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4. (SWE) Let $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ be a sequence of real numbers such that $0 \leq a_{n} \leq 1$ and $a_{n}-2 a_{n+1}+a_{n+2} \geq 0$ for $n=1,2,3, \ldots$. Prove that $$ 0 \leq(n+1)\left(a_{n}-a_{n+1}\right) \leq 2 \quad \text { for } n=1,2,3, \ldots $$
|
4. Put $\Delta a_{n}=a_{n}-a_{n+1}$. By the imposed condition, $\Delta a_{n}>\Delta a_{n+1}$. Suppose that for some $n, \Delta a_{n}<0$ : Then for each $k \geq n, \Delta a_{k}<\Delta a_{n}$; hence $a_{n}-a_{n+m}=\Delta a_{n}+\cdots+\Delta a_{n+m-1}<m \Delta a_{n}$. Thus for sufficiently large $m$ it holds that $a_{n}-a_{n+m}<-1$, which is impossible. This proves the first part of the inequality. Next one observes that $n \geq \sum_{k=1}^{n} a_{k}=n a_{n+1}+\sum_{k=1}^{n} k \Delta a_{k} \geq(1+2+\cdots+n) \Delta a_{n}=\frac{n(n+1)}{2} \Delta a_{n}$. Hence $(n+1) \Delta a_{n} \leq 2$.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
4. (SWE) Let $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ be a sequence of real numbers such that $0 \leq a_{n} \leq 1$ and $a_{n}-2 a_{n+1}+a_{n+2} \geq 0$ for $n=1,2,3, \ldots$. Prove that $$ 0 \leq(n+1)\left(a_{n}-a_{n+1}\right) \leq 2 \quad \text { for } n=1,2,3, \ldots $$
|
4. Put $\Delta a_{n}=a_{n}-a_{n+1}$. By the imposed condition, $\Delta a_{n}>\Delta a_{n+1}$. Suppose that for some $n, \Delta a_{n}<0$ : Then for each $k \geq n, \Delta a_{k}<\Delta a_{n}$; hence $a_{n}-a_{n+m}=\Delta a_{n}+\cdots+\Delta a_{n+m-1}<m \Delta a_{n}$. Thus for sufficiently large $m$ it holds that $a_{n}-a_{n+m}<-1$, which is impossible. This proves the first part of the inequality. Next one observes that $n \geq \sum_{k=1}^{n} a_{k}=n a_{n+1}+\sum_{k=1}^{n} k \Delta a_{k} \geq(1+2+\cdots+n) \Delta a_{n}=\frac{n(n+1)}{2} \Delta a_{n}$. Hence $(n+1) \Delta a_{n} \leq 2$.
|
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|
ae1d04f1-5c31-5baa-9c38-e184266086d6
| 25,062
|
5. (SWE) Let $M$ be the set of all positive integers that do not contain the digit 9 (base 10). If $x_{1}, \ldots, x_{n}$ are arbitrary but distinct elements in $M$, prove that $$ \sum_{j=1}^{n} \frac{1}{x_{j}}<80 $$
|
5. There are exactly $8 \cdot 9^{k-1} k$-digit numbers in $M$ (the first digit can be chosen in 8 ways, while any other position admits 9 possibilities). The least of them is $10^{k}$, and hence $$ \begin{aligned} \sum_{x_{j}<10^{k}} \frac{1}{x_{j}} & =\sum_{i=1}^{k} \sum_{10^{i-1} \leq x_{j}<10^{i}} \frac{1}{x_{j}}<\sum_{i=1}^{k} \sum_{10^{i-1} \leq x_{j}<10^{i}} \frac{1}{10^{i-1}} \\ & =\sum_{i=1}^{k} \frac{8 \cdot 9^{i-1}}{10^{i-1}}=80\left(1-\frac{9^{k}}{10^{k}}\right)<80 . \end{aligned} $$
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
5. (SWE) Let $M$ be the set of all positive integers that do not contain the digit 9 (base 10). If $x_{1}, \ldots, x_{n}$ are arbitrary but distinct elements in $M$, prove that $$ \sum_{j=1}^{n} \frac{1}{x_{j}}<80 $$
|
5. There are exactly $8 \cdot 9^{k-1} k$-digit numbers in $M$ (the first digit can be chosen in 8 ways, while any other position admits 9 possibilities). The least of them is $10^{k}$, and hence $$ \begin{aligned} \sum_{x_{j}<10^{k}} \frac{1}{x_{j}} & =\sum_{i=1}^{k} \sum_{10^{i-1} \leq x_{j}<10^{i}} \frac{1}{x_{j}}<\sum_{i=1}^{k} \sum_{10^{i-1} \leq x_{j}<10^{i}} \frac{1}{10^{i-1}} \\ & =\sum_{i=1}^{k} \frac{8 \cdot 9^{i-1}}{10^{i-1}}=80\left(1-\frac{9^{k}}{10^{k}}\right)<80 . \end{aligned} $$
|
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|
f2cf52f3-5640-5bf3-9ab5-0b3b274a0775
| 25,065
|
7. (GDR) Prove that from $x+y=1(x, y \in \mathbb{R})$ it follows that $$ x^{m+1} \sum_{j=0}^{n}\binom{m+j}{j} y^{j}+y^{n+1} \sum_{i=0}^{m}\binom{n+i}{i} x^{i}=1 \quad(m, n=0,1,2, \ldots) . $$
|
7. We use induction on $m$. Denote by $S_{m}$ the left-hand side of the equality to be proved. First $S_{0}=(1-y)\left(1+y+\cdots+y^{n}\right)+y^{n+1}=1$, since $x=1-y$. Furthermore, $$ \begin{aligned} & S_{m+1}-S_{m} \\ = & \binom{m+n+1}{m+1} x^{m+1} y^{n+1}+x^{m+1} \sum_{j=0}^{n}\left(\binom{m+1+j}{j} x y^{j}-\binom{m+j}{j} y^{j}\right) \\ = & \binom{m+n+1}{m+1} x^{m+1} y^{n+1} \\ & +x^{m+1} \sum_{j=0}^{n}\left(\binom{m+1+j}{j} y^{j}-\binom{m+j}{j} y^{j}-\binom{m+1+j}{j} y^{j+1}\right) \\ = & x^{m+1}\left[\binom{m+n+1}{n} y^{n+1}+\sum_{j=0}^{n}\left(\binom{m+j}{j-1} y^{j}-\binom{m+j+1}{j} y^{j+1}\right)\right] \\ = & 0 \end{aligned} $$ i.e., $S_{m+1}=S_{m}=1$ for every $m$. Second solution. Let us be given an unfair coin that, when tossed, shows heads with probability $x$ and tails with probability $y$. Note that $x^{m+1}\binom{m+j}{j} y^{j}$ is the probability that until the moment when the $(m+1)$ th head appears, exactly $j$ tails $(j<n+1)$ have appeared. Similarly, $y^{n+1}\binom{n+i}{i} x^{i}$ is the probability that exactly $i$ heads will appear before the $(n+1)$ th tail occurs. Therefore, the above sum is the probability that either $m+1$ heads will appear before $n+1$ tails, or vice versa, and this probability is clearly 1.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
7. (GDR) Prove that from $x+y=1(x, y \in \mathbb{R})$ it follows that $$ x^{m+1} \sum_{j=0}^{n}\binom{m+j}{j} y^{j}+y^{n+1} \sum_{i=0}^{m}\binom{n+i}{i} x^{i}=1 \quad(m, n=0,1,2, \ldots) . $$
|
7. We use induction on $m$. Denote by $S_{m}$ the left-hand side of the equality to be proved. First $S_{0}=(1-y)\left(1+y+\cdots+y^{n}\right)+y^{n+1}=1$, since $x=1-y$. Furthermore, $$ \begin{aligned} & S_{m+1}-S_{m} \\ = & \binom{m+n+1}{m+1} x^{m+1} y^{n+1}+x^{m+1} \sum_{j=0}^{n}\left(\binom{m+1+j}{j} x y^{j}-\binom{m+j}{j} y^{j}\right) \\ = & \binom{m+n+1}{m+1} x^{m+1} y^{n+1} \\ & +x^{m+1} \sum_{j=0}^{n}\left(\binom{m+1+j}{j} y^{j}-\binom{m+j}{j} y^{j}-\binom{m+1+j}{j} y^{j+1}\right) \\ = & x^{m+1}\left[\binom{m+n+1}{n} y^{n+1}+\sum_{j=0}^{n}\left(\binom{m+j}{j-1} y^{j}-\binom{m+j+1}{j} y^{j+1}\right)\right] \\ = & 0 \end{aligned} $$ i.e., $S_{m+1}=S_{m}=1$ for every $m$. Second solution. Let us be given an unfair coin that, when tossed, shows heads with probability $x$ and tails with probability $y$. Note that $x^{m+1}\binom{m+j}{j} y^{j}$ is the probability that until the moment when the $(m+1)$ th head appears, exactly $j$ tails $(j<n+1)$ have appeared. Similarly, $y^{n+1}\binom{n+i}{i} x^{i}$ is the probability that exactly $i$ heads will appear before the $(n+1)$ th tail occurs. Therefore, the above sum is the probability that either $m+1$ heads will appear before $n+1$ tails, or vice versa, and this probability is clearly 1.
|
{
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|
c559f05d-11d3-5d20-88eb-826c5dafcdd9
| 25,070
|
8. (NET) ${ }^{\mathrm{IMO} 3}$ On the sides of an arbitrary triangle $A B C$, triangles $B P C$, $C Q A$, and $A R B$ are externally erected such that $\measuredangle P B C=\measuredangle C A Q=45^{\circ}$, $\measuredangle B C P=\measuredangle Q C A=30^{\circ}$, $\measuredangle A B R=\measuredangle B A R=15^{\circ}$. Prove that $\measuredangle Q R P=90^{\circ}$ and $Q R=R P$.
|
8. Let $K$ and $L$ be the feet of perpendiculars from $P$ and $Q$ to $B C$ and $A C$ respectively. Let $M, N$ be points on $A B$ (ordered $A-N-M-B$ ) such that $R M N$ is a right isosceles triangle with $\angle R=90^{\circ}$. By sine theorem we have $\frac{B M}{B A}=\frac{B M}{B R} \cdot \frac{B R}{B A}=\frac{\sin 15^{\circ}}{\sin 45^{\circ}}$. Since $\frac{B K}{B C}=\frac{\sin 45^{\circ} \sin 30^{\circ}}{\cos 15^{\circ}}=\frac{\sin 15^{\circ}}{\sin 45^{\circ}}$, we deduce that $M K \| A C$ and $M K=A L$. Similarly, $N L \| B C$  and $N L=B K$. It follows that the vectors $\overrightarrow{R N}, \overrightarrow{N L}, \overrightarrow{L Q}$ are the images of $\overrightarrow{R M}, \overrightarrow{K P}, \overrightarrow{M K}$ respectively under a rotation of $90^{\circ}$, and consequently the same holds for their sums $\overrightarrow{R Q}$ and $\overrightarrow{R P}$. Therefore, $Q R=R P$ and $\angle Q R P=90^{\circ}$. Second solution. Let $A B S$ be the equilateral triangle constructed in the exterior of $\triangle A B C$. Obviously, the triangles $B P C, B R S, A R S, A Q C$ are similar. Let $f$ be the rotational homothety centered at $B$ that maps $P$ onto $C$, and let $g$ be the rotational homothety about $A$ that maps $C$ onto $Q$. The composition $h=g \circ f$ is also a rotational homothety; its angle is $\angle P B C+\angle C A Q=90^{\circ}$, and the coefficient is $\frac{B C}{B P} \cdot \frac{A Q}{A C}=1$. Moreover, $R$ is a fixed point of $h$ because $f(R)=S$ and $g(S)=R$. Hence $R$ is the center of $h$, and the statement follows from $h(P)=Q$. Remark. There are two more possible approaches: One includes using complex numbers and the other one is mere calculating of $R P, R Q, P Q$ by the cosine theorem. Second remark. The problem allows a generalization: Given that $\angle C B P=$ $\angle C A Q=\alpha, \angle B C P=\angle A C Q=\beta$, and $\angle R A B=\angle R B A=90^{\circ}-\alpha-\beta$, show that $R P=R Q$ and $\angle P R Q=2 \alpha$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
8. (NET) ${ }^{\mathrm{IMO} 3}$ On the sides of an arbitrary triangle $A B C$, triangles $B P C$, $C Q A$, and $A R B$ are externally erected such that $\measuredangle P B C=\measuredangle C A Q=45^{\circ}$, $\measuredangle B C P=\measuredangle Q C A=30^{\circ}$, $\measuredangle A B R=\measuredangle B A R=15^{\circ}$. Prove that $\measuredangle Q R P=90^{\circ}$ and $Q R=R P$.
|
8. Let $K$ and $L$ be the feet of perpendiculars from $P$ and $Q$ to $B C$ and $A C$ respectively. Let $M, N$ be points on $A B$ (ordered $A-N-M-B$ ) such that $R M N$ is a right isosceles triangle with $\angle R=90^{\circ}$. By sine theorem we have $\frac{B M}{B A}=\frac{B M}{B R} \cdot \frac{B R}{B A}=\frac{\sin 15^{\circ}}{\sin 45^{\circ}}$. Since $\frac{B K}{B C}=\frac{\sin 45^{\circ} \sin 30^{\circ}}{\cos 15^{\circ}}=\frac{\sin 15^{\circ}}{\sin 45^{\circ}}$, we deduce that $M K \| A C$ and $M K=A L$. Similarly, $N L \| B C$  and $N L=B K$. It follows that the vectors $\overrightarrow{R N}, \overrightarrow{N L}, \overrightarrow{L Q}$ are the images of $\overrightarrow{R M}, \overrightarrow{K P}, \overrightarrow{M K}$ respectively under a rotation of $90^{\circ}$, and consequently the same holds for their sums $\overrightarrow{R Q}$ and $\overrightarrow{R P}$. Therefore, $Q R=R P$ and $\angle Q R P=90^{\circ}$. Second solution. Let $A B S$ be the equilateral triangle constructed in the exterior of $\triangle A B C$. Obviously, the triangles $B P C, B R S, A R S, A Q C$ are similar. Let $f$ be the rotational homothety centered at $B$ that maps $P$ onto $C$, and let $g$ be the rotational homothety about $A$ that maps $C$ onto $Q$. The composition $h=g \circ f$ is also a rotational homothety; its angle is $\angle P B C+\angle C A Q=90^{\circ}$, and the coefficient is $\frac{B C}{B P} \cdot \frac{A Q}{A C}=1$. Moreover, $R$ is a fixed point of $h$ because $f(R)=S$ and $g(S)=R$. Hence $R$ is the center of $h$, and the statement follows from $h(P)=Q$. Remark. There are two more possible approaches: One includes using complex numbers and the other one is mere calculating of $R P, R Q, P Q$ by the cosine theorem. Second remark. The problem allows a generalization: Given that $\angle C B P=$ $\angle C A Q=\alpha, \angle B C P=\angle A C Q=\beta$, and $\angle R A B=\angle R B A=90^{\circ}-\alpha-\beta$, show that $R P=R Q$ and $\angle P R Q=2 \alpha$.
|
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|
d528ce24-647b-543c-9610-1c15d15bf9d7
| 25,073
|
1. (BUL 1) Let $A B C$ be a triangle with bisectors $A A_{1}, B B_{1}, C C_{1}\left(A_{1} \in\right.$ $B C$, etc.) and $M$ their common point. Consider the triangles $M B_{1} A$, $M C_{1} A, M C_{1} B, M A_{1} B, M A_{1} C, M B_{1} C$, and their inscribed circles. Prove that if four of these six inscribed circles have equal radii, then $A B=$ $B C=C A$.
|
1. Let $r$ denote the common inradius. Some two of the four triangles with the inradii $\rho$ have cross angles at $M$ : Suppose these are $\triangle A M B_{1}$ and $\triangle B M A_{1}$. We shall show that $\triangle A M B_{1} \cong \triangle B M A_{1}$. Indeed, the altitudes of these two triangles are both equal to $r$, the inradius of $\triangle A B C$, and their interior angles at $M$ are equal to some angle $\varphi$. If $P$ is the point of tangency of the incircle of $\triangle A_{1} M B$ with $M B$, then $\frac{r}{\rho}=\frac{A_{1} M+B M+A_{1} B}{A_{1} B}$, which also implies $\frac{r-2 \rho}{\rho}=\frac{A_{1} M+B M-A_{1} B}{A_{1} B}=\frac{2 M P}{A_{1} B}=\frac{2 r \cot (\varphi / 2)}{A_{1} B}$. Since similarly $\frac{r-2 \rho}{\rho}=\frac{2 r \cot (\varphi / 2)}{B_{1} A}$, we obtain $A_{1} B=B_{1} A$ and consequently $\triangle A M B_{1} \cong \triangle B M A_{1}$. Thus $\angle B A C=\angle A B C$ and $C C_{1} \perp A B$. There are two alternatives for the other two incircles: (i) If the inradii of $A M C_{1}$ and $A M B_{1}$ are equal to $r$, it is easy to obtain that $\triangle A M C_{1} \cong \triangle A M B_{1}$. Hence $\angle A B_{1} M=\angle A C_{1} M=90^{\circ}$, and $\triangle A B C$ is equilateral. (ii) The inradii of $A M B_{1}$ and $C M B_{1}$ are equal to $r$. Put $x=\angle M A C_{1}=$ $\angle M B C_{1}$. In this case $\varphi=2 x$ and $\angle B_{1} M C=90^{\circ}-x$. Now we have $\frac{A B_{1}}{C B_{1}}=\frac{S_{A M B_{1}}}{S_{C M B_{1}}}=\frac{A M+M B_{1}+A B_{1}}{C M+M B_{1}+C B_{1}}=\frac{A M+M B_{1}-A B_{1}}{C M+M B_{1}-C B_{1}}=\frac{\mathrm{cot} x}{\cot \left(45^{\circ}-x / 2\right)}$. On the other hand, $\frac{A B_{1}}{C B_{1}}=\frac{A B}{B C}=2 \cos 2 x$. Thus we have an equation for $x: \tan \left(45^{\circ}-x / 2\right)=2 \cos 2 x \tan x$, or equivalently $$ 2 \tan \left(45^{\circ}-\frac{x}{2}\right) \sin \left(45^{\circ}-\frac{x}{2}\right) \cos \left(45^{\circ}-\frac{x}{2}\right)=2 \cos 2 x \sin x . $$ Hence $\sin 3 x-\sin x=2 \sin ^{2}\left(45^{\circ}-\frac{x}{2}\right)=1-\sin x$, implying $\sin 3 x=1$, i.e., $x=30^{\circ}$. Therefore $\triangle A B C$ is equilateral.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
1. (BUL 1) Let $A B C$ be a triangle with bisectors $A A_{1}, B B_{1}, C C_{1}\left(A_{1} \in\right.$ $B C$, etc.) and $M$ their common point. Consider the triangles $M B_{1} A$, $M C_{1} A, M C_{1} B, M A_{1} B, M A_{1} C, M B_{1} C$, and their inscribed circles. Prove that if four of these six inscribed circles have equal radii, then $A B=$ $B C=C A$.
|
1. Let $r$ denote the common inradius. Some two of the four triangles with the inradii $\rho$ have cross angles at $M$ : Suppose these are $\triangle A M B_{1}$ and $\triangle B M A_{1}$. We shall show that $\triangle A M B_{1} \cong \triangle B M A_{1}$. Indeed, the altitudes of these two triangles are both equal to $r$, the inradius of $\triangle A B C$, and their interior angles at $M$ are equal to some angle $\varphi$. If $P$ is the point of tangency of the incircle of $\triangle A_{1} M B$ with $M B$, then $\frac{r}{\rho}=\frac{A_{1} M+B M+A_{1} B}{A_{1} B}$, which also implies $\frac{r-2 \rho}{\rho}=\frac{A_{1} M+B M-A_{1} B}{A_{1} B}=\frac{2 M P}{A_{1} B}=\frac{2 r \cot (\varphi / 2)}{A_{1} B}$. Since similarly $\frac{r-2 \rho}{\rho}=\frac{2 r \cot (\varphi / 2)}{B_{1} A}$, we obtain $A_{1} B=B_{1} A$ and consequently $\triangle A M B_{1} \cong \triangle B M A_{1}$. Thus $\angle B A C=\angle A B C$ and $C C_{1} \perp A B$. There are two alternatives for the other two incircles: (i) If the inradii of $A M C_{1}$ and $A M B_{1}$ are equal to $r$, it is easy to obtain that $\triangle A M C_{1} \cong \triangle A M B_{1}$. Hence $\angle A B_{1} M=\angle A C_{1} M=90^{\circ}$, and $\triangle A B C$ is equilateral. (ii) The inradii of $A M B_{1}$ and $C M B_{1}$ are equal to $r$. Put $x=\angle M A C_{1}=$ $\angle M B C_{1}$. In this case $\varphi=2 x$ and $\angle B_{1} M C=90^{\circ}-x$. Now we have $\frac{A B_{1}}{C B_{1}}=\frac{S_{A M B_{1}}}{S_{C M B_{1}}}=\frac{A M+M B_{1}+A B_{1}}{C M+M B_{1}+C B_{1}}=\frac{A M+M B_{1}-A B_{1}}{C M+M B_{1}-C B_{1}}=\frac{\mathrm{cot} x}{\cot \left(45^{\circ}-x / 2\right)}$. On the other hand, $\frac{A B_{1}}{C B_{1}}=\frac{A B}{B C}=2 \cos 2 x$. Thus we have an equation for $x: \tan \left(45^{\circ}-x / 2\right)=2 \cos 2 x \tan x$, or equivalently $$ 2 \tan \left(45^{\circ}-\frac{x}{2}\right) \sin \left(45^{\circ}-\frac{x}{2}\right) \cos \left(45^{\circ}-\frac{x}{2}\right)=2 \cos 2 x \sin x . $$ Hence $\sin 3 x-\sin x=2 \sin ^{2}\left(45^{\circ}-\frac{x}{2}\right)=1-\sin x$, implying $\sin 3 x=1$, i.e., $x=30^{\circ}$. Therefore $\triangle A B C$ is equilateral.
|
{
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d3abe3fd-c97f-5e9c-9ce8-ce3958ab3e06
| 25,079
|
11. (VIE 1) Prove that there exist infinitely many positive integers $n$ such that the decimal representation of $5^{n}$ contains a block of 1976 consecutive zeros.
|
11. We shall show by induction that $5^{2^{k}}-1=2^{k+2} q_{k}$ for each $k=0,1, \ldots$, where $q_{k} \in \mathbb{N}$. Indeed, the statement is true for $k=0$, and if it holds for some $k$ then $5^{2^{k+1}}-1=\left(5^{2^{k}}+1\right)\left(5^{2^{k}}-1\right)=2^{k+3} d_{k+1}$ where $d_{k+1}=$ $\left(5^{2^{k}}+1\right) d_{k} / 2$ is an integer by the inductive hypothesis. Let us now choose $n=2^{k}+k+2$. We have $5^{n}=10^{k+2} q_{k}+5^{k+2}$. It follows from $5^{4}<10^{3}$ that $5^{k+2}$ has at most $[3(k+2) / 4]+2$ nonzero digits, while $10^{k+2} q_{k}$ ends in $k+2$ zeros. Hence the decimal representation of $5^{n}$ contains at least $[(k+2) / 4]-2$ consecutive zeros. Now it suffices to take $k>4 \cdot 1978$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
11. (VIE 1) Prove that there exist infinitely many positive integers $n$ such that the decimal representation of $5^{n}$ contains a block of 1976 consecutive zeros.
|
11. We shall show by induction that $5^{2^{k}}-1=2^{k+2} q_{k}$ for each $k=0,1, \ldots$, where $q_{k} \in \mathbb{N}$. Indeed, the statement is true for $k=0$, and if it holds for some $k$ then $5^{2^{k+1}}-1=\left(5^{2^{k}}+1\right)\left(5^{2^{k}}-1\right)=2^{k+3} d_{k+1}$ where $d_{k+1}=$ $\left(5^{2^{k}}+1\right) d_{k} / 2$ is an integer by the inductive hypothesis. Let us now choose $n=2^{k}+k+2$. We have $5^{n}=10^{k+2} q_{k}+5^{k+2}$. It follows from $5^{4}<10^{3}$ that $5^{k+2}$ has at most $[3(k+2) / 4]+2$ nonzero digits, while $10^{k+2} q_{k}$ ends in $k+2$ zeros. Hence the decimal representation of $5^{n}$ contains at least $[(k+2) / 4]-2$ consecutive zeros. Now it suffices to take $k>4 \cdot 1978$.
|
{
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|
d9cc1944-cdf3-58c3-8cd6-79982439065b
| 25,083
|
2. (BUL 3) Let $a_{0}, a_{1}, \ldots, a_{n}, a_{n+1}$ be a sequence of real numbers satisfying the following conditions: $$ \begin{aligned} a_{0} & =a_{n+1}=0 \\ \left|a_{k-1}-2 a_{k}+a_{k+1}\right| & \leq 1 \quad(k=1,2, \ldots, n) \end{aligned} $$ Prove that $\left|a_{k}\right| \leq \frac{k(n+1-k)}{2}(k=0,1, \ldots, n+1)$.
|
2. Let us put $b_{i}=i(n+1-i) / 2$, and let $c_{i}=a_{i}-b_{i}, i=0,1, \ldots, n+1$. It is easy to verify that $b_{0}=b_{n+1}=0$ and $b_{i-1}-2 b_{i}+b_{i+1}=-1$. Subtracting this inequality from $a_{i-1}-2 a_{i}+a_{i+1} \geq-1$, we obtain $c_{i-1}-2 c_{i}+c_{i+1} \geq 0$, i.e., $2 c_{i} \leq c_{i-1}+c_{i+1}$. We also have $c_{0}=c_{n+1}=0$. Suppose that there exists $i \in\{1, \ldots, n\}$ for which $c_{i}>0$, and let $c_{k}$ be the maximal such $c_{i}$. Assuming w.l.o.g. that $c_{k-1}<c_{k}$, we obtain $c_{k-1}+c_{k+1}<2 c_{k}$, which is a contradiction. Hence $c_{i} \leq 0$ for all $i$; i.e., $a_{i} \leq b_{i}$. Similarly, considering the sequence $c_{i}^{\prime}=a_{i}+b_{i}$ one can show that $c_{i}^{\prime} \geq 0$, i.e., $a_{i} \geq-b_{i}$ for all $i$. This completes the proof.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
2. (BUL 3) Let $a_{0}, a_{1}, \ldots, a_{n}, a_{n+1}$ be a sequence of real numbers satisfying the following conditions: $$ \begin{aligned} a_{0} & =a_{n+1}=0 \\ \left|a_{k-1}-2 a_{k}+a_{k+1}\right| & \leq 1 \quad(k=1,2, \ldots, n) \end{aligned} $$ Prove that $\left|a_{k}\right| \leq \frac{k(n+1-k)}{2}(k=0,1, \ldots, n+1)$.
|
2. Let us put $b_{i}=i(n+1-i) / 2$, and let $c_{i}=a_{i}-b_{i}, i=0,1, \ldots, n+1$. It is easy to verify that $b_{0}=b_{n+1}=0$ and $b_{i-1}-2 b_{i}+b_{i+1}=-1$. Subtracting this inequality from $a_{i-1}-2 a_{i}+a_{i+1} \geq-1$, we obtain $c_{i-1}-2 c_{i}+c_{i+1} \geq 0$, i.e., $2 c_{i} \leq c_{i-1}+c_{i+1}$. We also have $c_{0}=c_{n+1}=0$. Suppose that there exists $i \in\{1, \ldots, n\}$ for which $c_{i}>0$, and let $c_{k}$ be the maximal such $c_{i}$. Assuming w.l.o.g. that $c_{k-1}<c_{k}$, we obtain $c_{k-1}+c_{k+1}<2 c_{k}$, which is a contradiction. Hence $c_{i} \leq 0$ for all $i$; i.e., $a_{i} \leq b_{i}$. Similarly, considering the sequence $c_{i}^{\prime}=a_{i}+b_{i}$ one can show that $c_{i}^{\prime} \geq 0$, i.e., $a_{i} \geq-b_{i}$ for all $i$. This completes the proof.
|
{
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|
6a850af9-9228-50e9-8428-9d038998985b
| 25,087
|
4. (GBR 1a) ${ }^{\mathrm{IMO} 6}$ For all positive integral $n$, $u_{n+1}=u_{n}\left(u_{n-1}^{2}-2\right)-u_{1}$, $u_{0}=2$, and $u_{1}=5 / 2$. Prove that $$ 3 \log _{2}\left[u_{n}\right]=2^{n}-(-1)^{n} $$ where $[x]$ is the integral part of $x$.
|
4. The first few values are easily verified to be $2^{r_{n}}+2^{-r_{n}}$, where $r_{0}=0$, $r_{1}=r_{2}=1, r_{3}=3, r_{4}=5, r_{5}=11, \ldots$. Let us put $u_{n}=2^{r_{n}}+2^{-r_{n}}$ (we will show that $r_{n}$ exists and is integer for each $n$ ). A simple calculation gives us $u_{n}\left(u_{n-1}^{2}-2\right)=2^{r_{n}+2 r_{n-1}}+2^{-r_{n}-2 r_{n-1}}+2^{r_{n}-2 r_{n-1}}+2^{-r_{n}+2 r_{n-1}}$. If an array $q_{n}$, with $q_{0}=0$ and $q_{1}=1$, is set so as to satisfy the linear recurrence $q_{n+1}=q_{n}+2 q_{n-1}$, then it also satisfies $q_{n}-2 q_{n-1}=-\left(q_{n-1}-\right.$ $\left.2 q_{n-2}\right)=\cdots=(-1)^{n-1}\left(q_{1}-2 q_{0}\right)=(-1)^{n-1}$. Assuming inductively up to $n r_{i}=q_{i}$, the expression for $u_{n}\left(u_{n-1}^{2}-2\right)=u_{n+1}+u_{1}$ reduces to $2^{q_{n+1}}+2^{-q_{n+1}}+u_{1}$. Therefore, $r_{n+1}=q_{n+1}$. The solution to this linear recurrence with $r_{0}=0, r_{1}=1$ is $r_{n}=q_{n}=\frac{2^{n}-(-1)^{n}}{3}$, and since $\left[u_{n}\right]=2^{r_{n}}$ for $n \geq 0$, the result follows. Remark. One could simply guess that $u_{n}=2^{r_{n}}+2^{-r_{n}}$ for $r_{n}=\frac{2^{n}-(-1)^{n}}{3}$, and then prove this result by induction.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
4. (GBR 1a) ${ }^{\mathrm{IMO} 6}$ For all positive integral $n$, $u_{n+1}=u_{n}\left(u_{n-1}^{2}-2\right)-u_{1}$, $u_{0}=2$, and $u_{1}=5 / 2$. Prove that $$ 3 \log _{2}\left[u_{n}\right]=2^{n}-(-1)^{n} $$ where $[x]$ is the integral part of $x$.
|
4. The first few values are easily verified to be $2^{r_{n}}+2^{-r_{n}}$, where $r_{0}=0$, $r_{1}=r_{2}=1, r_{3}=3, r_{4}=5, r_{5}=11, \ldots$. Let us put $u_{n}=2^{r_{n}}+2^{-r_{n}}$ (we will show that $r_{n}$ exists and is integer for each $n$ ). A simple calculation gives us $u_{n}\left(u_{n-1}^{2}-2\right)=2^{r_{n}+2 r_{n-1}}+2^{-r_{n}-2 r_{n-1}}+2^{r_{n}-2 r_{n-1}}+2^{-r_{n}+2 r_{n-1}}$. If an array $q_{n}$, with $q_{0}=0$ and $q_{1}=1$, is set so as to satisfy the linear recurrence $q_{n+1}=q_{n}+2 q_{n-1}$, then it also satisfies $q_{n}-2 q_{n-1}=-\left(q_{n-1}-\right.$ $\left.2 q_{n-2}\right)=\cdots=(-1)^{n-1}\left(q_{1}-2 q_{0}\right)=(-1)^{n-1}$. Assuming inductively up to $n r_{i}=q_{i}$, the expression for $u_{n}\left(u_{n-1}^{2}-2\right)=u_{n+1}+u_{1}$ reduces to $2^{q_{n+1}}+2^{-q_{n+1}}+u_{1}$. Therefore, $r_{n+1}=q_{n+1}$. The solution to this linear recurrence with $r_{0}=0, r_{1}=1$ is $r_{n}=q_{n}=\frac{2^{n}-(-1)^{n}}{3}$, and since $\left[u_{n}\right]=2^{r_{n}}$ for $n \geq 0$, the result follows. Remark. One could simply guess that $u_{n}=2^{r_{n}}+2^{-r_{n}}$ for $r_{n}=\frac{2^{n}-(-1)^{n}}{3}$, and then prove this result by induction.
|
{
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|
0d83d22e-b32e-5129-b152-b8a98f546c18
| 25,091
|
5. (NET 3) ${ }^{\mathrm{IMO} 5}$ Let a set of $p$ equations be given, $$ \begin{gathered} a_{11} x_{1}+\cdots+a_{1 q} x_{q}=0 \\ a_{21} x_{1}+\cdots+a_{2 q} x_{q}=0 \\ \vdots \\ a_{p 1} x_{1}+\cdots+a_{p q} x_{q}=0, \end{gathered} $$ with coefficients $a_{i j}$ satisfying $a_{i j}=-1,0$, or +1 for all $i=1, \ldots, p$ and $j=1, \ldots, q$. Prove that if $q=2 p$, there exists a solution $x_{1}, \ldots, x_{q}$ of this system such that all $x_{j}(j=1, \ldots, q)$ are integers satisfying $\left|x_{j}\right| \leq q$ and $x_{j} \neq 0$ for at least one value of $j$.
|
5. If one substitutes an integer $q$-tuple $\left(x_{1}, \ldots, x_{q}\right)$ satisfying $\left|x_{i}\right| \leq p$ for all $i$ in an equation of the given system, the absolute value of the right-hand member never exceeds $p q$. So for the right-hand member of the system there are $(2 p q+1)^{p}$ possibilities There are $(2 p+1)^{q}$ possible $q$-tuples $\left(x_{1}, \ldots, x_{q}\right)$. Since $(2 p+1)^{q} \geq(2 p q+1)^{p}$, there are at least two $q$-tuples $\left(y_{1}, \ldots, y_{q}\right)$ and $\left(z_{1}, \ldots, z_{q}\right)$ giving the same right-hand members in the given system. The difference $\left(x_{1}, \ldots, x_{q}\right)=\left(y_{1}-z_{1}, \ldots, y_{q}-z_{q}\right)$ thus satisfies all the requirements of the problem.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
5. (NET 3) ${ }^{\mathrm{IMO} 5}$ Let a set of $p$ equations be given, $$ \begin{gathered} a_{11} x_{1}+\cdots+a_{1 q} x_{q}=0 \\ a_{21} x_{1}+\cdots+a_{2 q} x_{q}=0 \\ \vdots \\ a_{p 1} x_{1}+\cdots+a_{p q} x_{q}=0, \end{gathered} $$ with coefficients $a_{i j}$ satisfying $a_{i j}=-1,0$, or +1 for all $i=1, \ldots, p$ and $j=1, \ldots, q$. Prove that if $q=2 p$, there exists a solution $x_{1}, \ldots, x_{q}$ of this system such that all $x_{j}(j=1, \ldots, q)$ are integers satisfying $\left|x_{j}\right| \leq q$ and $x_{j} \neq 0$ for at least one value of $j$.
|
5. If one substitutes an integer $q$-tuple $\left(x_{1}, \ldots, x_{q}\right)$ satisfying $\left|x_{i}\right| \leq p$ for all $i$ in an equation of the given system, the absolute value of the right-hand member never exceeds $p q$. So for the right-hand member of the system there are $(2 p q+1)^{p}$ possibilities There are $(2 p+1)^{q}$ possible $q$-tuples $\left(x_{1}, \ldots, x_{q}\right)$. Since $(2 p+1)^{q} \geq(2 p q+1)^{p}$, there are at least two $q$-tuples $\left(y_{1}, \ldots, y_{q}\right)$ and $\left(z_{1}, \ldots, z_{q}\right)$ giving the same right-hand members in the given system. The difference $\left(x_{1}, \ldots, x_{q}\right)=\left(y_{1}-z_{1}, \ldots, y_{q}-z_{q}\right)$ thus satisfies all the requirements of the problem.
|
{
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|
bc99d88c-f992-520a-88f7-cb6a5f33dfc1
| 25,093
|
7. (POL 1b) Let $I=(0,1]$ be the unit interval of the real line. For a given number $a \in(0,1)$ we define a map $T: I \rightarrow I$ by the formula $$ T(x, y)= \begin{cases}x+(1-a) & \text { if } 0<x \leq a \\ x-a & \text { if } a<x \leq 1\end{cases} $$ Show that for every interval $J \subset I$ there exists an integer $n>0$ such that $T^{n}(J) \cap J \neq \emptyset$.
|
7. The map $T$ transforms the interval $(0, a]$ onto $(1-a, 1]$ and the interval $(a, 1]$ onto $(0,1-a]$. Clearly $T$ preserves the measure. Since the measure of the interval $[0,1]$ is finite, there exist two positive integers $k, l>k$ such that $T^{k}(J)$ and $T^{l}(J)$ are not disjoint. But the map $T$ is bijective; hence $T^{l-k}(J)$ and $J$ are not disjoint.
|
proof
|
Yes
|
Yes
|
proof
|
Calculus
|
7. (POL 1b) Let $I=(0,1]$ be the unit interval of the real line. For a given number $a \in(0,1)$ we define a map $T: I \rightarrow I$ by the formula $$ T(x, y)= \begin{cases}x+(1-a) & \text { if } 0<x \leq a \\ x-a & \text { if } a<x \leq 1\end{cases} $$ Show that for every interval $J \subset I$ there exists an integer $n>0$ such that $T^{n}(J) \cap J \neq \emptyset$.
|
7. The map $T$ transforms the interval $(0, a]$ onto $(1-a, 1]$ and the interval $(a, 1]$ onto $(0,1-a]$. Clearly $T$ preserves the measure. Since the measure of the interval $[0,1]$ is finite, there exist two positive integers $k, l>k$ such that $T^{k}(J)$ and $T^{l}(J)$ are not disjoint. But the map $T$ is bijective; hence $T^{l-k}(J)$ and $J$ are not disjoint.
|
{
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|
9459f2dc-0cb3-5756-b3f7-dd3363cf63f9
| 25,097
|
8. (SWE 3) Let $P$ be a polynomial with real coefficients such that $P(x)>0$ if $x>0$. Prove that there exist polynomials $Q$ and $R$ with nonnegative coefficients such that $P(x)=\frac{Q(x)}{R(x)}$ if $x>0$.
|
8. Every polynomial with real coefficients can be factored as a product of linear and quadratic polynomials with real coefficients. Thus it suffices to prove the result only for a quadratic polynomial $P(x)=x^{2}-2 a x+b^{2}$, with $a>0$ and $b^{2}>a^{2}$. Using the identity $$ \left(x^{2}+b^{2}\right)^{2 n}-(2 a x)^{2 n}=\left(x^{2}-2 a x+b^{2}\right) \sum_{k=0}^{2 n-1}\left(x^{2}+b^{2}\right)^{k}(2 a x)^{2 n-k-1} $$ we have solved the problem if we can choose $n$ such that $b^{2 n}\binom{2 n}{n}>2^{2 n} a^{2 n}$. However, it is is easy to show that $2 n\binom{2 n}{n}<2^{2 n}$; hence it is enough to take $n$ such that $(b / a)^{2 n}>2 n$. Since $\lim _{n \rightarrow \infty}(2 n)^{1 /(2 n)}=1<b / a$, such an $n$ always exists.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
8. (SWE 3) Let $P$ be a polynomial with real coefficients such that $P(x)>0$ if $x>0$. Prove that there exist polynomials $Q$ and $R$ with nonnegative coefficients such that $P(x)=\frac{Q(x)}{R(x)}$ if $x>0$.
|
8. Every polynomial with real coefficients can be factored as a product of linear and quadratic polynomials with real coefficients. Thus it suffices to prove the result only for a quadratic polynomial $P(x)=x^{2}-2 a x+b^{2}$, with $a>0$ and $b^{2}>a^{2}$. Using the identity $$ \left(x^{2}+b^{2}\right)^{2 n}-(2 a x)^{2 n}=\left(x^{2}-2 a x+b^{2}\right) \sum_{k=0}^{2 n-1}\left(x^{2}+b^{2}\right)^{k}(2 a x)^{2 n-k-1} $$ we have solved the problem if we can choose $n$ such that $b^{2 n}\binom{2 n}{n}>2^{2 n} a^{2 n}$. However, it is is easy to show that $2 n\binom{2 n}{n}<2^{2 n}$; hence it is enough to take $n$ such that $(b / a)^{2 n}>2 n$. Since $\lim _{n \rightarrow \infty}(2 n)^{1 /(2 n)}=1<b / a$, such an $n$ always exists.
|
{
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|
1440c300-2dda-56b2-b26d-15d53bcaf61b
| 25,099
|
9. (FIN 2) ${ }^{\mathrm{IMO} 2}$ Let $P_{1}(x)=x^{2}-2, P_{j}(x)=P_{1}\left(P_{j-1}(x)\right), j=2,3, \ldots$. Show that for arbitrary $n$ the roots of the equation $P_{n}(x)=x$ are real and different from one another.
|
9. The equation $P_{n}(x)=x$ is of degree $2^{n}$, and has at most $2^{n}$ distinct roots. If $x>2$, then by simple induction $P_{n}(x)>x$ for all $n$. Similarly, if $x<-1$, then $P_{1}(x)>2$, which implies $P_{n}(x)>2$ for all $n$. It follows that all real roots of the equation $P_{n}(x)=x$ lie in the interval $[-2,2]$, and thus have the form $x=2 \cos t$. Now we observe that $P_{1}(2 \cos t)=4 \cos ^{2} t-2=2 \cos 2 t$, and in general $P_{n}(2 \cos t)=2 \cos 2^{n} t$. Our equation becomes $$ \cos 2^{n} t=\cos t $$ which indeed has $2^{n}$ different solutions $t=\frac{2 \pi m}{2^{n}-1}\left(m=0,1, \ldots, 2^{n-1}-1\right)$ and $t=\frac{2 \pi m}{2^{n}+1}\left(m=1,2, \ldots, 2^{n-1}\right)$.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
9. (FIN 2) ${ }^{\mathrm{IMO} 2}$ Let $P_{1}(x)=x^{2}-2, P_{j}(x)=P_{1}\left(P_{j-1}(x)\right), j=2,3, \ldots$. Show that for arbitrary $n$ the roots of the equation $P_{n}(x)=x$ are real and different from one another.
|
9. The equation $P_{n}(x)=x$ is of degree $2^{n}$, and has at most $2^{n}$ distinct roots. If $x>2$, then by simple induction $P_{n}(x)>x$ for all $n$. Similarly, if $x<-1$, then $P_{1}(x)>2$, which implies $P_{n}(x)>2$ for all $n$. It follows that all real roots of the equation $P_{n}(x)=x$ lie in the interval $[-2,2]$, and thus have the form $x=2 \cos t$. Now we observe that $P_{1}(2 \cos t)=4 \cos ^{2} t-2=2 \cos 2 t$, and in general $P_{n}(2 \cos t)=2 \cos 2^{n} t$. Our equation becomes $$ \cos 2^{n} t=\cos t $$ which indeed has $2^{n}$ different solutions $t=\frac{2 \pi m}{2^{n}-1}\left(m=0,1, \ldots, 2^{n-1}-1\right)$ and $t=\frac{2 \pi m}{2^{n}+1}\left(m=1,2, \ldots, 2^{n-1}\right)$.
|
{
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|
55cd0c96-bee3-58d9-b9ea-147a21f1ab00
| 25,101
|
1. (BUL 1) A pentagon $A B C D E$ inscribed in a circle for which $B C<C D$ and $A B<D E$ is the base of a pyramid with vertex $S$. If $A S$ is the longest edge starting from $S$, prove that $B S>C S$.
|
1. Let $P$ be the projection of $S$ onto the plane $A B C D E$. Obviously $B S>C S$ is equivalent to $B P>C P$. The conditions of the problem imply that $P A>P B$ and $P A>P E$. The locus of such points $P$ is the region of the plane that is determined by the perpendicular bisectors of segments $A B$ and $A E$ and that contains the point diametrically opposite $A$. But since $A B<D E$, the whole of this region lies on one side of the perpendicular bisector of $B C$. The result follows immediately. Remark. The assumption $B C<C D$ is redundant.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
1. (BUL 1) A pentagon $A B C D E$ inscribed in a circle for which $B C<C D$ and $A B<D E$ is the base of a pyramid with vertex $S$. If $A S$ is the longest edge starting from $S$, prove that $B S>C S$.
|
1. Let $P$ be the projection of $S$ onto the plane $A B C D E$. Obviously $B S>C S$ is equivalent to $B P>C P$. The conditions of the problem imply that $P A>P B$ and $P A>P E$. The locus of such points $P$ is the region of the plane that is determined by the perpendicular bisectors of segments $A B$ and $A E$ and that contains the point diametrically opposite $A$. But since $A B<D E$, the whole of this region lies on one side of the perpendicular bisector of $B C$. The result follows immediately. Remark. The assumption $B C<C D$ is redundant.
|
{
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|
329ab199-532c-5332-83b8-92144f42e18d
| 25,103
|
11. (FRG 2) Let $n$ and $z$ be integers greater than 1 and $(n, z)=1$. Prove: (a) At least one of the numbers $z_{i}=1+z+z^{2}+\cdots+z^{i}, i=0,1, \ldots, n-1$, is divisible by $n$. (b) If $(z-1, n)=1$, then at least one of the numbers $z_{i}, i=0,1, \ldots, n-2$, is divisible by $n$.
|
11. (a) Suppose to the contrary that none of the numbers $z_{0}, z_{1}, \ldots, z_{n-1}$ is divisible by $n$. Then two of these numbers, say $z_{k}$ and $z_{l}(0 \leq k<l \leq$ $n-1$ ), are congruent modulo $n$, and thus $n \mid z_{l}-z_{k}=z^{k+1} z_{l-k-1}$. But since $(n, z)=1$, this implies $n \mid z_{l-k-1}$, which is a contradiction. (b) Again suppose the contrary, that none of $z_{0}, z_{1}, \ldots, z_{n-2}$ is divisible by $n$. Since $(z-1, n)=1$, this is equivalent to $n \nmid(z-1) z_{j}$, i.e., $z^{k} \not \equiv 1$ $(\bmod n)$ for all $k=1,2, \ldots, n-1$. But since $(z, n)=1$, we also have that $z^{k} \not \equiv 0(\bmod n)$. It follows that there exist $k, l, 1 \leq k<l \leq n-1$ such that $z^{k} \equiv z^{l}$, i.e., $z^{l-k} \equiv 1(\bmod n)$, which is a contradiction.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
11. (FRG 2) Let $n$ and $z$ be integers greater than 1 and $(n, z)=1$. Prove: (a) At least one of the numbers $z_{i}=1+z+z^{2}+\cdots+z^{i}, i=0,1, \ldots, n-1$, is divisible by $n$. (b) If $(z-1, n)=1$, then at least one of the numbers $z_{i}, i=0,1, \ldots, n-2$, is divisible by $n$.
|
11. (a) Suppose to the contrary that none of the numbers $z_{0}, z_{1}, \ldots, z_{n-1}$ is divisible by $n$. Then two of these numbers, say $z_{k}$ and $z_{l}(0 \leq k<l \leq$ $n-1$ ), are congruent modulo $n$, and thus $n \mid z_{l}-z_{k}=z^{k+1} z_{l-k-1}$. But since $(n, z)=1$, this implies $n \mid z_{l-k-1}$, which is a contradiction. (b) Again suppose the contrary, that none of $z_{0}, z_{1}, \ldots, z_{n-2}$ is divisible by $n$. Since $(z-1, n)=1$, this is equivalent to $n \nmid(z-1) z_{j}$, i.e., $z^{k} \not \equiv 1$ $(\bmod n)$ for all $k=1,2, \ldots, n-1$. But since $(z, n)=1$, we also have that $z^{k} \not \equiv 0(\bmod n)$. It follows that there exist $k, l, 1 \leq k<l \leq n-1$ such that $z^{k} \equiv z^{l}$, i.e., $z^{l-k} \equiv 1(\bmod n)$, which is a contradiction.
|
{
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|
3251b732-4c2f-5ff1-87d8-f596f019db88
| 25,107
|
15. (GDR 1) Let $n$ be an integer greater than 1 . In the Cartesian coordinate system we consider all squares with integer vertices $(x, y)$ such that $1 \leq$ $x, y \leq n$. Denote by $p_{k}(k=0,1,2, \ldots)$ the number of pairs of points that are vertices of exactly $k$ such squares. Prove that $\sum_{k}(k-1) p_{k}=0$.
|
15. Each segment is an edge of at most two squares and a diagonal of at most one square. Therefore $p_{k}=0$ for $k>3$, and we have to prove that $$ p_{0}=p_{2}+2 p_{3} $$ Let us calculate the number $q(n)$ of considered squares. Each of these squares is inscribed in a square with integer vertices and sides parallel to the coordinate axes. There are $(n-s)^{2}$ squares of side $s$ with integer vertices and sides parallel to the coordinate axes, and each of them circumscribes exactly $s$ of the considered squares. It follows that $q(n)=\sum_{s=1}^{n-1}(n-s)^{2} s=n^{2}\left(n^{2}-1\right) / 12$. Computing the number of edges and diagonals of the considered squares in two ways, we obtain that $$ p_{1}+2 p_{2}+3 p_{3}=6 q(n) $$ On the other hand, the total number of segments with endpoints in the considered integer points is given by $$ p_{0}+p_{1}+p_{2}+p_{3}=\binom{n^{2}}{2}=\frac{n^{2}\left(n^{2}-1\right)}{2}=6 q(n) . $$ Now (1) follows immediately from (2) and (3).
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
15. (GDR 1) Let $n$ be an integer greater than 1 . In the Cartesian coordinate system we consider all squares with integer vertices $(x, y)$ such that $1 \leq$ $x, y \leq n$. Denote by $p_{k}(k=0,1,2, \ldots)$ the number of pairs of points that are vertices of exactly $k$ such squares. Prove that $\sum_{k}(k-1) p_{k}=0$.
|
15. Each segment is an edge of at most two squares and a diagonal of at most one square. Therefore $p_{k}=0$ for $k>3$, and we have to prove that $$ p_{0}=p_{2}+2 p_{3} $$ Let us calculate the number $q(n)$ of considered squares. Each of these squares is inscribed in a square with integer vertices and sides parallel to the coordinate axes. There are $(n-s)^{2}$ squares of side $s$ with integer vertices and sides parallel to the coordinate axes, and each of them circumscribes exactly $s$ of the considered squares. It follows that $q(n)=\sum_{s=1}^{n-1}(n-s)^{2} s=n^{2}\left(n^{2}-1\right) / 12$. Computing the number of edges and diagonals of the considered squares in two ways, we obtain that $$ p_{1}+2 p_{2}+3 p_{3}=6 q(n) $$ On the other hand, the total number of segments with endpoints in the considered integer points is given by $$ p_{0}+p_{1}+p_{2}+p_{3}=\binom{n^{2}}{2}=\frac{n^{2}\left(n^{2}-1\right)}{2}=6 q(n) . $$ Now (1) follows immediately from (2) and (3).
|
{
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|
9bb1fdef-ed93-5376-8123-a8f3a3a86c56
| 23,527
|
19. (GBR 1) Given any integer $m>1$ prove that there exist infinitely many positive integers $n$ such that the last $m$ digits of $5^{n}$ are a sequence $a_{m}, a_{m-1}, \ldots, a_{1}=5\left(0 \leq a_{j}<10\right)$ in which each digit except the last is of opposite parity to its successor (i.e., if $a_{i}$ is even, then $a_{i-1}$ is odd, and if $a_{i}$ is odd, then $a_{i-1}$ is even).
|
19. We shall prove the statement by induction on $m$. For $m=2$ it is trivial, since each power of 5 greater than 5 ends in 25 . Suppose that the statement is true for some $m \geq 2$, and that the last $m$ digits of $5^{n}$ alternate in parity. It can be shown by induction that the maximum power of 2 that divides $5^{2^{m-2}}-1$ is $2^{m}$, and consequently the difference $5^{n+2^{m-2}}-5^{n}$ is divisible by $10^{m}$ but not by $2 \cdot 10^{m}$. It follows that the last $m$ digits of the numbers $5^{n+2^{m-2}}$ and $5^{n}$ coincide, but the digits at the position $m+1$ have opposite parity. Hence the last $m+1$ digits of one of these two powers of 5 alternate in parity. The inductive proof is completed.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
19. (GBR 1) Given any integer $m>1$ prove that there exist infinitely many positive integers $n$ such that the last $m$ digits of $5^{n}$ are a sequence $a_{m}, a_{m-1}, \ldots, a_{1}=5\left(0 \leq a_{j}<10\right)$ in which each digit except the last is of opposite parity to its successor (i.e., if $a_{i}$ is even, then $a_{i-1}$ is odd, and if $a_{i}$ is odd, then $a_{i-1}$ is even).
|
19. We shall prove the statement by induction on $m$. For $m=2$ it is trivial, since each power of 5 greater than 5 ends in 25 . Suppose that the statement is true for some $m \geq 2$, and that the last $m$ digits of $5^{n}$ alternate in parity. It can be shown by induction that the maximum power of 2 that divides $5^{2^{m-2}}-1$ is $2^{m}$, and consequently the difference $5^{n+2^{m-2}}-5^{n}$ is divisible by $10^{m}$ but not by $2 \cdot 10^{m}$. It follows that the last $m$ digits of the numbers $5^{n+2^{m-2}}$ and $5^{n}$ coincide, but the digits at the position $m+1$ have opposite parity. Hence the last $m+1$ digits of one of these two powers of 5 alternate in parity. The inductive proof is completed.
|
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|
3357d13d-7361-5097-a0c6-0c899b85dab6
| 23,536
|
25. (HUN 3) Prove the identity $$ (z+a)^{n}=z^{n}+a \sum_{k=1}^{n}\binom{n}{k}(a-k b)^{k-1}(z+k b)^{n-k} . $$
|
25. Let $$ f_{n}(z)=z^{n}+a \sum_{k=1}^{n}\binom{n}{k}(a-k b)^{k-1}(z+k b)^{n-k} . $$ We shall prove by induction on $n$ that $f_{n}(z)=(z+a)^{n}$. This is trivial for $n=1$. Suppose that the statement is true for some positive integer $n-1$. Then $$ \begin{aligned} f_{n}^{\prime}(z) & =n z^{n-1}+a \sum_{k=1}^{n-1}\binom{n}{k}(n-k)(a-k b)^{k-1}(z+k b)^{n-k-1} \\ & =n z^{n-1}+n a \sum_{k=1}^{n-1}\binom{n-1}{k}(a-k b)^{k-1}(z+k b)^{n-k-1} \\ & =n f_{n-1}(z)=n(z+a)^{n-1} \end{aligned} $$ It remains to prove that $f_{n}(-a)=0$. For $z=-a$ we have by the lemma of (SL81-13), $$ \begin{aligned} f_{n}(-a) & =(-a)^{n}+a \sum_{k=1}^{n}\binom{n}{k}(-1)^{n-k}(a-k b)^{n-1} \\ & =a \sum_{k=0}^{n}\binom{n}{k}(-1)^{n-k}(a-k b)^{n-1}=0 . \end{aligned} $$
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
25. (HUN 3) Prove the identity $$ (z+a)^{n}=z^{n}+a \sum_{k=1}^{n}\binom{n}{k}(a-k b)^{k-1}(z+k b)^{n-k} . $$
|
25. Let $$ f_{n}(z)=z^{n}+a \sum_{k=1}^{n}\binom{n}{k}(a-k b)^{k-1}(z+k b)^{n-k} . $$ We shall prove by induction on $n$ that $f_{n}(z)=(z+a)^{n}$. This is trivial for $n=1$. Suppose that the statement is true for some positive integer $n-1$. Then $$ \begin{aligned} f_{n}^{\prime}(z) & =n z^{n-1}+a \sum_{k=1}^{n-1}\binom{n}{k}(n-k)(a-k b)^{k-1}(z+k b)^{n-k-1} \\ & =n z^{n-1}+n a \sum_{k=1}^{n-1}\binom{n-1}{k}(a-k b)^{k-1}(z+k b)^{n-k-1} \\ & =n f_{n-1}(z)=n(z+a)^{n-1} \end{aligned} $$ It remains to prove that $f_{n}(-a)=0$. For $z=-a$ we have by the lemma of (SL81-13), $$ \begin{aligned} f_{n}(-a) & =(-a)^{n}+a \sum_{k=1}^{n}\binom{n}{k}(-1)^{n-k}(a-k b)^{n-1} \\ & =a \sum_{k=0}^{n}\binom{n}{k}(-1)^{n-k}(a-k b)^{n-1}=0 . \end{aligned} $$
|
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|
01f824d5-bd16-5ed8-8851-5981492e97bf
| 23,553
|
26. (NET 1) Let $p$ be a prime number greater than 5 . Let $V$ be the collection of all positive integers $n$ that can be written in the form $n=k p+1$ or $n=k p-1(k=1,2, \ldots)$. A number $n \in V$ is called indecomposable in $V$ if it is impossible to find $k, l \in V$ such that $n=k l$. Prove that there exists a number $N \in V$ that can be factorized into indecomposable factors in $V$ in more than one way.
|
26. The result is an immediate consequence (for $G=\{-1,1\}$ ) of the following generalization. (1) Let $G$ be a proper subgroup of $\mathbb{Z}_{n}^{*}$ (the multiplicative group of residue classes modulo $n$ coprime to $n$ ), and let $V$ be the union of elements of $G$. A number $m \in V$ is called indecomposable in $V$ if there do not exist numbers $p, q \in V, p, q \notin\{-1,1\}$, such that $p q=m$. There exists a number $r \in V$ that can be expressed as a product of elements indecomposable in $V$ in more than one way. First proof. We shall start by proving the following lemma. Lemma. There are infinitely many primes not in $V$ that do not divide $n$. Proof. There is at least one such prime: In fact, any number other than $\pm 1$ not in $V$ must have a prime factor not in $V$, since $V$ is closed under multiplication. If there were a finite number of such primes, say $p_{1}, p_{2}, \ldots, p_{k}$, then one of the numbers $p_{1} p_{2} \cdots p_{k}+n, p_{1}^{2} p_{2} \cdots p_{k}+n$ is not in $V$ and is coprime to $n$ and $p_{1}, \ldots, p_{k}$, which is a contradiction. [This lemma is actually a direct consequence of Dirichlet's theorem.] Let us consider two such primes $p, q$ that are congruent modulo $n$. Let $p^{k}$ be the least power of $p$ that is in $V$. Then $p^{k}, q^{k}, p^{k-1} q, p q^{k-1}$ belong to $V$ and are indecomposable in $V$. It follows that $$ r=p^{k} \cdot q^{k}=p^{k-1} q \cdot p q^{k-1} $$ has the desired property. Second proof. Let $p$ be any prime not in $V$ that does not divide $n$, and let $p^{k}$ be the least power of $p$ that is in $V$. Obviously $p^{k}$ is indecomposable in $V$. Then the number $$ r=p^{k} \cdot\left(p^{k-1}+n\right)(p+n)=p\left(p^{k-1}+n\right) \cdot p^{k-1}(p+n) $$ has at least two different factorizations into indecomposable factors.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
26. (NET 1) Let $p$ be a prime number greater than 5 . Let $V$ be the collection of all positive integers $n$ that can be written in the form $n=k p+1$ or $n=k p-1(k=1,2, \ldots)$. A number $n \in V$ is called indecomposable in $V$ if it is impossible to find $k, l \in V$ such that $n=k l$. Prove that there exists a number $N \in V$ that can be factorized into indecomposable factors in $V$ in more than one way.
|
26. The result is an immediate consequence (for $G=\{-1,1\}$ ) of the following generalization. (1) Let $G$ be a proper subgroup of $\mathbb{Z}_{n}^{*}$ (the multiplicative group of residue classes modulo $n$ coprime to $n$ ), and let $V$ be the union of elements of $G$. A number $m \in V$ is called indecomposable in $V$ if there do not exist numbers $p, q \in V, p, q \notin\{-1,1\}$, such that $p q=m$. There exists a number $r \in V$ that can be expressed as a product of elements indecomposable in $V$ in more than one way. First proof. We shall start by proving the following lemma. Lemma. There are infinitely many primes not in $V$ that do not divide $n$. Proof. There is at least one such prime: In fact, any number other than $\pm 1$ not in $V$ must have a prime factor not in $V$, since $V$ is closed under multiplication. If there were a finite number of such primes, say $p_{1}, p_{2}, \ldots, p_{k}$, then one of the numbers $p_{1} p_{2} \cdots p_{k}+n, p_{1}^{2} p_{2} \cdots p_{k}+n$ is not in $V$ and is coprime to $n$ and $p_{1}, \ldots, p_{k}$, which is a contradiction. [This lemma is actually a direct consequence of Dirichlet's theorem.] Let us consider two such primes $p, q$ that are congruent modulo $n$. Let $p^{k}$ be the least power of $p$ that is in $V$. Then $p^{k}, q^{k}, p^{k-1} q, p q^{k-1}$ belong to $V$ and are indecomposable in $V$. It follows that $$ r=p^{k} \cdot q^{k}=p^{k-1} q \cdot p q^{k-1} $$ has the desired property. Second proof. Let $p$ be any prime not in $V$ that does not divide $n$, and let $p^{k}$ be the least power of $p$ that is in $V$. Obviously $p^{k}$ is indecomposable in $V$. Then the number $$ r=p^{k} \cdot\left(p^{k-1}+n\right)(p+n)=p\left(p^{k-1}+n\right) \cdot p^{k-1}(p+n) $$ has at least two different factorizations into indecomposable factors.
|
{
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|
f8138263-eee7-5113-ba57-38ed11c9e824
| 23,555
|
30. (NET 5) A triangle $A B C$ with $\angle A=30^{\circ}$ and $\angle C=54^{\circ}$ is given. On $B C$ a point $D$ is chosen such that $\angle C A D=12^{\circ}$. On $A B$ a point $E$ is chosen such that $\angle A C E=6^{\circ}$. Let $S$ be the point of intersection of $A D$ and $C E$. Prove that $B S=B C$.
|
30. Suppose $\angle S B A=x$. By the trigonometric form of Ceva's theorem we have $$ \frac{\sin \left(96^{\circ}-x\right)}{\sin x} \frac{\sin 18^{\circ}}{\sin 12^{\circ}} \frac{\sin 6^{\circ}}{\sin 48^{\circ}}=1 $$ We claim that $x=12^{\circ}$ is a solution of this equation. To prove this, it is enough to show that $\sin 84^{\circ} \sin 6^{\circ} \sin 18^{\circ}=\sin 48^{\circ} \sin 12^{\circ} \sin 12^{\circ}$, which is equivalent to $\sin 18^{\circ}=2 \sin 48^{\circ} \sin 12^{\circ}=\cos 36^{\circ}-\cos 60^{\circ}$. The last equality can be checked directly. Since the equation is equivalent to $\left(\sin 96^{\circ} \cot x-\cos 96^{\circ}\right) \sin 6^{\circ} \sin 18^{\circ}=$ $\sin 48^{\circ} \sin 12^{\circ}$, the solution $x \in[0, \pi)$ is unique. Hence $x=12^{\circ}$. Second solution. We know that if $a, b, c, a^{\prime}, b^{\prime}, c^{\prime}$ are points on the unit circle in the complex plane, the lines $a a^{\prime}, b b^{\prime}, c c^{\prime}$ are concurrent if and only if $$ \left(a-b^{\prime}\right)\left(b-c^{\prime}\right)\left(c-a^{\prime}\right)=\left(a-c^{\prime}\right)\left(b-a^{\prime}\right)\left(c-b^{\prime}\right) $$ We shall prove that $x=12^{\circ}$. We may suppose that $A B C$ is the triangle in the complex plane with vertices $a=1, b=\epsilon^{9}, c=\epsilon^{14}$, where $\epsilon=$ $\cos \frac{\pi}{15}+i \sin \frac{\pi}{15}$. If $a^{\prime}=\epsilon^{12}, b^{\prime}=\epsilon^{28}, c^{\prime}=\epsilon$, our task is the same as proving that lines $a a^{\prime}, b b^{\prime}, c c^{\prime}$ are concurrent, or by (1) that $$ \left(1-\epsilon^{28}\right)\left(\epsilon^{9}-\epsilon\right)\left(\epsilon^{14}-\epsilon^{12}\right)-(1-\epsilon)\left(\epsilon^{9}-\epsilon^{12}\right)\left(\epsilon^{14}-\epsilon^{28}\right)=0 . $$ The last equality holds, since the left-hand side is divisible by the minimum polynomial of $\epsilon: z^{8}+z^{7}-z^{5}-z^{4}-z^{3}+z+1$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
30. (NET 5) A triangle $A B C$ with $\angle A=30^{\circ}$ and $\angle C=54^{\circ}$ is given. On $B C$ a point $D$ is chosen such that $\angle C A D=12^{\circ}$. On $A B$ a point $E$ is chosen such that $\angle A C E=6^{\circ}$. Let $S$ be the point of intersection of $A D$ and $C E$. Prove that $B S=B C$.
|
30. Suppose $\angle S B A=x$. By the trigonometric form of Ceva's theorem we have $$ \frac{\sin \left(96^{\circ}-x\right)}{\sin x} \frac{\sin 18^{\circ}}{\sin 12^{\circ}} \frac{\sin 6^{\circ}}{\sin 48^{\circ}}=1 $$ We claim that $x=12^{\circ}$ is a solution of this equation. To prove this, it is enough to show that $\sin 84^{\circ} \sin 6^{\circ} \sin 18^{\circ}=\sin 48^{\circ} \sin 12^{\circ} \sin 12^{\circ}$, which is equivalent to $\sin 18^{\circ}=2 \sin 48^{\circ} \sin 12^{\circ}=\cos 36^{\circ}-\cos 60^{\circ}$. The last equality can be checked directly. Since the equation is equivalent to $\left(\sin 96^{\circ} \cot x-\cos 96^{\circ}\right) \sin 6^{\circ} \sin 18^{\circ}=$ $\sin 48^{\circ} \sin 12^{\circ}$, the solution $x \in[0, \pi)$ is unique. Hence $x=12^{\circ}$. Second solution. We know that if $a, b, c, a^{\prime}, b^{\prime}, c^{\prime}$ are points on the unit circle in the complex plane, the lines $a a^{\prime}, b b^{\prime}, c c^{\prime}$ are concurrent if and only if $$ \left(a-b^{\prime}\right)\left(b-c^{\prime}\right)\left(c-a^{\prime}\right)=\left(a-c^{\prime}\right)\left(b-a^{\prime}\right)\left(c-b^{\prime}\right) $$ We shall prove that $x=12^{\circ}$. We may suppose that $A B C$ is the triangle in the complex plane with vertices $a=1, b=\epsilon^{9}, c=\epsilon^{14}$, where $\epsilon=$ $\cos \frac{\pi}{15}+i \sin \frac{\pi}{15}$. If $a^{\prime}=\epsilon^{12}, b^{\prime}=\epsilon^{28}, c^{\prime}=\epsilon$, our task is the same as proving that lines $a a^{\prime}, b b^{\prime}, c c^{\prime}$ are concurrent, or by (1) that $$ \left(1-\epsilon^{28}\right)\left(\epsilon^{9}-\epsilon\right)\left(\epsilon^{14}-\epsilon^{12}\right)-(1-\epsilon)\left(\epsilon^{9}-\epsilon^{12}\right)\left(\epsilon^{14}-\epsilon^{28}\right)=0 . $$ The last equality holds, since the left-hand side is divisible by the minimum polynomial of $\epsilon: z^{8}+z^{7}-z^{5}-z^{4}-z^{3}+z+1$.
|
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|
60b7178f-4afb-5426-b27b-55ab01d7ac15
| 23,568
|
31. (POL 1) Let $f$ be a function defined on the set of pairs of nonzero rational numbers whose values are positive real numbers. Suppose that $f$ satisfies the following conditions: (1) $f(a b, c)=f(a, c) f(b, c), f(c, a b)=f(c, a) f(c, b)$; (2) $f(a, 1-a)=1$. Prove that $f(a, a)=f(a,-a)=1, f(a, b) f(b, a)=1$.
|
31. We obtain from (1) that $f(1, c)=f(1, c) f(1, c)$; hence $f(1, c)=1$ and consequently $f(-1, c) f(-1, c)=f(1, c)=1$, i.e. $f(-1, c)=1$. Analogously, $f(c, 1)=f(c,-1)=1$. Clearly $f(1,1)=f(-1,1)=f(1,-1)=1$. Now let us assume that $a \neq 1$. Observe that $f\left(x^{-1}, y\right)=f\left(x, y^{-1}\right)=f(x, y)^{-1}$. Thus by (1) and (2) we get $$ \begin{aligned} 1 & =f(a, 1-a) f(1 / a, 1-1 / a) \\ & =f(a, 1-a) f\left(a, \frac{1}{1-1 / a}\right)=f\left(a, \frac{1-a}{1-1 / a}\right)=f(a,-a) \end{aligned} $$ We now have $f(a, a)=f(a,-1) f(a,-a)=1 \cdot 1=1$ and $1=f(a b, a b)=$ $f(a, a b) f(b, a b)=f(a, a) f(a, b) f(b, a) f(b, b)=f(a, b) f(b, a)$.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
31. (POL 1) Let $f$ be a function defined on the set of pairs of nonzero rational numbers whose values are positive real numbers. Suppose that $f$ satisfies the following conditions: (1) $f(a b, c)=f(a, c) f(b, c), f(c, a b)=f(c, a) f(c, b)$; (2) $f(a, 1-a)=1$. Prove that $f(a, a)=f(a,-a)=1, f(a, b) f(b, a)=1$.
|
31. We obtain from (1) that $f(1, c)=f(1, c) f(1, c)$; hence $f(1, c)=1$ and consequently $f(-1, c) f(-1, c)=f(1, c)=1$, i.e. $f(-1, c)=1$. Analogously, $f(c, 1)=f(c,-1)=1$. Clearly $f(1,1)=f(-1,1)=f(1,-1)=1$. Now let us assume that $a \neq 1$. Observe that $f\left(x^{-1}, y\right)=f\left(x, y^{-1}\right)=f(x, y)^{-1}$. Thus by (1) and (2) we get $$ \begin{aligned} 1 & =f(a, 1-a) f(1 / a, 1-1 / a) \\ & =f(a, 1-a) f\left(a, \frac{1}{1-1 / a}\right)=f\left(a, \frac{1-a}{1-1 / a}\right)=f(a,-a) \end{aligned} $$ We now have $f(a, a)=f(a,-1) f(a,-a)=1 \cdot 1=1$ and $1=f(a b, a b)=$ $f(a, a b) f(b, a b)=f(a, a) f(a, b) f(b, a) f(b, b)=f(a, b) f(b, a)$.
|
{
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|
d4d12b5b-c2d2-57be-b7f9-4946d66e28cd
| 23,571
|
32. (POL 2) In a room there are nine men. Among every three of them there are two mutually acquainted. Prove that some four of them are mutually acquainted.
|
32. It is a known result that among six persons there are 3 mutually acquainted or 3 mutually unacquainted. By the condition of the problem the last case is excluded. If there is a man in the room who is not acquainted with four of the others, then these four men are mutually acquainted. Otherwise, each man is acquainted with at least five others, and since the sum of numbers of acquaintances of all men in the room is even, one of the men is acquainted with at least six men. Among these six there are three mutually acquainted, and they together with the first one make a group of four mutually acquainted men.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
32. (POL 2) In a room there are nine men. Among every three of them there are two mutually acquainted. Prove that some four of them are mutually acquainted.
|
32. It is a known result that among six persons there are 3 mutually acquainted or 3 mutually unacquainted. By the condition of the problem the last case is excluded. If there is a man in the room who is not acquainted with four of the others, then these four men are mutually acquainted. Otherwise, each man is acquainted with at least five others, and since the sum of numbers of acquaintances of all men in the room is even, one of the men is acquainted with at least six men. Among these six there are three mutually acquainted, and they together with the first one make a group of four mutually acquainted men.
|
{
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|
e7d937b8-0a76-5e5b-94a1-99e58595128c
| 23,573
|
33. (POL 3) A circle $K$ centered at $(0,0)$ is given. Prove that for every vector $\left(a_{1}, a_{2}\right)$ there is a positive integer $n$ such that the circle $K$ translated by the vector $n\left(a_{1}, a_{2}\right)$ contains a lattice point (i.e., a point both of whose coordinates are integers).
|
33. Let $r$ be the radius of $K$ and $s>\sqrt{2} / r$ an integer. Consider the points $A_{k}\left(k a_{1}-\left[k a_{1}\right], k a_{2}-\left[k a_{2}\right]\right)$, where $k=0,1,2, \ldots, s^{2}$. Since all these points are in the unit square, two of them, say $A_{p}, A_{q}, q>p$, are in a small square with side $1 / s$, and consequently $A_{p} A_{q} \leq \sqrt{2} / s<r$. Therefore, for $n=q-p, m_{1}=\left[q a_{1}\right]-\left[p a_{1}\right]$ and $m_{2}=\left[q a_{2}\right]-\left[p a_{2}\right]$ the distance between the points $n\left(a_{1}, a_{2}\right)$ and $\left(m_{1}, m_{2}\right)$ is less then $r$, i.e., the point $\left(m_{1}, m_{2}\right)$ is in the circle $K+n\left(a_{1}, a_{2}\right)$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
33. (POL 3) A circle $K$ centered at $(0,0)$ is given. Prove that for every vector $\left(a_{1}, a_{2}\right)$ there is a positive integer $n$ such that the circle $K$ translated by the vector $n\left(a_{1}, a_{2}\right)$ contains a lattice point (i.e., a point both of whose coordinates are integers).
|
33. Let $r$ be the radius of $K$ and $s>\sqrt{2} / r$ an integer. Consider the points $A_{k}\left(k a_{1}-\left[k a_{1}\right], k a_{2}-\left[k a_{2}\right]\right)$, where $k=0,1,2, \ldots, s^{2}$. Since all these points are in the unit square, two of them, say $A_{p}, A_{q}, q>p$, are in a small square with side $1 / s$, and consequently $A_{p} A_{q} \leq \sqrt{2} / s<r$. Therefore, for $n=q-p, m_{1}=\left[q a_{1}\right]-\left[p a_{1}\right]$ and $m_{2}=\left[q a_{2}\right]-\left[p a_{2}\right]$ the distance between the points $n\left(a_{1}, a_{2}\right)$ and $\left(m_{1}, m_{2}\right)$ is less then $r$, i.e., the point $\left(m_{1}, m_{2}\right)$ is in the circle $K+n\left(a_{1}, a_{2}\right)$.
|
{
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|
3d82e074-3655-52aa-a928-6df6ed735901
| 23,576
|
36. (ROM 2) Consider a sequence of numbers $\left(a_{1}, a_{2}, \ldots, a_{2^{n}}\right)$. Define the operation $$ S\left(\left(a_{1}, a_{2}, \ldots, a_{2^{n}}\right)\right)=\left(a_{1} a_{2}, a_{2} a_{3}, \ldots, a_{2^{n}-1} a_{2^{n}}, a_{2^{n}} a_{1}\right) $$ Prove that whatever the sequence $\left(a_{1}, a_{2}, \ldots, a_{2^{n}}\right)$ is, with $a_{i} \in\{-1,1\}$ for $i=1,2, \ldots, 2^{n}$, after finitely many applications of the operation we get the sequence $(1,1, \ldots, 1)$.
|
36. It can be shown by simple induction that $S^{m}\left(a_{1}, \ldots, a_{2^{n}}\right)=\left(b_{1}, \ldots, b_{2^{n}}\right)$, where $$ \left.b_{k}=\prod_{i=0}^{m} a_{k+i}^{\binom{m}{i}} \text { (assuming that } a_{k+2^{n}}=a_{k}\right) $$ If we take $m=2^{n}$ all the binomial coefficients $\binom{m}{i}$ apart from $i=0$ and $i=m$ will be even, and thus $b_{k}=a_{k} a_{k+m}=1$ for all $k$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
36. (ROM 2) Consider a sequence of numbers $\left(a_{1}, a_{2}, \ldots, a_{2^{n}}\right)$. Define the operation $$ S\left(\left(a_{1}, a_{2}, \ldots, a_{2^{n}}\right)\right)=\left(a_{1} a_{2}, a_{2} a_{3}, \ldots, a_{2^{n}-1} a_{2^{n}}, a_{2^{n}} a_{1}\right) $$ Prove that whatever the sequence $\left(a_{1}, a_{2}, \ldots, a_{2^{n}}\right)$ is, with $a_{i} \in\{-1,1\}$ for $i=1,2, \ldots, 2^{n}$, after finitely many applications of the operation we get the sequence $(1,1, \ldots, 1)$.
|
36. It can be shown by simple induction that $S^{m}\left(a_{1}, \ldots, a_{2^{n}}\right)=\left(b_{1}, \ldots, b_{2^{n}}\right)$, where $$ \left.b_{k}=\prod_{i=0}^{m} a_{k+i}^{\binom{m}{i}} \text { (assuming that } a_{k+2^{n}}=a_{k}\right) $$ If we take $m=2^{n}$ all the binomial coefficients $\binom{m}{i}$ apart from $i=0$ and $i=m$ will be even, and thus $b_{k}=a_{k} a_{k+m}=1$ for all $k$.
|
{
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|
5361b7b1-cdc2-52a9-8dfd-58e03ed142f9
| 23,583
|
37. (ROM 3) Let $A_{1}, A_{2}, \ldots, A_{n+1}$ be positive integers such that $\left(A_{i}, A_{n+1}\right)$ $=1$ for every $i=1,2, \ldots, n$. Show that the equation $$ x_{1}^{A_{1}}+x_{2}^{A_{2}}+\cdots+x_{n}^{A_{n}}=x_{n+1}^{A_{n+1}} $$ has an infinite set of solutions $\left(x_{1}, x_{2}, \ldots, x_{n+1}\right)$ in positive integers.
|
37. We look for a solution with $x_{1}^{A_{1}}=\cdots=x_{n}^{A_{n}}=n^{A_{1} A_{2} \cdots A_{n} x}$ and $x_{n+1}=$ $n^{y}$. In order for this to be a solution we must have $A_{1} A_{2} \cdots A_{n} x+1=$ $A_{n+1} y$. This equation has infinitely many solutions $(x, y)$ in $\mathbb{N}$, since $A_{1} A_{2} \cdots A_{n}$ and $A_{n+1}$ are coprime.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
37. (ROM 3) Let $A_{1}, A_{2}, \ldots, A_{n+1}$ be positive integers such that $\left(A_{i}, A_{n+1}\right)$ $=1$ for every $i=1,2, \ldots, n$. Show that the equation $$ x_{1}^{A_{1}}+x_{2}^{A_{2}}+\cdots+x_{n}^{A_{n}}=x_{n+1}^{A_{n+1}} $$ has an infinite set of solutions $\left(x_{1}, x_{2}, \ldots, x_{n+1}\right)$ in positive integers.
|
37. We look for a solution with $x_{1}^{A_{1}}=\cdots=x_{n}^{A_{n}}=n^{A_{1} A_{2} \cdots A_{n} x}$ and $x_{n+1}=$ $n^{y}$. In order for this to be a solution we must have $A_{1} A_{2} \cdots A_{n} x+1=$ $A_{n+1} y$. This equation has infinitely many solutions $(x, y)$ in $\mathbb{N}$, since $A_{1} A_{2} \cdots A_{n}$ and $A_{n+1}$ are coprime.
|
{
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|
24c2defc-aa0c-54d0-b79e-d81c7965e141
| 23,585
|
38. (ROM 4) Let $m_{j}>0$ for $j=1,2, \ldots, n$ and $a_{1} \leq \cdots \leq a_{n}<b_{1} \leq \cdots \leq$ $b_{n}<c_{1} \leq \cdots \leq c_{n}$ be real numbers. Prove: $$ \left[\sum_{j=1}^{n} m_{j}\left(a_{j}+b_{j}+c_{j}\right)\right]^{2}>3\left(\sum_{j=1}^{n} m_{j}\right)\left[\sum_{j=1}^{n} m_{j}\left(a_{j} b_{j}+b_{j} c_{j}+c_{j} a_{j}\right)\right] . $$
|
38. The condition says that the quadratic equation $f(x)=0$ has distinct real solutions, where $$ f(x)=3 x^{2} \sum_{j=1}^{n} m_{j}-2 x \sum_{j=1}^{n} m_{j}\left(a_{j}+b_{j}+c_{j}\right)+\sum_{j=1}^{n} m_{j}\left(a_{j} b_{j}+b_{j} c_{j}+c_{j} a_{j}\right) $$ It is easy to verify that the function $f$ is the derivative of $$ F(x)=\sum_{j=1}^{n} m_{j}\left(x-a_{j}\right)\left(x-b_{j}\right)\left(x-c_{j}\right) $$ Since $F\left(a_{1}\right) \leq 0 \leq F\left(a_{n}\right), F\left(b_{1}\right) \leq 0 \leq F\left(b_{n}\right)$ and $F\left(c_{1}\right) \leq 0 \leq F\left(c_{n}\right)$, $F(x)$ has three distinct real roots, and hence by Rolle's theorem its derivative $f(x)$ has two distinct real roots.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
38. (ROM 4) Let $m_{j}>0$ for $j=1,2, \ldots, n$ and $a_{1} \leq \cdots \leq a_{n}<b_{1} \leq \cdots \leq$ $b_{n}<c_{1} \leq \cdots \leq c_{n}$ be real numbers. Prove: $$ \left[\sum_{j=1}^{n} m_{j}\left(a_{j}+b_{j}+c_{j}\right)\right]^{2}>3\left(\sum_{j=1}^{n} m_{j}\right)\left[\sum_{j=1}^{n} m_{j}\left(a_{j} b_{j}+b_{j} c_{j}+c_{j} a_{j}\right)\right] . $$
|
38. The condition says that the quadratic equation $f(x)=0$ has distinct real solutions, where $$ f(x)=3 x^{2} \sum_{j=1}^{n} m_{j}-2 x \sum_{j=1}^{n} m_{j}\left(a_{j}+b_{j}+c_{j}\right)+\sum_{j=1}^{n} m_{j}\left(a_{j} b_{j}+b_{j} c_{j}+c_{j} a_{j}\right) $$ It is easy to verify that the function $f$ is the derivative of $$ F(x)=\sum_{j=1}^{n} m_{j}\left(x-a_{j}\right)\left(x-b_{j}\right)\left(x-c_{j}\right) $$ Since $F\left(a_{1}\right) \leq 0 \leq F\left(a_{n}\right), F\left(b_{1}\right) \leq 0 \leq F\left(b_{n}\right)$ and $F\left(c_{1}\right) \leq 0 \leq F\left(c_{n}\right)$, $F(x)$ has three distinct real roots, and hence by Rolle's theorem its derivative $f(x)$ has two distinct real roots.
|
{
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|
2b8083c8-731e-5268-8f96-459114bd37ab
| 23,587
|
39. (ROM 5) Consider 37 distinct points in space, all with integer coordinates. Prove that we may find among them three distinct points such that their barycenter has integers coordinates.
|
39. By the pigeonhole principle, we can find 5 distinct points among the given 37 such that their $x$-coordinates are congruent and their $y$-coordinates are congruent modulo 3 . Now among these 5 points either there exist three with $z$-coordinates congruent modulo 3 , or there exist three whose $z$ coordinates are congruent to $0,1,2$ modulo 3 . These three points are the desired ones. Remark. The minimum number $n$ such that among any $n$ integer points in space one can find three points whose barycenter is an integer point is $n=19$. Each proof of this result seems to consist in studying a great number of cases.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
39. (ROM 5) Consider 37 distinct points in space, all with integer coordinates. Prove that we may find among them three distinct points such that their barycenter has integers coordinates.
|
39. By the pigeonhole principle, we can find 5 distinct points among the given 37 such that their $x$-coordinates are congruent and their $y$-coordinates are congruent modulo 3 . Now among these 5 points either there exist three with $z$-coordinates congruent modulo 3 , or there exist three whose $z$ coordinates are congruent to $0,1,2$ modulo 3 . These three points are the desired ones. Remark. The minimum number $n$ such that among any $n$ integer points in space one can find three points whose barycenter is an integer point is $n=19$. Each proof of this result seems to consist in studying a great number of cases.
|
{
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|
ce2c955d-6ca9-5076-8cde-66d087a9fbe4
| 23,589
|
4. (BUL 4) We are given $n$ points in space. Some pairs of these points are connected by line segments so that the number of segments equals $\left[n^{2} / 4\right]$, and a connected triangle exists. Prove that any point from which the maximal number of segments starts is a vertex of a connected triangle.
|
4. Consider any vertex $v_{n}$ from which the maximal number $d$ of segments start, and suppose it is not a vertex of a triangle. Let $\mathcal{A}=$ $\left\{v_{1}, v_{2}, \ldots, v_{d}\right\}$ be the set of points that are connected to $v_{n}$, and let $\mathcal{B}=\left\{v_{d+1}, v_{d+2}, \ldots, v_{n}\right\}$ be the set of the other points. Since $v_{n}$ is not a vertex of a triangle, there is no segment both of whose vertices lie in $\mathcal{A}$; i.e., each segment has an end in $\mathcal{B}$. Thus, if $d_{j}$ denotes the number of segments at $v_{j}$ and $m$ denotes the total number of segments, we have $$ m \leq d_{d+1}+d_{d+2}+\cdots+d_{n} \leq d(n-d) \leq\left[\frac{n^{2}}{4}\right]=m $$ This means that each inequality must be equality, implying that each point in $\mathcal{B}$ is a vertex of $d$ segments, and each of these segments has the other end in $\mathcal{A}$. Then there is no triangle at all, which is a contradiction.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
4. (BUL 4) We are given $n$ points in space. Some pairs of these points are connected by line segments so that the number of segments equals $\left[n^{2} / 4\right]$, and a connected triangle exists. Prove that any point from which the maximal number of segments starts is a vertex of a connected triangle.
|
4. Consider any vertex $v_{n}$ from which the maximal number $d$ of segments start, and suppose it is not a vertex of a triangle. Let $\mathcal{A}=$ $\left\{v_{1}, v_{2}, \ldots, v_{d}\right\}$ be the set of points that are connected to $v_{n}$, and let $\mathcal{B}=\left\{v_{d+1}, v_{d+2}, \ldots, v_{n}\right\}$ be the set of the other points. Since $v_{n}$ is not a vertex of a triangle, there is no segment both of whose vertices lie in $\mathcal{A}$; i.e., each segment has an end in $\mathcal{B}$. Thus, if $d_{j}$ denotes the number of segments at $v_{j}$ and $m$ denotes the total number of segments, we have $$ m \leq d_{d+1}+d_{d+2}+\cdots+d_{n} \leq d(n-d) \leq\left[\frac{n^{2}}{4}\right]=m $$ This means that each inequality must be equality, implying that each point in $\mathcal{B}$ is a vertex of $d$ segments, and each of these segments has the other end in $\mathcal{A}$. Then there is no triangle at all, which is a contradiction.
|
{
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|
ac50a06b-fc86-5345-9e1d-73028d012e42
| 23,592
|
40. (SWE 1) The numbers $1,2,3, \ldots, 64$ are placed on a chessboard, one number in each square. Consider all squares on the chessboard of size $2 \times 2$. Prove that there are at least three such squares for which the sum of the 4 numbers contained exceeds 100.
|
40. Let us divide the chessboard into 16 squares $Q_{1}, Q_{2}, \ldots, Q_{16}$ of size $2 \times 2$. Let $s_{k}$ be the sum of numbers in $Q_{k}$, and let us assume that $s_{1} \geq s_{2} \geq$ $\cdots \geq s_{16}$. Since $s_{4}+s_{5}+\cdots+s_{16} \geq 1+2+\cdots+52=1378$, we must have $s_{4} \geq 100$ and hence $s_{1}, s_{2}, s_{3} \geq 100$ as well.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
40. (SWE 1) The numbers $1,2,3, \ldots, 64$ are placed on a chessboard, one number in each square. Consider all squares on the chessboard of size $2 \times 2$. Prove that there are at least three such squares for which the sum of the 4 numbers contained exceeds 100.
|
40. Let us divide the chessboard into 16 squares $Q_{1}, Q_{2}, \ldots, Q_{16}$ of size $2 \times 2$. Let $s_{k}$ be the sum of numbers in $Q_{k}$, and let us assume that $s_{1} \geq s_{2} \geq$ $\cdots \geq s_{16}$. Since $s_{4}+s_{5}+\cdots+s_{16} \geq 1+2+\cdots+52=1378$, we must have $s_{4} \geq 100$ and hence $s_{1}, s_{2}, s_{3} \geq 100$ as well.
|
{
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|
733d785c-fd81-52b9-8946-c5708b69553f
| 23,595
|
42. (SWE 3) The sequence $a_{n, k}, k=1,2,3, \ldots, 2^{n}, n=0,1,2, \ldots$, is defined by the following recurrence formula: $$ \begin{aligned} a_{1} & =2, \quad a_{n, k}=2 a_{n-1, k}^{3}, \quad a_{n, k+2^{n-1}}=\frac{1}{2} a_{n-1, k}^{3} \\ \text { for } k & =1,2,3, \ldots, 2^{n-1}, n=0,1,2, \ldots . \end{aligned} $$ Prove that the numbers $a_{n, k}$ are all different.
|
42. It can be proved by induction on $n$ that $\left\{a_{n, k} \mid 1 \leq k \leq 2^{n}\right\}=\left\{2^{m} \mid m=3^{n}+3^{n-1} s_{1}+\cdots+3^{1} s_{n-1}+s_{n}\left(s_{i}= \pm 1\right)\right\}$. Thus the result is an immediate consequence of the following lemma. Lemma. Each positive integer $s$ can be uniquely represented in the form $$ s=3^{n}+3^{n-1} s_{1}+\cdots+3^{1} s_{n-1}+s_{n}, \quad \text { where } s_{i} \in\{-1,0,1\} $$ Proof. Both the existence and the uniqueness can be shown by simple induction on $s$. The statement is trivial for $s=1$, while for $s>1$ there exist $q \in \mathbb{N}, r \in\{-1,0,1\}$ such that $s=3 q+r$, and $q$ has a unique representation of the form (1).
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
42. (SWE 3) The sequence $a_{n, k}, k=1,2,3, \ldots, 2^{n}, n=0,1,2, \ldots$, is defined by the following recurrence formula: $$ \begin{aligned} a_{1} & =2, \quad a_{n, k}=2 a_{n-1, k}^{3}, \quad a_{n, k+2^{n-1}}=\frac{1}{2} a_{n-1, k}^{3} \\ \text { for } k & =1,2,3, \ldots, 2^{n-1}, n=0,1,2, \ldots . \end{aligned} $$ Prove that the numbers $a_{n, k}$ are all different.
|
42. It can be proved by induction on $n$ that $\left\{a_{n, k} \mid 1 \leq k \leq 2^{n}\right\}=\left\{2^{m} \mid m=3^{n}+3^{n-1} s_{1}+\cdots+3^{1} s_{n-1}+s_{n}\left(s_{i}= \pm 1\right)\right\}$. Thus the result is an immediate consequence of the following lemma. Lemma. Each positive integer $s$ can be uniquely represented in the form $$ s=3^{n}+3^{n-1} s_{1}+\cdots+3^{1} s_{n-1}+s_{n}, \quad \text { where } s_{i} \in\{-1,0,1\} $$ Proof. Both the existence and the uniqueness can be shown by simple induction on $s$. The statement is trivial for $s=1$, while for $s>1$ there exist $q \in \mathbb{N}, r \in\{-1,0,1\}$ such that $s=3 q+r$, and $q$ has a unique representation of the form (1).
|
{
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|
e9a2d9cf-6604-500b-9b5d-9b8899302a81
| 23,600
|
44. (FIN 2) Let $E$ be a finite set of points in space such that $E$ is not contained in a plane and no three points of $E$ are collinear. Show that $E$ contains the vertices of a tetrahedron $T=A B C D$ such that $T \cap E=$ $\{A, B, C, D\}$ (including interior points of $T$ ) and such that the projection of $A$ onto the plane $B C D$ is inside a triangle that is similar to the triangle $B C D$ and whose sides have midpoints $B, C, D$.
|
44. Let $d(X, \sigma)$ denote the distance from a point $X$ to a plane $\sigma$. Let us consider the pair $(A, \pi)$ where $A \in E$ and $\pi$ is a plane containing some three points $B, C, D \in E$ such that $d(A, \pi)$ is the smallest possible. We may suppose that $B, C, D$ are selected such that $\triangle B C D$ contains no other points of $E$. Let $A^{\prime}$ be the projection of $A$ on $\pi$, and let $l_{b}, l_{c}, l_{d}$ be lines through $B, C, D$ parallel to $C D, D B, B C$ respectively. If $A^{\prime}$ is in the half-plane determined by $l_{d}$ not containing $B C$, then $d(D, A B C) \leq d\left(A^{\prime}, A B C\right)<d(A, B C D)$, which is impossible. Similarly, $A^{\prime}$ lies in the half-planes determined by $l_{b}, l_{c}$ that contain $D$, and hence $A^{\prime}$ is inside the triangle bordered by $l_{b}, l_{c}, l_{d}$. The minimality property of $(A, \pi)$ and the way in which $B C D$ was selected guarantee that $E \cap T=\{A, B, C, D\}$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
44. (FIN 2) Let $E$ be a finite set of points in space such that $E$ is not contained in a plane and no three points of $E$ are collinear. Show that $E$ contains the vertices of a tetrahedron $T=A B C D$ such that $T \cap E=$ $\{A, B, C, D\}$ (including interior points of $T$ ) and such that the projection of $A$ onto the plane $B C D$ is inside a triangle that is similar to the triangle $B C D$ and whose sides have midpoints $B, C, D$.
|
44. Let $d(X, \sigma)$ denote the distance from a point $X$ to a plane $\sigma$. Let us consider the pair $(A, \pi)$ where $A \in E$ and $\pi$ is a plane containing some three points $B, C, D \in E$ such that $d(A, \pi)$ is the smallest possible. We may suppose that $B, C, D$ are selected such that $\triangle B C D$ contains no other points of $E$. Let $A^{\prime}$ be the projection of $A$ on $\pi$, and let $l_{b}, l_{c}, l_{d}$ be lines through $B, C, D$ parallel to $C D, D B, B C$ respectively. If $A^{\prime}$ is in the half-plane determined by $l_{d}$ not containing $B C$, then $d(D, A B C) \leq d\left(A^{\prime}, A B C\right)<d(A, B C D)$, which is impossible. Similarly, $A^{\prime}$ lies in the half-planes determined by $l_{b}, l_{c}$ that contain $D$, and hence $A^{\prime}$ is inside the triangle bordered by $l_{b}, l_{c}, l_{d}$. The minimality property of $(A, \pi)$ and the way in which $B C D$ was selected guarantee that $E \cap T=\{A, B, C, D\}$.
|
{
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|
210f27e9-eb2a-5ede-9230-32375e005e9b
| 23,605
|
48. (USS 2) The intersection of a plane with a regular tetrahedron with edge $a$ is a quadrilateral with perimeter $P$. Prove that $2 a \leq P \leq 3 a$.
|
48. Let a plane cut the edges $A B, B C, C D, D A$ at points $K, L, M, N$ respectively. Let $D^{\prime}, A^{\prime}, B^{\prime}$ be distinct points in the plane $A B C$ such that the triangles $B C D^{\prime}, C D^{\prime} A^{\prime}, D^{\prime} A^{\prime} B^{\prime}$ are equilateral, and $M^{\prime} \in\left[C D^{\prime}\right], N^{\prime} \in\left[D^{\prime} A^{\prime}\right]$, and $K^{\prime} \in\left[A^{\prime} B^{\prime}\right]$ such that $C M^{\prime}=C M$, $A^{\prime} N^{\prime}=A N$, and $A^{\prime} K^{\prime}=A K$. The perimeter $P$ of the quadrilateral $K L M N$ is equal to the length of the polygonal line $K L M^{\prime} N^{\prime} K^{\prime}$, which is not less than $K K^{\prime}$. It follows that $P \geq 2 a$.  Let us consider all quadrilaterals $K L M N$ that are obtained by intersecting the tetrahedron by a plane parallel to a fixed plane $\alpha$. The lengths of the segments $K L, L M, M N, N K$ are linear functions in $A K$, and so is $P$. Thus $P$ takes its maximum at an endpoint of the interval, i.e., when the plane $K L M N$ passes through one of the vertices $A, B, C, D$, and it is easy to see that in this case $P \leq 3 a$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
48. (USS 2) The intersection of a plane with a regular tetrahedron with edge $a$ is a quadrilateral with perimeter $P$. Prove that $2 a \leq P \leq 3 a$.
|
48. Let a plane cut the edges $A B, B C, C D, D A$ at points $K, L, M, N$ respectively. Let $D^{\prime}, A^{\prime}, B^{\prime}$ be distinct points in the plane $A B C$ such that the triangles $B C D^{\prime}, C D^{\prime} A^{\prime}, D^{\prime} A^{\prime} B^{\prime}$ are equilateral, and $M^{\prime} \in\left[C D^{\prime}\right], N^{\prime} \in\left[D^{\prime} A^{\prime}\right]$, and $K^{\prime} \in\left[A^{\prime} B^{\prime}\right]$ such that $C M^{\prime}=C M$, $A^{\prime} N^{\prime}=A N$, and $A^{\prime} K^{\prime}=A K$. The perimeter $P$ of the quadrilateral $K L M N$ is equal to the length of the polygonal line $K L M^{\prime} N^{\prime} K^{\prime}$, which is not less than $K K^{\prime}$. It follows that $P \geq 2 a$.  Let us consider all quadrilaterals $K L M N$ that are obtained by intersecting the tetrahedron by a plane parallel to a fixed plane $\alpha$. The lengths of the segments $K L, L M, M N, N K$ are linear functions in $A K$, and so is $P$. Thus $P$ takes its maximum at an endpoint of the interval, i.e., when the plane $K L M N$ passes through one of the vertices $A, B, C, D$, and it is easy to see that in this case $P \leq 3 a$.
|
{
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|
5e196012-b7b4-50f4-83db-fe12b85fb55a
| 23,616
|
51. (USS 5) Several segments, which we shall call white, are given, and the sum of their lengths is 1 . Several other segments, which we shall call black, are given, and the sum of their lengths is 1 . Prove that every such system of segments can be distributed on the segment that is 1.51 long in the following way: Segments of the same color are disjoint, and segments of different colors are either disjoint or one is inside the other. Prove that there exists a system that cannot be distributed in that way on the segment that is 1.49 long.
|
51. We shall use the following algorithm: Choose a segment of maximum length ("basic" segment) and put on it unused segments of the opposite color without overlapping, each time of the maximum possible length, as long as it is possible. Repeat the procedure with remaining segments until all the segments are used. Let us suppose that the last basic segment is black. Then the length of the used part of any white basic segment is greater than the free part, and consequently at least one-half of the length of the white segments has been used more than once. Therefore all basic segments have total length at most 1.5 and can be distributed on a segment of length 1.51. On the other hand, if we are given two white segments of lengths 0.5 and two black segments of lengths 0.999 and 0.001 , we cannot distribute them on a segment of length less than 1.499.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
51. (USS 5) Several segments, which we shall call white, are given, and the sum of their lengths is 1 . Several other segments, which we shall call black, are given, and the sum of their lengths is 1 . Prove that every such system of segments can be distributed on the segment that is 1.51 long in the following way: Segments of the same color are disjoint, and segments of different colors are either disjoint or one is inside the other. Prove that there exists a system that cannot be distributed in that way on the segment that is 1.49 long.
|
51. We shall use the following algorithm: Choose a segment of maximum length ("basic" segment) and put on it unused segments of the opposite color without overlapping, each time of the maximum possible length, as long as it is possible. Repeat the procedure with remaining segments until all the segments are used. Let us suppose that the last basic segment is black. Then the length of the used part of any white basic segment is greater than the free part, and consequently at least one-half of the length of the white segments has been used more than once. Therefore all basic segments have total length at most 1.5 and can be distributed on a segment of length 1.51. On the other hand, if we are given two white segments of lengths 0.5 and two black segments of lengths 0.999 and 0.001 , we cannot distribute them on a segment of length less than 1.499.
|
{
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|
4b90eb12-cb9a-54ae-a242-eec374c05510
| 23,626
|
54. (USA 3) If $0 \leq a \leq b \leq c \leq d$, prove that $$ a^{b} b^{c} c^{d} d^{a} \geq b^{a} c^{b} d^{c} a^{d} $$
|
54. We shall use the following lemma. Lemma. If a real function $f$ is convex on the interval $I$ and $x, y, z \in I$, $x \leq y \leq z$, then $$ (y-z) f(x)+(z-x) f(y)+(x-y) f(z) \leq 0 . $$ Proof. The inequality is obvious for $x=y=z$. If $x<z$, then there exist $p, r$ such that $p+r=1$ and $y=p x+r z$. Then by Jensen's inequality $f(p x+r z) \leq p f(x)+r f(z)$, which is equivalent to the statement of the lemma. By applying the lemma to the convex function $-\ln x$ we obtain $x^{y} y^{z} z^{x} \geq$ $y^{x} z^{y} x^{z}$ for any $0<x \leq y \leq z$. Multiplying the inequalities $a^{b} b^{c} c^{a} \geq b^{a} c^{b} a^{c}$ and $a^{c} c^{d} d^{a} \geq c^{a} d^{c} a^{d}$ we get the desired inequality. Remark. Similarly, for $0<a_{1} \leq a_{2} \leq \cdots \leq a_{n}$ it holds that $a_{1}^{a_{2}} a_{2}^{a_{3}} \cdots a_{n}^{a_{1}} \geq a_{2}^{a_{1}} a_{3}^{a_{2}} \cdots a_{1}^{a_{n}}$.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
54. (USA 3) If $0 \leq a \leq b \leq c \leq d$, prove that $$ a^{b} b^{c} c^{d} d^{a} \geq b^{a} c^{b} d^{c} a^{d} $$
|
54. We shall use the following lemma. Lemma. If a real function $f$ is convex on the interval $I$ and $x, y, z \in I$, $x \leq y \leq z$, then $$ (y-z) f(x)+(z-x) f(y)+(x-y) f(z) \leq 0 . $$ Proof. The inequality is obvious for $x=y=z$. If $x<z$, then there exist $p, r$ such that $p+r=1$ and $y=p x+r z$. Then by Jensen's inequality $f(p x+r z) \leq p f(x)+r f(z)$, which is equivalent to the statement of the lemma. By applying the lemma to the convex function $-\ln x$ we obtain $x^{y} y^{z} z^{x} \geq$ $y^{x} z^{y} x^{z}$ for any $0<x \leq y \leq z$. Multiplying the inequalities $a^{b} b^{c} c^{a} \geq b^{a} c^{b} a^{c}$ and $a^{c} c^{d} d^{a} \geq c^{a} d^{c} a^{d}$ we get the desired inequality. Remark. Similarly, for $0<a_{1} \leq a_{2} \leq \cdots \leq a_{n}$ it holds that $a_{1}^{a_{2}} a_{2}^{a_{3}} \cdots a_{n}^{a_{1}} \geq a_{2}^{a_{1}} a_{3}^{a_{2}} \cdots a_{1}^{a_{n}}$.
|
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082d9a5e-6ceb-5268-a563-78276f052c3f
| 23,632
|
55. (USA 4) Through a point $O$ on the diagonal $B D$ of a parallelogram $A B C D$, segments $M N$ parallel to $A B$, and $P Q$ parallel to $A D$, are drawn, with $M$ on $A D$, and $Q$ on $A B$. Prove that diagonals $A O, B P, D N$ (extended if necessary) will be concurrent.
|
55. The statement is true without the assumption that $O \in B D$. Let $B P \cap$ $D N=\{K\}$. If we denote $\overrightarrow{A B}=a, \overrightarrow{A D}=b$ and $\overrightarrow{A O}=\alpha a+\beta b$ for some $\alpha, \beta \in \mathbb{R}, 1 / \alpha+1 / \beta \neq 1$, by straightforward calculation we obtain that $$ \overrightarrow{A K}=\frac{\alpha}{\alpha+\beta-\alpha \beta} a+\frac{\beta}{\alpha+\beta-\alpha \beta} b=\frac{1}{\alpha+\beta-\alpha \beta} \overrightarrow{A O} $$ Hence $A, K, O$ are collinear.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
55. (USA 4) Through a point $O$ on the diagonal $B D$ of a parallelogram $A B C D$, segments $M N$ parallel to $A B$, and $P Q$ parallel to $A D$, are drawn, with $M$ on $A D$, and $Q$ on $A B$. Prove that diagonals $A O, B P, D N$ (extended if necessary) will be concurrent.
|
55. The statement is true without the assumption that $O \in B D$. Let $B P \cap$ $D N=\{K\}$. If we denote $\overrightarrow{A B}=a, \overrightarrow{A D}=b$ and $\overrightarrow{A O}=\alpha a+\beta b$ for some $\alpha, \beta \in \mathbb{R}, 1 / \alpha+1 / \beta \neq 1$, by straightforward calculation we obtain that $$ \overrightarrow{A K}=\frac{\alpha}{\alpha+\beta-\alpha \beta} a+\frac{\beta}{\alpha+\beta-\alpha \beta} b=\frac{1}{\alpha+\beta-\alpha \beta} \overrightarrow{A O} $$ Hence $A, K, O$ are collinear.
|
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|
7acae887-fda2-521e-a48e-34339be11e17
| 23,634
|
58. (VIE 2) Prove that for every triangle the following inequality holds: $$ \frac{a b+b c+c a}{4 S} \geq \cot \frac{\pi}{6} $$ where $a, b, c$ are lengths of the sides and $S$ is the area of the triangle.
|
58. The following inequality (Finsler and Hadwiger, 1938) is sharper than the one we have to prove: $$ 2 a b+2 b c+2 c a-a^{2}-b^{2}-c^{2} \geq 4 S \sqrt{3} $$ First proof. Let us set $2 x=b+c-a, 2 y=c+a-b, 2 z=a+b-c$. Then $x, y, z>0$ and the inequality (1) becomes $$ y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2} \geq x y z(x+y+z) $$ which is equivalent to the obvious inequality $(x y-y z)^{2}+(y z-z x)^{2}+$ $(z x-x y)^{2} \geq 0$. Second proof. Using the known relations for a triangle $$ \begin{aligned} a^{2}+b^{2}+c^{2} & =2 s^{2}-2 r^{2}-8 r R, \\ a b+b c+c a & =s^{2}+r^{2}+4 r R, \\ S & =r s, \end{aligned} $$ where $r$ and $R$ are the radii of the incircle and the circumcircle, $s$ the semiperimeter and $S$ the area, we can transform (1) into $$ s \sqrt{3} \leq 4 R+r . $$ The last inequality is a consequence of the inequalities $2 r \leq R$ and $s^{2} \leq$ $4 R^{2}+4 R r+3 r^{2}$, where the last one follows from the equality $H I^{2}=$ $4 R^{2}+4 R r+3 r^{2}-s^{2}$ ( $H$ and $I$ being the orthocenter and the incenter of the triangle).
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
58. (VIE 2) Prove that for every triangle the following inequality holds: $$ \frac{a b+b c+c a}{4 S} \geq \cot \frac{\pi}{6} $$ where $a, b, c$ are lengths of the sides and $S$ is the area of the triangle.
|
58. The following inequality (Finsler and Hadwiger, 1938) is sharper than the one we have to prove: $$ 2 a b+2 b c+2 c a-a^{2}-b^{2}-c^{2} \geq 4 S \sqrt{3} $$ First proof. Let us set $2 x=b+c-a, 2 y=c+a-b, 2 z=a+b-c$. Then $x, y, z>0$ and the inequality (1) becomes $$ y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2} \geq x y z(x+y+z) $$ which is equivalent to the obvious inequality $(x y-y z)^{2}+(y z-z x)^{2}+$ $(z x-x y)^{2} \geq 0$. Second proof. Using the known relations for a triangle $$ \begin{aligned} a^{2}+b^{2}+c^{2} & =2 s^{2}-2 r^{2}-8 r R, \\ a b+b c+c a & =s^{2}+r^{2}+4 r R, \\ S & =r s, \end{aligned} $$ where $r$ and $R$ are the radii of the incircle and the circumcircle, $s$ the semiperimeter and $S$ the area, we can transform (1) into $$ s \sqrt{3} \leq 4 R+r . $$ The last inequality is a consequence of the inequalities $2 r \leq R$ and $s^{2} \leq$ $4 R^{2}+4 R r+3 r^{2}$, where the last one follows from the equality $H I^{2}=$ $4 R^{2}+4 R r+3 r^{2}-s^{2}$ ( $H$ and $I$ being the orthocenter and the incenter of the triangle).
|
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|
86fb6fbf-78c6-524b-981c-b20e0216777b
| 23,641
|
60. (VIE 4) Suppose $x_{0}, x_{1}, \ldots, x_{n}$ are integers and $x_{0}>x_{1}>\cdots>x_{n}$. Prove that at least one of the numbers $\left|F\left(x_{0}\right)\right|,\left|F\left(x_{1}\right)\right|,\left|F\left(x_{2}\right)\right|, \ldots$, $\left|F\left(x_{n}\right)\right|$, where $$ F(x)=x^{n}+a_{1} x^{n-1}+\cdots+a_{n}, \quad a_{i} \in \mathbb{R}, \quad i=1, \ldots, n $$ is greater than $\frac{n!}{2^{n}}$.
|
60. By Lagrange's interpolation formula we have $$ F(x)=\sum_{j=0}^{n} F\left(x_{j}\right) \frac{\prod_{i \neq j}\left(x-x_{j}\right)}{\prod_{i \neq j}\left(x_{i}-x_{j}\right)} . $$ Since the leading coefficient in $F(x)$ is 1 , it follows that $$ 1=\sum_{j=0}^{n} \frac{F\left(x_{j}\right)}{\prod_{i \neq j}\left(x_{i}-x_{j}\right)} $$ Since $$ \left|\prod_{i \neq j}\left(x_{i}-x_{j}\right)\right|=\prod_{i=0}^{j-1}\left|x_{i}-x_{j}\right| \prod_{i=j+1}^{n}\left|x_{i}-x_{j}\right| \geq j!(n-j)! $$ we have $$ 1 \leq \sum_{j=0}^{n} \frac{\left|F\left(x_{j}\right)\right|}{\left|\prod_{i \neq j}\left(x_{i}-x_{j}\right)\right|} \leq \frac{1}{n!} \sum_{j=0}^{n}\binom{n}{j}\left|F\left(x_{j}\right)\right| \leq \frac{2^{n}}{n!} \max \left|F\left(x_{j}\right)\right| . $$ Now the required inequality follows immediately.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
60. (VIE 4) Suppose $x_{0}, x_{1}, \ldots, x_{n}$ are integers and $x_{0}>x_{1}>\cdots>x_{n}$. Prove that at least one of the numbers $\left|F\left(x_{0}\right)\right|,\left|F\left(x_{1}\right)\right|,\left|F\left(x_{2}\right)\right|, \ldots$, $\left|F\left(x_{n}\right)\right|$, where $$ F(x)=x^{n}+a_{1} x^{n-1}+\cdots+a_{n}, \quad a_{i} \in \mathbb{R}, \quad i=1, \ldots, n $$ is greater than $\frac{n!}{2^{n}}$.
|
60. By Lagrange's interpolation formula we have $$ F(x)=\sum_{j=0}^{n} F\left(x_{j}\right) \frac{\prod_{i \neq j}\left(x-x_{j}\right)}{\prod_{i \neq j}\left(x_{i}-x_{j}\right)} . $$ Since the leading coefficient in $F(x)$ is 1 , it follows that $$ 1=\sum_{j=0}^{n} \frac{F\left(x_{j}\right)}{\prod_{i \neq j}\left(x_{i}-x_{j}\right)} $$ Since $$ \left|\prod_{i \neq j}\left(x_{i}-x_{j}\right)\right|=\prod_{i=0}^{j-1}\left|x_{i}-x_{j}\right| \prod_{i=j+1}^{n}\left|x_{i}-x_{j}\right| \geq j!(n-j)! $$ we have $$ 1 \leq \sum_{j=0}^{n} \frac{\left|F\left(x_{j}\right)\right|}{\left|\prod_{i \neq j}\left(x_{i}-x_{j}\right)\right|} \leq \frac{1}{n!} \sum_{j=0}^{n}\binom{n}{j}\left|F\left(x_{j}\right)\right| \leq \frac{2^{n}}{n!} \max \left|F\left(x_{j}\right)\right| . $$ Now the required inequality follows immediately.
|
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|
542dc560-3089-54eb-beb2-b701f1f953aa
| 23,648
|
7. (CZS 3) Prove the following assertion: If $c_{1}, c_{2}, \ldots, c_{n}(n \geq 2)$ are real numbers such that $$ (n-1)\left(c_{1}^{2}+c_{2}^{2}+\cdots+c_{n}^{2}\right)=\left(c_{1}+c_{2}+\cdots+c_{n}\right)^{2} $$ then either all these numbers are nonnegative or all these numbers are nonpositive.
|
7. Let us suppose that $c_{1} \leq c_{2} \leq \cdots \leq c_{n}$ and that $c_{1}<0<c_{n}$. There exists $k, 1 \leq k<n$, such that $c_{k} \leq 0<c_{k+1}$. Then we have $$ \begin{aligned} (n-1)\left(c_{1}^{2}+c_{2}^{2}+\cdots+c_{n}^{2}\right) \geq & k\left(c_{1}^{2}+\cdots+c_{k}^{2}\right)+(n-k)\left(c_{k+1}^{2}+\cdots+c_{n}^{2}\right) \\ \geq & \left(c_{1}+\cdots+c_{k}\right)^{2}+\left(c_{k+1}+\cdots+c_{n}\right)^{2} \\ = & \left(c_{1}+\cdots+c_{n}\right)^{2} \\ & -2\left(c_{1}+\cdots+c_{k}\right)\left(c_{k+1}+\cdots+c_{n}\right), \end{aligned} $$ from which we obtain $\left(c_{1}+\cdots+c_{k}\right)\left(c_{k+1}+\cdots+c_{n}\right) \geq 0$, a contradiction. Second solution. By the given condition and the inequality between arithmetic and quadratic mean we have $$ \begin{aligned} \left(c_{1}+\cdots+c_{n}\right)^{2} & =(n-1)\left(c_{1}^{2}+\cdots+c_{n-1}^{2}\right)+(n-1) c_{n}^{2} \\ & \geq\left(c_{1}+\cdots+c_{n-1}\right)^{2}+(n-1) c_{n}^{2} \end{aligned} $$ which is equivalent to $2\left(c_{1}+c_{2}+\cdots+c_{n}\right) c_{n} \geq n c_{n}^{2}$. Similarly, $2\left(c_{1}+c_{2}+\right.$ $\left.\cdots+c_{n}\right) c_{i} \geq n c_{i}^{2}$ for all $i=1, \ldots, n$. Hence all $c_{i}$ are of the same sign.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
7. (CZS 3) Prove the following assertion: If $c_{1}, c_{2}, \ldots, c_{n}(n \geq 2)$ are real numbers such that $$ (n-1)\left(c_{1}^{2}+c_{2}^{2}+\cdots+c_{n}^{2}\right)=\left(c_{1}+c_{2}+\cdots+c_{n}\right)^{2} $$ then either all these numbers are nonnegative or all these numbers are nonpositive.
|
7. Let us suppose that $c_{1} \leq c_{2} \leq \cdots \leq c_{n}$ and that $c_{1}<0<c_{n}$. There exists $k, 1 \leq k<n$, such that $c_{k} \leq 0<c_{k+1}$. Then we have $$ \begin{aligned} (n-1)\left(c_{1}^{2}+c_{2}^{2}+\cdots+c_{n}^{2}\right) \geq & k\left(c_{1}^{2}+\cdots+c_{k}^{2}\right)+(n-k)\left(c_{k+1}^{2}+\cdots+c_{n}^{2}\right) \\ \geq & \left(c_{1}+\cdots+c_{k}\right)^{2}+\left(c_{k+1}+\cdots+c_{n}\right)^{2} \\ = & \left(c_{1}+\cdots+c_{n}\right)^{2} \\ & -2\left(c_{1}+\cdots+c_{k}\right)\left(c_{k+1}+\cdots+c_{n}\right), \end{aligned} $$ from which we obtain $\left(c_{1}+\cdots+c_{k}\right)\left(c_{k+1}+\cdots+c_{n}\right) \geq 0$, a contradiction. Second solution. By the given condition and the inequality between arithmetic and quadratic mean we have $$ \begin{aligned} \left(c_{1}+\cdots+c_{n}\right)^{2} & =(n-1)\left(c_{1}^{2}+\cdots+c_{n-1}^{2}\right)+(n-1) c_{n}^{2} \\ & \geq\left(c_{1}+\cdots+c_{n-1}\right)^{2}+(n-1) c_{n}^{2} \end{aligned} $$ which is equivalent to $2\left(c_{1}+c_{2}+\cdots+c_{n}\right) c_{n} \geq n c_{n}^{2}$. Similarly, $2\left(c_{1}+c_{2}+\right.$ $\left.\cdots+c_{n}\right) c_{i} \geq n c_{i}^{2}$ for all $i=1, \ldots, n$. Hence all $c_{i}$ are of the same sign.
|
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8f03fe8b-75d5-5fda-bfb5-a12dabe97780
| 23,649
|
1. (BUL 1) The set $M=\{1,2, \ldots, 2 n\}$ is partitioned into $k$ nonintersecting subsets $M_{1}, M_{2}, \ldots, M_{k}$, where $n \geq k^{3}+k$. Prove that there exist even numbers $2 j_{1}, 2 j_{2}, \ldots, 2 j_{k+1}$ in $M$ that are in one and the same subset $M_{i}$ $(1 \leq i \leq k)$ such that the numbers $2 j_{1}-1,2 j_{2}-1, \ldots, 2 j_{k+1}-1$ are also in one and the same subset $M_{j}(1 \leq j \leq k)$.
|
1. There exists an $M_{s}$ that contains at least $2 n / k=2\left(k^{2}+1\right)$ elements. It follows that $M_{s}$ contains either at least $k^{2}+1$ even numbers or at least $k^{2}+1$ odd numbers. In the former case, consider the predecessors of those $k^{2}+1$ numbers: among them, at least $\frac{k^{2}+1}{k+1}>k$, i.e., at least $k+1$, belong to the same subset, say $M_{t}$. Then we choose $s, t$. The latter case is similar. Second solution. For all $i, j \in\{1,2, \ldots, k\}$, consider the set $N_{i j}=\{r \mid$ $\left.2 r \in M_{i}, 2 r-1 \in M_{j}\right\}$. Then $\left\{N_{i j} \mid i, j\right\}$ is a partition of $\{1,2, \ldots, n\}$ into $k^{2}$ subsets. For $n \geq k^{3}+1$ one of these subsets contains at least $k+1$ elements, and the statement follows. Remark. The statement is not necessarily true when $n=k^{3}$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
1. (BUL 1) The set $M=\{1,2, \ldots, 2 n\}$ is partitioned into $k$ nonintersecting subsets $M_{1}, M_{2}, \ldots, M_{k}$, where $n \geq k^{3}+k$. Prove that there exist even numbers $2 j_{1}, 2 j_{2}, \ldots, 2 j_{k+1}$ in $M$ that are in one and the same subset $M_{i}$ $(1 \leq i \leq k)$ such that the numbers $2 j_{1}-1,2 j_{2}-1, \ldots, 2 j_{k+1}-1$ are also in one and the same subset $M_{j}(1 \leq j \leq k)$.
|
1. There exists an $M_{s}$ that contains at least $2 n / k=2\left(k^{2}+1\right)$ elements. It follows that $M_{s}$ contains either at least $k^{2}+1$ even numbers or at least $k^{2}+1$ odd numbers. In the former case, consider the predecessors of those $k^{2}+1$ numbers: among them, at least $\frac{k^{2}+1}{k+1}>k$, i.e., at least $k+1$, belong to the same subset, say $M_{t}$. Then we choose $s, t$. The latter case is similar. Second solution. For all $i, j \in\{1,2, \ldots, k\}$, consider the set $N_{i j}=\{r \mid$ $\left.2 r \in M_{i}, 2 r-1 \in M_{j}\right\}$. Then $\left\{N_{i j} \mid i, j\right\}$ is a partition of $\{1,2, \ldots, n\}$ into $k^{2}$ subsets. For $n \geq k^{3}+1$ one of these subsets contains at least $k+1$ elements, and the statement follows. Remark. The statement is not necessarily true when $n=k^{3}$.
|
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d6c26b3e-550c-57c5-8ed7-eec738c792a3
| 23,656
|
10. (NET 1) ${ }^{\text {IMO6 }}$ An international society has its members in 6 different countries. The list of members contains 1978 names, numbered $1,2, \ldots$, 1978. Prove that there is at least one member whose number is the sum of the numbers of two, not necessarily distinct, of his compatriots.
|
10. Assume the opposite. One of the countries, say $A$, contains at least 330 members $a_{1}, a_{2}, \ldots, a_{330}$ of the society $(6 \cdot 329=1974)$. Consider the differences $a_{330}-a_{i},=1,2, \ldots, 329$ : the members with these numbers are not in $A$, so at least 66 of them, $a_{330}-a_{i_{1}}, \ldots, a_{330}-a_{i_{66}}$, belong to the same country, say $B$. Then the differences $\left(a_{i_{66}}-a_{330}\right)-\left(a_{i_{j}}-a_{330}\right)=$ $a_{i_{66}}-a_{i_{j}}, j=1,2, \ldots, 65$, are neither in $A$ nor in $B$. Continuing this procedure, we find that 17 of these differences are in the same country, say $C$, then 6 among 16 differences of themselves in a country $D$, and 3 among 5 differences of themselves in $E$; finally, one among two differences of these 3 differences belong to country $F$, so that the difference of themselves cannot be in any country. This is a contradiction. Remark. The following stronger $([6!e]=1957)$ statement can be proved in the same way. Schurr's lemma. If $n$ is a natural number and $e$ the logarithm base, then for every partition of the set $\{1,2, \ldots,[e n!]\}$ into $n$ subsets one of these subsets contains some two elements and their difference.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
10. (NET 1) ${ }^{\text {IMO6 }}$ An international society has its members in 6 different countries. The list of members contains 1978 names, numbered $1,2, \ldots$, 1978. Prove that there is at least one member whose number is the sum of the numbers of two, not necessarily distinct, of his compatriots.
|
10. Assume the opposite. One of the countries, say $A$, contains at least 330 members $a_{1}, a_{2}, \ldots, a_{330}$ of the society $(6 \cdot 329=1974)$. Consider the differences $a_{330}-a_{i},=1,2, \ldots, 329$ : the members with these numbers are not in $A$, so at least 66 of them, $a_{330}-a_{i_{1}}, \ldots, a_{330}-a_{i_{66}}$, belong to the same country, say $B$. Then the differences $\left(a_{i_{66}}-a_{330}\right)-\left(a_{i_{j}}-a_{330}\right)=$ $a_{i_{66}}-a_{i_{j}}, j=1,2, \ldots, 65$, are neither in $A$ nor in $B$. Continuing this procedure, we find that 17 of these differences are in the same country, say $C$, then 6 among 16 differences of themselves in a country $D$, and 3 among 5 differences of themselves in $E$; finally, one among two differences of these 3 differences belong to country $F$, so that the difference of themselves cannot be in any country. This is a contradiction. Remark. The following stronger $([6!e]=1957)$ statement can be proved in the same way. Schurr's lemma. If $n$ is a natural number and $e$ the logarithm base, then for every partition of the set $\{1,2, \ldots,[e n!]\}$ into $n$ subsets one of these subsets contains some two elements and their difference.
|
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|
178c7166-00ec-50cc-a629-195aed743739
| 23,658
|
11. (SWE 2) A function $f: I \rightarrow \mathbb{R}$, defined on an interval $I$, is called concave if $f(\theta x+(1-\theta) y) \geq \theta f(x)+(1-\theta) f(y)$ for all $x, y \in I$ and $0 \leq \theta \leq 1$. Assume that the functions $f_{1}, \ldots, f_{n}$, having all nonnegative values, are concave. Prove that the function $\left(f_{1} f_{2} \ldots f_{n}\right)^{1 / n}$ is concave.
|
11. Set $F(x)=f_{1}(x) f_{2}(x) \cdots f_{n}(x)$ : we must prove concavity of $F^{1 / n}$. By the assumption, $$ \begin{aligned} F(\theta x+(1-\theta) y) & \geq \prod_{i=1}^{n}\left[\theta f_{i}(x)+(1-\theta) f(y)\right] \\ & =\sum_{k=0}^{n} \theta^{k}(1-\theta)^{n-k} \sum f_{i_{1}}(x) \ldots f_{i_{k}}(x) f_{i_{k+1}}(y) f_{i_{n}}(y) \end{aligned} $$ where the second sum goes through all $\binom{n}{k} k$-subsets $\left\{i_{1}, \ldots, i_{k}\right\}$ of $\{1, \ldots, n\}$. The inequality between the arithmetic and geometric means now gives us $$ \sum f_{i_{1}}(x) f_{i_{2}}(x) \cdots f_{i_{k}}(x) f_{i_{k+1}}(y) f_{i_{n}}(y) \geq\binom{ n}{k} F(x)^{k / n} F(y)^{(n-k) / n} $$ Inserting this in the above inequality and using the binomial formula, we finally obtain $$ \begin{aligned} F(\theta x+(1-\theta) y) & \geq \sum_{k=0}^{n} \theta^{k}(1-\theta)^{n-k}\binom{n}{k} F(x)^{k / n} F(y)^{(n-k) / n} \\ & =\left(\theta F(x)^{1 / n}+(1-\theta) F(y)^{1 / n}\right)^{n}, \end{aligned} $$ which proves the assertion.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
11. (SWE 2) A function $f: I \rightarrow \mathbb{R}$, defined on an interval $I$, is called concave if $f(\theta x+(1-\theta) y) \geq \theta f(x)+(1-\theta) f(y)$ for all $x, y \in I$ and $0 \leq \theta \leq 1$. Assume that the functions $f_{1}, \ldots, f_{n}$, having all nonnegative values, are concave. Prove that the function $\left(f_{1} f_{2} \ldots f_{n}\right)^{1 / n}$ is concave.
|
11. Set $F(x)=f_{1}(x) f_{2}(x) \cdots f_{n}(x)$ : we must prove concavity of $F^{1 / n}$. By the assumption, $$ \begin{aligned} F(\theta x+(1-\theta) y) & \geq \prod_{i=1}^{n}\left[\theta f_{i}(x)+(1-\theta) f(y)\right] \\ & =\sum_{k=0}^{n} \theta^{k}(1-\theta)^{n-k} \sum f_{i_{1}}(x) \ldots f_{i_{k}}(x) f_{i_{k+1}}(y) f_{i_{n}}(y) \end{aligned} $$ where the second sum goes through all $\binom{n}{k} k$-subsets $\left\{i_{1}, \ldots, i_{k}\right\}$ of $\{1, \ldots, n\}$. The inequality between the arithmetic and geometric means now gives us $$ \sum f_{i_{1}}(x) f_{i_{2}}(x) \cdots f_{i_{k}}(x) f_{i_{k+1}}(y) f_{i_{n}}(y) \geq\binom{ n}{k} F(x)^{k / n} F(y)^{(n-k) / n} $$ Inserting this in the above inequality and using the binomial formula, we finally obtain $$ \begin{aligned} F(\theta x+(1-\theta) y) & \geq \sum_{k=0}^{n} \theta^{k}(1-\theta)^{n-k}\binom{n}{k} F(x)^{k / n} F(y)^{(n-k) / n} \\ & =\left(\theta F(x)^{1 / n}+(1-\theta) F(y)^{1 / n}\right)^{n}, \end{aligned} $$ which proves the assertion.
|
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6421fbb1-8546-5279-9c1b-baa7fcae777f
| 23,662
|
12. (USA 1) ${ }^{\mathrm{IMO} 4}$ In a triangle $A B C$ we have $A B=A C$. A circle is tangent internally to the circumcircle of $A B C$ and also to the sides $A B, A C$, at $P, Q$ respectively. Prove that the midpoint of $P Q$ is the center of the incircle of $A B C$.
|
12. Let $O$ be the center of the smaller circle, $T$ its contact point with the circumcircle of $A B C$, and $J$ the midpoint of segment $B C$. The figure is symmetric with respect to the line through $A, O, J, T$. A homothety centered at $A$ taking $T$ into $J$ will take the smaller circle into the incircle of $A B C$, hence will take $O$ into the incenter $I$. On the other hand, $\angle A B T=\angle A C T=90^{\circ}$ implies that the quadrilaterals $A B T C$ and $A P O Q$ are similar. Hence the above homothety also maps $O$ to the midpoint of $P Q$. This finishes the proof. Remark. The assertion is true for a nonisosceles triangle $A B C$ as well, and this (more difficult) case is a matter of SL93-3.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
12. (USA 1) ${ }^{\mathrm{IMO} 4}$ In a triangle $A B C$ we have $A B=A C$. A circle is tangent internally to the circumcircle of $A B C$ and also to the sides $A B, A C$, at $P, Q$ respectively. Prove that the midpoint of $P Q$ is the center of the incircle of $A B C$.
|
12. Let $O$ be the center of the smaller circle, $T$ its contact point with the circumcircle of $A B C$, and $J$ the midpoint of segment $B C$. The figure is symmetric with respect to the line through $A, O, J, T$. A homothety centered at $A$ taking $T$ into $J$ will take the smaller circle into the incircle of $A B C$, hence will take $O$ into the incenter $I$. On the other hand, $\angle A B T=\angle A C T=90^{\circ}$ implies that the quadrilaterals $A B T C$ and $A P O Q$ are similar. Hence the above homothety also maps $O$ to the midpoint of $P Q$. This finishes the proof. Remark. The assertion is true for a nonisosceles triangle $A B C$ as well, and this (more difficult) case is a matter of SL93-3.
|
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e40b06c1-f824-538b-b8f7-2f24bbfbce10
| 23,665
|
14. (VIE 2) Prove that it is possible to place $2 n(2 n+1)$ parallelepipedic (rectangular) pieces of soap of dimensions $1 \times 2 \times(n+1)$ in a cubic box with edge $2 n+1$ if and only if $n$ is even or $n=1$. Remark. It is assumed that the edges of the pieces of soap are parallel to the edges of the box.
|
14. We label the cells of the cube by $\left(a_{1}, a_{2}, a_{3}\right), a_{i} \in\{1,2, \ldots, 2 n+1\}$, in a natural way: for example, as Cartesian coordinates of centers of the cells $\left((1,1,1)\right.$ is one corner, etc.). Notice that there should be $(2 n+1)^{3}-$ $2 n(2 n+1) \cdot 2(n+1)=2 n+1$ void cells, i.e., those not covered by any piece of soap. $n=1$. In this case, six pieces of soap $1 \times 2 \times 2$ can be placed on the following positions: $[(1,1,1),(2,2,1)],[(3,1,1),(3,2,2)],[(2,3,1),(3,3,2)]$ and the symmetric ones with respect to the center of the box. (Here $[A, B]$ denotes the rectangle with opposite corners at $A, B$.) $n$ is even. Each of the $2 n+1$ planes $P_{k}=\left\{\left(a_{1}, a_{2}, k\right) \mid a_{i}=1, \ldots, 2 n+1\right\}$ can receive $2 n$ pieces of soap: In fact, $P_{k}$ can be partitioned into four $n \times(n+1)$ rectangles at the corners and the central cell, while an $n \times(n+1)$ rectangle can receive $n / 2$ pieces of soap. $n$ is odd, $n>1$. Let us color a cell $\left(a_{1}, a_{2}, a_{3}\right)$ blue, red, or yellow if exactly three, two or one $a_{i}$ respectively is equal to $n+1$. Thus there are 1 blue, $6 n$ red, and $12 n^{2}$ yellow cells. We notice that each piece of soap must contain at least one colored cell (because $2(n+1)>2 n+1)$. Also, every piece of soap contains an even number (actually, $1 \cdot 2,1(n+1)$, or $2(n+1)$ ) of cells in $P_{k}$. On the other hand, $2 n+1$ cells are void, i.e., one in each plane. There are several cases for a piece of soap $S$ : (i) $S$ consists of 1 blue, $n+1$ red and $n$ yellow cells; (ii) $S$ consists of 2 red and $2 n$ yellow cells (and no blue cells); (iii) $S$ contains 1 red cell, $n+1$ yellow cells, and the are rest uncolored; (iv) $S$ contains 2 yellow cells and no blue or red ones. From the descriptions of the last three cases, we can deduce that if $S$ contains $r$ red cells and no blue, then it contains exactly $2+(n-1) r$ red ones. $\quad(*)$ Now, let $B_{1}, \ldots, B_{k}$ be all boxes put in the cube, with a possible exception for the one covering the blue cell: thus $k=2 n(2 n+1)$ if the blue cell is void, or $k=2 n(2 n+1)-1$ otherwise. Let $r_{i}$ and $y_{i}$ respectively be the numbers of red and yellow cells inside $B_{i}$. By (*) we have $y_{1}+\cdots+y_{k}=2 k+(n-1)\left(r_{1}+\cdots+r_{k}\right)$. If the blue cell is void, then $r_{1}+\cdots+r_{k}=6 n$ and consequently $y_{1}+\cdots+y_{k}=$ $4 n(2 n+1)+6 n(n-1)=14 n^{2}-2 n$, which is impossible because there are only $12 n^{2}<14 n^{2}-2 n$ yellow cells. Otherwise, $r_{1}+\cdots+r_{k} \geq 5 n-2$ (because $n+1$ red cells are covered by the box containing the blue cell, and one can be void) and consequently $y_{1}+\cdots+y_{k} \geq 4 n(2 n+$ $1)-2+(n-1)(5 n-2)=13 n^{2}-3 n$; since there are $n$ more yellow cells in the box containing the blue one, this counts for $13 n^{2}-2 n>12 n^{2}$ ( $n \geq 3$ ), again impossible. Remark. The following solution of the case $n$ odd is simpler, but does not work for $n=3$. For $k=1,2,3$, let $m_{k}$ be the number of pieces whose long sides are perpendicular to the plane $\pi_{k}\left(a_{k}=n+1\right)$. Each of these $m_{k}$ pieces covers exactly 2 cells of $\pi_{k}$, while any other piece covers $n+1$, $2(n+1)$, or none. It follows that $4 n^{2}+4 n-2 m_{k}$ is divisible by $n+1$, and so is $2 m_{k}$. This further implies that $2 m_{1}+2 m_{2}+2 m_{3}=4 n(2 n+1)$ is a multiple of $n+1$, which is impossible for each odd $n$ except $n=1$ and $n=3$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
14. (VIE 2) Prove that it is possible to place $2 n(2 n+1)$ parallelepipedic (rectangular) pieces of soap of dimensions $1 \times 2 \times(n+1)$ in a cubic box with edge $2 n+1$ if and only if $n$ is even or $n=1$. Remark. It is assumed that the edges of the pieces of soap are parallel to the edges of the box.
|
14. We label the cells of the cube by $\left(a_{1}, a_{2}, a_{3}\right), a_{i} \in\{1,2, \ldots, 2 n+1\}$, in a natural way: for example, as Cartesian coordinates of centers of the cells $\left((1,1,1)\right.$ is one corner, etc.). Notice that there should be $(2 n+1)^{3}-$ $2 n(2 n+1) \cdot 2(n+1)=2 n+1$ void cells, i.e., those not covered by any piece of soap. $n=1$. In this case, six pieces of soap $1 \times 2 \times 2$ can be placed on the following positions: $[(1,1,1),(2,2,1)],[(3,1,1),(3,2,2)],[(2,3,1),(3,3,2)]$ and the symmetric ones with respect to the center of the box. (Here $[A, B]$ denotes the rectangle with opposite corners at $A, B$.) $n$ is even. Each of the $2 n+1$ planes $P_{k}=\left\{\left(a_{1}, a_{2}, k\right) \mid a_{i}=1, \ldots, 2 n+1\right\}$ can receive $2 n$ pieces of soap: In fact, $P_{k}$ can be partitioned into four $n \times(n+1)$ rectangles at the corners and the central cell, while an $n \times(n+1)$ rectangle can receive $n / 2$ pieces of soap. $n$ is odd, $n>1$. Let us color a cell $\left(a_{1}, a_{2}, a_{3}\right)$ blue, red, or yellow if exactly three, two or one $a_{i}$ respectively is equal to $n+1$. Thus there are 1 blue, $6 n$ red, and $12 n^{2}$ yellow cells. We notice that each piece of soap must contain at least one colored cell (because $2(n+1)>2 n+1)$. Also, every piece of soap contains an even number (actually, $1 \cdot 2,1(n+1)$, or $2(n+1)$ ) of cells in $P_{k}$. On the other hand, $2 n+1$ cells are void, i.e., one in each plane. There are several cases for a piece of soap $S$ : (i) $S$ consists of 1 blue, $n+1$ red and $n$ yellow cells; (ii) $S$ consists of 2 red and $2 n$ yellow cells (and no blue cells); (iii) $S$ contains 1 red cell, $n+1$ yellow cells, and the are rest uncolored; (iv) $S$ contains 2 yellow cells and no blue or red ones. From the descriptions of the last three cases, we can deduce that if $S$ contains $r$ red cells and no blue, then it contains exactly $2+(n-1) r$ red ones. $\quad(*)$ Now, let $B_{1}, \ldots, B_{k}$ be all boxes put in the cube, with a possible exception for the one covering the blue cell: thus $k=2 n(2 n+1)$ if the blue cell is void, or $k=2 n(2 n+1)-1$ otherwise. Let $r_{i}$ and $y_{i}$ respectively be the numbers of red and yellow cells inside $B_{i}$. By (*) we have $y_{1}+\cdots+y_{k}=2 k+(n-1)\left(r_{1}+\cdots+r_{k}\right)$. If the blue cell is void, then $r_{1}+\cdots+r_{k}=6 n$ and consequently $y_{1}+\cdots+y_{k}=$ $4 n(2 n+1)+6 n(n-1)=14 n^{2}-2 n$, which is impossible because there are only $12 n^{2}<14 n^{2}-2 n$ yellow cells. Otherwise, $r_{1}+\cdots+r_{k} \geq 5 n-2$ (because $n+1$ red cells are covered by the box containing the blue cell, and one can be void) and consequently $y_{1}+\cdots+y_{k} \geq 4 n(2 n+$ $1)-2+(n-1)(5 n-2)=13 n^{2}-3 n$; since there are $n$ more yellow cells in the box containing the blue one, this counts for $13 n^{2}-2 n>12 n^{2}$ ( $n \geq 3$ ), again impossible. Remark. The following solution of the case $n$ odd is simpler, but does not work for $n=3$. For $k=1,2,3$, let $m_{k}$ be the number of pieces whose long sides are perpendicular to the plane $\pi_{k}\left(a_{k}=n+1\right)$. Each of these $m_{k}$ pieces covers exactly 2 cells of $\pi_{k}$, while any other piece covers $n+1$, $2(n+1)$, or none. It follows that $4 n^{2}+4 n-2 m_{k}$ is divisible by $n+1$, and so is $2 m_{k}$. This further implies that $2 m_{1}+2 m_{2}+2 m_{3}=4 n(2 n+1)$ is a multiple of $n+1$, which is impossible for each odd $n$ except $n=1$ and $n=3$.
|
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|
6676d9f4-ed24-5eed-bd36-03d5be977993
| 23,670
|
15. (YUG 1) Let $p$ be a prime and $A=\left\{a_{1}, \ldots, a_{p-1}\right\}$ an arbitrary subset of the set of natural numbers such that none of its elements is divisible by $p$. Let us define a mapping $f$ from $\mathcal{P}(A)$ (the set of all subsets of $A$ ) to the set $P=\{0,1, \ldots, p-1\}$ in the following way: (i) if $B=\left\{a_{i_{1}}, \ldots, a_{i_{k}}\right\} \subset A$ and $\sum_{j=1}^{k} a_{i_{j}} \equiv n(\bmod p)$, then $f(B)=n$, (ii) $f(\emptyset)=0, \emptyset$ being the empty set. Prove that for each $n \in P$ there exists $B \subset A$ such that $f(B)=n$.
|
15. Let $C_{n}=\left\{a_{1}, \ldots, a_{n}\right\}\left(C_{0}=\emptyset\right)$ and $P_{n}=\left\{f(B) \mid B \subseteq C_{n}\right\}$. We claim that $P_{n}$ contains at least $n+1$ distinct elements. First note that $P_{0}=\{0\}$ contains one element. Suppose that $P_{n+1}=P_{n}$ for some $n$. Since $P_{n+1}=$ $\left\{a_{n+1}+r \mid r \in P_{n}\right\}$, it follows that for each $r \in P_{n}$, also $r+b_{n} \in P_{n}$. Then obviously $0 \in P_{n}$ implies $k b_{n} \in P_{n}$ for all $k$; therefore $P_{n}=P$ has at least $p \geq n+1$ elements. Otherwise, if $P_{n+1} \supset P_{n}$ for all $n$, then $\left|P_{n+1}\right| \geq\left|P_{n}\right|+1$ and hence $\left|P_{n}\right| \geq n+1$, as claimed. Consequently, $\left|P_{p-1}\right| \geq p$. (All the operations here are performed modulo $p$.)
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
15. (YUG 1) Let $p$ be a prime and $A=\left\{a_{1}, \ldots, a_{p-1}\right\}$ an arbitrary subset of the set of natural numbers such that none of its elements is divisible by $p$. Let us define a mapping $f$ from $\mathcal{P}(A)$ (the set of all subsets of $A$ ) to the set $P=\{0,1, \ldots, p-1\}$ in the following way: (i) if $B=\left\{a_{i_{1}}, \ldots, a_{i_{k}}\right\} \subset A$ and $\sum_{j=1}^{k} a_{i_{j}} \equiv n(\bmod p)$, then $f(B)=n$, (ii) $f(\emptyset)=0, \emptyset$ being the empty set. Prove that for each $n \in P$ there exists $B \subset A$ such that $f(B)=n$.
|
15. Let $C_{n}=\left\{a_{1}, \ldots, a_{n}\right\}\left(C_{0}=\emptyset\right)$ and $P_{n}=\left\{f(B) \mid B \subseteq C_{n}\right\}$. We claim that $P_{n}$ contains at least $n+1$ distinct elements. First note that $P_{0}=\{0\}$ contains one element. Suppose that $P_{n+1}=P_{n}$ for some $n$. Since $P_{n+1}=$ $\left\{a_{n+1}+r \mid r \in P_{n}\right\}$, it follows that for each $r \in P_{n}$, also $r+b_{n} \in P_{n}$. Then obviously $0 \in P_{n}$ implies $k b_{n} \in P_{n}$ for all $k$; therefore $P_{n}=P$ has at least $p \geq n+1$ elements. Otherwise, if $P_{n+1} \supset P_{n}$ for all $n$, then $\left|P_{n+1}\right| \geq\left|P_{n}\right|+1$ and hence $\left|P_{n}\right| \geq n+1$, as claimed. Consequently, $\left|P_{p-1}\right| \geq p$. (All the operations here are performed modulo $p$.)
|
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|
3a87058d-1bc2-5265-ab4c-a294679a93cd
| 23,673
|
17. (FRA 3) Prove that for any positive integers $x, y, z$ with $x y-z^{2}=1$ one can find nonnegative integers $a, b, c, d$ such that $x=a^{2}+b^{2}, y=c^{2}+d^{2}$, $z=a c+b d$. Set $z=(2 q)$ ! to deduce that for any prime number $p=4 q+1, p$ can be represented as the sum of squares of two integers.
|
17. Let $z_{0} \geq 1$ be a positive integer. Supposing that the statement is true for all triples $(x, y, z)$ with $z<z_{0}$, we shall prove that it is true for $z=z_{0}$ too. If $z_{0}=1$, verification is trivial, while $x_{0}=y_{0}$ is obviously impossible. So let there be given a triple $\left(x_{0}, y_{0}, z_{0}\right)$ with $z_{0}>1$ and $x_{0}<y_{0}$, and define another triple $(x, y, z)$ by $$ x=z_{0}, \quad y=x_{0}+y_{0}-2 z_{0}, \quad \text { and } \quad z=z_{0}-x_{0} $$ Then $x, y, z$ are positive integers. This is clear for $x, z$, while $y=x_{0}+y_{0}-$ $2 z_{0} \geq 2\left(\sqrt{x_{0} y_{0}}-z_{0}\right)>2\left(z_{0}-z_{0}\right)=0$. Moreover, $x y-z^{2}=x_{0}\left(x_{0}+y_{0}-\right.$ $\left.2 z_{0}\right)-\left(z_{0}-x_{0}\right)^{2}=x_{0} y_{0}-z_{0}^{2}=1$ and $z<z_{0}$, so that by the assumption, the statement holds for $x, y, z$. Thus for some nonnegative integers $a, b, c, d$ we have $$ x=a^{2}+b^{2}, \quad y=c^{2}+d^{2}, \quad z=a c+b d $$ But then we obtain representations of this sort for $x_{0}, y_{0}, z_{0}$ too: $$ x_{0}=a^{2}+b^{2}, \quad y_{0}=(a+c)^{2}+(b+d)^{2}, \quad z_{0}=a(a+c)+b(b+d) $$ For the second part of the problem, we note that for $z=(2 q)!$, $$ \begin{aligned} z^{2} & =(2 q)!(2 q)(2 q-1) \cdots 1 \equiv(2 q)!\cdot(-(2 q+1))(-(2 q+2)) \cdots(-4 q) \\ & =(-1)^{2 q}(4 q)!\equiv-1(\bmod p) \end{aligned} $$ by Wilson's theorem. Hence $p \mid z^{2}+1=p y$ for some positive integer $y>0$. Now it follows from the first part that there exist integers $a, b$ such that $x=p=a^{2}+b^{2}$. Second solution. Another possibility is using arithmetic of Gaussian integers. Lemma. Suppose $m, n, p, q$ are elements of $\mathbb{Z}$ or any other unique factorization domain, with $m n=p q$. then there exist elements $a, b, c, d$ such that $m=a b, n=c d, p=a c, q=b d$. Proof is direct, for example using factorization of $a, b, c, d$ into primes. We now apply this lemma to the Gaussian integers in our case (because $\mathbb{Z}[i]$ has the unique factorization property), having in mind that $x y=$ $z^{2}+1=(z+i)(z-i)$. We obtain (1) $x=a b$, (2) $y=c d$, (3) $z+i=a c$, (4) $z-i=b d$ for some $a, b, c, d \in \mathbb{Z}[i]$. Let $a=a_{1}+a_{2} i$, etc. By (3) and (4), $\operatorname{gcd}\left(a_{1}, a_{2}\right)=$ $\cdots=\operatorname{gcd}\left(d_{1}, d_{2}\right)$. Then (1) and (2) give us $b=\bar{a}, c=\bar{d}$. The statement follows at once: $x=a b=a \bar{a}=a_{1}^{2}+a_{2}^{2}, y=d \bar{d}=d_{1}^{2}+d_{2}^{2}$ and $z+i=$ $\left(a_{1} d_{1}+a_{2} d_{2}\right)+\imath\left(a_{2} d_{1}-a_{1} d_{2}\right) \Rightarrow z=a_{1} d_{1}+a_{2} d_{2}$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
17. (FRA 3) Prove that for any positive integers $x, y, z$ with $x y-z^{2}=1$ one can find nonnegative integers $a, b, c, d$ such that $x=a^{2}+b^{2}, y=c^{2}+d^{2}$, $z=a c+b d$. Set $z=(2 q)$ ! to deduce that for any prime number $p=4 q+1, p$ can be represented as the sum of squares of two integers.
|
17. Let $z_{0} \geq 1$ be a positive integer. Supposing that the statement is true for all triples $(x, y, z)$ with $z<z_{0}$, we shall prove that it is true for $z=z_{0}$ too. If $z_{0}=1$, verification is trivial, while $x_{0}=y_{0}$ is obviously impossible. So let there be given a triple $\left(x_{0}, y_{0}, z_{0}\right)$ with $z_{0}>1$ and $x_{0}<y_{0}$, and define another triple $(x, y, z)$ by $$ x=z_{0}, \quad y=x_{0}+y_{0}-2 z_{0}, \quad \text { and } \quad z=z_{0}-x_{0} $$ Then $x, y, z$ are positive integers. This is clear for $x, z$, while $y=x_{0}+y_{0}-$ $2 z_{0} \geq 2\left(\sqrt{x_{0} y_{0}}-z_{0}\right)>2\left(z_{0}-z_{0}\right)=0$. Moreover, $x y-z^{2}=x_{0}\left(x_{0}+y_{0}-\right.$ $\left.2 z_{0}\right)-\left(z_{0}-x_{0}\right)^{2}=x_{0} y_{0}-z_{0}^{2}=1$ and $z<z_{0}$, so that by the assumption, the statement holds for $x, y, z$. Thus for some nonnegative integers $a, b, c, d$ we have $$ x=a^{2}+b^{2}, \quad y=c^{2}+d^{2}, \quad z=a c+b d $$ But then we obtain representations of this sort for $x_{0}, y_{0}, z_{0}$ too: $$ x_{0}=a^{2}+b^{2}, \quad y_{0}=(a+c)^{2}+(b+d)^{2}, \quad z_{0}=a(a+c)+b(b+d) $$ For the second part of the problem, we note that for $z=(2 q)!$, $$ \begin{aligned} z^{2} & =(2 q)!(2 q)(2 q-1) \cdots 1 \equiv(2 q)!\cdot(-(2 q+1))(-(2 q+2)) \cdots(-4 q) \\ & =(-1)^{2 q}(4 q)!\equiv-1(\bmod p) \end{aligned} $$ by Wilson's theorem. Hence $p \mid z^{2}+1=p y$ for some positive integer $y>0$. Now it follows from the first part that there exist integers $a, b$ such that $x=p=a^{2}+b^{2}$. Second solution. Another possibility is using arithmetic of Gaussian integers. Lemma. Suppose $m, n, p, q$ are elements of $\mathbb{Z}$ or any other unique factorization domain, with $m n=p q$. then there exist elements $a, b, c, d$ such that $m=a b, n=c d, p=a c, q=b d$. Proof is direct, for example using factorization of $a, b, c, d$ into primes. We now apply this lemma to the Gaussian integers in our case (because $\mathbb{Z}[i]$ has the unique factorization property), having in mind that $x y=$ $z^{2}+1=(z+i)(z-i)$. We obtain (1) $x=a b$, (2) $y=c d$, (3) $z+i=a c$, (4) $z-i=b d$ for some $a, b, c, d \in \mathbb{Z}[i]$. Let $a=a_{1}+a_{2} i$, etc. By (3) and (4), $\operatorname{gcd}\left(a_{1}, a_{2}\right)=$ $\cdots=\operatorname{gcd}\left(d_{1}, d_{2}\right)$. Then (1) and (2) give us $b=\bar{a}, c=\bar{d}$. The statement follows at once: $x=a b=a \bar{a}=a_{1}^{2}+a_{2}^{2}, y=d \bar{d}=d_{1}^{2}+d_{2}^{2}$ and $z+i=$ $\left(a_{1} d_{1}+a_{2} d_{2}\right)+\imath\left(a_{2} d_{1}-a_{1} d_{2}\right) \Rightarrow z=a_{1} d_{1}+a_{2} d_{2}$.
|
{
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|
91a2625a-013e-5034-bd20-ad8ddf57f56b
| 23,676
|
2. (BUL 4) Two identically oriented equilateral triangles, $A B C$ with center $S$ and $A^{\prime} B^{\prime} C$, are given in the plane. We also have $A^{\prime} \neq S$ and $B^{\prime} \neq S$. If $M$ is the midpoint of $A^{\prime} B$ and $N$ the midpoint of $A B^{\prime}$, prove that the triangles $S B^{\prime} M$ and $S A^{\prime} N$ are similar.
|
2. Consider the transformation $\phi$ of the plane defined as the homothety $\mathcal{H}$ with center $B$ and coefficient 2 followed by the rotation $\mathcal{R}$ about the center $O$ through an angle of $60^{\circ}$. Being direct, this mapping must be a rotational homothety. We also see that $\mathcal{H}$ maps $S$ into the point symmetric to $S$ with respect to $O A$, and $\mathcal{R}$ takes it back to $S$. Hence $S$ is a fixed point, and is consequently also the center of $\phi$. Therefore $\phi$ is the rotational homothety about $S$ with the angle $60^{\circ}$  and coefficient 2. (In fact, this could also be seen from the fact that $\phi$ preserves angles of triangles and maps the segment $S R$ onto $S B$, where $R$ is the midpoint of $A B$.) Since $\phi(M)=B^{\prime}$, we conclude that $\angle M S B^{\prime}=60^{\circ}$ and $S B^{\prime} / S M=2$. Similarly, $\angle N S A^{\prime}=60^{\circ}$ and $S A^{\prime} / S N=2$, so triangles $M S B^{\prime}$ and $N S A^{\prime}$ are indeed similar. Second solution. Probably the simplest way here is using complex numbers. Put the origin at $O$ and complex numbers $a, a^{\prime}$ at points $A, A^{\prime}$, and denote the primitive sixth root of 1 by $\omega$. Then the numbers at $B, B^{\prime}$, $S$ and $N$ are $\omega a, \omega a^{\prime},(a+\omega a) / 3$, and $\left(a+\omega a^{\prime}\right) / 2$ respectively. Now it is easy to verify that $(n-s)=\omega\left(a^{\prime}-s\right) / 2$, i.e., that $\angle N S A^{\prime}=60^{\circ}$ and $S A^{\prime} / S N=2$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
2. (BUL 4) Two identically oriented equilateral triangles, $A B C$ with center $S$ and $A^{\prime} B^{\prime} C$, are given in the plane. We also have $A^{\prime} \neq S$ and $B^{\prime} \neq S$. If $M$ is the midpoint of $A^{\prime} B$ and $N$ the midpoint of $A B^{\prime}$, prove that the triangles $S B^{\prime} M$ and $S A^{\prime} N$ are similar.
|
2. Consider the transformation $\phi$ of the plane defined as the homothety $\mathcal{H}$ with center $B$ and coefficient 2 followed by the rotation $\mathcal{R}$ about the center $O$ through an angle of $60^{\circ}$. Being direct, this mapping must be a rotational homothety. We also see that $\mathcal{H}$ maps $S$ into the point symmetric to $S$ with respect to $O A$, and $\mathcal{R}$ takes it back to $S$. Hence $S$ is a fixed point, and is consequently also the center of $\phi$. Therefore $\phi$ is the rotational homothety about $S$ with the angle $60^{\circ}$  and coefficient 2. (In fact, this could also be seen from the fact that $\phi$ preserves angles of triangles and maps the segment $S R$ onto $S B$, where $R$ is the midpoint of $A B$.) Since $\phi(M)=B^{\prime}$, we conclude that $\angle M S B^{\prime}=60^{\circ}$ and $S B^{\prime} / S M=2$. Similarly, $\angle N S A^{\prime}=60^{\circ}$ and $S A^{\prime} / S N=2$, so triangles $M S B^{\prime}$ and $N S A^{\prime}$ are indeed similar. Second solution. Probably the simplest way here is using complex numbers. Put the origin at $O$ and complex numbers $a, a^{\prime}$ at points $A, A^{\prime}$, and denote the primitive sixth root of 1 by $\omega$. Then the numbers at $B, B^{\prime}$, $S$ and $N$ are $\omega a, \omega a^{\prime},(a+\omega a) / 3$, and $\left(a+\omega a^{\prime}\right) / 2$ respectively. Now it is easy to verify that $(n-s)=\omega\left(a^{\prime}-s\right) / 2$, i.e., that $\angle N S A^{\prime}=60^{\circ}$ and $S A^{\prime} / S N=2$.
|
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|
8f9212cb-ab60-57a6-9cdc-15efec4ad141
| 23,679
|
4. (CZS 2) Let $T_{1}$ be a triangle having $a, b, c$ as lengths of its sides and let $T_{2}$ be another triangle having $u, v, w$ as lengths of its sides. If $P, Q$ are the areas of the two triangles, prove that $$ 16 P Q \leq a^{2}\left(-u^{2}+v^{2}+w^{2}\right)+b^{2}\left(u^{2}-v^{2}+w^{2}\right)+c^{2}\left(u^{2}+v^{2}-w^{2}\right) $$ When does equality hold?
|
4. Let $\gamma, \varphi$ be the angles of $T_{1}$ and $T_{2}$ opposite to $c$ and $w$ respectively. By the cosine theorem, the inequality is transformed into $$ \begin{aligned} & a^{2}\left(2 v^{2}-2 u v \cos \varphi\right)+b^{2}\left(2 u^{2}-2 u v \cos \varphi\right) \\ & \quad+2\left(a^{2}+b^{2}-2 a b \cos \gamma\right) u v \cos \varphi \geq 4 a b u v \sin \gamma \sin \varphi \end{aligned} $$ This is equivalent to $2\left(a^{2} v^{2}+b^{2} u^{2}\right)-4 a b u v(\cos \gamma \cos \varphi+\sin \gamma \sin \varphi) \geq 0$, i.e., to $$ 2(a v-b u)^{2}+4 a b u v(1-\cos (\gamma-\varphi)) \geq 0 $$ which is clearly satisfied. Equality holds if and only if $\gamma=\varphi$ and $a / b=$ $u / v$, i.e., when the triangles are similar, $a$ corresponding to $u$ and $b$ to $v$.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
4. (CZS 2) Let $T_{1}$ be a triangle having $a, b, c$ as lengths of its sides and let $T_{2}$ be another triangle having $u, v, w$ as lengths of its sides. If $P, Q$ are the areas of the two triangles, prove that $$ 16 P Q \leq a^{2}\left(-u^{2}+v^{2}+w^{2}\right)+b^{2}\left(u^{2}-v^{2}+w^{2}\right)+c^{2}\left(u^{2}+v^{2}-w^{2}\right) $$ When does equality hold?
|
4. Let $\gamma, \varphi$ be the angles of $T_{1}$ and $T_{2}$ opposite to $c$ and $w$ respectively. By the cosine theorem, the inequality is transformed into $$ \begin{aligned} & a^{2}\left(2 v^{2}-2 u v \cos \varphi\right)+b^{2}\left(2 u^{2}-2 u v \cos \varphi\right) \\ & \quad+2\left(a^{2}+b^{2}-2 a b \cos \gamma\right) u v \cos \varphi \geq 4 a b u v \sin \gamma \sin \varphi \end{aligned} $$ This is equivalent to $2\left(a^{2} v^{2}+b^{2} u^{2}\right)-4 a b u v(\cos \gamma \cos \varphi+\sin \gamma \sin \varphi) \geq 0$, i.e., to $$ 2(a v-b u)^{2}+4 a b u v(1-\cos (\gamma-\varphi)) \geq 0 $$ which is clearly satisfied. Equality holds if and only if $\gamma=\varphi$ and $a / b=$ $u / v$, i.e., when the triangles are similar, $a$ corresponding to $u$ and $b$ to $v$.
|
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|
6fd1345a-6ede-5a9d-b468-af7343d20cef
| 23,685
|
5. (GDR 2) For every integer $d \geq 1$, let $M_{d}$ be the set of all positive integers that cannot be written as a sum of an arithmetic progression with difference $d$, having at least two terms and consisting of positive integers. Let $A=M_{1}, B=M_{2} \backslash\{2\}, C=M_{3}$. Prove that every $c \in C$ may be written in a unique way as $c=a b$ with $a \in A, b \in B$.
|
5. We first explicitly describe the elements of the sets $M_{1}, M_{2}$. $x \notin M_{1}$ is equivalent to $x=a+(a+1)+\cdots+(a+n-1)=n(2 a+n-1) / 2$ for some natural numbers $n, a, n \geq 2$. Among $n$ and $2 a+n-1$, one is odd and the other even, and both are greater than 1 ; so $x$ has an odd factor $\geq 3$. On the other hand, for every $x$ with an odd divisor $p>3$ it is easy to see that there exist corresponding $a, n$. Therefore $M_{1}=\left\{2^{k} \mid k=0,1,2, \ldots\right\}$. $x \notin M_{2}$ is equivalent to $x=a+(a+2)+\cdots+(a+2(n-1))=n(a+n-1)$, where $n \geq 2$, i.e. to $x$ being composite. Therefore $M_{2}=\{1\} \cup\{p \mid$ $p=$ prime $\}$. $x \notin M_{3}$ is equivalent to $x=a+(a+3)+\cdots+(a+3(n-1))=$ $n(2 a+3(n-1)) / 2$. It remains to show that every $c \in M_{3}$ can be written as $c=2^{k} p$ with $p$ prime. Suppose the opposite, that $c=2^{k} p q$, where $p, q$ are odd and $q \geq p \geq 3$. Then there exist positive integers $a, n(n \geq 2)$ such that $c=n(2 a+3(n-1)) / 2$ and hence $c \notin M_{3}$. Indeed, if $k=0$, then $n=2$ and $2 a+3=p q$ work; otherwise, setting $n=p$ one obtains $a=2^{k} q-$ $3(p-1) / 2 \geq 2 q-3(p-1) / 2 \geq(p+3) / 2>1$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
5. (GDR 2) For every integer $d \geq 1$, let $M_{d}$ be the set of all positive integers that cannot be written as a sum of an arithmetic progression with difference $d$, having at least two terms and consisting of positive integers. Let $A=M_{1}, B=M_{2} \backslash\{2\}, C=M_{3}$. Prove that every $c \in C$ may be written in a unique way as $c=a b$ with $a \in A, b \in B$.
|
5. We first explicitly describe the elements of the sets $M_{1}, M_{2}$. $x \notin M_{1}$ is equivalent to $x=a+(a+1)+\cdots+(a+n-1)=n(2 a+n-1) / 2$ for some natural numbers $n, a, n \geq 2$. Among $n$ and $2 a+n-1$, one is odd and the other even, and both are greater than 1 ; so $x$ has an odd factor $\geq 3$. On the other hand, for every $x$ with an odd divisor $p>3$ it is easy to see that there exist corresponding $a, n$. Therefore $M_{1}=\left\{2^{k} \mid k=0,1,2, \ldots\right\}$. $x \notin M_{2}$ is equivalent to $x=a+(a+2)+\cdots+(a+2(n-1))=n(a+n-1)$, where $n \geq 2$, i.e. to $x$ being composite. Therefore $M_{2}=\{1\} \cup\{p \mid$ $p=$ prime $\}$. $x \notin M_{3}$ is equivalent to $x=a+(a+3)+\cdots+(a+3(n-1))=$ $n(2 a+3(n-1)) / 2$. It remains to show that every $c \in M_{3}$ can be written as $c=2^{k} p$ with $p$ prime. Suppose the opposite, that $c=2^{k} p q$, where $p, q$ are odd and $q \geq p \geq 3$. Then there exist positive integers $a, n(n \geq 2)$ such that $c=n(2 a+3(n-1)) / 2$ and hence $c \notin M_{3}$. Indeed, if $k=0$, then $n=2$ and $2 a+3=p q$ work; otherwise, setting $n=p$ one obtains $a=2^{k} q-$ $3(p-1) / 2 \geq 2 q-3(p-1) / 2 \geq(p+3) / 2>1$.
|
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|
750d4052-da0a-58b0-bd92-8f49796e8f11
| 23,688
|
6. (FRA 2) $)^{\mathrm{IMO} 5}$ Let $\varphi:\{1,2,3, \ldots\} \rightarrow\{1,2,3, \ldots\}$ be injective. Prove that for all $n$, $$ \sum_{k=1}^{n} \frac{\varphi(k)}{k^{2}} \geq \sum_{k=1}^{n} \frac{1}{k} $$
|
6. For fixed $n$ and the set $\{\varphi(1), \ldots, \varphi(n)\}$, there are finitely many possibilities for mapping $\varphi$ to $\{1, \ldots, n\}$. Suppose $\varphi$ is the one among these for which $\sum_{k=1}^{n} \varphi(k) / k^{2}$ is minimal. If $i<j$ and $\varphi(i)<\varphi(j)$ for some $i, j \in\{1, \ldots, n\}$, define $\psi$ as $\psi(i)=\varphi(j), \psi(j)=\varphi(i)$, and $\psi(k)=\varphi(k)$ for all other $k$. Then $$ \begin{aligned} \sum \frac{\varphi(k)}{k^{2}}-\sum \frac{\psi(k)}{k^{2}} & =\left(\frac{\varphi(i)}{i^{2}}+\frac{\varphi(j)}{j^{2}}\right)-\left(\frac{\varphi(i)}{j^{2}}+\frac{\varphi(j)}{i^{2}}\right) \\ & =(i-j)(\varphi(j)-\varphi(i)) \frac{i+j}{i^{2} j^{2}}>0 \end{aligned} $$ which contradicts the assumption. This shows that $\varphi(1)<\cdots<\varphi(n)$, and consequently $\varphi(k) \geq k$ for all $k$. Hence $$ \sum_{k=1}^{n} \frac{\varphi(k)}{k^{2}} \geq \sum_{k=1}^{n} \frac{k}{k^{2}}=\sum_{k=1}^{n} \frac{1}{k} $$
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
6. (FRA 2) $)^{\mathrm{IMO} 5}$ Let $\varphi:\{1,2,3, \ldots\} \rightarrow\{1,2,3, \ldots\}$ be injective. Prove that for all $n$, $$ \sum_{k=1}^{n} \frac{\varphi(k)}{k^{2}} \geq \sum_{k=1}^{n} \frac{1}{k} $$
|
6. For fixed $n$ and the set $\{\varphi(1), \ldots, \varphi(n)\}$, there are finitely many possibilities for mapping $\varphi$ to $\{1, \ldots, n\}$. Suppose $\varphi$ is the one among these for which $\sum_{k=1}^{n} \varphi(k) / k^{2}$ is minimal. If $i<j$ and $\varphi(i)<\varphi(j)$ for some $i, j \in\{1, \ldots, n\}$, define $\psi$ as $\psi(i)=\varphi(j), \psi(j)=\varphi(i)$, and $\psi(k)=\varphi(k)$ for all other $k$. Then $$ \begin{aligned} \sum \frac{\varphi(k)}{k^{2}}-\sum \frac{\psi(k)}{k^{2}} & =\left(\frac{\varphi(i)}{i^{2}}+\frac{\varphi(j)}{j^{2}}\right)-\left(\frac{\varphi(i)}{j^{2}}+\frac{\varphi(j)}{i^{2}}\right) \\ & =(i-j)(\varphi(j)-\varphi(i)) \frac{i+j}{i^{2} j^{2}}>0 \end{aligned} $$ which contradicts the assumption. This shows that $\varphi(1)<\cdots<\varphi(n)$, and consequently $\varphi(k) \geq k$ for all $k$. Hence $$ \sum_{k=1}^{n} \frac{\varphi(k)}{k^{2}} \geq \sum_{k=1}^{n} \frac{k}{k^{2}}=\sum_{k=1}^{n} \frac{1}{k} $$
|
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|
97e3bcee-8f61-5fa8-af65-b51a75393991
| 23,689
|
7. (FRA 5) We consider three distinct half-lines $O x, O y, O z$ in a plane. Prove the existence and uniqueness of three points $A \in O x, B \in O y$, $C \in O z$ such that the perimeters of the triangles $O A B, O B C, O C A$ are all equal to a given number $2 p>0$.
|
7. Let $x=O A, y=O B, z=O C, \alpha=\angle B O C, \beta=\angle C O A, \gamma=\angle A O B$. The conditions yield the equation $x+y+\sqrt{x^{2}+y^{2}-2 x y \cos \gamma}=2 p$, which transforms to $(2 p-x-y)^{2}=x^{2}+y^{2}-2 x y \cos \gamma$, i.e. $(p-x)(p-y)=$ $x y(1-\cos \gamma)$. Thus $$ \frac{p-x}{x} \cdot \frac{p-y}{y}=1-\cos \gamma $$ and analogously $\frac{p-y}{y} \cdot \frac{p-z}{z}=1-\cos \alpha, \frac{p-z}{z} \cdot \frac{p-x}{x}=1-\cos \beta$. Setting $u=\frac{p-x}{x}, v=\frac{p-y}{y}, w=\frac{p-z}{z}$, the above system becomes $$ u v=1-\cos \gamma, \quad v w=1-\cos \alpha, \quad w u=1-\cos \beta $$ This system has a unique solution in positive real numbers $u, v, w$ : $u=\sqrt{\frac{(1-\cos \beta)(1-\cos \gamma)}{1-\cos \alpha}}$, etc. Finally, the values of $x, y, z$ are uniquely determined from $u, v, w$. Remark. It is not necessary that the three lines be in the same plane. Also, there could be any odd number of lines instead of three.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
7. (FRA 5) We consider three distinct half-lines $O x, O y, O z$ in a plane. Prove the existence and uniqueness of three points $A \in O x, B \in O y$, $C \in O z$ such that the perimeters of the triangles $O A B, O B C, O C A$ are all equal to a given number $2 p>0$.
|
7. Let $x=O A, y=O B, z=O C, \alpha=\angle B O C, \beta=\angle C O A, \gamma=\angle A O B$. The conditions yield the equation $x+y+\sqrt{x^{2}+y^{2}-2 x y \cos \gamma}=2 p$, which transforms to $(2 p-x-y)^{2}=x^{2}+y^{2}-2 x y \cos \gamma$, i.e. $(p-x)(p-y)=$ $x y(1-\cos \gamma)$. Thus $$ \frac{p-x}{x} \cdot \frac{p-y}{y}=1-\cos \gamma $$ and analogously $\frac{p-y}{y} \cdot \frac{p-z}{z}=1-\cos \alpha, \frac{p-z}{z} \cdot \frac{p-x}{x}=1-\cos \beta$. Setting $u=\frac{p-x}{x}, v=\frac{p-y}{y}, w=\frac{p-z}{z}$, the above system becomes $$ u v=1-\cos \gamma, \quad v w=1-\cos \alpha, \quad w u=1-\cos \beta $$ This system has a unique solution in positive real numbers $u, v, w$ : $u=\sqrt{\frac{(1-\cos \beta)(1-\cos \gamma)}{1-\cos \alpha}}$, etc. Finally, the values of $x, y, z$ are uniquely determined from $u, v, w$. Remark. It is not necessary that the three lines be in the same plane. Also, there could be any odd number of lines instead of three.
|
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d03adb4b-e062-5141-b91a-3ba5577b6bd2
| 23,692
|
1. (BEL 1) Prove that in the Euclidean plane every regular polygon having an even number of sides can be dissected into lozenges. (A lozenge is a quadrilateral whose four sides are all of equal length).
|
1. We prove more generally, by induction on $n$, that any $2 n$-gon with equal edges and opposite edges parallel to each other can be dissected. For $n=2$ the only possible such $2 n$-gon is a single lozenge, so our theorem holds in this case. We will now show that it holds for general $n$. Assume by induction that it holds for $n-1$. Let $A_{1} A_{2} \ldots A_{2 n}$ be an arbitrary $2 n$-gon with equal edges and opposite edges parallel to each other. Then we can construct points $B_{i}$ for $i=3,4, \ldots, n$ such that $\overrightarrow{A_{i} B_{i}}=\overrightarrow{A_{2} A_{1}}=\overrightarrow{A_{n+1} A_{n+2}}$. We set $B_{2}=A_{2 n+1}=A_{1}$ and $B_{n+1}=A_{n+2}$. It follows that $A_{i} B_{i} B_{i+1} A_{i+1}$ for $i=2,3,4, \ldots, n$ are all lozenges. It also follows that $B_{i} B_{i+1}$ for $i=2,3,4, \ldots, n$ are equal to the edges of $A_{1} A_{2} \ldots A_{2 n}$ and parallel to $A_{i} A_{i+1}$ and hence to $A_{n+i} A_{n+i+1}$. Thus $B_{2} \ldots B_{n+1} A_{n+3} \ldots A_{2 n}$ is a $2(n-1)$-gon with equal edges and opposite sides parallel and hence, by the induction hypothesis, can be dissected into lozenges. We have thus provided a dissection for $A_{1} A_{2} \ldots A_{2 n}$. This completes the proof.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
1. (BEL 1) Prove that in the Euclidean plane every regular polygon having an even number of sides can be dissected into lozenges. (A lozenge is a quadrilateral whose four sides are all of equal length).
|
1. We prove more generally, by induction on $n$, that any $2 n$-gon with equal edges and opposite edges parallel to each other can be dissected. For $n=2$ the only possible such $2 n$-gon is a single lozenge, so our theorem holds in this case. We will now show that it holds for general $n$. Assume by induction that it holds for $n-1$. Let $A_{1} A_{2} \ldots A_{2 n}$ be an arbitrary $2 n$-gon with equal edges and opposite edges parallel to each other. Then we can construct points $B_{i}$ for $i=3,4, \ldots, n$ such that $\overrightarrow{A_{i} B_{i}}=\overrightarrow{A_{2} A_{1}}=\overrightarrow{A_{n+1} A_{n+2}}$. We set $B_{2}=A_{2 n+1}=A_{1}$ and $B_{n+1}=A_{n+2}$. It follows that $A_{i} B_{i} B_{i+1} A_{i+1}$ for $i=2,3,4, \ldots, n$ are all lozenges. It also follows that $B_{i} B_{i+1}$ for $i=2,3,4, \ldots, n$ are equal to the edges of $A_{1} A_{2} \ldots A_{2 n}$ and parallel to $A_{i} A_{i+1}$ and hence to $A_{n+i} A_{n+i+1}$. Thus $B_{2} \ldots B_{n+1} A_{n+3} \ldots A_{2 n}$ is a $2(n-1)$-gon with equal edges and opposite sides parallel and hence, by the induction hypothesis, can be dissected into lozenges. We have thus provided a dissection for $A_{1} A_{2} \ldots A_{2 n}$. This completes the proof.
|
{
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|
b3b0e57d-5900-5685-849c-130347e94621
| 23,699
|
13. (GRE 1) Show that $\frac{20}{60}<\sin 20^{\circ}<\frac{21}{60}$.
|
13. From elementary trigonometry we have $\sin 3 t=3 \sin t-4 \sin ^{3} t$. Hence, if we denote $y=\sin 20^{\circ}$, we have $\sqrt{3} / 2=\sin 60^{\circ}=3 y-4 y^{3}$. Obviously $0<y<1 / 2=\sin 30^{\circ}$. The function $f(x)=3 x-4 x^{3}$ is strictly increasing on $[0,1 / 2)$ because $f^{\prime}(x)=3-12 x^{2}>0$ for $0 \leq x<1 / 2$. Now the desired inequality $\frac{20}{60}=\frac{1}{3}<\sin 20^{\circ}<\frac{21}{60}=\frac{7}{20}$ follows from $$ f\left(\frac{1}{3}\right)<\frac{\sqrt{3}}{2}<f\left(\frac{7}{20}\right) $$ which is directly verified.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
13. (GRE 1) Show that $\frac{20}{60}<\sin 20^{\circ}<\frac{21}{60}$.
|
13. From elementary trigonometry we have $\sin 3 t=3 \sin t-4 \sin ^{3} t$. Hence, if we denote $y=\sin 20^{\circ}$, we have $\sqrt{3} / 2=\sin 60^{\circ}=3 y-4 y^{3}$. Obviously $0<y<1 / 2=\sin 30^{\circ}$. The function $f(x)=3 x-4 x^{3}$ is strictly increasing on $[0,1 / 2)$ because $f^{\prime}(x)=3-12 x^{2}>0$ for $0 \leq x<1 / 2$. Now the desired inequality $\frac{20}{60}=\frac{1}{3}<\sin 20^{\circ}<\frac{21}{60}=\frac{7}{20}$ follows from $$ f\left(\frac{1}{3}\right)<\frac{\sqrt{3}}{2}<f\left(\frac{7}{20}\right) $$ which is directly verified.
|
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|
1be405c9-6db2-527a-b8b2-da3af2512cda
| 23,710
|
16. (ISR 4) Let $K$ denote the set $\{a, b, c, d, e\} . F$ is a collection of 16 different subsets of $K$, and it is known that any three members of $F$ have at least one element in common. Show that all 16 members of $F$ have exactly one element in common.
|
16. Obviously, no two elements of $F$ can be complements of each other. If one of the sets has one element, then the conclusion is trivial. If there exist two different 2-element sets, then they must contain a common element, which in turn must then be contained in all other sets. Thus we can assume that there exists at most one 2 -element subset of $K$ in $F$. Since there can be at most 6 subsets of more than 3 elements of a 5 -element set, it follows that at least 9 out of 10 possible 3 -element subsets of $K$ belong to $F$. Let us assume, without loss of generality, that all sets but $\{c, d, e\}$ belong to $F$. Then sets $\{a, b, c\},\{a, d, e\}$, and $\{b, c, d\}$ have no common element, which is a contradiction. Hence it follows that all sets have a common element.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
16. (ISR 4) Let $K$ denote the set $\{a, b, c, d, e\} . F$ is a collection of 16 different subsets of $K$, and it is known that any three members of $F$ have at least one element in common. Show that all 16 members of $F$ have exactly one element in common.
|
16. Obviously, no two elements of $F$ can be complements of each other. If one of the sets has one element, then the conclusion is trivial. If there exist two different 2-element sets, then they must contain a common element, which in turn must then be contained in all other sets. Thus we can assume that there exists at most one 2 -element subset of $K$ in $F$. Since there can be at most 6 subsets of more than 3 elements of a 5 -element set, it follows that at least 9 out of 10 possible 3 -element subsets of $K$ belong to $F$. Let us assume, without loss of generality, that all sets but $\{c, d, e\}$ belong to $F$. Then sets $\{a, b, c\},\{a, d, e\}$, and $\{b, c, d\}$ have no common element, which is a contradiction. Hence it follows that all sets have a common element.
|
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|
c66b7cb3-b0b7-5728-8ed0-5d9d56f1dfd7
| 23,717
|
18. (POL 1) Let $m$ positive integers $a_{1}, \ldots, a_{m}$ be given. Prove that there exist fewer than $2^{m}$ positive integers $b_{1}, \ldots, b_{n}$ such that all sums of distinct $b_{k}$ 's are distinct and all $a_{i}(i \leq m)$ occur among them.
|
18. Let us write all $a_{i}$ in binary representation. For $S \subseteq\{1,2, \ldots, m\}$ let us define $b(S)$ as the number in whose binary representation ones appear in exactly the slots where ones appear in all $a_{i}$ where $i \subseteq S$ and don't appear in any other $a_{i}$. Some $b(S)$, including $b(\emptyset)$, will equal 0 , and hence there are fewer than $2^{m}$ different positive $b(S)$. We note that no two positive $b\left(S_{1}\right)$ and $b\left(S_{2}\right)\left(S_{1} \neq S_{2}\right)$ have ones in the same decimal places. Hence sums of distinct $b(S)$ 's are distinct. Moreover $$ a_{i}=\sum_{i \in S} b(S) $$ and hence the positive $b(S)$ are indeed the numbers $b_{1}, \ldots, b_{n}$ whose existence we had to prove.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
18. (POL 1) Let $m$ positive integers $a_{1}, \ldots, a_{m}$ be given. Prove that there exist fewer than $2^{m}$ positive integers $b_{1}, \ldots, b_{n}$ such that all sums of distinct $b_{k}$ 's are distinct and all $a_{i}(i \leq m)$ occur among them.
|
18. Let us write all $a_{i}$ in binary representation. For $S \subseteq\{1,2, \ldots, m\}$ let us define $b(S)$ as the number in whose binary representation ones appear in exactly the slots where ones appear in all $a_{i}$ where $i \subseteq S$ and don't appear in any other $a_{i}$. Some $b(S)$, including $b(\emptyset)$, will equal 0 , and hence there are fewer than $2^{m}$ different positive $b(S)$. We note that no two positive $b\left(S_{1}\right)$ and $b\left(S_{2}\right)\left(S_{1} \neq S_{2}\right)$ have ones in the same decimal places. Hence sums of distinct $b(S)$ 's are distinct. Moreover $$ a_{i}=\sum_{i \in S} b(S) $$ and hence the positive $b(S)$ are indeed the numbers $b_{1}, \ldots, b_{n}$ whose existence we had to prove.
|
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1d6d7f48-51dd-5c3b-b375-5977897fc8ef
| 23,722
|
21. (USS 1) Let $N$ be the number of integral solutions of the equation $$ x^{2}-y^{2}=z^{3}-t^{3} $$ satisfying the condition $0 \leq x, y, z, t \leq 10^{6}$, and let $M$ be the number of integral solutions of the equation $$ x^{2}-y^{2}=z^{3}-t^{3}+1 $$ satisfying the condition $0 \leq x, y, z, t \leq 10^{6}$. Prove that $N>M$.
|
21. Let $f(n)$ be the number of different ways $n \in \mathbb{N}$ can be expressed as $x^{2}+y^{3}$ where $x, y \in\left\{0,1, \ldots, 10^{6}\right\}$. Clearly $f(n)=0$ for $n<0$ or $n>10^{12}+10^{18}$. The first equation can be written as $x^{2}+t^{3}=y^{2}+z^{3}=n$, whereas the second equation can be written as $x^{2}+t^{3}=n+1, y^{2}+z^{3}=n$. Hence we obtain the following formulas for $M$ and $N$ : $$ M=\sum_{i=0}^{m} f(i)^{2}, \quad N=\sum_{i=0}^{m-1} f(i) f(i+1) . $$ Using the AM-GM inequality we get $$ \begin{aligned} N & =\sum_{i=0}^{m-1} f(i) f(i+1) \\ & \leq \sum_{i=0}^{m-1} \frac{f(i)^{2}+f(i+1)^{2}}{2}=\frac{f(0)^{2}}{2}+\sum_{i=1}^{m-1} f(i)^{2}+\frac{f(m)^{2}}{2}<M . \end{aligned} $$ The last inequality is strong, since $f(0)=1>0$. This completes our proof.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
21. (USS 1) Let $N$ be the number of integral solutions of the equation $$ x^{2}-y^{2}=z^{3}-t^{3} $$ satisfying the condition $0 \leq x, y, z, t \leq 10^{6}$, and let $M$ be the number of integral solutions of the equation $$ x^{2}-y^{2}=z^{3}-t^{3}+1 $$ satisfying the condition $0 \leq x, y, z, t \leq 10^{6}$. Prove that $N>M$.
|
21. Let $f(n)$ be the number of different ways $n \in \mathbb{N}$ can be expressed as $x^{2}+y^{3}$ where $x, y \in\left\{0,1, \ldots, 10^{6}\right\}$. Clearly $f(n)=0$ for $n<0$ or $n>10^{12}+10^{18}$. The first equation can be written as $x^{2}+t^{3}=y^{2}+z^{3}=n$, whereas the second equation can be written as $x^{2}+t^{3}=n+1, y^{2}+z^{3}=n$. Hence we obtain the following formulas for $M$ and $N$ : $$ M=\sum_{i=0}^{m} f(i)^{2}, \quad N=\sum_{i=0}^{m-1} f(i) f(i+1) . $$ Using the AM-GM inequality we get $$ \begin{aligned} N & =\sum_{i=0}^{m-1} f(i) f(i+1) \\ & \leq \sum_{i=0}^{m-1} \frac{f(i)^{2}+f(i+1)^{2}}{2}=\frac{f(0)^{2}}{2}+\sum_{i=1}^{m-1} f(i)^{2}+\frac{f(m)^{2}}{2}<M . \end{aligned} $$ The last inequality is strong, since $f(0)=1>0$. This completes our proof.
|
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ba8fdbb3-fb12-5ea1-b1c1-7c3baa0d495c
| 23,732
|
22. (USS 3) ${ }^{\mathrm{IMO} 3}$ There are two circles in the plane. Let a point $A$ be one of the points of intersection of these circles. Two points begin moving simultaneously with constant speeds from the point $A$, each point along its own circle. The two points return to the point $A$ at the same time. Prove that there is a point $P$ in the plane such that at every moment of time the distances from the point $P$ to the moving points are equal.
|
22. Let the centers of the two circles be denoted by $O$ and $O_{1}$ and their respective radii by $r$ and $r_{1}$, and let the positions of the points on the circles at time $t$ be denoted by $M(t)$ and $N(t)$. Let $Q$ be the point such that $O A O_{1} Q$ is a parallelogram. We will show that $Q$ is the point $P$ we are looking for, i.e., that $Q M(t)=$ $Q N(t)$ for all $t$. We note that $O Q=$ $O_{1} A=r_{1}, O_{1} Q=O A=r$ and  $\angle Q O A=\angle Q O_{1} A=\phi$. Since the two points return to $A$ at the same time, it follows that $\angle M(t) O A=\angle N(t) O_{1} A=\omega t$. Therefore $\angle Q O M(t)=$ $\angle Q O_{1} N(t)=\phi+\omega t$, from which it follows that $\triangle Q O M(t) \cong \triangle Q O_{1} N(t)$. Hence $Q M(t)=Q N(t)$, as we claimed.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
22. (USS 3) ${ }^{\mathrm{IMO} 3}$ There are two circles in the plane. Let a point $A$ be one of the points of intersection of these circles. Two points begin moving simultaneously with constant speeds from the point $A$, each point along its own circle. The two points return to the point $A$ at the same time. Prove that there is a point $P$ in the plane such that at every moment of time the distances from the point $P$ to the moving points are equal.
|
22. Let the centers of the two circles be denoted by $O$ and $O_{1}$ and their respective radii by $r$ and $r_{1}$, and let the positions of the points on the circles at time $t$ be denoted by $M(t)$ and $N(t)$. Let $Q$ be the point such that $O A O_{1} Q$ is a parallelogram. We will show that $Q$ is the point $P$ we are looking for, i.e., that $Q M(t)=$ $Q N(t)$ for all $t$. We note that $O Q=$ $O_{1} A=r_{1}, O_{1} Q=O A=r$ and  $\angle Q O A=\angle Q O_{1} A=\phi$. Since the two points return to $A$ at the same time, it follows that $\angle M(t) O A=\angle N(t) O_{1} A=\omega t$. Therefore $\angle Q O M(t)=$ $\angle Q O_{1} N(t)=\phi+\omega t$, from which it follows that $\triangle Q O M(t) \cong \triangle Q O_{1} N(t)$. Hence $Q M(t)=Q N(t)$, as we claimed.
|
{
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|
3dde2cdb-a6ca-5ec1-be93-79d670aef3d4
| 23,733
|
24. (USA 5) A circle O with center $O$ on base $B C$ of an isosceles triangle $A B C$ is tangent to the equal sides $A B, A C$. If point $P$ on $A B$ and point $Q$ on $A C$ are selected such that $P B \times C Q=(B C / 2)^{2}$, prove that line segment $P Q$ is tangent to circle O , and prove the converse.
|
24. Clearly $O$ is the midpoint of $B C$. Let $M$ and $N$ be the points of tangency of the circle with $A B$ and $A C$, respectively, and let $\angle B A C=2 \varphi$. Then $\angle B O M=\angle C O N=\varphi$. Let us assume that $P Q$ touches the circle in $X$. If we set $\angle P O M=$ $\angle P O X=x$ and $\angle Q O N=\angle Q O X=y$, then $2 x+2 y=\angle M O N=$ $180^{\circ}-2 \varphi$, i.e., $y=90^{\circ}-\varphi-x$. It follows that $\angle O Q C=180^{\circ}-\angle Q O C-$ $\angle O C Q=180^{\circ}-(\varphi+y)-\left(90^{\circ}-\varphi\right)=90^{\circ}-y=x+\varphi=\angle B O P$. Hence the triangles $B O P$ and $C Q O$ are similar, and consequently $B P \cdot C Q=$ $B O \cdot C O=(B C / 2)^{2}$. Conversely, let $B P \cdot C Q=(B C / 2)^{2}$ and let $Q^{\prime}$ be the point on $(A C)$ such that $P Q^{\prime}$ is tangent to the circle. Then $B P \cdot C Q^{\prime}=(B C / 2)^{2}$, which implies $Q \equiv Q^{\prime}$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
24. (USA 5) A circle O with center $O$ on base $B C$ of an isosceles triangle $A B C$ is tangent to the equal sides $A B, A C$. If point $P$ on $A B$ and point $Q$ on $A C$ are selected such that $P B \times C Q=(B C / 2)^{2}$, prove that line segment $P Q$ is tangent to circle O , and prove the converse.
|
24. Clearly $O$ is the midpoint of $B C$. Let $M$ and $N$ be the points of tangency of the circle with $A B$ and $A C$, respectively, and let $\angle B A C=2 \varphi$. Then $\angle B O M=\angle C O N=\varphi$. Let us assume that $P Q$ touches the circle in $X$. If we set $\angle P O M=$ $\angle P O X=x$ and $\angle Q O N=\angle Q O X=y$, then $2 x+2 y=\angle M O N=$ $180^{\circ}-2 \varphi$, i.e., $y=90^{\circ}-\varphi-x$. It follows that $\angle O Q C=180^{\circ}-\angle Q O C-$ $\angle O C Q=180^{\circ}-(\varphi+y)-\left(90^{\circ}-\varphi\right)=90^{\circ}-y=x+\varphi=\angle B O P$. Hence the triangles $B O P$ and $C Q O$ are similar, and consequently $B P \cdot C Q=$ $B O \cdot C O=(B C / 2)^{2}$. Conversely, let $B P \cdot C Q=(B C / 2)^{2}$ and let $Q^{\prime}$ be the point on $(A C)$ such that $P Q^{\prime}$ is tangent to the circle. Then $B P \cdot C Q^{\prime}=(B C / 2)^{2}$, which implies $Q \equiv Q^{\prime}$.
|
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0ce38a29-cb8f-526e-ab40-3cf8ac013c4d
| 23,739
|
25. (USA 6) ${ }^{\mathrm{IMO} 4}$ Given a point $P$ in a given plane $\pi$ and also a given point $Q$ not in $\pi$, show how to determine a point $R$ in $\pi$ such that $\frac{Q P+P R}{Q R}$ is a maximum.
|
25. Let us first look for such a point $R$ on a line $l$ in $\pi$ going through $P$. Let $\angle Q P R=2 \theta$. Consider a point $Q^{\prime}$ on $l$ such that $Q^{\prime} P=Q P$. Then we have $$ \frac{Q P+P R}{Q R}=\frac{R Q^{\prime}}{Q R}=\frac{\sin Q^{\prime} Q R}{\sin Q Q^{\prime} R} $$ Since $Q Q^{\prime} P$ is fixed, the maximal value of the expression occurs when $\angle Q Q^{\prime} R=90^{\circ}$. In this case $(Q P+P R) / Q R=1 / \sin \theta$. Looking at all possible lines $l$, we see that $\theta$ is minimized when $l$ equals the projection of $P Q$ onto $\pi$. Hence, the point $R$ is the intersection of the projection of $P Q$ onto $\pi$ and the plane through $Q$ perpendicular to $P Q$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
25. (USA 6) ${ }^{\mathrm{IMO} 4}$ Given a point $P$ in a given plane $\pi$ and also a given point $Q$ not in $\pi$, show how to determine a point $R$ in $\pi$ such that $\frac{Q P+P R}{Q R}$ is a maximum.
|
25. Let us first look for such a point $R$ on a line $l$ in $\pi$ going through $P$. Let $\angle Q P R=2 \theta$. Consider a point $Q^{\prime}$ on $l$ such that $Q^{\prime} P=Q P$. Then we have $$ \frac{Q P+P R}{Q R}=\frac{R Q^{\prime}}{Q R}=\frac{\sin Q^{\prime} Q R}{\sin Q Q^{\prime} R} $$ Since $Q Q^{\prime} P$ is fixed, the maximal value of the expression occurs when $\angle Q Q^{\prime} R=90^{\circ}$. In this case $(Q P+P R) / Q R=1 / \sin \theta$. Looking at all possible lines $l$, we see that $\theta$ is minimized when $l$ equals the projection of $P Q$ onto $\pi$. Hence, the point $R$ is the intersection of the projection of $P Q$ onto $\pi$ and the plane through $Q$ perpendicular to $P Q$.
|
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405dfbfe-1767-5043-8569-67b186826a8e
| 23,741
|
26. (YUG 4) Prove that the functional equations $$ \begin{aligned} f(x+y) & =f(x)+f(y) \\ \text { and } \quad f(x+y+x y) & =f(x)+f(y)+f(x y) \quad(x, y \in \mathbb{R}) \end{aligned} $$ are equivalent.
|
26. Let us assume that $f(x+y)=f(x)+f(y)$ for all reals. In this case we trivially apply the equation to get $f(x+y+x y)=f(x+y)+f(x y)=$ $f(x)+f(y)+f(x y)$. Hence the equivalence is proved in the first direction. Now let us assume that $f(x+y+x y)=f(x)+f(y)+f(x y)$ for all reals. Plugging in $x=y=0$ we get $f(0)=0$. Plugging in $y=-1$ we get $f(x)=-f(-x)$. Plugging in $y=1$ we get $f(2 x+1)=2 f(x)+f(1)$ and hence $f(2(u+v+u v)+1)=2 f(u+v+u v)+f(1)=2 f(u v)+$ $2 f(u)+2 f(v)+f(1)$ for all real $u$ and $v$. On the other hand, plugging in $x=u$ and $y=2 v+1$ we get $f(2(u+v+u v)+1)=f(u+(2 v+$ 1) $+u(2 v+1))=f(u)+2 f(v)+f(1)+f(2 u v+u)$. Hence it follows that $2 f(u v)+2 f(u)+2 f(v)+f(1)=f(u)+2 f(v)+f(1)+f(2 u v+u)$, i.e., $$ f(2 u v+u)=2 f(u v)+f(u) $$ Plugging in $v=-1 / 2$ we get $0=2 f(-u / 2)+f(u)=-2 f(u / 2)+f(u)$. Hence, $f(u)=2 f(u / 2)$ and consequently $f(2 x)=2 f(x)$ for all reals. Now (1) reduces to $f(2 u v+u)=f(2 u v)+f(u)$. Plugging in $u=y$ and $x=2 u v$, we obtain $f(x)+f(y)=f(x+y)$ for all nonzero reals $x$ and $y$. Since $f(0)=0$, it trivially holds that $f(x+y)=f(x)+f(y)$ when one of $x$ and $y$ is 0 .
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
26. (YUG 4) Prove that the functional equations $$ \begin{aligned} f(x+y) & =f(x)+f(y) \\ \text { and } \quad f(x+y+x y) & =f(x)+f(y)+f(x y) \quad(x, y \in \mathbb{R}) \end{aligned} $$ are equivalent.
|
26. Let us assume that $f(x+y)=f(x)+f(y)$ for all reals. In this case we trivially apply the equation to get $f(x+y+x y)=f(x+y)+f(x y)=$ $f(x)+f(y)+f(x y)$. Hence the equivalence is proved in the first direction. Now let us assume that $f(x+y+x y)=f(x)+f(y)+f(x y)$ for all reals. Plugging in $x=y=0$ we get $f(0)=0$. Plugging in $y=-1$ we get $f(x)=-f(-x)$. Plugging in $y=1$ we get $f(2 x+1)=2 f(x)+f(1)$ and hence $f(2(u+v+u v)+1)=2 f(u+v+u v)+f(1)=2 f(u v)+$ $2 f(u)+2 f(v)+f(1)$ for all real $u$ and $v$. On the other hand, plugging in $x=u$ and $y=2 v+1$ we get $f(2(u+v+u v)+1)=f(u+(2 v+$ 1) $+u(2 v+1))=f(u)+2 f(v)+f(1)+f(2 u v+u)$. Hence it follows that $2 f(u v)+2 f(u)+2 f(v)+f(1)=f(u)+2 f(v)+f(1)+f(2 u v+u)$, i.e., $$ f(2 u v+u)=2 f(u v)+f(u) $$ Plugging in $v=-1 / 2$ we get $0=2 f(-u / 2)+f(u)=-2 f(u / 2)+f(u)$. Hence, $f(u)=2 f(u / 2)$ and consequently $f(2 x)=2 f(x)$ for all reals. Now (1) reduces to $f(2 u v+u)=f(2 u v)+f(u)$. Plugging in $u=y$ and $x=2 u v$, we obtain $f(x)+f(y)=f(x+y)$ for all nonzero reals $x$ and $y$. Since $f(0)=0$, it trivially holds that $f(x+y)=f(x)+f(y)$ when one of $x$ and $y$ is 0 .
|
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3a9c055f-3e8f-5d62-af88-4507561e8d81
| 23,743
|
4. (BUL 3) ${ }^{\mathrm{IMO} 2} \mathrm{~A}$ pentagonal prism $A_{1} A_{2} \ldots A_{5} B_{1} B_{2} \ldots B_{5}$ is given. The edges, the diagonals of the lateral walls and the internal diagonals of the prism are each colored either red or green in such a way that no triangle whose vertices are vertices of the prism has its three edges of the same color. Prove that all edges of the bases are of the same color.
|
4. Let us prove first that the edges $A_{1} A_{2}, A_{2} A_{3}, \ldots, A_{5} A_{1}$ are of the same color. Assume the contrary, and let w.l.o.g. $A_{1} A_{2}$ be red and $A_{2} A_{3}$ be green. Three of the segments $A_{2} B_{l}(l=1,2,3,4,5)$, say $A_{2} B_{i}, A_{2} B_{j}, A_{2} B_{k}$, have to be of the same color, let it w.l.o.g. be red. Then $A_{1} B_{i}, A_{1} B_{j}, A_{1} B_{k}$ must be green. At least one of the sides of triangle $B_{i} B_{j} B_{k}$, say $B_{i} B_{j}$, must be an edge of the prism. Then looking at the triangles $A_{1} B_{i} B_{j}$ and $A_{2} B_{i} B_{j}$ we deduce that $B_{i} B_{j}$ can be neither green nor red, which is a contradiction. Hence all five edges of the pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$ have the same color. Similarly, all five edges of $B_{1} B_{2} B_{3} B_{4} B_{5}$ have the same color. We now show that the two colors are the same. Assume otherwise, i.e., that w.l.o.g. the $A$ edges are painted red and the $B$ edges green. Let us call segments of the form $A_{i} B_{j}$ diagonal ( $i$ and $j$ may be equal). We now count the diagonal segments by grouping the red segments based on their $A$ point, and the green segments based on their $B$ point. As above, the assumption that three of $A_{i} B_{j}$ for fixed $i$ are red leads to a contradiction. Hence at most two diagonal segments out of each $A_{i}$ may be red, which counts up to at most 10 red segments. Similarly, at most 10 diagonal segments can be green. But then we can paint at most 20 diagonal segments out of 25 , which is a contradiction. Hence all edges in the pentagons $A_{1} A_{2} A_{3} A_{4} A_{5}$ and $B_{1} B_{2} B_{3} B_{4} B_{5}$ have the same color.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
4. (BUL 3) ${ }^{\mathrm{IMO} 2} \mathrm{~A}$ pentagonal prism $A_{1} A_{2} \ldots A_{5} B_{1} B_{2} \ldots B_{5}$ is given. The edges, the diagonals of the lateral walls and the internal diagonals of the prism are each colored either red or green in such a way that no triangle whose vertices are vertices of the prism has its three edges of the same color. Prove that all edges of the bases are of the same color.
|
4. Let us prove first that the edges $A_{1} A_{2}, A_{2} A_{3}, \ldots, A_{5} A_{1}$ are of the same color. Assume the contrary, and let w.l.o.g. $A_{1} A_{2}$ be red and $A_{2} A_{3}$ be green. Three of the segments $A_{2} B_{l}(l=1,2,3,4,5)$, say $A_{2} B_{i}, A_{2} B_{j}, A_{2} B_{k}$, have to be of the same color, let it w.l.o.g. be red. Then $A_{1} B_{i}, A_{1} B_{j}, A_{1} B_{k}$ must be green. At least one of the sides of triangle $B_{i} B_{j} B_{k}$, say $B_{i} B_{j}$, must be an edge of the prism. Then looking at the triangles $A_{1} B_{i} B_{j}$ and $A_{2} B_{i} B_{j}$ we deduce that $B_{i} B_{j}$ can be neither green nor red, which is a contradiction. Hence all five edges of the pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$ have the same color. Similarly, all five edges of $B_{1} B_{2} B_{3} B_{4} B_{5}$ have the same color. We now show that the two colors are the same. Assume otherwise, i.e., that w.l.o.g. the $A$ edges are painted red and the $B$ edges green. Let us call segments of the form $A_{i} B_{j}$ diagonal ( $i$ and $j$ may be equal). We now count the diagonal segments by grouping the red segments based on their $A$ point, and the green segments based on their $B$ point. As above, the assumption that three of $A_{i} B_{j}$ for fixed $i$ are red leads to a contradiction. Hence at most two diagonal segments out of each $A_{i}$ may be red, which counts up to at most 10 red segments. Similarly, at most 10 diagonal segments can be green. But then we can paint at most 20 diagonal segments out of 25 , which is a contradiction. Hence all edges in the pentagons $A_{1} A_{2} A_{3} A_{4} A_{5}$ and $B_{1} B_{2} B_{3} B_{4} B_{5}$ have the same color.
|
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74a6452b-9e57-544b-90f3-b916a77bc83d
| 23,749
|
7. (FRG 1) ${ }^{\mathrm{IMO} 1}$ Given that $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots-\frac{1}{1318}+\frac{1}{1319}=\frac{p}{q}$, where $p$ and $q$ are natural numbers having no common factor, prove that $p$ is divisible by 1979.
|
7. We denote the sum mentioned above by $S$. We have the following equalities: $$ \begin{aligned} S & =1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots-\frac{1}{1318}+\frac{1}{1319} \\ & =1+\frac{1}{2}+\cdots+\frac{1}{1319}-2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{1318}\right) \\ & =1+\frac{1}{2}+\cdots+\frac{1}{1319}-\left(1+\frac{1}{2}+\cdots+\frac{1}{659}\right) \\ & =\frac{1}{660}+\frac{1}{661}+\cdots+\frac{1}{1319} \\ & =\sum_{i=660}^{989} \frac{1}{i}+\frac{1}{1979-i}=\sum_{i=660}^{989} \frac{1979}{i \cdot(1979-i)} \end{aligned} $$ Since no term in the sum contains a denominator divisible by 1979 (1979 is a prime number), it follows that when $S$ is represented as $p / q$ the numerator $p$ will have to be divisible by 1979.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
7. (FRG 1) ${ }^{\mathrm{IMO} 1}$ Given that $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots-\frac{1}{1318}+\frac{1}{1319}=\frac{p}{q}$, where $p$ and $q$ are natural numbers having no common factor, prove that $p$ is divisible by 1979.
|
7. We denote the sum mentioned above by $S$. We have the following equalities: $$ \begin{aligned} S & =1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots-\frac{1}{1318}+\frac{1}{1319} \\ & =1+\frac{1}{2}+\cdots+\frac{1}{1319}-2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{1318}\right) \\ & =1+\frac{1}{2}+\cdots+\frac{1}{1319}-\left(1+\frac{1}{2}+\cdots+\frac{1}{659}\right) \\ & =\frac{1}{660}+\frac{1}{661}+\cdots+\frac{1}{1319} \\ & =\sum_{i=660}^{989} \frac{1}{i}+\frac{1}{1979-i}=\sum_{i=660}^{989} \frac{1979}{i \cdot(1979-i)} \end{aligned} $$ Since no term in the sum contains a denominator divisible by 1979 (1979 is a prime number), it follows that when $S$ is represented as $p / q$ the numerator $p$ will have to be divisible by 1979.
|
{
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1260db52-95f5-50e7-976b-00609bb84237
| 23,756
|
11. (NET) On a semicircle with unit radius four consecutive chords $A B, B C$, $C D, D E$ with lengths $a, b, c, d$, respectively, are given. Prove that $$ a^{2}+b^{2}+c^{2}+d^{2}+a b c+b c d<4 $$
|
11. Let us denote the center of the semicircle by $O$, and $\angle A O B=2 \alpha$, $\angle B O C=2 \beta, A C=m, C E=n$. We claim that $a^{2}+b^{2}+n^{2}+a b n=4$. Indeed, since $a=2 \sin \alpha, b=2 \sin \beta$, $n=2 \cos (\alpha+\beta)$, we have $$ \begin{aligned} a^{2} & +b^{2}+n^{2}+a b n \\ & =4\left(\sin ^{2} \alpha+\sin ^{2} \beta+\cos ^{2}(\alpha+\beta)+2 \sin \alpha \sin \beta \cos (\alpha+\beta)\right) \\ & =4+4\left(-\frac{\cos 2 \alpha}{2}-\frac{\cos 2 \beta}{2}+\cos (\alpha+\beta) \cos (\alpha-\beta)\right) \\ & =4+4(\cos (\alpha+\beta) \cos (\alpha-\beta)-\cos (\alpha+\beta) \cos (\alpha-\beta))=4 \end{aligned} $$ Analogously, $c^{2}+d^{2}+m^{2}+c d m=4$. By adding both equalities and subtracting $m^{2}+n^{2}=4$ we obtain $$ a^{2}+b^{2}+c^{2}+d^{2}+a b n+c d m=4 $$ Since $n>c$ and $m>b$, the desired inequality follows.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
11. (NET) On a semicircle with unit radius four consecutive chords $A B, B C$, $C D, D E$ with lengths $a, b, c, d$, respectively, are given. Prove that $$ a^{2}+b^{2}+c^{2}+d^{2}+a b c+b c d<4 $$
|
11. Let us denote the center of the semicircle by $O$, and $\angle A O B=2 \alpha$, $\angle B O C=2 \beta, A C=m, C E=n$. We claim that $a^{2}+b^{2}+n^{2}+a b n=4$. Indeed, since $a=2 \sin \alpha, b=2 \sin \beta$, $n=2 \cos (\alpha+\beta)$, we have $$ \begin{aligned} a^{2} & +b^{2}+n^{2}+a b n \\ & =4\left(\sin ^{2} \alpha+\sin ^{2} \beta+\cos ^{2}(\alpha+\beta)+2 \sin \alpha \sin \beta \cos (\alpha+\beta)\right) \\ & =4+4\left(-\frac{\cos 2 \alpha}{2}-\frac{\cos 2 \beta}{2}+\cos (\alpha+\beta) \cos (\alpha-\beta)\right) \\ & =4+4(\cos (\alpha+\beta) \cos (\alpha-\beta)-\cos (\alpha+\beta) \cos (\alpha-\beta))=4 \end{aligned} $$ Analogously, $c^{2}+d^{2}+m^{2}+c d m=4$. By adding both equalities and subtracting $m^{2}+n^{2}=4$ we obtain $$ a^{2}+b^{2}+c^{2}+d^{2}+a b n+c d m=4 $$ Since $n>c$ and $m>b$, the desired inequality follows.
|
{
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|
1dade684-86ef-5686-8320-67b30cfdd7da
| 23,767
|
14. (ROM) Prove that a convex pentagon (a five-sided polygon) $A B C D E$ with equal sides and for which the interior angles satisfy the condition $\angle A \geq \angle B \geq \angle C \geq \angle D \geq \angle E$ is a regular pentagon.
|
14. We need the following lemma. Lemma. If a convex quadrilateral $P Q R S$ satisfies $P S=Q R$ and $\angle S P Q \geq$ $\angle R Q P$, then $\angle Q R S \geq \angle P S R$. Proof. If the lines $P S$ and $Q R$ are parallel, then this quadrilateral is a parallelogram, and the statement is trivial. Otherwise, let $X$ be the point of intersection of lines $P S$ and $Q R$. Assume that $\angle S P Q+\angle R Q P>180^{\circ}$. Then $\angle X P Q \leq \angle X Q P$ implies that $X P \geq X Q$, and consequently $X S \geq X R$. Hence, $\angle Q R S=$ $\angle X R S \geq \angle X S R=\angle P S R$. Similarly, if $\angle S P Q+\angle R Q P<180^{\circ}$, then $\angle X P Q \geq \angle X Q P$, from which it follows that $X P \leq X Q$, and thus $X S \leq X R$; hence $\angle Q R S=$ $180^{\circ}-\angle X R S \geq 180^{\circ}-\angle X S R=\angle P S R$. Now we apply the lemma to the quadrilateral $A B C D$. Since $\angle B \geq \angle C$ and $A B=C D$, it follows that $\angle C D A \geq \angle B A D$, which together with $\angle E D A=\angle E A D$ gives $\angle D \geq \angle A$. Thus $\angle A=\angle B=\angle C=\angle D$. Analogously, by applying the lemma to $B C D E$ we obtain $\angle E \geq \angle B$, and hence $\angle B=\angle C=\angle D=\angle E$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
14. (ROM) Prove that a convex pentagon (a five-sided polygon) $A B C D E$ with equal sides and for which the interior angles satisfy the condition $\angle A \geq \angle B \geq \angle C \geq \angle D \geq \angle E$ is a regular pentagon.
|
14. We need the following lemma. Lemma. If a convex quadrilateral $P Q R S$ satisfies $P S=Q R$ and $\angle S P Q \geq$ $\angle R Q P$, then $\angle Q R S \geq \angle P S R$. Proof. If the lines $P S$ and $Q R$ are parallel, then this quadrilateral is a parallelogram, and the statement is trivial. Otherwise, let $X$ be the point of intersection of lines $P S$ and $Q R$. Assume that $\angle S P Q+\angle R Q P>180^{\circ}$. Then $\angle X P Q \leq \angle X Q P$ implies that $X P \geq X Q$, and consequently $X S \geq X R$. Hence, $\angle Q R S=$ $\angle X R S \geq \angle X S R=\angle P S R$. Similarly, if $\angle S P Q+\angle R Q P<180^{\circ}$, then $\angle X P Q \geq \angle X Q P$, from which it follows that $X P \leq X Q$, and thus $X S \leq X R$; hence $\angle Q R S=$ $180^{\circ}-\angle X R S \geq 180^{\circ}-\angle X S R=\angle P S R$. Now we apply the lemma to the quadrilateral $A B C D$. Since $\angle B \geq \angle C$ and $A B=C D$, it follows that $\angle C D A \geq \angle B A D$, which together with $\angle E D A=\angle E A D$ gives $\angle D \geq \angle A$. Thus $\angle A=\angle B=\angle C=\angle D$. Analogously, by applying the lemma to $B C D E$ we obtain $\angle E \geq \angle B$, and hence $\angle B=\angle C=\angle D=\angle E$.
|
{
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|
67c6ceda-73c3-5175-9cd6-3cadfb268091
| 23,773
|
16. (GBR) A sequence of real numbers $u_{1}, u_{2}, u_{3}, \ldots$ is determined by $u_{1}$ and the following recurrence relation for $n \geq 1$ : $$ 4 u_{n+1}=\sqrt[3]{64 u_{n}+15} $$ Describe, with proof, the behavior of $u_{n}$ as $n \rightarrow \infty$.
|
16. The sequence $\left\{u_{n}\right\}$ is bounded, whatever $u_{1}$ is. Indeed, assume the opposite, and let $u_{m}$ be the first member of the sequence such that $\left|u_{m}\right|>\max \left\{2,\left|u_{1}\right|\right\}$. Then $\left|u_{m-1}\right|=\left|u_{m}^{3}-15 / 64\right|>\left|u_{m}\right|$, which is impossible. Next, let us see for what values of $u_{m}, u_{m+1}$ is greater, equal, or smaller, respectively. If $u_{m+1}=u_{m}$, then $u_{m}=u_{m+1}^{3}-15 / 64=u_{m}^{3}-15 / 64$; i.e., $u_{m}$ is a root of $x^{3}-x-15 / 64=0$. This equation factors as $(x+1 / 4)\left(x^{2}-x / 4-\right.$ $15 / 16)=0$, and hence $u_{m}$ is equal to $x_{1}=(1-\sqrt{61}) / 8, x_{2}=-1 / 4$, or $x_{3}=(1+\sqrt{61}) / 8$, and these are the only possible limits of the sequence. Each of $u_{m+1}>u_{m}, u_{m+1}<u_{m}$ is equivalent to $u_{m}^{3}-u_{m}-15 / 64<0$ and $u_{m}^{3}-u_{m}-15 / 64>0$ respectively. Thus the former is satisfied for $u_{m}$ in the interval $I_{1}=\left(-\infty, x_{1}\right)$ or $I_{3}=\left(x_{2}, x_{3}\right)$, while the latter is satisfied for $u_{m}$ in $I_{2}=\left(x_{1}, x_{2}\right)$ or $I_{4}=\left(x_{3}, \infty\right)$. Moreover, since the function $f(x)=\sqrt[3]{x+15 / 64}$ is strictly increasing with fixed points $x_{1}, x_{2}, x_{3}$, it follows that $u_{m}$ will never escape from the interval $I_{1}, I_{2}, I_{3}$, or $I_{4}$ to which it belongs initially. Therefore: (1) if $u_{1}$ is one of $x_{1}, x_{2}, x_{3}$, the sequence $\left\{u_{m}\right\}$ is constant; (2) if $u_{1} \in I_{1}$, then the sequence is strictly increasing and tends to $x_{1}$; (3) if $u_{1} \in I_{2}$, then the sequence is strictly decreasing and tends to $x_{1}$; (4) if $u_{1} \in I_{3}$, then the sequence is strictly increasing and tends to $x_{3}$; (5) if $u_{1} \in I_{4}$, then the sequence is strictly decreasing and tends to $x_{3}$.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
16. (GBR) A sequence of real numbers $u_{1}, u_{2}, u_{3}, \ldots$ is determined by $u_{1}$ and the following recurrence relation for $n \geq 1$ : $$ 4 u_{n+1}=\sqrt[3]{64 u_{n}+15} $$ Describe, with proof, the behavior of $u_{n}$ as $n \rightarrow \infty$.
|
16. The sequence $\left\{u_{n}\right\}$ is bounded, whatever $u_{1}$ is. Indeed, assume the opposite, and let $u_{m}$ be the first member of the sequence such that $\left|u_{m}\right|>\max \left\{2,\left|u_{1}\right|\right\}$. Then $\left|u_{m-1}\right|=\left|u_{m}^{3}-15 / 64\right|>\left|u_{m}\right|$, which is impossible. Next, let us see for what values of $u_{m}, u_{m+1}$ is greater, equal, or smaller, respectively. If $u_{m+1}=u_{m}$, then $u_{m}=u_{m+1}^{3}-15 / 64=u_{m}^{3}-15 / 64$; i.e., $u_{m}$ is a root of $x^{3}-x-15 / 64=0$. This equation factors as $(x+1 / 4)\left(x^{2}-x / 4-\right.$ $15 / 16)=0$, and hence $u_{m}$ is equal to $x_{1}=(1-\sqrt{61}) / 8, x_{2}=-1 / 4$, or $x_{3}=(1+\sqrt{61}) / 8$, and these are the only possible limits of the sequence. Each of $u_{m+1}>u_{m}, u_{m+1}<u_{m}$ is equivalent to $u_{m}^{3}-u_{m}-15 / 64<0$ and $u_{m}^{3}-u_{m}-15 / 64>0$ respectively. Thus the former is satisfied for $u_{m}$ in the interval $I_{1}=\left(-\infty, x_{1}\right)$ or $I_{3}=\left(x_{2}, x_{3}\right)$, while the latter is satisfied for $u_{m}$ in $I_{2}=\left(x_{1}, x_{2}\right)$ or $I_{4}=\left(x_{3}, \infty\right)$. Moreover, since the function $f(x)=\sqrt[3]{x+15 / 64}$ is strictly increasing with fixed points $x_{1}, x_{2}, x_{3}$, it follows that $u_{m}$ will never escape from the interval $I_{1}, I_{2}, I_{3}$, or $I_{4}$ to which it belongs initially. Therefore: (1) if $u_{1}$ is one of $x_{1}, x_{2}, x_{3}$, the sequence $\left\{u_{m}\right\}$ is constant; (2) if $u_{1} \in I_{1}$, then the sequence is strictly increasing and tends to $x_{1}$; (3) if $u_{1} \in I_{2}$, then the sequence is strictly decreasing and tends to $x_{1}$; (4) if $u_{1} \in I_{3}$, then the sequence is strictly increasing and tends to $x_{3}$; (5) if $u_{1} \in I_{4}$, then the sequence is strictly decreasing and tends to $x_{3}$.
|
{
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|
9695c209-2a5c-52f4-bd2f-61d7efdcbd56
| 23,778
|
17. (USS) ${ }^{\text {IMO5 }}$ Three equal circles touch the sides of a triangle and have one common point $O$. Show that the center of the circle inscribed in and of the circle circumscribed about the triangle $A B C$ and the point $O$ are collinear.
|
17. Let us denote by $S_{A}, S_{B}, S_{C}$ the centers of the given circles, where $S_{A}$ lies on the bisector of $\angle A$, etc. Then $S_{A} S_{B}\left\|A B, S_{B} S_{C}\right\| B C, S_{C} S_{A} \| C A$, so that the inner bisectors of the angles of triangle $A B C$ are also inner bisectors of the angles of $\triangle S_{A} S_{B} S_{C}$. These two triangles thus have a common incenter $S$, which is also the center of the homothety $\chi$ mapping $\triangle S_{A} S_{B} S_{C}$ onto $\triangle A B C$. The point $O$ is the circumcenter of triangle $S_{A} S_{B} S_{C}$, and so is mapped by $\chi$ onto the circumcenter $P$ of $A B C$. This means that $O, P$, and the center $S$ of $\chi$ are collinear.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
17. (USS) ${ }^{\text {IMO5 }}$ Three equal circles touch the sides of a triangle and have one common point $O$. Show that the center of the circle inscribed in and of the circle circumscribed about the triangle $A B C$ and the point $O$ are collinear.
|
17. Let us denote by $S_{A}, S_{B}, S_{C}$ the centers of the given circles, where $S_{A}$ lies on the bisector of $\angle A$, etc. Then $S_{A} S_{B}\left\|A B, S_{B} S_{C}\right\| B C, S_{C} S_{A} \| C A$, so that the inner bisectors of the angles of triangle $A B C$ are also inner bisectors of the angles of $\triangle S_{A} S_{B} S_{C}$. These two triangles thus have a common incenter $S$, which is also the center of the homothety $\chi$ mapping $\triangle S_{A} S_{B} S_{C}$ onto $\triangle A B C$. The point $O$ is the circumcenter of triangle $S_{A} S_{B} S_{C}$, and so is mapped by $\chi$ onto the circumcenter $P$ of $A B C$. This means that $O, P$, and the center $S$ of $\chi$ are collinear.
|
{
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|
cdd2da4a-57bd-567b-9dbe-b317c3dcae2e
| 23,780
|
18. (USS) Several equal spherical planets are given in outer space. On the surface of each planet there is a set of points that is invisible from any of the remaining planets. Prove that the sum of the areas of all these sets is equal to the area of the surface of one planet.
|
18. Let $C$ be the convex hull of the set of the planets: its border consists of parts of planes, parts of cylinders, and parts of the surfaces of some planets. These parts of planets consist exactly of all the invisible points; any point on a planet that is inside $C$ is visible. Thus it remains to show that the areas of all the parts of planets lying on the border of $C$ add up to the area of one planet. As we have seen, an invisible part of a planet is bordered by some main spherical arcs, parallel two by two. Now fix any planet $P$, and translate these arcs onto arcs on the surface of $P$. All these arcs partition the surface of $P$ into several parts, each of which corresponds to the invisible part of one of the planets. This correspondence is bijective, and therefore the statement follows.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
18. (USS) Several equal spherical planets are given in outer space. On the surface of each planet there is a set of points that is invisible from any of the remaining planets. Prove that the sum of the areas of all these sets is equal to the area of the surface of one planet.
|
18. Let $C$ be the convex hull of the set of the planets: its border consists of parts of planes, parts of cylinders, and parts of the surfaces of some planets. These parts of planets consist exactly of all the invisible points; any point on a planet that is inside $C$ is visible. Thus it remains to show that the areas of all the parts of planets lying on the border of $C$ add up to the area of one planet. As we have seen, an invisible part of a planet is bordered by some main spherical arcs, parallel two by two. Now fix any planet $P$, and translate these arcs onto arcs on the surface of $P$. All these arcs partition the surface of $P$ into several parts, each of which corresponds to the invisible part of one of the planets. This correspondence is bijective, and therefore the statement follows.
|
{
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|
7520e280-b011-54d8-9c8d-258dba3854b2
| 23,782
|
19. (YUG) A finite set of unit circles is given in a plane such that the area of their union $U$ is $S$. Prove that there exists a subset of mutually disjoint circles such that the area of their union is greater that $\frac{2 S}{9}$.
|
19. Consider the partition of plane $\pi$ into regular hexagons, each having inradius 2. Fix one of these hexagons, denoted by $\gamma$. For any other hexagon $x$ in the partition, there exists a unique translation $\tau_{x}$ taking it onto $\gamma$. Define the mapping $\varphi: \pi \rightarrow \gamma$ as follows: If $A$ belongs to the interior of a hexagon $x$, then $\varphi(A)=\tau_{x}(A)$ (if $A$ is on the border of some hexagon, it does not actually matter where its image is). The total area of the images of the union of the given circles equals $S$, while the area of the hexagon $\gamma$ is $8 \sqrt{3}$. Thus there exists a point $B$ of $\gamma$ that is covered at least $\frac{S}{8 \sqrt{3}}$ times, i.e., such that $\varphi^{-1}(B)$ consists of at least $\frac{S}{8 \sqrt{3}}$ distinct points of the plane that belong to some of the circles. For any of these points, take a circle that contains it. All these circles are disjoint, with total area not less than $\frac{\pi}{8 \sqrt{3}} S \geq 2 S / 9$. Remark. The statement becomes false if the constant $2 / 9$ is replaced by any number greater than $1 / 4$. In that case a counterexample is, for example, a set of unit circles inside a circle of radius 2 covering a sufficiently large part of its area.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
19. (YUG) A finite set of unit circles is given in a plane such that the area of their union $U$ is $S$. Prove that there exists a subset of mutually disjoint circles such that the area of their union is greater that $\frac{2 S}{9}$.
|
19. Consider the partition of plane $\pi$ into regular hexagons, each having inradius 2. Fix one of these hexagons, denoted by $\gamma$. For any other hexagon $x$ in the partition, there exists a unique translation $\tau_{x}$ taking it onto $\gamma$. Define the mapping $\varphi: \pi \rightarrow \gamma$ as follows: If $A$ belongs to the interior of a hexagon $x$, then $\varphi(A)=\tau_{x}(A)$ (if $A$ is on the border of some hexagon, it does not actually matter where its image is). The total area of the images of the union of the given circles equals $S$, while the area of the hexagon $\gamma$ is $8 \sqrt{3}$. Thus there exists a point $B$ of $\gamma$ that is covered at least $\frac{S}{8 \sqrt{3}}$ times, i.e., such that $\varphi^{-1}(B)$ consists of at least $\frac{S}{8 \sqrt{3}}$ distinct points of the plane that belong to some of the circles. For any of these points, take a circle that contains it. All these circles are disjoint, with total area not less than $\frac{\pi}{8 \sqrt{3}} S \geq 2 S / 9$. Remark. The statement becomes false if the constant $2 / 9$ is replaced by any number greater than $1 / 4$. In that case a counterexample is, for example, a set of unit circles inside a circle of radius 2 covering a sufficiently large part of its area.
|
{
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|
d915778c-ee59-5416-9268-d436cc45f9b0
| 23,784
|
2. (BUL) A sphere $S$ is tangent to the edges $A B, B C, C D, D A$ of a tetrahedron $A B C D$ at the points $E, F, G, H$ respectively. The points $E, F, G, H$ are the vertices of a square. Prove that if the sphere is tangent to the edge $A C$, then it is also tangent to the edge $B D$.
|
2. Lemma. Let $E, F, G, H, I$, and $K$ be points on edges $A B, B C, C D, D A$, $A C$, and $B D$ of a tetrahedron. Then there is a sphere that touches the edges at these points if and only if $$ \begin{aligned} & A E=A H=A I, \quad B E=B F=B K, \\ & C F=C G=C I, \quad D G=D H=D K . \end{aligned} $$ Proof. The "only if" side of the equivalence is obvious. We now assume (*). Denote by $\epsilon, \phi, \gamma, \eta, \iota$, and $\kappa$ planes through $E, F, G, H, I, K$ perpendicular to $A B, B C, C D, D A, A C$ and $B D$ respectively. Since the three planes $\epsilon, \eta$, and $\iota$ are not mutually parallel, they intersect in a common point $O$. Clearly, $\triangle A E O \cong$  $\triangle A H O \cong \triangle A I O$; hence $O E=O H=O I=r$, and the sphere $\sigma(O, r)$ is tangent to $A B, A D, A C$. To prove that $\sigma$ is also tangent to $B C, C D, B D$ it suffices to show that planes $\phi, \gamma$, and $\kappa$ also pass through $O$. Without loss of generality we can prove this for just $\phi$. By the conditions for $E, F, I$, these are exactly the points of tangency of the incircle of $\triangle A B C$ and its sides, and if $S$ is the incenter, then $S E \perp A B, S F \perp B C, S I \perp A C$. Hence $\epsilon, \iota$, and $\phi$ all pass through $S$ and are perpendicular to the plane $A B C$, and consequently all share the line $l$ through $S$ perpendicular to $A B C$. Since $l=\epsilon \cap \iota$, the point $O$ will be situated on $l$, and hence $\phi$ will also contain $O$. This completes our proof of the lemma. Let $A H=A E=x, B E=B F=y, C F=C G=z$, and $D G=D H=w$. If the sphere is also tangent to $A C$ at some point $I$, then $A I=x$ and $I C=z$. Using the stated lemma it suffices to prove that if $A C=x+z$, then $B D=y+w$. Let $E F=F G=G H=H I=t, \angle B A D=\alpha, \angle A B C=\beta, \angle B C D=\gamma$, and $\angle A D C=\delta$. We get $$ t^{2}=E H^{2}=A E^{2}+A H^{2}-2 \cdot A E \cdot A H \cos \alpha=2 x^{2}(1-\cos \alpha) $$ We similarly conclude that $t^{2}=2 y^{2}(1-\cos \beta)=2 z^{2}(1-\cos \gamma)=2 w^{2}(1-$ $\cos \delta$ ). Further, using that $A B=x+y, B C=y+z, \cos \beta=1-t^{2} / 2 y^{2}$, we obtain $$ A C^{2}=A B^{2}+B C^{2}-2 A B \cdot B C \cos \beta=(x-z)^{2}+t^{2}\left(\frac{x}{y}+1\right)\left(\frac{z}{y}+1\right) $$ Analogously, from the triangle $A D C$ we get $A C^{2}=(x-z)^{2}+t^{2}(x / w+$ $1)(z / w+1)$, which gives $(x / y+1)(z / y+1)=(x / w+1)(z / w+1)$. Since $f(s)=(x / s+1)(z / s+1)$ is a decreasing function in $s$, it follows that $y=w$; similarly $x=z$. Hence $C F=C G=x$ and $D G=D H=y$. Hence $A C \| E F$ and $A C: t=$ $A C: E F=A B: E B=(x+y): y$; i.e., $A C=t(x+y) / y$. Similarly, from the triangle $A B D$, we get that $B D=t(x+y) / x$. Hence if $A C=x+z=2 x$, it follows that $2 x=t(x+y) / y \Rightarrow 2 x y=t(x+y) \Rightarrow B D=t(x+y) / x=$ $2 y=y+w$. This completes the proof. Second solution. Without loss of generality, assume that $E F=2$. Consider the Cartesian system in which points $O, E, F, G, H$ respectively have coordinates $(0,0,0),(-1,-1, a),(1,-1, a),(1,1, a),(-1,1, a)$. Line $A H$ is perpendicular to $O H$ and $A E$ is perpendicular to $O E$; hence from Pythagoras's theorem $A O^{2}=A H^{2}+H O^{2}=A E^{2}+E O^{2}=A E^{2}+H O^{2}$, which implies $A H=A E$. Therefore the $y$-coordinate of $A$ is zero; analogously the $x$-coordinates of $B$ and $D$ and the $y$-coordinate of $C$ are 0 . Let $A$ have coordinates $\left(x_{0}, 0, z_{1}\right)$ : then $\overrightarrow{E A}\left(x_{0}+1,1, z_{1}-a\right) \perp \overrightarrow{E O}(1,1,-a)$, i.e., $\overrightarrow{E A} \cdot \overrightarrow{E O}=x_{0}+2+a\left(a-z_{1}\right)=0$. Similarly, for $B\left(0, y_{0}, z_{2}\right)$ we have $y_{0}+2+a\left(a-z_{2}\right)=0$. This gives us $$ z_{1}=\frac{x_{0}+a^{2}+2}{a}, \quad z_{2}=\frac{y_{0}+a^{2}+2}{a} $$ We haven't used yet that $A\left(x_{0}, 0, z_{1}\right), E(-1,-1, a)$ and $B\left(0, y_{0}, z_{2}\right)$ are collinear, so let $A^{\prime}, B^{\prime}, E^{\prime}$ be the feet of perpendiculars from $A, B, E$ to the plane $x y$. The line $A^{\prime} B^{\prime}$, given by $y_{0} x+x_{0} y=x_{0} y_{0}, z=0$, contains the point $E^{\prime}(-1,-1,0)$, from which we obtain $$ \left(x_{0}+1\right)\left(y_{0}+1\right)=1 $$ In the same way, from the points $B$ and $C$ we get relations similar to (1) and (2) and conclude that $C$ has the coordinates $C\left(-x_{0}, 0, z_{1}\right)$. Similarly we get $D\left(0,-y_{0}, z_{2}\right)$. The condition that $A C$ is tangent to the sphere $\sigma(O, O E)$ is equivalent to $z_{1}=\sqrt{a^{2}+2}$, i.e., to $x_{0}=a \sqrt{a^{2}+2}-\left(a^{2}+2\right)$. But then (2) implies that $y_{0}=-a \sqrt{a^{2}+2}-\left(a^{2}+2\right)$ and $z_{2}=-\sqrt{a^{2}+2}$, which means that the sphere $\sigma$ is tangent to $B D$ as well. This finishes the proof.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
2. (BUL) A sphere $S$ is tangent to the edges $A B, B C, C D, D A$ of a tetrahedron $A B C D$ at the points $E, F, G, H$ respectively. The points $E, F, G, H$ are the vertices of a square. Prove that if the sphere is tangent to the edge $A C$, then it is also tangent to the edge $B D$.
|
2. Lemma. Let $E, F, G, H, I$, and $K$ be points on edges $A B, B C, C D, D A$, $A C$, and $B D$ of a tetrahedron. Then there is a sphere that touches the edges at these points if and only if $$ \begin{aligned} & A E=A H=A I, \quad B E=B F=B K, \\ & C F=C G=C I, \quad D G=D H=D K . \end{aligned} $$ Proof. The "only if" side of the equivalence is obvious. We now assume (*). Denote by $\epsilon, \phi, \gamma, \eta, \iota$, and $\kappa$ planes through $E, F, G, H, I, K$ perpendicular to $A B, B C, C D, D A, A C$ and $B D$ respectively. Since the three planes $\epsilon, \eta$, and $\iota$ are not mutually parallel, they intersect in a common point $O$. Clearly, $\triangle A E O \cong$  $\triangle A H O \cong \triangle A I O$; hence $O E=O H=O I=r$, and the sphere $\sigma(O, r)$ is tangent to $A B, A D, A C$. To prove that $\sigma$ is also tangent to $B C, C D, B D$ it suffices to show that planes $\phi, \gamma$, and $\kappa$ also pass through $O$. Without loss of generality we can prove this for just $\phi$. By the conditions for $E, F, I$, these are exactly the points of tangency of the incircle of $\triangle A B C$ and its sides, and if $S$ is the incenter, then $S E \perp A B, S F \perp B C, S I \perp A C$. Hence $\epsilon, \iota$, and $\phi$ all pass through $S$ and are perpendicular to the plane $A B C$, and consequently all share the line $l$ through $S$ perpendicular to $A B C$. Since $l=\epsilon \cap \iota$, the point $O$ will be situated on $l$, and hence $\phi$ will also contain $O$. This completes our proof of the lemma. Let $A H=A E=x, B E=B F=y, C F=C G=z$, and $D G=D H=w$. If the sphere is also tangent to $A C$ at some point $I$, then $A I=x$ and $I C=z$. Using the stated lemma it suffices to prove that if $A C=x+z$, then $B D=y+w$. Let $E F=F G=G H=H I=t, \angle B A D=\alpha, \angle A B C=\beta, \angle B C D=\gamma$, and $\angle A D C=\delta$. We get $$ t^{2}=E H^{2}=A E^{2}+A H^{2}-2 \cdot A E \cdot A H \cos \alpha=2 x^{2}(1-\cos \alpha) $$ We similarly conclude that $t^{2}=2 y^{2}(1-\cos \beta)=2 z^{2}(1-\cos \gamma)=2 w^{2}(1-$ $\cos \delta$ ). Further, using that $A B=x+y, B C=y+z, \cos \beta=1-t^{2} / 2 y^{2}$, we obtain $$ A C^{2}=A B^{2}+B C^{2}-2 A B \cdot B C \cos \beta=(x-z)^{2}+t^{2}\left(\frac{x}{y}+1\right)\left(\frac{z}{y}+1\right) $$ Analogously, from the triangle $A D C$ we get $A C^{2}=(x-z)^{2}+t^{2}(x / w+$ $1)(z / w+1)$, which gives $(x / y+1)(z / y+1)=(x / w+1)(z / w+1)$. Since $f(s)=(x / s+1)(z / s+1)$ is a decreasing function in $s$, it follows that $y=w$; similarly $x=z$. Hence $C F=C G=x$ and $D G=D H=y$. Hence $A C \| E F$ and $A C: t=$ $A C: E F=A B: E B=(x+y): y$; i.e., $A C=t(x+y) / y$. Similarly, from the triangle $A B D$, we get that $B D=t(x+y) / x$. Hence if $A C=x+z=2 x$, it follows that $2 x=t(x+y) / y \Rightarrow 2 x y=t(x+y) \Rightarrow B D=t(x+y) / x=$ $2 y=y+w$. This completes the proof. Second solution. Without loss of generality, assume that $E F=2$. Consider the Cartesian system in which points $O, E, F, G, H$ respectively have coordinates $(0,0,0),(-1,-1, a),(1,-1, a),(1,1, a),(-1,1, a)$. Line $A H$ is perpendicular to $O H$ and $A E$ is perpendicular to $O E$; hence from Pythagoras's theorem $A O^{2}=A H^{2}+H O^{2}=A E^{2}+E O^{2}=A E^{2}+H O^{2}$, which implies $A H=A E$. Therefore the $y$-coordinate of $A$ is zero; analogously the $x$-coordinates of $B$ and $D$ and the $y$-coordinate of $C$ are 0 . Let $A$ have coordinates $\left(x_{0}, 0, z_{1}\right)$ : then $\overrightarrow{E A}\left(x_{0}+1,1, z_{1}-a\right) \perp \overrightarrow{E O}(1,1,-a)$, i.e., $\overrightarrow{E A} \cdot \overrightarrow{E O}=x_{0}+2+a\left(a-z_{1}\right)=0$. Similarly, for $B\left(0, y_{0}, z_{2}\right)$ we have $y_{0}+2+a\left(a-z_{2}\right)=0$. This gives us $$ z_{1}=\frac{x_{0}+a^{2}+2}{a}, \quad z_{2}=\frac{y_{0}+a^{2}+2}{a} $$ We haven't used yet that $A\left(x_{0}, 0, z_{1}\right), E(-1,-1, a)$ and $B\left(0, y_{0}, z_{2}\right)$ are collinear, so let $A^{\prime}, B^{\prime}, E^{\prime}$ be the feet of perpendiculars from $A, B, E$ to the plane $x y$. The line $A^{\prime} B^{\prime}$, given by $y_{0} x+x_{0} y=x_{0} y_{0}, z=0$, contains the point $E^{\prime}(-1,-1,0)$, from which we obtain $$ \left(x_{0}+1\right)\left(y_{0}+1\right)=1 $$ In the same way, from the points $B$ and $C$ we get relations similar to (1) and (2) and conclude that $C$ has the coordinates $C\left(-x_{0}, 0, z_{1}\right)$. Similarly we get $D\left(0,-y_{0}, z_{2}\right)$. The condition that $A C$ is tangent to the sphere $\sigma(O, O E)$ is equivalent to $z_{1}=\sqrt{a^{2}+2}$, i.e., to $x_{0}=a \sqrt{a^{2}+2}-\left(a^{2}+2\right)$. But then (2) implies that $y_{0}=-a \sqrt{a^{2}+2}-\left(a^{2}+2\right)$ and $z_{2}=-\sqrt{a^{2}+2}$, which means that the sphere $\sigma$ is tangent to $B D$ as well. This finishes the proof.
|
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1cc01c72-fb9e-5db5-aefa-829b950f8aa6
| 23,786
|
6. (CUB) Let $P(z)$ and $Q(z)$ be complex-variable polynomials, with degree not less than 1. Let $$ P_{k}=\{z \in \mathbb{C} \mid P(z)=k\}, \quad Q_{k}=\{z \in \mathbb{C} \mid Q(z)=k\} $$ Let also $P_{0}=Q_{0}$ and $P_{1}=Q_{1}$. Prove that $P(z) \equiv Q(z)$.
|
6. Assume w.l.o.g. that $n=\operatorname{deg} P \geq \operatorname{deg} Q$, and let $P_{0}=\left\{z_{1}, z_{2}, \ldots, z_{k}\right\}$, $P_{1}=\left\{z_{k+1}, z_{k+2}, \ldots z_{k+m}\right\}$. The polynomials $P$ and $Q$ match at $k+m$ points $z_{1}, z_{2}, \ldots, z_{k+m}$; hence if we prove that $k+m>n$, the result will follow. By the assumption, $P(x)=\left(x-z_{1}\right)^{\alpha_{1}} \cdots\left(x-z_{k}\right)^{\alpha_{k}}=\left(x-z_{k+1}\right)^{\alpha_{k+1}} \cdots\left(x-z_{k+m}\right)^{\alpha_{k+m}}+1$ for some positive integers $\alpha_{1}, \ldots, \alpha_{k+m}$. Let us consider $P^{\prime}(x)$. As we know, it is divisible by $\left(x-z_{i}\right)^{\alpha_{i}-1}$ for $i=1,2, \ldots, k+m$; i.e., $$ \prod_{i=1}^{k+m}\left(x-z_{i}\right)^{\alpha_{i}-1} \mid P^{\prime}(x) $$ Therefore $2 n-k-m=\operatorname{deg} \prod_{i=1}^{k+m}\left(x-z_{i}\right)^{\alpha_{i}-1} \leq \operatorname{deg} P^{\prime}=n-1$, i.e., $k+m \geq n+1$, as we claimed.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
6. (CUB) Let $P(z)$ and $Q(z)$ be complex-variable polynomials, with degree not less than 1. Let $$ P_{k}=\{z \in \mathbb{C} \mid P(z)=k\}, \quad Q_{k}=\{z \in \mathbb{C} \mid Q(z)=k\} $$ Let also $P_{0}=Q_{0}$ and $P_{1}=Q_{1}$. Prove that $P(z) \equiv Q(z)$.
|
6. Assume w.l.o.g. that $n=\operatorname{deg} P \geq \operatorname{deg} Q$, and let $P_{0}=\left\{z_{1}, z_{2}, \ldots, z_{k}\right\}$, $P_{1}=\left\{z_{k+1}, z_{k+2}, \ldots z_{k+m}\right\}$. The polynomials $P$ and $Q$ match at $k+m$ points $z_{1}, z_{2}, \ldots, z_{k+m}$; hence if we prove that $k+m>n$, the result will follow. By the assumption, $P(x)=\left(x-z_{1}\right)^{\alpha_{1}} \cdots\left(x-z_{k}\right)^{\alpha_{k}}=\left(x-z_{k+1}\right)^{\alpha_{k+1}} \cdots\left(x-z_{k+m}\right)^{\alpha_{k+m}}+1$ for some positive integers $\alpha_{1}, \ldots, \alpha_{k+m}$. Let us consider $P^{\prime}(x)$. As we know, it is divisible by $\left(x-z_{i}\right)^{\alpha_{i}-1}$ for $i=1,2, \ldots, k+m$; i.e., $$ \prod_{i=1}^{k+m}\left(x-z_{i}\right)^{\alpha_{i}-1} \mid P^{\prime}(x) $$ Therefore $2 n-k-m=\operatorname{deg} \prod_{i=1}^{k+m}\left(x-z_{i}\right)^{\alpha_{i}-1} \leq \operatorname{deg} P^{\prime}=n-1$, i.e., $k+m \geq n+1$, as we claimed.
|
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d264eb7f-49ca-536c-a274-f9c8254a601d
| 23,796
|
13. C1 (NET 1) ${ }^{\mathrm{IMO} 2} \mathrm{~A}$ scalene triangle $A_{1} A_{2} A_{3}$ is given with sides $a_{1}, a_{2}, a_{3}$ ( $a_{i}$ is the side opposite to $A_{i}$ ). For all $i=1,2,3, M_{i}$ is the midpoint of side $a_{i}, T_{i}$ is the point where the incircle touches side $a_{i}$, and the reflection of $T_{i}$ in the interior bisector of $A_{i}$ yields the point $S_{i}$. Prove that the lines $M_{1} S_{1}, M_{2} S_{2}$, and $M_{3} S_{3}$ are concurrent.
|
13. Points $S_{1}, S_{2}, S_{3}$ clearly lie on the inscribed circle. Let $\widehat{X Y}$ denote the oriented arc $X Y$. The $\operatorname{arcs} \widehat{T_{2} S_{1}}$ and $\widehat{T_{1} T_{3}}$ are equal, since they are symmetric with respect to the bisector of $\angle A_{1}$. Similarly, $\widehat{T_{3} T_{2}}=\widehat{S_{2} T_{1}}$. Therefore $\widehat{T_{3} S_{1}}=$ $\widehat{T_{3} T_{2}}+\widehat{T_{2} S_{1}}=\widehat{S_{2} T_{1}}+\widehat{T_{1} T_{3}}=$ $S_{2} T_{3}$. It follows that $S_{1} S_{2}$ is parallel to $A_{1} A_{2}$, and consequently $S_{1} S_{2} \|$ $M_{1} M_{2}$. Analogously $S_{1} S_{3} \| M_{1} M_{3}$  and $S_{2} S_{3} \| M_{2} M_{3}$. Since the circumcircles of $\triangle M_{1} M_{2} M_{3}$ and $\triangle S_{1} S_{2} S_{3}$ are not equal, these triangles are not congruent and hence they must be homothetic. Then all the lines $M_{i} S_{i}$ pass through the center of homothety. Second solution. Set the complex plane so that the incenter of $\triangle A_{1} A_{2} A_{3}$ is the unit circle centered at the origin. Let $t_{i}, s_{i}$ respectively denote the complex numbers of modulus 1 corresponding to $T_{i}, S_{i}$. Clearly $t_{1} \overline{t_{1}}=$ $t_{2} \overline{t_{2}}=t_{3} \overline{t_{3}}=1$. Since $T_{2} T_{3}$ and $T_{1} S_{1}$ are parallel, we obtain $t_{2} t_{3}=t_{1} s_{1}$, or $s_{1}=t_{2} t_{3} \overline{t_{1}}$. Similarly $s_{2}=t_{1} t_{3} \overline{t_{2}}, s_{3}=t_{1} t_{2} \overline{t_{3}}$, from which it follows that $s_{2}-s_{3}=t_{1}\left(t_{3} \overline{t_{2}}-t_{2} \overline{t_{3}}\right)$. Since the number in parentheses is strictly imaginary, we conclude that $O T_{1} \perp S_{2} S_{3}$ and consequently $S_{2} S_{3} \| A_{2} A_{3}$. We proceed as in the first solution.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
13. C1 (NET 1) ${ }^{\mathrm{IMO} 2} \mathrm{~A}$ scalene triangle $A_{1} A_{2} A_{3}$ is given with sides $a_{1}, a_{2}, a_{3}$ ( $a_{i}$ is the side opposite to $A_{i}$ ). For all $i=1,2,3, M_{i}$ is the midpoint of side $a_{i}, T_{i}$ is the point where the incircle touches side $a_{i}$, and the reflection of $T_{i}$ in the interior bisector of $A_{i}$ yields the point $S_{i}$. Prove that the lines $M_{1} S_{1}, M_{2} S_{2}$, and $M_{3} S_{3}$ are concurrent.
|
13. Points $S_{1}, S_{2}, S_{3}$ clearly lie on the inscribed circle. Let $\widehat{X Y}$ denote the oriented arc $X Y$. The $\operatorname{arcs} \widehat{T_{2} S_{1}}$ and $\widehat{T_{1} T_{3}}$ are equal, since they are symmetric with respect to the bisector of $\angle A_{1}$. Similarly, $\widehat{T_{3} T_{2}}=\widehat{S_{2} T_{1}}$. Therefore $\widehat{T_{3} S_{1}}=$ $\widehat{T_{3} T_{2}}+\widehat{T_{2} S_{1}}=\widehat{S_{2} T_{1}}+\widehat{T_{1} T_{3}}=$ $S_{2} T_{3}$. It follows that $S_{1} S_{2}$ is parallel to $A_{1} A_{2}$, and consequently $S_{1} S_{2} \|$ $M_{1} M_{2}$. Analogously $S_{1} S_{3} \| M_{1} M_{3}$  and $S_{2} S_{3} \| M_{2} M_{3}$. Since the circumcircles of $\triangle M_{1} M_{2} M_{3}$ and $\triangle S_{1} S_{2} S_{3}$ are not equal, these triangles are not congruent and hence they must be homothetic. Then all the lines $M_{i} S_{i}$ pass through the center of homothety. Second solution. Set the complex plane so that the incenter of $\triangle A_{1} A_{2} A_{3}$ is the unit circle centered at the origin. Let $t_{i}, s_{i}$ respectively denote the complex numbers of modulus 1 corresponding to $T_{i}, S_{i}$. Clearly $t_{1} \overline{t_{1}}=$ $t_{2} \overline{t_{2}}=t_{3} \overline{t_{3}}=1$. Since $T_{2} T_{3}$ and $T_{1} S_{1}$ are parallel, we obtain $t_{2} t_{3}=t_{1} s_{1}$, or $s_{1}=t_{2} t_{3} \overline{t_{1}}$. Similarly $s_{2}=t_{1} t_{3} \overline{t_{2}}, s_{3}=t_{1} t_{2} \overline{t_{3}}$, from which it follows that $s_{2}-s_{3}=t_{1}\left(t_{3} \overline{t_{2}}-t_{2} \overline{t_{3}}\right)$. Since the number in parentheses is strictly imaginary, we conclude that $O T_{1} \perp S_{2} S_{3}$ and consequently $S_{2} S_{3} \| A_{2} A_{3}$. We proceed as in the first solution.
|
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62af918c-5e33-5eec-889a-6ce37e11d7ce
| 23,815
|
14. C2 (AUS 4) Let $A B C D$ be a convex plane quadrilateral and let $A_{1}$ denote the circumcenter of $\triangle B C D$. Define $B_{1}, C_{1}, D_{1}$ in a corresponding way. (a) Prove that either all of $A_{1}, B_{1}, C_{1}, D_{1}$ coincide in one point, or they are all distinct. Assuming the latter case, show that $A_{1}, C_{1}$ are on opposite sides of the line $B_{1} D_{1}$, and similarly, $B_{1}, D_{1}$ are on opposite sides of the line $A_{1} C_{1}$. (This establishes the convexity of the quadrilateral $A_{1} B_{1} C_{1} D_{1}$.) (b) Denote by $A_{2}$ the circumcenter of $B_{1} C_{1} D_{1}$, and define $B_{2}, C_{2}, D_{2}$ in an analogous way. Show that the quadrilateral $A_{2} B_{2} C_{2} D_{2}$ is similar to the quadrilateral $A B C D$.
|
14. (a) If any two of $A_{1}, B_{1}, C_{1}, D_{1}$ coincide, say $A_{1} \equiv B_{1}$, then $A B C D$ is inscribed in a circle centered at $A_{1}$ and hence all $A_{1}, B_{1}, C_{1}, D_{1}$ coincide. Assume now the opposite, and let w.l.o.g. $\angle D A B+\angle D C B<180^{\circ}$. Then $A$ is outside the circumcircle of $\triangle B C D$, so $A_{1} A>A_{1} C$. Similarly, $C_{1} C>C_{1} A$. Hence the perpendicular bisector $l_{A C}$ of $A C$ separates points $A_{1}$ and $C_{1}$. Since $B_{1}, D_{1}$ lie on $l_{A C}$, this means that $A_{1}$ and $C_{1}$ are on opposite sides $B_{1} D_{1}$. Similarly one can show that $B_{1}$ and $D_{1}$ are on opposite sides of $A_{1} C_{1}$. (b) Since $A_{2} B_{2} \perp C_{1} D_{1}$ and $C_{1} D_{1} \perp A B$, it follows that $A_{2} B_{2} \| A B$. Similarly $A_{2} C_{2}\left\|A C, A_{2} D_{2}\right\| A D, B_{2} C_{2}\left\|B C, B_{2} D_{2}\right\| B D$, and $C_{2} D_{2} \| C D$. Hence $\triangle A_{2} B_{2} C_{2} \sim \triangle A B C$ and $\triangle A_{2} D_{2} C_{2} \sim \triangle A D C$, and the result follows.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
14. C2 (AUS 4) Let $A B C D$ be a convex plane quadrilateral and let $A_{1}$ denote the circumcenter of $\triangle B C D$. Define $B_{1}, C_{1}, D_{1}$ in a corresponding way. (a) Prove that either all of $A_{1}, B_{1}, C_{1}, D_{1}$ coincide in one point, or they are all distinct. Assuming the latter case, show that $A_{1}, C_{1}$ are on opposite sides of the line $B_{1} D_{1}$, and similarly, $B_{1}, D_{1}$ are on opposite sides of the line $A_{1} C_{1}$. (This establishes the convexity of the quadrilateral $A_{1} B_{1} C_{1} D_{1}$.) (b) Denote by $A_{2}$ the circumcenter of $B_{1} C_{1} D_{1}$, and define $B_{2}, C_{2}, D_{2}$ in an analogous way. Show that the quadrilateral $A_{2} B_{2} C_{2} D_{2}$ is similar to the quadrilateral $A B C D$.
|
14. (a) If any two of $A_{1}, B_{1}, C_{1}, D_{1}$ coincide, say $A_{1} \equiv B_{1}$, then $A B C D$ is inscribed in a circle centered at $A_{1}$ and hence all $A_{1}, B_{1}, C_{1}, D_{1}$ coincide. Assume now the opposite, and let w.l.o.g. $\angle D A B+\angle D C B<180^{\circ}$. Then $A$ is outside the circumcircle of $\triangle B C D$, so $A_{1} A>A_{1} C$. Similarly, $C_{1} C>C_{1} A$. Hence the perpendicular bisector $l_{A C}$ of $A C$ separates points $A_{1}$ and $C_{1}$. Since $B_{1}, D_{1}$ lie on $l_{A C}$, this means that $A_{1}$ and $C_{1}$ are on opposite sides $B_{1} D_{1}$. Similarly one can show that $B_{1}$ and $D_{1}$ are on opposite sides of $A_{1} C_{1}$. (b) Since $A_{2} B_{2} \perp C_{1} D_{1}$ and $C_{1} D_{1} \perp A B$, it follows that $A_{2} B_{2} \| A B$. Similarly $A_{2} C_{2}\left\|A C, A_{2} D_{2}\right\| A D, B_{2} C_{2}\left\|B C, B_{2} D_{2}\right\| B D$, and $C_{2} D_{2} \| C D$. Hence $\triangle A_{2} B_{2} C_{2} \sim \triangle A B C$ and $\triangle A_{2} D_{2} C_{2} \sim \triangle A D C$, and the result follows.
|
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7168ac6e-7f3a-5913-89fe-895e17734829
| 23,818
|
15. C3 (CAN 5) Show that $$ \frac{1-s^{a}}{1-s} \leq(1+s)^{a-1} $$ holds for every $1 \neq s>0$ real and $0<a \leq 1$ rational.
|
15. Let $a=k / n$, where $n, k \in \mathbb{N}, n \geq k$. Putting $t^{n}=s$, the given inequality becomes $\frac{1-t^{k}}{1-t^{n}} \leq\left(1+t^{n}\right)^{k / n-1}$, or equivalently $$ \left(1+t+\cdots+t^{k-1}\right)^{n}\left(1+t^{n}\right)^{n-k} \leq\left(1+t+\cdots+t^{n-1}\right)^{n} $$ This is clearly true for $k=n$. Therefore it is enough to prove that the lefthand side of the above inequality is an increasing function of $k$. We are led to show that $\left(1+t+\cdots+t^{k-1}\right)^{n}\left(1+t^{n}\right)^{n-k} \leq\left(1+t+\cdots+t^{k}\right)^{n}\left(1+t^{n}\right)^{n-k-1}$. This is equivalent to $1+t^{n} \leq A^{n}$, where $A=\frac{1+t+\cdots+t^{k}}{1+t+\cdots+t^{k-1}}$. But this easily follows, since $$ \begin{aligned} A^{n}-t^{n} & =(A-t)\left(A^{n-1}+A^{n-2} t+\cdots+t^{n-1}\right) \\ & \geq(A-t)\left(1+t+\cdots+t^{n-1}\right)=\frac{1+t+\cdots+t^{n-1}}{1+t+\cdots+t^{k-1}} \geq 1 \end{aligned} $$ Remark. The original problem asked to prove the inequality for real $a$.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
15. C3 (CAN 5) Show that $$ \frac{1-s^{a}}{1-s} \leq(1+s)^{a-1} $$ holds for every $1 \neq s>0$ real and $0<a \leq 1$ rational.
|
15. Let $a=k / n$, where $n, k \in \mathbb{N}, n \geq k$. Putting $t^{n}=s$, the given inequality becomes $\frac{1-t^{k}}{1-t^{n}} \leq\left(1+t^{n}\right)^{k / n-1}$, or equivalently $$ \left(1+t+\cdots+t^{k-1}\right)^{n}\left(1+t^{n}\right)^{n-k} \leq\left(1+t+\cdots+t^{n-1}\right)^{n} $$ This is clearly true for $k=n$. Therefore it is enough to prove that the lefthand side of the above inequality is an increasing function of $k$. We are led to show that $\left(1+t+\cdots+t^{k-1}\right)^{n}\left(1+t^{n}\right)^{n-k} \leq\left(1+t+\cdots+t^{k}\right)^{n}\left(1+t^{n}\right)^{n-k-1}$. This is equivalent to $1+t^{n} \leq A^{n}$, where $A=\frac{1+t+\cdots+t^{k}}{1+t+\cdots+t^{k-1}}$. But this easily follows, since $$ \begin{aligned} A^{n}-t^{n} & =(A-t)\left(A^{n-1}+A^{n-2} t+\cdots+t^{n-1}\right) \\ & \geq(A-t)\left(1+t+\cdots+t^{n-1}\right)=\frac{1+t+\cdots+t^{n-1}}{1+t+\cdots+t^{k-1}} \geq 1 \end{aligned} $$ Remark. The original problem asked to prove the inequality for real $a$.
|
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1c6abac2-c636-5392-9906-22c90682f1cf
| 23,821
|
17. C5 (USS 5) The right triangles $A B C$ and $A B_{1} C_{1}$ are similar and have opposite orientation. The right angles are at $C$ and $C_{1}$, and we also have $\measuredangle C A B=\measuredangle C_{1} A B_{1}$. Let $M$ be the point of intersection of the lines $B C_{1}$ and $B_{1} C$. Prove that if the lines $A M$ and $C C_{1}$ exist, they are perpendicular.
|
17. Let $A$ be the origin of the Cartesian plane. Suppose that $B C: A C=k$ and that $(a, b)$ and $\left(a_{1}, b_{1}\right)$ are coordinates of the points $C$ and $C_{1}$, respectively. Then the coordinates of the point $B$ are $(a, b)+k(-b, a)=(a-k b, b+k a)$, while the coordinates of $B_{1}$ are $\left(a_{1}, b_{1}\right)+k\left(b_{1},-a_{1}\right)=\left(a+k b_{1}, b_{1}-k a_{1}\right)$. Thus the lines $B C_{1}$ and $C B_{1}$ are given by the equations $\frac{x-a_{1}}{y-b_{1}}=\frac{x-(a-k b)}{y-(b+k a)}$ and $\frac{x-a}{y-b}=\frac{x-\left(a_{1}+k b_{1}\right)}{y-\left(b_{1}-k a_{1}\right)}$ respectively. After multiplying, these equations transform into the forms $$ \begin{aligned} B C_{1}: & k a x+k b y & =k a a_{1}+k b b_{1}+b a_{1}-a b_{1}-\left(b-b_{1}\right) x+\left(a-a_{1}\right) y \\ C B_{1}: & k a_{1} x+k b_{1} y & =k a a_{1}+k b b_{1}+b a_{1}-a b_{1}-\left(b-b_{1}\right) x+\left(a-a_{1}\right) y . \end{aligned} $$ The coordinates $\left(x_{0}, y_{0}\right)$ of the point $M$ satisfy these equations, from which we deduce that $k a x_{0}+k b y_{0}=k a_{1} x_{0}+k b_{1} y_{0}$. This yields $\frac{x_{0}}{y_{0}}=-\frac{b_{1}-b}{a_{1}-a}$, implying that the lines $C C_{1}$ and $A M$ are perpendicular.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
17. C5 (USS 5) The right triangles $A B C$ and $A B_{1} C_{1}$ are similar and have opposite orientation. The right angles are at $C$ and $C_{1}$, and we also have $\measuredangle C A B=\measuredangle C_{1} A B_{1}$. Let $M$ be the point of intersection of the lines $B C_{1}$ and $B_{1} C$. Prove that if the lines $A M$ and $C C_{1}$ exist, they are perpendicular.
|
17. Let $A$ be the origin of the Cartesian plane. Suppose that $B C: A C=k$ and that $(a, b)$ and $\left(a_{1}, b_{1}\right)$ are coordinates of the points $C$ and $C_{1}$, respectively. Then the coordinates of the point $B$ are $(a, b)+k(-b, a)=(a-k b, b+k a)$, while the coordinates of $B_{1}$ are $\left(a_{1}, b_{1}\right)+k\left(b_{1},-a_{1}\right)=\left(a+k b_{1}, b_{1}-k a_{1}\right)$. Thus the lines $B C_{1}$ and $C B_{1}$ are given by the equations $\frac{x-a_{1}}{y-b_{1}}=\frac{x-(a-k b)}{y-(b+k a)}$ and $\frac{x-a}{y-b}=\frac{x-\left(a_{1}+k b_{1}\right)}{y-\left(b_{1}-k a_{1}\right)}$ respectively. After multiplying, these equations transform into the forms $$ \begin{aligned} B C_{1}: & k a x+k b y & =k a a_{1}+k b b_{1}+b a_{1}-a b_{1}-\left(b-b_{1}\right) x+\left(a-a_{1}\right) y \\ C B_{1}: & k a_{1} x+k b_{1} y & =k a a_{1}+k b b_{1}+b a_{1}-a b_{1}-\left(b-b_{1}\right) x+\left(a-a_{1}\right) y . \end{aligned} $$ The coordinates $\left(x_{0}, y_{0}\right)$ of the point $M$ satisfy these equations, from which we deduce that $k a x_{0}+k b y_{0}=k a_{1} x_{0}+k b_{1} y_{0}$. This yields $\frac{x_{0}}{y_{0}}=-\frac{b_{1}-b}{a_{1}-a}$, implying that the lines $C C_{1}$ and $A M$ are perpendicular.
|
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|
4db5dec3-ab28-5676-996d-b98addb12195
| 23,825
|
19. C7 (CZS 3) Let $M$ be the set of real numbers of the form $\frac{m+n}{\sqrt{m^{2}+n^{2}}}$, where $m$ and $n$ are positive integers. Prove that for every pair $x \in M$, $y \in M$ with $x<y$, there exists an element $z \in M$ such that $x<z<y$.
|
19. Let us set $x=m / n$. Since $f(x)=(m+n) / \sqrt{m^{2}+n^{2}}=(x+1) / \sqrt{1+x^{2}}$ is a continuous function of $x, f(x)$ takes all values between any two values of $f$; moreover, the corresponding $x$ can be rational. This completes the proof. Remark. Since $f$ is increasing for $x \geq 1,1 \leq x<z<y$ implies $f(x)<$ $f(z)<f(y)$.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
19. C7 (CZS 3) Let $M$ be the set of real numbers of the form $\frac{m+n}{\sqrt{m^{2}+n^{2}}}$, where $m$ and $n$ are positive integers. Prove that for every pair $x \in M$, $y \in M$ with $x<y$, there exists an element $z \in M$ such that $x<z<y$.
|
19. Let us set $x=m / n$. Since $f(x)=(m+n) / \sqrt{m^{2}+n^{2}}=(x+1) / \sqrt{1+x^{2}}$ is a continuous function of $x, f(x)$ takes all values between any two values of $f$; moreover, the corresponding $x$ can be rational. This completes the proof. Remark. Since $f$ is increasing for $x \geq 1,1 \leq x<z<y$ implies $f(x)<$ $f(z)<f(y)$.
|
{
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|
d75b223d-69e4-5242-ac27-8a00eca4c4aa
| 23,830
|
2. A2 (YUG 1) Let $K$ be a convex polygon in the plane and suppose that $K$ is positioned in the coordinate system in such a way that $$ \operatorname{area}\left(K \cap Q_{i}\right)=\frac{1}{4} \text { area } K(i=1,2,3,4,), $$ where the $Q_{i}$ denote the quadrants of the plane. Prove that if $K$ contains no nonzero lattice point, then the area of $K$ is less than 4.
|
2. Since $K$ does not contain a lattice point other than $O(0,0)$, it is bounded by four lines $u, v, w, x$ that pass through the points $U(1,0), V(0,1)$, $W(-1,0), X(0,-1)$ respectively. Let $P Q R S$ be the quadrilateral formed by these lines, where $U \in S P, V \in P Q, W \in Q R, X \in R S$. If one of the quadrants, say $Q_{1}$, contains no vertices of $P Q R S$, then $K \cap Q_{1}$ is contained in $\triangle O U V$ and hence has area less than $1 / 2$. Consequently the area of $K$ is less than 2 . Let us now suppose that $P, Q, R, S$ lie in different quadrants. One of the angles of $P Q R S$ is at least $90^{\circ}$ : let it be $\angle P$. Then $S_{U P V} \leq P U \cdot P V / 2 \leq$ $\left(P U^{2}+P V^{2}\right) / 4 \leq U V^{2} / 4=1 / 2$, which implies that $S_{K \cap Q_{1}}<S_{O U P V} \leq$ 1. Hence the area of $K$ is less than 4.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
2. A2 (YUG 1) Let $K$ be a convex polygon in the plane and suppose that $K$ is positioned in the coordinate system in such a way that $$ \operatorname{area}\left(K \cap Q_{i}\right)=\frac{1}{4} \text { area } K(i=1,2,3,4,), $$ where the $Q_{i}$ denote the quadrants of the plane. Prove that if $K$ contains no nonzero lattice point, then the area of $K$ is less than 4.
|
2. Since $K$ does not contain a lattice point other than $O(0,0)$, it is bounded by four lines $u, v, w, x$ that pass through the points $U(1,0), V(0,1)$, $W(-1,0), X(0,-1)$ respectively. Let $P Q R S$ be the quadrilateral formed by these lines, where $U \in S P, V \in P Q, W \in Q R, X \in R S$. If one of the quadrants, say $Q_{1}$, contains no vertices of $P Q R S$, then $K \cap Q_{1}$ is contained in $\triangle O U V$ and hence has area less than $1 / 2$. Consequently the area of $K$ is less than 2 . Let us now suppose that $P, Q, R, S$ lie in different quadrants. One of the angles of $P Q R S$ is at least $90^{\circ}$ : let it be $\angle P$. Then $S_{U P V} \leq P U \cdot P V / 2 \leq$ $\left(P U^{2}+P V^{2}\right) / 4 \leq U V^{2} / 4=1 / 2$, which implies that $S_{K \cap Q_{1}}<S_{O U P V} \leq$ 1. Hence the area of $K$ is less than 4.
|
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|
fd22b1b8-5585-5e9e-a4f3-9255d8eac227
| 23,832
|
20. C8 (TUN 3) Let $A B C D$ be a convex quadrilateral and draw regular triangles $A B M, C D P, B C N, A D Q$, the first two outward and the other two inward. Prove that $M N=A C$. What can be said about the quadrilateral $M N P Q$ ?
|
20. Since $M N$ is the image of $A C$ under rotation about $B$ for $60^{\circ}$, we have $M N=A C$. Similarly, $P Q$ is the image of $A C$ under rotation about $D$ through $60^{\circ}$, from which it follows that $P Q \| M N$. Hence either $M, N, P, Q$ are collinear or $M N P Q$ is a parallelogram.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
20. C8 (TUN 3) Let $A B C D$ be a convex quadrilateral and draw regular triangles $A B M, C D P, B C N, A D Q$, the first two outward and the other two inward. Prove that $M N=A C$. What can be said about the quadrilateral $M N P Q$ ?
|
20. Since $M N$ is the image of $A C$ under rotation about $B$ for $60^{\circ}$, we have $M N=A C$. Similarly, $P Q$ is the image of $A C$ under rotation about $D$ through $60^{\circ}$, from which it follows that $P Q \| M N$. Hence either $M, N, P, Q$ are collinear or $M N P Q$ is a parallelogram.
|
{
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|
ac03698c-0575-5f1a-a6e3-4ab7e11c00d4
| 23,834
|
3. A3 (USS 4) ${ }^{\mathrm{IMO} 3}$ Consider the infinite sequences $\left\{x_{n}\right\}$ of positive real numbers with the following properties: $$ x_{0}=1 \quad \text { and for all } i \geq 0, x_{i+1} \leq x_{i} $$ (a) Prove that for every such sequence there is an $n \geq 1$ such that $\frac{x_{0}^{2}}{x_{1}}+$ $\frac{x_{1}^{2}}{x_{2}}+\cdots+\frac{x_{n-1}^{2}}{x_{n}} \geq 3.999$. (b) Find such a sequence for which $\frac{x_{0}^{2}}{x_{1}}+\frac{x_{1}^{2}}{x_{2}}+\cdots+\frac{x_{n-1}^{2}}{x_{n}}<4$ for all $n$.
|
3. (a) By the Cauchy-Schwarz inequality we have $\left(x_{0}^{2} / x_{1}+\cdots+x_{n-1}^{2} / x_{n}\right)$. $\left(x_{1}+\cdots+x_{n}\right) \geq\left(x_{0}+\cdots+x_{n-1}\right)^{2}$. Let us set $X_{n-1}=x_{1}+x_{2}+$ $\cdots+x_{n-1}$. Using $x_{0}=1$, the last inequality can be rewritten as $$ \frac{x_{0}^{2}}{x_{1}}+\cdots+\frac{x_{n-1}^{2}}{x_{n}} \geq \frac{\left(1+X_{n-1}\right)^{2}}{X_{n-1}+x_{n}} \geq \frac{4 X_{n-1}}{X_{n-1}+x_{n}}=\frac{4}{1+x_{n} / X_{n-1}} $$ Since $x_{n} \leq x_{n-1} \leq \cdots \leq x_{1}$, it follows that $X_{n-1} \geq(n-1) x_{n}$. Now (1) yields $x_{0}^{2} / x_{1}+\cdots+x_{n-1}^{2} / x_{n} \geq 4(n-1) / n$, which exceeds 3.999 for $n>4000$. (b) The sequence $x_{n}=1 / 2^{n}$ obviously satisfies the required condition. Second solution to part (a). For each $n \in \mathbb{N}$, let us find a constant $c_{n}$ such that the inequality $x_{0}^{2} / x_{1}+\cdots+x_{n-1}^{2} / x_{n} \geq c_{n} x_{0}$ holds for any sequence $x_{0} \geq x_{1} \geq \cdots \geq x_{n}>0$. For $n=1$ we can take $c_{1}=1$. Assuming that $c_{n}$ exists, we have $$ \frac{x_{0}^{2}}{x_{1}}+\left(\frac{x_{1}^{2}}{x_{2}}+\cdots+\frac{x_{n}^{2}}{x_{n+1}}\right) \geq \frac{x_{0}^{2}}{x_{1}}+c_{n} x_{1} \geq 2 \sqrt{x_{0}^{2} c_{n}}=x_{0} \cdot 2 \sqrt{c_{n}} $$ Thus we can take $c_{n+1}=2 \sqrt{c_{n}}$. Then inductively $c_{n}=2^{2-1 / 2^{n-2}}$, and since $c_{n} \rightarrow 4$ as $n \rightarrow \infty$, the result follows. Third solution. Since $\left\{x_{n}\right\}$ is decreasing, there exists $\lim _{n \rightarrow \infty} x_{n}=x \geq 0$. If $x>0$, then $x_{n-1}^{2} / x_{n} \geq x_{n} \geq x$ holds for each $n$, and the result is trivial. If otherwise $x=0$, then we note that $x_{n-1}^{2} / x_{n} \geq 4\left(x_{n-1}-x_{n}\right)$ for each $n$, with equality if and only if $x_{n-1}=2 x_{n}$. Hence $$ \lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{x_{k-1}^{2}}{x_{k}} \geq \lim _{n \rightarrow \infty} \sum_{k=1}^{n} 4\left(x_{k-1}-x_{k}\right)=4 x_{0}=4 $$ Equality holds if and only if $x_{n-1}=2 x_{n}$ for all $n$, and consequently $x_{n}=1 / 2^{n}$.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
3. A3 (USS 4) ${ }^{\mathrm{IMO} 3}$ Consider the infinite sequences $\left\{x_{n}\right\}$ of positive real numbers with the following properties: $$ x_{0}=1 \quad \text { and for all } i \geq 0, x_{i+1} \leq x_{i} $$ (a) Prove that for every such sequence there is an $n \geq 1$ such that $\frac{x_{0}^{2}}{x_{1}}+$ $\frac{x_{1}^{2}}{x_{2}}+\cdots+\frac{x_{n-1}^{2}}{x_{n}} \geq 3.999$. (b) Find such a sequence for which $\frac{x_{0}^{2}}{x_{1}}+\frac{x_{1}^{2}}{x_{2}}+\cdots+\frac{x_{n-1}^{2}}{x_{n}}<4$ for all $n$.
|
3. (a) By the Cauchy-Schwarz inequality we have $\left(x_{0}^{2} / x_{1}+\cdots+x_{n-1}^{2} / x_{n}\right)$. $\left(x_{1}+\cdots+x_{n}\right) \geq\left(x_{0}+\cdots+x_{n-1}\right)^{2}$. Let us set $X_{n-1}=x_{1}+x_{2}+$ $\cdots+x_{n-1}$. Using $x_{0}=1$, the last inequality can be rewritten as $$ \frac{x_{0}^{2}}{x_{1}}+\cdots+\frac{x_{n-1}^{2}}{x_{n}} \geq \frac{\left(1+X_{n-1}\right)^{2}}{X_{n-1}+x_{n}} \geq \frac{4 X_{n-1}}{X_{n-1}+x_{n}}=\frac{4}{1+x_{n} / X_{n-1}} $$ Since $x_{n} \leq x_{n-1} \leq \cdots \leq x_{1}$, it follows that $X_{n-1} \geq(n-1) x_{n}$. Now (1) yields $x_{0}^{2} / x_{1}+\cdots+x_{n-1}^{2} / x_{n} \geq 4(n-1) / n$, which exceeds 3.999 for $n>4000$. (b) The sequence $x_{n}=1 / 2^{n}$ obviously satisfies the required condition. Second solution to part (a). For each $n \in \mathbb{N}$, let us find a constant $c_{n}$ such that the inequality $x_{0}^{2} / x_{1}+\cdots+x_{n-1}^{2} / x_{n} \geq c_{n} x_{0}$ holds for any sequence $x_{0} \geq x_{1} \geq \cdots \geq x_{n}>0$. For $n=1$ we can take $c_{1}=1$. Assuming that $c_{n}$ exists, we have $$ \frac{x_{0}^{2}}{x_{1}}+\left(\frac{x_{1}^{2}}{x_{2}}+\cdots+\frac{x_{n}^{2}}{x_{n+1}}\right) \geq \frac{x_{0}^{2}}{x_{1}}+c_{n} x_{1} \geq 2 \sqrt{x_{0}^{2} c_{n}}=x_{0} \cdot 2 \sqrt{c_{n}} $$ Thus we can take $c_{n+1}=2 \sqrt{c_{n}}$. Then inductively $c_{n}=2^{2-1 / 2^{n-2}}$, and since $c_{n} \rightarrow 4$ as $n \rightarrow \infty$, the result follows. Third solution. Since $\left\{x_{n}\right\}$ is decreasing, there exists $\lim _{n \rightarrow \infty} x_{n}=x \geq 0$. If $x>0$, then $x_{n-1}^{2} / x_{n} \geq x_{n} \geq x$ holds for each $n$, and the result is trivial. If otherwise $x=0$, then we note that $x_{n-1}^{2} / x_{n} \geq 4\left(x_{n-1}-x_{n}\right)$ for each $n$, with equality if and only if $x_{n-1}=2 x_{n}$. Hence $$ \lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{x_{k-1}^{2}}{x_{k}} \geq \lim _{n \rightarrow \infty} \sum_{k=1}^{n} 4\left(x_{k-1}-x_{k}\right)=4 x_{0}=4 $$ Equality holds if and only if $x_{n-1}=2 x_{n}$ for all $n$, and consequently $x_{n}=1 / 2^{n}$.
|
{
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ddd9f2b7-480d-550e-b20c-8ceb23d83a7d
| 23,837
|
6. A6 (VIE 1) ${ }^{\text {IMO6 }}$ Let $S$ be a square with sides of length 100 and let $L$ be a path within $S$ that does not meet itself and that is composed of linear segments $A_{0} A_{1}, A_{1} A_{2}, \ldots, A_{n-1} A_{n}$ with $A_{0} \neq A_{n}$. Suppose that for every point $P$ of the boundary of $S$ there is a point of $L$ at a distance from $P$ not greater than $\frac{1}{2}$. Prove that there are two points $X$ and $Y$ in $L$ such that the distance between $X$ and $Y$ is not greater than 1 and the length of that part of $L$ that lies between $X$ and $Y$ is not smaller than 198.
|
6. Denote by $d(U, V)$ the distance between points or sets of points $U$ and $V$. For $P, Q \in L$ we shall denote by $L_{P Q}$ the part of $L$ between points $P$ and $Q$ and by $l_{P Q}$ the length of this part. Let us denote by $S_{i}(i=1,2,3,4)$ the vertices of $S$ and by $T_{i}$ points of $L$ such that $S_{i} T_{i} \leq 1 / 2$ in such a way that $l_{A_{0} T_{1}}$ is the least of the $l_{A_{0} T_{i}}$ 's, $S_{2}$ and $S_{4}$ are neighbors of $S_{1}$, and $l_{A_{0} T_{2}}<l_{A_{0} T_{4}}$. Now we shall consider the points of the segment $S_{1} S_{4}$. Let $D$ and $E$ be the sets of points defined as follows: $D=\left\{X \in\left[S_{1} S_{4}\right] \mid d\left(X, L_{A_{0} T_{2}}\right) \leq 1 / 2\right\}$ and $E=\left\{X \in\left[S_{1} S_{4}\right] \mid d\left(X, L_{T_{2} A_{n}}\right) \leq 1 / 2\right\}$. Clearly $D$ and $E$ are closed, nonempty (indeed, $S_{1} \in D$ and $S_{4} \in E$ ) subsets of [ $\left.S_{1} S_{4}\right]$. Since their union is a connected set $S_{1} S_{4}$, it follows that they must have a nonempty intersection. Let $P \in D \cap E$. Then there exist points $X \in L_{A_{0} T_{2}}$ and $Y \in L_{T_{2} A_{n}}$ such that $d(P, X) \leq 1 / 2, d(P, Y) \leq 1 / 2$, and consequently $d(X, Y) \leq 1$. On the other hand, $T_{2}$ lies between $X$ and $Y$ on $L$, and thus $L_{X Y}=L_{X T_{2}}+L_{T_{2} Y} \geq X T_{2}+T_{2} Y \geq\left(P S_{2}-X P-S_{2} T_{2}\right)+\left(P S_{2}-Y P-\right.$ $\left.S_{2} T_{2}\right) \geq 99+99=198$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
6. A6 (VIE 1) ${ }^{\text {IMO6 }}$ Let $S$ be a square with sides of length 100 and let $L$ be a path within $S$ that does not meet itself and that is composed of linear segments $A_{0} A_{1}, A_{1} A_{2}, \ldots, A_{n-1} A_{n}$ with $A_{0} \neq A_{n}$. Suppose that for every point $P$ of the boundary of $S$ there is a point of $L$ at a distance from $P$ not greater than $\frac{1}{2}$. Prove that there are two points $X$ and $Y$ in $L$ such that the distance between $X$ and $Y$ is not greater than 1 and the length of that part of $L$ that lies between $X$ and $Y$ is not smaller than 198.
|
6. Denote by $d(U, V)$ the distance between points or sets of points $U$ and $V$. For $P, Q \in L$ we shall denote by $L_{P Q}$ the part of $L$ between points $P$ and $Q$ and by $l_{P Q}$ the length of this part. Let us denote by $S_{i}(i=1,2,3,4)$ the vertices of $S$ and by $T_{i}$ points of $L$ such that $S_{i} T_{i} \leq 1 / 2$ in such a way that $l_{A_{0} T_{1}}$ is the least of the $l_{A_{0} T_{i}}$ 's, $S_{2}$ and $S_{4}$ are neighbors of $S_{1}$, and $l_{A_{0} T_{2}}<l_{A_{0} T_{4}}$. Now we shall consider the points of the segment $S_{1} S_{4}$. Let $D$ and $E$ be the sets of points defined as follows: $D=\left\{X \in\left[S_{1} S_{4}\right] \mid d\left(X, L_{A_{0} T_{2}}\right) \leq 1 / 2\right\}$ and $E=\left\{X \in\left[S_{1} S_{4}\right] \mid d\left(X, L_{T_{2} A_{n}}\right) \leq 1 / 2\right\}$. Clearly $D$ and $E$ are closed, nonempty (indeed, $S_{1} \in D$ and $S_{4} \in E$ ) subsets of [ $\left.S_{1} S_{4}\right]$. Since their union is a connected set $S_{1} S_{4}$, it follows that they must have a nonempty intersection. Let $P \in D \cap E$. Then there exist points $X \in L_{A_{0} T_{2}}$ and $Y \in L_{T_{2} A_{n}}$ such that $d(P, X) \leq 1 / 2, d(P, Y) \leq 1 / 2$, and consequently $d(X, Y) \leq 1$. On the other hand, $T_{2}$ lies between $X$ and $Y$ on $L$, and thus $L_{X Y}=L_{X T_{2}}+L_{T_{2} Y} \geq X T_{2}+T_{2} Y \geq\left(P S_{2}-X P-S_{2} T_{2}\right)+\left(P S_{2}-Y P-\right.$ $\left.S_{2} T_{2}\right) \geq 99+99=198$.
|
{
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|
4105efbf-6626-5425-87d5-42f480a50dbf
| 23,844
|
7. B1 (CAN 2) Let $p(x)$ be a cubic polynomial with integer coefficients with leading coefficient 1 and with one of its roots equal to the product of the other two. Show that $2 p(-1)$ is a multiple of $p(1)+p(-1)-2(1+p(0))$.
|
7. Let $a, b, a b$ be the roots of the cubic polynomial $P(x)=(x-a)(x-b)(x-$ $a b)$. Observe that $$ \begin{aligned} 2 p(-1) & =-2(1+a)(1+b)(1+a b) \\ p(1)+p(-1)-2(1+p(0)) & =-2(1+a)(1+b) \end{aligned} $$ The statement of the problem is trivial if both the expressions are equal to zero. Otherwise, the quotient $\frac{2 p(-1)}{p(1)+p(-1)-2(1+p(0))}=1+a b$ is rational and consequently $a b$ is rational. But since $(a b)^{2}=-P(0)$ is an integer, it follows that $a b$ is also an integer. This completes the proof.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
7. B1 (CAN 2) Let $p(x)$ be a cubic polynomial with integer coefficients with leading coefficient 1 and with one of its roots equal to the product of the other two. Show that $2 p(-1)$ is a multiple of $p(1)+p(-1)-2(1+p(0))$.
|
7. Let $a, b, a b$ be the roots of the cubic polynomial $P(x)=(x-a)(x-b)(x-$ $a b)$. Observe that $$ \begin{aligned} 2 p(-1) & =-2(1+a)(1+b)(1+a b) \\ p(1)+p(-1)-2(1+p(0)) & =-2(1+a)(1+b) \end{aligned} $$ The statement of the problem is trivial if both the expressions are equal to zero. Otherwise, the quotient $\frac{2 p(-1)}{p(1)+p(-1)-2(1+p(0))}=1+a b$ is rational and consequently $a b$ is rational. But since $(a b)^{2}=-P(0)$ is an integer, it follows that $a b$ is also an integer. This completes the proof.
|
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|
e9b4375e-7957-51c2-951d-614281d080b6
| 23,847
|
8. B2 (POL 4) A convex, closed figure lies inside a given circle. The figure is seen from every point of the circumference at a right angle (that is, the two rays drawn from the point and supporting the convex figure are perpendicular). Prove that the center of the circle is a center of symmetry of the figure.
|
8. Let $\mathcal{F}$ be the given figure. Consider any chord $A B$ of the circumcircle $\gamma$ that supports $\mathcal{F}$. The other supporting lines to $\mathcal{F}$ from $A$ and $B$ intersect $\gamma$ again at $D$ and $C$ respectively so that $\angle D A B=\angle A B C=90^{\circ}$. Then $A B C D$ is a rectangle, and hence $C D$ must support $\mathcal{F}$ as well, from which it follows that $\mathcal{F}$ is inscribed in the rectangle $A B C D$ touching each of its sides. We easily conclude that $\mathcal{F}$ is the intersection of all such rectangles. Now, since the center $O$ of $\gamma$ is the center of symmetry of all these rectangles, it must be so for their intersection $\mathcal{F}$ as well.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
8. B2 (POL 4) A convex, closed figure lies inside a given circle. The figure is seen from every point of the circumference at a right angle (that is, the two rays drawn from the point and supporting the convex figure are perpendicular). Prove that the center of the circle is a center of symmetry of the figure.
|
8. Let $\mathcal{F}$ be the given figure. Consider any chord $A B$ of the circumcircle $\gamma$ that supports $\mathcal{F}$. The other supporting lines to $\mathcal{F}$ from $A$ and $B$ intersect $\gamma$ again at $D$ and $C$ respectively so that $\angle D A B=\angle A B C=90^{\circ}$. Then $A B C D$ is a rectangle, and hence $C D$ must support $\mathcal{F}$ as well, from which it follows that $\mathcal{F}$ is inscribed in the rectangle $A B C D$ touching each of its sides. We easily conclude that $\mathcal{F}$ is the intersection of all such rectangles. Now, since the center $O$ of $\gamma$ is the center of symmetry of all these rectangles, it must be so for their intersection $\mathcal{F}$ as well.
|
{
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|
1cfa1d54-ec3b-5baa-824e-56d5468ffdab
| 23,848
|
9. B3 (GBR 1) Let $A B C$ be a triangle, and let $P$ be a point inside it such that $\measuredangle P A C=\measuredangle P B C$. The perpendiculars from $P$ to $B C$ and $C A$ meet these lines at $L$ and $M$, respectively, and $D$ is the midpoint of $A B$. Prove that $D L=D M$.
|
9. Let $X$ and $Y$ be the midpoints of the segments $A P$ and $B P$. Then $D Y P X$ is a parallelogram. Since $X$ and $Y$ are the circumcenters of $\triangle A P M$ and $\triangle B P L$, it follows that $X M=$ $X P=D Y$ and $Y L=Y P=D X$. Furthermore, $\angle D X M=\angle D X P+$ $\angle P X M=\angle D X P+2 \angle P A M=$ $\angle D Y P+2 \angle P B L=\angle D Y P+$ $\angle P Y L=\angle D Y L$. Therefore, the triangles $D X M$ and $L Y D$ are con-  gruent, implying $D M=D L$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
9. B3 (GBR 1) Let $A B C$ be a triangle, and let $P$ be a point inside it such that $\measuredangle P A C=\measuredangle P B C$. The perpendiculars from $P$ to $B C$ and $C A$ meet these lines at $L$ and $M$, respectively, and $D$ is the midpoint of $A B$. Prove that $D L=D M$.
|
9. Let $X$ and $Y$ be the midpoints of the segments $A P$ and $B P$. Then $D Y P X$ is a parallelogram. Since $X$ and $Y$ are the circumcenters of $\triangle A P M$ and $\triangle B P L$, it follows that $X M=$ $X P=D Y$ and $Y L=Y P=D X$. Furthermore, $\angle D X M=\angle D X P+$ $\angle P X M=\angle D X P+2 \angle P A M=$ $\angle D Y P+2 \angle P B L=\angle D Y P+$ $\angle P Y L=\angle D Y L$. Therefore, the triangles $D X M$ and $L Y D$ are con-  gruent, implying $D M=D L$.
|
{
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|
1ec4d865-fb4d-54fa-909d-038cbbe1cb2c
| 23,850
|
1. (AUS 1) The localities $P_{1}, P_{2}, \ldots, P_{1983}$ are served by ten international airlines $A_{1}, A_{2}, \ldots, A_{10}$. It is noticed that there is direct service (without stops) between any two of these localities and that all airline schedules offer round-trip flights. Prove that at least one of the airlines can offer a round trip with an odd number of landings.
|
1. Suppose that there are $n$ airlines $A_{1}, \ldots, A_{n}$ and $N>2^{n}$ cities. We shall prove that there is a round trip by at least one $A_{i}$ containing an odd number of stops. For $n=1$ the statement is trivial, since one airline serves at least 3 cities and hence $P_{1} P_{2} P_{3} P_{1}$ is a round trip with 3 landings. We use induction on $n$, and assume that $n>1$. Suppose the contrary, that all round trips by $A_{n}$ consist of an even number of stops. Then we can separate the cities into two nonempty classes $Q=\left\{Q_{1}, \ldots, Q_{r}\right\}$ and $R=\left\{R_{1}, \ldots, R_{s}\right\}$ (where $r+s=N$ ), so that each flight by $A_{n}$ runs between a $Q$-city and an $R$-city. (Indeed, take any city $Q_{1}$ served by $A_{n}$; include each city linked to $Q_{1}$ by $A_{n}$ in $R$, then include in $Q$ each city linked by $A_{n}$ to any $R$-city, etc. Since all round trips are even, no contradiction can arise.) At least one of $r, s$ is larger than $2^{n-1}$, say $r>2^{n-1}$. But, only $A_{1}, \ldots, A_{n-1}$ run between cities in $\left\{Q_{1}, \ldots, Q_{r}\right\}$; hence by the induction hypothesis at least one of them flies a round trip with an odd number of landings, a contradiction. It only remains to notice that for $n=10,2^{n}=1024<1983$. Remark. If there are $N=2^{n}$ cities, there is a schedule with $n$ airlines that contain no odd round trip by any of the airlines. Let the cities be $P_{k}$, $k=0, \ldots, 2^{n}-1$, and write $k$ in the binary system as an $n$-digit number $\overline{a_{1} \ldots a_{n}}$ (e.g., $1=(0 \ldots 001)_{2}$ ). Link $P_{k}$ and $P_{l}$ by $A_{i}$ if the $i$ th digits $k$ and $l$ are distinct but the first $i-1$ digits are the same. All round trips under $A_{i}$ are even, since the $i$ th digit alternates.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
1. (AUS 1) The localities $P_{1}, P_{2}, \ldots, P_{1983}$ are served by ten international airlines $A_{1}, A_{2}, \ldots, A_{10}$. It is noticed that there is direct service (without stops) between any two of these localities and that all airline schedules offer round-trip flights. Prove that at least one of the airlines can offer a round trip with an odd number of landings.
|
1. Suppose that there are $n$ airlines $A_{1}, \ldots, A_{n}$ and $N>2^{n}$ cities. We shall prove that there is a round trip by at least one $A_{i}$ containing an odd number of stops. For $n=1$ the statement is trivial, since one airline serves at least 3 cities and hence $P_{1} P_{2} P_{3} P_{1}$ is a round trip with 3 landings. We use induction on $n$, and assume that $n>1$. Suppose the contrary, that all round trips by $A_{n}$ consist of an even number of stops. Then we can separate the cities into two nonempty classes $Q=\left\{Q_{1}, \ldots, Q_{r}\right\}$ and $R=\left\{R_{1}, \ldots, R_{s}\right\}$ (where $r+s=N$ ), so that each flight by $A_{n}$ runs between a $Q$-city and an $R$-city. (Indeed, take any city $Q_{1}$ served by $A_{n}$; include each city linked to $Q_{1}$ by $A_{n}$ in $R$, then include in $Q$ each city linked by $A_{n}$ to any $R$-city, etc. Since all round trips are even, no contradiction can arise.) At least one of $r, s$ is larger than $2^{n-1}$, say $r>2^{n-1}$. But, only $A_{1}, \ldots, A_{n-1}$ run between cities in $\left\{Q_{1}, \ldots, Q_{r}\right\}$; hence by the induction hypothesis at least one of them flies a round trip with an odd number of landings, a contradiction. It only remains to notice that for $n=10,2^{n}=1024<1983$. Remark. If there are $N=2^{n}$ cities, there is a schedule with $n$ airlines that contain no odd round trip by any of the airlines. Let the cities be $P_{k}$, $k=0, \ldots, 2^{n}-1$, and write $k$ in the binary system as an $n$-digit number $\overline{a_{1} \ldots a_{n}}$ (e.g., $1=(0 \ldots 001)_{2}$ ). Link $P_{k}$ and $P_{l}$ by $A_{i}$ if the $i$ th digits $k$ and $l$ are distinct but the first $i-1$ digits are the same. All round trips under $A_{i}$ are even, since the $i$ th digit alternates.
|
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|
4efccb15-7a41-53b1-8757-e8a77f14f90c
| 23,851
|
10. (FIN 1) Let $p$ and $q$ be integers. Show that there exists an interval $I$ of length $1 / q$ and a polynomial $P$ with integral coefficients such that $$ \left|P(x)-\frac{p}{q}\right|<\frac{1}{q^{2}} $$ for all $x \in I$.
|
10. Choose $P(x)=\frac{p}{q}\left((q x-1)^{2 n+1}+1\right), I=[1 / 2 q, 3 / 2 q]$. Then all the coefficients of $P$ are integers, and $$ \left|P(x)-\frac{p}{q}\right|=\left|\frac{p}{q}(q x-1)^{2 n+1}\right| \leq\left|\frac{p}{q}\right| \frac{1}{2^{2 n+1}}, $$ for $x \in I$. The desired inequality follows if $n$ is chosen large enough.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
10. (FIN 1) Let $p$ and $q$ be integers. Show that there exists an interval $I$ of length $1 / q$ and a polynomial $P$ with integral coefficients such that $$ \left|P(x)-\frac{p}{q}\right|<\frac{1}{q^{2}} $$ for all $x \in I$.
|
10. Choose $P(x)=\frac{p}{q}\left((q x-1)^{2 n+1}+1\right), I=[1 / 2 q, 3 / 2 q]$. Then all the coefficients of $P$ are integers, and $$ \left|P(x)-\frac{p}{q}\right|=\left|\frac{p}{q}(q x-1)^{2 n+1}\right| \leq\left|\frac{p}{q}\right| \frac{1}{2^{2 n+1}}, $$ for $x \in I$. The desired inequality follows if $n$ is chosen large enough.
|
{
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|
3c2f63ce-27ac-5379-aac6-f00f45a17f82
| 23,854
|
11. (FIN $\mathbf{2}^{\prime}$ ) Let $f:[0,1] \rightarrow \mathbb{R}$ be continuous and satisfy: $$ \begin{aligned} b f(2 x) & =f(x), & & 0 \leq x \leq 1 / 2 \\ f(x) & =b+(1-b) f(2 x-1), & & 1 / 2 \leq x \leq 1 \end{aligned} $$ where $b=\frac{1+c}{2+c}, c>0$. Show that $0<f(x)-x<c$ for every $x, 0<x<1$.
|
11. First suppose that the binary representation of $x$ is finite: $x=0, a_{1} a_{2} \ldots a_{n}$ $=\sum_{j=1}^{n} a_{j} 2^{-j}, a_{i} \in\{0,1\}$. We shall prove by induction on $n$ that $$ f(x)=\sum_{j=1}^{n} b_{0} \ldots b_{j-1} a_{j}, \quad \text { where } b_{k}=\left\{\begin{array}{lr} -b & \text { if } a_{k}=0 \\ 1-b & \text { if } a_{k}=1 \end{array}\right. $$ (Here $a_{0}=0$.) Indeed, by the recursion formula, $a_{1}=0 \Rightarrow f(x)=b f\left(\sum_{j=1}^{n-1} a_{j+1} 2^{-j}\right)=b \sum_{j=1}^{n-1} b_{1} \ldots b_{j} a_{j+1}$ hence $f(x)=$ $\sum_{j=0}^{n-1} b_{0} \ldots b_{j} a_{j+1}$ as $b_{0}=b_{1}=b ;$ $a_{1}=1 \Rightarrow f(x)=b+(1-b) f\left(\sum_{j=1}^{n-1} a_{j+1} 2^{-j}\right)=\sum_{j=0}^{n-1} b_{0} \ldots b_{j} a_{j+1}$, as $b_{0}=b, b_{1}=1-b$. Clearly, $f(0)=0, f(1)=1, f(1 / 2)=b>1 / 2$. Assume $x=\sum_{j=0}^{n} a_{j} 2^{-j}$, and for $k \geq 2, v=x+2^{-n-k+1}, u=x+2^{-n-k}=(v+x) / 2$. Then $f(v)=$ $f(x)+b_{0} \ldots b_{n} b^{k-2}$ and $f(u)=f(x)+b_{0} \ldots b_{n} b^{k-1}>(f(v)+f(x)) / 2$. This means that the point $(u, f(u))$ lies above the line joining $(x, f(x))$ and $(v, f(v))$. By induction, every $(x, f(x))$, where $x$ has a finite binary expansion, lies above the line joining $(0,0)$ and $(1 / 2, b)$ if $0<x<1 / 2$, or above the line joining $(1 / 2, b)$ and $(1,1)$ if $1 / 2<x<1$. It follows immediately that $f(x)>x$. For the second inequality, observe that $$ \begin{aligned} f(x)-x & =\sum_{j=1}^{\infty}\left(b_{0} \ldots b_{j-1}-2^{-j}\right) a_{j} \\ & <\sum_{j=1}^{\infty}\left(b^{j}-2^{-j}\right) a_{j}<\sum_{j=1}^{\infty}\left(b^{j}-2^{-j}\right)=\frac{b}{1-b}-1=c . \end{aligned} $$ By continuity, these inequalities also hold for $x$ with infinite binary representations.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
11. (FIN $\mathbf{2}^{\prime}$ ) Let $f:[0,1] \rightarrow \mathbb{R}$ be continuous and satisfy: $$ \begin{aligned} b f(2 x) & =f(x), & & 0 \leq x \leq 1 / 2 \\ f(x) & =b+(1-b) f(2 x-1), & & 1 / 2 \leq x \leq 1 \end{aligned} $$ where $b=\frac{1+c}{2+c}, c>0$. Show that $0<f(x)-x<c$ for every $x, 0<x<1$.
|
11. First suppose that the binary representation of $x$ is finite: $x=0, a_{1} a_{2} \ldots a_{n}$ $=\sum_{j=1}^{n} a_{j} 2^{-j}, a_{i} \in\{0,1\}$. We shall prove by induction on $n$ that $$ f(x)=\sum_{j=1}^{n} b_{0} \ldots b_{j-1} a_{j}, \quad \text { where } b_{k}=\left\{\begin{array}{lr} -b & \text { if } a_{k}=0 \\ 1-b & \text { if } a_{k}=1 \end{array}\right. $$ (Here $a_{0}=0$.) Indeed, by the recursion formula, $a_{1}=0 \Rightarrow f(x)=b f\left(\sum_{j=1}^{n-1} a_{j+1} 2^{-j}\right)=b \sum_{j=1}^{n-1} b_{1} \ldots b_{j} a_{j+1}$ hence $f(x)=$ $\sum_{j=0}^{n-1} b_{0} \ldots b_{j} a_{j+1}$ as $b_{0}=b_{1}=b ;$ $a_{1}=1 \Rightarrow f(x)=b+(1-b) f\left(\sum_{j=1}^{n-1} a_{j+1} 2^{-j}\right)=\sum_{j=0}^{n-1} b_{0} \ldots b_{j} a_{j+1}$, as $b_{0}=b, b_{1}=1-b$. Clearly, $f(0)=0, f(1)=1, f(1 / 2)=b>1 / 2$. Assume $x=\sum_{j=0}^{n} a_{j} 2^{-j}$, and for $k \geq 2, v=x+2^{-n-k+1}, u=x+2^{-n-k}=(v+x) / 2$. Then $f(v)=$ $f(x)+b_{0} \ldots b_{n} b^{k-2}$ and $f(u)=f(x)+b_{0} \ldots b_{n} b^{k-1}>(f(v)+f(x)) / 2$. This means that the point $(u, f(u))$ lies above the line joining $(x, f(x))$ and $(v, f(v))$. By induction, every $(x, f(x))$, where $x$ has a finite binary expansion, lies above the line joining $(0,0)$ and $(1 / 2, b)$ if $0<x<1 / 2$, or above the line joining $(1 / 2, b)$ and $(1,1)$ if $1 / 2<x<1$. It follows immediately that $f(x)>x$. For the second inequality, observe that $$ \begin{aligned} f(x)-x & =\sum_{j=1}^{\infty}\left(b_{0} \ldots b_{j-1}-2^{-j}\right) a_{j} \\ & <\sum_{j=1}^{\infty}\left(b^{j}-2^{-j}\right) a_{j}<\sum_{j=1}^{\infty}\left(b^{j}-2^{-j}\right)=\frac{b}{1-b}-1=c . \end{aligned} $$ By continuity, these inequalities also hold for $x$ with infinite binary representations.
|
{
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|
9e03eff3-f356-5304-9f27-aeed85dc3b3e
| 23,856
|
14. (POL 2) ${ }^{\mathrm{IMO} 5}$ Prove or disprove: From the interval $[1, \ldots, 30000]$ one can select a set of 1000 integers containing no arithmetic triple (three consecutive numbers of an arithmetic progression).
|
14. Let $T_{n}$ be the set of all nonnegative integers whose ternary representations consist of at most $n$ digits and do not contain a digit 2 . The cardinality of $T_{n}$ is $2^{n}$, and the greatest integer in $T_{n}$ is $11 \ldots 1=3^{0}+3^{1}+\cdots+3^{n-1}=$ $\left(3^{n}-1\right) / 2$. We claim that there is no arithmetic triple in $T_{n}$. To see this, suppose $x, y, z \in T_{n}$ and $2 y=x+z$. Then $2 y$ has only 0 's and 2's in its ternary representation, and a number of this form can be the sum of two integers $x, z \in T_{n}$ in only one way, namely $x=z=y$. But $\left|T_{10}\right|=2^{10}=1024$ and $\max T_{10}=\left(3^{10}-1\right) / 2=29524<30000$. Thus the answer is yes.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
14. (POL 2) ${ }^{\mathrm{IMO} 5}$ Prove or disprove: From the interval $[1, \ldots, 30000]$ one can select a set of 1000 integers containing no arithmetic triple (three consecutive numbers of an arithmetic progression).
|
14. Let $T_{n}$ be the set of all nonnegative integers whose ternary representations consist of at most $n$ digits and do not contain a digit 2 . The cardinality of $T_{n}$ is $2^{n}$, and the greatest integer in $T_{n}$ is $11 \ldots 1=3^{0}+3^{1}+\cdots+3^{n-1}=$ $\left(3^{n}-1\right) / 2$. We claim that there is no arithmetic triple in $T_{n}$. To see this, suppose $x, y, z \in T_{n}$ and $2 y=x+z$. Then $2 y$ has only 0 's and 2's in its ternary representation, and a number of this form can be the sum of two integers $x, z \in T_{n}$ in only one way, namely $x=z=y$. But $\left|T_{10}\right|=2^{10}=1024$ and $\max T_{10}=\left(3^{10}-1\right) / 2=29524<30000$. Thus the answer is yes.
|
{
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|
1996de62-1e8f-58a2-bfd0-43b96cbb00eb
| 23,862
|
15. (POL 3) Decide whether there exists a set $M$ of natural numbers satisfying the following conditions: (i) For any natural number $m>1$ there are $a, b \in M$ such that $a+b=m$. (ii) If $a, b, c, d \in M, a, b, c, d>10$ and $a+b=c+d$, then $a=c$ or $a=d$.
|
15. There is no such set. Suppose $M$ satisfies $(a)$ and $(b)$ and let $q_{n}=$ $|\{a \in M: a \leq n\}|$. Consider the differences $b-a$, where $a, b \in M$ and $10<a<b \leq k$. They are all positive and less than $k$, and (b) implies that they are $\left(\frac{q_{k}}{-} \frac{q_{10}}{2}\right)$ different integers. Hence $\binom{q_{k}-q_{10}}{2}<k$, so $q_{k} \leq \sqrt{2 k}+10$. It follows from (a) that among the numbers of the form $a+b$, where $a, b \in M, a \leq b \leq n$, or $a \leq n<b \leq 2 n$, there are all integers from the interval $[2,2 n+1]$. Thus $\binom{q_{n}+1}{2}+q_{n}\left(q_{2 n}-q_{n}\right) \geq 2 n$ for every $n \in \mathbb{N}$. Set $Q_{k}=\sqrt{2 k}+10$. We have $$ \begin{aligned} \binom{q_{n}+1}{2}+q_{n}\left(q_{2 n}-q_{n}\right) & =\frac{1}{2} q_{n}+\frac{1}{2} q_{n}\left(2 q_{2 n}-q_{n}\right) \\ & \leq \frac{1}{2} q_{n}+\frac{1}{2} q_{n}\left(2 Q_{2 n}-q_{n}\right) \\ & \leq \frac{1}{2} Q_{n}+\frac{1}{2} Q_{n}\left(2 Q_{2 n}-Q_{n}\right) \\ & \leq 2(\sqrt{2}-1) n+(20+\sqrt{2} / 2) \sqrt{n}+55 \end{aligned} $$ which is less than $n$ for $n$ large enough, a contradiction.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
15. (POL 3) Decide whether there exists a set $M$ of natural numbers satisfying the following conditions: (i) For any natural number $m>1$ there are $a, b \in M$ such that $a+b=m$. (ii) If $a, b, c, d \in M, a, b, c, d>10$ and $a+b=c+d$, then $a=c$ or $a=d$.
|
15. There is no such set. Suppose $M$ satisfies $(a)$ and $(b)$ and let $q_{n}=$ $|\{a \in M: a \leq n\}|$. Consider the differences $b-a$, where $a, b \in M$ and $10<a<b \leq k$. They are all positive and less than $k$, and (b) implies that they are $\left(\frac{q_{k}}{-} \frac{q_{10}}{2}\right)$ different integers. Hence $\binom{q_{k}-q_{10}}{2}<k$, so $q_{k} \leq \sqrt{2 k}+10$. It follows from (a) that among the numbers of the form $a+b$, where $a, b \in M, a \leq b \leq n$, or $a \leq n<b \leq 2 n$, there are all integers from the interval $[2,2 n+1]$. Thus $\binom{q_{n}+1}{2}+q_{n}\left(q_{2 n}-q_{n}\right) \geq 2 n$ for every $n \in \mathbb{N}$. Set $Q_{k}=\sqrt{2 k}+10$. We have $$ \begin{aligned} \binom{q_{n}+1}{2}+q_{n}\left(q_{2 n}-q_{n}\right) & =\frac{1}{2} q_{n}+\frac{1}{2} q_{n}\left(2 q_{2 n}-q_{n}\right) \\ & \leq \frac{1}{2} q_{n}+\frac{1}{2} q_{n}\left(2 Q_{2 n}-q_{n}\right) \\ & \leq \frac{1}{2} Q_{n}+\frac{1}{2} Q_{n}\left(2 Q_{2 n}-Q_{n}\right) \\ & \leq 2(\sqrt{2}-1) n+(20+\sqrt{2} / 2) \sqrt{n}+55 \end{aligned} $$ which is less than $n$ for $n$ large enough, a contradiction.
|
{
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|
47e66581-0b42-5bf7-a3e1-8f362c39dc0f
| 23,864
|
16. (GDR 1) Let $F(n)$ be the set of polynomials $P(x)=a_{0}+a_{1} x+\cdots+a_{n} x^{n}$, with $a_{0}, a_{1}, \ldots, a_{n} \in \mathbb{R}$ and $0 \leq a_{0}=a_{n} \leq a_{1}=a_{n-1} \leq \cdots \leq a_{[n / 2]}=$ $a_{[(n+1) / 2]}$. Prove that if $f \in F(m)$ and $g \in F(n)$, then $f g \in F(m+n)$.
|
16. Set $h_{n, i}(x)=x^{i}+\cdots+x^{n-i}, 2 i \leq n$. The set $F(n)$ is the set of linear combinations with nonnegative coefficients of the $h_{n, i}$ 's. This is a convex cone. Hence, it suffices to prove that $h_{n, i} h_{m, j} \in F(m+n)$. Indeed, setting $p=n-2 i$ and $q=m-2 j$ and assuming $p \leq q$ we obtain $$ h_{n, i}(x) h_{m, j}(x)=\left(x^{i}+\cdots+x^{i+p}\right)\left(x^{j}+\cdots+x^{j+q}\right)=\sum_{k=i+j}^{n-i+j} h_{m+n, k} $$ which proves the claim.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
16. (GDR 1) Let $F(n)$ be the set of polynomials $P(x)=a_{0}+a_{1} x+\cdots+a_{n} x^{n}$, with $a_{0}, a_{1}, \ldots, a_{n} \in \mathbb{R}$ and $0 \leq a_{0}=a_{n} \leq a_{1}=a_{n-1} \leq \cdots \leq a_{[n / 2]}=$ $a_{[(n+1) / 2]}$. Prove that if $f \in F(m)$ and $g \in F(n)$, then $f g \in F(m+n)$.
|
16. Set $h_{n, i}(x)=x^{i}+\cdots+x^{n-i}, 2 i \leq n$. The set $F(n)$ is the set of linear combinations with nonnegative coefficients of the $h_{n, i}$ 's. This is a convex cone. Hence, it suffices to prove that $h_{n, i} h_{m, j} \in F(m+n)$. Indeed, setting $p=n-2 i$ and $q=m-2 j$ and assuming $p \leq q$ we obtain $$ h_{n, i}(x) h_{m, j}(x)=\left(x^{i}+\cdots+x^{i+p}\right)\left(x^{j}+\cdots+x^{j+q}\right)=\sum_{k=i+j}^{n-i+j} h_{m+n, k} $$ which proves the claim.
|
{
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|
70dee911-2d86-56bc-8c39-085b29a02f60
| 23,868
|
18. (FRG 3) ${ }^{\mathrm{IMO} 3}$ Let $a, b, c$ be positive integers satisfying $(a, b)=(b, c)=$ $(c, a)=1$. Show that $2 a b c-a b-b c-c a$ is the largest integer not representable as $$ x b c+y c a+z a b $$ with nonnegative integers $x, y, z$.
|
18. Let $\left(x_{0}, y_{0}, z_{0}\right)$ be one solution of $b c x+c a y+a b z=n$ (not necessarily nonnegative). By subtracting $b c x_{0}+c a y_{0}+a b z_{0}=n$ we get $$ b c\left(x-x_{0}\right)+c a\left(y-y_{0}\right)+a b\left(z-z_{0}\right)=0 . $$ Since $(a, b)=(a, c)=1$, we must have $a \mid x-x_{0}$ or $x-x_{0}=a s$. Substituting this in the last equation gives $$ b c s+c\left(y-y_{0}\right)+b\left(z-z_{0}\right)=0 $$ Since $(b, c)=1$, we have $b \mid y-y_{0}$ or $y-y_{0}=b t$. If we substitute this in the last equation we get $b c s+b c t+b\left(z-z_{0}\right)=0$, or $c s+c t+z-z_{0}=0$, or $z-z_{0}=-c(s+t)$. In $x=x_{0}+a s$ and $y=y_{0}+b t$, we can choose $s$ and $t$ such that $0 \leq x \leq a-1$ and $0 \leq y \leq b-1$. If $n>2 a b c-b c-c a-a b$, then $a b z=n-b c x-a c y>2 a b c-a b-b c-c a-b c(a-1)-c a(b-1)=-a b$ or $z>-1$, i.e., $z \geq 0$. Hence, it is representable as $b c x+c a y+a b z$ with $x, y, z \geq 0$. Now we prove that $2 a b c-b c-c a-a b$ is not representable as $b c x+c a y+a b z$ with $x, y, z \geq 0$. Suppose that $b c x+c a y+a b z=2 a b c-a b-b c-c a$ with $x, y, z \geq 0$. Then $$ b c(x+1)+c a(y+1)+a b(z+1)=2 a b c $$ with $x+1, y+1, z+1 \geq 1$. Since $(a, b)=(a, c)=1$, we have $a \mid x+1$ and thus $a \leq x+1$. Similarly $b \leq y+1$ and $c \leq z+1$. Thus $b c a+c a b+a b c \leq 2 a b c$, a contradiction.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
18. (FRG 3) ${ }^{\mathrm{IMO} 3}$ Let $a, b, c$ be positive integers satisfying $(a, b)=(b, c)=$ $(c, a)=1$. Show that $2 a b c-a b-b c-c a$ is the largest integer not representable as $$ x b c+y c a+z a b $$ with nonnegative integers $x, y, z$.
|
18. Let $\left(x_{0}, y_{0}, z_{0}\right)$ be one solution of $b c x+c a y+a b z=n$ (not necessarily nonnegative). By subtracting $b c x_{0}+c a y_{0}+a b z_{0}=n$ we get $$ b c\left(x-x_{0}\right)+c a\left(y-y_{0}\right)+a b\left(z-z_{0}\right)=0 . $$ Since $(a, b)=(a, c)=1$, we must have $a \mid x-x_{0}$ or $x-x_{0}=a s$. Substituting this in the last equation gives $$ b c s+c\left(y-y_{0}\right)+b\left(z-z_{0}\right)=0 $$ Since $(b, c)=1$, we have $b \mid y-y_{0}$ or $y-y_{0}=b t$. If we substitute this in the last equation we get $b c s+b c t+b\left(z-z_{0}\right)=0$, or $c s+c t+z-z_{0}=0$, or $z-z_{0}=-c(s+t)$. In $x=x_{0}+a s$ and $y=y_{0}+b t$, we can choose $s$ and $t$ such that $0 \leq x \leq a-1$ and $0 \leq y \leq b-1$. If $n>2 a b c-b c-c a-a b$, then $a b z=n-b c x-a c y>2 a b c-a b-b c-c a-b c(a-1)-c a(b-1)=-a b$ or $z>-1$, i.e., $z \geq 0$. Hence, it is representable as $b c x+c a y+a b z$ with $x, y, z \geq 0$. Now we prove that $2 a b c-b c-c a-a b$ is not representable as $b c x+c a y+a b z$ with $x, y, z \geq 0$. Suppose that $b c x+c a y+a b z=2 a b c-a b-b c-c a$ with $x, y, z \geq 0$. Then $$ b c(x+1)+c a(y+1)+a b(z+1)=2 a b c $$ with $x+1, y+1, z+1 \geq 1$. Since $(a, b)=(a, c)=1$, we have $a \mid x+1$ and thus $a \leq x+1$. Similarly $b \leq y+1$ and $c \leq z+1$. Thus $b c a+c a b+a b c \leq 2 a b c$, a contradiction.
|
{
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|
5545b5ff-9213-5c2c-9312-ee3a82cb83a7
| 23,872
|
2. (BEL 1) Let $n$ be a positive integer. Let $\sigma(n)$ be the sum of the natural divisors $d$ of $n$ (including 1 and $n$ ). We say that an integer $m \geq 1$ is superabundant (P.Erdös, 1944) if $\forall k \in\{1,2, \ldots, m-1\}, \frac{\sigma(m)}{m}>\frac{\sigma(k)}{k}$. Prove that there exists an infinity of superabundant numbers.
|
2. By definition, $\sigma(n)=\sum_{d \mid n} d=\sum_{d \mid n} n / d=n \sum_{d \mid n} 1 / d$, hence $\sigma(n) / n=$ $\sum_{d \mid n} 1 / d$. In particular, $\sigma(n!) / n!=\sum_{d \mid n!} 1 / d \geq \sum_{k=1}^{n} 1 / k$. It follows that the sequence $\sigma(n) / n$ is unbounded, and consequently there exist an infinite number of integers $n$ such that $\sigma(n) / n$ is strictly greater than $\sigma(k) / k$ for $k<n$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
2. (BEL 1) Let $n$ be a positive integer. Let $\sigma(n)$ be the sum of the natural divisors $d$ of $n$ (including 1 and $n$ ). We say that an integer $m \geq 1$ is superabundant (P.Erdös, 1944) if $\forall k \in\{1,2, \ldots, m-1\}, \frac{\sigma(m)}{m}>\frac{\sigma(k)}{k}$. Prove that there exists an infinity of superabundant numbers.
|
2. By definition, $\sigma(n)=\sum_{d \mid n} d=\sum_{d \mid n} n / d=n \sum_{d \mid n} 1 / d$, hence $\sigma(n) / n=$ $\sum_{d \mid n} 1 / d$. In particular, $\sigma(n!) / n!=\sum_{d \mid n!} 1 / d \geq \sum_{k=1}^{n} 1 / k$. It follows that the sequence $\sigma(n) / n$ is unbounded, and consequently there exist an infinite number of integers $n$ such that $\sigma(n) / n$ is strictly greater than $\sigma(k) / k$ for $k<n$.
|
{
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|
1b7a1695-773a-5a5b-8d5a-fa9dfa38ac47
| 23,877
|
22. (SWE 4) Let $n$ be a positive integer having at least two different prime factors. Show that there exists a permutation $a_{1}, a_{2}, \ldots, a_{n}$ of the integers $1,2, \ldots, n$ such that $$ \sum_{k=1}^{n} k \cdot \cos \frac{2 \pi a_{k}}{n}=0 $$
|
22. Decompose $n$ into $n=s t$, where the greatest common divisor of $s$ and $t$ is 1 and where $s>1$ and $t>1$. For $1 \leq k \leq n$ put $k=v s+u$, where $0 \leq v \leq t-1$ and $1 \leq u \leq s$, and let $a_{k}=a_{v s+u}$ be the unique integer in the set $\{1,2,3, \ldots, n\}$ such that $v s+u t-a_{v s+u}$ is a multiple of $n$. To prove that this construction gives a permutation, assume that $a_{k_{1}}=a_{k_{2}}$, where $k_{i}=v_{i} s+u_{i}, i=1,2$. Then $\left(v_{1}-v_{2}\right) s+\left(u_{1}-u_{2}\right) t$ is a multiple of $n=s t$. It follows that $t$ divides $\left(v_{1}-v_{2}\right)$, while $\left|v_{1}-v_{2}\right| \leq t-1$, and that $s$ divides $\left(u_{1}-u_{2}\right)$, while $\left|u_{1}-u_{2}\right| \leq s-1$. Hence, $v_{1}=v_{2}, u_{1}=u_{2}$, and $k_{1}=k_{2}$. We have proved that $a_{1}, \ldots, a_{n}$ is a permutation of $\{1,2, \ldots, n\}$ and hence $$ \sum_{k=1}^{n} k \cos \frac{2 \pi a_{k}}{n}=\sum_{v=0}^{t-1}\left(\sum_{u=1}^{s}(v s+u) \cos \left(\frac{2 \pi v}{t}+\frac{2 \pi u}{s}\right)\right) $$ Using $\sum_{u=1}^{s} \cos (2 \pi u / s)=\sum_{u=1}^{s} \sin (2 \pi u / s)=0$ and the additive formulas for cosine, one finds that $$ \begin{aligned} \sum_{k=1}^{n} k \cos \frac{2 \pi a_{k}}{n}= & \sum_{v=0}^{t-1}\left(\cos \frac{2 \pi v}{t} \sum_{u=1}^{s} u \cos \frac{2 \pi u}{s}-\sin \frac{2 \pi v}{t} \sum_{u=1}^{s} u \sin \frac{2 \pi u}{s}\right) \\ = & \left(\sum_{u=1}^{s} u \cos \frac{2 \pi u}{s}\right)\left(\sum_{v=0}^{t-1} \cos \frac{2 \pi v}{t}\right) \\ & -\left(\sum_{u=1}^{s} u \sin \frac{2 \pi u}{s}\right)\left(\sum_{v=0}^{t-1} \sin \frac{2 \pi v}{t}\right)=0 \end{aligned} $$
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
22. (SWE 4) Let $n$ be a positive integer having at least two different prime factors. Show that there exists a permutation $a_{1}, a_{2}, \ldots, a_{n}$ of the integers $1,2, \ldots, n$ such that $$ \sum_{k=1}^{n} k \cdot \cos \frac{2 \pi a_{k}}{n}=0 $$
|
22. Decompose $n$ into $n=s t$, where the greatest common divisor of $s$ and $t$ is 1 and where $s>1$ and $t>1$. For $1 \leq k \leq n$ put $k=v s+u$, where $0 \leq v \leq t-1$ and $1 \leq u \leq s$, and let $a_{k}=a_{v s+u}$ be the unique integer in the set $\{1,2,3, \ldots, n\}$ such that $v s+u t-a_{v s+u}$ is a multiple of $n$. To prove that this construction gives a permutation, assume that $a_{k_{1}}=a_{k_{2}}$, where $k_{i}=v_{i} s+u_{i}, i=1,2$. Then $\left(v_{1}-v_{2}\right) s+\left(u_{1}-u_{2}\right) t$ is a multiple of $n=s t$. It follows that $t$ divides $\left(v_{1}-v_{2}\right)$, while $\left|v_{1}-v_{2}\right| \leq t-1$, and that $s$ divides $\left(u_{1}-u_{2}\right)$, while $\left|u_{1}-u_{2}\right| \leq s-1$. Hence, $v_{1}=v_{2}, u_{1}=u_{2}$, and $k_{1}=k_{2}$. We have proved that $a_{1}, \ldots, a_{n}$ is a permutation of $\{1,2, \ldots, n\}$ and hence $$ \sum_{k=1}^{n} k \cos \frac{2 \pi a_{k}}{n}=\sum_{v=0}^{t-1}\left(\sum_{u=1}^{s}(v s+u) \cos \left(\frac{2 \pi v}{t}+\frac{2 \pi u}{s}\right)\right) $$ Using $\sum_{u=1}^{s} \cos (2 \pi u / s)=\sum_{u=1}^{s} \sin (2 \pi u / s)=0$ and the additive formulas for cosine, one finds that $$ \begin{aligned} \sum_{k=1}^{n} k \cos \frac{2 \pi a_{k}}{n}= & \sum_{v=0}^{t-1}\left(\cos \frac{2 \pi v}{t} \sum_{u=1}^{s} u \cos \frac{2 \pi u}{s}-\sin \frac{2 \pi v}{t} \sum_{u=1}^{s} u \sin \frac{2 \pi u}{s}\right) \\ = & \left(\sum_{u=1}^{s} u \cos \frac{2 \pi u}{s}\right)\left(\sum_{v=0}^{t-1} \cos \frac{2 \pi v}{t}\right) \\ & -\left(\sum_{u=1}^{s} u \sin \frac{2 \pi u}{s}\right)\left(\sum_{v=0}^{t-1} \sin \frac{2 \pi v}{t}\right)=0 \end{aligned} $$
|
{
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|
66fdde35-355c-544d-9648-09d978e2f7d0
| 23,883
|
23. (USS 1) ${ }^{\mathrm{IMO} 2}$ Let $K$ be one of the two intersection points of the circles $W_{1}$ and $W_{2}$. Let $O_{1}$ and $O_{2}$ be the centers of $W_{1}$ and $W_{2}$. The two common tangents to the circles meet $W_{1}$ and $W_{2}$ respectively in $P_{1}$ and $P_{2}$, the first tangent, and $Q_{1}$ and $Q_{2}$, the second tangent. Let $M_{1}$ and $M_{2}$ be the midpoints of $P_{1} Q_{1}$ and $P_{2} Q_{2}$, respectively. Prove that $$ \angle O_{1} K O_{2}=\angle M_{1} K M_{2} $$
|
23. We note that $\angle O_{1} K O_{2}=\angle M_{1} K M_{2}$ is equivalent to $\angle O_{1} K M_{1}=$ $\angle O_{2} K M_{2}$. Let $S$ be the intersection point of the common tangents, and let $L$ be the second point of intersection of $S K$ and $W_{1}$. Since $\triangle S O_{1} P_{1} \sim \triangle S P_{1} M_{1}$, we have $S K$. $S L=S P_{1}^{2}=S O_{1} \cdot S M_{1}$ which implies that points $O_{1}, L, K, M_{1}$ lie on a circle. Hence $\angle O_{1} K M_{1}=$ $\angle O_{1} L M_{1}=\angle O_{2} K M_{2}$. 
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
23. (USS 1) ${ }^{\mathrm{IMO} 2}$ Let $K$ be one of the two intersection points of the circles $W_{1}$ and $W_{2}$. Let $O_{1}$ and $O_{2}$ be the centers of $W_{1}$ and $W_{2}$. The two common tangents to the circles meet $W_{1}$ and $W_{2}$ respectively in $P_{1}$ and $P_{2}$, the first tangent, and $Q_{1}$ and $Q_{2}$, the second tangent. Let $M_{1}$ and $M_{2}$ be the midpoints of $P_{1} Q_{1}$ and $P_{2} Q_{2}$, respectively. Prove that $$ \angle O_{1} K O_{2}=\angle M_{1} K M_{2} $$
|
23. We note that $\angle O_{1} K O_{2}=\angle M_{1} K M_{2}$ is equivalent to $\angle O_{1} K M_{1}=$ $\angle O_{2} K M_{2}$. Let $S$ be the intersection point of the common tangents, and let $L$ be the second point of intersection of $S K$ and $W_{1}$. Since $\triangle S O_{1} P_{1} \sim \triangle S P_{1} M_{1}$, we have $S K$. $S L=S P_{1}^{2}=S O_{1} \cdot S M_{1}$ which implies that points $O_{1}, L, K, M_{1}$ lie on a circle. Hence $\angle O_{1} K M_{1}=$ $\angle O_{1} L M_{1}=\angle O_{2} K M_{2}$. 
|
{
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|
0c2b2a43-d2cf-5373-8453-877727fc85c7
| 23,885
|
25. (USS 3) Prove that every partition of 3-dimensional space into three disjoint subsets has the following property: One of these subsets contains all possible distances; i.e., for every $a \in \mathbb{R}_{+}$, there are points $M$ and $N$ inside that subset such that distance between $M$ and $N$ is exactly $a$.
|
25. Suppose the contrary, that $\mathbb{R}^{3}=P_{1} \cup P_{2} \cup P_{3}$ is a partition such that $a_{1} \in \mathbb{R}^{+}$is not realized by $P_{1}, a_{2} \in \mathbb{R}^{+}$is not realized by $P_{2}$ and $a_{3} \in \mathbb{R}^{+}$ not realized by $P_{3}$, where w.l.o.g. $a_{1} \geq a_{2} \geq a_{3}$. If $P_{1}=\emptyset=P_{2}$, then $P_{3}=\mathbb{R}^{3}$, which is impossible. If $P_{1}=\emptyset$, and $X \in P_{2}$, the sphere centered at $X$ with radius $a_{2}$ is included in $P_{3}$ and $a_{3} \leq a_{2}$ is realized, which is impossible. If $P_{1} \neq \emptyset$, let $X_{1} \in P_{1}$. The sphere $S$ centered in $X_{1}$, of radius $a_{1}$ is included in $P_{2} \cap P_{3}$. Since $a_{1} \geq a_{3}, S \not \subset P_{3}$. Let $X_{2} \in P_{2} \cap S$. The circle $\left\{Y \in S \mid d\left(X_{2}, Y\right)=a_{2}\right\}$ is included in $P_{3}$, but $a_{2} \leq a_{1}$; hence it has radius $r=a_{2} \sqrt{1-a_{2}^{2} /\left(4 a_{1}^{2}\right)} \geq a_{2} \sqrt{3} / 2$ and $a_{3} \leq a_{2} \leq a_{2} \sqrt{3}<2 r$; hence $a_{3}$ is realized by $P_{3}$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
25. (USS 3) Prove that every partition of 3-dimensional space into three disjoint subsets has the following property: One of these subsets contains all possible distances; i.e., for every $a \in \mathbb{R}_{+}$, there are points $M$ and $N$ inside that subset such that distance between $M$ and $N$ is exactly $a$.
|
25. Suppose the contrary, that $\mathbb{R}^{3}=P_{1} \cup P_{2} \cup P_{3}$ is a partition such that $a_{1} \in \mathbb{R}^{+}$is not realized by $P_{1}, a_{2} \in \mathbb{R}^{+}$is not realized by $P_{2}$ and $a_{3} \in \mathbb{R}^{+}$ not realized by $P_{3}$, where w.l.o.g. $a_{1} \geq a_{2} \geq a_{3}$. If $P_{1}=\emptyset=P_{2}$, then $P_{3}=\mathbb{R}^{3}$, which is impossible. If $P_{1}=\emptyset$, and $X \in P_{2}$, the sphere centered at $X$ with radius $a_{2}$ is included in $P_{3}$ and $a_{3} \leq a_{2}$ is realized, which is impossible. If $P_{1} \neq \emptyset$, let $X_{1} \in P_{1}$. The sphere $S$ centered in $X_{1}$, of radius $a_{1}$ is included in $P_{2} \cap P_{3}$. Since $a_{1} \geq a_{3}, S \not \subset P_{3}$. Let $X_{2} \in P_{2} \cap S$. The circle $\left\{Y \in S \mid d\left(X_{2}, Y\right)=a_{2}\right\}$ is included in $P_{3}$, but $a_{2} \leq a_{1}$; hence it has radius $r=a_{2} \sqrt{1-a_{2}^{2} /\left(4 a_{1}^{2}\right)} \geq a_{2} \sqrt{3} / 2$ and $a_{3} \leq a_{2} \leq a_{2} \sqrt{3}<2 r$; hence $a_{3}$ is realized by $P_{3}$.
|
{
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|
453d5589-d68f-515c-919c-06ca88b7ef19
| 23,889
|
4. (BEL 5) On the sides of the triangle $A B C$, three similar isosceles triangles $A B P(A P=P B), A Q C(A Q=Q C)$, and $B R C(B R=R C)$ are constructed. The first two are constructed externally to the triangle $A B C$, but the third is placed in the same half-plane determined by the line $B C$ as the triangle $A B C$. Prove that $A P R Q$ is a parallelogram.
|
4. The rotational homothety centered at $C$ that sends $B$ to $R$ also sends $A$ to $Q$; hence the triangles $A B C$ and $Q R C$ are similar. For the same reason, $\triangle A B C$ and $\triangle P B R$ are similar. Moreover, $B R=C R$; hence $\triangle C R Q \cong$ $\triangle R B P$. Thus $P R=Q C=A Q$ and $Q R=P B=P A$, so $A P Q R$ is a parallelogram.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
4. (BEL 5) On the sides of the triangle $A B C$, three similar isosceles triangles $A B P(A P=P B), A Q C(A Q=Q C)$, and $B R C(B R=R C)$ are constructed. The first two are constructed externally to the triangle $A B C$, but the third is placed in the same half-plane determined by the line $B C$ as the triangle $A B C$. Prove that $A P R Q$ is a parallelogram.
|
4. The rotational homothety centered at $C$ that sends $B$ to $R$ also sends $A$ to $Q$; hence the triangles $A B C$ and $Q R C$ are similar. For the same reason, $\triangle A B C$ and $\triangle P B R$ are similar. Moreover, $B R=C R$; hence $\triangle C R Q \cong$ $\triangle R B P$. Thus $P R=Q C=A Q$ and $Q R=P B=P A$, so $A P Q R$ is a parallelogram.
|
{
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|
d450821d-6ed3-562a-8fd4-cdac402f03a6
| 23,894
|
6. (CAN 2) Suppose that $\left\{x_{1}, x_{2}, \ldots, x_{n}\right\}$ are positive integers for which $x_{1}+x_{2}+\cdots+x_{n}=2(n+1)$. Show that there exists an integer $r$ with $0 \leq r \leq n-1$ for which the following $n-1$ inequalities hold: $$ \begin{aligned} x_{r+1}+\cdots+x_{r+i} & \leq 2 i+1 & & \forall i, 1 \leq i \leq n-r ; \\ x_{r+1}+\cdots+x_{n}+x_{1}+\cdots+x_{i} & \leq 2(n-r+i)+1 & & \forall i, 1 \leq i \leq r-1 . \end{aligned} $$ Prove that if all the inequalities are strict, then $r$ is unique and that otherwise there are exactly two such $r$.
|
6. The existence of $r$ : Let $S=\left\{x_{1}+x_{2}+\cdots+x_{i}-2 i \mid i=1,2, \ldots, n\right\}$. Let $\max S$ be attained for the first time at $r^{\prime}$. If $r^{\prime}=n$, then $x_{1}+x_{2}+\cdots+x_{i}-2 i<2$ for $1 \leq i \leq n-1$, so one can take $r=r^{\prime}$. Suppose that $r^{\prime}<n$. Then for $l<n-r^{\prime}$ we have $x_{r^{\prime}+1}+x_{r^{\prime}+2}+\cdots+x_{r^{\prime}+l}=$ $\left(x_{1}+\cdots+x_{r^{\prime}+l}-2\left(r^{\prime}+l\right)\right)-\left(x_{1}+\cdots+x_{r^{\prime}}-2 r^{\prime}\right)+2 l \leq 2 l$; also, for $i<r^{\prime}$ we have $\left(x_{r^{\prime}+1}+\cdots+x_{n}\right)+\left(x_{1}+\cdots+x_{i}-2 i\right)<\left(x_{r^{\prime}+1}+\cdots+\right.$ $\left.x_{n}\right)+\left(x_{1}+\cdots+x_{r^{\prime}}-2 r^{\prime}\right)=\left(x_{1}+\cdots+x_{n}\right)-2 r^{\prime}=2\left(n-r^{\prime}\right)+2 \Rightarrow$ $x_{r^{\prime}+1}+\cdots+x_{n}+x_{1}+\cdots+x_{i} \leq 2\left(n+i-r^{\prime}\right)+1$, so we can again take $r=r^{\prime}$ 。 For the second part of the problem, we relabel the sequence so that $r=0$ works. Suppose that the inequalities are strict. We have $x_{1}+x_{2}+\cdots+x_{k} \leq 2 k$, $k=1, \ldots, n-1$. Now, $2 n+2=\left(x_{1}+\cdots+x_{k}\right)+\left(x_{k+1}+\cdots+x_{n}\right) \leq$ $2 k+x_{k+1}+\cdots+x_{n} \Rightarrow x_{k+1}+\cdots+x_{n} \geq 2(n-k)+2>2(n-k)+1$. So we cannot begin with $x_{k+1}$ for any $k>0$. Now assume that there is an equality for some $k$. There are two cases: (i) Suppose $x_{1}+x_{2}+\cdots+x_{i} \leq 2 i(i=1, \ldots, k)$ and $x_{1}+\cdots+x_{k}=2 k+1$, $x_{1}+\cdots+x_{k+l} \leq 2(k+l)+1(1 \leq l \leq n-1-k)$. For $i \leq k-1$ we have $x_{i+1}+\cdots+x_{n}=2(n+1)-\left(x_{1}+\cdots+x_{i}\right)>2(n-i)+1$, so we cannot take $r=i$. If there is a $j \geq 1$ such that $x_{1}+x_{2}+\cdots+x_{k+j} \leq 2(k+j)$, then also $x_{k+j+1}+\cdots+x_{n}>2(n-k-j)+1$. If $(\forall j \geq 1) x_{1}+\cdots+x_{k+j}=$ $2(k+j)+1$, then $x_{n}=3$ and $x_{k+1}=\cdots=x_{n-1}=2$. In this case we directly verify that we cannot take $r=k+j$. However, we can also take $r=k$ : for $k+l \leq n-1, x_{k+1}+\cdots+x_{k+l} \leq 2(k+l)+1-(2 k+1)=2 l$, also $x_{k+1}+\cdots+x_{n}=2(n-k)+1$, and moreover $x_{1} \leq 2, x_{1}+x_{2} \leq 4, \ldots$. (ii) Suppose $x_{1}+\cdots+x_{i} \leq 2 i(1 \leq i \leq n-2)$ and $x_{1}+\cdots+x_{n-1}=2 n-1$. Then we can obviously take $r=n-1$. On the other hand, for any $1 \leq i \leq n-2, x_{i+1}+\cdots+x_{n-1}+x_{n}=\left(x_{1}+\cdots+x_{n-1}\right)-\left(x_{1}+\cdots+\right.$ $\left.x_{i}\right)+3>2(n-i)+1$, so we cannot take another $r \neq 0$.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
6. (CAN 2) Suppose that $\left\{x_{1}, x_{2}, \ldots, x_{n}\right\}$ are positive integers for which $x_{1}+x_{2}+\cdots+x_{n}=2(n+1)$. Show that there exists an integer $r$ with $0 \leq r \leq n-1$ for which the following $n-1$ inequalities hold: $$ \begin{aligned} x_{r+1}+\cdots+x_{r+i} & \leq 2 i+1 & & \forall i, 1 \leq i \leq n-r ; \\ x_{r+1}+\cdots+x_{n}+x_{1}+\cdots+x_{i} & \leq 2(n-r+i)+1 & & \forall i, 1 \leq i \leq r-1 . \end{aligned} $$ Prove that if all the inequalities are strict, then $r$ is unique and that otherwise there are exactly two such $r$.
|
6. The existence of $r$ : Let $S=\left\{x_{1}+x_{2}+\cdots+x_{i}-2 i \mid i=1,2, \ldots, n\right\}$. Let $\max S$ be attained for the first time at $r^{\prime}$. If $r^{\prime}=n$, then $x_{1}+x_{2}+\cdots+x_{i}-2 i<2$ for $1 \leq i \leq n-1$, so one can take $r=r^{\prime}$. Suppose that $r^{\prime}<n$. Then for $l<n-r^{\prime}$ we have $x_{r^{\prime}+1}+x_{r^{\prime}+2}+\cdots+x_{r^{\prime}+l}=$ $\left(x_{1}+\cdots+x_{r^{\prime}+l}-2\left(r^{\prime}+l\right)\right)-\left(x_{1}+\cdots+x_{r^{\prime}}-2 r^{\prime}\right)+2 l \leq 2 l$; also, for $i<r^{\prime}$ we have $\left(x_{r^{\prime}+1}+\cdots+x_{n}\right)+\left(x_{1}+\cdots+x_{i}-2 i\right)<\left(x_{r^{\prime}+1}+\cdots+\right.$ $\left.x_{n}\right)+\left(x_{1}+\cdots+x_{r^{\prime}}-2 r^{\prime}\right)=\left(x_{1}+\cdots+x_{n}\right)-2 r^{\prime}=2\left(n-r^{\prime}\right)+2 \Rightarrow$ $x_{r^{\prime}+1}+\cdots+x_{n}+x_{1}+\cdots+x_{i} \leq 2\left(n+i-r^{\prime}\right)+1$, so we can again take $r=r^{\prime}$ 。 For the second part of the problem, we relabel the sequence so that $r=0$ works. Suppose that the inequalities are strict. We have $x_{1}+x_{2}+\cdots+x_{k} \leq 2 k$, $k=1, \ldots, n-1$. Now, $2 n+2=\left(x_{1}+\cdots+x_{k}\right)+\left(x_{k+1}+\cdots+x_{n}\right) \leq$ $2 k+x_{k+1}+\cdots+x_{n} \Rightarrow x_{k+1}+\cdots+x_{n} \geq 2(n-k)+2>2(n-k)+1$. So we cannot begin with $x_{k+1}$ for any $k>0$. Now assume that there is an equality for some $k$. There are two cases: (i) Suppose $x_{1}+x_{2}+\cdots+x_{i} \leq 2 i(i=1, \ldots, k)$ and $x_{1}+\cdots+x_{k}=2 k+1$, $x_{1}+\cdots+x_{k+l} \leq 2(k+l)+1(1 \leq l \leq n-1-k)$. For $i \leq k-1$ we have $x_{i+1}+\cdots+x_{n}=2(n+1)-\left(x_{1}+\cdots+x_{i}\right)>2(n-i)+1$, so we cannot take $r=i$. If there is a $j \geq 1$ such that $x_{1}+x_{2}+\cdots+x_{k+j} \leq 2(k+j)$, then also $x_{k+j+1}+\cdots+x_{n}>2(n-k-j)+1$. If $(\forall j \geq 1) x_{1}+\cdots+x_{k+j}=$ $2(k+j)+1$, then $x_{n}=3$ and $x_{k+1}=\cdots=x_{n-1}=2$. In this case we directly verify that we cannot take $r=k+j$. However, we can also take $r=k$ : for $k+l \leq n-1, x_{k+1}+\cdots+x_{k+l} \leq 2(k+l)+1-(2 k+1)=2 l$, also $x_{k+1}+\cdots+x_{n}=2(n-k)+1$, and moreover $x_{1} \leq 2, x_{1}+x_{2} \leq 4, \ldots$. (ii) Suppose $x_{1}+\cdots+x_{i} \leq 2 i(1 \leq i \leq n-2)$ and $x_{1}+\cdots+x_{n-1}=2 n-1$. Then we can obviously take $r=n-1$. On the other hand, for any $1 \leq i \leq n-2, x_{i+1}+\cdots+x_{n-1}+x_{n}=\left(x_{1}+\cdots+x_{n-1}\right)-\left(x_{1}+\cdots+\right.$ $\left.x_{i}\right)+3>2(n-i)+1$, so we cannot take another $r \neq 0$.
|
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|
5b4f9753-f7af-530c-a25a-13def9f4bcea
| 23,896
|
7. (CAN 5) Let $a$ be a positive integer and let $\left\{a_{n}\right\}$ be defined by $a_{0}=0$ and $$ a_{n+1}=\left(a_{n}+1\right) a+(a+1) a_{n}+2 \sqrt{a(a+1) a_{n}\left(a_{n}+1\right)} \quad(n=1,2 \ldots) $$ Show that for each positive integer $n, a_{n}$ is a positive integer.
|
7. Clearly, each $a_{n}$ is positive and $\sqrt{a_{n+1}}=\sqrt{a_{n}} \sqrt{a+1}+\sqrt{a_{n}+1} \sqrt{a}$. Notice that $\sqrt{a_{n+1}+1}=\sqrt{a+1} \sqrt{a_{n}+1}+\sqrt{a} \sqrt{a_{n}}$. Therefore $$ \begin{aligned} & (\sqrt{a+1}-\sqrt{a})\left(\sqrt{a_{n}+1}-\sqrt{a_{n}}\right) \\ & \quad=\left(\sqrt{a+1} \sqrt{a_{n}+1}+\sqrt{a} \sqrt{a_{n}}\right)-\left(\sqrt{a_{n}} \sqrt{a+1}+\sqrt{a_{n}+1} \sqrt{a}\right) \\ & \quad=\sqrt{a_{n+1}+1}-\sqrt{a_{n+1}} \end{aligned} $$ By induction, $\sqrt{a_{n+1}}-\sqrt{a_{n}}=(\sqrt{a+1}-\sqrt{a})^{n}$. Similarly, $\sqrt{a_{n+1}}+\sqrt{a_{n}}=$ $(\sqrt{a+1}+\sqrt{a})^{n}$. Hence, $$ \sqrt{a_{n}}=\frac{1}{2}\left[(\sqrt{a+1}+\sqrt{a})^{n}-(\sqrt{a+1}-\sqrt{a})^{n}\right] $$ from which the result follows.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
7. (CAN 5) Let $a$ be a positive integer and let $\left\{a_{n}\right\}$ be defined by $a_{0}=0$ and $$ a_{n+1}=\left(a_{n}+1\right) a+(a+1) a_{n}+2 \sqrt{a(a+1) a_{n}\left(a_{n}+1\right)} \quad(n=1,2 \ldots) $$ Show that for each positive integer $n, a_{n}$ is a positive integer.
|
7. Clearly, each $a_{n}$ is positive and $\sqrt{a_{n+1}}=\sqrt{a_{n}} \sqrt{a+1}+\sqrt{a_{n}+1} \sqrt{a}$. Notice that $\sqrt{a_{n+1}+1}=\sqrt{a+1} \sqrt{a_{n}+1}+\sqrt{a} \sqrt{a_{n}}$. Therefore $$ \begin{aligned} & (\sqrt{a+1}-\sqrt{a})\left(\sqrt{a_{n}+1}-\sqrt{a_{n}}\right) \\ & \quad=\left(\sqrt{a+1} \sqrt{a_{n}+1}+\sqrt{a} \sqrt{a_{n}}\right)-\left(\sqrt{a_{n}} \sqrt{a+1}+\sqrt{a_{n}+1} \sqrt{a}\right) \\ & \quad=\sqrt{a_{n+1}+1}-\sqrt{a_{n+1}} \end{aligned} $$ By induction, $\sqrt{a_{n+1}}-\sqrt{a_{n}}=(\sqrt{a+1}-\sqrt{a})^{n}$. Similarly, $\sqrt{a_{n+1}}+\sqrt{a_{n}}=$ $(\sqrt{a+1}+\sqrt{a})^{n}$. Hence, $$ \sqrt{a_{n}}=\frac{1}{2}\left[(\sqrt{a+1}+\sqrt{a})^{n}-(\sqrt{a+1}-\sqrt{a})^{n}\right] $$ from which the result follows.
|
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|
090a17ca-979a-5435-8f2d-d34ceec6687a
| 23,898
|
9. (USA 3) ${ }^{\mathrm{IMO} 6}$ If $a, b$, and $c$ are sides of a triangle, prove that $$ a^{2} b(a-b)+b^{2} c(b-c)+c^{2} a(c-a) \geq 0 $$ Determine when there is equality.
|
9. For any triangle of sides $a, b, c$ there exist 3 nonnegative numbers $x, y, z$ such that $a=y+z, b=z+x, c=x+y$ (these numbers correspond to the division of the sides of a triangle by the point of contact of the incircle). The inequality becomes $$ (y+z)^{2}(z+x)(y-x)+(z+x)^{2}(x+y)(z-y)+(x+y)^{2}(y+z)(x-z) \geq 0 $$ Expanding, we get $x y^{3}+y z^{3}+z x^{3} \geq x y z(x+y+z)$. This follows from Cauchy's inequality $\left(x y^{3}+y z^{3}+z x^{3}\right)(z+x+y) \geq(\sqrt{x y z}(x+y+z))^{2}$ with equality if and only if $x y^{3} / z=y z^{3} / x=z x^{3} / y$, or equivalently $x=y=z$, i.e., $a=b=c$.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
9. (USA 3) ${ }^{\mathrm{IMO} 6}$ If $a, b$, and $c$ are sides of a triangle, prove that $$ a^{2} b(a-b)+b^{2} c(b-c)+c^{2} a(c-a) \geq 0 $$ Determine when there is equality.
|
9. For any triangle of sides $a, b, c$ there exist 3 nonnegative numbers $x, y, z$ such that $a=y+z, b=z+x, c=x+y$ (these numbers correspond to the division of the sides of a triangle by the point of contact of the incircle). The inequality becomes $$ (y+z)^{2}(z+x)(y-x)+(z+x)^{2}(x+y)(z-y)+(x+y)^{2}(y+z)(x-z) \geq 0 $$ Expanding, we get $x y^{3}+y z^{3}+z x^{3} \geq x y z(x+y+z)$. This follows from Cauchy's inequality $\left(x y^{3}+y z^{3}+z x^{3}\right)(z+x+y) \geq(\sqrt{x y z}(x+y+z))^{2}$ with equality if and only if $x y^{3} / z=y z^{3} / x=z x^{3} / y$, or equivalently $x=y=z$, i.e., $a=b=c$.
|
{
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|
d8790b10-927d-547c-9952-3f2304fd3385
| 23,904
|
10. (GBR 1) Prove that the product of five consecutive positive integers cannot be the square of an integer.
|
10. Suppose that the product of some five consecutive numbers is a square. It is easily seen that among them at least one, say $n$, is divisible neither by 2 nor 3 . Since $n$ is coprime to the remaining four numbers, it is itself a square of a number $m$ of the form $6 k \pm 1$. Thus $n=(6 k \pm 1)^{2}=24 r+1$, where $r=k(3 k \pm 1) / 2$. Note that neither of the numbers $24 r-1,24 r+5$ is one of our five consecutive numbers because it is not a square. Hence the five numbers must be $24 r, 24 r+1, \ldots, 24 r+4$. However, the number $24 r+4=(6 k \pm 1)^{2}+3$ is divisible by $6 r+1$, which implies that it is a square as well. It follows that these two squares are 1 and 4 , which is impossible.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
10. (GBR 1) Prove that the product of five consecutive positive integers cannot be the square of an integer.
|
10. Suppose that the product of some five consecutive numbers is a square. It is easily seen that among them at least one, say $n$, is divisible neither by 2 nor 3 . Since $n$ is coprime to the remaining four numbers, it is itself a square of a number $m$ of the form $6 k \pm 1$. Thus $n=(6 k \pm 1)^{2}=24 r+1$, where $r=k(3 k \pm 1) / 2$. Note that neither of the numbers $24 r-1,24 r+5$ is one of our five consecutive numbers because it is not a square. Hence the five numbers must be $24 r, 24 r+1, \ldots, 24 r+4$. However, the number $24 r+4=(6 k \pm 1)^{2}+3$ is divisible by $6 r+1$, which implies that it is a square as well. It follows that these two squares are 1 and 4 , which is impossible.
|
{
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|
a8b802dc-646e-51ea-8cb1-fb24be2f7f9d
| 23,909
|
13. (BUL 5) Prove that the volume of a tetrahedron inscribed in a right circular cylinder of volume 1 does not exceed $\frac{2}{3 \pi}$.
|
13. Let $Z$ be the given cylinder of radius $r$, altitude $h$, and volume $\pi r^{2} h=1, k_{1}$ and $k_{2}$ the circles surrounding its bases, and $V$ the volume of an inscribed tetrahedron $A B C D$. We claim that there is no loss of generality in assuming that $A, B, C, D$ all lie on $k_{1} \cup k_{2}$. Indeed, if the vertices $A, B, C$ are fixed and $D$ moves along a segment $E F$ parallel to the axis of the cylinder $\left(E \in k_{1}, F \in k_{2}\right)$, the maximum distance of $D$ from the plane $A B C$ (and consequently the maximum value of $V$ ) is achieved either at $E$ or at $F$. Hence we shall consider only the following two cases: (i) $A, B \in k_{1}$ and $C, D \in k_{2}$. Let $P, Q$ be the projections of $A, B$ on the plane of $k_{2}$, and $R, S$ the projections of $C, D$ on the plane of $k_{1}$, respectively. Then $V$ is one-third of the volume $V^{\prime}$ of the prism $A R B S C P D Q$ with bases $A R B S$ and $C P D Q$. The area of the quadrilateral $A R B S$ inscribed in $k_{1}$ does not exceed the area of the square inscribed therein, which is $2 r^{2}$. Hence $3 V=V^{\prime} \leq 2 r^{2} h=2 / \pi$. (ii) $A, B, C \in k_{1}$ and $D \in k_{2}$. The area of the triangle $A B C$ does not exceed the area of an equilateral triangle inscribed in $k_{1}$, which is $3 \sqrt{3} r^{2} / 4$. Consequently, $V \leq \frac{\sqrt{3}}{4} r^{2} h=\frac{\sqrt{3}}{4 \pi}<\frac{2}{3 \pi}$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
13. (BUL 5) Prove that the volume of a tetrahedron inscribed in a right circular cylinder of volume 1 does not exceed $\frac{2}{3 \pi}$.
|
13. Let $Z$ be the given cylinder of radius $r$, altitude $h$, and volume $\pi r^{2} h=1, k_{1}$ and $k_{2}$ the circles surrounding its bases, and $V$ the volume of an inscribed tetrahedron $A B C D$. We claim that there is no loss of generality in assuming that $A, B, C, D$ all lie on $k_{1} \cup k_{2}$. Indeed, if the vertices $A, B, C$ are fixed and $D$ moves along a segment $E F$ parallel to the axis of the cylinder $\left(E \in k_{1}, F \in k_{2}\right)$, the maximum distance of $D$ from the plane $A B C$ (and consequently the maximum value of $V$ ) is achieved either at $E$ or at $F$. Hence we shall consider only the following two cases: (i) $A, B \in k_{1}$ and $C, D \in k_{2}$. Let $P, Q$ be the projections of $A, B$ on the plane of $k_{2}$, and $R, S$ the projections of $C, D$ on the plane of $k_{1}$, respectively. Then $V$ is one-third of the volume $V^{\prime}$ of the prism $A R B S C P D Q$ with bases $A R B S$ and $C P D Q$. The area of the quadrilateral $A R B S$ inscribed in $k_{1}$ does not exceed the area of the square inscribed therein, which is $2 r^{2}$. Hence $3 V=V^{\prime} \leq 2 r^{2} h=2 / \pi$. (ii) $A, B, C \in k_{1}$ and $D \in k_{2}$. The area of the triangle $A B C$ does not exceed the area of an equilateral triangle inscribed in $k_{1}$, which is $3 \sqrt{3} r^{2} / 4$. Consequently, $V \leq \frac{\sqrt{3}}{4} r^{2} h=\frac{\sqrt{3}}{4 \pi}<\frac{2}{3 \pi}$.
|
{
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|
d3cb56d0-9478-515b-b60c-02eaf3a2976c
| 23,914
|
15. (LUX 2) Angles of a given triangle $A B C$ are all smaller than $120^{\circ}$. Equilateral triangles $A F B, B D C$ and $C E A$ are constructed in the exterior of $\triangle A B C$. (a) Prove that the lines $A D, B E$, and $C F$ pass through one point $S$. (b) Prove that $S D+S E+S F=2(S A+S B+S C)$.
|
15. (a) Since rotation by $60^{\circ}$ around $A$ transforms the triangle $C A F$ into $\triangle E A B$, it follows that $\measuredangle(C F, E B)=60^{\circ}$. We similarly deduce that $\measuredangle(E B, A D)=\measuredangle(A D, F C)=60^{\circ}$. Let $S$ be the intersection point of $B E$ and $A D$. Since $\measuredangle C S E=\measuredangle C A E=60^{\circ}$, it follows that $E A S C$ is cyclic. Therefore $\measuredangle(A S, S C)=60^{\circ}=\measuredangle(A D, F C)$, which implies that $S$ lies on $C F$ as well. (b) A rotation of $E A S C$ around $E$ by $60^{\circ}$ transforms $A$ into $C$ and $S$ into a point $T$ for which $S E=S T=S C+C T=S C+S A$. Summing the equality $S E=S C+S A$ and the analogous equalities $S D=S B+S C$ and $S F=S A+S B$ yields the result.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
15. (LUX 2) Angles of a given triangle $A B C$ are all smaller than $120^{\circ}$. Equilateral triangles $A F B, B D C$ and $C E A$ are constructed in the exterior of $\triangle A B C$. (a) Prove that the lines $A D, B E$, and $C F$ pass through one point $S$. (b) Prove that $S D+S E+S F=2(S A+S B+S C)$.
|
15. (a) Since rotation by $60^{\circ}$ around $A$ transforms the triangle $C A F$ into $\triangle E A B$, it follows that $\measuredangle(C F, E B)=60^{\circ}$. We similarly deduce that $\measuredangle(E B, A D)=\measuredangle(A D, F C)=60^{\circ}$. Let $S$ be the intersection point of $B E$ and $A D$. Since $\measuredangle C S E=\measuredangle C A E=60^{\circ}$, it follows that $E A S C$ is cyclic. Therefore $\measuredangle(A S, S C)=60^{\circ}=\measuredangle(A D, F C)$, which implies that $S$ lies on $C F$ as well. (b) A rotation of $E A S C$ around $E$ by $60^{\circ}$ transforms $A$ into $C$ and $S$ into a point $T$ for which $S E=S T=S C+C T=S C+S A$. Summing the equality $S E=S C+S A$ and the analogous equalities $S D=S B+S C$ and $S F=S A+S B$ yields the result.
|
{
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|
a924a0b5-20db-5cf9-b8bf-97c7be9240d6
| 23,918
|
2. (CAN 2) Prove: (a) There are infinitely many triples of positive integers $m, n, p$ such that $4 m n-m-n=p^{2}-1$. (b) There are no positive integers $m, n, p$ such that $4 m n-m-n=p^{2}$.
|
2. (a) For $m=t(t-1) / 2$ and $n=t(t+1) / 2$ we have $4 m n-m-n=$ $\left(t^{2}-1\right)^{2}-1$ (b) Suppose that $4 m n-m-n=p^{2}$, or equivalently, $(4 m-1)(4 n-1)=$ $4 p^{2}+1$. The number $4 m-1$ has at least one prime divisor, say $q$, that is of the form $4 k+3$. Then $4 p^{2} \equiv-1(\bmod q)$. However, by Fermat's theorem we have $$ 1 \equiv(2 p)^{q-1}=\left(4 p^{2}\right)^{\frac{q-1}{2}} \equiv(-1)^{\frac{q-1}{2}}(\bmod q) $$ which is impossible since $(q-1) / 2=2 k+1$ is odd.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
2. (CAN 2) Prove: (a) There are infinitely many triples of positive integers $m, n, p$ such that $4 m n-m-n=p^{2}-1$. (b) There are no positive integers $m, n, p$ such that $4 m n-m-n=p^{2}$.
|
2. (a) For $m=t(t-1) / 2$ and $n=t(t+1) / 2$ we have $4 m n-m-n=$ $\left(t^{2}-1\right)^{2}-1$ (b) Suppose that $4 m n-m-n=p^{2}$, or equivalently, $(4 m-1)(4 n-1)=$ $4 p^{2}+1$. The number $4 m-1$ has at least one prime divisor, say $q$, that is of the form $4 k+3$. Then $4 p^{2} \equiv-1(\bmod q)$. However, by Fermat's theorem we have $$ 1 \equiv(2 p)^{q-1}=\left(4 p^{2}\right)^{\frac{q-1}{2}} \equiv(-1)^{\frac{q-1}{2}}(\bmod q) $$ which is impossible since $(q-1) / 2=2 k+1$ is odd.
|
{
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87ce569b-b9f5-5f80-ade2-860310eee076
| 23,928
|
4. (MON 1) ${ }^{\mathrm{IMO} 5}$ Let $d$ be the sum of the lengths of all diagonals of a convex polygon of $n(n>3)$ vertices and let $p$ be its perimeter. Prove that $$ \frac{n-3}{2}<\frac{d}{p}<\frac{1}{2}\left(\left[\frac{n}{2}\right]\left[\frac{n+1}{2}\right]-2\right) . $$
|
4. Consider the convex $n$-gon $A_{1} A_{2} \ldots A_{n}$ (the indices are considered modulo $n)$. For any diagonal $A_{i} A_{j}$ we have $A_{i} A_{j}+A_{i+1} A_{j+1}>A_{i} A_{i+1}+A_{j} A_{j+1}$. Summing all such $n(n-3) / 2$ inequalities, we obtain $2 d>(n-3) p$, proving the first inequality. Let us now prove the second inequality. We notice that for each diagonal $A_{i} A_{i+j}$ (we may assume w.l.o.g. that $j \leq[n / 2]$ ) the following relation holds: $$ A_{i} A_{i+j}<A_{i} A_{i+1}+\cdots+A_{i+j-1} A_{i+j} . $$ If $n=2 k+1$, then summing the inequalities (1) for $j=2,3, \ldots, k$ and $i=1,2, \ldots, n$ yields $d<(2+3+\cdots+k) p=([n / 2][n+1 / 2]-2) p / 2$. If $n=2 k$, then summing the inequalities (1) for $j=2,3, \ldots, k-1$, $i=1,2, \ldots, n$ and for $j=k, i=1,2, \ldots, k$ again yields $d<(2+3+\cdots+$ $(k-1)+k / 2) p=\frac{1}{2}([n / 2][n+1 / 2]-2) p$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
4. (MON 1) ${ }^{\mathrm{IMO} 5}$ Let $d$ be the sum of the lengths of all diagonals of a convex polygon of $n(n>3)$ vertices and let $p$ be its perimeter. Prove that $$ \frac{n-3}{2}<\frac{d}{p}<\frac{1}{2}\left(\left[\frac{n}{2}\right]\left[\frac{n+1}{2}\right]-2\right) . $$
|
4. Consider the convex $n$-gon $A_{1} A_{2} \ldots A_{n}$ (the indices are considered modulo $n)$. For any diagonal $A_{i} A_{j}$ we have $A_{i} A_{j}+A_{i+1} A_{j+1}>A_{i} A_{i+1}+A_{j} A_{j+1}$. Summing all such $n(n-3) / 2$ inequalities, we obtain $2 d>(n-3) p$, proving the first inequality. Let us now prove the second inequality. We notice that for each diagonal $A_{i} A_{i+j}$ (we may assume w.l.o.g. that $j \leq[n / 2]$ ) the following relation holds: $$ A_{i} A_{i+j}<A_{i} A_{i+1}+\cdots+A_{i+j-1} A_{i+j} . $$ If $n=2 k+1$, then summing the inequalities (1) for $j=2,3, \ldots, k$ and $i=1,2, \ldots, n$ yields $d<(2+3+\cdots+k) p=([n / 2][n+1 / 2]-2) p / 2$. If $n=2 k$, then summing the inequalities (1) for $j=2,3, \ldots, k-1$, $i=1,2, \ldots, n$ and for $j=k, i=1,2, \ldots, k$ again yields $d<(2+3+\cdots+$ $(k-1)+k / 2) p=\frac{1}{2}([n / 2][n+1 / 2]-2) p$.
|
{
"resource_path": "IMO/segmented/en-compendium.jsonl",
"problem_match": null,
"solution_match": null
}
|
3f042e42-6739-5323-bab4-41f96abf953f
| 23,934
|
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