{x_{n}^{2}+x_{1} x_{2}} \leq n-1 . $$ Supplementary Problems
|
18. Set $y_{i}=\frac{x_{i}^{2}}{x_{i+1} x_{i+2}}$, where $x_{n+i}=x_{i}$. Then $\prod_{i=1}^{n} y_{i}=1$ and the inequality to be proved becomes $\sum_{i=1}^{n} \frac{y_{i}}{1+y_{i}} \leq n-1$, or equivalently $$ \sum_{i=1}^{n} \frac{1}{1+y_{i}} \geq 1 $$ We prove this inequality by induction on $n$. Since $\frac{1}{1+y}+\frac{1}{1+y^{-1}}=1$, the inequality is true for $n=2$. Assume that it is true for $n-1$, and let there be given $y_{1}, \ldots, y_{n}>0$ with $\prod_{i=1}^{n} y_{i}=1$. Then $\frac{1}{1+y_{n-1}}+\frac{1}{1+y_{n}}>\frac{1}{1+y_{n-1} y_{n}}$, which is equivalent to $1+y_{n} y_{n-1}(1+$ $\left.y_{n}+y_{n-1}\right)>0$. Hence by the inductive hypothesis $$ \sum_{i=1}^{n} \frac{1}{1+y_{i}} \geq \sum_{i=1}^{n-2} \frac{1}{1+y_{i}}+\frac{1}{1+y_{n-1} y_{n}} \geq 1 $$ Remark. The constant $n-1$ is best possible (take for example $x_{i}=a^{i}$ with $a$ arbitrarily large).
|
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|
cfe04ad4-f3ed-5a5b-9a19-d17cdca9a671
| 23,970
|
21. (IRE 1) The tangents at $B$ and $C$ to the circumcircle of the acute-angled triangle $A B C$ meet at $X$. Let $M$ be the midpoint of $B C$. Prove that (a) $\angle B A M=\angle C A X$, and (b) $\frac{A M}{A X}=\cos \angle B A C$.
|
21. Let $B_{1}$ and $C_{1}$ be the points on the rays $A C$ and $A B$ respectively such that $X B_{1}=X C=X B=X C_{1}$. Then $\angle X B_{1} C=\angle X C B_{1}=\angle A B C$ and $\angle X C_{1} B=\angle X B C_{1}=\angle A C B$, which imply that $B_{1}, X, C_{1}$ are collinear and $\triangle A B_{1} C_{1} \sim \triangle A B C$. Moreover, $X$ is the midpoint of $B_{1} C_{1}$ because $X B_{1}=X C=X B=X C_{1}$, from which we conclude that $\triangle A X C_{1} \sim$ $\triangle A M C$. Therefore $\angle B A X=\angle C A M$ and $$ \frac{A M}{A X}=\frac{C M}{X C_{1}}=\frac{C M}{X C}=\cos \alpha $$
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
21. (IRE 1) The tangents at $B$ and $C$ to the circumcircle of the acute-angled triangle $A B C$ meet at $X$. Let $M$ be the midpoint of $B C$. Prove that (a) $\angle B A M=\angle C A X$, and (b) $\frac{A M}{A X}=\cos \angle B A C$.
|
21. Let $B_{1}$ and $C_{1}$ be the points on the rays $A C$ and $A B$ respectively such that $X B_{1}=X C=X B=X C_{1}$. Then $\angle X B_{1} C=\angle X C B_{1}=\angle A B C$ and $\angle X C_{1} B=\angle X B C_{1}=\angle A C B$, which imply that $B_{1}, X, C_{1}$ are collinear and $\triangle A B_{1} C_{1} \sim \triangle A B C$. Moreover, $X$ is the midpoint of $B_{1} C_{1}$ because $X B_{1}=X C=X B=X C_{1}$, from which we conclude that $\triangle A X C_{1} \sim$ $\triangle A M C$. Therefore $\angle B A X=\angle C A M$ and $$ \frac{A M}{A X}=\frac{C M}{X C_{1}}=\frac{C M}{X C}=\cos \alpha $$
|
{
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|
6eece399-06b6-53ab-b05a-48312b82e8ce
| 23,979
|
22. (USS 5) ${ }^{\mathrm{IMO}} \mathrm{A}$ circle with center $O$ passes through points $A$ and $C$ and intersects the sides $A B$ and $B C$ of the triangle $A B C$ at points $K$ and $N$, respectively. The circumscribed circles of the triangles $A B C$ and $K B N$ intersect at two distinct points $B$ and $M$. Prove that $\angle O M B=90^{\circ}$.
|
22. Assume that $\triangle A B C$ is acute (the case of an obtuse $\triangle A B C$ is similar). Let $S$ and $R$ be the centers of the circumcircles of $\triangle A B C$ and $\triangle K B N$, respectively. Since $\angle B N K=\angle B A C$, the triangles $B N K$ and $B A C$ are similar. Now we have $\angle C B R=\angle A B S=90^{\circ}-\angle A C B$, which gives us $B R \perp A C$ and consequently $B R \| O S$. Similarly $B S \perp K N$ implies that $B S \| O R$. Hence $B R O S$ is a parallelogram. Let $L$ be the point symmetric to $B$ with respect to $R$. Then $R L O S$ is also a parallelogram, and since $S R \perp B M$, we obtain $O L \perp B M$. However, we also have $L M \perp B M$, from which we conclude that $O, L, M$ are collinear and $O M \perp B M$. Second solution. The lines $B M, N K$, and $C A$ are the radical axes of pairs of the three circles, and hence they intersect at a single point $P$. Also, the quadrilateral $M N C P$ is cyclic. Let $O A=O C=O K=O N=r$. We then have $B M \cdot B P=B N \cdot B C=O B^{2}-r^{2}$, $P M \cdot P B=P N \cdot P K=O P^{2}-r^{2}$. It follows that $O B^{2}-O P^{2}=$ $B P(B M-P M)=B M^{2}-P M^{2}$, which implies that $O M \perp M B$. 
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
22. (USS 5) ${ }^{\mathrm{IMO}} \mathrm{A}$ circle with center $O$ passes through points $A$ and $C$ and intersects the sides $A B$ and $B C$ of the triangle $A B C$ at points $K$ and $N$, respectively. The circumscribed circles of the triangles $A B C$ and $K B N$ intersect at two distinct points $B$ and $M$. Prove that $\angle O M B=90^{\circ}$.
|
22. Assume that $\triangle A B C$ is acute (the case of an obtuse $\triangle A B C$ is similar). Let $S$ and $R$ be the centers of the circumcircles of $\triangle A B C$ and $\triangle K B N$, respectively. Since $\angle B N K=\angle B A C$, the triangles $B N K$ and $B A C$ are similar. Now we have $\angle C B R=\angle A B S=90^{\circ}-\angle A C B$, which gives us $B R \perp A C$ and consequently $B R \| O S$. Similarly $B S \perp K N$ implies that $B S \| O R$. Hence $B R O S$ is a parallelogram. Let $L$ be the point symmetric to $B$ with respect to $R$. Then $R L O S$ is also a parallelogram, and since $S R \perp B M$, we obtain $O L \perp B M$. However, we also have $L M \perp B M$, from which we conclude that $O, L, M$ are collinear and $O M \perp B M$. Second solution. The lines $B M, N K$, and $C A$ are the radical axes of pairs of the three circles, and hence they intersect at a single point $P$. Also, the quadrilateral $M N C P$ is cyclic. Let $O A=O C=O K=O N=r$. We then have $B M \cdot B P=B N \cdot B C=O B^{2}-r^{2}$, $P M \cdot P B=P N \cdot P K=O P^{2}-r^{2}$. It follows that $O B^{2}-O P^{2}=$ $B P(B M-P M)=B M^{2}-P M^{2}$, which implies that $O M \perp M B$. 
|
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|
e3169739-9f43-5527-adc8-4cf0ae682803
| 23,981
|
3. (NET 3) ${ }^{\mathrm{IMO} 3}$ The weight $w(p)$ of a polynomial $p, p(x)=\sum_{i=0}^{n} a_{i} x^{i}$, with integer coefficients $a_{i}$ is defined as the number of its odd coefficients. For $i=0,1,2, \ldots$, let $q_{i}(x)=(1+x)^{i}$. Prove that for any finite sequence $0 \leq i_{1}<i_{2}<\cdots<i_{n}$, the inequality $$ w\left(q_{i_{1}}+\cdots+q_{i_{n}}\right) \geq w\left(q_{i_{1}}\right) $$ holds.
|
3. We shall write $P \equiv Q$ for two polynomials $P$ and $Q$ if $P(x)-Q(x)$ has even coefficients. We observe that $(1+x)^{2^{m}} \equiv 1+x^{2^{m}}$ for every $m \in \mathbb{N}$. Consequently, for every polynomial $p$ with degree less than $k=2^{m}, w\left(p \cdot q_{k}\right)=2 w(p)$. Now we prove the inequality from the problem by induction on $i_{n}$. If $i_{n} \leq 1$, the inequality is trivial. Assume it is true for any sequence with $i_{1}<\cdots<i_{n}<2^{m}(m \geq 1)$, and let there be given a sequence with $k=2^{m} \leq i_{n}<2^{m+1}$. Consider two cases. (i) $i_{1} \geq k$. Then $w\left(q_{i_{1}}+\cdots+q_{i_{n}}\right)=2 w\left(q_{i_{1}-k}+\cdots+q_{i_{n}-k}\right) \geq 2 w\left(q_{i_{1}-k}\right)=$ $w\left(q_{i_{1}}\right)$. (ii) $i_{1}<k$. Then the polynomial $p=q_{i_{1}}+\cdots+q_{i_{n}}$ has the form $$ p=\sum_{i=0}^{k-1} a_{i} x^{i}+(1+x)^{k} \sum_{i=0}^{k-1} b_{i} x^{i} \equiv \sum_{i=0}^{k-1}\left[\left(a_{i}+b_{i}\right) x^{i}+b_{i} x^{i+k}\right] $$ Whenever some $a_{i}$ is odd, either $a_{i}+b_{i}$ or $b_{i}$ in the above sum will be odd. It follows that $w(p) \geq w\left(q_{i_{1}}\right)$, as claimed. The proof is complete.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
3. (NET 3) ${ }^{\mathrm{IMO} 3}$ The weight $w(p)$ of a polynomial $p, p(x)=\sum_{i=0}^{n} a_{i} x^{i}$, with integer coefficients $a_{i}$ is defined as the number of its odd coefficients. For $i=0,1,2, \ldots$, let $q_{i}(x)=(1+x)^{i}$. Prove that for any finite sequence $0 \leq i_{1}<i_{2}<\cdots<i_{n}$, the inequality $$ w\left(q_{i_{1}}+\cdots+q_{i_{n}}\right) \geq w\left(q_{i_{1}}\right) $$ holds.
|
3. We shall write $P \equiv Q$ for two polynomials $P$ and $Q$ if $P(x)-Q(x)$ has even coefficients. We observe that $(1+x)^{2^{m}} \equiv 1+x^{2^{m}}$ for every $m \in \mathbb{N}$. Consequently, for every polynomial $p$ with degree less than $k=2^{m}, w\left(p \cdot q_{k}\right)=2 w(p)$. Now we prove the inequality from the problem by induction on $i_{n}$. If $i_{n} \leq 1$, the inequality is trivial. Assume it is true for any sequence with $i_{1}<\cdots<i_{n}<2^{m}(m \geq 1)$, and let there be given a sequence with $k=2^{m} \leq i_{n}<2^{m+1}$. Consider two cases. (i) $i_{1} \geq k$. Then $w\left(q_{i_{1}}+\cdots+q_{i_{n}}\right)=2 w\left(q_{i_{1}-k}+\cdots+q_{i_{n}-k}\right) \geq 2 w\left(q_{i_{1}-k}\right)=$ $w\left(q_{i_{1}}\right)$. (ii) $i_{1}<k$. Then the polynomial $p=q_{i_{1}}+\cdots+q_{i_{n}}$ has the form $$ p=\sum_{i=0}^{k-1} a_{i} x^{i}+(1+x)^{k} \sum_{i=0}^{k-1} b_{i} x^{i} \equiv \sum_{i=0}^{k-1}\left[\left(a_{i}+b_{i}\right) x^{i}+b_{i} x^{i+k}\right] $$ Whenever some $a_{i}$ is odd, either $a_{i}+b_{i}$ or $b_{i}$ in the above sum will be odd. It follows that $w(p) \geq w\left(q_{i_{1}}\right)$, as claimed. The proof is complete.
|
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366abe28-f6cf-598d-87d5-6b9a640fbed4
| 23,983
|
4. (AUS 1) ${ }^{\mathrm{IMO} 2}$ Each of the numbers in the set $N=\{1,2,3, \ldots, n-1\}$, where $n \geq 3$, is colored with one of two colors, say red or black, so that: (i) $i$ and $n-i$ always receive the same color, and (ii) for some $j \in N$, relatively prime to $n, i$ and $|j-i|$ receive the same color for all $i \in N, i \neq j$. Prove that all numbers in $N$ must receive the same color.
|
4. Let $\langle x\rangle$ denote the residue of an integer $x$ modulo $n$. Also, we write $a \sim b$ if $a$ and $b$ receive the same color. We claim that all the numbers $\langle i j\rangle$, $i=1,2, \ldots, n-1$, are of the same color. Since $j$ and $n$ are coprime, this will imply the desired result. We use induction on $i$. For $i=1$ the statement is trivial. Assume now that the statement is true for $i=1, \ldots, k-1$. For $1<k<n$ we have $\langle k j\rangle \neq j$. If $\langle k j\rangle>j$, then by (ii), $\langle k j\rangle \sim\langle k j\rangle-j=\langle(k-1) j\rangle$. If otherwise $\langle k j\rangle<j$, then by (ii) and (i), $\langle k j\rangle \sim j-\langle k j\rangle \sim n-j+\langle k j\rangle=\langle(k-1) j\rangle$. This completes the induction.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
4. (AUS 1) ${ }^{\mathrm{IMO} 2}$ Each of the numbers in the set $N=\{1,2,3, \ldots, n-1\}$, where $n \geq 3$, is colored with one of two colors, say red or black, so that: (i) $i$ and $n-i$ always receive the same color, and (ii) for some $j \in N$, relatively prime to $n, i$ and $|j-i|$ receive the same color for all $i \in N, i \neq j$. Prove that all numbers in $N$ must receive the same color.
|
4. Let $\langle x\rangle$ denote the residue of an integer $x$ modulo $n$. Also, we write $a \sim b$ if $a$ and $b$ receive the same color. We claim that all the numbers $\langle i j\rangle$, $i=1,2, \ldots, n-1$, are of the same color. Since $j$ and $n$ are coprime, this will imply the desired result. We use induction on $i$. For $i=1$ the statement is trivial. Assume now that the statement is true for $i=1, \ldots, k-1$. For $1<k<n$ we have $\langle k j\rangle \neq j$. If $\langle k j\rangle>j$, then by (ii), $\langle k j\rangle \sim\langle k j\rangle-j=\langle(k-1) j\rangle$. If otherwise $\langle k j\rangle<j$, then by (ii) and (i), $\langle k j\rangle \sim j-\langle k j\rangle \sim n-j+\langle k j\rangle=\langle(k-1) j\rangle$. This completes the induction.
|
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2c89fecc-31fe-581f-8b6e-25a13c37fcf1
| 23,987
|
5. (ROM 1) Let $D$ be the interior of the circle $C$ and let $A \in C$. Show that the function $f: D \rightarrow \mathbb{R}, f(M)=\frac{|M A|}{\left|M M^{\prime}\right|}$, where $M^{\prime}=(A M \cap C$, is strictly convex; i.e., $f(P)<\frac{f\left(M_{1}\right)+f\left(M_{2}\right)}{2}, \forall M_{1}, M_{2} \in D, M_{1} \neq M_{2}$, where $P$ is the midpoint of the segment $M_{1} M_{2}$.
|
5. Let w.l.o.g. circle $C$ have unit radius. For each $m \in \mathbb{R}$, the locus of points $M$ such that $f(M)=m$ is the circle $C_{m}$ with radius $r_{m}=m /(m+1)$, that is tangent to $C$ at $A$. Let $O_{m}$ be the center of $C_{m}$. We have to show that if $M \in C_{m}$ and $N \in C_{n}$, where $m, n>0$, then the midpoint $P$ of $M N$ lies inside the circle $C_{(m+n) / 2}$. This is trivial if $m=n$, so let $m \neq n$. For fixed $M, P$ is in the image $C_{n}^{\prime}$ of $C_{n}$ under the homothety with center $M$ and coefficient $1 / 2$. The center of the circle $C_{n}^{\prime}$ is at the midpoint of $O_{n} M$. If we let both $M$ and $N$ vary, $P$ will be on the union of circles with radius $r_{n} / 2$ and centers in the image of $C_{m}$ under the homothety with center $O_{n}$ and coefficient $1 / 2$. Hence $P$ is not outside the circle centered at the midpoint $O_{m} O_{n}$ and with radius $\left(r_{m}+r_{n}\right) / 2$. It remains to show that $r_{(m+n) / 2}>\left(r_{m}+r_{n}\right) / 2$. But this inequality is easily reduced to $(m-n)^{2}>0$, which is true.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
5. (ROM 1) Let $D$ be the interior of the circle $C$ and let $A \in C$. Show that the function $f: D \rightarrow \mathbb{R}, f(M)=\frac{|M A|}{\left|M M^{\prime}\right|}$, where $M^{\prime}=(A M \cap C$, is strictly convex; i.e., $f(P)<\frac{f\left(M_{1}\right)+f\left(M_{2}\right)}{2}, \forall M_{1}, M_{2} \in D, M_{1} \neq M_{2}$, where $P$ is the midpoint of the segment $M_{1} M_{2}$.
|
5. Let w.l.o.g. circle $C$ have unit radius. For each $m \in \mathbb{R}$, the locus of points $M$ such that $f(M)=m$ is the circle $C_{m}$ with radius $r_{m}=m /(m+1)$, that is tangent to $C$ at $A$. Let $O_{m}$ be the center of $C_{m}$. We have to show that if $M \in C_{m}$ and $N \in C_{n}$, where $m, n>0$, then the midpoint $P$ of $M N$ lies inside the circle $C_{(m+n) / 2}$. This is trivial if $m=n$, so let $m \neq n$. For fixed $M, P$ is in the image $C_{n}^{\prime}$ of $C_{n}$ under the homothety with center $M$ and coefficient $1 / 2$. The center of the circle $C_{n}^{\prime}$ is at the midpoint of $O_{n} M$. If we let both $M$ and $N$ vary, $P$ will be on the union of circles with radius $r_{n} / 2$ and centers in the image of $C_{m}$ under the homothety with center $O_{n}$ and coefficient $1 / 2$. Hence $P$ is not outside the circle centered at the midpoint $O_{m} O_{n}$ and with radius $\left(r_{m}+r_{n}\right) / 2$. It remains to show that $r_{(m+n) / 2}>\left(r_{m}+r_{n}\right) / 2$. But this inequality is easily reduced to $(m-n)^{2}>0$, which is true.
|
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|
d7d260c6-b19f-5840-a984-4b530151b043
| 23,991
|
6. (POL 1) Let $x_{n}=\sqrt[2]{2+\sqrt[3]{3+\ldots+\sqrt[n]{n}}}$. Prove that $$ x_{n+1}-x_{n}<\frac{1}{n!}, \quad n=2,3, \ldots $$ ## Alternatives
|
6. Let us set $$ \begin{aligned} & x_{n, i}=\sqrt[i]{i+\sqrt[i+1]{i+1+\cdots+\sqrt[n]{n}}} \\ & y_{n, i}=x_{n+1, i}^{i-1}+x_{n+1, i}^{i-2} x_{n, i}+\cdots+x_{n, i}^{i-1} \end{aligned} $$ In particular, $x_{n, 2}=x_{n}$ and $x_{n, i}=0$ for $i>n$. We observe that for $n>i>2$, $$ x_{n+1, i}-x_{n, i}=\frac{x_{n+1, i}^{i}-x_{n, i}^{i}}{y_{n, i}}=\frac{x_{n+1, i+1}-x_{n, i+1}}{y_{n, i}} . $$ Since $y_{n, i}>i x_{n, i}^{i-1} \geq i^{1+(i-1) / i} \geq i^{3 / 2}$ and $x_{n+1, n+1}-x_{n, n+1}=\sqrt[n+1]{n+1}$, simple induction gives $$ x_{n+1}-x_{n} \leq \frac{\sqrt[n+1]{n+1}}{(n!)^{3 / 2}}<\frac{1}{n!} \quad \text { for } n>2 $$ The inequality for $n=2$ is directly verified.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
6. (POL 1) Let $x_{n}=\sqrt[2]{2+\sqrt[3]{3+\ldots+\sqrt[n]{n}}}$. Prove that $$ x_{n+1}-x_{n}<\frac{1}{n!}, \quad n=2,3, \ldots $$ ## Alternatives
|
6. Let us set $$ \begin{aligned} & x_{n, i}=\sqrt[i]{i+\sqrt[i+1]{i+1+\cdots+\sqrt[n]{n}}} \\ & y_{n, i}=x_{n+1, i}^{i-1}+x_{n+1, i}^{i-2} x_{n, i}+\cdots+x_{n, i}^{i-1} \end{aligned} $$ In particular, $x_{n, 2}=x_{n}$ and $x_{n, i}=0$ for $i>n$. We observe that for $n>i>2$, $$ x_{n+1, i}-x_{n, i}=\frac{x_{n+1, i}^{i}-x_{n, i}^{i}}{y_{n, i}}=\frac{x_{n+1, i+1}-x_{n, i+1}}{y_{n, i}} . $$ Since $y_{n, i}>i x_{n, i}^{i-1} \geq i^{1+(i-1) / i} \geq i^{3 / 2}$ and $x_{n+1, n+1}-x_{n, n+1}=\sqrt[n+1]{n+1}$, simple induction gives $$ x_{n+1}-x_{n} \leq \frac{\sqrt[n+1]{n+1}}{(n!)^{3 / 2}}<\frac{1}{n!} \quad \text { for } n>2 $$ The inequality for $n=2$ is directly verified.
|
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|
8a8b46a2-f638-5a76-99ae-51ac853788fb
| 23,994
|
7. 1a.(CZS 3) The positive integers $x_{1}, \ldots, x_{n}, n \geq 3$, satisfy $x_{1}<x_{2}<$ $\cdots<x_{n}<2 x_{1}$. Set $P=x_{1} x_{2} \cdots x_{n}$. Prove that if $p$ is a prime number, $k$ a positive integer, and $P$ is divisible by $p^{k}$, then $\frac{P}{p^{k}} \geq n!$.
|
7. Let $k_{i} \geq 0$ be the largest integer such that $p^{k_{i}} \mid x_{i}, i=1, \ldots, n$, and $y_{i}=x_{i} / p^{k_{i}}$. We may assume that $k=k_{1}+\cdots+k_{n}$. All the $y_{i}$ must be distinct. Indeed, if $y_{i}=y_{j}$ and $k_{i}>k_{j}$, then $x_{i} \geq p x_{j} \geq 2 x_{i} \geq 2 x_{1}$, which is impossible. Thus $y_{1} y_{2} \ldots y_{n}=P / p^{k} \geq n!$. If equality holds, we must have $y_{i}=1, y_{j}=2$ and $y_{k}=3$ for some $i, j, k$. Thus $p \geq 5$, which implies that either $y_{i} / y_{j} \leq 1 / 2$ or $y_{i} / y_{j} \geq 5 / 2$, which is impossible. Hence the inequality is strict.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
7. 1a.(CZS 3) The positive integers $x_{1}, \ldots, x_{n}, n \geq 3$, satisfy $x_{1}<x_{2}<$ $\cdots<x_{n}<2 x_{1}$. Set $P=x_{1} x_{2} \cdots x_{n}$. Prove that if $p$ is a prime number, $k$ a positive integer, and $P$ is divisible by $p^{k}$, then $\frac{P}{p^{k}} \geq n!$.
|
7. Let $k_{i} \geq 0$ be the largest integer such that $p^{k_{i}} \mid x_{i}, i=1, \ldots, n$, and $y_{i}=x_{i} / p^{k_{i}}$. We may assume that $k=k_{1}+\cdots+k_{n}$. All the $y_{i}$ must be distinct. Indeed, if $y_{i}=y_{j}$ and $k_{i}>k_{j}$, then $x_{i} \geq p x_{j} \geq 2 x_{i} \geq 2 x_{1}$, which is impossible. Thus $y_{1} y_{2} \ldots y_{n}=P / p^{k} \geq n!$. If equality holds, we must have $y_{i}=1, y_{j}=2$ and $y_{k}=3$ for some $i, j, k$. Thus $p \geq 5$, which implies that either $y_{i} / y_{j} \leq 1 / 2$ or $y_{i} / y_{j} \geq 5 / 2$, which is impossible. Hence the inequality is strict.
|
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|
16292d68-d3a1-551f-a783-4b20a8aec5f4
| 23,996
|
1. (GBR 3) ${ }^{\text {IMO5 }}$ Find, with proof, all functions $f$ defined on the nonnegative real numbers and taking nonnegative real values such that (i) $f[x f(y)] f(y)=f(x+y)$, (ii) $f(2)=0$ but $f(x) \neq 0$ for $0 \leq x<2$.
|
1. If $w>2$, then setting in (i) $x=w-2, y=2$, we get $f(w)=f((w-$ 2) $f(w)) f(2)=0$. Thus $$ f(x)=0 \quad \text { if and only if } \quad x \geq 2 $$ Now let $0 \leq y<2$ and $x \geq 0$. The LHS in (i) is zero if and only if $x f(y) \geq 2$, while the RHS is zero if and only if $x+y \geq 2$. It follows that $x \geq 2 / f(y)$ if and only if $x \geq 2-y$. Therefore $$ f(y)=\left\{\begin{array}{cl} \frac{2}{2-y} & \text { for } 0 \leq y<2 \\ 0 & \text { for } y \geq 2 \end{array}\right. $$ The confirmation that $f$ satisfies the given conditions is straightforward.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
1. (GBR 3) ${ }^{\text {IMO5 }}$ Find, with proof, all functions $f$ defined on the nonnegative real numbers and taking nonnegative real values such that (i) $f[x f(y)] f(y)=f(x+y)$, (ii) $f(2)=0$ but $f(x) \neq 0$ for $0 \leq x<2$.
|
1. If $w>2$, then setting in (i) $x=w-2, y=2$, we get $f(w)=f((w-$ 2) $f(w)) f(2)=0$. Thus $$ f(x)=0 \quad \text { if and only if } \quad x \geq 2 $$ Now let $0 \leq y<2$ and $x \geq 0$. The LHS in (i) is zero if and only if $x f(y) \geq 2$, while the RHS is zero if and only if $x+y \geq 2$. It follows that $x \geq 2 / f(y)$ if and only if $x \geq 2-y$. Therefore $$ f(y)=\left\{\begin{array}{cl} \frac{2}{2-y} & \text { for } 0 \leq y<2 \\ 0 & \text { for } y \geq 2 \end{array}\right. $$ The confirmation that $f$ satisfies the given conditions is straightforward.
|
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|
479302f7-de2e-5c45-aeb0-e61b3e096ef6
| 24,003
|
12. (GDR 3) ${ }^{\mathrm{IMO} 3}$ To each vertex $P_{i}(i=1, \ldots, 5)$ of a pentagon an integer $x_{i}$ is assigned, the sum $s=\sum x_{i}$ being positive. The following operation is allowed, provided at least one of the $x_{i}$ 's is negative: Choose a negative $x_{i}$, replace it by $-x_{i}$, and add the former value of $x_{i}$ to the integers assigned to the two neighboring vertices of $P_{i}$ (the remaining two integers are left unchanged). This operation is to be performed repeatedly until all negative integers disappear. Decide whether this procedure must eventually terminate.
|
12. We define $f\left(x_{1}, \ldots, x_{5}\right)=\sum_{i=1}^{5}\left(x_{i+1}-x_{i-1}\right)^{2}\left(x_{0}=x_{5}, x_{6}=x_{1}\right)$. Assuming that $x_{3}<0$, according to the rules the lattice vector $X=$ $\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)$ changes into $Y=\left(x_{1}, x_{2}+x_{3},-x_{3}, x_{4}+x_{3}, x_{5}\right)$. Then $$ \begin{aligned} f(Y)-f(X)= & \left(x_{2}+x_{3}-x_{5}\right)^{2}+\left(x_{1}+x_{3}\right)^{2}+\left(x_{2}-x_{4}\right)^{2} \\ & +\left(x_{3}+x_{5}\right)^{2}+\left(x_{1}-x_{3}-x_{4}\right)^{2}-\left(x_{2}-x_{5}\right)^{2} \\ & -\left(x_{3}-x_{1}\right)^{2}-\left(x_{4}-x_{2}\right)^{2}-\left(x_{5}-x_{3}\right)^{2}-\left(x_{1}-x_{4}\right)^{2} \\ = & 2 x_{3}\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right)=2 x_{3} S<0 \end{aligned} $$ Thus $f$ strictly decreases after each step, and since it takes only positive integer values, the number of steps must be finite. Remark. One could inspect the behavior of $g(x)=\sum_{i=1}^{5} \sum_{j=1}^{5} \mid x_{i}+x_{i+1}+$ $\cdots+x_{j-1} \mid$ instead. Then $g(Y)-g(X)=\left|S+x_{3}\right|-\left|S-x_{3}\right|>0$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
12. (GDR 3) ${ }^{\mathrm{IMO} 3}$ To each vertex $P_{i}(i=1, \ldots, 5)$ of a pentagon an integer $x_{i}$ is assigned, the sum $s=\sum x_{i}$ being positive. The following operation is allowed, provided at least one of the $x_{i}$ 's is negative: Choose a negative $x_{i}$, replace it by $-x_{i}$, and add the former value of $x_{i}$ to the integers assigned to the two neighboring vertices of $P_{i}$ (the remaining two integers are left unchanged). This operation is to be performed repeatedly until all negative integers disappear. Decide whether this procedure must eventually terminate.
|
12. We define $f\left(x_{1}, \ldots, x_{5}\right)=\sum_{i=1}^{5}\left(x_{i+1}-x_{i-1}\right)^{2}\left(x_{0}=x_{5}, x_{6}=x_{1}\right)$. Assuming that $x_{3}<0$, according to the rules the lattice vector $X=$ $\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)$ changes into $Y=\left(x_{1}, x_{2}+x_{3},-x_{3}, x_{4}+x_{3}, x_{5}\right)$. Then $$ \begin{aligned} f(Y)-f(X)= & \left(x_{2}+x_{3}-x_{5}\right)^{2}+\left(x_{1}+x_{3}\right)^{2}+\left(x_{2}-x_{4}\right)^{2} \\ & +\left(x_{3}+x_{5}\right)^{2}+\left(x_{1}-x_{3}-x_{4}\right)^{2}-\left(x_{2}-x_{5}\right)^{2} \\ & -\left(x_{3}-x_{1}\right)^{2}-\left(x_{4}-x_{2}\right)^{2}-\left(x_{5}-x_{3}\right)^{2}-\left(x_{1}-x_{4}\right)^{2} \\ = & 2 x_{3}\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right)=2 x_{3} S<0 \end{aligned} $$ Thus $f$ strictly decreases after each step, and since it takes only positive integer values, the number of steps must be finite. Remark. One could inspect the behavior of $g(x)=\sum_{i=1}^{5} \sum_{j=1}^{5} \mid x_{i}+x_{i+1}+$ $\cdots+x_{j-1} \mid$ instead. Then $g(Y)-g(X)=\left|S+x_{3}\right|-\left|S-x_{3}\right|>0$.
|
{
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|
01152158-3321-54a1-845c-af957b6a3927
| 24,012
|
14. (IRE 1) The circle inscribed in a triangle $A B C$ touches the sides $B C, C A, A B$ in $D, E, F$, respectively, and $X, Y, Z$ are the midpoints of $E F, F D, D E$, respectively. Prove that the centers of the inscribed circle and of the circles around $X Y Z$ and $A B C$ are collinear.
|
14. We shall use the following simple fact. Lemma. If $\widehat{k}$ is the image of a circle $k$ under an inversion centered at a point $Z$, and $O_{1}, O_{2}$ are centers of $k$ and $\widehat{k}$, then $O_{1}, O_{2}$, and $Z$ are collinear. Proof. The result follows immediately from the symmetry with respect to the line $Z O_{1}$. Let $I$ be the center of the inscribed circle $i$. Since $I X \cdot I A=I E^{2}$, the inversion with respect to $i$ takes points $A$ into $X$, and analogously $B, C$ into $Y, Z$ respectively. It follows from the lemma that the center of circle $A B C$, the center of circle $X Y Z$, and point $I$ are collinear.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
14. (IRE 1) The circle inscribed in a triangle $A B C$ touches the sides $B C, C A, A B$ in $D, E, F$, respectively, and $X, Y, Z$ are the midpoints of $E F, F D, D E$, respectively. Prove that the centers of the inscribed circle and of the circles around $X Y Z$ and $A B C$ are collinear.
|
14. We shall use the following simple fact. Lemma. If $\widehat{k}$ is the image of a circle $k$ under an inversion centered at a point $Z$, and $O_{1}, O_{2}$ are centers of $k$ and $\widehat{k}$, then $O_{1}, O_{2}$, and $Z$ are collinear. Proof. The result follows immediately from the symmetry with respect to the line $Z O_{1}$. Let $I$ be the center of the inscribed circle $i$. Since $I X \cdot I A=I E^{2}$, the inversion with respect to $i$ takes points $A$ into $X$, and analogously $B, C$ into $Y, Z$ respectively. It follows from the lemma that the center of circle $A B C$, the center of circle $X Y Z$, and point $I$ are collinear.
|
{
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|
77b8897e-276e-5c3a-9c67-493d80644a48
| 24,018
|
17. ( $\mathbf{C H N} 3)^{\mathrm{IMO} 2}$ Let $A, B, C$ be fixed points in the plane. A man starts from a certain point $P_{0}$ and walks directly to $A$. At $A$ he turns his direction by $60^{\circ}$ to the left and walks to $P_{1}$ such that $P_{0} A=A P_{1}$. After he does the same action 1986 times successively around the points $A, B, C, A, B, C, \ldots$, he returns to the starting point. Prove that $\triangle A B C$ is equilateral and that the vertices $A, B, C$ are arranged counterclockwise.
|
17. We use complex numbers to represent the position of a point in the plane. For convenience, let $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, \ldots$ be $A, B, C, A, B, \ldots$ respectively, and let $P_{0}$ be the origin. After the $k$ th step, the position of $P_{k}$ will be $P_{k}=A_{k}+\left(P_{k-1}-A_{k}\right) u, k=1,2,3, \ldots$, where $u=e^{4 \pi \tau / 3}$. We easily obtain $$ P_{k}=(1-u)\left(A_{k}+u A_{k-1}+u^{2} A_{k-2}+\cdots+u^{k-1} A_{1}\right) $$ The condition $P_{0} \equiv P_{1986}$ is equivalent to $A_{1986}+u A_{1985}+\cdots+u^{1984} A_{2}+$ $u^{1985} A_{1}=0$, which, having in mind that $A_{1}=A_{4}=A_{7}=\cdots, A_{2}=A_{5}=$ $A_{8}=\cdots, A_{3}=A_{6}=A_{9}=\cdots$, reduces to $$ 662\left(A_{3}+u A_{2}+u^{2} A_{1}\right)=\left(1+u^{3}+\cdots+u^{1983}\right)\left(A_{3}+u A_{2}+u^{2} A_{1}\right)=0 $$ It follows that $A_{3}-A_{1}=u\left(A_{1}-A_{2}\right)$, and the assertion follows. Second solution. Let $f_{P}$ denote the rotation with center $P$ through $120^{\circ}$ clockwise. Let $f_{1}=f_{A}$. Then $f_{1}\left(P_{0}\right)=P_{1}$. Let $B^{\prime}=f_{1}(B), C^{\prime}=f_{1}(C)$, and $f_{2}=f_{B^{\prime}}$. Then $f_{2}\left(P_{1}\right)=P_{2}$ and $f_{2}\left(A B^{\prime} C^{\prime}\right)=A^{\prime} B^{\prime} C^{\prime \prime}$. Finally, let $f_{3}=f_{C^{\prime \prime}}$ and $f_{3}\left(A^{\prime} B^{\prime} C^{\prime \prime}\right)=A^{\prime \prime} B^{\prime \prime} C^{\prime \prime}$. Then $g=f_{3} f_{2} f_{1}$ is a translation sending $P_{0}$ to $P_{3}$ and $C$ to $C^{\prime \prime}$. Now $P_{1986}=P_{0}$ implies that $g^{662}$ is the identity, and thus $C=C^{\prime \prime}$. Let $K$ be such that $A B K$ is equilateral and positively oriented. We observe that $f_{2} f_{1}(K)=K$; therefore the rotation $f_{2} f_{1}$ satisfies $f_{2} f_{1}(P) \neq P$ for $P \neq K$. Hence $f_{2} f_{1}(C)=C^{\prime \prime}=C$ implies $K=C$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
17. ( $\mathbf{C H N} 3)^{\mathrm{IMO} 2}$ Let $A, B, C$ be fixed points in the plane. A man starts from a certain point $P_{0}$ and walks directly to $A$. At $A$ he turns his direction by $60^{\circ}$ to the left and walks to $P_{1}$ such that $P_{0} A=A P_{1}$. After he does the same action 1986 times successively around the points $A, B, C, A, B, C, \ldots$, he returns to the starting point. Prove that $\triangle A B C$ is equilateral and that the vertices $A, B, C$ are arranged counterclockwise.
|
17. We use complex numbers to represent the position of a point in the plane. For convenience, let $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, \ldots$ be $A, B, C, A, B, \ldots$ respectively, and let $P_{0}$ be the origin. After the $k$ th step, the position of $P_{k}$ will be $P_{k}=A_{k}+\left(P_{k-1}-A_{k}\right) u, k=1,2,3, \ldots$, where $u=e^{4 \pi \tau / 3}$. We easily obtain $$ P_{k}=(1-u)\left(A_{k}+u A_{k-1}+u^{2} A_{k-2}+\cdots+u^{k-1} A_{1}\right) $$ The condition $P_{0} \equiv P_{1986}$ is equivalent to $A_{1986}+u A_{1985}+\cdots+u^{1984} A_{2}+$ $u^{1985} A_{1}=0$, which, having in mind that $A_{1}=A_{4}=A_{7}=\cdots, A_{2}=A_{5}=$ $A_{8}=\cdots, A_{3}=A_{6}=A_{9}=\cdots$, reduces to $$ 662\left(A_{3}+u A_{2}+u^{2} A_{1}\right)=\left(1+u^{3}+\cdots+u^{1983}\right)\left(A_{3}+u A_{2}+u^{2} A_{1}\right)=0 $$ It follows that $A_{3}-A_{1}=u\left(A_{1}-A_{2}\right)$, and the assertion follows. Second solution. Let $f_{P}$ denote the rotation with center $P$ through $120^{\circ}$ clockwise. Let $f_{1}=f_{A}$. Then $f_{1}\left(P_{0}\right)=P_{1}$. Let $B^{\prime}=f_{1}(B), C^{\prime}=f_{1}(C)$, and $f_{2}=f_{B^{\prime}}$. Then $f_{2}\left(P_{1}\right)=P_{2}$ and $f_{2}\left(A B^{\prime} C^{\prime}\right)=A^{\prime} B^{\prime} C^{\prime \prime}$. Finally, let $f_{3}=f_{C^{\prime \prime}}$ and $f_{3}\left(A^{\prime} B^{\prime} C^{\prime \prime}\right)=A^{\prime \prime} B^{\prime \prime} C^{\prime \prime}$. Then $g=f_{3} f_{2} f_{1}$ is a translation sending $P_{0}$ to $P_{3}$ and $C$ to $C^{\prime \prime}$. Now $P_{1986}=P_{0}$ implies that $g^{662}$ is the identity, and thus $C=C^{\prime \prime}$. Let $K$ be such that $A B K$ is equilateral and positively oriented. We observe that $f_{2} f_{1}(K)=K$; therefore the rotation $f_{2} f_{1}$ satisfies $f_{2} f_{1}(P) \neq P$ for $P \neq K$. Hence $f_{2} f_{1}(C)=C^{\prime \prime}=C$ implies $K=C$.
|
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|
1c1d1843-f3d4-576c-8173-100a65f88570
| 24,024
|
18. (TUR 1) Let $A X, B Y, C Z$ be three cevians concurrent at an interior point $D$ of a triangle $A B C$. Prove that if two of the quadrangles $D Y A Z, D Z B X, D X C Y$ are circumscribable, so is the third.
|
18. We shall use the following criterion for a quadrangle to be circumscribable. Lemma. The quadrangle $A Y D Z$ is circumscribable if and only if $D B-$ $D C=A B-A C$. Proof. Suppose that $A Y D Z$ is circumscribable and that the incircle is tangent to $A Z, Z D, D Y, Y A$ at $M, N, P, Q$ respectively. Then $D B-D C=P B-N C=M B-Q C=A B-A C$. Conversely, assume that $D B-D C=A B-A C$ and let a tangent from $D$ to the incircle of the triangle $A C Z$ meet $C Z$ and $C A$ at $D^{\prime} \neq Z$ and $Y^{\prime} \neq A$ respectively. According to the first part we have $D^{\prime} B-D^{\prime} C=A B-A C$. It follows that $\left|D^{\prime} B-D B\right|=$  $\left|D^{\prime} C-D C\right|=D D^{\prime}$, implying that $D^{\prime} \equiv D$. Let us assume that $D Z B X$ and $D X C Y$ are circumscribable. Using the lemma we obtain $D C-D A=B C-B A$ and $D A-D B=C A-C B$. Adding these two inequalities yields $D C-D B=A C-A B$, and the statement follows from the lemma.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
18. (TUR 1) Let $A X, B Y, C Z$ be three cevians concurrent at an interior point $D$ of a triangle $A B C$. Prove that if two of the quadrangles $D Y A Z, D Z B X, D X C Y$ are circumscribable, so is the third.
|
18. We shall use the following criterion for a quadrangle to be circumscribable. Lemma. The quadrangle $A Y D Z$ is circumscribable if and only if $D B-$ $D C=A B-A C$. Proof. Suppose that $A Y D Z$ is circumscribable and that the incircle is tangent to $A Z, Z D, D Y, Y A$ at $M, N, P, Q$ respectively. Then $D B-D C=P B-N C=M B-Q C=A B-A C$. Conversely, assume that $D B-D C=A B-A C$ and let a tangent from $D$ to the incircle of the triangle $A C Z$ meet $C Z$ and $C A$ at $D^{\prime} \neq Z$ and $Y^{\prime} \neq A$ respectively. According to the first part we have $D^{\prime} B-D^{\prime} C=A B-A C$. It follows that $\left|D^{\prime} B-D B\right|=$  $\left|D^{\prime} C-D C\right|=D D^{\prime}$, implying that $D^{\prime} \equiv D$. Let us assume that $D Z B X$ and $D X C Y$ are circumscribable. Using the lemma we obtain $D C-D A=B C-B A$ and $D A-D B=C A-C B$. Adding these two inequalities yields $D C-D B=A C-A B$, and the statement follows from the lemma.
|
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|
e2a7165e-de7e-54a2-b9f4-9471e6134739
| 24,027
|
20. (CAN 3) Prove that the sum of the face angles at each vertex of a tetrahedron is a straight angle if and only if the faces are congruent triangles.
|
20. If the faces of the tetrahedron $A B C D$ are congruent triangles, we must have $A B=C D, A C=B D$, and $A D=B C$. Then the sum of angles at $A$ is $\angle B A C+\angle C A D+\angle D A B=\angle B D C+\angle C B D+\angle D C B=180^{\circ}$. We now assume that the sum of angles at each vertex is $180^{\circ}$. Let us construct triangles $B C D^{\prime}, C A D^{\prime \prime}, A B D^{\prime \prime \prime}$ in the plane $A B C$, exterior to $\triangle A B C$, such that $\triangle B C D^{\prime} \cong \triangle B C D, \triangle C A D^{\prime \prime} \cong \triangle C A D$, and $\triangle A B D^{\prime \prime \prime} \cong \triangle A B D$. Then by the assumption, $A \in D^{\prime \prime} D^{\prime \prime \prime}, B \in D^{\prime \prime \prime} D^{\prime}$, and $C \in D^{\prime} D^{\prime \prime}$. Since also $D^{\prime \prime} A=D^{\prime \prime \prime} A=D A$, etc., $A, B, C$ are the mid- points of segments $D^{\prime \prime} D^{\prime \prime \prime}, D^{\prime \prime \prime} D^{\prime}, D^{\prime} D^{\prime \prime}$ respectively. Thus the triangles $A B C, B C D^{\prime}, C A D^{\prime \prime}, A B D^{\prime \prime \prime}$ are congruent, and the statement follows.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
20. (CAN 3) Prove that the sum of the face angles at each vertex of a tetrahedron is a straight angle if and only if the faces are congruent triangles.
|
20. If the faces of the tetrahedron $A B C D$ are congruent triangles, we must have $A B=C D, A C=B D$, and $A D=B C$. Then the sum of angles at $A$ is $\angle B A C+\angle C A D+\angle D A B=\angle B D C+\angle C B D+\angle D C B=180^{\circ}$. We now assume that the sum of angles at each vertex is $180^{\circ}$. Let us construct triangles $B C D^{\prime}, C A D^{\prime \prime}, A B D^{\prime \prime \prime}$ in the plane $A B C$, exterior to $\triangle A B C$, such that $\triangle B C D^{\prime} \cong \triangle B C D, \triangle C A D^{\prime \prime} \cong \triangle C A D$, and $\triangle A B D^{\prime \prime \prime} \cong \triangle A B D$. Then by the assumption, $A \in D^{\prime \prime} D^{\prime \prime \prime}, B \in D^{\prime \prime \prime} D^{\prime}$, and $C \in D^{\prime} D^{\prime \prime}$. Since also $D^{\prime \prime} A=D^{\prime \prime \prime} A=D A$, etc., $A, B, C$ are the mid- points of segments $D^{\prime \prime} D^{\prime \prime \prime}, D^{\prime \prime \prime} D^{\prime}, D^{\prime} D^{\prime \prime}$ respectively. Thus the triangles $A B C, B C D^{\prime}, C A D^{\prime \prime}, A B D^{\prime \prime \prime}$ are congruent, and the statement follows.
|
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|
7ffdba7e-6f49-5fcc-9e75-05c0e70fbb19
| 24,032
|
21. (TUR 2) Let $A B C D$ be a tetrahedron having each sum of opposite sides equal to 1. Prove that $$ r_{A}+r_{B}+r_{C}+r_{D} \leq \frac{\sqrt{3}}{3} $$ where $r_{A}, r_{B}, r_{C}, r_{D}$ are the inradii of the faces, equality holding only if $A B C D$ is regular.
|
21. Since the sum of all edges of $A B C D$ is 3 , the statement of the problem is an immediate consequence of the following statement: Lemma. Let $r$ be the inradius of a triangle with sides $a, b, c$. Then $a+$ $b+c \geq 6 \sqrt{3} \cdot r$, with equality if and only if the triangle is equilateral. Proof. If $S$ and $p$ denotes the area and semiperimeter of the triangle, by Heron's formula and the AM-GM inequality we have $$ \begin{aligned} p r & =S=\sqrt{p(p-a)(p-b)(p-c)} \\ & \leq \sqrt{p\left(\frac{(p-a)+(p-b)+(p-c)}{3}\right)^{3}}=\sqrt{\frac{p^{4}}{27}}=\frac{p^{2}}{3 \sqrt{3}}, \end{aligned} $$ i.e., $p \geq 3 \sqrt{3} \cdot r$, which is equivalent to the claim.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
21. (TUR 2) Let $A B C D$ be a tetrahedron having each sum of opposite sides equal to 1. Prove that $$ r_{A}+r_{B}+r_{C}+r_{D} \leq \frac{\sqrt{3}}{3} $$ where $r_{A}, r_{B}, r_{C}, r_{D}$ are the inradii of the faces, equality holding only if $A B C D$ is regular.
|
21. Since the sum of all edges of $A B C D$ is 3 , the statement of the problem is an immediate consequence of the following statement: Lemma. Let $r$ be the inradius of a triangle with sides $a, b, c$. Then $a+$ $b+c \geq 6 \sqrt{3} \cdot r$, with equality if and only if the triangle is equilateral. Proof. If $S$ and $p$ denotes the area and semiperimeter of the triangle, by Heron's formula and the AM-GM inequality we have $$ \begin{aligned} p r & =S=\sqrt{p(p-a)(p-b)(p-c)} \\ & \leq \sqrt{p\left(\frac{(p-a)+(p-b)+(p-c)}{3}\right)^{3}}=\sqrt{\frac{p^{4}}{27}}=\frac{p^{2}}{3 \sqrt{3}}, \end{aligned} $$ i.e., $p \geq 3 \sqrt{3} \cdot r$, which is equivalent to the claim.
|
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|
a96be212-2423-5f27-b9f5-dd3480252a3e
| 24,035
|
5. (FRG 1) ${ }^{\mathrm{IMO1}}$ The set $S=\{2,5,13\}$ has the property that for every $a, b \in S, a \neq b$, the number $a b-1$ is a perfect square. Show that for every positive integer $d$ not in $S$, the set $S \cup\{d\}$ does not have the above property.
|
5. Suppose that for every $a, b \in\{2,5,13, d\}, a \neq b$, the number $a b-1$ is a perfect square. In particular, for some integers $x, y, z$ we have $$ 2 d-1=x^{2}, \quad 5 d-1=y^{2}, \quad 13 d-1=z^{2} $$ Since $x$ is clearly odd, $d=\left(x^{2}+1\right) / 2$ is also odd because $4 \nmid x^{2}+1$. It follows that $y$ and $z$ are even, say $y=2 y_{1}$ and $z=2 z_{1}$. Hence $\left(z_{1}-\right.$ $\left.y_{1}\right)\left(z_{1}+y_{1}\right)=\left(z^{2}-y^{2}\right) / 4=2 d$. But in this case one of the factors $z_{1}-y_{1}$, $z_{1}+y_{1}$ is odd and the other one is even, which is impossible.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
5. (FRG 1) ${ }^{\mathrm{IMO1}}$ The set $S=\{2,5,13\}$ has the property that for every $a, b \in S, a \neq b$, the number $a b-1$ is a perfect square. Show that for every positive integer $d$ not in $S$, the set $S \cup\{d\}$ does not have the above property.
|
5. Suppose that for every $a, b \in\{2,5,13, d\}, a \neq b$, the number $a b-1$ is a perfect square. In particular, for some integers $x, y, z$ we have $$ 2 d-1=x^{2}, \quad 5 d-1=y^{2}, \quad 13 d-1=z^{2} $$ Since $x$ is clearly odd, $d=\left(x^{2}+1\right) / 2$ is also odd because $4 \nmid x^{2}+1$. It follows that $y$ and $z$ are even, say $y=2 y_{1}$ and $z=2 z_{1}$. Hence $\left(z_{1}-\right.$ $\left.y_{1}\right)\left(z_{1}+y_{1}\right)=\left(z^{2}-y^{2}\right) / 4=2 d$. But in this case one of the factors $z_{1}-y_{1}$, $z_{1}+y_{1}$ is odd and the other one is even, which is impossible.
|
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|
5dfb2371-db87-50bc-9881-157e710865b7
| 24,042
|
7. (FRA 5) Let real numbers $x_{1}, x_{2}, \ldots, x_{n}$ satisfy $0<x_{1}<x_{2}<\cdots<$ $x_{n}<1$ and set $x_{0}=0, x_{n+1}=1$. Suppose that these numbers satisfy the following system of equations: $$ \sum_{j=0, j \neq i}^{n+1} \frac{1}{x_{i}-x_{j}}=0 \quad \text { where } i=1,2, \ldots, n $$ Prove that $x_{n+1-i}=1-x_{i}$ for $i=1,2, \ldots, n$.
|
7. Let $P(x)=\left(x-x_{0}\right)\left(x-x_{1}\right) \cdots\left(x-x_{n}\right)\left(x-x_{n+1}\right)$. Then $$ P^{\prime}(x)=\sum_{j=0}^{n+1} \frac{P(x)}{x-x_{j}} \quad \text { and } \quad P^{\prime \prime}(x)=\sum_{j=0}^{n+1} \sum_{k \neq j} \frac{P(x)}{\left(x-x_{j}\right)\left(x-x_{k}\right)} . $$ Therefore $$ P^{\prime \prime}\left(x_{i}\right)=2 P^{\prime}\left(x_{i}\right) \sum_{j \neq i} \frac{1}{\left(x_{i}-x_{j}\right)} $$ for $i=0,1, \ldots, n+1$, and the given condition implies $P^{\prime \prime}\left(x_{i}\right)=0$ for $i=1,2, \ldots, n$. Consequently, $$ x(x-1) P^{\prime \prime}(x)=(n+2)(n+1) P(x) . $$ It is easy to observe that there is a unique monic polynomial of degree $n+2$ satisfying differential equation (1). On the other hand, the polynomial $Q(x)=(-1)^{n} P(1-x)$ also satisfies this equation, is monic, and $\operatorname{deg} Q=$ $n+2$. Therefore $(-1)^{n} P(1-x)=P(x)$, and the result follows.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
7. (FRA 5) Let real numbers $x_{1}, x_{2}, \ldots, x_{n}$ satisfy $0<x_{1}<x_{2}<\cdots<$ $x_{n}<1$ and set $x_{0}=0, x_{n+1}=1$. Suppose that these numbers satisfy the following system of equations: $$ \sum_{j=0, j \neq i}^{n+1} \frac{1}{x_{i}-x_{j}}=0 \quad \text { where } i=1,2, \ldots, n $$ Prove that $x_{n+1-i}=1-x_{i}$ for $i=1,2, \ldots, n$.
|
7. Let $P(x)=\left(x-x_{0}\right)\left(x-x_{1}\right) \cdots\left(x-x_{n}\right)\left(x-x_{n+1}\right)$. Then $$ P^{\prime}(x)=\sum_{j=0}^{n+1} \frac{P(x)}{x-x_{j}} \quad \text { and } \quad P^{\prime \prime}(x)=\sum_{j=0}^{n+1} \sum_{k \neq j} \frac{P(x)}{\left(x-x_{j}\right)\left(x-x_{k}\right)} . $$ Therefore $$ P^{\prime \prime}\left(x_{i}\right)=2 P^{\prime}\left(x_{i}\right) \sum_{j \neq i} \frac{1}{\left(x_{i}-x_{j}\right)} $$ for $i=0,1, \ldots, n+1$, and the given condition implies $P^{\prime \prime}\left(x_{i}\right)=0$ for $i=1,2, \ldots, n$. Consequently, $$ x(x-1) P^{\prime \prime}(x)=(n+2)(n+1) P(x) . $$ It is easy to observe that there is a unique monic polynomial of degree $n+2$ satisfying differential equation (1). On the other hand, the polynomial $Q(x)=(-1)^{n} P(1-x)$ also satisfies this equation, is monic, and $\operatorname{deg} Q=$ $n+2$. Therefore $(-1)^{n} P(1-x)=P(x)$, and the result follows.
|
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|
922a4a41-8ed1-55e8-bd79-aa75aa27b0d7
| 24,047
|
8. (USA 1) From a collection of $n$ persons $q$ distinct two-member teams are selected and ranked $1, \ldots, q$ (no ties). Let $m$ be the least integer larger than or equal to $2 q / n$. Show that there are $m$ distinct teams that may be listed so that (i) each pair of consecutive teams on the list have one member in common and (ii) the chain of teams on the list are in rank order. Alternative formulation. Given a graph with $n$ vertices and $q$ edges numbered $1, \ldots, q$, show that there exists a chain of $m$ edges, $m \geq \frac{2 q}{n}$, each two consecutive edges having a common vertex, arranged monotonically with respect to the numbering.
|
8. We shall solve the problem in the alternative formulation. Let $L_{G}(v)$ denote the length of the longest directed chain of edges in the given graph $G$ that begins in a vertex $v$ and is arranged decreasingly relative to the numbering. By the pigeonhole principle it suffices to show that $\sum_{v} L(v) \geq 2 q$ in every such graph. We do this by induction on $q$. For $q=1$ the claim is obvious. We assume that it is true for $q-1$ and consider a graph $G$ with $q$ edges numbered $1, \ldots, q$. Let the edge number $q$ connect vertices $u$ and $w$. Removing this edge, we get a graph $G^{\prime}$ with $q-1$ edges. We then have $$ L_{G}(u) \geq L_{G^{\prime}}(w)+1, L_{G}(w) \geq L_{G^{\prime}}(u)+1, L_{G}(v) \geq L_{G^{\prime}}(v) \text { for other } v $$ Since $\sum L_{G^{\prime}}(v) \geq 2(q-1)$ by inductive assumption, it follows that $\sum L_{G}(v) \geq 2(q-1)+2=2 q$ as desired. Second solution. Let us place a spider at each vertex of the graph. Let us now interchange the positions of the two spiders at the endpoints of each edge, listing the edges increasingly with respect to the numbering. This way we will move spiders exactly $2 q$ times (two for each edge). Hence there is a spider that will be moved at least $2 q / n$ times. All that remains is to notice that the path of each spider consists of edges numbered in increasing order. Remark. A chain of the stated length having all vertices distinct does not necessarily exist. An example is $n=4, q=6$ with the numbering following the order $a b, c d, a c, b d, a d, b c$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
8. (USA 1) From a collection of $n$ persons $q$ distinct two-member teams are selected and ranked $1, \ldots, q$ (no ties). Let $m$ be the least integer larger than or equal to $2 q / n$. Show that there are $m$ distinct teams that may be listed so that (i) each pair of consecutive teams on the list have one member in common and (ii) the chain of teams on the list are in rank order. Alternative formulation. Given a graph with $n$ vertices and $q$ edges numbered $1, \ldots, q$, show that there exists a chain of $m$ edges, $m \geq \frac{2 q}{n}$, each two consecutive edges having a common vertex, arranged monotonically with respect to the numbering.
|
8. We shall solve the problem in the alternative formulation. Let $L_{G}(v)$ denote the length of the longest directed chain of edges in the given graph $G$ that begins in a vertex $v$ and is arranged decreasingly relative to the numbering. By the pigeonhole principle it suffices to show that $\sum_{v} L(v) \geq 2 q$ in every such graph. We do this by induction on $q$. For $q=1$ the claim is obvious. We assume that it is true for $q-1$ and consider a graph $G$ with $q$ edges numbered $1, \ldots, q$. Let the edge number $q$ connect vertices $u$ and $w$. Removing this edge, we get a graph $G^{\prime}$ with $q-1$ edges. We then have $$ L_{G}(u) \geq L_{G^{\prime}}(w)+1, L_{G}(w) \geq L_{G^{\prime}}(u)+1, L_{G}(v) \geq L_{G^{\prime}}(v) \text { for other } v $$ Since $\sum L_{G^{\prime}}(v) \geq 2(q-1)$ by inductive assumption, it follows that $\sum L_{G}(v) \geq 2(q-1)+2=2 q$ as desired. Second solution. Let us place a spider at each vertex of the graph. Let us now interchange the positions of the two spiders at the endpoints of each edge, listing the edges increasingly with respect to the numbering. This way we will move spiders exactly $2 q$ times (two for each edge). Hence there is a spider that will be moved at least $2 q / n$ times. All that remains is to notice that the path of each spider consists of edges numbered in increasing order. Remark. A chain of the stated length having all vertices distinct does not necessarily exist. An example is $n=4, q=6$ with the numbering following the order $a b, c d, a c, b d, a d, b c$.
|
{
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c0c64e16-d003-54db-8d18-bac411f61c21
| 24,050
|
9. (GDR 1) ${ }^{\mathrm{IMO} 6}$ Prove or disprove: Given a finite set of points with integer coordinates in the plane, it is possible to color some of these points red and the remaining ones white in such a way that for any straight line $L$ parallel to one of the coordinate axes, the number of red colored points and the number of white colored points on $L$ differ by at most 1.
|
9. We shall use induction on the number $n$ of points. The case $n=1$ is trivial. Let us suppose that the statement is true for all $1,2, \ldots, n-1$, and that we are given a set $T$ of $n$ points. If there exists a point $P \in T$ and a line $l$ that is parallel to an axis and contains $P$ and no other points of $T$, then by the inductive hypothesis we can color the set $T \backslash\{P\}$ and then use a suitable color for $P$. Let us now suppose that whenever a line parallel to an axis contains a point of $T$, it contains another point of $T$. It follows that for an arbitrary point $P_{0} \in T$ we can choose points $P_{1}, P_{2}, \ldots$ such that $P_{k} P_{k+1}$ is parallel to the $x$-axis for $k$ even, and to the $y$-axis for $k$ odd. We eventually come to a pair of integers $(r, s)$ of the same parity, $0 \leq r<s$, such that lines $P_{r} P_{r+1}$ and $P_{s} P_{s+1}$ coincide. Hence the closed polygonal line $P_{r+1} P_{r+2} \ldots P_{s} P_{r+1}$ is of even length. Thus we may color the points of this polygonal line alternately and then apply the inductive assumption for the rest of the set $T$. The induction is complete. Second solution. Let $P_{1}, P_{2}, \ldots, P_{k}$ be the points lying on a line $l$ parallel to an axis, going from left to right or from up to down. We draw segments joining $P_{1}$ with $P_{2}, P_{3}$ with $P_{4}$, and generally $P_{2 i-1}$ with $P_{2 i}$. Having this done for every such line $l$, we obtain a set of segments forming certain polygonal lines. If one of these polygonal lines is closed, then it must have an even number of vertices. Thus, we can color the vertices on each of the polygonal lines alternately (a point not lying on any of the polygonal lines may be colored arbitrarily). The obtained coloring satisfies the conditions.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
9. (GDR 1) ${ }^{\mathrm{IMO} 6}$ Prove or disprove: Given a finite set of points with integer coordinates in the plane, it is possible to color some of these points red and the remaining ones white in such a way that for any straight line $L$ parallel to one of the coordinate axes, the number of red colored points and the number of white colored points on $L$ differ by at most 1.
|
9. We shall use induction on the number $n$ of points. The case $n=1$ is trivial. Let us suppose that the statement is true for all $1,2, \ldots, n-1$, and that we are given a set $T$ of $n$ points. If there exists a point $P \in T$ and a line $l$ that is parallel to an axis and contains $P$ and no other points of $T$, then by the inductive hypothesis we can color the set $T \backslash\{P\}$ and then use a suitable color for $P$. Let us now suppose that whenever a line parallel to an axis contains a point of $T$, it contains another point of $T$. It follows that for an arbitrary point $P_{0} \in T$ we can choose points $P_{1}, P_{2}, \ldots$ such that $P_{k} P_{k+1}$ is parallel to the $x$-axis for $k$ even, and to the $y$-axis for $k$ odd. We eventually come to a pair of integers $(r, s)$ of the same parity, $0 \leq r<s$, such that lines $P_{r} P_{r+1}$ and $P_{s} P_{s+1}$ coincide. Hence the closed polygonal line $P_{r+1} P_{r+2} \ldots P_{s} P_{r+1}$ is of even length. Thus we may color the points of this polygonal line alternately and then apply the inductive assumption for the rest of the set $T$. The induction is complete. Second solution. Let $P_{1}, P_{2}, \ldots, P_{k}$ be the points lying on a line $l$ parallel to an axis, going from left to right or from up to down. We draw segments joining $P_{1}$ with $P_{2}, P_{3}$ with $P_{4}$, and generally $P_{2 i-1}$ with $P_{2 i}$. Having this done for every such line $l$, we obtain a set of segments forming certain polygonal lines. If one of these polygonal lines is closed, then it must have an even number of vertices. Thus, we can color the vertices on each of the polygonal lines alternately (a point not lying on any of the polygonal lines may be colored arbitrarily). The obtained coloring satisfies the conditions.
|
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|
2af3b98c-f4e8-557e-b115-e0c2fdf08389
| 24,052
|
13. (GDR 2) ${ }^{\mathrm{IMO} 5}$ Is it possible to put 1987 points in the Euclidean plane such that the distance between each pair of points is irrational and each three points determine a nondegenerate triangle with rational area?
|
13. We claim that the points $P_{i}\left(i, i^{2}\right), i=1,2, \ldots, 1987$, satisfy the conditions. In fact: (i) $\overline{P_{i} P_{j}}=\sqrt{(i-j)^{2}+\left(i^{2}-j^{2}\right)^{2}}=|i-j| \sqrt{1+(i+j)^{2}}$. It is known that for each positive integer $n, \sqrt{n}$ is either an integer or an irrational number. Since $i+j<\sqrt{1+(i+j)^{2}}<i+j+1$, $\sqrt{1+(i+j)^{2}}$ is not an integer, it is irrational, and so is $\overline{P_{i} P_{j}}$. (ii) The area $A$ of the triangle $P_{i} P_{j} P_{k}$, for distinct $i, j, k$, is given by $$ \begin{aligned} A & =\left|\frac{i^{2}+j^{2}}{2}(i-j)+\frac{j^{2}+k^{2}}{2}(j-k)+\frac{k^{2}+i^{2}}{2}(k-i)\right| \\ & =\left|\frac{(i-j)(j-k)(k-i)}{2}\right| \in \mathbb{Q} \backslash\{0\}, \end{aligned} $$ also showing that this triangle is nondegenerate.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
13. (GDR 2) ${ }^{\mathrm{IMO} 5}$ Is it possible to put 1987 points in the Euclidean plane such that the distance between each pair of points is irrational and each three points determine a nondegenerate triangle with rational area?
|
13. We claim that the points $P_{i}\left(i, i^{2}\right), i=1,2, \ldots, 1987$, satisfy the conditions. In fact: (i) $\overline{P_{i} P_{j}}=\sqrt{(i-j)^{2}+\left(i^{2}-j^{2}\right)^{2}}=|i-j| \sqrt{1+(i+j)^{2}}$. It is known that for each positive integer $n, \sqrt{n}$ is either an integer or an irrational number. Since $i+j<\sqrt{1+(i+j)^{2}}<i+j+1$, $\sqrt{1+(i+j)^{2}}$ is not an integer, it is irrational, and so is $\overline{P_{i} P_{j}}$. (ii) The area $A$ of the triangle $P_{i} P_{j} P_{k}$, for distinct $i, j, k$, is given by $$ \begin{aligned} A & =\left|\frac{i^{2}+j^{2}}{2}(i-j)+\frac{j^{2}+k^{2}}{2}(j-k)+\frac{k^{2}+i^{2}}{2}(k-i)\right| \\ & =\left|\frac{(i-j)(j-k)(k-i)}{2}\right| \in \mathbb{Q} \backslash\{0\}, \end{aligned} $$ also showing that this triangle is nondegenerate.
|
{
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|
b5062a67-4ea9-5211-a6f6-f2e8c73951f8
| 24,063
|
15. (FRG 2) $)^{\mathrm{IMO} 3}$ Suppose $x_{1}, x_{2}, \ldots, x_{n}$ are real numbers with $x_{1}^{2}+x_{2}^{2}+$ $\cdots+x_{n}^{2}=1$. Prove that for any integer $k>1$ there are integers $e_{i}$ not all 0 and with $\left|e_{i}\right|<k$ such that $$ \left|e_{1} x_{1}+e_{2} x_{2}+\cdots+e_{n} x_{n}\right| \leq \frac{(k-1) \sqrt{n}}{k^{n}-1} $$
|
15. Since $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$, we get by the Cauchy-Schwarz inequality $$ \left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\right| \leq \sqrt{n\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)}=\sqrt{n} $$ Hence all $k^{n}$ sums of the form $e_{1} x_{1}+e_{2} x_{2}+\cdots+e_{n} x_{n}$, with $e_{i} \in$ $\{0,1,2, \ldots, k-1\}$, must lie in some closed interval $\Im$ of length $(k-1) \sqrt{n}$. This interval can be covered with $k^{n}-1$ closed subintervals of length $\frac{k-1}{k^{n}-1} \sqrt{n}$. By the pigeonhole principle there must be two of these sums lying in the same subinterval. Their difference, which is of the form $e_{1} x_{1}+e_{2} x_{2}+\cdots+e_{n} x_{n}$ where $e_{i} \in\{0, \pm 1, \ldots, \pm(k-1)\}$, satisfies $$ \left|e_{1} x_{1}+e_{2} x_{2}+\cdots+e_{n} x_{n}\right| \leq \frac{(k-1) \sqrt{n}}{k^{n}-1} $$
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
15. (FRG 2) $)^{\mathrm{IMO} 3}$ Suppose $x_{1}, x_{2}, \ldots, x_{n}$ are real numbers with $x_{1}^{2}+x_{2}^{2}+$ $\cdots+x_{n}^{2}=1$. Prove that for any integer $k>1$ there are integers $e_{i}$ not all 0 and with $\left|e_{i}\right|<k$ such that $$ \left|e_{1} x_{1}+e_{2} x_{2}+\cdots+e_{n} x_{n}\right| \leq \frac{(k-1) \sqrt{n}}{k^{n}-1} $$
|
15. Since $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$, we get by the Cauchy-Schwarz inequality $$ \left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\right| \leq \sqrt{n\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)}=\sqrt{n} $$ Hence all $k^{n}$ sums of the form $e_{1} x_{1}+e_{2} x_{2}+\cdots+e_{n} x_{n}$, with $e_{i} \in$ $\{0,1,2, \ldots, k-1\}$, must lie in some closed interval $\Im$ of length $(k-1) \sqrt{n}$. This interval can be covered with $k^{n}-1$ closed subintervals of length $\frac{k-1}{k^{n}-1} \sqrt{n}$. By the pigeonhole principle there must be two of these sums lying in the same subinterval. Their difference, which is of the form $e_{1} x_{1}+e_{2} x_{2}+\cdots+e_{n} x_{n}$ where $e_{i} \in\{0, \pm 1, \ldots, \pm(k-1)\}$, satisfies $$ \left|e_{1} x_{1}+e_{2} x_{2}+\cdots+e_{n} x_{n}\right| \leq \frac{(k-1) \sqrt{n}}{k^{n}-1} $$
|
{
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|
b4018ac1-1683-5d8b-96dc-a9986f26073c
| 24,068
|
16. (FRG 3) ${ }^{\mathrm{IMO1}}$ Let $S$ be a set of $n$ elements. We denote the number of all permutations of $S$ that have exactly $k$ fixed points by $p_{n}(k)$. Prove: (a) $\sum_{k=0}^{n} k p_{n}(k)=n$ !; (b) $\sum_{k=0}^{n}(k-1)^{2} p_{n}(k)=n$ !.
|
16. We assume that $S=\{1,2, \ldots, n\}$, and use the obvious fact $$ \sum_{k=0}^{n} p_{n}(k)=n! $$ (a) To each permutation $\pi$ of $S$ we assign an $n$-vector $\left(e_{1}, e_{2}, \ldots, e_{n}\right)$, where $e_{i}$ is 1 if $i$ is a fixed point of $\pi$, and 0 otherwise. Since exactly $p_{n}(k)$ of the assigned vectors contain exactly $k$ " 1 "s, the considered sum $\sum_{k=0}^{n} k p_{n}(k)$ counts all the " 1 "s occurring in all the $n$ ! assigned vectors. But for each $i, 1 \leq i \leq n$, there are exactly $(n-1)$ ! permutations that fix $i$; i.e., exactly $(n-1)$ ! of the vectors have $e_{i}=1$. Therefore the total number of " 1 "s is $n \cdot(n-1)$ ! = n!, implying $$ \sum_{k=0}^{n} k p_{n}(k)=n! $$ (b) In this case, to each permutation $\pi$ of $S$ we assign a vector $\left(d_{1}, \ldots, d_{n}\right)$ instead, with $d_{i}=k$ if $i$ is a fixed point of $\pi$, and $d_{i}=0$ otherwise, where $k$ is the number of fixed points of $\pi$. Let us count the sum $Z$ of all components $d_{i}$ for all the $n$ ! permutations. There are $p_{n}(k)$ such vectors with exactly $k$ components equal to $k$, and sums of components equal to $k^{2}$. Thus, $Z=\sum_{k=0}^{n} k^{2} p_{n}(k)$. On the other hand, we may first calculate the sum of all components $d_{i}$ for fixed $i$. In fact, the value $d_{i}=k>0$ will occur exactly $p_{n-1}(k-1)$ times, so that the sum of the $d_{i}$ 's is $\sum_{k=1}^{n} k p_{n-1}(k-1)=\sum_{k=0}^{n-1}(k+$ 1) $p_{n-1}(k)=2(n-1)$ !. Summation over $i$ yields $$ Z=\sum_{k=0}^{n} k^{2} p_{n}(k)=2 n!. $$ From (0), (1), and (2), we conclude that $$ \sum_{k=0}^{n}(k-1)^{2} p_{n}(k)=\sum_{k=0}^{n} k^{2} p_{n}(k)-2 \sum_{k=0}^{n} k p_{n}(k)+\sum_{k=0}^{n} p_{n}(k)=n! $$ Remark. Only the first part of this problem was given on the IMO.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
16. (FRG 3) ${ }^{\mathrm{IMO1}}$ Let $S$ be a set of $n$ elements. We denote the number of all permutations of $S$ that have exactly $k$ fixed points by $p_{n}(k)$. Prove: (a) $\sum_{k=0}^{n} k p_{n}(k)=n$ !; (b) $\sum_{k=0}^{n}(k-1)^{2} p_{n}(k)=n$ !.
|
16. We assume that $S=\{1,2, \ldots, n\}$, and use the obvious fact $$ \sum_{k=0}^{n} p_{n}(k)=n! $$ (a) To each permutation $\pi$ of $S$ we assign an $n$-vector $\left(e_{1}, e_{2}, \ldots, e_{n}\right)$, where $e_{i}$ is 1 if $i$ is a fixed point of $\pi$, and 0 otherwise. Since exactly $p_{n}(k)$ of the assigned vectors contain exactly $k$ " 1 "s, the considered sum $\sum_{k=0}^{n} k p_{n}(k)$ counts all the " 1 "s occurring in all the $n$ ! assigned vectors. But for each $i, 1 \leq i \leq n$, there are exactly $(n-1)$ ! permutations that fix $i$; i.e., exactly $(n-1)$ ! of the vectors have $e_{i}=1$. Therefore the total number of " 1 "s is $n \cdot(n-1)$ ! = n!, implying $$ \sum_{k=0}^{n} k p_{n}(k)=n! $$ (b) In this case, to each permutation $\pi$ of $S$ we assign a vector $\left(d_{1}, \ldots, d_{n}\right)$ instead, with $d_{i}=k$ if $i$ is a fixed point of $\pi$, and $d_{i}=0$ otherwise, where $k$ is the number of fixed points of $\pi$. Let us count the sum $Z$ of all components $d_{i}$ for all the $n$ ! permutations. There are $p_{n}(k)$ such vectors with exactly $k$ components equal to $k$, and sums of components equal to $k^{2}$. Thus, $Z=\sum_{k=0}^{n} k^{2} p_{n}(k)$. On the other hand, we may first calculate the sum of all components $d_{i}$ for fixed $i$. In fact, the value $d_{i}=k>0$ will occur exactly $p_{n-1}(k-1)$ times, so that the sum of the $d_{i}$ 's is $\sum_{k=1}^{n} k p_{n-1}(k-1)=\sum_{k=0}^{n-1}(k+$ 1) $p_{n-1}(k)=2(n-1)$ !. Summation over $i$ yields $$ Z=\sum_{k=0}^{n} k^{2} p_{n}(k)=2 n!. $$ From (0), (1), and (2), we conclude that $$ \sum_{k=0}^{n}(k-1)^{2} p_{n}(k)=\sum_{k=0}^{n} k^{2} p_{n}(k)-2 \sum_{k=0}^{n} k p_{n}(k)+\sum_{k=0}^{n} p_{n}(k)=n! $$ Remark. Only the first part of this problem was given on the IMO.
|
{
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|
e0210168-76d1-5c4b-a8b3-0845d0988cd0
| 24,070
|
17. (ROM 1) Prove that there exists a four-coloring of the set $M=$ $\{1,2, \ldots, 1987\}$ such that any arithmetic progression with 10 terms in the set $M$ is not monochromatic. Alternative formulation. Let $M=\{1,2, \ldots, 1987\}$. Prove that there is a function $f: M \rightarrow\{1,2,3,4\}$ that is not constant on every set of 10 terms from $M$ that form an arithmetic progression.
|
17. The number of 4 -colorings of the set $M$ is equal to $4^{1987}$. Let $A$ be the number of arithmetic progressions in $M$ with 10 terms. The number of colorings containing a monochromatic arithmetic progression with 10 terms is less than $4 A \cdot 4^{1977}$. So, if $A<4^{9}$, then there exist 4 -colorings with the required property. Now we estimate the value of $A$. If the first term of a 10-term progression is $k$ and the difference is $d$, then $1 \leq k \leq 1978$ and $d \leq\left[\frac{1987-k}{9}\right]$; hence $$ A=\sum_{k=1}^{1978}\left[\frac{1987-k}{9}\right]<\frac{1986+1985+\cdots+9}{9}=\frac{1995 \cdot 1978}{18}<4^{9} $$
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
17. (ROM 1) Prove that there exists a four-coloring of the set $M=$ $\{1,2, \ldots, 1987\}$ such that any arithmetic progression with 10 terms in the set $M$ is not monochromatic. Alternative formulation. Let $M=\{1,2, \ldots, 1987\}$. Prove that there is a function $f: M \rightarrow\{1,2,3,4\}$ that is not constant on every set of 10 terms from $M$ that form an arithmetic progression.
|
17. The number of 4 -colorings of the set $M$ is equal to $4^{1987}$. Let $A$ be the number of arithmetic progressions in $M$ with 10 terms. The number of colorings containing a monochromatic arithmetic progression with 10 terms is less than $4 A \cdot 4^{1977}$. So, if $A<4^{9}$, then there exist 4 -colorings with the required property. Now we estimate the value of $A$. If the first term of a 10-term progression is $k$ and the difference is $d$, then $1 \leq k \leq 1978$ and $d \leq\left[\frac{1987-k}{9}\right]$; hence $$ A=\sum_{k=1}^{1978}\left[\frac{1987-k}{9}\right]<\frac{1986+1985+\cdots+9}{9}=\frac{1995 \cdot 1978}{18}<4^{9} $$
|
{
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|
e0a856c0-a82e-5ffb-a7f1-92c9f1e88d5f
| 24,073
|
19. (USS 2) Let $\alpha, \beta, \gamma$ be positive real numbers such that $\alpha+\beta+\gamma<\pi$, $\alpha+\beta>\gamma, \beta+\gamma>\alpha, \gamma+\alpha>\beta$. Prove that with the segments of lengths $\sin \alpha, \sin \beta, \sin \gamma$ we can construct a triangle and that its area is not greater than $$ \frac{1}{8}(\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma) $$
|
19. The facts given in the problem allow us to draw a triangular pyramid with angles $2 \alpha, 2 \beta, 2 \gamma$ at the top and lateral edges of length $1 / 2$. At the base there is a triangle whose side lengths are exactly $\sin \alpha, \sin \beta, \sin \gamma$. The area of this triangle does not exceed the sum of areas of the lateral sides, which equals $(\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma) / 8$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
19. (USS 2) Let $\alpha, \beta, \gamma$ be positive real numbers such that $\alpha+\beta+\gamma<\pi$, $\alpha+\beta>\gamma, \beta+\gamma>\alpha, \gamma+\alpha>\beta$. Prove that with the segments of lengths $\sin \alpha, \sin \beta, \sin \gamma$ we can construct a triangle and that its area is not greater than $$ \frac{1}{8}(\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma) $$
|
19. The facts given in the problem allow us to draw a triangular pyramid with angles $2 \alpha, 2 \beta, 2 \gamma$ at the top and lateral edges of length $1 / 2$. At the base there is a triangle whose side lengths are exactly $\sin \alpha, \sin \beta, \sin \gamma$. The area of this triangle does not exceed the sum of areas of the lateral sides, which equals $(\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma) / 8$.
|
{
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|
9603fcc7-6507-553b-8939-3b786953c058
| 24,078
|
2. (USA 3) At a party attended by $n$ married couples, each person talks to everyone else at the party except his or her spouse. The conversations involve sets of persons or cliques $C_{1}, C_{2}, \ldots, C_{k}$ with the following property: no couple are members of the same clique, but for every other pair of persons there is exactly one clique to which both members belong. Prove that if $n \geq 4$, then $k \geq 2 n$.
|
2. Let $d_{i}$ denote the number of cliques of which person $i$ is a member. Clearly $d_{i} \geq 2$. We now distinguish two cases: (i) For some $i, d_{i}=2$. Suppose that $i$ is a member of two cliques, $C_{p}$ and $C_{q}$. Then $\left|C_{p}\right|=\left|C_{q}\right|=n$, since for each couple other than $i$ and his/her spouse, one member is in $C_{p}$ and one in $C_{q}$. There are thus $(n-1)(n-2)$ pairs $(r, s)$ of nonspouse persons distinct from $i$, where $r \in C_{p}, s \in C_{q}$. We observe that each such pair accounts for a different clique. Otherwise, we find two members of $C_{p}$ or $C_{q}$ who belong to one other clique. It follows that $k \geq 2+(n-1)(n-2) \geq 2 n$ for $n \geq 4$. (ii) For every $i, d_{i} \geq 3$. Suppose that $k<2 n$. For $i=1,2, \ldots, 2 n$ assign to person $i$ an indeterminant $x_{i}$, and for $j=1,2, \ldots, k$ set $y=\sum_{i \in C_{j}} x_{i}$. From linear algebra, we know that if $k<2 n$, then there exist $x_{1}, x_{2}, \ldots, x_{2 n}$, not all zero, such that $y_{1}=y_{2}=\cdots=y_{k}=0$. On the other hand, suppose that $y_{1}=y_{2}=\cdots=y_{k}=0$. Let $M$ be the set of the couples and $M^{\prime}$ the set of all other pairs of persons. Then $$ \begin{aligned} 0 & =\sum_{j=1}^{k} y_{j}^{2}=\sum_{i=1}^{2 n} d_{i} x_{i}^{2}+2 \sum_{(i, j) \in M^{\prime}} x_{i} x_{j} \\ & =\sum_{i=1}^{2 n}\left(d_{i}-2\right) x_{i}^{2}+\left(x_{1}+x_{2}+\cdots+x_{2 n}\right)^{2}+\sum_{(i, j) \in M}\left(x_{i}-x_{j}\right)^{2} \\ & \geq \sum_{i=1}^{2 n} x_{i}^{2}>0 \end{aligned} $$ if not all $x_{1}, x_{2}, \ldots, x_{2 n}$ are zero, which is a contradiction. Hence $k \geq$ $2 n$. Remark. The condition $n \geq 4$ is essential. For a party attended by 3 couples $\{(1,4),(2,5),(3,6)\}$, there is a collection of 4 cliques satisfying the conditions: $\{(1,2,3),(3,4,5),(5,6,1),(2,4,6)\}$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
2. (USA 3) At a party attended by $n$ married couples, each person talks to everyone else at the party except his or her spouse. The conversations involve sets of persons or cliques $C_{1}, C_{2}, \ldots, C_{k}$ with the following property: no couple are members of the same clique, but for every other pair of persons there is exactly one clique to which both members belong. Prove that if $n \geq 4$, then $k \geq 2 n$.
|
2. Let $d_{i}$ denote the number of cliques of which person $i$ is a member. Clearly $d_{i} \geq 2$. We now distinguish two cases: (i) For some $i, d_{i}=2$. Suppose that $i$ is a member of two cliques, $C_{p}$ and $C_{q}$. Then $\left|C_{p}\right|=\left|C_{q}\right|=n$, since for each couple other than $i$ and his/her spouse, one member is in $C_{p}$ and one in $C_{q}$. There are thus $(n-1)(n-2)$ pairs $(r, s)$ of nonspouse persons distinct from $i$, where $r \in C_{p}, s \in C_{q}$. We observe that each such pair accounts for a different clique. Otherwise, we find two members of $C_{p}$ or $C_{q}$ who belong to one other clique. It follows that $k \geq 2+(n-1)(n-2) \geq 2 n$ for $n \geq 4$. (ii) For every $i, d_{i} \geq 3$. Suppose that $k<2 n$. For $i=1,2, \ldots, 2 n$ assign to person $i$ an indeterminant $x_{i}$, and for $j=1,2, \ldots, k$ set $y=\sum_{i \in C_{j}} x_{i}$. From linear algebra, we know that if $k<2 n$, then there exist $x_{1}, x_{2}, \ldots, x_{2 n}$, not all zero, such that $y_{1}=y_{2}=\cdots=y_{k}=0$. On the other hand, suppose that $y_{1}=y_{2}=\cdots=y_{k}=0$. Let $M$ be the set of the couples and $M^{\prime}$ the set of all other pairs of persons. Then $$ \begin{aligned} 0 & =\sum_{j=1}^{k} y_{j}^{2}=\sum_{i=1}^{2 n} d_{i} x_{i}^{2}+2 \sum_{(i, j) \in M^{\prime}} x_{i} x_{j} \\ & =\sum_{i=1}^{2 n}\left(d_{i}-2\right) x_{i}^{2}+\left(x_{1}+x_{2}+\cdots+x_{2 n}\right)^{2}+\sum_{(i, j) \in M}\left(x_{i}-x_{j}\right)^{2} \\ & \geq \sum_{i=1}^{2 n} x_{i}^{2}>0 \end{aligned} $$ if not all $x_{1}, x_{2}, \ldots, x_{2 n}$ are zero, which is a contradiction. Hence $k \geq$ $2 n$. Remark. The condition $n \geq 4$ is essential. For a party attended by 3 couples $\{(1,4),(2,5),(3,6)\}$, there is a collection of 4 cliques satisfying the conditions: $\{(1,2,3),(3,4,5),(5,6,1),(2,4,6)\}$.
|
{
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|
20c7d614-3e32-55a9-aa63-1cff86c407d1
| 24,079
|
20. (USS 3) ${ }^{\text {IMO6 }}$ Let $f(x)=x^{2}+x+p, p \in \mathbb{N}$. Prove that if the numbers $f(0), f(1), \ldots, f([\sqrt{p / 3}])$ are primes, then all the numbers $f(0), f(1), \ldots$, $f(p-2)$ are primes.
|
20. Let $y$ be the smallest nonnegative integer with $y \leq p-2$ for which $f(y)$ is a composite number. Denote by $q$ the smallest prime divisor of $f(y)$. We claim that $y<q$. Suppose the contrary, that $y \geq q$. Let $r$ be a positive integer such that $y \equiv r(\bmod q)$. Then $f(y) \equiv f(r) \equiv 0(\bmod q)$, and since $q \leq y \leq p-2 \leq$ $f(r)$, we conclude that $q \mid f(r)$, which is a contradiction to the minimality of $y$. Now, we will prove that $q>2 y$. Suppose the contrary, that $q \leq 2 y$. Since $$ f(y)-f(x)=(y-x)(y+x+1) $$ we observe that $f(y)-f(q-1-y)=(2 y-q+1) q$, from which it follows that $f(q-1-y)$ is divisible by $q$. But by the assumptions, $q-1-y<y$, implying that $f(q-1-y)$ is prime and therefore equal to $q$. This is impossible, because $$ f(q-1-y)=(q-1-y)^{2}+(q-1-y)+p>q+p-y-1 \geq q . $$ Therefore $q \geq 2 y+1$. Now, since $f(y)$, being composite, cannot be equal to $q$, and $q$ is its smallest prime divisor, we obtain that $f(y) \geq q^{2}$. Consequently, $$ y^{2}+y+p \geq q^{2} \geq(2 y+1)^{2}=4 y^{2}+4 y+1 \Rightarrow 3\left(y^{2}+y\right) \leq p-1 $$ and from this we easily conclude that $y<\sqrt{p / 3}$, which contradicts the condition of the problem. In this way, all the numbers $$ f(0), f(1), \ldots, f(p-2) $$ must be prime.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
20. (USS 3) ${ }^{\text {IMO6 }}$ Let $f(x)=x^{2}+x+p, p \in \mathbb{N}$. Prove that if the numbers $f(0), f(1), \ldots, f([\sqrt{p / 3}])$ are primes, then all the numbers $f(0), f(1), \ldots$, $f(p-2)$ are primes.
|
20. Let $y$ be the smallest nonnegative integer with $y \leq p-2$ for which $f(y)$ is a composite number. Denote by $q$ the smallest prime divisor of $f(y)$. We claim that $y<q$. Suppose the contrary, that $y \geq q$. Let $r$ be a positive integer such that $y \equiv r(\bmod q)$. Then $f(y) \equiv f(r) \equiv 0(\bmod q)$, and since $q \leq y \leq p-2 \leq$ $f(r)$, we conclude that $q \mid f(r)$, which is a contradiction to the minimality of $y$. Now, we will prove that $q>2 y$. Suppose the contrary, that $q \leq 2 y$. Since $$ f(y)-f(x)=(y-x)(y+x+1) $$ we observe that $f(y)-f(q-1-y)=(2 y-q+1) q$, from which it follows that $f(q-1-y)$ is divisible by $q$. But by the assumptions, $q-1-y<y$, implying that $f(q-1-y)$ is prime and therefore equal to $q$. This is impossible, because $$ f(q-1-y)=(q-1-y)^{2}+(q-1-y)+p>q+p-y-1 \geq q . $$ Therefore $q \geq 2 y+1$. Now, since $f(y)$, being composite, cannot be equal to $q$, and $q$ is its smallest prime divisor, we obtain that $f(y) \geq q^{2}$. Consequently, $$ y^{2}+y+p \geq q^{2} \geq(2 y+1)^{2}=4 y^{2}+4 y+1 \Rightarrow 3\left(y^{2}+y\right) \leq p-1 $$ and from this we easily conclude that $y<\sqrt{p / 3}$, which contradicts the condition of the problem. In this way, all the numbers $$ f(0), f(1), \ldots, f(p-2) $$ must be prime.
|
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|
306bb4d0-b2c0-5e05-a1c2-9a04af542bf3
| 24,081
|
21. (USS 4) ${ }^{\mathrm{IMO} 2}$ The prolongation of the bisector $A L(L \in B C)$ in the acuteangled triangle $A B C$ intersects the circumscribed circle at point $N$. From point $L$ to the sides $A B$ and $A C$ are drawn the perpendiculars $L K$ and $L M$ respectively. Prove that the area of the triangle $A B C$ is equal to the area of the quadrilateral $A K N M$.
|
21. Let $P$ be the second point of intersection of segment $B C$ and the circle circumscribed about quadrilateral $A K L M$. Denote by $E$ the intersection point of the lines $K N$ and $B C$ and by $F$ the intersection point of the lines $M N$ and $B C$. Then $\angle B C N=\angle B A N$ and $\angle M A L=$ $\angle M P L$, as angles on the same arc. Since $A L$ is a bisector, $\angle B C N=$ $\angle B A L=\angle M A L=\angle M P L$, and  consequently $P M \| N C$. Similarly we prove $K P \| B N$. Then the quadrilaterals $B K P N$ and $N P M C$ are trapezoids; hence $$ S_{B K E}=S_{N P E} \quad \text { and } \quad S_{N P F}=S_{C M F} $$ Therefore $S_{A B C}=S_{A K N M}$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
21. (USS 4) ${ }^{\mathrm{IMO} 2}$ The prolongation of the bisector $A L(L \in B C)$ in the acuteangled triangle $A B C$ intersects the circumscribed circle at point $N$. From point $L$ to the sides $A B$ and $A C$ are drawn the perpendiculars $L K$ and $L M$ respectively. Prove that the area of the triangle $A B C$ is equal to the area of the quadrilateral $A K N M$.
|
21. Let $P$ be the second point of intersection of segment $B C$ and the circle circumscribed about quadrilateral $A K L M$. Denote by $E$ the intersection point of the lines $K N$ and $B C$ and by $F$ the intersection point of the lines $M N$ and $B C$. Then $\angle B C N=\angle B A N$ and $\angle M A L=$ $\angle M P L$, as angles on the same arc. Since $A L$ is a bisector, $\angle B C N=$ $\angle B A L=\angle M A L=\angle M P L$, and  consequently $P M \| N C$. Similarly we prove $K P \| B N$. Then the quadrilaterals $B K P N$ and $N P M C$ are trapezoids; hence $$ S_{B K E}=S_{N P E} \quad \text { and } \quad S_{N P F}=S_{C M F} $$ Therefore $S_{A B C}=S_{A K N M}$.
|
{
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|
cffe60d7-2711-5ea0-b87c-555ca6c50b28
| 24,084
|
22. (VIE 3) ${ }^{\mathrm{IMO} 4}$ Does there exist a function $f: \mathbb{N} \rightarrow \mathbb{N}$, such that $f(f(n))=$ $n+1987$ for every natural number $n$ ?
|
22. Suppose that there exists such function $f$. Then we obtain $$ f(n+1987)=f(f(f(n)))=f(n)+1987 \quad \text { for all } n \in \mathbb{N} $$ and from here, by induction, $f(n+1987 t)=f(n)+1987 t$ for all $n, t \in \mathbb{N}$. Further, for any $r \in\{0,1, \ldots, 1986\}$, let $f(r)=1987 k+l, k, l \in \mathbb{N}$, $l \leq 1986$. We have $$ r+1987=f(f(r))=f(l+1987 k)=f(l)+1987 k, $$ and consequently there are two possibilities: (i) $k=1 \Rightarrow f(r)=l+1987$ and $f(l)=r$; (ii) $k=0 \Rightarrow f(r)=l$ and $f(l)=r+1987$; in both cases, $r \neq l$. In this way, the set $\{0,1, \ldots, 1986\}$ decomposes to pairs $\{a, b\}$ such that $$ f(a)=b \text { and } f(b)=a+1987, \quad \text { or } \quad f(b)=a \text { and } f(a)=b+1987 . $$ But the set $\{0,1, \ldots, 1986\}$ has an odd number of elements, and cannot be decomposed into pairs. Contradiction.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
22. (VIE 3) ${ }^{\mathrm{IMO} 4}$ Does there exist a function $f: \mathbb{N} \rightarrow \mathbb{N}$, such that $f(f(n))=$ $n+1987$ for every natural number $n$ ?
|
22. Suppose that there exists such function $f$. Then we obtain $$ f(n+1987)=f(f(f(n)))=f(n)+1987 \quad \text { for all } n \in \mathbb{N} $$ and from here, by induction, $f(n+1987 t)=f(n)+1987 t$ for all $n, t \in \mathbb{N}$. Further, for any $r \in\{0,1, \ldots, 1986\}$, let $f(r)=1987 k+l, k, l \in \mathbb{N}$, $l \leq 1986$. We have $$ r+1987=f(f(r))=f(l+1987 k)=f(l)+1987 k, $$ and consequently there are two possibilities: (i) $k=1 \Rightarrow f(r)=l+1987$ and $f(l)=r$; (ii) $k=0 \Rightarrow f(r)=l$ and $f(l)=r+1987$; in both cases, $r \neq l$. In this way, the set $\{0,1, \ldots, 1986\}$ decomposes to pairs $\{a, b\}$ such that $$ f(a)=b \text { and } f(b)=a+1987, \quad \text { or } \quad f(b)=a \text { and } f(a)=b+1987 . $$ But the set $\{0,1, \ldots, 1986\}$ has an odd number of elements, and cannot be decomposed into pairs. Contradiction.
|
{
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|
f88523b7-5be5-5a2c-947e-21519531b30f
| 24,087
|
23. (YUG 2) Prove that for every natural number $k(k \geq 2)$ there exists an irrational number $r$ such that for every natural number $m$, $$ \left[r^{m}\right] \equiv-1 \quad(\bmod k) $$ Remark. An easier variant: Find $r$ as a root of a polynomial of second degree with integer coefficients.
|
23. If we prove the existence of $p, q \in \mathbb{N}$ such that the roots $r, s$ of $$ f(x)=x^{2}-k p \cdot x+k q=0 $$ are irrational real numbers with $0<s<1$ (and consequently $r>1$ ), then we are done, because from $r+s, r s \equiv 0(\bmod k)$ we get $r^{m}+s^{m} \equiv 0$ $(\bmod k)$, and $0<s^{m}<1$ yields the assertion. To prove the existence of such natural numbers $p$ and $q$, we can take them such that $f(0)>0>f(1)$, i.e., $$ k q>0>k(q-p)+1 \quad \Rightarrow \quad p>q>0 $$ The irrationality of $r$ can be obtained by taking $q=p-1$, because the discriminant $D=(k p)^{2}-4 k p+4 k$, for $(k p-2)^{2}<D<(k p-1)^{2}$, is not a perfect square for $p \geq 2$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
23. (YUG 2) Prove that for every natural number $k(k \geq 2)$ there exists an irrational number $r$ such that for every natural number $m$, $$ \left[r^{m}\right] \equiv-1 \quad(\bmod k) $$ Remark. An easier variant: Find $r$ as a root of a polynomial of second degree with integer coefficients.
|
23. If we prove the existence of $p, q \in \mathbb{N}$ such that the roots $r, s$ of $$ f(x)=x^{2}-k p \cdot x+k q=0 $$ are irrational real numbers with $0<s<1$ (and consequently $r>1$ ), then we are done, because from $r+s, r s \equiv 0(\bmod k)$ we get $r^{m}+s^{m} \equiv 0$ $(\bmod k)$, and $0<s^{m}<1$ yields the assertion. To prove the existence of such natural numbers $p$ and $q$, we can take them such that $f(0)>0>f(1)$, i.e., $$ k q>0>k(q-p)+1 \quad \Rightarrow \quad p>q>0 $$ The irrationality of $r$ can be obtained by taking $q=p-1$, because the discriminant $D=(k p)^{2}-4 k p+4 k$, for $(k p-2)^{2}<D<(k p-1)^{2}$, is not a perfect square for $p \geq 2$.
|
{
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|
57d3bfa8-091e-58d7-b2ec-0efdd008ab6c
| 24,088
|
5. (GBR 1) Find, with proof, the point $P$ in the interior of an acute-angled triangle $A B C$ for which $B L^{2}+C M^{2}+A N^{2}$ is a minimum, where $L, M, N$ are the feet of the perpendiculars from $P$ to $B C, C A, A B$ respectively.
|
5. Assuming the notation $a=\overline{B C}, b=\overline{A C}, c=\overline{A B} ; x=\overline{B L}, y=\overline{C M}$, $z=\overline{A N}$, from the Pythagorean theorem we obtain $$ \begin{aligned} (a-x)^{2}+(b-y)^{2} & +(c-z)^{2}=x^{2}+y^{2}+z^{2} \\ & =\frac{x^{2}+(a-x)^{2}+y^{2}+(b-y)^{2}+z^{2}+(c-z)^{2}}{2} \end{aligned} $$ Since $x^{2}+(a-x)^{2}=a^{2} / 2+(a-2 x)^{2} / 2 \geq a^{2} / 2$ and similarly $y^{2}+(b-y)^{2} \geq$ $b^{2} / 2$ and $z^{2}+(c-z)^{2} \geq c^{2} / 2$, we get $$ x^{2}+y^{2}+z^{2} \geq \frac{a^{2}+b^{2}+c^{2}}{4} $$ Equality holds if and only if $P$ is the circumcenter of the triangle $A B C$, i.e., when $x=a / 2, y=b / 2, z=c / 2$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
5. (GBR 1) Find, with proof, the point $P$ in the interior of an acute-angled triangle $A B C$ for which $B L^{2}+C M^{2}+A N^{2}$ is a minimum, where $L, M, N$ are the feet of the perpendiculars from $P$ to $B C, C A, A B$ respectively.
|
5. Assuming the notation $a=\overline{B C}, b=\overline{A C}, c=\overline{A B} ; x=\overline{B L}, y=\overline{C M}$, $z=\overline{A N}$, from the Pythagorean theorem we obtain $$ \begin{aligned} (a-x)^{2}+(b-y)^{2} & +(c-z)^{2}=x^{2}+y^{2}+z^{2} \\ & =\frac{x^{2}+(a-x)^{2}+y^{2}+(b-y)^{2}+z^{2}+(c-z)^{2}}{2} \end{aligned} $$ Since $x^{2}+(a-x)^{2}=a^{2} / 2+(a-2 x)^{2} / 2 \geq a^{2} / 2$ and similarly $y^{2}+(b-y)^{2} \geq$ $b^{2} / 2$ and $z^{2}+(c-z)^{2} \geq c^{2} / 2$, we get $$ x^{2}+y^{2}+z^{2} \geq \frac{a^{2}+b^{2}+c^{2}}{4} $$ Equality holds if and only if $P$ is the circumcenter of the triangle $A B C$, i.e., when $x=a / 2, y=b / 2, z=c / 2$.
|
{
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|
e84d1f4a-ef18-5ab3-a282-f684bc11c8a8
| 24,095
|
7. (NET 1) Given five real numbers $u_{0}, u_{1}, u_{2}, u_{3}, u_{4}$, prove that it is always possible to find five real numbers $v_{0}, v_{1}, v_{2}, v_{3}, v_{4}$ that satisfy the following conditions: (i) $u_{i}-v_{i} \in \mathbb{N}$. (ii) $\sum_{0 \leq i<j \leq 4}\left(v_{i}-v_{j}\right)^{2}<4$.
|
7. For all real numbers $v$ the following inequality holds: $$ \sum_{0 \leq i<j \leq 4}\left(v_{i}-v_{j}\right)^{2} \leq 5 \sum_{i=0}^{4}\left(v_{i}-v\right)^{2} $$ Indeed, $$ \begin{aligned} \sum_{0 \leq i<j \leq 4}\left(v_{i}-v_{j}\right)^{2} & =\sum_{0 \leq i<j \leq 4}\left[\left(v_{i}-v\right)-\left(v_{j}-v\right)\right]^{2} \\ & =5 \sum_{i=0}^{4}\left(v_{i}-v\right)^{2}-\left(\sum_{i=0}^{4}\left(v_{i}-v\right)\right)^{2} \leq 5 \sum_{i=0}^{4}\left(v_{i}-v\right)^{2} . \end{aligned} $$ Let us first take $v_{i}$ 's, satisfying condition (1), so that w.l.o.g. $v_{0} \leq v_{1} \leq$ $v_{2} \leq v_{3} \leq v_{4} \leq 1+v_{0}$. Defining $v_{5}=1+v_{0}$, we see that one of the differences $v_{j+1}-v_{j}, j=0, \ldots, 4$, is at most $1 / 5$. Take $v=\left(v_{j+1}+v_{j}\right) / 2$, and then place the other three $v_{j}$ 's in the segment $[v-1 / 2, v+1 / 2]$. Now we have $\left|v-v_{j}\right| \leq 1 / 10,\left|v-v_{j+1}\right| \leq 1 / 10$, and $\left|v-v_{k}\right| \leq 1 / 2$, for any $k$ different from $j, j+1$. The $v_{i}$ 's thus obtained have the required property. In fact, using the inequality (1), we obtain $$ \sum_{0 \leq i<j \leq 4}\left(v_{i}-v_{j}\right)^{2} \leq 5\left(2\left(\frac{1}{10}\right)^{2}+3\left(\frac{1}{2}\right)^{2}\right)=3.85<4 $$ Remark. The best possible estimate for the right-hand side is 2 .
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
7. (NET 1) Given five real numbers $u_{0}, u_{1}, u_{2}, u_{3}, u_{4}$, prove that it is always possible to find five real numbers $v_{0}, v_{1}, v_{2}, v_{3}, v_{4}$ that satisfy the following conditions: (i) $u_{i}-v_{i} \in \mathbb{N}$. (ii) $\sum_{0 \leq i<j \leq 4}\left(v_{i}-v_{j}\right)^{2}<4$.
|
7. For all real numbers $v$ the following inequality holds: $$ \sum_{0 \leq i<j \leq 4}\left(v_{i}-v_{j}\right)^{2} \leq 5 \sum_{i=0}^{4}\left(v_{i}-v\right)^{2} $$ Indeed, $$ \begin{aligned} \sum_{0 \leq i<j \leq 4}\left(v_{i}-v_{j}\right)^{2} & =\sum_{0 \leq i<j \leq 4}\left[\left(v_{i}-v\right)-\left(v_{j}-v\right)\right]^{2} \\ & =5 \sum_{i=0}^{4}\left(v_{i}-v\right)^{2}-\left(\sum_{i=0}^{4}\left(v_{i}-v\right)\right)^{2} \leq 5 \sum_{i=0}^{4}\left(v_{i}-v\right)^{2} . \end{aligned} $$ Let us first take $v_{i}$ 's, satisfying condition (1), so that w.l.o.g. $v_{0} \leq v_{1} \leq$ $v_{2} \leq v_{3} \leq v_{4} \leq 1+v_{0}$. Defining $v_{5}=1+v_{0}$, we see that one of the differences $v_{j+1}-v_{j}, j=0, \ldots, 4$, is at most $1 / 5$. Take $v=\left(v_{j+1}+v_{j}\right) / 2$, and then place the other three $v_{j}$ 's in the segment $[v-1 / 2, v+1 / 2]$. Now we have $\left|v-v_{j}\right| \leq 1 / 10,\left|v-v_{j+1}\right| \leq 1 / 10$, and $\left|v-v_{k}\right| \leq 1 / 2$, for any $k$ different from $j, j+1$. The $v_{i}$ 's thus obtained have the required property. In fact, using the inequality (1), we obtain $$ \sum_{0 \leq i<j \leq 4}\left(v_{i}-v_{j}\right)^{2} \leq 5\left(2\left(\frac{1}{10}\right)^{2}+3\left(\frac{1}{2}\right)^{2}\right)=3.85<4 $$ Remark. The best possible estimate for the right-hand side is 2 .
|
{
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|
4eba5e10-b0fd-5637-9120-be5798676ecb
| 24,102
|
8. (HUN 1) (a) Let $(m, k)=1$. Prove that there exist integers $a_{1}, a_{2}, \ldots, a_{m}$ and $b_{1}, b_{2}, \ldots, b_{k}$ such that each product $a_{i} b_{j}(i=1,2, \ldots, m ; j=$ $1,2, \ldots, k)$ gives a different residue when divided by $m k$. (b) Let $(m, k)>1$. Prove that for any integers $a_{1}, a_{2}, \ldots, a_{m}$ and $b_{1}, b_{2}$, $\ldots, b_{k}$ there must be two products $a_{i} b_{j}$ and $a_{s} b_{t}((i, j) \neq(s, t))$ that give the same residue when divided by $m k$.
|
8. (a) Consider $$ a_{i}=i k+1, \quad i=1,2, \ldots, m ; \quad b_{j}=j m+1, \quad j=1,2, \ldots, k $$ Assume that $m k \mid a_{i} b_{j}-a_{s} b_{t}=(i k+1)(j m+1)-(s k+1)(t m+1)=$ $k m(i j-s t)+m(j-t)+k(i-s)$. Since $m$ divides this sum, we get that $m \mid k(i-s)$, or, together with $\operatorname{gcd}(k, m)=1$, that $i=s$. Similarly $j=t$, which proves part (a). (b) Suppose the opposite, i.e., that all the residues are distinct. Then the residue 0 must also occur, say at $a_{1} b_{1}: m k \mid a_{1} b_{1}$; so, for some $a^{\prime}$ and $b^{\prime}, a^{\prime}\left|a_{1}, b^{\prime}\right| b_{1}$, and $a^{\prime} b^{\prime}=m k$. Assuming that for some $i, s \neq i$, $a^{\prime} \mid a_{i}-a_{s}$, we obtain $m k=a^{\prime} b^{\prime} \mid a_{i} b_{1}-a_{s} b_{1}$, a contradiction. This shows that $a^{\prime} \geq m$ and similarly $b^{\prime} \geq k$, and thus from $a^{\prime} b^{\prime}=m k$ we have $a^{\prime}=m, b^{\prime}=k$. We also get (1): all $a_{i}$ 's give distinct residues modulo $m=a^{\prime}$, and all $b_{j}$ 's give distinct residues modulo $k=b^{\prime}$. Now let $p$ be a common prime divisor of $m$ and $k$. By $(*)$, exactly $\frac{p-1}{p} m$ of $a_{i}$ 's and exactly $\frac{p-1}{p} k$ of $b_{j}$ 's are not divisible by $p$. Therefore there are precisely $\frac{(p-1)^{2}}{p^{2}} m k$ products $a_{i} b_{j}$ that are not divisible by $p$, although from the assumption that they all give distinct residues it follows that the number of such products is $\frac{p-1}{p} m k \neq \frac{(p-1)^{2}}{p^{2}} m k$. We have arrived at a contradiction, thus proving (b).
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
8. (HUN 1) (a) Let $(m, k)=1$. Prove that there exist integers $a_{1}, a_{2}, \ldots, a_{m}$ and $b_{1}, b_{2}, \ldots, b_{k}$ such that each product $a_{i} b_{j}(i=1,2, \ldots, m ; j=$ $1,2, \ldots, k)$ gives a different residue when divided by $m k$. (b) Let $(m, k)>1$. Prove that for any integers $a_{1}, a_{2}, \ldots, a_{m}$ and $b_{1}, b_{2}$, $\ldots, b_{k}$ there must be two products $a_{i} b_{j}$ and $a_{s} b_{t}((i, j) \neq(s, t))$ that give the same residue when divided by $m k$.
|
8. (a) Consider $$ a_{i}=i k+1, \quad i=1,2, \ldots, m ; \quad b_{j}=j m+1, \quad j=1,2, \ldots, k $$ Assume that $m k \mid a_{i} b_{j}-a_{s} b_{t}=(i k+1)(j m+1)-(s k+1)(t m+1)=$ $k m(i j-s t)+m(j-t)+k(i-s)$. Since $m$ divides this sum, we get that $m \mid k(i-s)$, or, together with $\operatorname{gcd}(k, m)=1$, that $i=s$. Similarly $j=t$, which proves part (a). (b) Suppose the opposite, i.e., that all the residues are distinct. Then the residue 0 must also occur, say at $a_{1} b_{1}: m k \mid a_{1} b_{1}$; so, for some $a^{\prime}$ and $b^{\prime}, a^{\prime}\left|a_{1}, b^{\prime}\right| b_{1}$, and $a^{\prime} b^{\prime}=m k$. Assuming that for some $i, s \neq i$, $a^{\prime} \mid a_{i}-a_{s}$, we obtain $m k=a^{\prime} b^{\prime} \mid a_{i} b_{1}-a_{s} b_{1}$, a contradiction. This shows that $a^{\prime} \geq m$ and similarly $b^{\prime} \geq k$, and thus from $a^{\prime} b^{\prime}=m k$ we have $a^{\prime}=m, b^{\prime}=k$. We also get (1): all $a_{i}$ 's give distinct residues modulo $m=a^{\prime}$, and all $b_{j}$ 's give distinct residues modulo $k=b^{\prime}$. Now let $p$ be a common prime divisor of $m$ and $k$. By $(*)$, exactly $\frac{p-1}{p} m$ of $a_{i}$ 's and exactly $\frac{p-1}{p} k$ of $b_{j}$ 's are not divisible by $p$. Therefore there are precisely $\frac{(p-1)^{2}}{p^{2}} m k$ products $a_{i} b_{j}$ that are not divisible by $p$, although from the assumption that they all give distinct residues it follows that the number of such products is $\frac{p-1}{p} m k \neq \frac{(p-1)^{2}}{p^{2}} m k$. We have arrived at a contradiction, thus proving (b).
|
{
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|
3fb74acc-b0d4-5f2c-afe0-09bee5ed9e28
| 24,104
|
9. (HUN 2) Does there exist a set $M$ in usual Euclidean space such that for every plane $\lambda$ the intersection $M \cap \lambda$ is finite and nonempty?
|
9. The answer is yes. Consider the curve $$ C=\left\{(x, y, z) \mid x=t, y=t^{3}, z=t^{5}, \quad t \in \mathbb{R}\right\} $$ Any plane defined by an equation of the form $a x+b y+c z+d=0$ intersects the curve $C$ at points $\left(t, t^{3}, t^{5}\right)$ with $t$ satisfying $c t^{5}+b t^{3}+a t+d=0$. This last equation has at least one but only finitely many solutions.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
9. (HUN 2) Does there exist a set $M$ in usual Euclidean space such that for every plane $\lambda$ the intersection $M \cap \lambda$ is finite and nonempty?
|
9. The answer is yes. Consider the curve $$ C=\left\{(x, y, z) \mid x=t, y=t^{3}, z=t^{5}, \quad t \in \mathbb{R}\right\} $$ Any plane defined by an equation of the form $a x+b y+c z+d=0$ intersects the curve $C$ at points $\left(t, t^{3}, t^{5}\right)$ with $t$ satisfying $c t^{5}+b t^{3}+a t+d=0$. This last equation has at least one but only finitely many solutions.
|
{
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|
d80f0abe-625a-530f-8fdf-77df2fa8b390
| 24,107
|
1. (BUL 1) An integer sequence is defined by $$ a_{n}=2 a_{n-1}+a_{n-2} \quad(n>1), \quad a_{0}=0, \quad a_{1}=1 $$ Prove that $2^{k}$ divides $a_{n}$ if and only if $2^{k}$ divides $n$.
|
1. Assume that $p$ and $q$ are real and $b_{0}, b_{1}, b_{2}, \ldots$ is a sequence such that $b_{n}=p b_{n-1}+q b_{n-2}$ for all $n>1$. From the equalities $b_{n}=p b_{n-1}+q b_{n-2}$, $b_{n+1}=p b_{n}+q b_{n-1}, b_{n+2}=p b_{n+1}+q b_{n}$, eliminating $b_{n+1}$ and $b_{n-1}$ we obtain that $b_{n+2}=\left(p^{2}+2 q\right) b_{n}-q^{2} b_{n-2}$. So the sequence $b_{0}, b_{2}, b_{4}, \ldots$ has the property $$ b_{2 n}=P b_{2 n-2}+Q b_{2 n-4}, \quad P=p^{2}+2 q, \quad Q=-q^{2} . $$ We shall solve the problem by induction. The sequence $a_{n}$ has $p=2$, $q=1$, and hence $P=6, Q=-1$. Let $k=1$. Then $a_{0}=0, a_{1}=1$, and $a_{n}$ is of the same parity as $a_{n-2}$; i.e., it is even if and only if $n$ is even. Let $k \geq 1$. We assume that for $n=2^{k} m$, the numbers $a_{n}$ are divisible by $2^{k}$, but divisible by $2^{k+1}$ if and only if $m$ is even. We assume also that the sequence $c_{0}, c_{1}, \ldots$, with $c_{m}=a_{m \cdot 2^{k}}$, satisfies the condition $c_{n}=$ $p c_{n-1}-c_{n-2}$, where $p \equiv 2(\bmod 4)($ for $k=1$ it is true). We shall prove the same statement for $k+1$. According to (1), $c_{2 n}=P c_{2 n-2}-c_{2 n-4}$, where $P=p^{2}-2$. Obviously $P \equiv 2(\bmod 4)$. Since $P=4 s+2$ for some integer $s$, and $c_{2 n}=2^{k+1} d_{2 n}, c_{0}=0, c_{1} \equiv 2^{k}\left(\bmod 2^{k+1}\right)$, and $c_{2}=p c_{1} \equiv 2^{k+1}$ $\left(\bmod 2^{k+2}\right)$, we have $$ c_{2 n}=(4 s+2) 2^{k+1} d_{2 n-2}-c_{2 n-4} \equiv c_{2 n-4}\left(\bmod 2^{k+2}\right) $$ i.e., $0 \equiv c_{0} \equiv c_{4} \equiv c_{8} \equiv \cdots$ and $2^{k+1} \equiv c_{2} \equiv c_{6} \equiv \cdots\left(\bmod 2^{k+2}\right)$, which proves the statement. Second solution. The recursion is solved by $$ a_{n}=\frac{1}{2 \sqrt{2}}\left((1+\sqrt{2})^{n}-(1-\sqrt{2})^{n}\right)=\binom{n}{1}+2\binom{n}{3}+2^{2}\binom{n}{5}+\cdots . $$ Let $n=2^{k} m$ with $m$ odd; then for $p>0$ the summand $$ 2^{p}\binom{n}{2 p+1}=2^{k+p} m \frac{(n-1) \ldots(n-2 p)}{(2 p+1)!}=2^{k+p} \frac{m}{2 p+1}\binom{n-1}{2 p} $$ is divisible by $2^{k+p}$, because the denominator $2 p+1$ is odd. Hence $$ a_{n}=n+\sum_{p>0} 2^{p}\binom{n}{2 p+1}=2^{k} m+2^{k+1} N $$ for some integer $N$, so that $a_{n}$ is exactly divisible by $2^{k}$. Third solution. It can be proven by induction that $a_{2 n}=2 a_{n}\left(a_{n}+a_{n+1}\right)$. The required result follows easily, again by induction on $k$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
1. (BUL 1) An integer sequence is defined by $$ a_{n}=2 a_{n-1}+a_{n-2} \quad(n>1), \quad a_{0}=0, \quad a_{1}=1 $$ Prove that $2^{k}$ divides $a_{n}$ if and only if $2^{k}$ divides $n$.
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1. Assume that $p$ and $q$ are real and $b_{0}, b_{1}, b_{2}, \ldots$ is a sequence such that $b_{n}=p b_{n-1}+q b_{n-2}$ for all $n>1$. From the equalities $b_{n}=p b_{n-1}+q b_{n-2}$, $b_{n+1}=p b_{n}+q b_{n-1}, b_{n+2}=p b_{n+1}+q b_{n}$, eliminating $b_{n+1}$ and $b_{n-1}$ we obtain that $b_{n+2}=\left(p^{2}+2 q\right) b_{n}-q^{2} b_{n-2}$. So the sequence $b_{0}, b_{2}, b_{4}, \ldots$ has the property $$ b_{2 n}=P b_{2 n-2}+Q b_{2 n-4}, \quad P=p^{2}+2 q, \quad Q=-q^{2} . $$ We shall solve the problem by induction. The sequence $a_{n}$ has $p=2$, $q=1$, and hence $P=6, Q=-1$. Let $k=1$. Then $a_{0}=0, a_{1}=1$, and $a_{n}$ is of the same parity as $a_{n-2}$; i.e., it is even if and only if $n$ is even. Let $k \geq 1$. We assume that for $n=2^{k} m$, the numbers $a_{n}$ are divisible by $2^{k}$, but divisible by $2^{k+1}$ if and only if $m$ is even. We assume also that the sequence $c_{0}, c_{1}, \ldots$, with $c_{m}=a_{m \cdot 2^{k}}$, satisfies the condition $c_{n}=$ $p c_{n-1}-c_{n-2}$, where $p \equiv 2(\bmod 4)($ for $k=1$ it is true). We shall prove the same statement for $k+1$. According to (1), $c_{2 n}=P c_{2 n-2}-c_{2 n-4}$, where $P=p^{2}-2$. Obviously $P \equiv 2(\bmod 4)$. Since $P=4 s+2$ for some integer $s$, and $c_{2 n}=2^{k+1} d_{2 n}, c_{0}=0, c_{1} \equiv 2^{k}\left(\bmod 2^{k+1}\right)$, and $c_{2}=p c_{1} \equiv 2^{k+1}$ $\left(\bmod 2^{k+2}\right)$, we have $$ c_{2 n}=(4 s+2) 2^{k+1} d_{2 n-2}-c_{2 n-4} \equiv c_{2 n-4}\left(\bmod 2^{k+2}\right) $$ i.e., $0 \equiv c_{0} \equiv c_{4} \equiv c_{8} \equiv \cdots$ and $2^{k+1} \equiv c_{2} \equiv c_{6} \equiv \cdots\left(\bmod 2^{k+2}\right)$, which proves the statement. Second solution. The recursion is solved by $$ a_{n}=\frac{1}{2 \sqrt{2}}\left((1+\sqrt{2})^{n}-(1-\sqrt{2})^{n}\right)=\binom{n}{1}+2\binom{n}{3}+2^{2}\binom{n}{5}+\cdots . $$ Let $n=2^{k} m$ with $m$ odd; then for $p>0$ the summand $$ 2^{p}\binom{n}{2 p+1}=2^{k+p} m \frac{(n-1) \ldots(n-2 p)}{(2 p+1)!}=2^{k+p} \frac{m}{2 p+1}\binom{n-1}{2 p} $$ is divisible by $2^{k+p}$, because the denominator $2 p+1$ is odd. Hence $$ a_{n}=n+\sum_{p>0} 2^{p}\binom{n}{2 p+1}=2^{k} m+2^{k+1} N $$ for some integer $N$, so that $a_{n}$ is exactly divisible by $2^{k}$. Third solution. It can be proven by induction that $a_{2 n}=2 a_{n}\left(a_{n}+a_{n+1}\right)$. The required result follows easily, again by induction on $k$.
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22c01ce9-daf5-5f85-8f8b-6c8684bfa342
| 24,110
|
12. (GRE 2) In a triangle $A B C$, choose any points $K \in B C, L \in A C$, $M \in A B, N \in L M, R \in M K$, and $F \in K L$. If $E_{1}, E_{2}, E_{3}, E_{4}, E_{5}$, $E_{6}$, and $E$ denote the areas of the triangles $A M R, C K R, B K F, A L F$, $B N M, C L N$, and $A B C$ respectively, show that $$ E \geq 8 \sqrt[6]{E_{1} E_{2} E_{3} E_{4} E_{5} E_{6}} $$ Remark. Points $K, L, M, N, R, F$ lie on segments $B C, A C, A B, L M$, $M K, K L$ respectively.
|
12. Let $E(X Y Z)$ stand for the area of a triangle $X Y Z$. We have $$ \begin{gathered} \frac{E_{1}}{E}=\frac{E(A M R)}{E(A M K)} \cdot \frac{E(A M K)}{E(A B K)} \cdot \frac{E(A B K)}{E(A B C)}=\frac{M R}{M K} \cdot \frac{A M}{A B} \cdot \frac{B K}{B C} \Rightarrow \\ \left(\frac{E_{1}}{E}\right)^{1 / 3} \leq \frac{1}{3}\left(\frac{M R}{M K}+\frac{A M}{A B}+\frac{B K}{B C}\right) \end{gathered} $$ We similarly obtain $$ \left(\frac{E_{2}}{E}\right)^{1 / 3} \leq \frac{1}{3}\left(\frac{K R}{M K}+\frac{B M}{A B}+\frac{C K}{B C}\right) $$ Therefore $\left(E_{1} / E\right)^{1 / 3}+\left(E_{2} / E\right)^{1 / 3} \leq 1$, i.e., $\sqrt[3]{E_{1}}+\sqrt[3]{E_{2}} \leq \sqrt[3]{E}$. Analogously, $\sqrt[3]{E_{3}}+\sqrt[3]{E_{4}} \leq \sqrt[3]{E}$ and $\sqrt[3]{E_{5}}+\sqrt[3]{E_{6}} \leq \sqrt[3]{E}$; hence $$ \begin{aligned} & 8 \sqrt[6]{E_{1} E_{2} E_{3} E_{4} E_{5} E_{6}} \\ & \quad=2\left(\sqrt[3]{E_{1}} \sqrt[3]{E_{2}}\right)^{1 / 2} \cdot 2\left(\sqrt[3]{E_{3}} \sqrt[3]{E_{4}}\right)^{1 / 2} \cdot 2\left(\sqrt[3]{E_{5}} \sqrt[3]{E_{6}}\right)^{1 / 2} \\ & \quad \leq\left(\sqrt[3]{E_{1}}+\sqrt[3]{E_{2}}\right) \cdot\left(\sqrt[3]{E_{3}}+\sqrt[3]{E_{4}}\right) \cdot\left(\sqrt[3]{E_{5}}+\sqrt[3]{E_{6}}\right) \leq E \end{aligned} $$
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
12. (GRE 2) In a triangle $A B C$, choose any points $K \in B C, L \in A C$, $M \in A B, N \in L M, R \in M K$, and $F \in K L$. If $E_{1}, E_{2}, E_{3}, E_{4}, E_{5}$, $E_{6}$, and $E$ denote the areas of the triangles $A M R, C K R, B K F, A L F$, $B N M, C L N$, and $A B C$ respectively, show that $$ E \geq 8 \sqrt[6]{E_{1} E_{2} E_{3} E_{4} E_{5} E_{6}} $$ Remark. Points $K, L, M, N, R, F$ lie on segments $B C, A C, A B, L M$, $M K, K L$ respectively.
|
12. Let $E(X Y Z)$ stand for the area of a triangle $X Y Z$. We have $$ \begin{gathered} \frac{E_{1}}{E}=\frac{E(A M R)}{E(A M K)} \cdot \frac{E(A M K)}{E(A B K)} \cdot \frac{E(A B K)}{E(A B C)}=\frac{M R}{M K} \cdot \frac{A M}{A B} \cdot \frac{B K}{B C} \Rightarrow \\ \left(\frac{E_{1}}{E}\right)^{1 / 3} \leq \frac{1}{3}\left(\frac{M R}{M K}+\frac{A M}{A B}+\frac{B K}{B C}\right) \end{gathered} $$ We similarly obtain $$ \left(\frac{E_{2}}{E}\right)^{1 / 3} \leq \frac{1}{3}\left(\frac{K R}{M K}+\frac{B M}{A B}+\frac{C K}{B C}\right) $$ Therefore $\left(E_{1} / E\right)^{1 / 3}+\left(E_{2} / E\right)^{1 / 3} \leq 1$, i.e., $\sqrt[3]{E_{1}}+\sqrt[3]{E_{2}} \leq \sqrt[3]{E}$. Analogously, $\sqrt[3]{E_{3}}+\sqrt[3]{E_{4}} \leq \sqrt[3]{E}$ and $\sqrt[3]{E_{5}}+\sqrt[3]{E_{6}} \leq \sqrt[3]{E}$; hence $$ \begin{aligned} & 8 \sqrt[6]{E_{1} E_{2} E_{3} E_{4} E_{5} E_{6}} \\ & \quad=2\left(\sqrt[3]{E_{1}} \sqrt[3]{E_{2}}\right)^{1 / 2} \cdot 2\left(\sqrt[3]{E_{3}} \sqrt[3]{E_{4}}\right)^{1 / 2} \cdot 2\left(\sqrt[3]{E_{5}} \sqrt[3]{E_{6}}\right)^{1 / 2} \\ & \quad \leq\left(\sqrt[3]{E_{1}}+\sqrt[3]{E_{2}}\right) \cdot\left(\sqrt[3]{E_{3}}+\sqrt[3]{E_{4}}\right) \cdot\left(\sqrt[3]{E_{5}}+\sqrt[3]{E_{6}}\right) \leq E \end{aligned} $$
|
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44a21cd8-3079-5a09-888e-ad96876021bb
| 24,115
|
15. (ICE 1) Let $A B C$ be an acute-angled triangle. Three lines $L_{A}, L_{B}$, and $L_{C}$ are constructed through the vertices $A, B$, and $C$ respectively according to the following prescription: Let $H$ be the foot of the altitude drawn from the vertex $A$ to the side $B C$; let $S_{A}$ be the circle with diameter $A H$; let $S_{A}$ meet the sides $A B$ and $A C$ at $M$ and $N$ respectively, where $M$ and $N$ are distinct from $A$; then $L_{A}$ is the line through $A$ perpendicular to $M N$. The lines $L_{B}$ and $L_{C}$ are constructed similarly. Prove that $L_{A}$, $L_{B}$, and $L_{C}$ are concurrent.
|
15. Referring to the description of $L_{A}$, we have $\angle A M N=\angle A H N=90^{\circ}-$ $\angle H A C=\angle C$, and similarly $\angle A N M=\angle B$. Since the triangle $A B C$ is acute-angled, the line $L_{A}$ lies inside the angle $A$. Hence if $P=L_{A} \cap B C$ and $Q=L_{B} \cap A C$, we get $\angle B A P=90^{\circ}-\angle C$; hence $A P$ passes through the circumcenter $O$ of $\triangle A B C$. Similarly we prove that $L_{B}$ and $L_{C}$ contains the circumcenter $O$ also. It follows that $L_{A}, L_{B}$ and $L_{C}$ intersect at the point $O$. Remark. Without identifying the point of intersection, one can prove the concurrence of the three lines using Ceva's theorem, in usual or trigonometric form.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
15. (ICE 1) Let $A B C$ be an acute-angled triangle. Three lines $L_{A}, L_{B}$, and $L_{C}$ are constructed through the vertices $A, B$, and $C$ respectively according to the following prescription: Let $H$ be the foot of the altitude drawn from the vertex $A$ to the side $B C$; let $S_{A}$ be the circle with diameter $A H$; let $S_{A}$ meet the sides $A B$ and $A C$ at $M$ and $N$ respectively, where $M$ and $N$ are distinct from $A$; then $L_{A}$ is the line through $A$ perpendicular to $M N$. The lines $L_{B}$ and $L_{C}$ are constructed similarly. Prove that $L_{A}$, $L_{B}$, and $L_{C}$ are concurrent.
|
15. Referring to the description of $L_{A}$, we have $\angle A M N=\angle A H N=90^{\circ}-$ $\angle H A C=\angle C$, and similarly $\angle A N M=\angle B$. Since the triangle $A B C$ is acute-angled, the line $L_{A}$ lies inside the angle $A$. Hence if $P=L_{A} \cap B C$ and $Q=L_{B} \cap A C$, we get $\angle B A P=90^{\circ}-\angle C$; hence $A P$ passes through the circumcenter $O$ of $\triangle A B C$. Similarly we prove that $L_{B}$ and $L_{C}$ contains the circumcenter $O$ also. It follows that $L_{A}, L_{B}$ and $L_{C}$ intersect at the point $O$. Remark. Without identifying the point of intersection, one can prove the concurrence of the three lines using Ceva's theorem, in usual or trigonometric form.
|
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437b63b9-4434-5dd2-9830-75f0de56e2ff
| 24,122
|
17. (ISR 2) In the convex pentagon $A B C D E$, the sides $B C, C D, D E$ have the same length. Moreover, each diagonal of the pentagon is parallel to a side ( $A C$ is parallel to $D E, B D$ is parallel to $A E$, etc.). Prove that $A B C D E$ is a regular pentagon.
|
17. Let $A C$ and $A D$ meet $B E$ in $R, S$, respectively. Then by the conditions of the problem, $$ \begin{aligned} & \angle A E B=\angle E B D=\angle B D C=\angle D B C=\angle A D B=\angle E A D=\alpha, \\ & \angle A B E=\angle B E C=\angle E C D=\angle C E D=\angle A C E=\angle B A C=\beta \\ & \angle B C A=\angle C A D=\angle A D E=\gamma \end{aligned} $$ Since $\angle S A E=\angle S E A$, it follows that $A S=S E$, and analogously $B R=$ $R A$. But $B S D C$ and $R E D C$ are parallelograms; hence $B S=C D=R E$, giving us $B R=S E$ and $A R=A S$. Then also $A C=A D$, because $R S \|$ $C D$. We deduce that $2 \beta=\angle A C D=\angle A D C=2 \alpha$, i.e., $\alpha=\beta$. It will be sufficient to show that $\alpha=\gamma$, since that will imply $\alpha=\beta=\gamma=$ $36^{\circ}$. We have that the sum of the interior angles of $A C D$ is $4 \alpha+\gamma=180^{\circ}$. We have $$ \frac{\sin \gamma}{\sin \alpha}=\frac{A E}{D E}=\frac{A E}{C D}=\frac{A E}{R E}=\frac{\sin (2 \alpha+\gamma)}{\sin (\alpha+\gamma)} $$ i.e., $\cos \alpha-\cos (\alpha+2 \gamma)=2 \sin \gamma \sin (\alpha+\gamma)=2 \sin \alpha \sin (2 \alpha+\gamma)=\cos (\alpha+$ $\gamma)-\cos (3 \alpha+\gamma)$. From $4 \alpha+\gamma=180^{\circ}$ we obtain $-\cos (3 \alpha+\gamma)=\cos \alpha$. Hence $$ \cos (\alpha+\gamma)+\cos (\alpha+2 \gamma)=2 \cos \frac{\gamma}{2} \cos \frac{2 \alpha+3 \gamma}{2}=0 $$ so that $2 \alpha+3 \gamma=180^{\circ}$. It follows that $\alpha=\gamma$. Second solution. We have $\angle B E C=\angle E C D=\angle D E C=\angle E C A=$ $\angle C A B$, and hence the trapezoid $B A E C$ is cyclic; consequently, $A E=$ $B C$. Similarly $A B=E D$, and $A B C D$ is cyclic as well. Thus $A B C D E$ is cyclic and has all sides equal; i.e., it is regular.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
17. (ISR 2) In the convex pentagon $A B C D E$, the sides $B C, C D, D E$ have the same length. Moreover, each diagonal of the pentagon is parallel to a side ( $A C$ is parallel to $D E, B D$ is parallel to $A E$, etc.). Prove that $A B C D E$ is a regular pentagon.
|
17. Let $A C$ and $A D$ meet $B E$ in $R, S$, respectively. Then by the conditions of the problem, $$ \begin{aligned} & \angle A E B=\angle E B D=\angle B D C=\angle D B C=\angle A D B=\angle E A D=\alpha, \\ & \angle A B E=\angle B E C=\angle E C D=\angle C E D=\angle A C E=\angle B A C=\beta \\ & \angle B C A=\angle C A D=\angle A D E=\gamma \end{aligned} $$ Since $\angle S A E=\angle S E A$, it follows that $A S=S E$, and analogously $B R=$ $R A$. But $B S D C$ and $R E D C$ are parallelograms; hence $B S=C D=R E$, giving us $B R=S E$ and $A R=A S$. Then also $A C=A D$, because $R S \|$ $C D$. We deduce that $2 \beta=\angle A C D=\angle A D C=2 \alpha$, i.e., $\alpha=\beta$. It will be sufficient to show that $\alpha=\gamma$, since that will imply $\alpha=\beta=\gamma=$ $36^{\circ}$. We have that the sum of the interior angles of $A C D$ is $4 \alpha+\gamma=180^{\circ}$. We have $$ \frac{\sin \gamma}{\sin \alpha}=\frac{A E}{D E}=\frac{A E}{C D}=\frac{A E}{R E}=\frac{\sin (2 \alpha+\gamma)}{\sin (\alpha+\gamma)} $$ i.e., $\cos \alpha-\cos (\alpha+2 \gamma)=2 \sin \gamma \sin (\alpha+\gamma)=2 \sin \alpha \sin (2 \alpha+\gamma)=\cos (\alpha+$ $\gamma)-\cos (3 \alpha+\gamma)$. From $4 \alpha+\gamma=180^{\circ}$ we obtain $-\cos (3 \alpha+\gamma)=\cos \alpha$. Hence $$ \cos (\alpha+\gamma)+\cos (\alpha+2 \gamma)=2 \cos \frac{\gamma}{2} \cos \frac{2 \alpha+3 \gamma}{2}=0 $$ so that $2 \alpha+3 \gamma=180^{\circ}$. It follows that $\alpha=\gamma$. Second solution. We have $\angle B E C=\angle E C D=\angle D E C=\angle E C A=$ $\angle C A B$, and hence the trapezoid $B A E C$ is cyclic; consequently, $A E=$ $B C$. Similarly $A B=E D$, and $A B C D$ is cyclic as well. Thus $A B C D E$ is cyclic and has all sides equal; i.e., it is regular.
|
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32244f60-60ba-5889-b01e-078823a8ea72
| 24,127
|
21. (POL 4) Forty-nine students solve a set of three problems. The score for each problem is a whole number of points from 0 to 7 . Prove that there exist two students $A$ and $B$ such that for each problem, $A$ will score at least as many points as $B$.
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21. Let $X$ be the set of all ordered triples $a=\left(a_{1}, a_{2}, a_{3}\right)$ for $a_{i} \in\{0,1, \ldots, 7\}$. Write $a \prec b$ if $a_{i} \leq b_{i}$ for $i=1,2,3$ and $a \neq b$. Call a subset $Y \subset X$ independent if there are no $a, b \in Y$ with $a \prec b$. We shall prove that an independent set contains at most 48 elements. For $j=0,1, \ldots, 21$ let $X_{j}=\left\{\left(a_{1}, a_{2}, a_{3}\right) \in X \mid a_{1}+a_{2}+a_{3}=j\right\}$. If $x \prec y$ and $x \in X_{j}, y \in X_{j+1}$ for some $j$, then we say that $y$ is a successor of $x$, and $x$ a predecessor of $y$. Lemma. If $A$ is an $m$-element subset of $X_{j}$ and $j \leq 10$, then there are at least $m$ distinct successors of the elements of $A$. Proof. For $k=0,1,2,3$ let $X_{j, k}=\left\{\left(a_{1}, a_{2}, a_{3}\right) \in X_{j} \mid \min \left(a_{1}, a_{2}, a_{3}, 7-\right.\right.$ $\left.\left.a_{1}, 7-a_{2}, 7-a_{3}\right)=k\right\}$. It is easy to verify that every element of $X_{j, k}$ has at least two successors in $X_{j+1, k}$ and every element of $X_{j+1, k}$ has at most two predecessors in $X_{j, k}$. Therefore the number of elements of $A \cap X_{j, k}$ is not greater than the number of their successors. Since $X_{j}$ is a disjoint union of $X_{j, k}, k=0,1,2,3$, the lemma follows. Similarly, elements of an $m$-element subset of $X_{j}, j \geq 11$, have at least $m$ predecessors. Let $Y$ be an independent set, and let $p, q$ be integers such that $p<10<q$. We can transform $Y$ by replacing all the elements of $Y \cap X_{p}$ with their successors, and all the elements of $Y \cap X_{q}$ with their predecessors. After this transformation $Y$ will still be independent, and by the lemma its size will not be reduced. Every independent set can be eventually transformed in this way into a subset of $X_{10}$, and $X_{10}$ has exactly 48 elements.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
21. (POL 4) Forty-nine students solve a set of three problems. The score for each problem is a whole number of points from 0 to 7 . Prove that there exist two students $A$ and $B$ such that for each problem, $A$ will score at least as many points as $B$.
|
21. Let $X$ be the set of all ordered triples $a=\left(a_{1}, a_{2}, a_{3}\right)$ for $a_{i} \in\{0,1, \ldots, 7\}$. Write $a \prec b$ if $a_{i} \leq b_{i}$ for $i=1,2,3$ and $a \neq b$. Call a subset $Y \subset X$ independent if there are no $a, b \in Y$ with $a \prec b$. We shall prove that an independent set contains at most 48 elements. For $j=0,1, \ldots, 21$ let $X_{j}=\left\{\left(a_{1}, a_{2}, a_{3}\right) \in X \mid a_{1}+a_{2}+a_{3}=j\right\}$. If $x \prec y$ and $x \in X_{j}, y \in X_{j+1}$ for some $j$, then we say that $y$ is a successor of $x$, and $x$ a predecessor of $y$. Lemma. If $A$ is an $m$-element subset of $X_{j}$ and $j \leq 10$, then there are at least $m$ distinct successors of the elements of $A$. Proof. For $k=0,1,2,3$ let $X_{j, k}=\left\{\left(a_{1}, a_{2}, a_{3}\right) \in X_{j} \mid \min \left(a_{1}, a_{2}, a_{3}, 7-\right.\right.$ $\left.\left.a_{1}, 7-a_{2}, 7-a_{3}\right)=k\right\}$. It is easy to verify that every element of $X_{j, k}$ has at least two successors in $X_{j+1, k}$ and every element of $X_{j+1, k}$ has at most two predecessors in $X_{j, k}$. Therefore the number of elements of $A \cap X_{j, k}$ is not greater than the number of their successors. Since $X_{j}$ is a disjoint union of $X_{j, k}, k=0,1,2,3$, the lemma follows. Similarly, elements of an $m$-element subset of $X_{j}, j \geq 11$, have at least $m$ predecessors. Let $Y$ be an independent set, and let $p, q$ be integers such that $p<10<q$. We can transform $Y$ by replacing all the elements of $Y \cap X_{p}$ with their successors, and all the elements of $Y \cap X_{q}$ with their predecessors. After this transformation $Y$ will still be independent, and by the lemma its size will not be reduced. Every independent set can be eventually transformed in this way into a subset of $X_{10}$, and $X_{10}$ has exactly 48 elements.
|
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a6feec24-35d5-54dc-9d9d-d1d8a93a5ae8
| 24,139
|
22. (KOR 2) Let $p$ be the product of two consecutive integers greater than 2. Show that there are no integers $x_{1}, x_{2}, \ldots, x_{p}$ satisfying the equation $$ \sum_{i=1}^{p} x_{i}^{2}-\frac{4}{4 p+1}\left(\sum_{i=1}^{p} x_{i}\right)^{2}=1 $$ Alternative formulation. Show that there are only two values of $p$ for which there are integers $x_{1}, x_{2}, \ldots, x_{p}$ satisfying the above inequality.
|
22. Set $X=\sum_{i=1}^{p} x_{i}$ and w.l.o.g. assume that $X \geq 0$ (if $\left(x_{1}, \ldots, x_{p}\right)$ is a solution, then $\left(-x_{1}, \ldots,-x_{p}\right)$ is a solution too). Since $x^{2} \geq x$ for all integers $x$, it follows that $\sum_{i=1}^{p} x_{i}^{2} \geq X$. If the last inequality is an equality, then all $x_{i}$ 's are 0 or 1 ; then, taking that there are $a 1$ 's, the equation becomes $4 p+1=4(a+1)+\frac{4}{a-1}$, which forces $p=6$ and $a=5$. Otherwise, we have $X+1 \leq \sum_{i=1}^{p} x_{i}^{2}=\frac{4}{4 p+1} X^{2}+1$, so $X \geq p+1$. Also, by the Cauchy-Schwarz inequality, $X^{2} \leq p \sum_{i=1}^{p} x_{i}^{2}=\frac{4 p}{4 p+1} X^{2}+p$, so $X^{2} \leq 4 p^{2}+p$ and $X \leq 2 p$. Thus $1 \leq X / p \leq 2$. However, $$ \begin{aligned} \sum_{i=1}^{p}\left(x_{i}-\frac{X}{p}\right)^{2} & =\sum x_{i}^{2}-\frac{2 X}{p} \sum x_{i}+\frac{X^{2}}{p} \\ & =\sum x_{i}^{2}-p \frac{X^{2}}{p^{2}}=1-\frac{X^{2}}{p(4 p+1)}<1 \end{aligned} $$ and we deduce that $-1<x_{i}-X / p<1$ for all $i$. This finally gives $x_{i} \in\{1,2\}$. Suppose there are $b 2$ 's. Then $3 b+p=4(b+p)^{2} /(4 p+1)+1$, so $p=b+1 /(4 b-3)$, which leads to $p=2, b=1$. Thus there are no solutions for any $p \notin\{2,6\}$. Remark. The condition $p=n(n+1), n \geq 3$, was unnecessary in the official solution, too (its only role was to simplify showing that $X \neq p-1$ ).
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
22. (KOR 2) Let $p$ be the product of two consecutive integers greater than 2. Show that there are no integers $x_{1}, x_{2}, \ldots, x_{p}$ satisfying the equation $$ \sum_{i=1}^{p} x_{i}^{2}-\frac{4}{4 p+1}\left(\sum_{i=1}^{p} x_{i}\right)^{2}=1 $$ Alternative formulation. Show that there are only two values of $p$ for which there are integers $x_{1}, x_{2}, \ldots, x_{p}$ satisfying the above inequality.
|
22. Set $X=\sum_{i=1}^{p} x_{i}$ and w.l.o.g. assume that $X \geq 0$ (if $\left(x_{1}, \ldots, x_{p}\right)$ is a solution, then $\left(-x_{1}, \ldots,-x_{p}\right)$ is a solution too). Since $x^{2} \geq x$ for all integers $x$, it follows that $\sum_{i=1}^{p} x_{i}^{2} \geq X$. If the last inequality is an equality, then all $x_{i}$ 's are 0 or 1 ; then, taking that there are $a 1$ 's, the equation becomes $4 p+1=4(a+1)+\frac{4}{a-1}$, which forces $p=6$ and $a=5$. Otherwise, we have $X+1 \leq \sum_{i=1}^{p} x_{i}^{2}=\frac{4}{4 p+1} X^{2}+1$, so $X \geq p+1$. Also, by the Cauchy-Schwarz inequality, $X^{2} \leq p \sum_{i=1}^{p} x_{i}^{2}=\frac{4 p}{4 p+1} X^{2}+p$, so $X^{2} \leq 4 p^{2}+p$ and $X \leq 2 p$. Thus $1 \leq X / p \leq 2$. However, $$ \begin{aligned} \sum_{i=1}^{p}\left(x_{i}-\frac{X}{p}\right)^{2} & =\sum x_{i}^{2}-\frac{2 X}{p} \sum x_{i}+\frac{X^{2}}{p} \\ & =\sum x_{i}^{2}-p \frac{X^{2}}{p^{2}}=1-\frac{X^{2}}{p(4 p+1)}<1 \end{aligned} $$ and we deduce that $-1<x_{i}-X / p<1$ for all $i$. This finally gives $x_{i} \in\{1,2\}$. Suppose there are $b 2$ 's. Then $3 b+p=4(b+p)^{2} /(4 p+1)+1$, so $p=b+1 /(4 b-3)$, which leads to $p=2, b=1$. Thus there are no solutions for any $p \notin\{2,6\}$. Remark. The condition $p=n(n+1), n \geq 3$, was unnecessary in the official solution, too (its only role was to simplify showing that $X \neq p-1$ ).
|
{
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|
35098702-1c43-5607-a909-489e09577d47
| 24,141
|
23. (SIN 2) Let $Q$ be the center of the inscribed circle of a triangle $A B C$. Prove that for any point $P$, $a(P A)^{2}+b(P B)^{2}+c(P C)^{2}=a(Q A)^{2}+b(Q B)^{2}+c(Q C)^{2}+(a+b+c)(Q P)^{2}$, where $a=B C, b=C A$, and $c=A B$.
|
23. Denote by $R$ the intersection point of lines $A Q$ and $B C$. We know that $B R: R C=c: b$ and $A Q: Q R=(b+c): a$. By applying Stewart's theorem to $\triangle P B C$ and $\triangle P A R$ we obtain $$ \begin{aligned} a \cdot A P^{2} & +b \cdot B P^{2}+c \cdot C P^{2}=a P A^{2}+(b+c) P R^{2}+(b+c) R B \cdot R C \\ & =(a+b+c) Q P^{2}+(b+c) R B \cdot R C+(a+b+c) Q A \cdot Q R \end{aligned} $$ On the other hand, putting $P=Q$ into (1), we get that $$ a \cdot A Q^{2}+b \cdot B Q^{2}+c \cdot C Q^{2}=(b+c) R B \cdot R C+(a+b+c) Q A \cdot Q R, $$ and the required statement follows. Second solution. At vertices $A, B, C$ place weights equal to $a, b, c$ in some units respectively, so that $Q$ is the center of gravity of the system. The left side of the equality to be proved is in fact the moment of inertia of the system about the axis through $P$ and perpendicular to the plane $A B C$. On the other side, the right side expresses the same, due to the parallel axes theorem. Alternative approach. Analytical geometry. The fact that all the variable segments appear squared usually implies that this is a good approach. Assign coordinates $A\left(x_{a}, y_{a}\right), B\left(x_{b}, y_{b}\right), C\left(x_{c}, y_{c}\right)$, and $P(x, y)$, use that $(a+b+c) \mathbf{Q}=a \mathbf{A}+b \mathbf{B}+c \mathbf{C}$, and calculate. Alternatively, differentiate $f(x, y)=a \cdot A P^{2}+b \cdot B P^{2}+c \cdot C P^{2}-(a+b+c) Q P^{2}$ and show that it is constant.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
23. (SIN 2) Let $Q$ be the center of the inscribed circle of a triangle $A B C$. Prove that for any point $P$, $a(P A)^{2}+b(P B)^{2}+c(P C)^{2}=a(Q A)^{2}+b(Q B)^{2}+c(Q C)^{2}+(a+b+c)(Q P)^{2}$, where $a=B C, b=C A$, and $c=A B$.
|
23. Denote by $R$ the intersection point of lines $A Q$ and $B C$. We know that $B R: R C=c: b$ and $A Q: Q R=(b+c): a$. By applying Stewart's theorem to $\triangle P B C$ and $\triangle P A R$ we obtain $$ \begin{aligned} a \cdot A P^{2} & +b \cdot B P^{2}+c \cdot C P^{2}=a P A^{2}+(b+c) P R^{2}+(b+c) R B \cdot R C \\ & =(a+b+c) Q P^{2}+(b+c) R B \cdot R C+(a+b+c) Q A \cdot Q R \end{aligned} $$ On the other hand, putting $P=Q$ into (1), we get that $$ a \cdot A Q^{2}+b \cdot B Q^{2}+c \cdot C Q^{2}=(b+c) R B \cdot R C+(a+b+c) Q A \cdot Q R, $$ and the required statement follows. Second solution. At vertices $A, B, C$ place weights equal to $a, b, c$ in some units respectively, so that $Q$ is the center of gravity of the system. The left side of the equality to be proved is in fact the moment of inertia of the system about the axis through $P$ and perpendicular to the plane $A B C$. On the other side, the right side expresses the same, due to the parallel axes theorem. Alternative approach. Analytical geometry. The fact that all the variable segments appear squared usually implies that this is a good approach. Assign coordinates $A\left(x_{a}, y_{a}\right), B\left(x_{b}, y_{b}\right), C\left(x_{c}, y_{c}\right)$, and $P(x, y)$, use that $(a+b+c) \mathbf{Q}=a \mathbf{A}+b \mathbf{B}+c \mathbf{C}$, and calculate. Alternatively, differentiate $f(x, y)=a \cdot A P^{2}+b \cdot B P^{2}+c \cdot C P^{2}-(a+b+c) Q P^{2}$ and show that it is constant.
|
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|
1663474f-666d-585b-93ca-fc772db6705f
| 24,143
|
24. (SWE 2) Let $\left\{a_{k}\right\}_{1}^{\infty}$ be a sequence of nonnegative real numbers such that $a_{k}-2 a_{k+1}+a_{k+2} \geq 0$ and $\sum_{j=1}^{k} a_{j} \leq 1$ for all $k=1,2, \ldots$. Prove that $0 \leq\left(a_{k}-a_{k+1}\right)<\frac{2}{k^{2}}$ for all $k=1,2, \ldots$.
|
24. The first condition means in fact that $a_{k}-a_{k+1}$ is decreasing. In particular, if $a_{k}-a_{k+1}=-\delta<0$, then $a_{k}-a_{k+m}=\left(a_{k}-a_{k+1}\right)+\cdots+\left(a_{k+m-1}-\right.$ $\left.a_{k+m}\right)<-m \delta$, which implies that $a_{k+m}>a_{k}+m \delta$, and consequently $a_{k+m}>1$ for large enough $m$, a contradiction. Thus $a_{k}-a_{k+1} \geq 0$ for all $k$. Suppose that $a_{k}-a_{k+1}>2 / k^{2}$. Then for all $i<k, a_{i}-a_{i+1}>2 / k^{2}$, so that $a_{i}-a_{k+1}>2(k+1-i) / k^{2}$, i.e., $a_{i}>2(k+1-i) / k^{2}, i=1,2, \ldots, k$. But this implies $a_{1}+a_{2}+\cdots+a_{k}>2 / k^{2}+4 / k^{2}+\cdots+2 k / k^{2}=k(k+1) / k^{2}$, which is impossible. Therefore $a_{k}-a_{k+1} \leq 2 / k^{2}$ for all $k$.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
24. (SWE 2) Let $\left\{a_{k}\right\}_{1}^{\infty}$ be a sequence of nonnegative real numbers such that $a_{k}-2 a_{k+1}+a_{k+2} \geq 0$ and $\sum_{j=1}^{k} a_{j} \leq 1$ for all $k=1,2, \ldots$. Prove that $0 \leq\left(a_{k}-a_{k+1}\right)<\frac{2}{k^{2}}$ for all $k=1,2, \ldots$.
|
24. The first condition means in fact that $a_{k}-a_{k+1}$ is decreasing. In particular, if $a_{k}-a_{k+1}=-\delta<0$, then $a_{k}-a_{k+m}=\left(a_{k}-a_{k+1}\right)+\cdots+\left(a_{k+m-1}-\right.$ $\left.a_{k+m}\right)<-m \delta$, which implies that $a_{k+m}>a_{k}+m \delta$, and consequently $a_{k+m}>1$ for large enough $m$, a contradiction. Thus $a_{k}-a_{k+1} \geq 0$ for all $k$. Suppose that $a_{k}-a_{k+1}>2 / k^{2}$. Then for all $i<k, a_{i}-a_{i+1}>2 / k^{2}$, so that $a_{i}-a_{k+1}>2(k+1-i) / k^{2}$, i.e., $a_{i}>2(k+1-i) / k^{2}, i=1,2, \ldots, k$. But this implies $a_{1}+a_{2}+\cdots+a_{k}>2 / k^{2}+4 / k^{2}+\cdots+2 k / k^{2}=k(k+1) / k^{2}$, which is impossible. Therefore $a_{k}-a_{k+1} \leq 2 / k^{2}$ for all $k$.
|
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|
981f28b7-43d8-59b6-9841-581bab9a60b2
| 24,146
|
25. (GBR 1) A positive integer is called a double number if its decimal representation consists of a block of digits, not commencing with 0 , followed immediately by an identical block. For instance, 360360 is a double number, but 36036 is not. Show that there are infinitely many double numbers that are perfect squares.
|
25. Observe that $1001=7 \cdot 143$, i.e., $10^{3}=-1+7 a, a=143$. Then by the binomial theorem, $10^{21}=(-1+7 a)^{7}=-1+7^{2} b$ for some integer $b$, so that we also have $10^{21 n} \equiv-1(\bmod 49)$ for any odd integer $n>0$. Hence $N=\frac{9}{49}\left(10^{21 n}+1\right)$ is an integer of $21 n$ digits, and $N\left(10^{21 n}+1\right)=$ $\left(\frac{3}{7}\left(10^{21 n}+1\right)\right)^{2}$ is a double number that is a perfect square.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
25. (GBR 1) A positive integer is called a double number if its decimal representation consists of a block of digits, not commencing with 0 , followed immediately by an identical block. For instance, 360360 is a double number, but 36036 is not. Show that there are infinitely many double numbers that are perfect squares.
|
25. Observe that $1001=7 \cdot 143$, i.e., $10^{3}=-1+7 a, a=143$. Then by the binomial theorem, $10^{21}=(-1+7 a)^{7}=-1+7^{2} b$ for some integer $b$, so that we also have $10^{21 n} \equiv-1(\bmod 49)$ for any odd integer $n>0$. Hence $N=\frac{9}{49}\left(10^{21 n}+1\right)$ is an integer of $21 n$ digits, and $N\left(10^{21 n}+1\right)=$ $\left(\frac{3}{7}\left(10^{21 n}+1\right)\right)^{2}$ is a double number that is a perfect square.
|
{
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|
cc488605-6756-5071-9c40-b178afd33fe5
| 24,148
|
27. (GBR 4) The triangle $A B C$ is acute-angled. Let $L$ be any line in the plane of the triangle and let $u, v, w$ be the lengths of the perpendiculars from $A, B, C$ respectively to $L$. Prove that $$ u^{2} \tan A+v^{2} \tan B+w^{2} \tan C \geq 2 \Delta $$ where $\Delta$ is the area of the triangle, and determine the lines $L$ for which equality holds.
|
27. Consider a Cartesian system with the $x$-axis on the line $B C$ and origin at the foot of the perpendicular from $A$ to $B C$, so that $A$ lies on the $y$-axis. Let $A$ be $(0, \alpha), B(-\beta, 0), C(\gamma, 0)$, where $\alpha, \beta, \gamma>0$ (because $A B C$ is acute-angled). Then $\tan B=\frac{\alpha}{\beta}, \quad \tan C=\frac{\alpha}{\gamma} \quad$ and $\quad \tan A=-\tan (B+C)=\frac{\alpha(\beta+\gamma)}{\alpha^{2}-\beta \gamma} ;$ here $\tan A>0$, so $\alpha^{2}>\beta \gamma$. Let $L$ have equation $x \cos \theta+y \sin \theta+p=0$. Then $$ \begin{aligned} & u^{2} \tan A+v^{2} \tan B+w^{2} \tan C \\ & =\frac{\alpha(\beta+\gamma)}{\alpha^{2}-\beta \gamma}(\alpha \sin \theta+p)^{2}+\frac{\alpha}{\beta}(-\beta \cos \theta+p)^{2}+\frac{\alpha}{\gamma}(\gamma \cos \theta+p)^{2} \\ & =\left(\alpha^{2} \sin ^{2} \theta+2 \alpha p \sin \theta+p^{2}\right) \frac{\alpha(\beta+\gamma)}{\alpha^{2}-\beta \gamma}+\alpha(\beta+\gamma) \cos ^{2} \theta+\frac{\alpha(\beta+\gamma)}{\beta \gamma} p^{2} \\ & =\frac{\alpha(\beta+\gamma)}{\beta \gamma\left(\alpha^{2}-\beta \gamma\right)}\left(\alpha^{2} p^{2}+2 \alpha p \beta \gamma \sin \theta+\alpha^{2} \beta \gamma \sin ^{2} \theta+\beta \gamma\left(\alpha^{2}-\beta \gamma\right) \cos ^{2} \theta\right) \\ & =\frac{\alpha(\beta+\gamma)}{\beta \gamma\left(\alpha^{2}-\beta \gamma\right)}\left[(\alpha p+\beta \gamma \sin \theta)^{2}+\beta \gamma\left(\alpha^{2}-\beta \gamma\right)\right] \geq \alpha(\beta+\gamma)=2 \Delta \end{aligned} $$ with equality when $\alpha p+\beta \gamma \sin \theta=0$, i.e., if and only if $L$ passes through $(0, \beta \gamma / \alpha)$, which is the orthocenter of the triangle.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
27. (GBR 4) The triangle $A B C$ is acute-angled. Let $L$ be any line in the plane of the triangle and let $u, v, w$ be the lengths of the perpendiculars from $A, B, C$ respectively to $L$. Prove that $$ u^{2} \tan A+v^{2} \tan B+w^{2} \tan C \geq 2 \Delta $$ where $\Delta$ is the area of the triangle, and determine the lines $L$ for which equality holds.
|
27. Consider a Cartesian system with the $x$-axis on the line $B C$ and origin at the foot of the perpendicular from $A$ to $B C$, so that $A$ lies on the $y$-axis. Let $A$ be $(0, \alpha), B(-\beta, 0), C(\gamma, 0)$, where $\alpha, \beta, \gamma>0$ (because $A B C$ is acute-angled). Then $\tan B=\frac{\alpha}{\beta}, \quad \tan C=\frac{\alpha}{\gamma} \quad$ and $\quad \tan A=-\tan (B+C)=\frac{\alpha(\beta+\gamma)}{\alpha^{2}-\beta \gamma} ;$ here $\tan A>0$, so $\alpha^{2}>\beta \gamma$. Let $L$ have equation $x \cos \theta+y \sin \theta+p=0$. Then $$ \begin{aligned} & u^{2} \tan A+v^{2} \tan B+w^{2} \tan C \\ & =\frac{\alpha(\beta+\gamma)}{\alpha^{2}-\beta \gamma}(\alpha \sin \theta+p)^{2}+\frac{\alpha}{\beta}(-\beta \cos \theta+p)^{2}+\frac{\alpha}{\gamma}(\gamma \cos \theta+p)^{2} \\ & =\left(\alpha^{2} \sin ^{2} \theta+2 \alpha p \sin \theta+p^{2}\right) \frac{\alpha(\beta+\gamma)}{\alpha^{2}-\beta \gamma}+\alpha(\beta+\gamma) \cos ^{2} \theta+\frac{\alpha(\beta+\gamma)}{\beta \gamma} p^{2} \\ & =\frac{\alpha(\beta+\gamma)}{\beta \gamma\left(\alpha^{2}-\beta \gamma\right)}\left(\alpha^{2} p^{2}+2 \alpha p \beta \gamma \sin \theta+\alpha^{2} \beta \gamma \sin ^{2} \theta+\beta \gamma\left(\alpha^{2}-\beta \gamma\right) \cos ^{2} \theta\right) \\ & =\frac{\alpha(\beta+\gamma)}{\beta \gamma\left(\alpha^{2}-\beta \gamma\right)}\left[(\alpha p+\beta \gamma \sin \theta)^{2}+\beta \gamma\left(\alpha^{2}-\beta \gamma\right)\right] \geq \alpha(\beta+\gamma)=2 \Delta \end{aligned} $$ with equality when $\alpha p+\beta \gamma \sin \theta=0$, i.e., if and only if $L$ passes through $(0, \beta \gamma / \alpha)$, which is the orthocenter of the triangle.
|
{
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|
3d9f8670-6d8b-5fed-bfb1-68354b7f9c18
| 24,153
|
28. (GBR 5) The sequence $\left\{a_{n}\right\}$ of integers is defined by $a_{1}=2, a_{2}=7$, and $$ -\frac{1}{2}<a_{n+1}-\frac{a_{n}^{2}}{a_{n-1}} \leq \frac{1}{2}, \quad \text { for } n \geq 2 $$ Prove that $a_{n}$ is odd for all $n>1$.
|
28. The sequence is uniquely determined by the conditions, and $a_{1}=2, a_{2}=$ $7, a_{3}=25, a_{4}=89, a_{5}=317, \ldots$; it satisfies $a_{n}=3 a_{n-1}+2 a_{n-2}$ for $n=3,4,5$. We show that the sequence $b_{n}$ given by $b_{1}=2, b_{2}=7$, $b_{n}=3 b_{n-1}+2 b_{n-2}$ has the same inequality property, i.e., that $b_{n}=a_{n}$ : $b_{n+1} b_{n-1}-b_{n}^{2}=\left(3 b_{n}+2 b_{n-1}\right) b_{n-1}-b_{n}\left(3 b_{n-1}+2 b_{n-2}\right)=-2\left(b_{n} b_{n-2}-b_{n-1}^{2}\right)$ for $n>2$ gives that $b_{n+1} b_{n-1}-b_{n}^{2}=(-2)^{n-2}$ for all $n \geq 2$. But then $$ \left|b_{n+1}-\frac{b_{n}^{2}}{b_{n-1}}\right|=\frac{2^{n-2}}{b_{n-1}}<\frac{1}{2}, $$ since it is easily shown that $b_{n-1}>2^{n-1}$ for all $n$. It is obvious that $a_{n}=b_{n}$ are odd for $n>1$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
28. (GBR 5) The sequence $\left\{a_{n}\right\}$ of integers is defined by $a_{1}=2, a_{2}=7$, and $$ -\frac{1}{2}<a_{n+1}-\frac{a_{n}^{2}}{a_{n-1}} \leq \frac{1}{2}, \quad \text { for } n \geq 2 $$ Prove that $a_{n}$ is odd for all $n>1$.
|
28. The sequence is uniquely determined by the conditions, and $a_{1}=2, a_{2}=$ $7, a_{3}=25, a_{4}=89, a_{5}=317, \ldots$; it satisfies $a_{n}=3 a_{n-1}+2 a_{n-2}$ for $n=3,4,5$. We show that the sequence $b_{n}$ given by $b_{1}=2, b_{2}=7$, $b_{n}=3 b_{n-1}+2 b_{n-2}$ has the same inequality property, i.e., that $b_{n}=a_{n}$ : $b_{n+1} b_{n-1}-b_{n}^{2}=\left(3 b_{n}+2 b_{n-1}\right) b_{n-1}-b_{n}\left(3 b_{n-1}+2 b_{n-2}\right)=-2\left(b_{n} b_{n-2}-b_{n-1}^{2}\right)$ for $n>2$ gives that $b_{n+1} b_{n-1}-b_{n}^{2}=(-2)^{n-2}$ for all $n \geq 2$. But then $$ \left|b_{n+1}-\frac{b_{n}^{2}}{b_{n-1}}\right|=\frac{2^{n-2}}{b_{n-1}}<\frac{1}{2}, $$ since it is easily shown that $b_{n-1}>2^{n-1}$ for all $n$. It is obvious that $a_{n}=b_{n}$ are odd for $n>1$.
|
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e7b6c498-aea3-5e16-90b4-23e7bc16a208
| 24,156
|
29. (USA 3) A number of signal lights are equally spaced along a one-way railroad track, labeled in order $1,2, \ldots, N(N \geq 2)$. As a safety rule, a train is not allowed to pass a signal if any other train is in motion on the length of track between it and the following signal. However, there is no limit to the number of trains that can be parked motionless at a signal, one behind the other. (Assume that the trains have zero length.) A series of $K$ freight trains must be driven from Signal 1 to Signal $N$. Each train travels at a distinct but constant speed (i.e., the speed is fixed and different from that of each of the other trains) at all times when it is not blocked by the safety rule. Show that regardless of the order in which the trains are arranged, the same time will elapse between the first train's departure from Signal 1 and the last train's arrival at Signal $N$.
|
29. Let the first train start from Signal 1 at time 0, and let $t_{j}$ be the time it takes for the $j$ th train in the series to travel from one signal to the next. By induction on $k$, we show that Train $k$ arrives at signal $n$ at time $s_{k}+(n-2) m_{k}$, where $s_{k}=t_{1}+\cdots+t_{k}$ and $m_{k}=\max _{j=1, \ldots, k} t_{j}$. For $k=1$ the statement is clear. We now suppose that it is true for $k$ trains and for every $n$, and add a $(k+1)$ th train behind the others at Signal 1. There are two cases to consider: (i) $t_{k+1} \geq m_{k}$, i.e., $m_{k+1}=t_{k+1}$. Then Train $k+1$ leaves Signal 1 when all the others reach Signal 2, which by the induction happens at time $s_{k}$. Since by the induction hypothesis Train $k$ arrives at Signal $i+1$ at time $s_{k}+(i-1) m_{k} \leq s_{k}+(i-1) t_{k+1}$, Train $k+1$ is never forced to stop. The journey finishes at time $s_{k}+(n-1) t_{k+1}=s_{k+1}+(n-2) m_{k+1}$. (ii) $t_{k+1}<m_{k}$, i.e., $m_{k+1}=m_{k}$. Train $k+1$ leaves Signal 1 at time $s_{k}$, and reaches Signal 2 at time $s_{k}+t_{k+1}$, but must wait there until all the other trains get to Signal 3, i.e., until time $s_{k}+m_{k}$ (by the induction hypothesis). So it reaches Signal 3 only at time $s_{k}+m_{k}+$ $t_{k+1}$. Similarly, it gets to Signal 4 at time $s_{k}+2 m_{k}+t_{k+1}$, etc. Thus the entire schedule finishes at time $s_{k}+(n-2) m_{k}+t_{k+1}=s_{k+1}+$ $(n-2) m_{k+1}$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
29. (USA 3) A number of signal lights are equally spaced along a one-way railroad track, labeled in order $1,2, \ldots, N(N \geq 2)$. As a safety rule, a train is not allowed to pass a signal if any other train is in motion on the length of track between it and the following signal. However, there is no limit to the number of trains that can be parked motionless at a signal, one behind the other. (Assume that the trains have zero length.) A series of $K$ freight trains must be driven from Signal 1 to Signal $N$. Each train travels at a distinct but constant speed (i.e., the speed is fixed and different from that of each of the other trains) at all times when it is not blocked by the safety rule. Show that regardless of the order in which the trains are arranged, the same time will elapse between the first train's departure from Signal 1 and the last train's arrival at Signal $N$.
|
29. Let the first train start from Signal 1 at time 0, and let $t_{j}$ be the time it takes for the $j$ th train in the series to travel from one signal to the next. By induction on $k$, we show that Train $k$ arrives at signal $n$ at time $s_{k}+(n-2) m_{k}$, where $s_{k}=t_{1}+\cdots+t_{k}$ and $m_{k}=\max _{j=1, \ldots, k} t_{j}$. For $k=1$ the statement is clear. We now suppose that it is true for $k$ trains and for every $n$, and add a $(k+1)$ th train behind the others at Signal 1. There are two cases to consider: (i) $t_{k+1} \geq m_{k}$, i.e., $m_{k+1}=t_{k+1}$. Then Train $k+1$ leaves Signal 1 when all the others reach Signal 2, which by the induction happens at time $s_{k}$. Since by the induction hypothesis Train $k$ arrives at Signal $i+1$ at time $s_{k}+(i-1) m_{k} \leq s_{k}+(i-1) t_{k+1}$, Train $k+1$ is never forced to stop. The journey finishes at time $s_{k}+(n-1) t_{k+1}=s_{k+1}+(n-2) m_{k+1}$. (ii) $t_{k+1}<m_{k}$, i.e., $m_{k+1}=m_{k}$. Train $k+1$ leaves Signal 1 at time $s_{k}$, and reaches Signal 2 at time $s_{k}+t_{k+1}$, but must wait there until all the other trains get to Signal 3, i.e., until time $s_{k}+m_{k}$ (by the induction hypothesis). So it reaches Signal 3 only at time $s_{k}+m_{k}+$ $t_{k+1}$. Similarly, it gets to Signal 4 at time $s_{k}+2 m_{k}+t_{k+1}$, etc. Thus the entire schedule finishes at time $s_{k}+(n-2) m_{k}+t_{k+1}=s_{k+1}+$ $(n-2) m_{k+1}$.
|
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|
ceb62031-26c1-508c-a392-fec6877b581e
| 24,159
|
3. (CAN 1) The triangle $A B C$ is inscribed in a circle. The interior bisectors of the angles $A, B$, and $C$ meet the circle again at $A^{\prime}, B^{\prime}$, and $C^{\prime}$ respectively. Prove that the area of triangle $A^{\prime} B^{\prime} C^{\prime}$ is greater than or equal to the area of triangle $A B C$.
|
3. Let $R$ be the circumradius, $r$ the inradius, $s$ the semiperimeter, $\Delta$ the area of $A B C$ and $\Delta^{\prime}$ the area of $A^{\prime} B^{\prime} C^{\prime}$. The angles of triangle $A^{\prime} B^{\prime} C^{\prime}$ are $A^{\prime}=90^{\circ}-A / 2, B^{\prime}=90^{\circ}-B / 2$, and $C^{\prime}=90^{\circ}-C / 2$, and hence $$ \Delta=2 R^{2} \sin A \sin B \sin C $$ and $\Delta^{\prime}=2 R^{2} \sin A^{\prime} \sin B^{\prime} \sin C^{\prime}=2 R^{2} \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$. Hence, $$ \frac{\Delta}{\Delta^{\prime}}=\frac{\sin A \sin B \sin C}{\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}=8 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}=\frac{2 r}{R} $$ where we have used that $r=A I \sin (A / 2)=\cdots=4 R \sin (A / 2) \cdot \sin (B / 2)$. $\sin (C / 2)$. Euler's inequality $2 r \leq R$ shows that $\Delta \leq \Delta^{\prime}$. Second solution. Let $H$ be orthocenter of triangle $A B C$, and $H_{a}, H_{b}, H_{c}$ points symmetric to $H$ with respect to $B C, C A, A B$, respectively. Since $\angle B H_{a} C=\angle B H C=180^{\circ}-\angle A$, points $H_{a}, H_{b}, H_{c}$ lie on the circumcircle of $A B C$, and the area of the hexagon $A H_{c} B H_{a} C H_{b}$ is double the area of $A B C$. (1) Let us apply the analogous result for the triangle $A^{\prime} B^{\prime} C^{\prime}$. Since its orthocenter is the incenter $I$ of $A B C$, and the point symmetric to $I$ with respect to $B^{\prime} C^{\prime}$ is the point $A$, we find by (1) that the area of the hexagon $A C^{\prime} B A^{\prime} C B^{\prime}$ is double the area of $A^{\prime} B^{\prime} C^{\prime}$. But it is clear that the area of $\Delta C H_{a} B$ is less than or equal to the area of $\Delta C A^{\prime} B$ etc.; hence, the area of $A H_{c} B H_{a} C H_{b}$ does not exceed the area of $A C^{\prime} B A^{\prime} C B^{\prime}$. The statement follows immediately.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
3. (CAN 1) The triangle $A B C$ is inscribed in a circle. The interior bisectors of the angles $A, B$, and $C$ meet the circle again at $A^{\prime}, B^{\prime}$, and $C^{\prime}$ respectively. Prove that the area of triangle $A^{\prime} B^{\prime} C^{\prime}$ is greater than or equal to the area of triangle $A B C$.
|
3. Let $R$ be the circumradius, $r$ the inradius, $s$ the semiperimeter, $\Delta$ the area of $A B C$ and $\Delta^{\prime}$ the area of $A^{\prime} B^{\prime} C^{\prime}$. The angles of triangle $A^{\prime} B^{\prime} C^{\prime}$ are $A^{\prime}=90^{\circ}-A / 2, B^{\prime}=90^{\circ}-B / 2$, and $C^{\prime}=90^{\circ}-C / 2$, and hence $$ \Delta=2 R^{2} \sin A \sin B \sin C $$ and $\Delta^{\prime}=2 R^{2} \sin A^{\prime} \sin B^{\prime} \sin C^{\prime}=2 R^{2} \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$. Hence, $$ \frac{\Delta}{\Delta^{\prime}}=\frac{\sin A \sin B \sin C}{\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}=8 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}=\frac{2 r}{R} $$ where we have used that $r=A I \sin (A / 2)=\cdots=4 R \sin (A / 2) \cdot \sin (B / 2)$. $\sin (C / 2)$. Euler's inequality $2 r \leq R$ shows that $\Delta \leq \Delta^{\prime}$. Second solution. Let $H$ be orthocenter of triangle $A B C$, and $H_{a}, H_{b}, H_{c}$ points symmetric to $H$ with respect to $B C, C A, A B$, respectively. Since $\angle B H_{a} C=\angle B H C=180^{\circ}-\angle A$, points $H_{a}, H_{b}, H_{c}$ lie on the circumcircle of $A B C$, and the area of the hexagon $A H_{c} B H_{a} C H_{b}$ is double the area of $A B C$. (1) Let us apply the analogous result for the triangle $A^{\prime} B^{\prime} C^{\prime}$. Since its orthocenter is the incenter $I$ of $A B C$, and the point symmetric to $I$ with respect to $B^{\prime} C^{\prime}$ is the point $A$, we find by (1) that the area of the hexagon $A C^{\prime} B A^{\prime} C B^{\prime}$ is double the area of $A^{\prime} B^{\prime} C^{\prime}$. But it is clear that the area of $\Delta C H_{a} B$ is less than or equal to the area of $\Delta C A^{\prime} B$ etc.; hence, the area of $A H_{c} B H_{a} C H_{b}$ does not exceed the area of $A C^{\prime} B A^{\prime} C B^{\prime}$. The statement follows immediately.
|
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|
9986d417-954f-5532-b18e-73a8ee60712a
| 24,162
|
31. (USS 2) Around a circular table an even number of persons have a discussion. After a break they sit again around the circular table in a different order. Prove that there are at least two people such that the number of participants sitting between them before and after the break is the same.
|
31. Denote the number of participants by $2 n$, and assign to each seat one of the numbers $1,2, \ldots, 2 n$. Let the participant who was sitting at the seat $k$ before the break move to seat $\pi(k)$. It suffices to prove that for every permutation $\pi$ of the set $\{1,2, \ldots, 2 n\}$, there exist distinct $i, j$ such that $\pi(i)-\pi(j)= \pm(i-j)$, the differences being calculated modulo $2 n$. If there are distinct $i$ and $j$ such that $\pi(i)-i=\pi(j)-j$ modulo $2 n$, then we are done. Suppose that all the differences $\pi(i)-i$ are distinct modulo $2 n$. Then they take values $0,1, \ldots, 2 n-1$ in some order, and consequently $$ \sum_{i=1}^{2 n}(\pi(i)-i)=0+1+\cdots+(2 n-1) \equiv n(2 n-1)(\bmod 2 n) $$ On the other hand, $\sum_{i=1}^{2 n}(\pi(i)-i)=\sum \pi(i)-\sum i=0$, which is a contradiction because $n(2 n-1)$ is not divisible by $2 n$. Remark. For an odd number of participants, the statement is false. For example, the permutation $(a, 2 a, \ldots,(2 n+1) a)$ of $(1,2, \ldots, 2 n+1)$ modulo $2 n+1$ does not satisfy the statement when $\operatorname{gcd}\left(a^{2}-1,2 n+1\right)=1$. Check that such an $a$ always exists.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
31. (USS 2) Around a circular table an even number of persons have a discussion. After a break they sit again around the circular table in a different order. Prove that there are at least two people such that the number of participants sitting between them before and after the break is the same.
|
31. Denote the number of participants by $2 n$, and assign to each seat one of the numbers $1,2, \ldots, 2 n$. Let the participant who was sitting at the seat $k$ before the break move to seat $\pi(k)$. It suffices to prove that for every permutation $\pi$ of the set $\{1,2, \ldots, 2 n\}$, there exist distinct $i, j$ such that $\pi(i)-\pi(j)= \pm(i-j)$, the differences being calculated modulo $2 n$. If there are distinct $i$ and $j$ such that $\pi(i)-i=\pi(j)-j$ modulo $2 n$, then we are done. Suppose that all the differences $\pi(i)-i$ are distinct modulo $2 n$. Then they take values $0,1, \ldots, 2 n-1$ in some order, and consequently $$ \sum_{i=1}^{2 n}(\pi(i)-i)=0+1+\cdots+(2 n-1) \equiv n(2 n-1)(\bmod 2 n) $$ On the other hand, $\sum_{i=1}^{2 n}(\pi(i)-i)=\sum \pi(i)-\sum i=0$, which is a contradiction because $n(2 n-1)$ is not divisible by $2 n$. Remark. For an odd number of participants, the statement is false. For example, the permutation $(a, 2 a, \ldots,(2 n+1) a)$ of $(1,2, \ldots, 2 n+1)$ modulo $2 n+1$ does not satisfy the statement when $\operatorname{gcd}\left(a^{2}-1,2 n+1\right)=1$. Check that such an $a$ always exists.
|
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d7a6420c-f654-5065-8970-e4d565fb9563
| 24,166
|
4. (CZS 1) An $n \times n$ chessboard $(n \geq 2)$ is numbered by the numbers $1,2, \ldots, n^{2}$ (every number occurs once). Prove that there exist two neighboring (which share a common edge) squares such that their numbers differ by at least $n$.
|
4. Suppose that the numbers of any two neighboring squares differ by at most $n-1$. For $k=1,2, \ldots, n^{2}-n$, let $A_{k}, B_{k}$, and $C_{k}$ denote, respectively, the sets of squares numbered by $1,2, \ldots, k$; of squares numbered by $k+$ $n, k+n+1, \ldots, n^{2}$; and of squares numbered by $k+1, \ldots, k+n-1$. By the assumption, the squares from $A_{k}$ and $B_{k}$ have no edge in common; $C_{k}$ has $n-1$ elements only. Consequently, for each $k$ there exists a row and a column all belonging either to $A_{k}$, or to $B_{k}$. For $k=1$, it must belong to $B_{k}$, while for $k=n^{2}-n$ it belongs to $A_{k}$. Let $k$ be the smallest index such that $A_{k}$ contains a whole row and a whole column. Since $B_{k-1}$ has that property too, it must have at least two squares in common with $A_{k}$, which is impossible.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
4. (CZS 1) An $n \times n$ chessboard $(n \geq 2)$ is numbered by the numbers $1,2, \ldots, n^{2}$ (every number occurs once). Prove that there exist two neighboring (which share a common edge) squares such that their numbers differ by at least $n$.
|
4. Suppose that the numbers of any two neighboring squares differ by at most $n-1$. For $k=1,2, \ldots, n^{2}-n$, let $A_{k}, B_{k}$, and $C_{k}$ denote, respectively, the sets of squares numbered by $1,2, \ldots, k$; of squares numbered by $k+$ $n, k+n+1, \ldots, n^{2}$; and of squares numbered by $k+1, \ldots, k+n-1$. By the assumption, the squares from $A_{k}$ and $B_{k}$ have no edge in common; $C_{k}$ has $n-1$ elements only. Consequently, for each $k$ there exists a row and a column all belonging either to $A_{k}$, or to $B_{k}$. For $k=1$, it must belong to $B_{k}$, while for $k=n^{2}-n$ it belongs to $A_{k}$. Let $k$ be the smallest index such that $A_{k}$ contains a whole row and a whole column. Since $B_{k-1}$ has that property too, it must have at least two squares in common with $A_{k}$, which is impossible.
|
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|
9c093677-de4d-52ef-bb65-6da8f9cf843f
| 24,170
|
6. (CZS 3) In a given tetrahedron $A B C D$ let $K$ and $L$ be the centers of edges $A B$ and $C D$ respectively. Prove that every plane that contains the line $K L$ divides the tetrahedron into two parts of equal volume.
|
6. Let $\omega$ be the plane through $A B$, parallel to $C D$. Define the point transformation $f: X \mapsto X^{\prime}$ in space as follows. If $X \in K L$, then $X^{\prime}=X$; otherwise, let $\omega_{X}$ be the plane through $X$ parallel to $\omega$ : then $X^{\prime}$ is the point symmetric to $X$ with respect to the intersection point of $K L$ with $\omega_{X}$. Clearly, $f(A)=B, f(B)=A, f(C)=D, f(D)=C$; hence $f$ maps the tetrahedron onto itself. We shall show that $f$ preserves volumes. Let $s: X \mapsto X^{\prime \prime}$ denote the symmetry with respect to $K L$, and $g$ the transformation mapping $X^{\prime \prime}$ into $X^{\prime}$; then $f=g \circ s$. If points $X_{1}^{\prime \prime}=s\left(X_{1}\right)$ and $X_{2}^{\prime \prime}=s\left(X_{2}\right)$ have the property that $X_{1}^{\prime \prime} X_{2}^{\prime \prime}$ is parallel to $K L$, then the segments $X_{1}^{\prime \prime} X_{2}^{\prime \prime}$ and $X_{1}^{\prime} X_{2}^{\prime}$ have the same length and lie on the same line. Then by Cavalieri's principle $g$ preserves volume, and so does $f$. Now, if $\alpha$ is any plane containing the line $K L$, the two parts of the tetrahedron on which it is partitioned by $\alpha$ are transformed into each other by $f$, and therefore have the same volumes. Second solution. Suppose w.l.o.g. that the plane $\alpha$ through $K L$ meets the interiors of edges $A C$ and $B D$ at $X$ and $Y$. Let $\overrightarrow{A X}=\lambda \overrightarrow{A C}$ and $\overrightarrow{B Y}=\mu \overrightarrow{B D}$, for $0 \leq \lambda, \mu \leq 1$. Then the vectors $\overrightarrow{K X}=\lambda \overrightarrow{A C}-\overrightarrow{A B} / 2$, $\overrightarrow{K Y}=\mu \overrightarrow{B D}+\overrightarrow{A B} / 2, \overrightarrow{K L}=\overrightarrow{A C} / 2+$ $\overrightarrow{B D} / 2$ are coplanar; i.e., there exist real numbers $a, b, c$, not all zero, such that  $$ \overrightarrow{0}=a \overrightarrow{K X}+b \overrightarrow{K Y}+c \overrightarrow{K L}=(\lambda a+c / 2) \overrightarrow{A C}+(\mu b+c / 2) \overrightarrow{B D}+\frac{b-a}{2} \overrightarrow{A B} $$ Since $\overrightarrow{A C}, \overrightarrow{B D}, \overrightarrow{A B}$ are linearly independent, we must have $a=b$ and $\lambda=\mu$. We need to prove that the volume of the polyhedron $K X L Y B C$, which is one of the parts of the tetrahedron $A B C D$ partitioned by $\alpha$, equals half of the volume $V$ of $A B C D$. Indeed, we obtain $$ V_{K X L Y B C}=V_{K X L C}+V_{K B Y L C}=\frac{1}{4}(1-\lambda) V+\frac{1}{4}(1+\mu) V=\frac{1}{2} V . $$
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
6. (CZS 3) In a given tetrahedron $A B C D$ let $K$ and $L$ be the centers of edges $A B$ and $C D$ respectively. Prove that every plane that contains the line $K L$ divides the tetrahedron into two parts of equal volume.
|
6. Let $\omega$ be the plane through $A B$, parallel to $C D$. Define the point transformation $f: X \mapsto X^{\prime}$ in space as follows. If $X \in K L$, then $X^{\prime}=X$; otherwise, let $\omega_{X}$ be the plane through $X$ parallel to $\omega$ : then $X^{\prime}$ is the point symmetric to $X$ with respect to the intersection point of $K L$ with $\omega_{X}$. Clearly, $f(A)=B, f(B)=A, f(C)=D, f(D)=C$; hence $f$ maps the tetrahedron onto itself. We shall show that $f$ preserves volumes. Let $s: X \mapsto X^{\prime \prime}$ denote the symmetry with respect to $K L$, and $g$ the transformation mapping $X^{\prime \prime}$ into $X^{\prime}$; then $f=g \circ s$. If points $X_{1}^{\prime \prime}=s\left(X_{1}\right)$ and $X_{2}^{\prime \prime}=s\left(X_{2}\right)$ have the property that $X_{1}^{\prime \prime} X_{2}^{\prime \prime}$ is parallel to $K L$, then the segments $X_{1}^{\prime \prime} X_{2}^{\prime \prime}$ and $X_{1}^{\prime} X_{2}^{\prime}$ have the same length and lie on the same line. Then by Cavalieri's principle $g$ preserves volume, and so does $f$. Now, if $\alpha$ is any plane containing the line $K L$, the two parts of the tetrahedron on which it is partitioned by $\alpha$ are transformed into each other by $f$, and therefore have the same volumes. Second solution. Suppose w.l.o.g. that the plane $\alpha$ through $K L$ meets the interiors of edges $A C$ and $B D$ at $X$ and $Y$. Let $\overrightarrow{A X}=\lambda \overrightarrow{A C}$ and $\overrightarrow{B Y}=\mu \overrightarrow{B D}$, for $0 \leq \lambda, \mu \leq 1$. Then the vectors $\overrightarrow{K X}=\lambda \overrightarrow{A C}-\overrightarrow{A B} / 2$, $\overrightarrow{K Y}=\mu \overrightarrow{B D}+\overrightarrow{A B} / 2, \overrightarrow{K L}=\overrightarrow{A C} / 2+$ $\overrightarrow{B D} / 2$ are coplanar; i.e., there exist real numbers $a, b, c$, not all zero, such that  $$ \overrightarrow{0}=a \overrightarrow{K X}+b \overrightarrow{K Y}+c \overrightarrow{K L}=(\lambda a+c / 2) \overrightarrow{A C}+(\mu b+c / 2) \overrightarrow{B D}+\frac{b-a}{2} \overrightarrow{A B} $$ Since $\overrightarrow{A C}, \overrightarrow{B D}, \overrightarrow{A B}$ are linearly independent, we must have $a=b$ and $\lambda=\mu$. We need to prove that the volume of the polyhedron $K X L Y B C$, which is one of the parts of the tetrahedron $A B C D$ partitioned by $\alpha$, equals half of the volume $V$ of $A B C D$. Indeed, we obtain $$ V_{K X L Y B C}=V_{K X L C}+V_{K B Y L C}=\frac{1}{4}(1-\lambda) V+\frac{1}{4}(1+\mu) V=\frac{1}{2} V . $$
|
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03c7bf52-c5b7-5a6b-a90e-f9e70d0d067f
| 24,175
|
7. (FRA 2) Let $a$ be the greatest positive root of the equation $x^{3}-3 x^{2}+1=$ 0 . Show that $\left[a^{1788}\right]$ and $\left[a^{1988}\right]$ are both divisible by 17 . ( $[x]$ denotes the integer part of $x$.)
|
7. The algebraic equation $x^{3}-3 x^{2}+1=0$ admits three real roots $\beta, \gamma, a$, with $$ -0.6<\beta<-0.5, \quad 0.6<\gamma<0.7, \quad \sqrt{8}<a<3 . $$ Define, for all integers $n$, $$ u_{n}=\beta^{n}+\gamma^{n}+a^{n} . $$ It holds that $u_{n+3}=3 u_{n+2}-u_{n}$. Obviously, $0<\beta^{n}+\gamma^{n}<1$ for all $n \geq 2$, and we see that $u_{n}-1=\left[a^{n}\right]$ for $n \geq 2$. It is now a question whether $u_{1788}-1$ and $u_{1988}-1$ are divisible by 17 . Working modulo 17 , we get $u_{0} \equiv 3, u_{1} \equiv 3, u_{2} \equiv 9, u_{3} \equiv 7, u_{4} \equiv$ $1, \ldots, u_{16}=3, u_{17}=3, u_{18}=9$. Thus, $u_{n}$ is periodic modulo 17 , with period 16. Since $1788=16 \cdot 111+12,1988=16 \cdot 124+4$, it follows that $u_{1788} \equiv u_{12} \equiv 1$ and $u_{1988} \equiv u_{4}=1$. So, $\left[a^{1788}\right]$ and $\left[a^{1988}\right]$ are divisible by 17 . Second solution. The polynomial $x^{3}-3 x^{2}+1$ allows the factorization modulo 17 as $(x-4)(x-5)(x+6)$. Hence it is easily seen that $u_{n} \equiv$ $4^{n}+5^{n}+(-6)^{n}$. Fermat's theorem gives us $4^{n} \equiv 5^{n} \equiv(-6)^{n} \equiv 1$ for $16 \mid n$, and the rest follows easily. Remark. In fact, the roots of $x^{3}-3 x^{2}+1=0$ are $\frac{1}{2 \sin 10^{\circ}}, \frac{1}{2 \sin 50^{\circ}}$, and $-\frac{1}{2 \sin 70^{\circ}}$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
7. (FRA 2) Let $a$ be the greatest positive root of the equation $x^{3}-3 x^{2}+1=$ 0 . Show that $\left[a^{1788}\right]$ and $\left[a^{1988}\right]$ are both divisible by 17 . ( $[x]$ denotes the integer part of $x$.)
|
7. The algebraic equation $x^{3}-3 x^{2}+1=0$ admits three real roots $\beta, \gamma, a$, with $$ -0.6<\beta<-0.5, \quad 0.6<\gamma<0.7, \quad \sqrt{8}<a<3 . $$ Define, for all integers $n$, $$ u_{n}=\beta^{n}+\gamma^{n}+a^{n} . $$ It holds that $u_{n+3}=3 u_{n+2}-u_{n}$. Obviously, $0<\beta^{n}+\gamma^{n}<1$ for all $n \geq 2$, and we see that $u_{n}-1=\left[a^{n}\right]$ for $n \geq 2$. It is now a question whether $u_{1788}-1$ and $u_{1988}-1$ are divisible by 17 . Working modulo 17 , we get $u_{0} \equiv 3, u_{1} \equiv 3, u_{2} \equiv 9, u_{3} \equiv 7, u_{4} \equiv$ $1, \ldots, u_{16}=3, u_{17}=3, u_{18}=9$. Thus, $u_{n}$ is periodic modulo 17 , with period 16. Since $1788=16 \cdot 111+12,1988=16 \cdot 124+4$, it follows that $u_{1788} \equiv u_{12} \equiv 1$ and $u_{1988} \equiv u_{4}=1$. So, $\left[a^{1788}\right]$ and $\left[a^{1988}\right]$ are divisible by 17 . Second solution. The polynomial $x^{3}-3 x^{2}+1$ allows the factorization modulo 17 as $(x-4)(x-5)(x+6)$. Hence it is easily seen that $u_{n} \equiv$ $4^{n}+5^{n}+(-6)^{n}$. Fermat's theorem gives us $4^{n} \equiv 5^{n} \equiv(-6)^{n} \equiv 1$ for $16 \mid n$, and the rest follows easily. Remark. In fact, the roots of $x^{3}-3 x^{2}+1=0$ are $\frac{1}{2 \sin 10^{\circ}}, \frac{1}{2 \sin 50^{\circ}}$, and $-\frac{1}{2 \sin 70^{\circ}}$.
|
{
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|
a623dfad-2098-57a6-b699-3ac1de818bb9
| 24,177
|
8. (FRA 3) Let $u_{1}, u_{2}, \ldots, u_{m}$ be $m$ vectors in the plane, each of length less than or equal to 1 , which add up to zero. Show that one can rearrange $u_{1}, u_{2}, \ldots, u_{m}$ as a sequence $v_{1}, v_{2}, \ldots, v_{m}$ such that each partial sum $v_{1}, v_{1}+v_{2}, v_{1}+v_{2}+v_{3}, \ldots, v_{1}+v_{2}+\cdots+v_{m}$ has length less than or equal to $\sqrt{5}$.
|
8. Consider first the case that the vectors are on the same line. Then if $e$ is a unit vector, we can write $u_{1}=x_{1} e, \ldots, u_{n}=x_{n} e$ for scalars $x_{i},\left|x_{i}\right| \leq 1$, with zero sum. It is now easy to permute $x_{1}, x_{2}, \ldots, x_{n}$ into $z_{1}, z_{2}, \ldots z_{n}$ so that $\left|z_{1}\right| \leq 1,\left|z_{1}+z_{2}\right| \leq 1, \ldots,\left|z_{1}+z_{2}+\cdots+z_{n-1}\right| \leq 1$. Indeed, suppose w.l.o.g. that $z_{1}=x_{1} \geq 0$; then we choose $z_{2}, \ldots, z_{r}$ from the $x_{i}$ 's to be negative, until we get to the first $r$ with $x_{1}+x_{2}+\cdots+x_{r} \leq 0$; we continue successively choosing positive $z_{j}$ 's from the remaining $x_{i}$ 's until we get the first partial sum that is positive, and so on. It is easy to verify that $\left|z_{1}+z_{2}+\cdots+z_{j}\right| \leq 1$ for all $j=1,2, \ldots, n$. Now we pass to the general case. Let $s$ be the longest vector that can be obtained by summing a subset of $u_{1}, \ldots, u_{m}$, and assume w.l.o.g. that $s=u_{1}+\cdots+u_{p}$. Further, let $\delta$ and $\delta^{\prime}$ respectively be the lines through the origin $O$ in the direction of $s$ and perpendicular to $s$, and $e, e^{\prime}$ respectively the unit vectors on $\delta$ and $\delta^{\prime}$. Put $u_{i}=x_{i} e+y_{i} e^{\prime}$, $i=1,2, \ldots, m$. By the definition of $\delta$ and $\delta^{\prime}$, we have $\left|x_{i}\right|,\left|y_{i}\right| \leq 1$; $x_{1}+\cdots+x_{m}=y_{1}+\cdots+y_{m}=0 ; y_{1}+\cdots+y_{p}=y_{p+1}+\cdots+y_{m}=0$; we also have $x_{p+1}, \ldots, x_{m} \leq 0$ (otherwise, if $x_{i}>0$ for some $i$, then $\left|s+v_{i}\right|>|s|$ ), and similarly $x_{1}, \ldots, x_{p} \geq 0$. Finally, suppose by the one-dimensional case that $y_{1}, \ldots, y_{p}$ and $y_{p+1}, \ldots, y_{m}$ are permuted in such a way that all the sums $y_{1}+\cdots+y_{i}$ and $y_{p+1}+\cdots+y_{p+i}$ are $\leq 1$ in absolute value. We apply the construction of the one-dimensional case to $x_{1}, \ldots, x_{m}$ taking, as described above, positive $z_{i}$ 's from $x_{1}, x_{2}, \ldots, x_{p}$ and negative ones from $x_{p+1}, \ldots, x_{m}$, but so that the order is preserved; this way we get a permutation $x_{\sigma_{1}}, x_{\sigma_{2}}, \ldots, x_{\sigma_{m}}$. It is then clear that each sum $y_{\sigma_{1}}+y_{\sigma_{2}}+$ $\cdots+y_{\sigma_{k}}$ decomposes into the sum $\left(y_{1}+y_{2}+\cdots+y_{l}\right)+\left(y_{p+1}+\cdots+y_{p+n}\right)$ (because of the preservation of order), and that each of these sums is less than or equal to 1 in absolute value. Thus each sum $u_{\sigma_{1}}+\cdots+u_{\sigma_{k}}$ is composed of a vector of length at most 2 and an orthogonal vector of length at most 1 , and so is itself of length at most $\sqrt{5}$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
8. (FRA 3) Let $u_{1}, u_{2}, \ldots, u_{m}$ be $m$ vectors in the plane, each of length less than or equal to 1 , which add up to zero. Show that one can rearrange $u_{1}, u_{2}, \ldots, u_{m}$ as a sequence $v_{1}, v_{2}, \ldots, v_{m}$ such that each partial sum $v_{1}, v_{1}+v_{2}, v_{1}+v_{2}+v_{3}, \ldots, v_{1}+v_{2}+\cdots+v_{m}$ has length less than or equal to $\sqrt{5}$.
|
8. Consider first the case that the vectors are on the same line. Then if $e$ is a unit vector, we can write $u_{1}=x_{1} e, \ldots, u_{n}=x_{n} e$ for scalars $x_{i},\left|x_{i}\right| \leq 1$, with zero sum. It is now easy to permute $x_{1}, x_{2}, \ldots, x_{n}$ into $z_{1}, z_{2}, \ldots z_{n}$ so that $\left|z_{1}\right| \leq 1,\left|z_{1}+z_{2}\right| \leq 1, \ldots,\left|z_{1}+z_{2}+\cdots+z_{n-1}\right| \leq 1$. Indeed, suppose w.l.o.g. that $z_{1}=x_{1} \geq 0$; then we choose $z_{2}, \ldots, z_{r}$ from the $x_{i}$ 's to be negative, until we get to the first $r$ with $x_{1}+x_{2}+\cdots+x_{r} \leq 0$; we continue successively choosing positive $z_{j}$ 's from the remaining $x_{i}$ 's until we get the first partial sum that is positive, and so on. It is easy to verify that $\left|z_{1}+z_{2}+\cdots+z_{j}\right| \leq 1$ for all $j=1,2, \ldots, n$. Now we pass to the general case. Let $s$ be the longest vector that can be obtained by summing a subset of $u_{1}, \ldots, u_{m}$, and assume w.l.o.g. that $s=u_{1}+\cdots+u_{p}$. Further, let $\delta$ and $\delta^{\prime}$ respectively be the lines through the origin $O$ in the direction of $s$ and perpendicular to $s$, and $e, e^{\prime}$ respectively the unit vectors on $\delta$ and $\delta^{\prime}$. Put $u_{i}=x_{i} e+y_{i} e^{\prime}$, $i=1,2, \ldots, m$. By the definition of $\delta$ and $\delta^{\prime}$, we have $\left|x_{i}\right|,\left|y_{i}\right| \leq 1$; $x_{1}+\cdots+x_{m}=y_{1}+\cdots+y_{m}=0 ; y_{1}+\cdots+y_{p}=y_{p+1}+\cdots+y_{m}=0$; we also have $x_{p+1}, \ldots, x_{m} \leq 0$ (otherwise, if $x_{i}>0$ for some $i$, then $\left|s+v_{i}\right|>|s|$ ), and similarly $x_{1}, \ldots, x_{p} \geq 0$. Finally, suppose by the one-dimensional case that $y_{1}, \ldots, y_{p}$ and $y_{p+1}, \ldots, y_{m}$ are permuted in such a way that all the sums $y_{1}+\cdots+y_{i}$ and $y_{p+1}+\cdots+y_{p+i}$ are $\leq 1$ in absolute value. We apply the construction of the one-dimensional case to $x_{1}, \ldots, x_{m}$ taking, as described above, positive $z_{i}$ 's from $x_{1}, x_{2}, \ldots, x_{p}$ and negative ones from $x_{p+1}, \ldots, x_{m}$, but so that the order is preserved; this way we get a permutation $x_{\sigma_{1}}, x_{\sigma_{2}}, \ldots, x_{\sigma_{m}}$. It is then clear that each sum $y_{\sigma_{1}}+y_{\sigma_{2}}+$ $\cdots+y_{\sigma_{k}}$ decomposes into the sum $\left(y_{1}+y_{2}+\cdots+y_{l}\right)+\left(y_{p+1}+\cdots+y_{p+n}\right)$ (because of the preservation of order), and that each of these sums is less than or equal to 1 in absolute value. Thus each sum $u_{\sigma_{1}}+\cdots+u_{\sigma_{k}}$ is composed of a vector of length at most 2 and an orthogonal vector of length at most 1 , and so is itself of length at most $\sqrt{5}$.
|
{
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|
ed2738fb-cbf4-52ba-a4e6-b9d41b0b37fd
| 24,180
|
9. (FRG 1) ${ }^{\text {IMO6 }}$ Let $a$ and $b$ be two positive integers such that $a b+1$ divides $a^{2}+b^{2}$. Show that $\frac{a^{2}+b^{2}}{a b+1}$ is a perfect square.
|
9. Let us assume $\frac{a^{2}+b^{2}}{a b+1}=k \in \mathbb{N}$. We then have $a^{2}-k a b+b^{2}=k$. Let us assume that $k$ is not an integer square, which implies $k \geq 2$. Now we observe the minimal pair $(a, b)$ such that $a^{2}-k a b+b^{2}=k$ holds. We may assume w.l.o.g. that $a \geq b$. For $a=b$ we get $k=(2-k) a^{2} \leq 0$; hence we must have $a>b$. Let us observe the quadratic equation $x^{2}-k b x+b^{2}-k=0$, which has solutions $a$ and $a_{1}$. Since $a+a_{1}=k b$, it follows that $a_{1} \in \mathbb{Z}$. Since $a>k b$ implies $k>a+b^{2}>k b$ and $a=k b$ implies $k=b^{2}$, it follows that $a<k b$ and thus $b^{2}>k$. Since $a a_{1}=b^{2}-k>0$ and $a>0$, it follows that $a_{1} \in \mathbb{N}$ and $a_{1}=\frac{b^{2}-k}{a}<\frac{a^{2}-1}{a}<a$. We have thus found an integer pair $\left(a_{1}, b\right)$ with $0<a_{1}<a$ that satisfies the original equation. This is a contradiction of the initial assumption that $(a, b)$ is minimal. Hence $k$ must be an integer square.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
9. (FRG 1) ${ }^{\text {IMO6 }}$ Let $a$ and $b$ be two positive integers such that $a b+1$ divides $a^{2}+b^{2}$. Show that $\frac{a^{2}+b^{2}}{a b+1}$ is a perfect square.
|
9. Let us assume $\frac{a^{2}+b^{2}}{a b+1}=k \in \mathbb{N}$. We then have $a^{2}-k a b+b^{2}=k$. Let us assume that $k$ is not an integer square, which implies $k \geq 2$. Now we observe the minimal pair $(a, b)$ such that $a^{2}-k a b+b^{2}=k$ holds. We may assume w.l.o.g. that $a \geq b$. For $a=b$ we get $k=(2-k) a^{2} \leq 0$; hence we must have $a>b$. Let us observe the quadratic equation $x^{2}-k b x+b^{2}-k=0$, which has solutions $a$ and $a_{1}$. Since $a+a_{1}=k b$, it follows that $a_{1} \in \mathbb{Z}$. Since $a>k b$ implies $k>a+b^{2}>k b$ and $a=k b$ implies $k=b^{2}$, it follows that $a<k b$ and thus $b^{2}>k$. Since $a a_{1}=b^{2}-k>0$ and $a>0$, it follows that $a_{1} \in \mathbb{N}$ and $a_{1}=\frac{b^{2}-k}{a}<\frac{a^{2}-1}{a}<a$. We have thus found an integer pair $\left(a_{1}, b\right)$ with $0<a_{1}<a$ that satisfies the original equation. This is a contradiction of the initial assumption that $(a, b)$ is minimal. Hence $k$ must be an integer square.
|
{
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|
ba236fa5-5e34-54b7-b4b0-8f9c7fe84d9e
| 24,183
|
1. (AUS 2) ${ }^{\mathrm{IMO} 2}$ Let $A B C$ be a triangle. The bisector of angle $A$ meets the circumcircle of triangle $A B C$ in $A_{1}$. Points $B_{1}$ and $C_{1}$ are defined similarly. Let $A A_{1}$ meet the lines that bisect the two external angles at $B$ and $C$ in point $A^{0}$. Define $B^{0}$ and $C^{0}$ similarly. If $S_{X_{1} X_{2} \ldots X_{n}}$ denotes the area of the polygon $X_{1} X_{2} \ldots X_{n}$, prove that $$ S_{A^{0} B^{0} C^{0}}=2 S_{A C_{1} B A_{1} C B_{1}} \geq 4 S_{A B C} . $$
|
1. Let $I$ denote the intersection of the three internal bisectors. Then $I A_{1}=A_{1} A^{0}$. One way proving this is to realize that the circumcircle of $A B C$ is the nine-point circle of $A^{0} B^{0} C^{0}$, hence it bisects $I A^{0}$, since $I$ is the orthocenter of $A^{0} B^{0} C^{0}$. Another way is through noting that $I A_{1}=A_{1} B$, which follows from $\angle A_{1} I B=\angle I B A_{1}=(\angle A+\angle B) / 2$, and $A_{1} B=A_{1} A^{0}$ which follows from $\angle A_{1} A^{0} B=\angle A_{1} B A^{0}=90^{\circ}-$  $\angle I B A_{1}$. Hence, we obtain $S_{I A_{1} B}=S_{A^{0} A_{1} B}$. Repeating this argument for the six triangles that have a vertex at $I$ and adding them up gives us $S_{A^{0} B^{0} C^{0}}=2 S_{A C_{1} B A_{1} C B_{1}}$. To prove $S_{A C_{1} B A_{1} C B_{1}} \geq 2 S_{A B C}$, draw the three altitudes in triangle $A B C$ intersecting in $H$. Let $X, Y$, and $Z$ be the symmetric points of $H$ with respect the sides $B C, C A$, and $A B$, respectively. Then $X, Y, Z$ are points on the circumcircle of $\triangle A B C$ (because $\left.\angle B X C=\angle B H C=180^{\circ}-\angle A\right)$. Since $A_{1}$ is the midpoint of the arc $B C$, we have $S_{B A_{1} C} \geq S_{B X C}$. Hence $$ S_{A C_{1} B A_{1} C B_{1}} \geq S_{A Z B X C Y}=2\left(S_{B H C}+S_{C H A}+S_{A H B}\right)=2 S_{A B C} $$
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
1. (AUS 2) ${ }^{\mathrm{IMO} 2}$ Let $A B C$ be a triangle. The bisector of angle $A$ meets the circumcircle of triangle $A B C$ in $A_{1}$. Points $B_{1}$ and $C_{1}$ are defined similarly. Let $A A_{1}$ meet the lines that bisect the two external angles at $B$ and $C$ in point $A^{0}$. Define $B^{0}$ and $C^{0}$ similarly. If $S_{X_{1} X_{2} \ldots X_{n}}$ denotes the area of the polygon $X_{1} X_{2} \ldots X_{n}$, prove that $$ S_{A^{0} B^{0} C^{0}}=2 S_{A C_{1} B A_{1} C B_{1}} \geq 4 S_{A B C} . $$
|
1. Let $I$ denote the intersection of the three internal bisectors. Then $I A_{1}=A_{1} A^{0}$. One way proving this is to realize that the circumcircle of $A B C$ is the nine-point circle of $A^{0} B^{0} C^{0}$, hence it bisects $I A^{0}$, since $I$ is the orthocenter of $A^{0} B^{0} C^{0}$. Another way is through noting that $I A_{1}=A_{1} B$, which follows from $\angle A_{1} I B=\angle I B A_{1}=(\angle A+\angle B) / 2$, and $A_{1} B=A_{1} A^{0}$ which follows from $\angle A_{1} A^{0} B=\angle A_{1} B A^{0}=90^{\circ}-$  $\angle I B A_{1}$. Hence, we obtain $S_{I A_{1} B}=S_{A^{0} A_{1} B}$. Repeating this argument for the six triangles that have a vertex at $I$ and adding them up gives us $S_{A^{0} B^{0} C^{0}}=2 S_{A C_{1} B A_{1} C B_{1}}$. To prove $S_{A C_{1} B A_{1} C B_{1}} \geq 2 S_{A B C}$, draw the three altitudes in triangle $A B C$ intersecting in $H$. Let $X, Y$, and $Z$ be the symmetric points of $H$ with respect the sides $B C, C A$, and $A B$, respectively. Then $X, Y, Z$ are points on the circumcircle of $\triangle A B C$ (because $\left.\angle B X C=\angle B H C=180^{\circ}-\angle A\right)$. Since $A_{1}$ is the midpoint of the arc $B C$, we have $S_{B A_{1} C} \geq S_{B X C}$. Hence $$ S_{A C_{1} B A_{1} C B_{1}} \geq S_{A Z B X C Y}=2\left(S_{B H C}+S_{C H A}+S_{A H B}\right)=2 S_{A B C} $$
|
{
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|
c656962e-13c3-508c-b7c1-a040a7f5c087
| 24,186
|
11. (HUN 1) Define sequence $a_{n}$ by $\sum_{d \mid n} a_{d}=2^{n}$. Show that $n \mid a_{n}$.
|
11. Call a binary sequence $S$ of length $n$ repeating if for some $d \mid n, d>1, S$ can be split into $d$ identical blocks. Let $x_{n}$ be the number of nonrepeating binary sequences of length $n$. The total number of binary sequences of length $n$ is obviously $2^{n}$. Any sequence of length $n$ can be produced by repeating its unique longest nonrepeating initial block according to need. Hence, we obtain the recursion relation $\sum_{d \mid n} x_{d}=2^{n}$. This, along with $x_{1}=2$, gives us $a_{n}=x_{n}$ for all $n$. We now have that the sequences counted by $x_{n}$ can be grouped into groups of $n$, the sequences in the same group being cyclic shifts of each other. Hence, $n \mid x_{n}=a_{n}$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
11. (HUN 1) Define sequence $a_{n}$ by $\sum_{d \mid n} a_{d}=2^{n}$. Show that $n \mid a_{n}$.
|
11. Call a binary sequence $S$ of length $n$ repeating if for some $d \mid n, d>1, S$ can be split into $d$ identical blocks. Let $x_{n}$ be the number of nonrepeating binary sequences of length $n$. The total number of binary sequences of length $n$ is obviously $2^{n}$. Any sequence of length $n$ can be produced by repeating its unique longest nonrepeating initial block according to need. Hence, we obtain the recursion relation $\sum_{d \mid n} x_{d}=2^{n}$. This, along with $x_{1}=2$, gives us $a_{n}=x_{n}$ for all $n$. We now have that the sequences counted by $x_{n}$ can be grouped into groups of $n$, the sequences in the same group being cyclic shifts of each other. Hence, $n \mid x_{n}=a_{n}$.
|
{
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|
121df7f2-6d61-5b13-b858-5de97e65034d
| 24,191
|
12. (HUN 3) At $n$ distinct points of a circular race course there are $n$ cars ready to start. Each car moves at a constant speed and covers the circle in an hour. On hearing the initial signal, each of them selects a direction and starts moving immediately. If two cars meet, both of them change directions and go on without loss of speed. Show that at a certain moment each car will be at its starting point.
|
12. Assume that each car starts with a unique ranking number. Suppose that while turning back at a meeting point two cars always exchanged their ranking numbers. We can observe that ranking numbers move at a constant speed and direction. One hour later, after several exchanges, each starting point will be occupied by a car of the same ranking number and proceeding in the same direction as the one that started from there one hour ago. We now give the cars back their original ranking numbers. Since the sequence of the cars along the track cannot be changed, the only possibility is that the original situation has been rotated, maybe onto itself. Hence for some $d \mid n$, after $d$ hours each car will be at its starting position and orientation.
|
proof
|
Yes
|
Yes
|
proof
|
Logic and Puzzles
|
12. (HUN 3) At $n$ distinct points of a circular race course there are $n$ cars ready to start. Each car moves at a constant speed and covers the circle in an hour. On hearing the initial signal, each of them selects a direction and starts moving immediately. If two cars meet, both of them change directions and go on without loss of speed. Show that at a certain moment each car will be at its starting point.
|
12. Assume that each car starts with a unique ranking number. Suppose that while turning back at a meeting point two cars always exchanged their ranking numbers. We can observe that ranking numbers move at a constant speed and direction. One hour later, after several exchanges, each starting point will be occupied by a car of the same ranking number and proceeding in the same direction as the one that started from there one hour ago. We now give the cars back their original ranking numbers. Since the sequence of the cars along the track cannot be changed, the only possibility is that the original situation has been rotated, maybe onto itself. Hence for some $d \mid n$, after $d$ hours each car will be at its starting position and orientation.
|
{
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|
3389ccde-e2d0-5173-9ca3-e2c6373fa72e
| 24,194
|
14. (IND 2) A bicentric quadrilateral is one that is both inscribable in and circumscribable about a circle. Show that for such a quadrilateral, the centers of the two associated circles are collinear with the point of intersection of the diagonals.
|
14. Lemma 1. In a quadrilateral $A B C D$ circumscribed about a circle, with points of tangency $P, Q, R, S$ on $D A, A B, B C, C D$ respectively, the lines $A C, B D, P R, Q S$ concur. Proof. Follows immediately, for example, from Brianchon's theorem. Lemma 2. Let a variable chord $X Y$ of a circle $C(I, r)$ subtend a right angle at a fixed point $Z$ within the circle. Then the locus of the midpoint $P$ of $X Y$ is a circle whose center is at the midpoint $M$ of $I Z$ and whose radius is $\sqrt{r^{2} / 2-I Z^{2} / 4}$. Proof. From $\angle X Z Y=90^{\circ}$ follows $\overrightarrow{Z X} \cdot \overrightarrow{Z Y}=(\overrightarrow{I X}-\overrightarrow{I Z}) \cdot(\overrightarrow{I Y}-\overrightarrow{I Z})=0$. Therefore, $$ \begin{aligned} \overrightarrow{M P}^{2} & =(\overrightarrow{M I}+\overrightarrow{I P})^{2}=\frac{1}{4}(-\overrightarrow{I Z}+\overrightarrow{I X}+\overrightarrow{I Y})^{2} \\ & =\frac{1}{4}\left(I X^{2}+I Y^{2}-I Z^{2}+2(\overrightarrow{I X}-\overrightarrow{I Z}) \cdot(\overrightarrow{I Y}-\overrightarrow{I Z})\right) \\ & =\frac{1}{2} r^{2}-\frac{1}{4} I Z^{2} \end{aligned} $$ Lemma 3. Using notation as in Lemma 1, if $A B C D$ is cyclic, $P R$ is perpendicular to $Q S$. Proof. Consider the inversion in $C(I, r)$, mapping $A$ to $A^{\prime}$ etc. $(P, Q, R, S$ are fixed). As is easily seen, $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$ will lie at the midpoints of $P Q, Q R, R S, S P$, respectively. $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is a parallelogram, but also cyclic, since inversion preserves circles; thus it must be a rectangle, and so $P R \perp Q S$. Now we return to the main result. Let $I$ and $O$ be the incenter and circumcenter, $Z$ the intersection of the diagonals, and $P, Q, R, S, A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$ points as defined in Lemmas 1 and 3. From Lemma 3, the chords $P Q, Q R, R S, S P$ subtend $90^{\circ}$ at $Z$. Therefore by Lemma 2 the points $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$ lie on a circle whose center is the midpoint $Y$ of $I Z$. Since this circle is the image of the circle $A B C D$ under the considered inversion (centered at $I$ ), it follows that $I, O, Y$ are collinear, and hence so are $I, O, Z$. Remark. This is the famous Newton's theorem for bicentric quadrilaterals.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
14. (IND 2) A bicentric quadrilateral is one that is both inscribable in and circumscribable about a circle. Show that for such a quadrilateral, the centers of the two associated circles are collinear with the point of intersection of the diagonals.
|
14. Lemma 1. In a quadrilateral $A B C D$ circumscribed about a circle, with points of tangency $P, Q, R, S$ on $D A, A B, B C, C D$ respectively, the lines $A C, B D, P R, Q S$ concur. Proof. Follows immediately, for example, from Brianchon's theorem. Lemma 2. Let a variable chord $X Y$ of a circle $C(I, r)$ subtend a right angle at a fixed point $Z$ within the circle. Then the locus of the midpoint $P$ of $X Y$ is a circle whose center is at the midpoint $M$ of $I Z$ and whose radius is $\sqrt{r^{2} / 2-I Z^{2} / 4}$. Proof. From $\angle X Z Y=90^{\circ}$ follows $\overrightarrow{Z X} \cdot \overrightarrow{Z Y}=(\overrightarrow{I X}-\overrightarrow{I Z}) \cdot(\overrightarrow{I Y}-\overrightarrow{I Z})=0$. Therefore, $$ \begin{aligned} \overrightarrow{M P}^{2} & =(\overrightarrow{M I}+\overrightarrow{I P})^{2}=\frac{1}{4}(-\overrightarrow{I Z}+\overrightarrow{I X}+\overrightarrow{I Y})^{2} \\ & =\frac{1}{4}\left(I X^{2}+I Y^{2}-I Z^{2}+2(\overrightarrow{I X}-\overrightarrow{I Z}) \cdot(\overrightarrow{I Y}-\overrightarrow{I Z})\right) \\ & =\frac{1}{2} r^{2}-\frac{1}{4} I Z^{2} \end{aligned} $$ Lemma 3. Using notation as in Lemma 1, if $A B C D$ is cyclic, $P R$ is perpendicular to $Q S$. Proof. Consider the inversion in $C(I, r)$, mapping $A$ to $A^{\prime}$ etc. $(P, Q, R, S$ are fixed). As is easily seen, $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$ will lie at the midpoints of $P Q, Q R, R S, S P$, respectively. $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is a parallelogram, but also cyclic, since inversion preserves circles; thus it must be a rectangle, and so $P R \perp Q S$. Now we return to the main result. Let $I$ and $O$ be the incenter and circumcenter, $Z$ the intersection of the diagonals, and $P, Q, R, S, A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$ points as defined in Lemmas 1 and 3. From Lemma 3, the chords $P Q, Q R, R S, S P$ subtend $90^{\circ}$ at $Z$. Therefore by Lemma 2 the points $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$ lie on a circle whose center is the midpoint $Y$ of $I Z$. Since this circle is the image of the circle $A B C D$ under the considered inversion (centered at $I$ ), it follows that $I, O, Y$ are collinear, and hence so are $I, O, Z$. Remark. This is the famous Newton's theorem for bicentric quadrilaterals.
|
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3e423520-deae-5bbd-96aa-5747a57b50e7
| 24,199
|
20. (NET 1) ${ }^{\text {IMO3 }}$ Given a set $S$ in the plane containing $n$ points and satisfying the conditions: (i) no three points of $S$ are collinear, (ii) for every point $P$ of $S$ there exist at least $k$ points in $S$ that have the same distance to $P$, prove that the following inequality holds: $$ k<\frac{1}{2}+\sqrt{2 n} $$
|
20. Suppose $k \geq 1 / 2+\sqrt{2 n}$. Consider a point $P$ in $S$. There are at least $k$ points in $S$ having all the same distance to $P$, so there are at least $\binom{k}{2}$ pairs of points $A, B$ with $A P=B P$. Since this is true for every point $P \in S$, there are at least $n\binom{k}{2}$ triples of points $(A, B, P)$ for which $A P=B P$ holds. However, $$ \begin{aligned} n\binom{k}{2} & =n \frac{k(k-1)}{2} \geq \frac{n}{2}\left(\sqrt{2 n}+\frac{1}{2}\right)\left(\sqrt{2 n}-\frac{1}{2}\right) \\ & =\frac{n}{2}\left(2 n-\frac{1}{4}\right)>n(n-1)=2\binom{n}{2} \end{aligned} $$ Since $\binom{n}{2}$ is the number of all possible pairs $(A, B)$ with $A, B \in S$, there must exist a pair of points $A, B$ with more than two points $P_{i}$ such that $A P_{i}=B P_{i}$. These points $P_{i}$ are collinear (they lie on the perpendicular bisector of $A B$ ), contradicting condition (1).
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
20. (NET 1) ${ }^{\text {IMO3 }}$ Given a set $S$ in the plane containing $n$ points and satisfying the conditions: (i) no three points of $S$ are collinear, (ii) for every point $P$ of $S$ there exist at least $k$ points in $S$ that have the same distance to $P$, prove that the following inequality holds: $$ k<\frac{1}{2}+\sqrt{2 n} $$
|
20. Suppose $k \geq 1 / 2+\sqrt{2 n}$. Consider a point $P$ in $S$. There are at least $k$ points in $S$ having all the same distance to $P$, so there are at least $\binom{k}{2}$ pairs of points $A, B$ with $A P=B P$. Since this is true for every point $P \in S$, there are at least $n\binom{k}{2}$ triples of points $(A, B, P)$ for which $A P=B P$ holds. However, $$ \begin{aligned} n\binom{k}{2} & =n \frac{k(k-1)}{2} \geq \frac{n}{2}\left(\sqrt{2 n}+\frac{1}{2}\right)\left(\sqrt{2 n}-\frac{1}{2}\right) \\ & =\frac{n}{2}\left(2 n-\frac{1}{4}\right)>n(n-1)=2\binom{n}{2} \end{aligned} $$ Since $\binom{n}{2}$ is the number of all possible pairs $(A, B)$ with $A, B \in S$, there must exist a pair of points $A, B$ with more than two points $P_{i}$ such that $A P_{i}=B P_{i}$. These points $P_{i}$ are collinear (they lie on the perpendicular bisector of $A B$ ), contradicting condition (1).
|
{
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|
b234fb9c-1ff2-5189-8afd-1d35d128ce6b
| 24,218
|
21. (NET 2) Prove that the intersection of a plane and a regular tetrahedron can be an obtuse-angled triangle and that the obtuse angle in any such triangle is always smaller than $120^{\circ}$.
|
21. In order to obtain a triangle as the intersection we must have three points $P, Q, R$ on three sides of the tetrahedron passing through one vertex, say $T$. It is clear that we may suppose w.l.o.g. that $P$ is a vertex, and $Q$ and $R$ lie on the edges $T P_{1}$ and $T P_{2}\left(P_{1}, P_{2}\right.$ are vertices) or on their extensions respectively. Suppose that $\overrightarrow{T Q}=\lambda \overrightarrow{T P_{1}}$ and $\overrightarrow{T R}=\mu \overrightarrow{T P_{2}}$, where $\lambda, \mu>0$. Then $$ \cos \angle Q P R=\frac{\overrightarrow{P Q} \cdot \overrightarrow{P R}}{\overline{P Q} \cdot \overline{P R}}=\frac{(\lambda-1)(\mu-1)+1}{2 \sqrt{\lambda^{2}-\lambda+1} \sqrt{\mu^{2}-\mu+1}} $$ In order to obtain an obtuse angle (with $\cos <0$ ) we must choose $\mu<1$ and $\lambda>\frac{2-\mu}{1-\mu}>1$. Since $\sqrt{\lambda^{2}-\lambda+1}>\lambda-1$ and $\sqrt{\mu^{2}-\mu+1}>1-\mu$, we get that for $(\lambda-1)(\mu-1)+1<0$, $$ \cos \angle Q P R>\frac{1-(1-\mu)(\lambda-1)}{2(1-\mu)(\lambda-1)}>-\frac{1}{2} ; \quad \text { hence } \angle Q P R<120^{\circ} $$ Remark. After obtaining the formula for $\cos \angle Q P R$, the official solution was as follows: For fixed $\mu_{0}<1$ and $\lambda>1, \cos \angle Q P R$ is a decreasing function of $\lambda$ : indeed, $$ \frac{\partial \cos \angle Q P R}{\partial \lambda}=\frac{\mu-(3-\mu) \lambda}{4\left(\lambda^{2}-\lambda+1\right)^{3 / 2}\left(\mu^{2}-\mu+1\right)^{1 / 2}}<0 $$ Similarly, for a fixed, sufficiently large $\lambda_{0}, \cos \angle Q P R$ is decreasing for $\mu$ decreasing to 0 . Since $\lim _{\lambda \rightarrow 0, \mu \rightarrow 0+} \cos \angle Q P R=-1 / 2$, we conclude that $\angle Q P R<120^{\circ}$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
21. (NET 2) Prove that the intersection of a plane and a regular tetrahedron can be an obtuse-angled triangle and that the obtuse angle in any such triangle is always smaller than $120^{\circ}$.
|
21. In order to obtain a triangle as the intersection we must have three points $P, Q, R$ on three sides of the tetrahedron passing through one vertex, say $T$. It is clear that we may suppose w.l.o.g. that $P$ is a vertex, and $Q$ and $R$ lie on the edges $T P_{1}$ and $T P_{2}\left(P_{1}, P_{2}\right.$ are vertices) or on their extensions respectively. Suppose that $\overrightarrow{T Q}=\lambda \overrightarrow{T P_{1}}$ and $\overrightarrow{T R}=\mu \overrightarrow{T P_{2}}$, where $\lambda, \mu>0$. Then $$ \cos \angle Q P R=\frac{\overrightarrow{P Q} \cdot \overrightarrow{P R}}{\overline{P Q} \cdot \overline{P R}}=\frac{(\lambda-1)(\mu-1)+1}{2 \sqrt{\lambda^{2}-\lambda+1} \sqrt{\mu^{2}-\mu+1}} $$ In order to obtain an obtuse angle (with $\cos <0$ ) we must choose $\mu<1$ and $\lambda>\frac{2-\mu}{1-\mu}>1$. Since $\sqrt{\lambda^{2}-\lambda+1}>\lambda-1$ and $\sqrt{\mu^{2}-\mu+1}>1-\mu$, we get that for $(\lambda-1)(\mu-1)+1<0$, $$ \cos \angle Q P R>\frac{1-(1-\mu)(\lambda-1)}{2(1-\mu)(\lambda-1)}>-\frac{1}{2} ; \quad \text { hence } \angle Q P R<120^{\circ} $$ Remark. After obtaining the formula for $\cos \angle Q P R$, the official solution was as follows: For fixed $\mu_{0}<1$ and $\lambda>1, \cos \angle Q P R$ is a decreasing function of $\lambda$ : indeed, $$ \frac{\partial \cos \angle Q P R}{\partial \lambda}=\frac{\mu-(3-\mu) \lambda}{4\left(\lambda^{2}-\lambda+1\right)^{3 / 2}\left(\mu^{2}-\mu+1\right)^{1 / 2}}<0 $$ Similarly, for a fixed, sufficiently large $\lambda_{0}, \cos \angle Q P R$ is decreasing for $\mu$ decreasing to 0 . Since $\lim _{\lambda \rightarrow 0, \mu \rightarrow 0+} \cos \angle Q P R=-1 / 2$, we conclude that $\angle Q P R<120^{\circ}$.
|
{
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|
9265bfc6-9615-5d9e-921e-f858fb81faf9
| 24,221
|
22. (PHI 1) ${ }^{\text {IMO1 }}$ Prove that the set $\{1,2, \ldots, 1989\}$ can be expressed as the disjoint union of 17 subsets $A_{1}, A_{2}, \ldots, A_{17}$ such that: (i) each $A_{i}$ contains the same number of elements; (ii) the sum of all elements of each $A_{i}$ is the same for $i=1,2, \ldots, 17$.
|
22. The statement remains valid if 17 is replaced by any divisor $k$ of $1989=3^{2}$. $13 \cdot 17,1<k<1989$, so let $k$ be one such divisor. The set $\{1,2, \ldots, 1989\}$ can be partitioned as $\{1,2, \ldots, 3 k\} \cup \bigcup_{j=1}^{L}\{(2 j+1) k+1,(2 j+1) k+$ $2, \ldots,(2 j+1) k+2 k\}=X \cup Y_{1} \cup \cdots \cup Y_{L}$, where $L=(1989-3 k) / 2 k$. The required statement will be an obvious consequence of the following two claims. Claim 1. The set $X=\{1,2, \ldots, 3 k\}$ can be partitioned into $k$ disjoint subsets, each having 3 elements and the same sum. Proof. Since $k$ is odd, let $t=k-1 / 2$ and $X=\{1,2, \ldots, 6 t+3\}$. For $l=1,2, \ldots, t$, define $$ \begin{aligned} X_{2 l-1} & =\{l, 3 t+1+l, 6 t+5-2 l\} \\ X_{2 l} & =\{t+1+l, 2 t+1+l, 6 t+4-2 l\} \\ X_{2 t+1} & =X_{k}=\{t+1,4 t+2,4 t+3\} \end{aligned} $$ It is easily seen that these three subsets are disjoint and that the sum of elements in each set is $9 t+6$. Claim 2. Each $Y_{j}=\{(2 j+1) k+1, \ldots,(2 j+1) k+2 k\}$ can be partitioned into $k$ disjoint subsets, each having 2 elements and the same sum. Proof. The obvious partitioning works: $$ Y_{j}=\{(2 j+1) k+1,(2 j+1) k+2 k\} \cup \cdots \cup\{(2 j+1) k+k,(2 j+1) k+(k+1)\} . $$
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
22. (PHI 1) ${ }^{\text {IMO1 }}$ Prove that the set $\{1,2, \ldots, 1989\}$ can be expressed as the disjoint union of 17 subsets $A_{1}, A_{2}, \ldots, A_{17}$ such that: (i) each $A_{i}$ contains the same number of elements; (ii) the sum of all elements of each $A_{i}$ is the same for $i=1,2, \ldots, 17$.
|
22. The statement remains valid if 17 is replaced by any divisor $k$ of $1989=3^{2}$. $13 \cdot 17,1<k<1989$, so let $k$ be one such divisor. The set $\{1,2, \ldots, 1989\}$ can be partitioned as $\{1,2, \ldots, 3 k\} \cup \bigcup_{j=1}^{L}\{(2 j+1) k+1,(2 j+1) k+$ $2, \ldots,(2 j+1) k+2 k\}=X \cup Y_{1} \cup \cdots \cup Y_{L}$, where $L=(1989-3 k) / 2 k$. The required statement will be an obvious consequence of the following two claims. Claim 1. The set $X=\{1,2, \ldots, 3 k\}$ can be partitioned into $k$ disjoint subsets, each having 3 elements and the same sum. Proof. Since $k$ is odd, let $t=k-1 / 2$ and $X=\{1,2, \ldots, 6 t+3\}$. For $l=1,2, \ldots, t$, define $$ \begin{aligned} X_{2 l-1} & =\{l, 3 t+1+l, 6 t+5-2 l\} \\ X_{2 l} & =\{t+1+l, 2 t+1+l, 6 t+4-2 l\} \\ X_{2 t+1} & =X_{k}=\{t+1,4 t+2,4 t+3\} \end{aligned} $$ It is easily seen that these three subsets are disjoint and that the sum of elements in each set is $9 t+6$. Claim 2. Each $Y_{j}=\{(2 j+1) k+1, \ldots,(2 j+1) k+2 k\}$ can be partitioned into $k$ disjoint subsets, each having 2 elements and the same sum. Proof. The obvious partitioning works: $$ Y_{j}=\{(2 j+1) k+1,(2 j+1) k+2 k\} \cup \cdots \cup\{(2 j+1) k+k,(2 j+1) k+(k+1)\} . $$
|
{
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|
9b7773be-9e9b-5abd-9629-fe39139c4c9e
| 24,223
|
25. (KOR 3) Let $a, b$ be integers that are not perfect squares. Prove that if $$ x^{2}-a y^{2}-b z^{2}+a b w^{2}=0 $$ has a nontrivial solution in integers, then so does $$ x^{2}-a y^{2}-b z^{2}=0 $$
|
25. We may assume w.l.o.g. that $a>0$ (because $a, b<0$ is impossible, and $a, b \neq 0$ from the condition of the problem). Let $\left(x_{0}, y_{0}, z_{0}, w_{0}\right) \neq$ $(0,0,0,0)$ be a solution of $x^{2}-a y^{2}-b z^{2}+a b w^{2}$. Then $$ x_{0}^{2}-a y_{0}^{2}=b\left(z_{0}^{2}-a w_{0}^{2}\right) $$ Multiplying both sides by $\left(z_{0}^{2}-a w_{0}^{2}\right)$, we get $$ \begin{gathered} \left(x_{0}^{2}-a y_{0}^{2}\right)\left(z_{0}^{2}-a w_{0}^{2}\right)-b\left(z_{0}^{2}-a w_{0}^{2}\right)^{2}=0 \\ \Leftrightarrow\left(x_{0} z_{0}-a y_{0} w_{0}\right)^{2}-a\left(y_{0} z_{0}-x_{0} w_{0}\right)^{2}-b\left(z_{0}^{2}-a w_{0}^{2}\right)^{2}=0 . \end{gathered} $$ Hence, for $x_{1}=x_{0} z_{0}-a y_{0} w_{0}, \quad y_{1}=y_{0} z_{0}-x_{0} w_{0}, \quad z_{1}=z_{0}^{2}-a w_{0}^{2}$, we have $$ x_{1}^{2}-a y_{1}^{2}-b z_{1}^{2}=0 . $$ If $\left(x_{1}, y_{1}, z_{1}\right)$ is the trivial solution, then $z_{1}=0$ implies $z_{0}=w_{0}=0$ and similarly $x_{0}=y_{0}=0$ because $a$ is not a perfect square. This contradicts the initial assumption.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
25. (KOR 3) Let $a, b$ be integers that are not perfect squares. Prove that if $$ x^{2}-a y^{2}-b z^{2}+a b w^{2}=0 $$ has a nontrivial solution in integers, then so does $$ x^{2}-a y^{2}-b z^{2}=0 $$
|
25. We may assume w.l.o.g. that $a>0$ (because $a, b<0$ is impossible, and $a, b \neq 0$ from the condition of the problem). Let $\left(x_{0}, y_{0}, z_{0}, w_{0}\right) \neq$ $(0,0,0,0)$ be a solution of $x^{2}-a y^{2}-b z^{2}+a b w^{2}$. Then $$ x_{0}^{2}-a y_{0}^{2}=b\left(z_{0}^{2}-a w_{0}^{2}\right) $$ Multiplying both sides by $\left(z_{0}^{2}-a w_{0}^{2}\right)$, we get $$ \begin{gathered} \left(x_{0}^{2}-a y_{0}^{2}\right)\left(z_{0}^{2}-a w_{0}^{2}\right)-b\left(z_{0}^{2}-a w_{0}^{2}\right)^{2}=0 \\ \Leftrightarrow\left(x_{0} z_{0}-a y_{0} w_{0}\right)^{2}-a\left(y_{0} z_{0}-x_{0} w_{0}\right)^{2}-b\left(z_{0}^{2}-a w_{0}^{2}\right)^{2}=0 . \end{gathered} $$ Hence, for $x_{1}=x_{0} z_{0}-a y_{0} w_{0}, \quad y_{1}=y_{0} z_{0}-x_{0} w_{0}, \quad z_{1}=z_{0}^{2}-a w_{0}^{2}$, we have $$ x_{1}^{2}-a y_{1}^{2}-b z_{1}^{2}=0 . $$ If $\left(x_{1}, y_{1}, z_{1}\right)$ is the trivial solution, then $z_{1}=0$ implies $z_{0}=w_{0}=0$ and similarly $x_{0}=y_{0}=0$ because $a$ is not a perfect square. This contradicts the initial assumption.
|
{
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|
5d79e9dd-8e6d-5ce5-9104-6a24c39adbc1
| 24,231
|
31. (SWE 3) Let $a_{1} \geq a_{2} \geq a_{3}$ be given positive integers and let $N\left(a_{1}, a_{2}, a_{3}\right)$ be the number of solutions $\left(x_{1}, x_{2}, x_{3}\right)$ of the equation $$ \frac{a_{1}}{x_{1}}+\frac{a_{2}}{x_{2}}+\frac{a_{3}}{x_{3}}=1 $$ where $x_{1}, x_{2}$, and $x_{3}$ are positive integers. Show that $$ N\left(a_{1}, a_{2}, a_{3}\right) \leq 6 a_{1} a_{2}\left(3+\ln \left(2 a_{1}\right)\right) $$
|
31. Let us denote by $N_{p q r}$ the number of solutions for which $a_{p} / x_{p} \geq a_{q} / x_{q} \geq$ $a_{r} / x_{r}$, where $(p, q, r)$ is one of six permutations of $(1,2,3)$. It is clearly enough to prove that $N_{p q r}+N_{q p r} \leq 2 a_{1} a_{2}\left(3+\ln \left(2 a_{1}\right)\right)$. First, from $$ \frac{3 a_{p}}{x_{p}} \geq \frac{a_{p}}{x_{p}}+\frac{a_{q}}{x_{q}}+\frac{a_{r}}{x_{r}}=1 \quad \text { and } \quad \frac{a_{p}}{x_{p}}<1 $$ we get $a_{p}+1 \leq x_{p} \leq 3 a_{p}$. Similarly, for fixed $x_{p}$ we have $$ \frac{2 a_{q}}{x_{q}} \geq \frac{a_{q}}{x_{q}}+\frac{a_{r}}{x_{r}}=1-\frac{a_{p}}{x_{p}} \quad \text { and } \quad \frac{a_{q}}{x_{q}} \leq \min \left(\frac{a_{p}}{x_{p}}, 1-\frac{a_{p}}{x_{p}}\right), $$ which gives $\max \left\{a_{q} \cdot x_{p} / a_{p}, a_{q} \cdot x_{p} /\left(x_{p}-a_{p}\right)\right\} \leq x_{q} \leq 2 a_{q} \cdot x_{p} /\left(x_{p}-a_{p}\right)$, i.e., if $a_{p}+1 \leq x_{p} \leq 2 a_{p}$ there are at most $a_{q} \cdot x_{p} /\left(x_{p}-a_{p}\right)+1 / 2$ possible values for $x_{q}$ (because there are $[2 x]-[x]=[x+1 / 2]$ integers between $x$ and $2 x)$, and if $2 a_{p}+1 \leq x_{p} \leq 3 a_{p}$, at most $2 a_{q} \cdot x_{p} /\left(x_{p}-a_{p}\right)-a_{q} \cdot x_{p} / a_{p}+$ 1 possible values. Given $x_{p}$ and $x_{q}, x_{r}$ is uniquely determined. Hence $$ \begin{aligned} N_{p q r} & \leq \sum_{x_{p}=a_{p}+1}^{2 a_{p}}\left(\frac{a_{q} \cdot x_{p}}{x_{p}-a_{p}}+\frac{1}{2}\right)+\sum_{x_{p}=2 a_{p}+1}^{3 a_{p}}\left(\frac{2 a_{q} \cdot x_{p}}{x_{p}-a_{p}}-\frac{a_{q} \cdot x_{p}}{a_{p}}+1\right) \\ & =\frac{3 a_{p}}{2}+a_{q} \sum_{k=1}^{a_{p}}\left[\frac{k+a_{p}}{k}+\left(\frac{2\left(k+2 a_{p}\right)}{k+a_{p}}-\frac{k+2 a_{p}}{a_{p}}\right)\right] \\ & =\frac{3 a_{p}}{2}+a_{q} \sum_{k=1}^{a_{p}}\left[1-\frac{k}{a_{p}}+a_{p}\left(\frac{1}{k}+\frac{2}{k+a_{p}}\right)\right] \\ & =\frac{3 a_{p}}{2}-\frac{a_{q}}{2}+a_{p} a_{q}\left(\frac{1}{2}+\sum_{k=1}^{a_{p}}\left(\frac{1}{k}+\frac{2}{k+a_{p}}\right)\right) \\ & \leq a_{p} a_{q}\left(\frac{3}{2 a_{q}}-\frac{1}{2 a_{p}}+\ln \left(2 a_{p}\right)+\frac{5}{2}-\ln 2\right), \end{aligned} $$ where we have used $\sum_{k=1}^{n}(1 / k+2 /(k+n)) \leq \ln (2 n)+2-\ln 2$ (this can be proved by induction). Hence, $N_{p q r}+N_{q p r} \leq 2 a_{p} a_{q}\left(1+0.5+\ln \left(2 a_{p}\right)+2-\ln 2\right)<2 a_{1} a_{2}\left(2.81+\ln \left(2 a_{1}\right)\right)$. Remark. The official solution was somewhat simpler, but used that the interval $(x, 2 x]$, for real $x$, cannot contain more than $x$ integers, which is false in general. Thus it could give only a weaker estimate $N \leq 6 a_{1} a_{2}\left(9 / 2-\ln 2+\ln \left(2 a_{1}\right)\right)$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
31. (SWE 3) Let $a_{1} \geq a_{2} \geq a_{3}$ be given positive integers and let $N\left(a_{1}, a_{2}, a_{3}\right)$ be the number of solutions $\left(x_{1}, x_{2}, x_{3}\right)$ of the equation $$ \frac{a_{1}}{x_{1}}+\frac{a_{2}}{x_{2}}+\frac{a_{3}}{x_{3}}=1 $$ where $x_{1}, x_{2}$, and $x_{3}$ are positive integers. Show that $$ N\left(a_{1}, a_{2}, a_{3}\right) \leq 6 a_{1} a_{2}\left(3+\ln \left(2 a_{1}\right)\right) $$
|
31. Let us denote by $N_{p q r}$ the number of solutions for which $a_{p} / x_{p} \geq a_{q} / x_{q} \geq$ $a_{r} / x_{r}$, where $(p, q, r)$ is one of six permutations of $(1,2,3)$. It is clearly enough to prove that $N_{p q r}+N_{q p r} \leq 2 a_{1} a_{2}\left(3+\ln \left(2 a_{1}\right)\right)$. First, from $$ \frac{3 a_{p}}{x_{p}} \geq \frac{a_{p}}{x_{p}}+\frac{a_{q}}{x_{q}}+\frac{a_{r}}{x_{r}}=1 \quad \text { and } \quad \frac{a_{p}}{x_{p}}<1 $$ we get $a_{p}+1 \leq x_{p} \leq 3 a_{p}$. Similarly, for fixed $x_{p}$ we have $$ \frac{2 a_{q}}{x_{q}} \geq \frac{a_{q}}{x_{q}}+\frac{a_{r}}{x_{r}}=1-\frac{a_{p}}{x_{p}} \quad \text { and } \quad \frac{a_{q}}{x_{q}} \leq \min \left(\frac{a_{p}}{x_{p}}, 1-\frac{a_{p}}{x_{p}}\right), $$ which gives $\max \left\{a_{q} \cdot x_{p} / a_{p}, a_{q} \cdot x_{p} /\left(x_{p}-a_{p}\right)\right\} \leq x_{q} \leq 2 a_{q} \cdot x_{p} /\left(x_{p}-a_{p}\right)$, i.e., if $a_{p}+1 \leq x_{p} \leq 2 a_{p}$ there are at most $a_{q} \cdot x_{p} /\left(x_{p}-a_{p}\right)+1 / 2$ possible values for $x_{q}$ (because there are $[2 x]-[x]=[x+1 / 2]$ integers between $x$ and $2 x)$, and if $2 a_{p}+1 \leq x_{p} \leq 3 a_{p}$, at most $2 a_{q} \cdot x_{p} /\left(x_{p}-a_{p}\right)-a_{q} \cdot x_{p} / a_{p}+$ 1 possible values. Given $x_{p}$ and $x_{q}, x_{r}$ is uniquely determined. Hence $$ \begin{aligned} N_{p q r} & \leq \sum_{x_{p}=a_{p}+1}^{2 a_{p}}\left(\frac{a_{q} \cdot x_{p}}{x_{p}-a_{p}}+\frac{1}{2}\right)+\sum_{x_{p}=2 a_{p}+1}^{3 a_{p}}\left(\frac{2 a_{q} \cdot x_{p}}{x_{p}-a_{p}}-\frac{a_{q} \cdot x_{p}}{a_{p}}+1\right) \\ & =\frac{3 a_{p}}{2}+a_{q} \sum_{k=1}^{a_{p}}\left[\frac{k+a_{p}}{k}+\left(\frac{2\left(k+2 a_{p}\right)}{k+a_{p}}-\frac{k+2 a_{p}}{a_{p}}\right)\right] \\ & =\frac{3 a_{p}}{2}+a_{q} \sum_{k=1}^{a_{p}}\left[1-\frac{k}{a_{p}}+a_{p}\left(\frac{1}{k}+\frac{2}{k+a_{p}}\right)\right] \\ & =\frac{3 a_{p}}{2}-\frac{a_{q}}{2}+a_{p} a_{q}\left(\frac{1}{2}+\sum_{k=1}^{a_{p}}\left(\frac{1}{k}+\frac{2}{k+a_{p}}\right)\right) \\ & \leq a_{p} a_{q}\left(\frac{3}{2 a_{q}}-\frac{1}{2 a_{p}}+\ln \left(2 a_{p}\right)+\frac{5}{2}-\ln 2\right), \end{aligned} $$ where we have used $\sum_{k=1}^{n}(1 / k+2 /(k+n)) \leq \ln (2 n)+2-\ln 2$ (this can be proved by induction). Hence, $N_{p q r}+N_{q p r} \leq 2 a_{p} a_{q}\left(1+0.5+\ln \left(2 a_{p}\right)+2-\ln 2\right)<2 a_{1} a_{2}\left(2.81+\ln \left(2 a_{1}\right)\right)$. Remark. The official solution was somewhat simpler, but used that the interval $(x, 2 x]$, for real $x$, cannot contain more than $x$ integers, which is false in general. Thus it could give only a weaker estimate $N \leq 6 a_{1} a_{2}\left(9 / 2-\ln 2+\ln \left(2 a_{1}\right)\right)$.
|
{
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|
3f08cc6b-af7e-59cb-8e03-1b7b5df5baf1
| 24,249
|
4. (BUL 3) Prove that for every integer $n>1$ the equation $$ \frac{x^{n}}{n!}+\frac{x^{n-1}}{(n-1)!}+\cdots+\frac{x^{2}}{2!}+\frac{x}{1!}+1=0 $$ has no rational roots.
|
4. First we note that for every integer $k>0$ and prime number $p, p^{k}$ doesn't divide $k!$. This follows from the fact that the highest exponent $r$ of $p$ for which $p^{r} \mid k$ ! is $$ r=\left[\frac{k}{p}\right]+\left[\frac{k}{p^{2}}\right]+\cdots<\frac{k}{p}+\frac{k}{p^{2}}+\cdots=\frac{k}{p-1}<k . $$ Now suppose that $\alpha$ is a rational root of the given equation. Then $$ \alpha^{n}+\frac{n!}{(n-1)!} \alpha^{n-1}+\cdots+\frac{n!}{2!} \alpha^{2}+\frac{n!}{1!} \alpha+n!=0 $$ from which we can conclude that $\alpha$ must be an integer, not equal to $\pm 1$. Let $p$ be a prime divisor of $n$ and let $r$ be the highest exponent of $p$ for which $p^{r} \mid n$ !. Then $p \mid \alpha$. Since $p^{k} \mid \alpha^{k}$ and $p^{k} \nmid k$ !, we obtain that $p^{r+1} \mid n!\alpha^{k} / k$ ! for $k=1,2, \ldots, n$. But then it follows from (1) that $p^{r+1} \mid n$ !, a contradiction.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
4. (BUL 3) Prove that for every integer $n>1$ the equation $$ \frac{x^{n}}{n!}+\frac{x^{n-1}}{(n-1)!}+\cdots+\frac{x^{2}}{2!}+\frac{x}{1!}+1=0 $$ has no rational roots.
|
4. First we note that for every integer $k>0$ and prime number $p, p^{k}$ doesn't divide $k!$. This follows from the fact that the highest exponent $r$ of $p$ for which $p^{r} \mid k$ ! is $$ r=\left[\frac{k}{p}\right]+\left[\frac{k}{p^{2}}\right]+\cdots<\frac{k}{p}+\frac{k}{p^{2}}+\cdots=\frac{k}{p-1}<k . $$ Now suppose that $\alpha$ is a rational root of the given equation. Then $$ \alpha^{n}+\frac{n!}{(n-1)!} \alpha^{n-1}+\cdots+\frac{n!}{2!} \alpha^{2}+\frac{n!}{1!} \alpha+n!=0 $$ from which we can conclude that $\alpha$ must be an integer, not equal to $\pm 1$. Let $p$ be a prime divisor of $n$ and let $r$ be the highest exponent of $p$ for which $p^{r} \mid n$ !. Then $p \mid \alpha$. Since $p^{k} \mid \alpha^{k}$ and $p^{k} \nmid k$ !, we obtain that $p^{r+1} \mid n!\alpha^{k} / k$ ! for $k=1,2, \ldots, n$. But then it follows from (1) that $p^{r+1} \mid n$ !, a contradiction.
|
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b73ca753-0b95-5002-b8f9-48904e5baf18
| 24,253
|
6. (CZS 1) For a triangle $A B C$, let $k$ be its circumcircle with radius $r$. The bisectors of the inner angles $A, B$, and $C$ of the triangle intersect respectively the circle $k$ again at points $A^{\prime}, B^{\prime}$, and $C^{\prime}$. Prove the inequality $$ 16 Q^{3} \geq 27 r^{4} P $$ where $Q$ and $P$ are the areas of the triangles $A^{\prime} B^{\prime} C^{\prime}$ and $A B C$ respectively.
|
6. Let us denote the measures of the inner angles of the triangle $A B C$ by $\alpha, \beta, \gamma$. Then $P=r^{2}(\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma) / 2$. Since the inner angles of the triangle $A^{\prime} B^{\prime} C^{\prime}$ are $(\beta+\gamma) / 2,(\gamma+\alpha) / 2,(\alpha+\beta) / 2$, we also have $Q=r^{2}[\sin (\beta+\gamma)+\sin (\gamma+\alpha)+\sin (\alpha+\beta)] / 2$. Applying the AM-GM mean inequality, we now obtain $$ \begin{aligned} 16 Q^{3} & =\frac{16}{8} r^{6}(\sin (\beta+\gamma)+\sin (\gamma+\alpha)+\sin (\alpha+\beta))^{3} \\ & \geq 54 r^{6} \sin (\beta+\gamma) \sin (\gamma+\alpha) \sin (\alpha+\beta) \\ & =27 r^{6}[\cos (\alpha-\beta)-\cos (\alpha+\beta+2 \gamma)] \sin (\alpha+\beta) \\ & =27 r^{6}[\cos (\alpha-\beta)+\cos \gamma] \sin (\alpha+\beta) \\ & =\frac{27}{2} r^{6}[\sin (\alpha+\beta+\gamma)+\sin (\alpha+\beta-\gamma)+\sin 2 \alpha+\sin 2 \beta] \\ & =\frac{27}{2} r^{6}[\sin (2 \gamma)+\sin 2 \alpha+\sin 2 \beta]=27 r^{4} P . \end{aligned} $$ This completes the proof.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
6. (CZS 1) For a triangle $A B C$, let $k$ be its circumcircle with radius $r$. The bisectors of the inner angles $A, B$, and $C$ of the triangle intersect respectively the circle $k$ again at points $A^{\prime}, B^{\prime}$, and $C^{\prime}$. Prove the inequality $$ 16 Q^{3} \geq 27 r^{4} P $$ where $Q$ and $P$ are the areas of the triangles $A^{\prime} B^{\prime} C^{\prime}$ and $A B C$ respectively.
|
6. Let us denote the measures of the inner angles of the triangle $A B C$ by $\alpha, \beta, \gamma$. Then $P=r^{2}(\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma) / 2$. Since the inner angles of the triangle $A^{\prime} B^{\prime} C^{\prime}$ are $(\beta+\gamma) / 2,(\gamma+\alpha) / 2,(\alpha+\beta) / 2$, we also have $Q=r^{2}[\sin (\beta+\gamma)+\sin (\gamma+\alpha)+\sin (\alpha+\beta)] / 2$. Applying the AM-GM mean inequality, we now obtain $$ \begin{aligned} 16 Q^{3} & =\frac{16}{8} r^{6}(\sin (\beta+\gamma)+\sin (\gamma+\alpha)+\sin (\alpha+\beta))^{3} \\ & \geq 54 r^{6} \sin (\beta+\gamma) \sin (\gamma+\alpha) \sin (\alpha+\beta) \\ & =27 r^{6}[\cos (\alpha-\beta)-\cos (\alpha+\beta+2 \gamma)] \sin (\alpha+\beta) \\ & =27 r^{6}[\cos (\alpha-\beta)+\cos \gamma] \sin (\alpha+\beta) \\ & =\frac{27}{2} r^{6}[\sin (\alpha+\beta+\gamma)+\sin (\alpha+\beta-\gamma)+\sin 2 \alpha+\sin 2 \beta] \\ & =\frac{27}{2} r^{6}[\sin (2 \gamma)+\sin 2 \alpha+\sin 2 \beta]=27 r^{4} P . \end{aligned} $$ This completes the proof.
|
{
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|
3af89abf-ee98-5f80-b709-2579537e68b8
| 24,257
|
7. (FIN 1) Show that any two points lying inside a regular $n$-gon $E$ can be joined by two circular arcs lying inside $E$ and meeting at an angle of at least $\left(1-\frac{2}{n}\right) \pi$.
|
7. Assume that $P_{1}$ and $P_{2}$ are points inside $E$, and that the line $P_{1} P_{2}$ intersects the perimeter of $E$ at $Q_{1}$ and $Q_{2}$. If we prove the statement for $Q_{1}$ and $Q_{2}$, we are done, since these arcs can be mapped homothetically to join $P_{1}$ and $P_{2}$. Let $V_{1}, V_{2}$ be two vertices of $E$. Then applying two homotheties to the inscribed circle of $E$ one can find two arcs (one of them may be a side of $E)$ joining these two points, both tangent to the sides of $E$ that meet at $V_{1}$ and $V_{2}$. If $A$ is any point of the side $V_{2} V_{3}$, two homotheties with center $V_{1}$ take the arcs joining $V_{1}$ to $V_{2}$ and $V_{3}$ into arcs joining $V_{1}$ to $A$; their angle of incidence at $A$ remains $(1-2 / n) \pi$. Next, for two arbitrary points $Q_{1}$ and $Q_{2}$ on two different sides $V_{1} V_{2}$ and $V_{3} V_{4}$, we join $V_{1}$ and $V_{2}$ to $Q_{2}$ with pairs of arcs that meet at $Q_{2}$ and have an angle of incidence $(1-2 / n) \pi$. The two arcs that meet the line $Q_{1} Q_{2}$ again outside $E$ meet at $Q_{2}$ at an angle greater than or equal to $(1-2 / n) \pi$. Two homotheties with center $Q_{2}$ carry these arcs to ones meeting also at $Q_{1}$ with the same angle of incidence.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
7. (FIN 1) Show that any two points lying inside a regular $n$-gon $E$ can be joined by two circular arcs lying inside $E$ and meeting at an angle of at least $\left(1-\frac{2}{n}\right) \pi$.
|
7. Assume that $P_{1}$ and $P_{2}$ are points inside $E$, and that the line $P_{1} P_{2}$ intersects the perimeter of $E$ at $Q_{1}$ and $Q_{2}$. If we prove the statement for $Q_{1}$ and $Q_{2}$, we are done, since these arcs can be mapped homothetically to join $P_{1}$ and $P_{2}$. Let $V_{1}, V_{2}$ be two vertices of $E$. Then applying two homotheties to the inscribed circle of $E$ one can find two arcs (one of them may be a side of $E)$ joining these two points, both tangent to the sides of $E$ that meet at $V_{1}$ and $V_{2}$. If $A$ is any point of the side $V_{2} V_{3}$, two homotheties with center $V_{1}$ take the arcs joining $V_{1}$ to $V_{2}$ and $V_{3}$ into arcs joining $V_{1}$ to $A$; their angle of incidence at $A$ remains $(1-2 / n) \pi$. Next, for two arbitrary points $Q_{1}$ and $Q_{2}$ on two different sides $V_{1} V_{2}$ and $V_{3} V_{4}$, we join $V_{1}$ and $V_{2}$ to $Q_{2}$ with pairs of arcs that meet at $Q_{2}$ and have an angle of incidence $(1-2 / n) \pi$. The two arcs that meet the line $Q_{1} Q_{2}$ again outside $E$ meet at $Q_{2}$ at an angle greater than or equal to $(1-2 / n) \pi$. Two homotheties with center $Q_{2}$ carry these arcs to ones meeting also at $Q_{1}$ with the same angle of incidence.
|
{
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d286b72c-2a77-547f-a58e-30fc6a3d3b2b
| 24,260
|
8. (FRA 2) Let $R$ be a rectangle that is the union of a finite number of rectangles $R_{i}, 1 \leq i \leq n$, satisfying the following conditions: (i) The sides of every rectangle $R_{i}$ are parallel to the sides of $R$. (ii) The interiors of any two different $R_{i}$ are disjoint. (iii) Every $R_{i}$ has at least one side of integral length. Prove that $R$ has at least one side of integral length.
|
8. Let $A, B, C, D$ denote the vertices of $R$. We consider the set $\mathcal{S}$ of all points $E$ of the plane that are vertices of at least one rectangle, and its subset $\mathcal{S}^{\prime}$ consisting of those points in $\mathcal{S}$ that have both coordinates integral in the orthonormal coordinate system with point $A$ as the origin and lines $A B, A D$ as axes. First, to each $E \in \mathcal{S}$ we can assign an integer $n_{E}$ as the number of rectangles $R_{i}$ with one vertex at $E$. It is easy to check that $n_{E}=1$ if $E$ is one of the vertices $A, B, C, D$; in all other cases $n_{E}$ is either 2 or 4. Furthermore, for each rectangle $R_{i}$ we define $f\left(R_{i}\right)$ as the number of vertices of $R_{i}$ that belong to $\mathcal{S}^{\prime}$. Since every $R_{i}$ has at least one side of integer length, $f\left(R_{i}\right)$ can take only values 0,2 , or 4 . Therefore we have $$ \sum_{i=1}^{n} f\left(R_{i}\right) \equiv 0(\bmod 2) $$ On the other hand, $\sum_{i=1}^{n} f\left(R_{i}\right)$ is equal to $\sum_{E \in \mathcal{S}^{\prime}} n_{E}$, implying that $$ \sum_{E \in \mathcal{S}^{\prime}} n_{E} \equiv 0(\bmod 2) $$ However, since $n_{A}=1$, at least one other $n_{E}$, where $E \in \mathcal{S}^{\prime}$, must be odd, and that can happen only for $E$ being $B, C$, or $D$. We conclude that at least one of the sides of $R$ has integral length. Second solution. Consider the coordinate system introduced above. If $D$ is a rectangle whose sides are parallel to the axes of the system, it is easy to prove that $$ \int_{D} \sin 2 \pi x \sin 2 \pi y d x d y=0 $$ if and only if at least one side of $D$ has integral length. This holds for all $R_{i}$ 's, so that adding up these equalities we get $\int_{R} \sin 2 \pi x \sin 2 \pi y d x d y=0$. Thus, $R$ also has a side of integral length.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
8. (FRA 2) Let $R$ be a rectangle that is the union of a finite number of rectangles $R_{i}, 1 \leq i \leq n$, satisfying the following conditions: (i) The sides of every rectangle $R_{i}$ are parallel to the sides of $R$. (ii) The interiors of any two different $R_{i}$ are disjoint. (iii) Every $R_{i}$ has at least one side of integral length. Prove that $R$ has at least one side of integral length.
|
8. Let $A, B, C, D$ denote the vertices of $R$. We consider the set $\mathcal{S}$ of all points $E$ of the plane that are vertices of at least one rectangle, and its subset $\mathcal{S}^{\prime}$ consisting of those points in $\mathcal{S}$ that have both coordinates integral in the orthonormal coordinate system with point $A$ as the origin and lines $A B, A D$ as axes. First, to each $E \in \mathcal{S}$ we can assign an integer $n_{E}$ as the number of rectangles $R_{i}$ with one vertex at $E$. It is easy to check that $n_{E}=1$ if $E$ is one of the vertices $A, B, C, D$; in all other cases $n_{E}$ is either 2 or 4. Furthermore, for each rectangle $R_{i}$ we define $f\left(R_{i}\right)$ as the number of vertices of $R_{i}$ that belong to $\mathcal{S}^{\prime}$. Since every $R_{i}$ has at least one side of integer length, $f\left(R_{i}\right)$ can take only values 0,2 , or 4 . Therefore we have $$ \sum_{i=1}^{n} f\left(R_{i}\right) \equiv 0(\bmod 2) $$ On the other hand, $\sum_{i=1}^{n} f\left(R_{i}\right)$ is equal to $\sum_{E \in \mathcal{S}^{\prime}} n_{E}$, implying that $$ \sum_{E \in \mathcal{S}^{\prime}} n_{E} \equiv 0(\bmod 2) $$ However, since $n_{A}=1$, at least one other $n_{E}$, where $E \in \mathcal{S}^{\prime}$, must be odd, and that can happen only for $E$ being $B, C$, or $D$. We conclude that at least one of the sides of $R$ has integral length. Second solution. Consider the coordinate system introduced above. If $D$ is a rectangle whose sides are parallel to the axes of the system, it is easy to prove that $$ \int_{D} \sin 2 \pi x \sin 2 \pi y d x d y=0 $$ if and only if at least one side of $D$ has integral length. This holds for all $R_{i}$ 's, so that adding up these equalities we get $\int_{R} \sin 2 \pi x \sin 2 \pi y d x d y=0$. Thus, $R$ also has a side of integral length.
|
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4e14febc-cea2-5729-860a-80358e52d45c
| 24,263
|
9. (FRA 4) For all integers $n, n \geq 0$, there exist uniquely determined integers $a_{n}, b_{n}, c_{n}$ such that $$ (1+4 \sqrt[3]{2}-4 \sqrt[3]{4})^{n}=a_{n}+b_{n} \sqrt[3]{2}+c_{n} \sqrt[3]{4} $$ Prove that $c_{n}=0$ implies $n=0$.
|
9. From $a_{n+1}+b_{n+1} \sqrt[3]{2}+c_{n+1} \sqrt[3]{4}=\left(a_{n}+b_{n} \sqrt[3]{2}+c_{n} \sqrt[3]{4}\right)(1+4 \sqrt[3]{2}-4 \sqrt[3]{4})$ we obtain $a_{n+1}=a_{n}-8 b_{n}+8 c_{n}$. Since $a_{0}=1, a_{n}$ is odd for all $n$. For an integer $k>0$, we can write $k=2^{l} k^{\prime}, k^{\prime}$ being odd and $l$ a nonnegative integer. Let us set $v(k)=l$, and define $\beta_{n}=v\left(b_{n}\right), \gamma_{n}=v\left(c_{n}\right)$. We prove the following lemmas: Lemma 1. For every integer $p \geq 0, b_{2^{p}}$ and $c_{2^{p}}$ are nonzero, and $\beta_{2^{p}}=$ $\gamma_{2^{p}}=p+2$. Proof. By induction on $p$. For $p=0, b_{1}=4$ and $c_{1}=-4$, so the assertion is true. Suppose that it holds for $p$. Then $$ (1+4 \sqrt[3]{2}-4 \sqrt[3]{4})^{2^{p+1}}=\left(a+2^{p+2}\left(b^{\prime} \sqrt[3]{2}+c^{\prime} \sqrt[3]{4}\right)\right)^{2} \text { with } a, b^{\prime}, \text { and } c^{\prime} \text { odd. } $$ Then we easily obtain that $(1+4 \sqrt[3]{2}-4 \sqrt[3]{4})^{2^{p+1}}=A+2^{p+3}(B \sqrt[3]{2}+$ $C \sqrt[3]{4}$ ), where $A, B=a b^{\prime}+2^{p+1} E, C=a c^{\prime}+2^{p+1} F$ are odd. Therefore Lemma 1 holds for $p+1$. Lemma 2. Suppose that for integers $n, m \geq 0, \beta_{n}=\gamma_{n}=\lambda>\beta_{m}=$ $\gamma_{m}=\mu$. Then $b_{n+m}, c_{n+m}$ are nonzero and $\beta_{n+m}=\gamma_{n+m}=\mu$. Proof. Calculating $\left(a^{\prime}+2^{\lambda}\left(b^{\prime} \sqrt[3]{2}+c^{\prime} \sqrt[3]{4}\right)\right)\left(a^{\prime \prime}+2^{\mu}\left(b^{\prime \prime} \sqrt[3]{2}+c^{\prime \prime} \sqrt[3]{4}\right)\right)$, with $a^{\prime}, b^{\prime}, c^{\prime}, a^{\prime \prime}, b^{\prime \prime}, c^{\prime \prime}$ odd, we easily obtain the product $A+2^{\mu}(B \sqrt[3]{2}+$ $C \sqrt[3]{4})$, where $A, B=a^{\prime} b^{\prime \prime}+2^{\lambda-\mu} E$, and $C=a^{\prime} c^{\prime \prime}+2^{\lambda-\mu} F$ are odd, which proves Lemma 2. Since every integer $n>0$ can be written as $n=2^{p_{r}}+\cdots+2^{p_{1}}$, with $0 \leq p_{1}<\cdots<p_{r}$, from Lemmas 1 and 2 it follows that $c_{n}$ is nonzero, and that $\gamma_{n}=p_{1}+2$. Remark. $b_{1989}$ and $c_{1989}$ are divisible by 4 , but not by 8 .
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
9. (FRA 4) For all integers $n, n \geq 0$, there exist uniquely determined integers $a_{n}, b_{n}, c_{n}$ such that $$ (1+4 \sqrt[3]{2}-4 \sqrt[3]{4})^{n}=a_{n}+b_{n} \sqrt[3]{2}+c_{n} \sqrt[3]{4} $$ Prove that $c_{n}=0$ implies $n=0$.
|
9. From $a_{n+1}+b_{n+1} \sqrt[3]{2}+c_{n+1} \sqrt[3]{4}=\left(a_{n}+b_{n} \sqrt[3]{2}+c_{n} \sqrt[3]{4}\right)(1+4 \sqrt[3]{2}-4 \sqrt[3]{4})$ we obtain $a_{n+1}=a_{n}-8 b_{n}+8 c_{n}$. Since $a_{0}=1, a_{n}$ is odd for all $n$. For an integer $k>0$, we can write $k=2^{l} k^{\prime}, k^{\prime}$ being odd and $l$ a nonnegative integer. Let us set $v(k)=l$, and define $\beta_{n}=v\left(b_{n}\right), \gamma_{n}=v\left(c_{n}\right)$. We prove the following lemmas: Lemma 1. For every integer $p \geq 0, b_{2^{p}}$ and $c_{2^{p}}$ are nonzero, and $\beta_{2^{p}}=$ $\gamma_{2^{p}}=p+2$. Proof. By induction on $p$. For $p=0, b_{1}=4$ and $c_{1}=-4$, so the assertion is true. Suppose that it holds for $p$. Then $$ (1+4 \sqrt[3]{2}-4 \sqrt[3]{4})^{2^{p+1}}=\left(a+2^{p+2}\left(b^{\prime} \sqrt[3]{2}+c^{\prime} \sqrt[3]{4}\right)\right)^{2} \text { with } a, b^{\prime}, \text { and } c^{\prime} \text { odd. } $$ Then we easily obtain that $(1+4 \sqrt[3]{2}-4 \sqrt[3]{4})^{2^{p+1}}=A+2^{p+3}(B \sqrt[3]{2}+$ $C \sqrt[3]{4}$ ), where $A, B=a b^{\prime}+2^{p+1} E, C=a c^{\prime}+2^{p+1} F$ are odd. Therefore Lemma 1 holds for $p+1$. Lemma 2. Suppose that for integers $n, m \geq 0, \beta_{n}=\gamma_{n}=\lambda>\beta_{m}=$ $\gamma_{m}=\mu$. Then $b_{n+m}, c_{n+m}$ are nonzero and $\beta_{n+m}=\gamma_{n+m}=\mu$. Proof. Calculating $\left(a^{\prime}+2^{\lambda}\left(b^{\prime} \sqrt[3]{2}+c^{\prime} \sqrt[3]{4}\right)\right)\left(a^{\prime \prime}+2^{\mu}\left(b^{\prime \prime} \sqrt[3]{2}+c^{\prime \prime} \sqrt[3]{4}\right)\right)$, with $a^{\prime}, b^{\prime}, c^{\prime}, a^{\prime \prime}, b^{\prime \prime}, c^{\prime \prime}$ odd, we easily obtain the product $A+2^{\mu}(B \sqrt[3]{2}+$ $C \sqrt[3]{4})$, where $A, B=a^{\prime} b^{\prime \prime}+2^{\lambda-\mu} E$, and $C=a^{\prime} c^{\prime \prime}+2^{\lambda-\mu} F$ are odd, which proves Lemma 2. Since every integer $n>0$ can be written as $n=2^{p_{r}}+\cdots+2^{p_{1}}$, with $0 \leq p_{1}<\cdots<p_{r}$, from Lemmas 1 and 2 it follows that $c_{n}$ is nonzero, and that $\gamma_{n}=p_{1}+2$. Remark. $b_{1989}$ and $c_{1989}$ are divisible by 4 , but not by 8 .
|
{
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0ce82997-869e-50c0-924c-a119d5b78a7b
| 24,266
|
14. (JAP 2) In the coordinate plane a rectangle with vertices $(0,0),(m, 0)$, $(0, n),(m, n)$ is given where both $m$ and $n$ are odd integers. The rectangle is partitioned into triangles in such a way that (i) each triangle in the partition has at least one side (to be called a "good" side) that lies on a line of the form $x=j$ or $y=k$, where $j$ and $k$ are integers, and the altitude on this side has length 1 ; (ii) each "bad" side (i.e., a side of any triangle in the partition that is not a "good" one) is a common side of two triangles in the partition. Prove that there exist at least two triangles in the partition each of which has two good sides.
|
14. Let $V$ be the set of all midpoints of bad sides, and $E$ the set of segments connecting two points in $V$ that belong to the same triangle. Each edge in $E$ is parallel to exactly one good side and thus is parallel to the coordinate grid and has half-integer coordinates. Thus, the edges of $E$ are a subset of the grid formed by joining the centers of the squares in the original grid to each other. Let $G$ be a graph whose set of vertices is $V$ and set of edges is $E$. The degree of each vertex $X$, denoted by $d(X)$, is 0 , 1 , or 2 . We observe the following cases: (i) $d(X)=0$ for some $X$. Then both triangles containing $X$ have two good sides. (ii) $d(X)=1$ for some $X$. Since $\sum_{X \in V} d(X)=2|E|$ is even, it follows that at least another vertex $Y$ has the degree 1. Hence both $X$ and $Y$ belong to triangles having two good sides. (iii) $d(X)=2$ for all $X \in V$. We will show that this case cannot occur. We prove first that centers of all the squares of the $m \times n$ board belong to $V \cup E$. A bad side contains no points with half-integer coordinates in its interior other than its midpoint. Therefore either a point $X$ is in $V$, or it lies on the segment connecting the midpoints of the two bad sides. Evidently, the graph $G$ can be partitioned into disjoint cycles. Each center of a square is passed exactly once in exactly one cycle. Let us color the board black and white in a standard chessboard fashion. Each cycle passes through centers that must alternate in color, and hence it contains an equal number of black and white centers. Consequently, the numbers of black and white squares on the entire board must be equal, contradicting the condition that $m$ and $n$ are odd. Our proof is thus completed.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
14. (JAP 2) In the coordinate plane a rectangle with vertices $(0,0),(m, 0)$, $(0, n),(m, n)$ is given where both $m$ and $n$ are odd integers. The rectangle is partitioned into triangles in such a way that (i) each triangle in the partition has at least one side (to be called a "good" side) that lies on a line of the form $x=j$ or $y=k$, where $j$ and $k$ are integers, and the altitude on this side has length 1 ; (ii) each "bad" side (i.e., a side of any triangle in the partition that is not a "good" one) is a common side of two triangles in the partition. Prove that there exist at least two triangles in the partition each of which has two good sides.
|
14. Let $V$ be the set of all midpoints of bad sides, and $E$ the set of segments connecting two points in $V$ that belong to the same triangle. Each edge in $E$ is parallel to exactly one good side and thus is parallel to the coordinate grid and has half-integer coordinates. Thus, the edges of $E$ are a subset of the grid formed by joining the centers of the squares in the original grid to each other. Let $G$ be a graph whose set of vertices is $V$ and set of edges is $E$. The degree of each vertex $X$, denoted by $d(X)$, is 0 , 1 , or 2 . We observe the following cases: (i) $d(X)=0$ for some $X$. Then both triangles containing $X$ have two good sides. (ii) $d(X)=1$ for some $X$. Since $\sum_{X \in V} d(X)=2|E|$ is even, it follows that at least another vertex $Y$ has the degree 1. Hence both $X$ and $Y$ belong to triangles having two good sides. (iii) $d(X)=2$ for all $X \in V$. We will show that this case cannot occur. We prove first that centers of all the squares of the $m \times n$ board belong to $V \cup E$. A bad side contains no points with half-integer coordinates in its interior other than its midpoint. Therefore either a point $X$ is in $V$, or it lies on the segment connecting the midpoints of the two bad sides. Evidently, the graph $G$ can be partitioned into disjoint cycles. Each center of a square is passed exactly once in exactly one cycle. Let us color the board black and white in a standard chessboard fashion. Each cycle passes through centers that must alternate in color, and hence it contains an equal number of black and white centers. Consequently, the numbers of black and white squares on the entire board must be equal, contradicting the condition that $m$ and $n$ are odd. Our proof is thus completed.
|
{
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96451e6d-93a4-5d09-bf0f-2874af21c15d
| 24,279
|
16. (NET 1) ${ }^{\text {IMO6 }}$ Is there a 1990-gon with the following properties (i) and (ii)? (i) All angles are equal; (ii) The lengths of the 1990 sides are a permutation of the numbers $1^{2}, 2^{2}, \ldots, 1989^{2}, 1990^{2}$.
|
16. Let $A_{0} A_{1} \ldots A_{1989}$ be the desired 1990-gon. We also define $A_{1990}=A_{0}$. Let $O$ be an arbitrary point. For $1 \leq i \leq 1990$ let $B_{i}$ be a point such that $\overrightarrow{O B_{i}}=\overrightarrow{A_{i-1} A_{i}}$. We define $B_{0}=B_{1990}$. The points $B_{i}$ must satisfy the following properties: $\angle B_{i} O B_{i+1}=\frac{2 \pi}{1990}, 0 \leq i \leq 1989$, lengths of $O B_{i}$ are a permutation of $1^{2}, 2^{2}, \ldots, 1989^{2}, 1990^{2}$, and $\sum_{i=0}^{1989} \overrightarrow{O B_{i}}=\overrightarrow{0}$. Conversely, any such set of points $B_{i}$ corresponds to a desired 1990-gon. Hence, our goal is to construct vectors $\overrightarrow{O B_{i}}$ satisfying all the stated properties. Let us group vectors of lengths $(2 n-1)^{2}$ and $(2 n)^{2}$ into pairs and put them diametrically opposite each other. The length of the resulting vectors is $4 n-1$. The problem thus reduces to arranging vectors of lengths $3,7,11, \ldots, 3979$ at mutual angles of $\frac{2 \pi}{995}$ such that their sum is $\overrightarrow{0}$. We partition the 995 directions into 199 sets of five directions at mutual angles $\frac{2 \pi}{5}$. The directions when intersected with a unit circle form a regular pentagon. We group the set of lengths of vectors $3,7, \ldots, 3979$ into 199 sets of five consecutive elements of the set. We place each group of lengths on directions belonging to the same group of directions, thus constructing five vectors. We use that $\overrightarrow{O C_{1}}+\cdots+\overrightarrow{O C_{n}}=0$ where $O$ is the center of a regular $n$-gon $C_{1} \ldots C_{n}$. In other words, vectors of equal lengths along directions that form a regular $n$-gon cancel each other out. Such are the groups of five directions. Hence, we can assume for each group of five lengths for its lengths to be $\{0,4,8,12,16\}$. We place these five lengths in a random fashion on a single group of directions. We then rotate the configuration clockwise by $\frac{2 \pi}{199}$ to cover other groups of directions and repeat until all groups of directions are exhausted. It follows that all vectors of each of the lengths $\{0,4,8,12,16\}$ will form a regular 199-gon and will thus cancel each other out. We have thus constructed a way of obtaining points $B_{i}$ and have hence shown the existence of the 1990-gon satisfying $(i)$ and (ii).
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
16. (NET 1) ${ }^{\text {IMO6 }}$ Is there a 1990-gon with the following properties (i) and (ii)? (i) All angles are equal; (ii) The lengths of the 1990 sides are a permutation of the numbers $1^{2}, 2^{2}, \ldots, 1989^{2}, 1990^{2}$.
|
16. Let $A_{0} A_{1} \ldots A_{1989}$ be the desired 1990-gon. We also define $A_{1990}=A_{0}$. Let $O$ be an arbitrary point. For $1 \leq i \leq 1990$ let $B_{i}$ be a point such that $\overrightarrow{O B_{i}}=\overrightarrow{A_{i-1} A_{i}}$. We define $B_{0}=B_{1990}$. The points $B_{i}$ must satisfy the following properties: $\angle B_{i} O B_{i+1}=\frac{2 \pi}{1990}, 0 \leq i \leq 1989$, lengths of $O B_{i}$ are a permutation of $1^{2}, 2^{2}, \ldots, 1989^{2}, 1990^{2}$, and $\sum_{i=0}^{1989} \overrightarrow{O B_{i}}=\overrightarrow{0}$. Conversely, any such set of points $B_{i}$ corresponds to a desired 1990-gon. Hence, our goal is to construct vectors $\overrightarrow{O B_{i}}$ satisfying all the stated properties. Let us group vectors of lengths $(2 n-1)^{2}$ and $(2 n)^{2}$ into pairs and put them diametrically opposite each other. The length of the resulting vectors is $4 n-1$. The problem thus reduces to arranging vectors of lengths $3,7,11, \ldots, 3979$ at mutual angles of $\frac{2 \pi}{995}$ such that their sum is $\overrightarrow{0}$. We partition the 995 directions into 199 sets of five directions at mutual angles $\frac{2 \pi}{5}$. The directions when intersected with a unit circle form a regular pentagon. We group the set of lengths of vectors $3,7, \ldots, 3979$ into 199 sets of five consecutive elements of the set. We place each group of lengths on directions belonging to the same group of directions, thus constructing five vectors. We use that $\overrightarrow{O C_{1}}+\cdots+\overrightarrow{O C_{n}}=0$ where $O$ is the center of a regular $n$-gon $C_{1} \ldots C_{n}$. In other words, vectors of equal lengths along directions that form a regular $n$-gon cancel each other out. Such are the groups of five directions. Hence, we can assume for each group of five lengths for its lengths to be $\{0,4,8,12,16\}$. We place these five lengths in a random fashion on a single group of directions. We then rotate the configuration clockwise by $\frac{2 \pi}{199}$ to cover other groups of directions and repeat until all groups of directions are exhausted. It follows that all vectors of each of the lengths $\{0,4,8,12,16\}$ will form a regular 199-gon and will thus cancel each other out. We have thus constructed a way of obtaining points $B_{i}$ and have hence shown the existence of the 1990-gon satisfying $(i)$ and (ii).
|
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|
c3875878-8e2e-585f-9513-140a533b9fa2
| 24,283
|
20. (POL 3) Prove that every integer $k$ greater than 1 has a multiple that is less than $k^{4}$ and can be written in the decimal system with at most four different digits.
|
20. Let $n$ be the unique integer such that $2^{n-1} \leq k<2^{n}$. Let $S(n)$ be the set of numbers less than $10^{n}$ that are written with only the digits $\{0,1\}$ in the decimal system. Evidently $|S(n)|=2^{n}>k$ and hence there exist two numbers $x, y \in S(n)$ such that $k \mid x-y$. Let us show that $w=|x-y|$ is the desired number. By definition $k \mid w$. We also have $$ w<1.2 \cdot 10^{n-1} \leq 1.2 \cdot\left(2^{3} \sqrt{2}\right)^{n-1} \leq 1.2 \cdot k^{3} \sqrt{k} \leq k^{4} $$ Finally, since $x, y \in S(n)$, it follows that $w=|x-y|$ can be written using only the digits $\{0,1,8,9\}$. This completes the proof.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
20. (POL 3) Prove that every integer $k$ greater than 1 has a multiple that is less than $k^{4}$ and can be written in the decimal system with at most four different digits.
|
20. Let $n$ be the unique integer such that $2^{n-1} \leq k<2^{n}$. Let $S(n)$ be the set of numbers less than $10^{n}$ that are written with only the digits $\{0,1\}$ in the decimal system. Evidently $|S(n)|=2^{n}>k$ and hence there exist two numbers $x, y \in S(n)$ such that $k \mid x-y$. Let us show that $w=|x-y|$ is the desired number. By definition $k \mid w$. We also have $$ w<1.2 \cdot 10^{n-1} \leq 1.2 \cdot\left(2^{3} \sqrt{2}\right)^{n-1} \leq 1.2 \cdot k^{3} \sqrt{k} \leq k^{4} $$ Finally, since $x, y \in S(n)$, it follows that $w=|x-y|$ can be written using only the digits $\{0,1,8,9\}$. This completes the proof.
|
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|
39d39de1-88b8-5478-9d6c-cf79bffc1597
| 24,296
|
22. (ROM 4) Ten localities are served by two international airlines such that there exists a direct service (without stops) between any two of these localities and all airline schedules offer round-trip service between the cities they serve. Prove that at least one of the airlines can offer two disjoint round trips each containing an odd number of landings.
|
22. We can assume without loss of generality that each connection is serviced by only one airline and the problem reduces to finding two disjoint monochromatic cycles of the same color and of odd length on a complete graph of 10 points colored by two colors. We use the following two standard lemmas: Lemma 1. Given a complete graph on six points whose edges are colored with two colors there exists a monochromatic triangle. $\operatorname{Proof}$. Let us denote the vertices by $c_{1}, c_{2}, c_{3}, c_{4}, c_{5}, c_{6}$. By the pigeonhole principle at least three vertices out of $c_{1}$, say $c_{2}, c_{3}, c_{4}$, are of the same color, let us call it red. Assuming that at least one of the edges connecting points $c_{2}, c_{3}, c_{4}$ is red, the connected points along with $c_{1}$ form a red triangle. Otherwise, edges connecting $c_{2}, c_{3}, c_{4}$ are all of the opposite color, let us call it blue, and hence in all cases we have a monochromatic triangle. Lemma 2. Given a complete graph on five points whose edges are colored two colors there exists a monochromatic triangle or a monochromatic cycle of length five. Proof. Let us denote the vertices by $c_{1}, c_{2}, c_{3}, c_{4}, c_{5}$. Assume that out of a point $c_{i}$ three vertices are of the same color. We can then proceed as in Lemma 1 to obtain a monochromatic triangle. Otherwise, each point is connected to other points with exactly two red and two blue vertices. Hence, we obtain monochromatic cycles starting from a single point and moving along the edges of the same color. Since each cycle must be of length at least three (i.e., we cannot have more than one cycle of one color), it follows that for both red and blue we must have one cycle of length five of that color. We now apply the lemmas. Let us denote the vertices by $c_{1}, c_{2}, \ldots, c_{10}$. We apply Lemma 1 to vertices $c_{1}, \ldots, c_{6}$ to obtain a monochromatic triangle. Out of the seven remaining vertices we select 6 and again apply Lemma 1 to obtain another monochromatic triangle. If they are of the same color, we are done. Otherwise, out of the nine edges connecting the two triangles of opposite color at least 5 are of the same color, we can assume blue w.l.o.g., and hence a vertex of a red triangle must contain at least two blue edges whose endpoints are connected with a blue edge. Hence there exist two triangles of different colors joined at a vertex. These take up five points. Applying Lemma 2 on the five remaining points, we obtain a monochromatic cycle of odd length that is of the same color as one of the two joined triangles and disjoint from both of them.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
22. (ROM 4) Ten localities are served by two international airlines such that there exists a direct service (without stops) between any two of these localities and all airline schedules offer round-trip service between the cities they serve. Prove that at least one of the airlines can offer two disjoint round trips each containing an odd number of landings.
|
22. We can assume without loss of generality that each connection is serviced by only one airline and the problem reduces to finding two disjoint monochromatic cycles of the same color and of odd length on a complete graph of 10 points colored by two colors. We use the following two standard lemmas: Lemma 1. Given a complete graph on six points whose edges are colored with two colors there exists a monochromatic triangle. $\operatorname{Proof}$. Let us denote the vertices by $c_{1}, c_{2}, c_{3}, c_{4}, c_{5}, c_{6}$. By the pigeonhole principle at least three vertices out of $c_{1}$, say $c_{2}, c_{3}, c_{4}$, are of the same color, let us call it red. Assuming that at least one of the edges connecting points $c_{2}, c_{3}, c_{4}$ is red, the connected points along with $c_{1}$ form a red triangle. Otherwise, edges connecting $c_{2}, c_{3}, c_{4}$ are all of the opposite color, let us call it blue, and hence in all cases we have a monochromatic triangle. Lemma 2. Given a complete graph on five points whose edges are colored two colors there exists a monochromatic triangle or a monochromatic cycle of length five. Proof. Let us denote the vertices by $c_{1}, c_{2}, c_{3}, c_{4}, c_{5}$. Assume that out of a point $c_{i}$ three vertices are of the same color. We can then proceed as in Lemma 1 to obtain a monochromatic triangle. Otherwise, each point is connected to other points with exactly two red and two blue vertices. Hence, we obtain monochromatic cycles starting from a single point and moving along the edges of the same color. Since each cycle must be of length at least three (i.e., we cannot have more than one cycle of one color), it follows that for both red and blue we must have one cycle of length five of that color. We now apply the lemmas. Let us denote the vertices by $c_{1}, c_{2}, \ldots, c_{10}$. We apply Lemma 1 to vertices $c_{1}, \ldots, c_{6}$ to obtain a monochromatic triangle. Out of the seven remaining vertices we select 6 and again apply Lemma 1 to obtain another monochromatic triangle. If they are of the same color, we are done. Otherwise, out of the nine edges connecting the two triangles of opposite color at least 5 are of the same color, we can assume blue w.l.o.g., and hence a vertex of a red triangle must contain at least two blue edges whose endpoints are connected with a blue edge. Hence there exist two triangles of different colors joined at a vertex. These take up five points. Applying Lemma 2 on the five remaining points, we obtain a monochromatic cycle of odd length that is of the same color as one of the two joined triangles and disjoint from both of them.
|
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a5d45b73-dc18-5563-a3ce-02e020a0aa2e
| 24,300
|
26. (USA 2) Let $P$ be a cubic polynomial with rational coefficients, and let $q_{1}, q_{2}, q_{3}, \ldots$ be a sequence of rational numbers such that $q_{n}=P\left(q_{n+1}\right)$ for all $n \geq 1$. Prove that there exists $k \geq 1$ such that for all $n \geq 1, q_{n+k}=q_{n}$.
|
26. We note that $|P(x) / x| \rightarrow \infty$. Hence, there exists an integer number $M$ such that $M>\left|q_{1}\right|$ and $|P(x)| \leq|x| \Rightarrow|x|<M$. It follows that $\left|q_{i}\right|<M$ for all $i \in \mathbb{N}$ because assuming $\left|q_{i}\right| \geq M$ for some $i$ we get $\left|q_{i-1}\right|=$ $\left|P\left(q_{i}\right)\right|>\left|q_{i}\right| \geq M$ and this ultimately contradicts $\left|q_{1}\right|<M$. Let us define $q_{1}=\frac{r}{s}$ and $P(x)=\frac{a x^{3}+b x^{2}+c x+d}{e}$ where $r, s, a, b, c, d, e$ are all integers. For $N=s a$ we shall prove by induction that $N q_{i}$ is an integer for all $i \in \mathbb{N}$. By definition $N \neq 0$. For $i=1$ this obviously holds. Assume it holds for some $i \in \mathbb{N}$. Then using $q_{i}=P\left(q_{i+1}\right)$ we have that $N q_{i+1}$ is a zero of the polynomial $$ \begin{aligned} Q(x) & =\frac{e}{a} N^{3}\left(P\left(\frac{x}{N}\right)-q_{i}\right) \\ & =x^{3}+(s b) x^{2}+\left(s^{2} a c\right) x+\left(s^{3} a^{2} d-s^{2} a e\left(N q_{i}\right)\right) . \end{aligned} $$ Since $Q(x)$ is a monic polynomial with integer coefficients (a conclusion for which we must assume the induction hypothesis) and $N q_{i+1}$ is rational it follows by the rational root theorem that $N q_{i+1}$ is an integer. It follows that all $q_{i}$ are multiples of $1 / N$. Since $-M<q_{i}<M$ we conclude that $q_{i}$ can take less than $T=2 M|N|$ distinct values. Therefore for each $j$ there are $m_{j}$ and $m_{j}+k_{j}\left(k_{j}>0\right)$ both belonging to the set $\{j T+1, j T+2, \ldots, j T+T\}$ such that $q_{m_{j}}=q_{m_{j}+k_{j}}$. Since $k_{j}<T$ for all $k_{j}$ it follows that there exists a positive integer $k$ which appears an infinite number of times in the sequence $k_{j}$, i.e. there exist infinitely many integers $m$ such that $q_{m}=q_{m+k}$. Moreover, $q_{m}=q_{m+k}$ clearly implies $q_{n}=q_{n+k}$ for all $n \leq m$. Hence $q_{n}=q_{n+k}$ holds for all $n$.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
26. (USA 2) Let $P$ be a cubic polynomial with rational coefficients, and let $q_{1}, q_{2}, q_{3}, \ldots$ be a sequence of rational numbers such that $q_{n}=P\left(q_{n+1}\right)$ for all $n \geq 1$. Prove that there exists $k \geq 1$ such that for all $n \geq 1, q_{n+k}=q_{n}$.
|
26. We note that $|P(x) / x| \rightarrow \infty$. Hence, there exists an integer number $M$ such that $M>\left|q_{1}\right|$ and $|P(x)| \leq|x| \Rightarrow|x|<M$. It follows that $\left|q_{i}\right|<M$ for all $i \in \mathbb{N}$ because assuming $\left|q_{i}\right| \geq M$ for some $i$ we get $\left|q_{i-1}\right|=$ $\left|P\left(q_{i}\right)\right|>\left|q_{i}\right| \geq M$ and this ultimately contradicts $\left|q_{1}\right|<M$. Let us define $q_{1}=\frac{r}{s}$ and $P(x)=\frac{a x^{3}+b x^{2}+c x+d}{e}$ where $r, s, a, b, c, d, e$ are all integers. For $N=s a$ we shall prove by induction that $N q_{i}$ is an integer for all $i \in \mathbb{N}$. By definition $N \neq 0$. For $i=1$ this obviously holds. Assume it holds for some $i \in \mathbb{N}$. Then using $q_{i}=P\left(q_{i+1}\right)$ we have that $N q_{i+1}$ is a zero of the polynomial $$ \begin{aligned} Q(x) & =\frac{e}{a} N^{3}\left(P\left(\frac{x}{N}\right)-q_{i}\right) \\ & =x^{3}+(s b) x^{2}+\left(s^{2} a c\right) x+\left(s^{3} a^{2} d-s^{2} a e\left(N q_{i}\right)\right) . \end{aligned} $$ Since $Q(x)$ is a monic polynomial with integer coefficients (a conclusion for which we must assume the induction hypothesis) and $N q_{i+1}$ is rational it follows by the rational root theorem that $N q_{i+1}$ is an integer. It follows that all $q_{i}$ are multiples of $1 / N$. Since $-M<q_{i}<M$ we conclude that $q_{i}$ can take less than $T=2 M|N|$ distinct values. Therefore for each $j$ there are $m_{j}$ and $m_{j}+k_{j}\left(k_{j}>0\right)$ both belonging to the set $\{j T+1, j T+2, \ldots, j T+T\}$ such that $q_{m_{j}}=q_{m_{j}+k_{j}}$. Since $k_{j}<T$ for all $k_{j}$ it follows that there exists a positive integer $k$ which appears an infinite number of times in the sequence $k_{j}$, i.e. there exist infinitely many integers $m$ such that $q_{m}=q_{m+k}$. Moreover, $q_{m}=q_{m+k}$ clearly implies $q_{n}=q_{n+k}$ for all $n \leq m$. Hence $q_{n}=q_{n+k}$ holds for all $n$.
|
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b8d6adfa-6760-511e-8704-727292d6d31f
| 24,308
|
28. (USS 3) Prove that on the coordinate plane it is impossible to draw a closed broken line such that (i) the coordinates of each vertex are rational; (ii) the length each of its edges is 1 ; (iii) the line has an odd number of vertices.
|
28. Let us first prove the following lemma. Lemma. Let $\left(b^{\prime} / a^{\prime}, d^{\prime} / c^{\prime}\right)$ and $\left(b^{\prime \prime} / a^{\prime \prime}, d^{\prime \prime} / c^{\prime \prime}\right)$ be two points with rational coordinates where the fractions given are irreducible. If both $a^{\prime}$ and $c^{\prime}$ are odd and the distance between the two points is 1 then it follows that $a^{\prime \prime}$ and $c^{\prime \prime}$ are odd, and that $b^{\prime}+d^{\prime}$ and $b^{\prime \prime}+d^{\prime \prime}$ are of a different parity. Proof. Let $b / a$ and $d / c$ be irreducible fractions such that $b^{\prime} / a^{\prime}-b^{\prime \prime} / a^{\prime \prime}=$ $b / a$ and $d^{\prime} / c^{\prime}-d^{\prime \prime} / c^{\prime \prime}=d / c$. Then it follows that $b^{2} / a^{2}+d^{2} / c^{2}=$ $1 \Rightarrow b^{2} c^{2}+a^{2} d^{2}=a^{2} c^{2}$. Since $(a, b)=1$ and $(c, d)=1$ it follows that $a|c, c| a$ and hence $a=c$. Consequently $b^{2}+d^{2}=a^{2}$. Since $a$ is mutually co-prime to $b$ and $d$ it follows that $a$ and $b+d$ are odd. From $b^{\prime \prime} / a^{\prime \prime}=b / a+b^{\prime} / a^{\prime}$ we get that $a^{\prime \prime} \mid a a^{\prime}$, so $a^{\prime \prime}$ is odd. Similarly, $c^{\prime \prime}$ is odd as well. Now it follows that $b^{\prime \prime} \equiv b+b^{\prime}$ and similarly $d^{\prime \prime} \equiv d+d^{\prime}$ $(\bmod 2)$. Hence $b^{\prime \prime}+d^{\prime \prime} \equiv b^{\prime}+d^{\prime}+b+d \equiv b^{\prime}+d^{\prime}+1(\bmod 2)$, from which it follows that $b^{\prime}+d^{\prime}$ and $b^{\prime \prime}+d^{\prime \prime}$ are of a different parity. Without loss of generality we start from the origin of the coordinate system $(0 / 1,0 / 1)$. Initially $b+d=0$ and after moving to each subsequent point along the broken line $b+d$ changes parity by the lemma. Hence it will not be possible to return to the origin after an odd number of steps since $b+d$ will be odd.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
28. (USS 3) Prove that on the coordinate plane it is impossible to draw a closed broken line such that (i) the coordinates of each vertex are rational; (ii) the length each of its edges is 1 ; (iii) the line has an odd number of vertices.
|
28. Let us first prove the following lemma. Lemma. Let $\left(b^{\prime} / a^{\prime}, d^{\prime} / c^{\prime}\right)$ and $\left(b^{\prime \prime} / a^{\prime \prime}, d^{\prime \prime} / c^{\prime \prime}\right)$ be two points with rational coordinates where the fractions given are irreducible. If both $a^{\prime}$ and $c^{\prime}$ are odd and the distance between the two points is 1 then it follows that $a^{\prime \prime}$ and $c^{\prime \prime}$ are odd, and that $b^{\prime}+d^{\prime}$ and $b^{\prime \prime}+d^{\prime \prime}$ are of a different parity. Proof. Let $b / a$ and $d / c$ be irreducible fractions such that $b^{\prime} / a^{\prime}-b^{\prime \prime} / a^{\prime \prime}=$ $b / a$ and $d^{\prime} / c^{\prime}-d^{\prime \prime} / c^{\prime \prime}=d / c$. Then it follows that $b^{2} / a^{2}+d^{2} / c^{2}=$ $1 \Rightarrow b^{2} c^{2}+a^{2} d^{2}=a^{2} c^{2}$. Since $(a, b)=1$ and $(c, d)=1$ it follows that $a|c, c| a$ and hence $a=c$. Consequently $b^{2}+d^{2}=a^{2}$. Since $a$ is mutually co-prime to $b$ and $d$ it follows that $a$ and $b+d$ are odd. From $b^{\prime \prime} / a^{\prime \prime}=b / a+b^{\prime} / a^{\prime}$ we get that $a^{\prime \prime} \mid a a^{\prime}$, so $a^{\prime \prime}$ is odd. Similarly, $c^{\prime \prime}$ is odd as well. Now it follows that $b^{\prime \prime} \equiv b+b^{\prime}$ and similarly $d^{\prime \prime} \equiv d+d^{\prime}$ $(\bmod 2)$. Hence $b^{\prime \prime}+d^{\prime \prime} \equiv b^{\prime}+d^{\prime}+b+d \equiv b^{\prime}+d^{\prime}+1(\bmod 2)$, from which it follows that $b^{\prime}+d^{\prime}$ and $b^{\prime \prime}+d^{\prime \prime}$ are of a different parity. Without loss of generality we start from the origin of the coordinate system $(0 / 1,0 / 1)$. Initially $b+d=0$ and after moving to each subsequent point along the broken line $b+d$ changes parity by the lemma. Hence it will not be possible to return to the origin after an odd number of steps since $b+d$ will be odd.
|
{
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5b15ea9f-7980-5e7e-8db8-08964398c27c
| 24,313
|
4. (CZS 2) Assume that the set of all positive integers is decomposed into $r$ (disjoint) subsets $A_{1} \cup A_{2} \cup \cdots A_{r}=\mathbb{N}$. Prove that one of them, say $A_{i}$, has the following property: There exists a positive $m$ such that for any $k$ one can find numbers $a_{1}, a_{2}, \ldots, a_{k}$ in $A_{i}$ with $0<a_{j+1}-a_{j} \leq m$ $(1 \leq j \leq k-1)$.
|
4. Assuming that $A_{1}$ is not such a set $A_{i}$, it follows that for every $m$ there exist $m$ consecutive numbers not in $A_{1}$. It follows that $A_{2} \cup A_{3} \cup \cdots \cup A_{r}$ contains arbitrarily long sequences of numbers. Inductively, let us assume that $A_{j} \cup A_{j+1} \cup \cdots \cup A_{r}$ contains arbitrarily long sequences of consecutive numbers and none of $A_{1}, A_{2}, \ldots, A_{j-1}$ is the desired set $A_{i}$. Let us assume that $A_{j}$ is also not $A_{i}$. Hence for each $m$ there exists $k(m)$ such that among $k(m)$ elements of $A_{j}$ there exist two consecutive elements that differ by at least $m$. Let us consider $m \cdot k(m)$ consecutive numbers in $A_{j} \cup \cdots \cup A_{r}$, which exist by the induction hypothesis. Then either $A_{j}$ contains fewer than $k(m)$ of these integers, in which case $A_{j+1} \cup \cdots \cup A_{r}$ contains $m$ consecutive integers by the pigeonhole principle or $A_{j}$ contains $k(m)$ integers among which there exists a gap of length $m$ of consecutive integers that belong to $A_{j+1} \cup \cdots \cup A_{r}$. Hence we have proven that $A_{j+1} \cup \cdots \cup A_{r}$ contains sequences of integers of arbitrary length. By induction, assuming that $A_{1}, A_{2}, \ldots, A_{r-1}$ do not satisfy the conditions to be the set $A_{i}$, it follows that $A_{r}$ contains sequences of consecutive integers of arbitrary length and hence satisfies the conditions necessary for it to be the set $A_{i}$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
4. (CZS 2) Assume that the set of all positive integers is decomposed into $r$ (disjoint) subsets $A_{1} \cup A_{2} \cup \cdots A_{r}=\mathbb{N}$. Prove that one of them, say $A_{i}$, has the following property: There exists a positive $m$ such that for any $k$ one can find numbers $a_{1}, a_{2}, \ldots, a_{k}$ in $A_{i}$ with $0<a_{j+1}-a_{j} \leq m$ $(1 \leq j \leq k-1)$.
|
4. Assuming that $A_{1}$ is not such a set $A_{i}$, it follows that for every $m$ there exist $m$ consecutive numbers not in $A_{1}$. It follows that $A_{2} \cup A_{3} \cup \cdots \cup A_{r}$ contains arbitrarily long sequences of numbers. Inductively, let us assume that $A_{j} \cup A_{j+1} \cup \cdots \cup A_{r}$ contains arbitrarily long sequences of consecutive numbers and none of $A_{1}, A_{2}, \ldots, A_{j-1}$ is the desired set $A_{i}$. Let us assume that $A_{j}$ is also not $A_{i}$. Hence for each $m$ there exists $k(m)$ such that among $k(m)$ elements of $A_{j}$ there exist two consecutive elements that differ by at least $m$. Let us consider $m \cdot k(m)$ consecutive numbers in $A_{j} \cup \cdots \cup A_{r}$, which exist by the induction hypothesis. Then either $A_{j}$ contains fewer than $k(m)$ of these integers, in which case $A_{j+1} \cup \cdots \cup A_{r}$ contains $m$ consecutive integers by the pigeonhole principle or $A_{j}$ contains $k(m)$ integers among which there exists a gap of length $m$ of consecutive integers that belong to $A_{j+1} \cup \cdots \cup A_{r}$. Hence we have proven that $A_{j+1} \cup \cdots \cup A_{r}$ contains sequences of integers of arbitrary length. By induction, assuming that $A_{1}, A_{2}, \ldots, A_{r-1}$ do not satisfy the conditions to be the set $A_{i}$, it follows that $A_{r}$ contains sequences of consecutive integers of arbitrary length and hence satisfies the conditions necessary for it to be the set $A_{i}$.
|
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|
9cccecc5-6d00-5e67-8d43-aeb31d9a5739
| 24,318
|
5. (FRA 1) Given $\triangle A B C$ with no side equal to another side, let $G, K$, and $H$ be its centroid, incenter, and orthocenter, respectively. Prove that $\angle G K H>90^{\circ}$.
|
5. Let $O$ be the circumcenter of $A B C, E$ the midpoint of $O H$, and $R$ and $r$ the radii of the circumcircle and incircle respectively. We use the following facts from elementary geometry: $\overrightarrow{O H}=3 \overrightarrow{O G}, O K^{2}=R^{2}-2 R r$, and $K E=\frac{R}{2}-r$. Hence $\overrightarrow{K H}=2 \overrightarrow{K E}-\overrightarrow{K O}$ and $\overrightarrow{K G}=\frac{2 \overrightarrow{K E}+\overrightarrow{K O}}{3}$. We then obtain $$ \overrightarrow{K H} \cdot \overrightarrow{K G}=\frac{1}{3}\left(4 K E^{2}-K O^{2}\right)=-\frac{2}{3} r(R-2 r)<0 . $$ Hence $\cos \angle G K H<0 \Rightarrow \angle G K H>90^{\circ}$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
5. (FRA 1) Given $\triangle A B C$ with no side equal to another side, let $G, K$, and $H$ be its centroid, incenter, and orthocenter, respectively. Prove that $\angle G K H>90^{\circ}$.
|
5. Let $O$ be the circumcenter of $A B C, E$ the midpoint of $O H$, and $R$ and $r$ the radii of the circumcircle and incircle respectively. We use the following facts from elementary geometry: $\overrightarrow{O H}=3 \overrightarrow{O G}, O K^{2}=R^{2}-2 R r$, and $K E=\frac{R}{2}-r$. Hence $\overrightarrow{K H}=2 \overrightarrow{K E}-\overrightarrow{K O}$ and $\overrightarrow{K G}=\frac{2 \overrightarrow{K E}+\overrightarrow{K O}}{3}$. We then obtain $$ \overrightarrow{K H} \cdot \overrightarrow{K G}=\frac{1}{3}\left(4 K E^{2}-K O^{2}\right)=-\frac{2}{3} r(R-2 r)<0 . $$ Hence $\cos \angle G K H<0 \Rightarrow \angle G K H>90^{\circ}$.
|
{
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|
380cde0d-084b-54cd-ab3d-6aca3eb9d8de
| 24,320
|
7. (GRE 2) Let $f(0)=f(1)=0$ and $$ f(n+2)=4^{n+2} f(n+1)-16^{n+1} f(n)+n \cdot 2^{n^{2}}, \quad n=0,1,2,3, \ldots $$ Show that the numbers $f(1989), f(1990), f(1991)$ are divisible by 13.
|
7. Let $f(n)=g(n) 2^{n^{2}}$ for all $n$. The recursion then transforms into $g(n+$ 2) $-2 g(n+1)+g(n)=n \cdot 16^{-n-1}$ for $n \in \mathbb{N}_{0}$. By summing this equation from 0 to $n-1$, we get $$ g(n+1)-g(n)=\frac{1}{15^{2}} \cdot\left(1-(15 n+1) 16^{-n}\right) $$ By summing up again from 0 to $n-1$ we get $g(n)=\frac{1}{15^{3}} \cdot(15 n-32+$ $\left.(15 n+2) 16^{-n+1}\right)$. Hence $$ f(n)=\frac{1}{15^{3}} \cdot\left(15 n+2+(15 n-32) 16^{n-1}\right) \cdot 2^{(n-2)^{2}} $$ Now let us look at the values of $f(n)$ modulo 13: $$ f(n) \equiv 15 n+2+(15 n-32) 16^{n-1} \equiv 2 n+2+(2 n-6) 3^{n-1} $$ We have $3^{3} \equiv 1(\bmod 13)$. Plugging in $n \equiv 1(\bmod 13)$ and $n \equiv 1(\bmod$ 3 ) for $n=1990$ gives us $f(1990) \equiv 0(\bmod 13)$. We similarly calculate $f(1989) \equiv 0$ and $f(1991) \equiv 0(\bmod 13)$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
7. (GRE 2) Let $f(0)=f(1)=0$ and $$ f(n+2)=4^{n+2} f(n+1)-16^{n+1} f(n)+n \cdot 2^{n^{2}}, \quad n=0,1,2,3, \ldots $$ Show that the numbers $f(1989), f(1990), f(1991)$ are divisible by 13.
|
7. Let $f(n)=g(n) 2^{n^{2}}$ for all $n$. The recursion then transforms into $g(n+$ 2) $-2 g(n+1)+g(n)=n \cdot 16^{-n-1}$ for $n \in \mathbb{N}_{0}$. By summing this equation from 0 to $n-1$, we get $$ g(n+1)-g(n)=\frac{1}{15^{2}} \cdot\left(1-(15 n+1) 16^{-n}\right) $$ By summing up again from 0 to $n-1$ we get $g(n)=\frac{1}{15^{3}} \cdot(15 n-32+$ $\left.(15 n+2) 16^{-n+1}\right)$. Hence $$ f(n)=\frac{1}{15^{3}} \cdot\left(15 n+2+(15 n-32) 16^{n-1}\right) \cdot 2^{(n-2)^{2}} $$ Now let us look at the values of $f(n)$ modulo 13: $$ f(n) \equiv 15 n+2+(15 n-32) 16^{n-1} \equiv 2 n+2+(2 n-6) 3^{n-1} $$ We have $3^{3} \equiv 1(\bmod 13)$. Plugging in $n \equiv 1(\bmod 13)$ and $n \equiv 1(\bmod$ 3 ) for $n=1990$ gives us $f(1990) \equiv 0(\bmod 13)$. We similarly calculate $f(1989) \equiv 0$ and $f(1991) \equiv 0(\bmod 13)$.
|
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|
d6c2e1d3-5fee-5167-b3f3-25ecf012d758
| 24,326
|
1. (PHI 3) Let $A B C$ be any triangle and $P$ any point in its interior. Let $P_{1}, P_{2}$ be the feet of the perpendiculars from $P$ to the two sides $A C$ and $B C$. Draw $A P$ and $B P$, and from $C$ drop perpendiculars to $A P$ and $B P$. Let $Q_{1}$ and $Q_{2}$ be the feet of these perpendiculars. Prove that the lines $Q_{1} P_{2}, Q_{2} P_{1}$, and $A B$ are concurrent.
|
1. All the angles $\angle P P_{1} C, \angle P P_{2} C, \angle P Q_{1} C, \angle P Q_{2} C$ are right, hence $P_{1}, P_{2}$, $Q_{1}, Q_{2}$ lie on the circle with diameter $P C$. The result now follows immediately from Pascal's theorem applied to the hexagon $P_{1} P P_{2} Q_{1} C Q_{2}$. It tells us that the points of intersection of the three pairs of lines $P_{1} C, P Q_{1}$ (intersection $A$ ), $P_{1} Q_{2}, P_{2} Q_{1}$ (intersection  $X)$ and $P Q_{2}, P_{2} C$ (intersection $B$ ) are collinear.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
1. (PHI 3) Let $A B C$ be any triangle and $P$ any point in its interior. Let $P_{1}, P_{2}$ be the feet of the perpendiculars from $P$ to the two sides $A C$ and $B C$. Draw $A P$ and $B P$, and from $C$ drop perpendiculars to $A P$ and $B P$. Let $Q_{1}$ and $Q_{2}$ be the feet of these perpendiculars. Prove that the lines $Q_{1} P_{2}, Q_{2} P_{1}$, and $A B$ are concurrent.
|
1. All the angles $\angle P P_{1} C, \angle P P_{2} C, \angle P Q_{1} C, \angle P Q_{2} C$ are right, hence $P_{1}, P_{2}$, $Q_{1}, Q_{2}$ lie on the circle with diameter $P C$. The result now follows immediately from Pascal's theorem applied to the hexagon $P_{1} P P_{2} Q_{1} C Q_{2}$. It tells us that the points of intersection of the three pairs of lines $P_{1} C, P Q_{1}$ (intersection $A$ ), $P_{1} Q_{2}, P_{2} Q_{1}$ (intersection  $X)$ and $P Q_{2}, P_{2} C$ (intersection $B$ ) are collinear.
|
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|
2ffed25a-997c-5dc6-9b73-2db5687141ce
| 24,332
|
10. (USA 5) ${ }^{\mathrm{IMO} 4}$ Suppose $G$ is a connected graph with $n$ edges. Prove that it is possible to label the edges of $G$ from 1 to $n$ in such a way that in every vertex $v$ of $G$ with two or more incident edges, the set of numbers labeling those edges has no common divisor greater than 1.
|
10. We start at some vertex $v_{0}$ and walk along distinct edges of the graph, numbering them $1,2, \ldots$ in the order of appearance, until this is no longer possible without reusing an edge. If there are still edges which are not numbered, one of them has a vertex which has already been visited (else $G$ would not be connected). Starting from this vertex, we continue to walk along unused edges resuming the numbering, until we eventually get stuck. Repeating this procedure as long as possible, we shall number all the edges. Let $v$ be a vertex which is incident with $e \geq 2$ edges. If $v=v_{0}$, then it is on the edge 1 , so the gcd at $v$ is 1 . If $v \neq v_{0}$, suppose that it was reached for the first time by the edge $r$. At that time there was at least one unused edge incident with $v$ (as $e \geq 2$ ), hence one of them was labelled by $r+1$. The gcd at $v$ again equals $\operatorname{gcd}(r, r+1)=1$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
10. (USA 5) ${ }^{\mathrm{IMO} 4}$ Suppose $G$ is a connected graph with $n$ edges. Prove that it is possible to label the edges of $G$ from 1 to $n$ in such a way that in every vertex $v$ of $G$ with two or more incident edges, the set of numbers labeling those edges has no common divisor greater than 1.
|
10. We start at some vertex $v_{0}$ and walk along distinct edges of the graph, numbering them $1,2, \ldots$ in the order of appearance, until this is no longer possible without reusing an edge. If there are still edges which are not numbered, one of them has a vertex which has already been visited (else $G$ would not be connected). Starting from this vertex, we continue to walk along unused edges resuming the numbering, until we eventually get stuck. Repeating this procedure as long as possible, we shall number all the edges. Let $v$ be a vertex which is incident with $e \geq 2$ edges. If $v=v_{0}$, then it is on the edge 1 , so the gcd at $v$ is 1 . If $v \neq v_{0}$, suppose that it was reached for the first time by the edge $r$. At that time there was at least one unused edge incident with $v$ (as $e \geq 2$ ), hence one of them was labelled by $r+1$. The gcd at $v$ again equals $\operatorname{gcd}(r, r+1)=1$.
|
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|
fc765b3d-88c1-5c56-9728-3131f3063214
| 24,335
|
13. (POL 4) Given any integer $n \geq 2$, assume that the integers $a_{1}, a_{2}, \ldots, a_{n}$ are not divisible by $n$ and, moreover, that $n$ does not divide $a_{1}+a_{2}+$ $\cdots+a_{n}$. Prove that there exist at least $n$ different sequences $\left(e_{1}, e_{2}, \cdots, e_{n}\right)$ consisting of zeros or ones such that $e_{1} a_{1}+e_{2} a_{2}+\cdots+e_{n} a_{n}$ is divisible by $n$.
|
13. Call a sequence $e_{1}, \ldots, e_{n}$ good if $e_{1} a_{1}+\cdots+e_{n} a_{n}$ is divisible by $n$. Among the sums $s_{0}=0, s_{1}=a_{1}, s_{2}=a_{1}+a_{2}, \ldots, s_{n}=a_{1}+\cdots+a_{n}$, two give the same remainder modulo $n$, and their difference corresponds to a good sequence. To show that, permuting the $a_{i}$ 's, we can find $n-1$ different sequences, we use the following Lemma. Let $A$ be a $k \times n(k \leq n-2)$ matrix of zeros and ones, whose every row contains at least one 0 and at least two 1 's. Then it is possible to permute columns of $A$ is such a way that in any row 1 's do not form a block. Proof. We will use the induction on $k$. The case $k=1$ and arbitrary $n \geq 3$ is trivial. Suppose that $k \geq 2$ and that for $k-1$ and any $n \geq k+1$ the lemma is true. Consider a $k \times n$ matrix $A, n \geq k+2$. We mark an element $a_{i j}$ if either it is the only zero in the $i$-th row, or one of the 1 's in the row if it contains exactly two 1 's. Since $n \geq 4$, every row contains at most two marked elements, which adds up to at most $2 k<2 n$ marked elements in total. It follows that there is a column with at most one marked element. Assume w.l.o.g. that it is the first column and that $a_{1 j}$ isn't marked for $j>1$. The matrix $B$, obtained by omitting the first row and first column from $A$, satisfies the conditions of the lemma. Therefore, we can permute columns of $B$ and get the required form. Considered as a permutation of column of $A$, this permutation may leave a block of 1's only in the first row of $A$. In the case that it is so, if $a_{11}=1$ we put the first column in the last place, otherwise we put it between any two columns having 1's in the first row. The obtained matrix has the required property. Suppose now that we have got $k$ different nontrivial good sequences $e_{1}^{i}, \ldots, e_{n}^{i}, i=1, \ldots, k$, and that $k \leq n-2$. The matrix $A=\left(e_{j}^{i}\right)$ fulfils the conditions of Lemma, hence there is a permutation $\sigma$ from Lemma. Now among the sums $s_{0}=0, s_{1}=a_{\sigma(1)}, s_{2}=a_{\sigma(1)}+a_{\sigma(2)}$, $\ldots, s_{n}=a_{\sigma(1)}+\cdots+a_{\sigma(n)}$, two give the same remainder modulo $n$. Let $s_{p} \equiv s_{q}(\bmod n), p<q$. Then $n \mid s_{q}-s_{p}=a_{\sigma(p+1)}+\cdots+a_{\sigma(q)}$, and this yields a good sequence $e_{1}, \ldots, e_{n}$ with $e_{\sigma(p+1)}=\cdots=e_{\sigma(q)}=1$ and other $e$ 's equal to zero. Since from the construction we see that none of the sequences $e_{\sigma(j)^{i}}$ has all 1's in a block, in this way we have got a new nontrivial good sequence, and we can continue this procedure until there are $n-1$ sequences. Together with the trivial $0, \ldots, 0$ sequence, we have found $n$ good sequences.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
13. (POL 4) Given any integer $n \geq 2$, assume that the integers $a_{1}, a_{2}, \ldots, a_{n}$ are not divisible by $n$ and, moreover, that $n$ does not divide $a_{1}+a_{2}+$ $\cdots+a_{n}$. Prove that there exist at least $n$ different sequences $\left(e_{1}, e_{2}, \cdots, e_{n}\right)$ consisting of zeros or ones such that $e_{1} a_{1}+e_{2} a_{2}+\cdots+e_{n} a_{n}$ is divisible by $n$.
|
13. Call a sequence $e_{1}, \ldots, e_{n}$ good if $e_{1} a_{1}+\cdots+e_{n} a_{n}$ is divisible by $n$. Among the sums $s_{0}=0, s_{1}=a_{1}, s_{2}=a_{1}+a_{2}, \ldots, s_{n}=a_{1}+\cdots+a_{n}$, two give the same remainder modulo $n$, and their difference corresponds to a good sequence. To show that, permuting the $a_{i}$ 's, we can find $n-1$ different sequences, we use the following Lemma. Let $A$ be a $k \times n(k \leq n-2)$ matrix of zeros and ones, whose every row contains at least one 0 and at least two 1 's. Then it is possible to permute columns of $A$ is such a way that in any row 1 's do not form a block. Proof. We will use the induction on $k$. The case $k=1$ and arbitrary $n \geq 3$ is trivial. Suppose that $k \geq 2$ and that for $k-1$ and any $n \geq k+1$ the lemma is true. Consider a $k \times n$ matrix $A, n \geq k+2$. We mark an element $a_{i j}$ if either it is the only zero in the $i$-th row, or one of the 1 's in the row if it contains exactly two 1 's. Since $n \geq 4$, every row contains at most two marked elements, which adds up to at most $2 k<2 n$ marked elements in total. It follows that there is a column with at most one marked element. Assume w.l.o.g. that it is the first column and that $a_{1 j}$ isn't marked for $j>1$. The matrix $B$, obtained by omitting the first row and first column from $A$, satisfies the conditions of the lemma. Therefore, we can permute columns of $B$ and get the required form. Considered as a permutation of column of $A$, this permutation may leave a block of 1's only in the first row of $A$. In the case that it is so, if $a_{11}=1$ we put the first column in the last place, otherwise we put it between any two columns having 1's in the first row. The obtained matrix has the required property. Suppose now that we have got $k$ different nontrivial good sequences $e_{1}^{i}, \ldots, e_{n}^{i}, i=1, \ldots, k$, and that $k \leq n-2$. The matrix $A=\left(e_{j}^{i}\right)$ fulfils the conditions of Lemma, hence there is a permutation $\sigma$ from Lemma. Now among the sums $s_{0}=0, s_{1}=a_{\sigma(1)}, s_{2}=a_{\sigma(1)}+a_{\sigma(2)}$, $\ldots, s_{n}=a_{\sigma(1)}+\cdots+a_{\sigma(n)}$, two give the same remainder modulo $n$. Let $s_{p} \equiv s_{q}(\bmod n), p<q$. Then $n \mid s_{q}-s_{p}=a_{\sigma(p+1)}+\cdots+a_{\sigma(q)}$, and this yields a good sequence $e_{1}, \ldots, e_{n}$ with $e_{\sigma(p+1)}=\cdots=e_{\sigma(q)}=1$ and other $e$ 's equal to zero. Since from the construction we see that none of the sequences $e_{\sigma(j)^{i}}$ has all 1's in a block, in this way we have got a new nontrivial good sequence, and we can continue this procedure until there are $n-1$ sequences. Together with the trivial $0, \ldots, 0$ sequence, we have found $n$ good sequences.
|
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0b3edf8d-f26d-5795-a4fa-5255ae8dbed5
| 24,343
|
14. (POL 3) Let $a, b, c$ be integers and $p$ an odd prime number. Prove that if $f(x)=a x^{2}+b x+c$ is a perfect square for $2 p-1$ consecutive integer values of $x$, then $p$ divides $b^{2}-4 a c$.
|
14. Suppose that $f\left(x_{0}\right), f\left(x_{0}+1\right), \ldots, f\left(x_{0}+2 p-2\right)$ are squares. If $p \mid a$ and $p \nmid b$, then $f(x) \equiv b x+c(\bmod p)$ for $x=x_{0}, \ldots, x_{0}+p-1$ form a complete system of residues modulo $p$. However, a square is always congruent to exactly one of the $\frac{p+1}{2}$ numbers $0,1^{2}, 2^{2}, \ldots,\left(\frac{p-1}{2}\right)^{2}$ and thus cannot give every residue modulo $p$. Also, if $p \mid a$ and $p \mid b$, then $p \mid b^{2}-4 a c$. We now assume $p \nmid a$. The following identities hold for any quadric polynomial: $$ 4 a \cdot f(x)=(2 a x+b)^{2}-\left(b^{2}-4 a c\right) $$ and $$ f(x+p)-f(x)=p(2 a x+b)+p^{2} a . $$ Suppose that there is an $y, x_{0} \leq y \leq x_{0}+p-2$, for which $f(y)$ is divisible by $p$. Then both $f(y)$ and $f(y+p)$ are squares divisible by $p$, and therefore both are divisible by $p^{2}$. But relation (2) implies that $p \mid 2 a y+b$, and hence by (1) $b^{2}-4 a c$ is divisible by $p$ as well. Therefore it suffices to show that such an $y$ exists, and for that aim we prove that there are two such $y$ in $\left[x_{0}, x_{0}+p-1\right]$. Assume the opposite. Since for $x=x_{0}, x_{0}+1, \ldots, x_{0}+p-1 f(x)$ is congruent modulo $p$ to one of the $\frac{p-1}{2}$ numbers $1^{2}, 2^{2}, \ldots,\left(\frac{p-1}{2}\right)^{2}$, it follows by the pigeon-hole principle that for some mutually distinct $u, v, w \in\left\{x_{0}, \ldots, x_{0}+p-1\right\}$ we have $f(u) \equiv f(v) \equiv f(w)(\bmod p)$. Consequently the difference $f(u)-f(v)=$ $(u-v)(a(u+v)+b)$ is divisible by $p$, but it is clear that $p \nmid u-v$, hence $a(u+v) \equiv-b(\bmod p)$. Similarly $a(u+w) \equiv-b(\bmod p)$, which together with the previous congruence yields $p|a(v-w) \Rightarrow p| v-w$ which is clearly impossible. It follows that $p \mid f\left(y_{1}\right)$ for at least one $y_{1}$, $x_{0} \leq y_{1}<x_{0}+p$. If $y_{2}, x_{0} \leq y_{2}<x_{0}+p$ is such that $a\left(y_{1}+y_{2}\right)+b \equiv 0(\bmod p)$, we have $p\left|f\left(y_{1}\right)-f\left(y_{2}\right) \Rightarrow p\right| f\left(y_{2}\right)$. If $y_{1}=y_{2}$, then by (1) $p \mid b^{2}-4 a c$. Otherwise, among $y_{1}, y_{2}$ one belongs to $\left[x_{0}, x_{0}+p-2\right]$ as required. Second solution. Using Legendre's symbols $\left(\frac{a}{p}\right)$ for quadratic residues we can prove a stronger statement for $p \geq 5$. It can be shown that $$ \sum_{x=0}^{p-1}\left(\frac{a x^{2}+b x+c}{p}\right)=-\left(\frac{a}{p}\right) \quad \text { if } \quad p \nmid b^{2}-4 a c $$ hence for at most $\frac{p+3}{2}$ values of $x$ between $x_{0}$ and $x_{0}+p-1$ inclusive, $a x^{2}+b x+c$ is a quadratic residue or 0 modulo $p$. Therefore, if $p \geq 5$ and $f(x)$ is a square for $\frac{p+5}{2}$ consecutive values, then $p \mid b^{2}-4 a c$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
14. (POL 3) Let $a, b, c$ be integers and $p$ an odd prime number. Prove that if $f(x)=a x^{2}+b x+c$ is a perfect square for $2 p-1$ consecutive integer values of $x$, then $p$ divides $b^{2}-4 a c$.
|
14. Suppose that $f\left(x_{0}\right), f\left(x_{0}+1\right), \ldots, f\left(x_{0}+2 p-2\right)$ are squares. If $p \mid a$ and $p \nmid b$, then $f(x) \equiv b x+c(\bmod p)$ for $x=x_{0}, \ldots, x_{0}+p-1$ form a complete system of residues modulo $p$. However, a square is always congruent to exactly one of the $\frac{p+1}{2}$ numbers $0,1^{2}, 2^{2}, \ldots,\left(\frac{p-1}{2}\right)^{2}$ and thus cannot give every residue modulo $p$. Also, if $p \mid a$ and $p \mid b$, then $p \mid b^{2}-4 a c$. We now assume $p \nmid a$. The following identities hold for any quadric polynomial: $$ 4 a \cdot f(x)=(2 a x+b)^{2}-\left(b^{2}-4 a c\right) $$ and $$ f(x+p)-f(x)=p(2 a x+b)+p^{2} a . $$ Suppose that there is an $y, x_{0} \leq y \leq x_{0}+p-2$, for which $f(y)$ is divisible by $p$. Then both $f(y)$ and $f(y+p)$ are squares divisible by $p$, and therefore both are divisible by $p^{2}$. But relation (2) implies that $p \mid 2 a y+b$, and hence by (1) $b^{2}-4 a c$ is divisible by $p$ as well. Therefore it suffices to show that such an $y$ exists, and for that aim we prove that there are two such $y$ in $\left[x_{0}, x_{0}+p-1\right]$. Assume the opposite. Since for $x=x_{0}, x_{0}+1, \ldots, x_{0}+p-1 f(x)$ is congruent modulo $p$ to one of the $\frac{p-1}{2}$ numbers $1^{2}, 2^{2}, \ldots,\left(\frac{p-1}{2}\right)^{2}$, it follows by the pigeon-hole principle that for some mutually distinct $u, v, w \in\left\{x_{0}, \ldots, x_{0}+p-1\right\}$ we have $f(u) \equiv f(v) \equiv f(w)(\bmod p)$. Consequently the difference $f(u)-f(v)=$ $(u-v)(a(u+v)+b)$ is divisible by $p$, but it is clear that $p \nmid u-v$, hence $a(u+v) \equiv-b(\bmod p)$. Similarly $a(u+w) \equiv-b(\bmod p)$, which together with the previous congruence yields $p|a(v-w) \Rightarrow p| v-w$ which is clearly impossible. It follows that $p \mid f\left(y_{1}\right)$ for at least one $y_{1}$, $x_{0} \leq y_{1}<x_{0}+p$. If $y_{2}, x_{0} \leq y_{2}<x_{0}+p$ is such that $a\left(y_{1}+y_{2}\right)+b \equiv 0(\bmod p)$, we have $p\left|f\left(y_{1}\right)-f\left(y_{2}\right) \Rightarrow p\right| f\left(y_{2}\right)$. If $y_{1}=y_{2}$, then by (1) $p \mid b^{2}-4 a c$. Otherwise, among $y_{1}, y_{2}$ one belongs to $\left[x_{0}, x_{0}+p-2\right]$ as required. Second solution. Using Legendre's symbols $\left(\frac{a}{p}\right)$ for quadratic residues we can prove a stronger statement for $p \geq 5$. It can be shown that $$ \sum_{x=0}^{p-1}\left(\frac{a x^{2}+b x+c}{p}\right)=-\left(\frac{a}{p}\right) \quad \text { if } \quad p \nmid b^{2}-4 a c $$ hence for at most $\frac{p+3}{2}$ values of $x$ between $x_{0}$ and $x_{0}+p-1$ inclusive, $a x^{2}+b x+c$ is a quadratic residue or 0 modulo $p$. Therefore, if $p \geq 5$ and $f(x)$ is a square for $\frac{p+5}{2}$ consecutive values, then $p \mid b^{2}-4 a c$.
|
{
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|
c11847e9-bb59-51d2-acdc-f3a26ffb29ac
| 24,345
|
15. (USS 2) Let $a_{n}$ be the last nonzero digit in the decimal representation of the number $n$ !. Does the sequence $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ become periodic after a finite number of terms?
|
15. Assume that the sequence has the period $T$. We can find integers $k>m>$ 0 , as large as we like, such that $10^{k} \equiv 10^{m}(\bmod T)$, using for example Euler's theorem. It is obvious that $a_{10^{k}-1}=a_{10^{k}}$ and hence, taking $k$ sufficiently large and using the periodicity, we see that $$ a_{2 \cdot 10^{k}-10^{m}-1}=a_{10^{k}-1}=a_{10^{k}}=a_{2 \cdot 10^{k}-10^{m}} $$ Since $\left(2 \cdot 10^{k}-10^{m}\right)!=\left(2 \cdot 10^{k}-10^{m}\right)\left(2 \cdot 10^{k}-10^{m}-1\right)!$ and the last nonzero digit of $2 \cdot 10^{k}-10^{m}$ is nine, we must have $a_{2 \cdot 10^{k}-10^{m}-1}=5$ (if $s$ is a digit, the last digit of $9 s$ is $s$ only if $s=5$ ). But this means that 5 divides $n$ ! with a greater power than 2 does, which is impossible. Indeed, if the exponents of these powers are $\alpha_{2}, \alpha_{5}$ respectively, then $\alpha_{5}=$ $[n / 5]+\left[n / 5^{2}\right]+\cdots \leq \alpha_{2}=[n / 2]+\left[n / 2^{2}\right]+\cdots$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
15. (USS 2) Let $a_{n}$ be the last nonzero digit in the decimal representation of the number $n$ !. Does the sequence $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ become periodic after a finite number of terms?
|
15. Assume that the sequence has the period $T$. We can find integers $k>m>$ 0 , as large as we like, such that $10^{k} \equiv 10^{m}(\bmod T)$, using for example Euler's theorem. It is obvious that $a_{10^{k}-1}=a_{10^{k}}$ and hence, taking $k$ sufficiently large and using the periodicity, we see that $$ a_{2 \cdot 10^{k}-10^{m}-1}=a_{10^{k}-1}=a_{10^{k}}=a_{2 \cdot 10^{k}-10^{m}} $$ Since $\left(2 \cdot 10^{k}-10^{m}\right)!=\left(2 \cdot 10^{k}-10^{m}\right)\left(2 \cdot 10^{k}-10^{m}-1\right)!$ and the last nonzero digit of $2 \cdot 10^{k}-10^{m}$ is nine, we must have $a_{2 \cdot 10^{k}-10^{m}-1}=5$ (if $s$ is a digit, the last digit of $9 s$ is $s$ only if $s=5$ ). But this means that 5 divides $n$ ! with a greater power than 2 does, which is impossible. Indeed, if the exponents of these powers are $\alpha_{2}, \alpha_{5}$ respectively, then $\alpha_{5}=$ $[n / 5]+\left[n / 5^{2}\right]+\cdots \leq \alpha_{2}=[n / 2]+\left[n / 2^{2}\right]+\cdots$.
|
{
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|
3592ca99-f0af-53b6-a19b-5daeb3e96e24
| 24,347
|
16. (ROM 1) ${ }^{\mathrm{IMO} 2}$ Let $n>6$ and $a_{1}<a_{2}<\cdots<a_{k}$ be all natural numbers that are less than $n$ and relatively prime to $n$. Show that if $a_{1}, a_{2}, \ldots, a_{k}$ is an arithmetic progression, then $n$ is a prime number or a natural power of two.
|
16. Let $p$ be the least prime number that does not divide $n$ : thus $a_{1}=1$ and $a_{2}=p$. Since $a_{2}-a_{1}=a_{3}-a_{2}=\cdots=r$, the $a_{i}$ 's are $1, p, 2 p-1,3 p-2, \ldots$ We have the following cases: $p=2$. Then $r=1$ and the numbers $1,2,3, \ldots, n-1$ are relatively prime to $n$, hence $n$ is a prime. $p=3$. Then $r=2$, so every odd number less than $n$ is relatively prime to $n$, from which we deduce that $n$ has no odd divisors. Therefore $n=2^{k}$ for some $k \in \mathbb{N}$. $p>3$. Then $r=p-1$ and $a_{k+1}=a_{1}+k(p-1)=1+k(p-1)$. Since $n-1$ also must belong to the progression, we have $p-1 \mid n-2$. Let $q$ be any prime divisor of $p-1$. Then also $q \mid n-2$. On the other hand, since $q<p$, it must divide $n$ too, therefore $q \mid 2$, i.e. $q=2$. This means that $p-1$ has no prime divisors other than 2 and thus $p=2^{l}+1$ for some $l \geq 2$. But in order for $p$ to be prime, $l$ must be even (because $3 \mid 2^{l}+1$ for $l$ odd). Now we recall that $2 p-1$ is also relatively prime to $n$; but $2 p-1=2^{l+1}+1$ is divisible by 3 , which is a contradiction because $3 \mid n$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
16. (ROM 1) ${ }^{\mathrm{IMO} 2}$ Let $n>6$ and $a_{1}<a_{2}<\cdots<a_{k}$ be all natural numbers that are less than $n$ and relatively prime to $n$. Show that if $a_{1}, a_{2}, \ldots, a_{k}$ is an arithmetic progression, then $n$ is a prime number or a natural power of two.
|
16. Let $p$ be the least prime number that does not divide $n$ : thus $a_{1}=1$ and $a_{2}=p$. Since $a_{2}-a_{1}=a_{3}-a_{2}=\cdots=r$, the $a_{i}$ 's are $1, p, 2 p-1,3 p-2, \ldots$ We have the following cases: $p=2$. Then $r=1$ and the numbers $1,2,3, \ldots, n-1$ are relatively prime to $n$, hence $n$ is a prime. $p=3$. Then $r=2$, so every odd number less than $n$ is relatively prime to $n$, from which we deduce that $n$ has no odd divisors. Therefore $n=2^{k}$ for some $k \in \mathbb{N}$. $p>3$. Then $r=p-1$ and $a_{k+1}=a_{1}+k(p-1)=1+k(p-1)$. Since $n-1$ also must belong to the progression, we have $p-1 \mid n-2$. Let $q$ be any prime divisor of $p-1$. Then also $q \mid n-2$. On the other hand, since $q<p$, it must divide $n$ too, therefore $q \mid 2$, i.e. $q=2$. This means that $p-1$ has no prime divisors other than 2 and thus $p=2^{l}+1$ for some $l \geq 2$. But in order for $p$ to be prime, $l$ must be even (because $3 \mid 2^{l}+1$ for $l$ odd). Now we recall that $2 p-1$ is also relatively prime to $n$; but $2 p-1=2^{l+1}+1$ is divisible by 3 , which is a contradiction because $3 \mid n$.
|
{
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|
c6ae07ce-791d-5674-bf66-a58a30f1dcb3
| 24,349
|
2. (JAP 5) For an acute triangle $A B C, M$ is the midpoint of the segment $B C, P$ is a point on the segment $A M$ such that $P M=B M, H$ is the foot of the perpendicular line from $P$ to $B C, Q$ is the point of intersection of segment $A B$ and the line passing through $H$ that is perpendicular to $P B$, and finally, $R$ is the point of intersection of the segment $A C$ and the line passing through $H$ that is perpendicular to $P C$. Show that the circumcircle of $\triangle Q H R$ is tangent to the side $B C$ at point $H$.
|
2. Let $H Q$ meet $P B$ at $Q^{\prime}$ and $H R$ meet $P C$ at $R^{\prime}$. From $M P=M B=M C$ we have $\angle B P C=90^{\circ}$. So $P R^{\prime} H Q^{\prime}$ is a rectangle. Since $P H$ is perpendicular to $B C$, it follows that the circle with diameter $P H$, through $P, R^{\prime}, H, Q^{\prime}$, is tangent to $B C$. It is now sufficient to show that $Q R$ is parallel to $Q^{\prime} R^{\prime}$. Let $C P$ meet $A B$ at $X$, and $B P$ meet $A C$ at $Y$. Since $P$ is on the median, it follows (for  example, by Ceva's theorem) that $A X / X B=A Y / Y C$, i.e. that $X Y$ is parallel to $B C$. Consequently, $P Y / B P=P X / C P$. Since $H Q$ is parallel to $C X$, we have $Q Q^{\prime} / H Q^{\prime}=$ $P X / C P$ and similarly $R R^{\prime} / H R^{\prime}=P Y / B P$. It follows that $Q Q^{\prime} / H Q^{\prime}=$ $R R^{\prime} / H R^{\prime}$, hence $Q R$ is parallel to $Q^{\prime} R^{\prime}$ as required. Second solution. It suffices to show that $\angle R H C=\angle R Q H$, or equivalently $R H: Q H=P C: P B$. We assume $P C: P B=1: x$. Let $X \in A B$ and $Y \in A C$ be points such that $M X \perp P B$ and $M Y \perp P C$. Since $M X$ bisects $\angle A M B$ and $M Y$ bisects $A M C$, we deduce $$ \begin{aligned} & A X: X B=A M: M B=A Y: Y C \Rightarrow X Y \| B C \Rightarrow \\ & \quad \Rightarrow \triangle X Y M \sim \triangle C B P \Rightarrow X M: M Y=1: x . \end{aligned} $$ Now from $C H: H B=1: x^{2}$ we obtain $R H: M Y=C H: C M=1: \frac{1+x^{2}}{2}$ and $Q H: M X=B H: B M=x^{2}: \frac{1+x^{2}}{2}$. Therefore $$ R H: Q H=\frac{2}{1+x^{2}} M Y: \frac{2 x^{2}}{1+x^{2}} M X=1: x $$
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
2. (JAP 5) For an acute triangle $A B C, M$ is the midpoint of the segment $B C, P$ is a point on the segment $A M$ such that $P M=B M, H$ is the foot of the perpendicular line from $P$ to $B C, Q$ is the point of intersection of segment $A B$ and the line passing through $H$ that is perpendicular to $P B$, and finally, $R$ is the point of intersection of the segment $A C$ and the line passing through $H$ that is perpendicular to $P C$. Show that the circumcircle of $\triangle Q H R$ is tangent to the side $B C$ at point $H$.
|
2. Let $H Q$ meet $P B$ at $Q^{\prime}$ and $H R$ meet $P C$ at $R^{\prime}$. From $M P=M B=M C$ we have $\angle B P C=90^{\circ}$. So $P R^{\prime} H Q^{\prime}$ is a rectangle. Since $P H$ is perpendicular to $B C$, it follows that the circle with diameter $P H$, through $P, R^{\prime}, H, Q^{\prime}$, is tangent to $B C$. It is now sufficient to show that $Q R$ is parallel to $Q^{\prime} R^{\prime}$. Let $C P$ meet $A B$ at $X$, and $B P$ meet $A C$ at $Y$. Since $P$ is on the median, it follows (for  example, by Ceva's theorem) that $A X / X B=A Y / Y C$, i.e. that $X Y$ is parallel to $B C$. Consequently, $P Y / B P=P X / C P$. Since $H Q$ is parallel to $C X$, we have $Q Q^{\prime} / H Q^{\prime}=$ $P X / C P$ and similarly $R R^{\prime} / H R^{\prime}=P Y / B P$. It follows that $Q Q^{\prime} / H Q^{\prime}=$ $R R^{\prime} / H R^{\prime}$, hence $Q R$ is parallel to $Q^{\prime} R^{\prime}$ as required. Second solution. It suffices to show that $\angle R H C=\angle R Q H$, or equivalently $R H: Q H=P C: P B$. We assume $P C: P B=1: x$. Let $X \in A B$ and $Y \in A C$ be points such that $M X \perp P B$ and $M Y \perp P C$. Since $M X$ bisects $\angle A M B$ and $M Y$ bisects $A M C$, we deduce $$ \begin{aligned} & A X: X B=A M: M B=A Y: Y C \Rightarrow X Y \| B C \Rightarrow \\ & \quad \Rightarrow \triangle X Y M \sim \triangle C B P \Rightarrow X M: M Y=1: x . \end{aligned} $$ Now from $C H: H B=1: x^{2}$ we obtain $R H: M Y=C H: C M=1: \frac{1+x^{2}}{2}$ and $Q H: M X=B H: B M=x^{2}: \frac{1+x^{2}}{2}$. Therefore $$ R H: Q H=\frac{2}{1+x^{2}} M Y: \frac{2 x^{2}}{1+x^{2}} M X=1: x $$
|
{
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|
1ed478eb-efea-538a-9d62-cba26b39d689
| 24,358
|
20. (IRE 3) Let $\alpha$ be the positive root of the equation $x^{2}=1991 x+1$. For natural numbers $m, n$ define $$ m * n=m n+[\alpha m][\alpha n] $$ where $[x]$ is the greatest integer not exceeding $x$. Prove that for all natural numbers $p, q, r$, $$ (p * q) * r=p *(q * r) $$
|
20. We prove the result with 1991 replaced by any positive integer $k$. For natural numbers $p, q$, let $\epsilon=(\alpha p-[\alpha p])(\alpha q-[\alpha q])$. Then $0<\epsilon<1$ and $$ \epsilon=\alpha^{2} p q-\alpha(p[\alpha q]+q[\alpha p])+[\alpha p][\alpha q] . $$ Multiplying this equality by $\alpha-k$ and using $\alpha^{2}=k \alpha+1$, i.e. $\alpha(\alpha-k)=1$, we get $$ (\alpha-k) \epsilon=\alpha(p q+[\alpha p][\alpha q])-(p[\alpha q]+q[\alpha p]+k[\alpha p][\alpha q]) $$ Since $0<(\alpha-k) \epsilon<1$, we have $[\alpha(p * q)]=p[\alpha q]+q[\alpha p]+k[\alpha p][\alpha q]$. Now $$ \begin{aligned} (p * q) * r & =(p * q) r+[\alpha(p * q)][\alpha r]= \\ & =p q r+[\alpha p][\alpha q] r+[\alpha q][\alpha r] p+[\alpha r][\alpha p] q+k[\alpha p][\alpha q][\alpha r] . \end{aligned} $$ Since the last expression is symmetric, the same formula is obtained for $p *(q * r)$.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
20. (IRE 3) Let $\alpha$ be the positive root of the equation $x^{2}=1991 x+1$. For natural numbers $m, n$ define $$ m * n=m n+[\alpha m][\alpha n] $$ where $[x]$ is the greatest integer not exceeding $x$. Prove that for all natural numbers $p, q, r$, $$ (p * q) * r=p *(q * r) $$
|
20. We prove the result with 1991 replaced by any positive integer $k$. For natural numbers $p, q$, let $\epsilon=(\alpha p-[\alpha p])(\alpha q-[\alpha q])$. Then $0<\epsilon<1$ and $$ \epsilon=\alpha^{2} p q-\alpha(p[\alpha q]+q[\alpha p])+[\alpha p][\alpha q] . $$ Multiplying this equality by $\alpha-k$ and using $\alpha^{2}=k \alpha+1$, i.e. $\alpha(\alpha-k)=1$, we get $$ (\alpha-k) \epsilon=\alpha(p q+[\alpha p][\alpha q])-(p[\alpha q]+q[\alpha p]+k[\alpha p][\alpha q]) $$ Since $0<(\alpha-k) \epsilon<1$, we have $[\alpha(p * q)]=p[\alpha q]+q[\alpha p]+k[\alpha p][\alpha q]$. Now $$ \begin{aligned} (p * q) * r & =(p * q) r+[\alpha(p * q)][\alpha r]= \\ & =p q r+[\alpha p][\alpha q] r+[\alpha q][\alpha r] p+[\alpha r][\alpha p] q+k[\alpha p][\alpha q][\alpha r] . \end{aligned} $$ Since the last expression is symmetric, the same formula is obtained for $p *(q * r)$.
|
{
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|
cf86e1e5-219f-5ce1-ad24-efa7cc9fae51
| 24,359
|
21. (HKG 6) Let $f(x)$ be a monic polynomial of degree 1991 with integer coefficients. Define $g(x)=f^{2}(x)-9$. Show that the number of distinct integer solutions of $g(x)=0$ cannot exceed 1995.
|
21. The polynomial $g(x)$ factorizes as $g(x)=f(x)^{2}-9=(f(x)-3)(f(x)+3)$. If one of the equations $f(x)+3=0$ and $f(x)-3=0$ has no integer solutions, then the number of integer solutions of $g(x)=0$ clearly does not exceed 1991. Suppose now that both $f(x)+3=0$ and $f(x)-3=0$ have integer solutions. Let $x_{1}, \ldots, x_{k}$ be distinct integer solutions of the former, and $x_{k+1}, \ldots, x_{k+l}$ be distinct integer solutions of the latter equation. There exist monic polynomials $p(x), q(x)$ with integer coefficients such that $f(x)+3=\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{k}\right) p(x)$ and $f(x)-3=$ $\left(x-x_{k+1}\right)\left(x-x_{k+2}\right) \ldots\left(x-x_{k+l}\right) q(x)$. Thus we obtain $\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{k}\right) p(x)-\left(x-x_{k+1}\right)\left(x-x_{k+2}\right) \ldots\left(x-x_{k+l}\right) q(x)=6$. Putting $x=x_{k+1}$ we get $\left(x_{k+1}-x_{1}\right)\left(x_{k+1}-x_{2}\right) \cdots\left(x_{k+1}-x_{k}\right) \mid 6$, and since the product of more than four distinct integers cannot divide 6 , this implies $k \leq 4$. Similarly $l \leq 4$; hence $g(x)=0$ has at most 8 distinct integer solutions. Remark. The proposer provided a solution for the upper bound of 1995 roots which was essentially the same as that of (IMO74-6).
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
21. (HKG 6) Let $f(x)$ be a monic polynomial of degree 1991 with integer coefficients. Define $g(x)=f^{2}(x)-9$. Show that the number of distinct integer solutions of $g(x)=0$ cannot exceed 1995.
|
21. The polynomial $g(x)$ factorizes as $g(x)=f(x)^{2}-9=(f(x)-3)(f(x)+3)$. If one of the equations $f(x)+3=0$ and $f(x)-3=0$ has no integer solutions, then the number of integer solutions of $g(x)=0$ clearly does not exceed 1991. Suppose now that both $f(x)+3=0$ and $f(x)-3=0$ have integer solutions. Let $x_{1}, \ldots, x_{k}$ be distinct integer solutions of the former, and $x_{k+1}, \ldots, x_{k+l}$ be distinct integer solutions of the latter equation. There exist monic polynomials $p(x), q(x)$ with integer coefficients such that $f(x)+3=\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{k}\right) p(x)$ and $f(x)-3=$ $\left(x-x_{k+1}\right)\left(x-x_{k+2}\right) \ldots\left(x-x_{k+l}\right) q(x)$. Thus we obtain $\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{k}\right) p(x)-\left(x-x_{k+1}\right)\left(x-x_{k+2}\right) \ldots\left(x-x_{k+l}\right) q(x)=6$. Putting $x=x_{k+1}$ we get $\left(x_{k+1}-x_{1}\right)\left(x_{k+1}-x_{2}\right) \cdots\left(x_{k+1}-x_{k}\right) \mid 6$, and since the product of more than four distinct integers cannot divide 6 , this implies $k \leq 4$. Similarly $l \leq 4$; hence $g(x)=0$ has at most 8 distinct integer solutions. Remark. The proposer provided a solution for the upper bound of 1995 roots which was essentially the same as that of (IMO74-6).
|
{
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|
3c6f1a4b-0ce9-5749-9d5a-b5a5b2c8b2d1
| 24,362
|
25. (USA 1) Suppose that $n \geq 2$ and $x_{1}, x_{2}, \ldots, x_{n}$ are real numbers between 0 and 1 (inclusive). Prove that for some index $i$ between 1 and $n-1$ the inequality $$ x_{i}\left(1-x_{i+1}\right) \geq \frac{1}{4} x_{1}\left(1-x_{n}\right) $$ holds.
|
25. Since replacing $x_{1}$ by 1 can only reduce the set of indices $i$ for which the desired inequality holds, we may assume $x_{1}=1$. Similarly we may assume $x_{n}=0$. Now we can let $i$ be the largest index such that $x_{i}>1 / 2$. Then $x_{i+1} \leq 1 / 2$, hence $$ x_{i}\left(1-x_{i+1}\right) \geq \frac{1}{4}=\frac{1}{4} x_{1}\left(1-x_{n}\right) . $$
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
25. (USA 1) Suppose that $n \geq 2$ and $x_{1}, x_{2}, \ldots, x_{n}$ are real numbers between 0 and 1 (inclusive). Prove that for some index $i$ between 1 and $n-1$ the inequality $$ x_{i}\left(1-x_{i+1}\right) \geq \frac{1}{4} x_{1}\left(1-x_{n}\right) $$ holds.
|
25. Since replacing $x_{1}$ by 1 can only reduce the set of indices $i$ for which the desired inequality holds, we may assume $x_{1}=1$. Similarly we may assume $x_{n}=0$. Now we can let $i$ be the largest index such that $x_{i}>1 / 2$. Then $x_{i+1} \leq 1 / 2$, hence $$ x_{i}\left(1-x_{i+1}\right) \geq \frac{1}{4}=\frac{1}{4} x_{1}\left(1-x_{n}\right) . $$
|
{
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|
f6ce68ef-e58e-5cf9-a5ec-2c61b2146f32
| 24,373
|
28. (NET 1) ${ }^{\mathrm{IMO}}$ Given a real number $a>1$, construct an infinite and bounded sequence $x_{0}, x_{1}, x_{2}, \ldots$ such that for all natural numbers $i$ and $j, i \neq j$, the following inequality holds: $$ \left|x_{i}-x_{j}\right||i-j|^{a} \geq 1 $$
|
28. Let $x_{n}=c(n \sqrt{2}-[n \sqrt{2}])$ for some constant $c>0$. For $i>j$, putting $p=[i \sqrt{2}]-[j \sqrt{2}]$, we have $\left|x_{i}-x_{j}\right|=c|(i-j) \sqrt{2}-p|=\frac{\left|2(i-j)^{2}-p^{2}\right| c}{(i-j) \sqrt{2}+p} \geq \frac{c}{(i-j) \sqrt{2}+p} \geq \frac{c}{4(i-j)}$, because $p<(i-j) \sqrt{2}+1$. Taking $c=4$, we obtain that for any $i>j$, $(i-j)\left|x_{i}-x_{j}\right| \geq 1$. Of course, this implies $(i-j)^{a}\left|x_{i}-x_{j}\right| \geq 1$ for any $a>1$ 。 Remark. The constant 4 can be replaced with $3 / 2+\sqrt{2}$. Second solution. Another example of a sequence $\left\{x_{n}\right\}$ is constructed in the following way: $x_{1}=0, x_{2}=1, x_{3}=2$ and $x_{3^{k} i+m}=x_{m}+\frac{i}{3^{k}}$ for $i=1,2$ and $1 \leq m \leq 3^{k}$. It is easily shown that $|i-j| \cdot\left|x_{i}-x_{j}\right| \geq 1 / 3$ for any $i \neq j$. Third solution. If $n=b_{0}+2 b_{1}+\cdots+2^{k} b_{k}, b_{i} \in\{0,1\}$, then one can set $x_{n}$ to be $=b_{0}+2^{-a} b_{1}+\cdots+2^{-k a} b_{k}$. In this case it holds that $|i-j|^{a}\left|x_{i}-x_{j}\right| \geq$ $\frac{2^{a}-2}{2^{a}-1}$.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
28. (NET 1) ${ }^{\mathrm{IMO}}$ Given a real number $a>1$, construct an infinite and bounded sequence $x_{0}, x_{1}, x_{2}, \ldots$ such that for all natural numbers $i$ and $j, i \neq j$, the following inequality holds: $$ \left|x_{i}-x_{j}\right||i-j|^{a} \geq 1 $$
|
28. Let $x_{n}=c(n \sqrt{2}-[n \sqrt{2}])$ for some constant $c>0$. For $i>j$, putting $p=[i \sqrt{2}]-[j \sqrt{2}]$, we have $\left|x_{i}-x_{j}\right|=c|(i-j) \sqrt{2}-p|=\frac{\left|2(i-j)^{2}-p^{2}\right| c}{(i-j) \sqrt{2}+p} \geq \frac{c}{(i-j) \sqrt{2}+p} \geq \frac{c}{4(i-j)}$, because $p<(i-j) \sqrt{2}+1$. Taking $c=4$, we obtain that for any $i>j$, $(i-j)\left|x_{i}-x_{j}\right| \geq 1$. Of course, this implies $(i-j)^{a}\left|x_{i}-x_{j}\right| \geq 1$ for any $a>1$ 。 Remark. The constant 4 can be replaced with $3 / 2+\sqrt{2}$. Second solution. Another example of a sequence $\left\{x_{n}\right\}$ is constructed in the following way: $x_{1}=0, x_{2}=1, x_{3}=2$ and $x_{3^{k} i+m}=x_{m}+\frac{i}{3^{k}}$ for $i=1,2$ and $1 \leq m \leq 3^{k}$. It is easily shown that $|i-j| \cdot\left|x_{i}-x_{j}\right| \geq 1 / 3$ for any $i \neq j$. Third solution. If $n=b_{0}+2 b_{1}+\cdots+2^{k} b_{k}, b_{i} \in\{0,1\}$, then one can set $x_{n}$ to be $=b_{0}+2^{-a} b_{1}+\cdots+2^{-k a} b_{k}$. In this case it holds that $|i-j|^{a}\left|x_{i}-x_{j}\right| \geq$ $\frac{2^{a}-2}{2^{a}-1}$.
|
{
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|
2e7bbc05-757e-5791-9779-44d6f3ea93f9
| 24,378
|
29. (FIN 2) We call a set $S$ on the real line $\mathbb{R}$ superinvariant if for any stretching $A$ of the set by the transformation taking $x$ to $A(x)=x_{0}+$ $a\left(x-x_{0}\right)$ there exists a translation $B, B(x)=x+b$, such that the images of $S$ under $A$ and $B$ agree; i.e., for any $x \in S$ there is a $y \in S$ such that $A(x)=B(y)$ and for any $t \in S$ there is a $u \in S$ such that $B(t)=A(u)$. Determine all superinvariant sets. Remark. It is assumed that $a>0$.
|
29. One easily observes that the following sets are super-invariant: one-point set, its complement, closed and open half-lines or their complements, and the whole real line. To show that these are the only possibilities, we first observe that $S$ is super-invariant if and only if for each $a>0$ there is a $b$ such that $x \in S \Leftrightarrow a x+b \in S$. (i) Suppose that for some $a$ there are two such $b$ 's: $b_{1}$ and $b_{2}$. Then $x \in$ $S \Leftrightarrow a x+b_{1} \in S$ and $x \in S \Leftrightarrow a x+b_{2} \in S$, which implies that $S$ is periodic: $y \in S \Leftrightarrow y+\frac{b_{1}-b_{2}}{a} \in S$. Since $S$ is identical to a translate of any stretching of $S$, all positive numbers are periods of $S$. Therefore $S \equiv \mathbb{R}$. (ii) Assume that, for each $a, b=f(a)$ is unique. Then for any $a_{1}$ and $a_{2}$, $$ \begin{aligned} x \in S & \Leftrightarrow a_{1} x+f\left(a_{1}\right) \in S \Leftrightarrow a_{1} a_{2} x+a_{2} f\left(a_{1}\right)+f\left(a_{2}\right) \in S \\ & \Leftrightarrow a_{2} x+f\left(a_{2}\right) \in S \Leftrightarrow a_{1} a_{2} x+a_{1} f\left(a_{2}\right)+f\left(a_{1}\right) \in S . \end{aligned} $$ As above it follows that $a_{1} f\left(a_{2}\right)+f\left(a_{1}\right)=a_{2} f\left(a_{1}\right)+f\left(a_{2}\right)$, or equivalently $f\left(a_{1}\right)\left(a_{2}-1\right)=f\left(a_{2}\right)\left(a_{1}-1\right)$. Hence (for some $\left.c\right), f(a)=c(a-1)$ for all $a$. Now $x \in S \Leftrightarrow a x+c(a-1) \in S$ actually means that $y-c \in S \Leftrightarrow a y-c \in S$ for all $a$. Then it is easy to conclude that $\{y-c \mid y \in S\}$ is either a half-line or the whole line, and so is $S$.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
29. (FIN 2) We call a set $S$ on the real line $\mathbb{R}$ superinvariant if for any stretching $A$ of the set by the transformation taking $x$ to $A(x)=x_{0}+$ $a\left(x-x_{0}\right)$ there exists a translation $B, B(x)=x+b$, such that the images of $S$ under $A$ and $B$ agree; i.e., for any $x \in S$ there is a $y \in S$ such that $A(x)=B(y)$ and for any $t \in S$ there is a $u \in S$ such that $B(t)=A(u)$. Determine all superinvariant sets. Remark. It is assumed that $a>0$.
|
29. One easily observes that the following sets are super-invariant: one-point set, its complement, closed and open half-lines or their complements, and the whole real line. To show that these are the only possibilities, we first observe that $S$ is super-invariant if and only if for each $a>0$ there is a $b$ such that $x \in S \Leftrightarrow a x+b \in S$. (i) Suppose that for some $a$ there are two such $b$ 's: $b_{1}$ and $b_{2}$. Then $x \in$ $S \Leftrightarrow a x+b_{1} \in S$ and $x \in S \Leftrightarrow a x+b_{2} \in S$, which implies that $S$ is periodic: $y \in S \Leftrightarrow y+\frac{b_{1}-b_{2}}{a} \in S$. Since $S$ is identical to a translate of any stretching of $S$, all positive numbers are periods of $S$. Therefore $S \equiv \mathbb{R}$. (ii) Assume that, for each $a, b=f(a)$ is unique. Then for any $a_{1}$ and $a_{2}$, $$ \begin{aligned} x \in S & \Leftrightarrow a_{1} x+f\left(a_{1}\right) \in S \Leftrightarrow a_{1} a_{2} x+a_{2} f\left(a_{1}\right)+f\left(a_{2}\right) \in S \\ & \Leftrightarrow a_{2} x+f\left(a_{2}\right) \in S \Leftrightarrow a_{1} a_{2} x+a_{1} f\left(a_{2}\right)+f\left(a_{1}\right) \in S . \end{aligned} $$ As above it follows that $a_{1} f\left(a_{2}\right)+f\left(a_{1}\right)=a_{2} f\left(a_{1}\right)+f\left(a_{2}\right)$, or equivalently $f\left(a_{1}\right)\left(a_{2}-1\right)=f\left(a_{2}\right)\left(a_{1}-1\right)$. Hence (for some $\left.c\right), f(a)=c(a-1)$ for all $a$. Now $x \in S \Leftrightarrow a x+c(a-1) \in S$ actually means that $y-c \in S \Leftrightarrow a y-c \in S$ for all $a$. Then it is easy to conclude that $\{y-c \mid y \in S\}$ is either a half-line or the whole line, and so is $S$.
|
{
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|
d74dae8f-8d2d-5642-ad5c-1696ae2902c5
| 24,380
|
30. (BUL 3) Two students $A$ and $B$ are playing the following game: Each of them writes down on a sheet of paper a positive integer and gives the sheet to the referee. The referee writes down on a blackboard two integers, one of which is the sum of the integers written by the players. After that, the referee asks student $A$ : "Can you tell the integer written by the other student?" If $A$ answers "no," the referee puts the same question to student $B$. If $B$ answers "no," the referee puts the question back to $A$, and so on. Assume that both students are intelligent and truthful. Prove that after a finite number of questions, one of the students will answer "yes."
|
30. Let $a$ and $b$ be the integers written by $A$ and $B$ respectively, and let $x<y$ be the two integers written by the referee. Suppose that none of $A$ and $B$ ever answers "yes". Initially, regardless of $a, A$ knows that $0 \leq b \leq y$ and answers "no". In the second step, $B$ knows that $A$ obtained $0 \leq b \leq y$, but if $a$ were greater than $x, A$ would know that $a+b=y$ and would thus answer "yes". So $B$ concludes $0 \leq a \leq x$ but answers "no". The process continues. Suppose that, in the $n$-th step, $A$ knows that $B$ obtained $r_{n-1} \leq a \leq s_{n-1}$. If $b>x-r_{n-1}, B$ would know that $a+b>x$ and hence $a+b=y$, while if $b<y-s_{n-1}, B$ would know that $a+b<y$, i.e. $a+b=x$ : in both cases he would be able to guess $a$. However, $B$ answered "no", from which $A$ concludes $y-s_{n-1} \leq b \leq x-r_{n-1}$. Put $r_{n}=y-s_{n-1}$ and $s_{n}=x-r_{n-1}$. Similarly, in the next step $B$ knows that $A$ obtained $r_{n} \leq b \leq s_{n}$ and, since $A$ answered "no", concludes $y-s_{n} \leq a \leq x-r_{n}$. Put $r_{n+1}=y-s_{n}$ and $s_{n+1}=x-r_{n}$. Notice that in both cases $s_{i+1}-r_{i+1}=s_{i}-r_{i}-(y-x)$. Since $y-x>0$, there exists an $m$ for which $s_{m}-r_{m}<0$, a contradiction.
|
proof
|
Yes
|
Yes
|
proof
|
Logic and Puzzles
|
30. (BUL 3) Two students $A$ and $B$ are playing the following game: Each of them writes down on a sheet of paper a positive integer and gives the sheet to the referee. The referee writes down on a blackboard two integers, one of which is the sum of the integers written by the players. After that, the referee asks student $A$ : "Can you tell the integer written by the other student?" If $A$ answers "no," the referee puts the same question to student $B$. If $B$ answers "no," the referee puts the question back to $A$, and so on. Assume that both students are intelligent and truthful. Prove that after a finite number of questions, one of the students will answer "yes."
|
30. Let $a$ and $b$ be the integers written by $A$ and $B$ respectively, and let $x<y$ be the two integers written by the referee. Suppose that none of $A$ and $B$ ever answers "yes". Initially, regardless of $a, A$ knows that $0 \leq b \leq y$ and answers "no". In the second step, $B$ knows that $A$ obtained $0 \leq b \leq y$, but if $a$ were greater than $x, A$ would know that $a+b=y$ and would thus answer "yes". So $B$ concludes $0 \leq a \leq x$ but answers "no". The process continues. Suppose that, in the $n$-th step, $A$ knows that $B$ obtained $r_{n-1} \leq a \leq s_{n-1}$. If $b>x-r_{n-1}, B$ would know that $a+b>x$ and hence $a+b=y$, while if $b<y-s_{n-1}, B$ would know that $a+b<y$, i.e. $a+b=x$ : in both cases he would be able to guess $a$. However, $B$ answered "no", from which $A$ concludes $y-s_{n-1} \leq b \leq x-r_{n-1}$. Put $r_{n}=y-s_{n-1}$ and $s_{n}=x-r_{n-1}$. Similarly, in the next step $B$ knows that $A$ obtained $r_{n} \leq b \leq s_{n}$ and, since $A$ answered "no", concludes $y-s_{n} \leq a \leq x-r_{n}$. Put $r_{n+1}=y-s_{n}$ and $s_{n+1}=x-r_{n}$. Notice that in both cases $s_{i+1}-r_{i+1}=s_{i}-r_{i}-(y-x)$. Since $y-x>0$, there exists an $m$ for which $s_{m}-r_{m}<0$, a contradiction.
|
{
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|
dd848260-7073-52d8-b853-d81194b2f27c
| 24,385
|
4. (FRA 2) ${ }^{\mathrm{IMO5}}$ Let $A B C$ be a triangle and $M$ an interior point in $A B C$. Show that at least one of the angles $\measuredangle M A B, \measuredangle M B C$, and $\measuredangle M C A$ is less than or equal to $30^{\circ}$.
|
4. Assume the contrary, that $\angle M A B, \angle M B C, \angle M C A$ are all greater than $30^{\circ}$. By the sine Ceva theorem, it holds that $$ \begin{aligned} & \sin \angle M A C \sin \angle M B A \sin \angle M C B \\ = & \sin \angle M A B \sin \angle M B C \sin \angle M C A>\sin ^{3} 30^{\circ}=\frac{1}{8} . \end{aligned} $$ On the other hand, since $\angle M A C+\angle M B A+\angle M C B<180^{\circ}-3 \cdot 30^{\circ}=90^{\circ}$, Jensen's inequality applied on the concave function $\ln \sin x(x \in[0, \pi])$ gives us $\sin \angle M A C \sin \angle M B A \sin \angle M C B<\sin ^{3} 30^{\circ}$, contradicting (*). Second solution. Denote the intersections of $P A, P B, P C$ with $B C, C A$, $A B$ by $A_{1}, B_{1}, C_{1}$, respectively. Suppose that each of the angles $\angle P A B$, $\angle P B C, \angle P C A$ is greater than $30^{\circ}$ and denote $P A=2 x, P B=2 y, P C=$ $2 z$. Then $P C_{1}>x, P A_{1}>y, P B_{1}>z$. On the other hand, we know that $$ \frac{P C_{1}}{P C+P C_{1}}+\frac{P A_{1}}{P A+P A_{1}}+\frac{P B_{1}}{P B+P B_{1}}=\frac{S_{A B P}}{S_{A B C}}+\frac{S_{P B C}}{S_{A B C}}+\frac{S_{A P C}}{S_{A B C}}=1 . $$ Since the function $\frac{t}{p+t}$ is increasing, we obtain $\frac{x}{2 z+x}+\frac{y}{2 x+y}+\frac{z}{2 y+z}<1$. But on the contrary, Cauchy-Schwartz inequality (or alternatively Jensen's inequality) yields $$ \frac{x}{2 z+x}+\frac{y}{2 x+y}+\frac{z}{2 y+z} \geq \frac{(x+y+z)^{2}}{x(2 z+x)+y(2 x+y)+z(2 y+z)}=1 . $$
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
4. (FRA 2) ${ }^{\mathrm{IMO5}}$ Let $A B C$ be a triangle and $M$ an interior point in $A B C$. Show that at least one of the angles $\measuredangle M A B, \measuredangle M B C$, and $\measuredangle M C A$ is less than or equal to $30^{\circ}$.
|
4. Assume the contrary, that $\angle M A B, \angle M B C, \angle M C A$ are all greater than $30^{\circ}$. By the sine Ceva theorem, it holds that $$ \begin{aligned} & \sin \angle M A C \sin \angle M B A \sin \angle M C B \\ = & \sin \angle M A B \sin \angle M B C \sin \angle M C A>\sin ^{3} 30^{\circ}=\frac{1}{8} . \end{aligned} $$ On the other hand, since $\angle M A C+\angle M B A+\angle M C B<180^{\circ}-3 \cdot 30^{\circ}=90^{\circ}$, Jensen's inequality applied on the concave function $\ln \sin x(x \in[0, \pi])$ gives us $\sin \angle M A C \sin \angle M B A \sin \angle M C B<\sin ^{3} 30^{\circ}$, contradicting (*). Second solution. Denote the intersections of $P A, P B, P C$ with $B C, C A$, $A B$ by $A_{1}, B_{1}, C_{1}$, respectively. Suppose that each of the angles $\angle P A B$, $\angle P B C, \angle P C A$ is greater than $30^{\circ}$ and denote $P A=2 x, P B=2 y, P C=$ $2 z$. Then $P C_{1}>x, P A_{1}>y, P B_{1}>z$. On the other hand, we know that $$ \frac{P C_{1}}{P C+P C_{1}}+\frac{P A_{1}}{P A+P A_{1}}+\frac{P B_{1}}{P B+P B_{1}}=\frac{S_{A B P}}{S_{A B C}}+\frac{S_{P B C}}{S_{A B C}}+\frac{S_{A P C}}{S_{A B C}}=1 . $$ Since the function $\frac{t}{p+t}$ is increasing, we obtain $\frac{x}{2 z+x}+\frac{y}{2 x+y}+\frac{z}{2 y+z}<1$. But on the contrary, Cauchy-Schwartz inequality (or alternatively Jensen's inequality) yields $$ \frac{x}{2 z+x}+\frac{y}{2 x+y}+\frac{z}{2 y+z} \geq \frac{(x+y+z)^{2}}{x(2 z+x)+y(2 x+y)+z(2 y+z)}=1 . $$
|
{
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|
aeb27d8b-fdd4-55a7-b7aa-ef2f3521a480
| 24,388
|
5. (SPA 4) In the triangle $A B C$, with $\measuredangle A=60^{\circ}$, a parallel $I F$ to $A C$ is drawn through the incenter $I$ of the triangle, where $F$ lies on the side $A B$. The point $P$ on the side $B C$ is such that $3 B P=B C$. Show that $\measuredangle B F P=\measuredangle B / 2$.
|
5. Let $P_{1}$ be the point on the side $B C$ such that $\angle B F P_{1}=\beta / 2$. Then $\angle B P_{1} F=180^{\circ}-3 \beta / 2$, and the sine law gives us $\frac{B F}{B P_{1}}=\frac{\sin (3 \beta / 2)}{\sin (\beta / 2)}=$ $3-4 \sin ^{2}(\beta / 2)=1+2 \cos \beta$. Now we calculate $\frac{B F}{B P}$. We have $\angle B I F=120^{\circ}-\beta / 2, \angle B F I=60^{\circ}$ and $\angle B I C=120^{\circ}, \angle B C I=\gamma / 2=60^{\circ}-\beta / 2$. By the sine law, $$ B F=B I \frac{\sin \left(120^{\circ}-\beta / 2\right)}{\sin 60^{\circ}}, \quad B P=\frac{1}{3} B C=B I \frac{\sin 120^{\circ}}{3 \sin \left(60^{\circ}-\beta / 2\right)} . $$ It follows that $\frac{B F}{B P}=\frac{3 \sin \left(60^{\circ}-\beta / 2\right) \sin \left(60^{\circ}+\beta / 2\right)}{\sin ^{2} 60^{\circ}}=4 \sin \left(60^{\circ}-\beta / 2\right) \sin \left(60^{\circ}+\right.$ $\beta / 2)=2\left(\cos \beta-\cos 120^{\circ}\right)=2 \cos \beta+1=\frac{B F}{B P_{1}}$. Therefore $P \equiv P_{1}$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
5. (SPA 4) In the triangle $A B C$, with $\measuredangle A=60^{\circ}$, a parallel $I F$ to $A C$ is drawn through the incenter $I$ of the triangle, where $F$ lies on the side $A B$. The point $P$ on the side $B C$ is such that $3 B P=B C$. Show that $\measuredangle B F P=\measuredangle B / 2$.
|
5. Let $P_{1}$ be the point on the side $B C$ such that $\angle B F P_{1}=\beta / 2$. Then $\angle B P_{1} F=180^{\circ}-3 \beta / 2$, and the sine law gives us $\frac{B F}{B P_{1}}=\frac{\sin (3 \beta / 2)}{\sin (\beta / 2)}=$ $3-4 \sin ^{2}(\beta / 2)=1+2 \cos \beta$. Now we calculate $\frac{B F}{B P}$. We have $\angle B I F=120^{\circ}-\beta / 2, \angle B F I=60^{\circ}$ and $\angle B I C=120^{\circ}, \angle B C I=\gamma / 2=60^{\circ}-\beta / 2$. By the sine law, $$ B F=B I \frac{\sin \left(120^{\circ}-\beta / 2\right)}{\sin 60^{\circ}}, \quad B P=\frac{1}{3} B C=B I \frac{\sin 120^{\circ}}{3 \sin \left(60^{\circ}-\beta / 2\right)} . $$ It follows that $\frac{B F}{B P}=\frac{3 \sin \left(60^{\circ}-\beta / 2\right) \sin \left(60^{\circ}+\beta / 2\right)}{\sin ^{2} 60^{\circ}}=4 \sin \left(60^{\circ}-\beta / 2\right) \sin \left(60^{\circ}+\right.$ $\beta / 2)=2\left(\cos \beta-\cos 120^{\circ}\right)=2 \cos \beta+1=\frac{B F}{B P_{1}}$. Therefore $P \equiv P_{1}$.
|
{
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|
5d8338c1-b000-5bff-869d-0f2cafa6206a
| 24,391
|
6. (USS 4) ${ }^{\mathrm{IMO}}$ Prove for each triangle $A B C$ the inequality $$ \frac{1}{4}<\frac{I A \cdot I B \cdot I C}{l_{A} l_{B} l_{C}} \leq \frac{8}{27} $$ where $I$ is the incenter and $l_{A}, l_{B}, l_{C}$ are the lengths of the angle bisectors of $A B C$.
|
6. Let $a, b, c$ be sides of the triangle. Let $A_{1}$ be the intersection of line $A I$ with $B C$. By the known fact, $B A_{1}: A_{1} C=c: b$ and $A I: I A_{1}=A B: B A_{1}$, hence $B A_{1}=\frac{a c}{b+c}$ and $\frac{A I}{I A_{1}}=\frac{A B}{B A_{1}}=\frac{b+c}{a}$. Consequently $\frac{A I}{l_{A}}=\frac{b+c}{a+b+c}$. Put $a=n+p, b=p+m, c=m+n$ : it is obvious that $m, n, p$ are positive. Our inequality becomes $$ 2<\frac{(2 m+n+p)(m+2 n+p)(m+n+2 p)}{(m+n+p)^{3}} \leq \frac{64}{27} $$ The right side inequality immediately follows from the inequality between arithmetic and geometric means applied on $2 m+n+p, m+2 n+p$ and $m+n+2 p$. For the left side inequality, denote by $T=m+n+p$. Then we can write $(2 m+n+p)(m+2 n+p)(m+n+2 p)=(T+m)(T+n)(T+p)$ and $(T+m)(T+n)(T+p)=T^{3}+(m+n+p) T^{2}+(m n+n p+p n) T+m n p>2 T^{3}$. Remark. The inequalities cannot be improved. In fact, $\frac{A I \cdot B I \cdot C I}{l_{A} l_{B} l_{C}}$ is equal to $8 / 27$ for $a=b=c$, while it can be arbitrarily close to $1 / 4$ if $a=b$ and $c$ is sufficiently small.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
6. (USS 4) ${ }^{\mathrm{IMO}}$ Prove for each triangle $A B C$ the inequality $$ \frac{1}{4}<\frac{I A \cdot I B \cdot I C}{l_{A} l_{B} l_{C}} \leq \frac{8}{27} $$ where $I$ is the incenter and $l_{A}, l_{B}, l_{C}$ are the lengths of the angle bisectors of $A B C$.
|
6. Let $a, b, c$ be sides of the triangle. Let $A_{1}$ be the intersection of line $A I$ with $B C$. By the known fact, $B A_{1}: A_{1} C=c: b$ and $A I: I A_{1}=A B: B A_{1}$, hence $B A_{1}=\frac{a c}{b+c}$ and $\frac{A I}{I A_{1}}=\frac{A B}{B A_{1}}=\frac{b+c}{a}$. Consequently $\frac{A I}{l_{A}}=\frac{b+c}{a+b+c}$. Put $a=n+p, b=p+m, c=m+n$ : it is obvious that $m, n, p$ are positive. Our inequality becomes $$ 2<\frac{(2 m+n+p)(m+2 n+p)(m+n+2 p)}{(m+n+p)^{3}} \leq \frac{64}{27} $$ The right side inequality immediately follows from the inequality between arithmetic and geometric means applied on $2 m+n+p, m+2 n+p$ and $m+n+2 p$. For the left side inequality, denote by $T=m+n+p$. Then we can write $(2 m+n+p)(m+2 n+p)(m+n+2 p)=(T+m)(T+n)(T+p)$ and $(T+m)(T+n)(T+p)=T^{3}+(m+n+p) T^{2}+(m n+n p+p n) T+m n p>2 T^{3}$. Remark. The inequalities cannot be improved. In fact, $\frac{A I \cdot B I \cdot C I}{l_{A} l_{B} l_{C}}$ is equal to $8 / 27$ for $a=b=c$, while it can be arbitrarily close to $1 / 4$ if $a=b$ and $c$ is sufficiently small.
|
{
"resource_path": "IMO/segmented/en-compendium.jsonl",
"problem_match": null,
"solution_match": null
}
|
11309cc1-d440-5a42-b60c-0638ce4f76f9
| 24,393
|
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