year stringdate 1961-01-01 00:00:00 2025-01-01 00:00:00 ⌀ | tier stringclasses 5
values | problem_label stringclasses 119
values | problem_type stringclasses 13
values | exam stringclasses 28
values | problem stringlengths 87 2.77k | solution stringlengths 834 13k | metadata dict | problem_tokens int64 50 903 | solution_tokens int64 500 3.93k |
|---|---|---|---|---|---|---|---|---|---|
2020 | T3 | 5 | null | INMO | Infinitely many equidistant parallel lines are drawn in the plane. A positive integer $n \geq 3$ is called frameable if it is possible to draw a regular polygon with $n$ sides all whose vertices lie on these lines and no line contains more than one vertex of the polygon.
(a) Show that $3,4,6$ are frameable.
(b) Show ... | For $n=3,4,6$ it is possible to draw regular polygons with vertices on the parallel lines (note that when we show a regular hexagon is a framed polygon, it includes the equilateral triangle case).
Figure 1:
Figure 2:
Figure 3:
We will prove that it is not possible for $n \geq 7$. In fact, we prove a stronger statem... | {
"problem_match": "\n5.",
"resource_path": "INMO/segmented/en-sol-inmo-20.jsonl",
"solution_match": "\nSolution:"
} | 106 | 600 |
2016 | T3 | 2 | null | INMO | For positive real numbers $a, b, c$, which of the following statements necessarily implies $a=b=c$ : (I) $a\left(b^{3}+c^{3}\right)=b\left(c^{3}+a^{3}\right)=c\left(a^{3}+b^{3}\right)$, (II) $a\left(a^{3}+b^{3}\right)=b\left(b^{3}+c^{3}\right)=c\left(c^{3}+a^{3}\right)$ ? Justify your answer. | We show that (I) need not imply that $a=b=c$ where as (II) always implies $a=b=c$.
Observe that $a\left(b^{3}+c^{3}\right)=b\left(c^{3}+a^{3}\right)$ gives $c^{3}(a-b)=a b\left(a^{2}-b^{2}\right)$. This gives either $a=b$ or $a b(a+b)=c^{3}$. Similarly, $b=c$ or $b c(b+c)=a^{3}$. If $a \neq b$ and $b \neq c$, we obtai... | {
"problem_match": "\n2.",
"resource_path": "INMO/segmented/en-sol-inmo16.jsonl",
"solution_match": "\nSolution:"
} | 123 | 1,279 |
2016 | T3 | 3 | null | INMO | Let $\mathbb{N}$ denote the set of all natural numbers. Define a function $T: \mathbb{N} \rightarrow \mathbb{N}$ by $T(2 k)=k$ and $T(2 k+1)=2 k+2$. We write $T^{2}(n)=T(T(n))$ and in general $T^{k}(n)=T^{k-1}(T(n))$ for any $k>1$.
(i) Show that for each $n \in \mathbb{N}$, there exists $k$ such that $T^{k}(n)=1$.
(i... | (i) For $n=1$, we have $T(1)=2$ and $T^{2}(1)=T(2)=1$. Hence we may assume that $n>1$.
Suppose $n>1$ is even. Then $T(n)=n / 2$. We observe that $(n / 2) \leq n-1$ for $n>1$.
Suppose $n>1$ is odd so that $n \geq 3$. Then $T(n)=n+1$ and $T^{2}(n)=(n+1) / 2$. Again we see that $(n+1) / 2 \leq(n-1)$ for $n \geq 3$.
Thu... | {
"problem_match": "\n3.",
"resource_path": "INMO/segmented/en-sol-inmo16.jsonl",
"solution_match": "## Solution:"
} | 211 | 729 |
2016 | T3 | 4 | null | INMO | Suppose 2016 points of the circumference of a circle are coloured red and the remaining points are coloured blue. Given any natural number $n \geq 3$, prove that there is a regular $n$-sided polygon all of whose vertices are blue. | Let $A_{1}, A_{2}, \ldots, A_{2016}$ be 2016 points on the circle which are coloured red and the remain-
ing blue. Let $n \geq 3$ and let $B_{1}, B_{2}, \ldots, B_{n}$ be a regular $n$-sided polygon inscribed in this circle with the vertices chosen in anti-clock-wise direction. We place $B_{1}$ at $A_{1}$. (It is possi... | {
"problem_match": "\n4.",
"resource_path": "INMO/segmented/en-sol-inmo16.jsonl",
"solution_match": "\nSolution:"
} | 57 | 566 |
2016 | T3 | 5 | null | INMO | Let $A B C$ be a right-angled triangle with $\angle B=90^{\circ}$. Let $D$ be a point on $A C$ such that the in-radii of the triangles $A B D$ and $C B D$ are equal. If this common value is $r^{\prime}$ and if $r$ is the in-radius of triangle $A B C$, prove that
$$
\frac{1}{r^{\prime}}=\frac{1}{r}+\frac{1}{B D}
$$ | Let $E$ and $F$ be the incentres of triangles $A B D$ and $C B D$ respectively. Let the incircles of triangles $A B D$ and $C B D$ touch $A C$ in $P$ and $Q$ respectively. If $\angle B D A=\theta$, we see that
$$
r^{\prime}=P D \tan (\theta / 2)=Q D \cot (\theta / 2)
$$
Hence
$$
P Q=P D+Q D=r^{\prime}\left(\cot \fra... | {
"problem_match": "\n5.",
"resource_path": "INMO/segmented/en-sol-inmo16.jsonl",
"solution_match": "\nSolution:"
} | 118 | 2,347 |
2016 | T3 | 6 | null | INMO | Consider a non-constant arithmetic progression $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ Suppose there exist relatively prime positive integers $p>1$ and $q>1$ such that $a_{1}^{2}, a_{p+1}^{2}$ and $a_{q+1}^{2}$ are also the terms of the same arithmetic progression. Prove that the terms of the arithmetic progression are a... | Let us take $a_{1}=a$. We have
$$
a^{2}=a+k d, \quad(a+p d)^{2}=a+l d, \quad(a+q d)^{2}=a+m d
$$
Thus we have
$$
a+l d=(a+p d)^{2}=a^{2}+2 p a d+p^{2} d^{2}=a+k d+2 p a d+p^{2} d^{2}
$$
Since we have non-constant AP, we see that $d \neq 0$. Hence we obtain $2 p a+p^{2} d=l-k$. Similarly, we get $2 q a+q^{2} d=m-k$.... | {
"problem_match": "\n6.",
"resource_path": "INMO/segmented/en-sol-inmo16.jsonl",
"solution_match": "\nSolution:"
} | 99 | 1,194 |
2017 | T3 | 2 | null | INMO | Suppose $n \geq 0$ is an integer and all the roots of $x^{3}+\alpha x+4-\left(2 \times 2016^{n}\right)=0$ are integers. Find all possible values of $\alpha$. | Let $a, b, c \in \mathbb{Z}$ be the roots of the given equation for some $n \in \mathbb{N}_{0}$. By Vieta Theorem, we know that
$$
\begin{gathered}
a+b+c=0 \\
a b+b c+c a=\alpha \\
a b c=2 \times 2016^{n}-4
\end{gathered}
$$
If possible, let us have $n \geq 1$. Since $7 \mid 2016$, we have that
$$
7|a b c+4 \Longrig... | {
"problem_match": "\n2.",
"resource_path": "INMO/segmented/en-sol-inmo_17.jsonl",
"solution_match": "\nSolution 2:"
} | 57 | 874 |
2017 | T3 | 3 | null | INMO | Find the number of triples $(x, a, b)$ where $x$ is a real number and $a, b$ belong to the set $\{1,2,3,4,5,6,7,8,9\}$ such that
$$
x^{2}-a\{x\}+b=0
$$
where $\{x\}$ denotes the fractional part of the real number $x$. (For example $\{1.1\}=0.1=$ $\{-0.9\}$. | Let us write $x=n+f$ where $n=[x]$ and $f=\{x\}$. Then
$$
f^{2}+(2 n-a) f+n^{2}+b=0
$$
Observe that the product of the roots of (1) is $n^{2}+b \geq 1$. If this equation has to have a solution $0 \leq f<1$, the larger root of (1) is greater 1 . We conclude that the equation (1) has a real root less than 1 only if $P(... | {
"problem_match": "\n3.",
"resource_path": "INMO/segmented/en-sol-inmo_17.jsonl",
"solution_match": "\nSolution:"
} | 112 | 650 |
2017 | T3 | 4 | null | INMO | Let $A B C D E$ be a convex pentagon in which $\angle A=\angle B=\angle C=\angle D=120^{\circ}$ and whose side lengths are 5 consecutive integers in some order. Find all possible values of $A B+B C+C D$. | Let $A B=a, B C=b$, and $C D=c$. By symmetry, we may assume that $c<a$. We show that $D E=a+b$ and $E A=b+c$.
Draw a line parallel to $B C$ through $D$. Extend $E A$ to meet this line at $F$. Draw a line parallel to $C D$ through $B$ and let it intersect $D F$ in $G$. Let $A B$ intersect $D F$ in $H$. We have $\angle ... | {
"problem_match": "\n4.",
"resource_path": "INMO/segmented/en-sol-inmo_17.jsonl",
"solution_match": "\nSolution 1:"
} | 61 | 711 |
2017 | T3 | 5 | null | INMO | Let $A B C$ be a triangle with $\angle A=90^{\circ}$ and $A B<A C$. Let $A D$ be the altitude from $A$ on to $B C$. Let $P, Q$ and $I$ denote respectively the incentres of triangles $A B D, A C D$ and $A B C$. Prove that $A I$ is perpendicular to $P Q$ and $A I=P Q$. | Draw $P S \| B C$ and $Q S \| A D$. Then $P S Q$ is a right-angled triangle with $\angle P S Q=90^{\circ}$. Observe that $P S=r_{1}+r_{2}$ and $S Q=r_{2}-r_{1}$, where $r_{1}$ and $r_{2}$ are the inradii of triangles $A B D$ and $A C D$, respectively. We observe that triangles $D A B$ and $D C A$ are similar to triangl... | {
"problem_match": "\n5.",
"resource_path": "INMO/segmented/en-sol-inmo_17.jsonl",
"solution_match": "\nSolution:"
} | 98 | 637 |
2017 | T3 | 5 | null | INMO | Let $A B C$ be a triangle with $\angle A=90^{\circ}$ and $A B<A C$. Let $A D$ be the altitude from $A$ on to $B C$. Let $P, Q$ and $I$ denote respectively the incentres of triangles $A B D, A C D$ and $A B C$. Prove that $A I$ is perpendicular to $P Q$ and $A I=P Q$. | In the figure, we have made the construction as mentioned in the hint. Since $P, Q$ are the incentres of $\triangle A B D, \triangle A C D, D P, D Q$ are the internal angle bisectors of $\angle A D B, \angle A D C$ respectively. Since $A D$ is the altitude on the hypotenuse $B C$ in $\triangle A B C$, we have that $\an... | {
"problem_match": "\n5.",
"resource_path": "INMO/segmented/en-sol-inmo_17.jsonl",
"solution_match": "\nSolution 2:"
} | 98 | 625 |
2017 | T3 | 5 | null | INMO | Let $A B C$ be a triangle with $\angle A=90^{\circ}$ and $A B<A C$. Let $A D$ be the altitude from $A$ on to $B C$. Let $P, Q$ and $I$ denote respectively the incentres of triangles $A B D, A C D$ and $A B C$. Prove that $A I$ is perpendicular to $P Q$ and $A I=P Q$. | We know that the angle bisector of $\angle B$ passes through $P, I$ which implies that $B, P, I$ are collinear. Similarly, $C, Q, I$ are also collinear. Since $I$ is the incentre of $\triangle A B C$, we know that
$$
\angle P I Q=\angle B I C=90^{\circ}+\frac{\angle A}{2}=135^{\circ}
$$
Join $A P, A Q$. We know that ... | {
"problem_match": "\n5.",
"resource_path": "INMO/segmented/en-sol-inmo_17.jsonl",
"solution_match": "\nSolution 3:"
} | 98 | 956 |
2017 | T3 | 6 | null | INMO | Let $n \geq 1$ be an integer and consider the sum
$$
x=\sum_{k \geq 0}\binom{n}{2 k} 2^{n-2 k} 3^{k}=\binom{n}{0} 2^{n}+\binom{n}{2} 2^{n-2} \cdot 3+\binom{n}{4} 2^{n-4} \cdot 3^{2}+\cdots
$$
Show that $2 x-1,2 x, 2 x+1$ form the sides of a triangle whose area and inradius are also integers. | We will first show that the numbers $2 x_{n}-1,2 x_{n}, 2 x_{n}+1$ form the sides of a triangle. To show that, it suffices to prove that $2 x_{n}-1+2 x_{n}>2 x_{n}+1$. If possible, let the converse hold. Then, we see that we must have $4 x_{n}-1 \leq 2 x_{n}+1$, which implies that $x_{n} \leq 1$. But we see that even f... | {
"problem_match": "\n6.",
"resource_path": "INMO/segmented/en-sol-inmo_17.jsonl",
"solution_match": "\nSolution 2:"
} | 139 | 1,052 |
2014 | T1 | 3 | null | IZho | There are given 100 distinct positive integers. We call a pair of integers among them good if the ratio of its elements is either 2 or 3 . What is the maximum number $g$ of good pairs that these 100 numbers can form? (A same number can be used in several pairs.) | Like so often in Russian problems, numbers are used instead of generic symbols. Let us therefore denote $10=n>1,2=k>1,3=\ell>1$, with the extra condition both $k$ and $\ell$ aren't powers of a same number. Consider the digraph $G$ whose set of vertices $V(G)$ is made of $v=n^{2}$ distinct positive integers, and whose s... | {
"problem_match": "\nProblem 3.",
"resource_path": "IZho/segmented/en-2014_zhautykov_resenja_e.jsonl",
"solution_match": "\nSolution."
} | 66 | 1,129 |
2014 | T1 | 5 | null | IZho | Let $U=\{1,2,3, \ldots, 2014\}$. For all $a, b, c \in \mathbb{N}$ let $f(a, b, c)$ be the number of ordered sextuplets $\left(X_{1}, X_{2}, X_{3}, Y_{1}, Y_{2}, Y_{3}\right)$ of subsets of $U$, satisfying the following conditions
(i) $Y_{1} \subseteq X_{1} \subseteq U$ and $\left|X_{1}\right|=a$;
(ii) $Y_{2} \subsete... | In order to avoid any confusion between the letters $a, b, c$ and their numerical values (as cardinalities of sets), the most convenient way will be to denote by $|\ell|$ the cardinality symbolized by any such letter $\ell$. We can now consider the true 3 -element set $\{a, b, c\}$, and the canonical bijection $\phi:\{... | {
"problem_match": "\nProblem 5.",
"resource_path": "IZho/segmented/en-2014_zhautykov_resenja_e.jsonl",
"solution_match": "\nSolution."
} | 242 | 1,220 |
2015 | T1 | 3 | null | IZho | Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f\left(x^{3}+y^{3}+x y\right)=x^{2} f(x)+y^{2} f(y)+f(x y)
$$
for all $x, y \in \mathbb{R}$. | (Ştefan Tudose on AoPS) Let $P(x, y)$ be the assertion that $f\left(x^{3}+y^{3}+x y\right)=x^{2} f(x)+y^{2} f(y)+f(x y)$. From $P(1,0)$ we get that $f(0)=0$, hence from $P(x, 0)$ we get $f\left(x^{3}\right)=x^{2} f(x) . P(x,-x)$ yields $f(x)=-f(-x)$.
From $P(x, y)-P(x,-y)$ we get
$$
f\left(y^{3}\right)+f(x y)=\frac{1... | {
"problem_match": "\nProblem 3.",
"resource_path": "IZho/segmented/en-2015_zhautykov_resenja_e.jsonl",
"solution_match": "\nSolution."
} | 79 | 534 |
2015 | T1 | 3 | null | IZho | Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f\left(x^{3}+y^{3}+x y\right)=x^{2} f(x)+y^{2} f(y)+f(x y)
$$
for all $x, y \in \mathbb{R}$. | (AoPS - user pco) Under the notations of above we similarly get $f(0)=0, f\left(x^{3}\right)=x^{2} f(x)$, and $f(x)$ odd.
Let then $u, v$ be such that $u+v \leq 0$. It's easy to see that the system $x^{3}+y^{3}+x y=u,-x^{3}-y^{3}+x y=v$ always has a solution, and then
$$
\begin{aligned}
& P(x, y) \text { yields } f(u... | {
"problem_match": "\nProblem 3.",
"resource_path": "IZho/segmented/en-2015_zhautykov_resenja_e.jsonl",
"solution_match": "\nAlternative Solution."
} | 79 | 620 |
2015 | T1 | 5 | null | IZho | Let $A_{n}$ be the set of partitions of the sequence $(1,2, \ldots, n)$ into several subsequences such that every two neighbouring terms of each subsequence have different parity, and let $B_{n}$ be the set of partitions of the sequence $(1,2, \ldots, n)$ into several subsequences such that all the terms of each subseq... | For each partition $\pi$ of $\{1,2, \ldots, n\}$, with the elements within each block written in ascending order, denote by $k(\pi)$ the number of blocks of $\pi$ ending in an even number and by $\ell(\pi)$ the number of blocks of $\pi$ ending in an odd number.
Also denote $f_{n}(x, y)=\sum_{\pi \in A_{n}} x^{k(\pi)} ... | {
"problem_match": "\nProblem 5.",
"resource_path": "IZho/segmented/en-2015_zhautykov_resenja_e.jsonl",
"solution_match": "\nSolution."
} | 122 | 1,256 |
2018 | T1 | 2 | null | IZho | Points $N, K, L$ lie on the sides $A B, B C, C A$ of a triangle $A B C$ respectively so that $A L=B K$ and $C N$ is the bisector of the angle $C$. The segments $A K$ and $B L$ meet at the point $P$. Let $I$ and $J$ be the incentres of the triangles $A P L$ and $B P K$ respectively. The lines $C N$ and $I J$ meet at poi... | The case $C A=C B$ is trivial. If $C A \neq C B$, we may suppose, without loss of generality, that $C N$ meets the segment $P K$.
Let the circumcircles $\omega_{1}$ and $\omega_{2}$ of the triangles $A P L$ and $B P K$ respectively meet again at point $T$. Then
$$
\angle L A T=\angle T P B=\angle T K B
$$
and $\angl... | {
"problem_match": "\n2.",
"resource_path": "IZho/segmented/en-2018_zhautykov_resenja_e.jsonl",
"solution_match": "\nSolution."
} | 127 | 575 |
2018 | T1 | 5 | null | IZho | Find all real $a$ for which there exists a function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x-f(y))=f(x)+a[y]$ for every real $x$ и $y$ ( $[y]$ denotes the integral part of $y$ ).
Answer: $a=-n^{2}$ for arbitrary integer $n$. | First note that $a=0$ satisfies the problem condition (for example, the equation is satisfied by the function $f(x) \equiv 0)$.
Now suppose $a \neq 0$.
Lemma. $f(y)=f(z)$ if and only if $[y]=[z]$.
Suppose $f(y)=f(z)$ for some $y, z$. Then the given equation implies $f(x)+a[y]=f(x-f(y))=f(x-f(z))=f(x)+a[z]$ whence $[... | {
"problem_match": "\n5.",
"resource_path": "IZho/segmented/en-2018_zhautykov_resenja_e.jsonl",
"solution_match": "\nSolution."
} | 83 | 782 |
2018 | T1 | 6 | null | IZho | A convex hexagon $A B C D E F$ is inscribed in a circle with radius $R$. Diagonals $A D$ and $B E, B E$ and $C F, A D$ and $C F$ of the hexagon meet at points $M, N, K$ respectively. Let $r_{1}, r_{2}, r_{3}, r_{4}, r_{5}, r_{6}$ be the inradii of the triangles $A B M$, $B C N, C D K, D E M, E F N, A F K$ respectively.... | We start with a lemma.
Lemma. Let $R$ be the circumradius of a quadrilateral $X Y Z T$, the diagonals of $X Y Z T$ meet at $U$, and $\varphi=\frac{1}{2} \angle X U Y$. Then the radii $r_{1}$ and $r_{2}$ of the incentres of $X Y U$ and $Z T U$ satisfy
$$
\frac{r_{1}+r_{2}}{R} \leqslant 2 \tan \varphi(1-\sin \varphi)
$... | {
"problem_match": "\n6.",
"resource_path": "IZho/segmented/en-2018_zhautykov_resenja_e.jsonl",
"solution_match": "## Solution."
} | 174 | 1,004 |
2019 | T1 | 3 | null | IZho | The extension of median $C M$ of the triangle $A B C$ intersects its circumcircle $\omega$ at $N$. Let $P$ and $Q$ be the points on the rays $C A$ and $C B$ respectively such that $P M \| B N$ and $Q M \| A N$. Let $X$ and $Y$ be the points on the segments $P M$ and $Q M$ respectively such that $P Y$ and $Q X$ are tang... | Lemma. The points $K$ and $L$ lie on the sides $B C$ and $A C$ of a triangle $A B C$. The segments $A K$ and $B L$ intersect at $D$. Then the quadrilateral $C K D L$ is circumscribed if and only if $A C-B C=A D-B D$.
Proof. Let $C K D L$ be circumscribed and its incircle touches $L C, C K, K D, D L$ at $X, Y, Z, T$ re... | {
"problem_match": "\n№3.",
"resource_path": "IZho/segmented/en-2019_zhautykov_resenja_e.jsonl",
"solution_match": "## Solution."
} | 142 | 674 |
2020 | T1 | 2 | null | IZho | In a set of 20 elements there are $2 k+1$ different subsets of 7 elements such that each of these subsets intersects exactly $k$ other subsets. Find the maximum $k$ for which this is possible.
The answer is $k=2$. | Let $M$ be the set of residues mod20. An example is given by the sets $A_{i}=\{4 i+1,4 i+$ $2,4 i+3,4 i+4,4 i+5,4 i+6,4 i+7\} \subset M, i=0,1,2,3,4$.
Let $k \geq 2$. Obviously among any three 7-element subsets there are two intersecting subsets.
Let $A$ be any of the $2 k+1$ subsets. It intersects $k$ other subsets ... | {
"problem_match": "\n2.",
"resource_path": "IZho/segmented/en-2020_zhautykov_resenja_e.jsonl",
"solution_match": "\nSolution."
} | 56 | 524 |
2020 | T1 | 3 | null | IZho | A convex hexagon $A B C D E F$ is inscribed in a circle. Prove the inequality
$$
A C \cdot B D \cdot C E \cdot D F \cdot A E \cdot B F \geq 27 A B \cdot B C \cdot C D \cdot D E \cdot E F \cdot F A
$$ | Let
$$
d_{1}=A B \cdot B C \cdot C D \cdot D E \cdot E F \cdot F A, d_{2}=A C \cdot B D \cdot C E \cdot D F \cdot A E \cdot B F, d_{3}=A D \cdot B E \cdot C F
$$
Applying Ptolemy's theorem to quadrilaterals $A B C D, B C D E, C D E F, D E F A, E F A B, F A B C$, we obtain six equations $A C \cdot B D-A B \cdot C D=B ... | {
"problem_match": "\n3.",
"resource_path": "IZho/segmented/en-2020_zhautykov_resenja_e.jsonl",
"solution_match": "\nSolution."
} | 78 | 726 |
2020 | T1 | 5 | null | IZho | Find all the functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that $f(4 x+3 y)=f(3 x+$ $+y)+f(x+2 y)$ for all integers $x$ and $y$.
Answer: $f(x)=\frac{a x}{5}$ for $x$ divisible by 5 and $f(x)=b x$ for $x$ not
=f(3 x+y)+f(x+2 y)
$$
we get
$$
f(3 y)=f(y)+f(2 y)
$$
Next, (1) for $y=-2 x$ gives us $f(-2 x)=f(x)+f(-3 x)=f(x)+f(-x)+f(-2 x)$ (in view of (2)). It follows that
$$
f(-x)=-f(x)
$$
Now, let $x=2 z-v, y=3 v-z$ in (1). Then
$$
f(5 z+5 v)=f(5 z)+f(5 v)
$$
for all ... | {
"problem_match": "\n№5.",
"resource_path": "IZho/segmented/en-2020_zhautykov_resenja_e.jsonl",
"solution_match": "\nSolution."
} | 186 | 891 |
2020 | T1 | 6 | null | IZho | Some squares of a $n \times n$ table $(n>2)$ are black, the rest are white. In every white square we write the number of all the black squares having at least one common vertex with it. Find the maximum possible sum of all these numbers.
The answer is $3 n^{2}-5 n+2$. | The sum attains this value when all squares in even rows are black and the rest are white. It remains to prove that this is the maximum value.
The sum in question is the number of pairs of differently coloured squares sharing at least one vertex. There are two kinds of such pairs: sharing a side and sharing only one v... | {
"problem_match": "\n№6.",
"resource_path": "IZho/segmented/en-2020_zhautykov_resenja_e.jsonl",
"solution_match": "\nSolution."
} | 70 | 803 |
2021 | T1 | 3 | null | IZho | Let $n \geq 2$ be an integer. Elwyn is given an $n \times n$ table filled with real numbers (each cell of the table contains exactly one number). We define a rook set as a set of $n$ cells of the table situated in $n$ distinct rows as well as in $n$ distinct columns. Assume that, for every rook set, the sum of $n$ numb... | We start with the following known consequence of Hall's lemma.
Lemma. Let $G=(U \sqcup V, E)$ be a bipartite multigraph with parts $U$ and $V$, both of size $n$. Assume that each vertex has degree $k$; then the edges can be partitioned into $k$ perfect matchings.
Proof. Induction on $k$; the base case $k=1$ is trivia... | {
"problem_match": "\n№3.",
"resource_path": "IZho/segmented/en-2021_zhautykov_resenja_e.jsonl",
"solution_match": "\nSolution 1."
} | 512 | 932 |
2021 | T1 | 3 | null | IZho | Let $n \geq 2$ be an integer. Elwyn is given an $n \times n$ table filled with real numbers (each cell of the table contains exactly one number). We define a rook set as a set of $n$ cells of the table situated in $n$ distinct rows as well as in $n$ distinct columns. Assume that, for every rook set, the sum of $n$ numb... | Say that the badness of a table is the sum of absolute values of all its negative entries. In Step 1, we will show that, whenever the badness of a good table is nonzero, Elwyn can make some moves decreasing the badness. In a (technical) Step 2, we will show that this claim yields the required result.
Step 1. Let $r$ b... | {
"problem_match": "\n№3.",
"resource_path": "IZho/segmented/en-2021_zhautykov_resenja_e.jsonl",
"solution_match": "\nSolution 2."
} | 512 | 2,388 |
2021 | T1 | 4 | null | IZho | A circle with radius $r$ is inscribed in the triangle $A B C$. Circles with radii $r_{1}, r_{2}, r_{3}\left(r_{1}, r_{2}, r_{3}<r\right)$ are inscribed in the angles $A, B, C$ so that each touches the incircle externally. Prove that $r_{1}+r_{2}+r_{3} \geq r$. | Let $A_{0}, B_{0}, C_{0}, A_{1}, B_{1}, C_{1}$ retain the meaning they had in the first solution. We have $\angle B_{1} I C_{1}=90^{\circ}+\frac{\angle A}{2}, \angle A_{1} I C_{1}=90^{\circ}+\frac{\angle B}{2}, \angle A_{1} I B_{1}=90^{\circ}+\frac{\angle C}{2}$. Obviously
$$
\left(\overrightarrow{I A_{1}}+\overrighta... | {
"problem_match": "\n№4.",
"resource_path": "IZho/segmented/en-2021_zhautykov_resenja_e.jsonl",
"solution_match": "\nSecond solution."
} | 98 | 780 |
2021 | T1 | 5 | null | IZho | On a party with 99 guests, hosts Ann and Bob play a game (the hosts are not regarded as guests). There are 99 chairs arranged in a circle; initially, all guests hang around those chairs. The hosts take turns alternately. By a turn, a host orders any standing guest to sit on an unoccupied chair $c$. If some chair adjace... | Preliminary notes. Let $F$ denote the number of occupied chairs at the current position in the game. Notice that, on any turn, $F$ does not decrease. Thus, we need to determine the maximal value of $F$ Ann can guarantee after an arbitrary move (either hers or her opponent's).
Say that the situation in the game is stab... | {
"problem_match": "\n№5.",
"resource_path": "IZho/segmented/en-2021_zhautykov_resenja_e.jsonl",
"solution_match": "\nSolution."
} | 179 | 992 |
1987 | T2 | A2 | null | IberoAmerican_MO | In the triangle $\mathrm{ABC}$, the midpoints of $\mathrm{AC}$ and $\mathrm{AB}$ are $\mathrm{M}$ and $\mathrm{N}$ respectively. $\mathrm{BM}$ and $\mathrm{CN}$ meet at $P$. Show that if it is possible to inscribe a circle in the quadrilateral AMPN (touching every side), then $\mathrm{ABC}$ is isosceles. | If the quadrilateral has an inscribed circle then $\mathrm{AM}+\mathrm{PN}=\mathrm{AN}+\mathrm{PM}$ (consider the tangents to the circle from $A, M, P, N$ ). But if $\mathrm{AB}>\mathrm{AC}$, then $\mathrm{BM}>\mathrm{CN}$ (see below). We have $\mathrm{AN}=$ $\mathrm{AB} / 2, \mathrm{PM}=\mathrm{BM} / 3, \mathrm{AM}=\m... | {
"problem_match": "\n## Problem A2",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 89 | 510 |
1989 | T2 | A2 | null | IberoAmerican_MO | Given positive real numbers $\mathrm{x}, \mathrm{y}, \mathrm{z}$ each less than $\pi / 2$, show that $\pi / 2+2 \sin \mathrm{x} \cos \mathrm{y}+2 \sin \mathrm{y}$ $\cos z>\sin 2 x+\sin 2 y+\sin 2 z$. | We have $\sin 2 x+\sin 2 y+\sin 2 z-2 \sin x \cos y-2 \sin y \cos z=2 \sin x(\cos x-\cos y)+2 \sin$ $y(\cos y-\cos z)+2 \sin z \cos z$, so we wish to show that $\sin x(\cos x-\cos y)+\sin y(\cos y-\cos$ $\mathrm{z})+\sin \mathrm{z} \cos \mathrm{z}<\pi / 2(*)$.
We have to consider six cases: (1) $\mathrm{x} \leq \mathr... | {
"problem_match": "\n## Problem A2",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 77 | 510 |
1989 | T2 | A3 | null | IberoAmerican_MO | If $\mathrm{a}, \mathrm{b}, \mathrm{c}$, are the sides of a triangle, show that $(\mathrm{a}-\mathrm{b}) /(\mathrm{a}+\mathrm{b})+(\mathrm{b}-\mathrm{c}) /(\mathrm{b}+\mathrm{c})+(\mathrm{c}-\mathrm{a}) /(\mathrm{a}+\mathrm{c})<$ $1 / 16$. | Put $\mathrm{f}(\mathrm{a}, \mathrm{b}, \mathrm{c})=(\mathrm{a}-\mathrm{b}) /(\mathrm{a}+\mathrm{b})+(\mathrm{b}-\mathrm{c}) /(\mathrm{b}+\mathrm{c})+(\mathrm{c}-\mathrm{a}) /(\mathrm{a}+\mathrm{c})$. Let $\mathrm{A}, \mathrm{B}, \mathrm{C}$ be the permutation of $\mathrm{a}, \mathrm{b}, \mathrm{c}$, with $\mathrm{A}<=... | {
"problem_match": "\n## Problem A3",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 98 | 951 |
1989 | T2 | B1 | null | IberoAmerican_MO | The incircle of the triangle $\mathrm{ABC}$ touches $\mathrm{AC}$ at $\mathrm{M}$ and $\mathrm{BC}$ at $\mathrm{N}$ and has center $\mathrm{O}$. $\mathrm{AO}$ meets $\mathrm{MN}$ at $\mathrm{P}$ and BO meets MN at $\mathrm{Q}$. Show that MP.OA $=\mathrm{BC} . \mathrm{OQ}$. | The key to getting started is to notice that angle $\mathrm{AQB}=90^{\circ}$.
Angle $\mathrm{BAQ}=90^{\circ}-\mathrm{B} / 2$, so angle $\mathrm{OAQ}=90^{\circ}-\mathrm{B} / 2-\mathrm{A} / 2=\mathrm{C} / 2$. So $\mathrm{OQ}=\mathrm{AO} \sin \mathrm{C} / 2$. Thus we have to show that $\mathrm{MP}=\mathrm{BC} \sin \mathr... | {
"problem_match": "\n## Problem B1",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 91 | 553 |
1990 | T2 | A2 | null | IberoAmerican_MO | I is the incenter of the triangle $\mathrm{ABC}$ and the incircle touches $\mathrm{BC}, \mathrm{CA}, \mathrm{AB}$ at $\mathrm{D}, \mathrm{E}, \mathrm{F}$ respectively. AD meets the incircle again at P. M is the midpoint of EF. Show that PMID is cyclic (or the points are collinear). | The first part is trivial. If $\mathrm{p}$ does not divide $(\mathrm{x}+\mathrm{b})$, then it does not divide $(\mathrm{x}+\mathrm{b})^{2}$, so it does not divide $(\mathrm{x}+\mathrm{b})^{2}+\mathrm{c}$. On the other hand, if $\mathrm{p}$ does divide $\mathrm{x}+\mathrm{b}$, then $\mathrm{p}^{2}$ divides $(\mathrm{x}+... | {
"problem_match": "\n## Problem A2",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 80 | 513 |
1990 | T2 | B1 | null | IberoAmerican_MO | The circle $\mathrm{C}$ has diameter $\mathrm{AB}$. The tangent at $\mathrm{B}$ is $\mathrm{T}$. For each point $\mathrm{M}$ (not equal to $\mathrm{A}$ ) on $\mathrm{C}$ there is a circle $\mathrm{C}^{\prime}$ which touches $\mathrm{T}$ and touches $\mathrm{C}$ at M. Find the point at which $\mathrm{C}^{\prime}$ touche... | Let $\mathrm{O}$ be the center of $\mathrm{C}$. Let the line $\mathrm{AM}$ meet $\mathrm{T}$ at $\mathrm{N}$. Let the perpendicular to $\mathrm{T}$ at $\mathrm{N}$ meet the line $\mathrm{OM}$ at $\mathrm{O}^{\prime}$. Then $\angle \mathrm{O}^{\prime} \mathrm{NM}=\angle \mathrm{MAB}$ ( $\mathrm{O}^{\prime} \mathrm{N}$ p... | {
"problem_match": "\n## Problem B1",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 218 | 525 |
1992 | T2 | A3 | null | IberoAmerican_MO | $\mathrm{ABC}$ is an equilateral triangle with side 2 . Show that any point $\mathrm{P}$ on the incircle satisfies $\mathrm{PA}^{2}$ $+\mathrm{PB}^{2}+\mathrm{PC}^{2}=5$. Show also that the triangle with side lengths $\mathrm{PA}, \mathrm{PB}, \mathrm{PC}$ has area $(\sqrt{3}) / 4$. | Take vectors centered at the center $\mathrm{O}$ of the triangle. Write the vector $\mathrm{OA}$ as $\mathbf{A}$ etc. Then $\mathrm{PA}^{2}$ $+\mathrm{PB}^{2}+\mathrm{PC}^{2}=(\mathbf{P}-\mathbf{A})^{2}+(\mathbf{P}-\mathbf{B})^{2}+(\mathbf{P}-\mathbf{C})^{2}=3 \mathrm{P}^{2}+\left(\mathrm{A}^{2}+\mathrm{B}^{2}+\mathrm{... | {
"problem_match": "\n## Problem A3",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 91 | 598 |
1994 | T2 | A1 | null | IberoAmerican_MO | Show that there is a number $1<\mathrm{b}<1993$ such that if 1994 is written in base $\mathrm{b}$ then all its digits are the same. Show that there is no number $1<b<1992$ such that if 1993 is written in base $\mathrm{b}$ then all its digits are the same. | Any even number $2 \mathrm{n}$ can be written as 22 in base $\mathrm{n}-1$. In particular $1994=22996$.
We have to show that we cannot write $1993=$ aaa ... $\mathrm{a}_{\mathrm{b}}$. If the number has $\mathrm{n}$ digits, then 1993 $=\mathrm{a}\left(1+\mathrm{b}+\ldots+\mathrm{b}^{\mathrm{n}-1}\right)=\mathrm{a}\left... | {
"problem_match": "\n## Problem A1",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 84 | 565 |
1994 | T2 | A2 | null | IberoAmerican_MO | $\mathrm{ABCD}$ is a cyclic quadrilateral. A circle whose center is on the side $\mathrm{AB}$ touches the other three sides. Show that $\mathrm{AB}=\mathrm{AD}+\mathrm{BC}$. What is the maximum possible area of $\mathrm{ABCD}$ in terms of $|\mathrm{AB}|$ and $|C D|$ ?
## Answer
![](https://cdn.mathpix.com/cropped/202... | Let the circle have center $\mathrm{O}$ on $\mathrm{AB}$ and radius $\mathrm{r}$. Let $\angle \mathrm{OAD}=\theta, \angle \mathrm{OBC}=\varphi$. Since $\mathrm{ABCD}$ is cyclic, $\angle \mathrm{ADC}=180^{\circ}-\varphi$, so $\angle \mathrm{ODA}=90^{\circ}-\varphi / 2$. If $\mathrm{AD}$ touches the circle at $\mathrm{X}... | {
"problem_match": "\n## Problem A2",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 154 | 1,053 |
1994 | T2 | A2 | null | IberoAmerican_MO | $\mathrm{ABCD}$ is a cyclic quadrilateral. A circle whose center is on the side $\mathrm{AB}$ touches the other three sides. Show that $\mathrm{AB}=\mathrm{AD}+\mathrm{BC}$. What is the maximum possible area of $\mathrm{ABCD}$ in terms of $|\mathrm{AB}|$ and $|C D|$ ?
## Answer
![](https://cdn.mathpix.com/cropped/202... | If $\mathrm{n}$ is odd, touch each bulb in the first column. Then bulbs in the first column are each switched $\mathrm{n}$ times, which is odd and so end up on. All other bulbs are switched just once, and so end up on. $\mathrm{n}$ is obviously minimal, because if $\mathrm{m}<\mathrm{n}$, then there is a bulb which is ... | {
"problem_match": "\n## Problem A2",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n# Solution"
} | 154 | 1,111 |
1994 | T2 | B2 | null | IberoAmerican_MO | $\mathrm{n}$ and $\mathrm{r}$ are positive integers. Find the smallest $\mathrm{k}$ for which we can construct $\mathrm{r}$ subsets $\mathrm{A}_{1}, \mathrm{~A}_{2}$, $\ldots, A_{r}$ of $\{0,1,2, \ldots, n-1\}$ each with $k$ elements such that each integer $0 \leq m<n$ can be written as a sum of one element from each o... | We can form at most $\mathrm{k}^{r}$ distinct sums, so $\mathrm{k}^{r}$ must be $\geq n$.
Now consider $\mathrm{A}_{1}=\{0,1,2, \ldots, \mathrm{k}-1\}, \mathrm{A}_{2}=\{0, \mathrm{k}, 2 \mathrm{k}, \ldots,(\mathrm{k}-1) \mathrm{k}\}, \mathrm{A}_{3}=\left\{0, \mathrm{k}^{2}, 2 \mathrm{k}^{2}, \ldots,(\mathrm{k}-\right.... | {
"problem_match": "\n## Problem B2",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 136 | 526 |
1996 | T2 | A3 | null | IberoAmerican_MO | $\mathrm{n}=\mathrm{k}^{2}-\mathrm{k}+1$, where $\mathrm{k}$ is a prime plus one. Show that we can color some squares of an $\mathrm{n} \mathrm{x}$ board black so that each row and column has exactly $\mathrm{k}$ black squares, but there is no rectangle with sides parallel to the sides of the board which has its four c... | We can regard the rows as lines and the columns as points. Black squares denote incidence. So line 3 contains point 4 iff square $(3,4)$ is black. The condition about rectangles then means that there is at most one line through two distinct points.
Suppose we take the points to be ( $\mathrm{a}, \mathrm{b}, \mathrm{c}... | {
"problem_match": "\n## Problem A3",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 85 | 1,388 |
1996 | T2 | B3 | null | IberoAmerican_MO | $A_{1}, A_{2}, \ldots, A_{n}$ are points in the plane. A non-zero real number $k_{i}$ is assigned to each point, so that the square of the distance between $A_{i}$ and $A_{j}($ for $i \neq j)$ is $k_{i}+k_{j}$. Show that $n$ is at most 4 and that if $n=4$, then $1 / k_{1}+1 / k_{2}+1 / k_{3}+1 / k_{4}=0$. | Suppose we have four points A, B, C, D with associated numbers $a, b, c, d$. Then $\mathrm{AB}^{2}=\mathrm{a}+\mathrm{b}$, $A C^{2}=a+c$, so $A B^{2}-A C^{2}=b-c$. Similarly, $D B^{2}-D C^{2}=b-c$, so $A B^{2}-A C^{2}=D B^{2}-D C^{2}$. Let $X$ be the foot of the perpendicular from $\mathrm{A}$ to $\mathrm{BC}$, and $\m... | {
"problem_match": "\n## Problem B3",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 128 | 783 |
1997 | T2 | B2 | null | IberoAmerican_MO | $\mathrm{ABC}$ is an acute-angled triangle with orthocenter $\mathrm{H} . \mathrm{AE}$ and $\mathrm{BF}$ are altitudes. $\mathrm{AE}$ is reflected in the angle bisector of angle $\mathrm{A}$ and $\mathrm{BF}$ is reflected in the angle bisector of angle $\mathrm{B}$. The two reflections intersect at $\mathrm{O}$. The ra... | We show first that $\mathrm{O}$ is the circumcenter of $\mathrm{ABC} . \angle \mathrm{ABF}=90^{\circ}-\mathrm{A}$. The line $\mathrm{BC}$ is the reflection in $B D$ of the line $B A$ and the line $\mathrm{BF}^{\prime}$ is the refection of $\mathrm{BF}$, so angle $\mathrm{CBF}^{\prime}=90^{\circ}$ A. But if $\mathrm{O}^... | {
"problem_match": "\n## Problem B2",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 190 | 529 |
1998 | T2 | A3 | null | IberoAmerican_MO | Find the smallest number $\mathrm{n}$ such that given any $\mathrm{n}$ distinct numbers from $\{1,2,3, \ldots, 999\}$, one can choose four different numbers $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ such that $\mathrm{a}+2 \mathrm{~b}+3 \mathrm{c}=\mathrm{d}$. | Answer: $\mathrm{n}=835$.
Consider the set $S=\{166,167, \ldots, 999\}$. The smallest possible value for $a+2 b+3 c$, for distinct $\mathrm{a}, \mathrm{b}, \mathrm{c}$ in $\mathrm{S}$ is $168+2.167+3.166=1000$. So we cannot find distinct $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ in $\mathrm{S}$ with $a+2 b+3 c=... | {
"problem_match": "\n## Problem A3",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 92 | 624 |
1999 | T2 | A2 | null | IberoAmerican_MO | Given two circles $\mathrm{C}$ and $\mathrm{C}^{\prime}$ we say that $\mathrm{C}$ bisects $\mathrm{C}^{\prime}$ if their common chord is a diameter of $\mathrm{C}^{\prime}$. Show that for any two circles which are not concentric, there are infinitely many circles which bisect them both. Find the locus of the centers of... | Let $\mathrm{C}, \mathrm{C}^{\prime}$ have center $\mathrm{O}, \mathrm{O}^{\prime}$ respectively and radius $\mathrm{r}, \mathrm{r}^{\prime}$ respectively. Let a circle center $\mathrm{P}$ bisect $C$. Suppose it meets $C$ at $A$ and $B$. Then $A B$ is perpendicular to $O P$ and is a diameter of C. Hence $\mathrm{PA}^{2... | {
"problem_match": "\n## Problem A2",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 92 | 999 |
1999 | T2 | B3 | null | IberoAmerican_MO | Given two points $\mathrm{A}$ and $\mathrm{B}$, take $\mathrm{C}$ on the perpendicular bisector of $\mathrm{AB}$. Define the sequence $C_{1}, C_{2}, C_{3}, \ldots$ as follows. $C_{1}=C$. If $C_{n}$ is not on $A B$, then $C_{n+1}$ is the circumcenter of the triangle $\mathrm{ABC}_{\mathrm{n}}$. If $\mathrm{C}_{\mathrm{n... | Answer: any $\mathrm{C}$ such that $\angle \mathrm{ACB}=180^{\circ} \mathrm{r} / \mathrm{s}$, with $\mathrm{r}$ and $\mathrm{s}$ relatively prime integers and $\mathrm{s}$ not a power of 2 .
Let $\angle \mathrm{AC}_{\mathrm{n}} \mathrm{B}=\mathrm{x}_{\mathrm{n}}$, where the angle is measured clockwise, so that $\mathr... | {
"problem_match": "\n## Problem B3",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 160 | 677 |
2000 | T2 | A3 | null | IberoAmerican_MO | Find all solutions to $(\mathrm{m}+1)^{\mathrm{a}}=\mathrm{m}^{\mathrm{b}}+1$ in integers greater than 1 .
## Answer
$(\mathrm{m}, \mathrm{a}, \mathrm{b})=(2,2,3)$. | Taking equation mod $\mathrm{m}+1$ we get $(-1)^{\mathrm{b}}=-1$, so $\mathrm{b}$ is odd. Hence we can divide the rhs by $\mathrm{m}+1$ to get $m^{b-1}-m^{b-2}+\ldots-m+1$. This has an odd number of terms. If $m$ is odd, then each term is odd and so the total is odd, but $(\mathrm{m}+1)^{\mathrm{a}-1}$ is even (note th... | {
"problem_match": "\n## Problem A3",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 66 | 536 |
2000 | T2 | B2 | null | IberoAmerican_MO | Given a pile of 2000 stones, two players take turns in taking stones from the pile. Each player must remove $1,2,3,4$, or 5 stones from the pile at each turn, but may not take the same number as his opponent took on his last move. The player who takes the last stone wins. Does the first or second player have a winning ... | The first player has a winning strategy. He takes 4 on his first move leaving 7 mod $13(2000=$ $153.13+7+4)$. Now we claim that the first player can always leave: (1) $0 \bmod 13$, (2) $3 \bmod$ 13 by taking away 3 , (3) $5 \bmod 13$ by taking away 5 , or (4) $7 \bmod 13$, and that the second player can never leave 0 m... | {
"problem_match": "\n## Problem B2",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 82 | 678 |
2000 | T2 | B3 | null | IberoAmerican_MO | A convex hexagon is called a unit if it has four diagonals of length 1, whose endpoints include all the vertices of the hexagon. Show that there is a unit of area $\mathrm{k}$ for any $0<\mathrm{k} \leq 1$. What is the largest possible area for a unit? | Answer: We can get arbitrarily close to (but not achieve) ( $3 \sqrt{3}$ )/4 (approx 1.3) by:
To prove the first part, consider the diagram below. Take $\mathrm{AB}=\mathrm{AC}=1$ and angle $\mathrm{BAC}=2 \theta$. Take $\mathrm{DE}=\mathrm{DF}=1$ and take the points of intersection $\mathrm{X}$ and $\mathrm{Y}$ such ... | {
"problem_match": "\n## Problem B3",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 67 | 895 |
2002 | T2 | B3 | null | IberoAmerican_MO | A game is played on a 2001 x 2001 board as follows. The first player's piece is the policeman, the second player's piece is the robber. Each piece can move one square south, one square east or one square northwest. In addition, the policeman (but not the robber) can move from the bottom right to the top left square in ... | Color the squares with three colors as follows:
```
0<ccccccccoc
1cccccccccoc
2cccccccccccollo
```
```
\(0 \begin{array}{llllllllll}0 & 1 & 2 & 0 & 1 & 2 & 0 & \ldots & 2\end{array}\)
\(\begin{array}{llllllllll}1 & 2 & 0 & 1 & 2 & 0 & 1 & \cdots & 0\end{array}\)
. .
\(\begin{array}{lllllllll}2 & 0 & 1 & 2 & 0 & 1 & 2... | {
"problem_match": "\n## Problem B3",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 175 | 2,075 |
2003 | T2 | B1 | null | IberoAmerican_MO | A $\square\{1,2,3, \ldots, 49\}$ does not contain six consecutive integers. Find the largest possible value of |A|. How many such subsets are there (of the maximum size)?
## Answer
$\max =41$; no. ways 495 | We must exclude at least one element of each of the 8 sets $\{1,2, \ldots, 6\},\{7, \ldots, 12\},\{13, \ldots$, $18\}, \ldots,\{43, \ldots, 48\}$. So $|\mathrm{A}| \leq 41$. But a value of 41 is certainly possible, for example, exclude $2,8,14, \ldots, 44$.
The largest excluded element must be at least 44 (or we have ... | {
"problem_match": "\n## Problem B1",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 65 | 750 |
2003 | T2 | B2 | null | IberoAmerican_MO | $\mathrm{ABCD}$ is a square. $\mathrm{P}, \mathrm{Q}$ are points on the sides $\mathrm{BC}, \mathrm{CD}$ respectively, distinct from the endpoints such that $\mathrm{BP}=\mathrm{CQ}$. $\mathrm{X}, \mathrm{Y}$ are points on $\mathrm{AP}, \mathrm{AQ}$ respectively. Show that there is a triangle with side lengths $B X, X ... | We have $\mathrm{DY}<\mathrm{BY} \leq \mathrm{BX}+\mathrm{XY}$ (this is almost obvious, but to prove formally use the cosine formula for BAY and DAY and notice that $\angle \mathrm{BAY}>\angle \mathrm{DAY}$ ). Similarly, $\mathrm{BX}<\mathrm{DX} \leq \mathrm{DY}+$ YX. So it remains to show that $X Y<B X+D Y$.
Take $\m... | {
"problem_match": "\n## Problem B2",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 99 | 560 |
2003 | T2 | B3 | null | IberoAmerican_MO | The sequences $\mathrm{a}_{0}, \mathrm{a}_{1}, \mathrm{a}_{2}, \ldots$ and $\mathrm{b}_{0}, \mathrm{~b}_{1}, \mathrm{~b}_{2}, \ldots$ are defined by $\mathrm{a}_{0}=1, \mathrm{~b}_{0}=4, \mathrm{a}_{\mathrm{n}+1}=\mathrm{a}_{\mathrm{n}}{ }^{2001}+\mathrm{b}_{\mathrm{n}}$, $b_{n+1}=b_{n}^{2001}+a_{n}$. Show that no memb... | 2003 is prime, so $\mathrm{a}^{2002}=1 \bmod 2003$ for any a not divisible by 2003. Thus $a_{n+1}=a_{n}^{-1}+b_{n}$ $\bmod 2003, b_{n+1}=b_{n}^{-1}+a_{n} \bmod 2003$. Put $c_{n}=a_{n} b_{n}$. Then $c_{n+1}=c_{n}+1 / c_{n}+2=\left(c_{n}+1\right)_{2} / c_{n}$ $\bmod 2003$. So if $\mathrm{c}_{\mathrm{n}} \neq 0 \bmod 2003... | {
"problem_match": "\n## Problem B3",
"resource_path": "IberoAmerican_MO/segmented/en-1985-2003-IberoamericanMO.jsonl",
"solution_match": "\n## Solution"
} | 155 | 890 |
2021 | T3 | 1 | null | JBMO | Let $n(n \geq 1)$ be an integer. Consider the equation
$$
2 \cdot\left\lfloor\frac{1}{2 x}\right\rfloor-n+1=(n+1)(1-n x)
$$
where $x$ is the unknown real variable.
(a) Solve the equation for $n=8$.
(b) Prove that there exists an integer $n$ for which the equation has at least 2021 solutions. (For any real number $y... | Let $k=\left[\frac{1}{2 x}\right], k \in \mathbb{Z}$.
(a) For $n=8$, the equation becomes
$$
k=\left[\frac{1}{2 x}\right]=8-36 x \Rightarrow x \neq 0 \text { and } x=\frac{8-k}{36}
$$
Since $x \neq 0$, we have $k \neq 8$, and the last relation implies $k=\left[\frac{1}{2 x}\right]=\left[\frac{18}{8-k}\right]$. Check... | {
"problem_match": "\nProblem 1.",
"resource_path": "JBMO/segmented/en-official/en-2021_jbmo_problems_and_sol.jsonl",
"solution_match": "\nSolution."
} | 137 | 502 |
2021 | T3 | 2 | null | JBMO | For any set $A=\left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right\}$ of five distinct positive integers denote by $S_{A}$ the sum of its elements, and denote by $T_{A}$ the number of triples $(i, j, k)$ with $1 \leqslant i<j<k \leqslant 5$ for which $x_{i}+x_{j}+x_{k}$ divides $S_{A}$.
Find the largest possible value of $... | We will prove that the maximum value that $T_{A}$ can attain is 4 . Let $A=$ $\left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right\}$ be a set of five positive integers such that $x_{1}<x_{2}<x_{3}<x_{4}<x_{5}$. Call a triple $(i, j, k)$ with $1 \leqslant i<j<k \leqslant 5$ good if $x_{i}+x_{j}+x_{k}$ divides $S_{A}$. None o... | {
"problem_match": "\nProblem 2.",
"resource_path": "JBMO/segmented/en-official/en-2021_jbmo_problems_and_sol.jsonl",
"solution_match": "\nSolution."
} | 124 | 1,021 |
2021 | T3 | 3 | null | JBMO | Let $A B C$ be an acute scalene triangle with circumcenter $O$. Let $D$ be the foot of the altitude from $A$ to the side $B C$. The lines $B C$ and $A O$ intersect at $E$. Let $s$ be the line through $E$ perpendicular to $A O$. The line $s$ intersects $A B$ and $A C$ at $K$ and $L$, respectively. Denote by $\omega$ the... | Let us denote angles of triangle $A B C$ with $\alpha, \beta, \gamma$ in a standard way. By basic anglechasing we have
$$
\angle B A D=90^{\circ}-\beta=\angle O A C \text { and } \angle C A D=\angle B A O=90^{\circ}-\gamma
$$
Using the fact that lines $A E$ and $A X$ are isogonal with respect to $\angle K A L$ we can... | {
"problem_match": "\nProblem 3.",
"resource_path": "JBMO/segmented/en-official/en-2021_jbmo_problems_and_sol.jsonl",
"solution_match": "## Solution."
} | 157 | 723 |
2021 | T3 | 4 | null | JBMO | Let $M$ be a subset of the set of 2021 integers $\{1,2,3, \ldots, 2021\}$ such that for any three elements (not necessarily distinct) $a, b, c$ of $M$ we have $|a+b-c|>10$. Determine the largest possible number of elements of $M$. | The set $M=\{1016,1017, \ldots, 2021\}$ has 1006 elements and satisfies the required property, since $a, b, c \in M$ implies that $a+b-c \geqslant 1016+1016-2021=11$. We will show that this is optimal.
Suppose $M$ satisfies the condition in the problem. Let $k$ be the minimal element of $M$. Then $k=|k+k-k|>10 \Righta... | {
"problem_match": "\nProblem 4.",
"resource_path": "JBMO/segmented/en-official/en-2021_jbmo_problems_and_sol.jsonl",
"solution_match": "\nSolution."
} | 81 | 701 |
2022 | T3 | 2 | null | JBMO | Let $A B C$ be an acute triangle such that $A H=H D$, where $H$ is the orthocenter of $A B C$ and $D \in B C$ is the foot of the altitude from the vertex $A$. Let $\ell$ denote the line through $H$ which is tangent to the circumcircle of the triangle $B H C$. Let $S$ and $T$ be the intersection points of $\ell$ with $A... | 1. In order to prove that $S M$ and $T N$ are parallel, it suffices to prove that both of them are perpendicular to $S T$. Due to symmetry, we will provide a detailed proof of $S M \perp S T$, whereas the proof of $T N \perp S T$ is analogous. In this solution we will use the following notation: $\angle B A C=\alpha, \... | {
"problem_match": "\nProblem 2.",
"resource_path": "JBMO/segmented/en-official/en-jbmo-2022-solutions.jsonl",
"solution_match": "\nSolution "
} | 150 | 819 |
2022 | T3 | 3 | null | JBMO | Find all quadruples of positive integers $(p, q, a, b)$, where $p$ and $q$ are prime numbers and $a>1$, such that
$$
p^{a}=1+5 q^{b}
$$ | 1. First of all, observe that if $p, q$ are both odd, then the left hand side of the given equation is odd and the right hand side is even so there are no solutions in this case. In other words, one of these numbers has to be equal to 2 so we can discuss the following two cases:
- $p=2$
In this case the given equatio... | {
"problem_match": "\nProblem 3.",
"resource_path": "JBMO/segmented/en-official/en-jbmo-2022-solutions.jsonl",
"solution_match": "\nSolution "
} | 51 | 2,322 |
2022 | T3 | 4 | null | JBMO | We call an even positive integer $n$ nice if the set $\{1,2, \ldots, n\}$ can be partitioned into $\frac{n}{2}$ two-element subsets, such that the sum of the elements in each subset is a power of 3 . For example, 6 is nice, because the set $\{1,2,3,4,5,6\}$ can be partitioned into subsets $\{1,2\},\{3,6\},\{4,5\}$. Fin... | For a nice number $n$ and a given partition of the set $\{1,2, \ldots, n\}$ into twoelement subsets such that the sum of the elements in each subset is a power of 3 , we say that $a, b \in\{1,2, \ldots, n\}$ are paired if both of them belong to the same subset.
Let $x$ be a nice number and $k$ be a (unique) non-negati... | {
"problem_match": "\nProblem 4.",
"resource_path": "JBMO/segmented/en-official/en-jbmo-2022-solutions.jsonl",
"solution_match": "\nSolution."
} | 132 | 898 |
2016 | T3 | 1 | null | JBMO | A trapezoid $A B C D(A B \| C D, A B>C D)$ is circumscribed. The incircle of the triangle $A B C$ touches the lines $A B$ and $A C$ at the points $M$ and $N$, respectively. Prove that the incenter of the trapezoid $A B C D$ lies on the line $M N$.
=90^{\circ}-\frac{1}{2} m(\widehat{M A N}) \quad \text { and } \quad m(\widehat{B I C})=90^{\circ}+\frac{1}{2} m(\widehat{M A N})
$$
the quadrilateral $I R N C$ is cyclic.
It fo... | {
"problem_match": "\nProblem 1.",
"resource_path": "JBMO/segmented/en-official/en-jbmo_2016_paper_eng.jsonl",
"solution_match": "## Solution."
} | 159 | 714 |
2016 | T3 | 3 | null | JBMO | Find all the triples of integers $(a, b, c)$ such that the number
$$
N=\frac{(a-b)(b-c)(c-a)}{2}+2
$$
is a power of 2016 .
(A power of 2016 is an integer of the form $2016^{n}$, where $n$ is a non-negative integer.) | Let $a, b, c$ be integers and $n$ be a positive integer such that
$$
(a-b)(b-c)(c-a)+4=2 \cdot 2016^{n}
$$
We set $a-b=-x, b-c=-y$ and we rewrite the equation as
$$
x y(x+y)+4=2 \cdot 2016^{n}
$$
If $n>0$, then the right hand side is divisible by 7 , so we have that
$$
x y(x+y)+4 \equiv 0 \quad(\bmod 7)
$$
or
$$... | {
"problem_match": "\nProblem 3.",
"resource_path": "JBMO/segmented/en-official/en-jbmo_2016_paper_eng.jsonl",
"solution_match": "\nSolution."
} | 84 | 511 |
2016 | T3 | 4 | null | JBMO | A $5 \times 5$ table is called regular if each of its cells contains one of four pairwise distinct real numbers, such that each of them occurs exactly once in every $2 \times 2$ subtable. The sum of all numbers of a regular table is called the total sum of the table. With any four numbers, one constructs all possible r... | We will prove that the maximum number of total sums is 60 .
The proof is based on the following claim.
Claim. In a regular table either each row contains exactly two of the numbers, or each column contains exactly two of the numbers.
Proof of the Claim. Indeed, let $R$ be a row containing at least three of the numbe... | {
"problem_match": "\nProblem 4.",
"resource_path": "JBMO/segmented/en-official/en-jbmo_2016_paper_eng.jsonl",
"solution_match": "\nSolution."
} | 92 | 1,589 |
2017 | T3 | 2 | null | JBMO | Let $x, y, z$ be positive integers such that $x \neq y \neq z \neq x$. Prove that
$$
(x+y+z)(x y+y z+z x-2) \geq 9 x y z
$$
When does the equality hold? | Since $x, y, z$ are distinct positive integers, the required inequality is symmetric and WLOG we can suppose that $x \geq y+1 \geq z+2$. We consider 2 possible cases:
Case 1. $y \geq z+2$. Since $x \geq y+1 \geq z+3$ it follows that
$$
(x-y)^{2} \geq 1, \quad(y-z)^{2} \geq 4, \quad(x-z)^{2} \geq 9
$$
which are equiv... | {
"problem_match": "\nProblem 2.",
"resource_path": "JBMO/segmented/en-official/en-jbmo_2017_english_solutions.jsonl",
"solution_match": "\nSolution."
} | 64 | 619 |
2017 | T3 | 3 | null | JBMO | Let $A B C$ be an acute triangle such that $A B \neq A C$, with circumcircle $\Gamma$ and circumcenter $O$. Let $M$ be the midpoint of $B C$ and $D$ be a point on $\Gamma$ such that $A D \perp B C$. Let $T$ be a point such that $B D C T$ is a parallelogram and $Q$ a point on the same side of $B C$ as $A$ such that
$$
... | Let $X^{\prime}$ be symmetric point to $Q$ in line $B C$. Now since $\angle C B A=\angle C Q M=\angle C X^{\prime} M$, $\angle B C A=\angle B Q M=\angle B X^{\prime} M$, we have
$$
\angle B X^{\prime} C=\angle B X^{\prime} M+\angle C X^{\prime} M=\angle C B A+\angle B C A=180^{\circ}-\angle B A C
$$
we have that $X^{... | {
"problem_match": "\nProblem 3.",
"resource_path": "JBMO/segmented/en-official/en-jbmo_2017_english_solutions.jsonl",
"solution_match": "\nSolution."
} | 207 | 920 |
2017 | T3 | 4 | null | JBMO | Consider a regular $2 n$-gon $P, A_{1} A_{2} \ldots A_{2 n}$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$, if the line segment $S E$ contains no other points that lie on the sides of $P$ except $S$. We color the si... | Answer: For $n=2$, the answer is 36 ; for $n=3$, the answer is 30 and for $n \geq 4$, the answer is $6 n$.
Lemma 1. Given a regular $2 n$-gon in the plane and a sequence of $n$ consecutive sides $s_{1}, s_{2}, \ldots, s_{n}$ there is an external point $Q$ in the plane, such that the color of each $s_{i}$ can be seen f... | {
"problem_match": "\nProblem 4.",
"resource_path": "JBMO/segmented/en-official/en-jbmo_2017_english_solutions.jsonl",
"solution_match": "\nSolution "
} | 205 | 1,429 |
2018 | T3 | 2 | null | JBMO | Let $n$ three-digit numbers satisfy the following properties:
(1) No number contains the digit 0 .
(2) The sum of the digits of each number is 9 .
(3) The units digits of any two numbers are different.
(4) The tens digits of any two numbers are different.
(5) The hundreds digits of any two numbers are different.
... | Let $S$ denote the set of three-digit numbers that have digit sum equal to 9 and no digit equal to 0 . We will first find the cardinality of $S$. We start from the number 111 and each element of $S$ can be obtained from 111 by a string of 6 A's (which means that we add 1 to the current digit) and $2 G$ 's (which means ... | {
"problem_match": "\nProblem 2.",
"resource_path": "JBMO/segmented/en-official/en-jbmo_2018_solutions.jsonl",
"solution_match": "\nSolution."
} | 86 | 644 |
2018 | T3 | 4 | null | JBMO | Let $A B C$ be an acute triangle, $A^{\prime}, B^{\prime}$ and $C^{\prime}$ be the reflections of the vertices $A, B$ and $C$ with respect to $B C, C A$, and $A B$, respectively, and let the circumcircles of triangles $A B B^{\prime}$ and $A C C^{\prime}$ meet again at $A_{1}$. Points $B_{1}$ and $C_{1}$ are defined si... | Let $O_{1}, O_{2}$ and $O$ be the circumcenters of triangles $A B B^{\prime}, A C C^{\prime}$ and $A B C$ respectively. As $A B$ is the perpendicular bisector of the line segment $C C^{\prime}, O_{2}$ is the intersection of the perpendicular bisector of $A C$ with $A B$. Similarly, $O_{1}$ is the intersection of the pe... | {
"problem_match": "\nProblem 4.",
"resource_path": "JBMO/segmented/en-official/en-jbmo_2018_solutions.jsonl",
"solution_match": "\nSolution."
} | 139 | 555 |
2020 | T3 | 1 | null | JBMO | Find all triples $(a, b, c)$ of real numbers such that the following system holds:
$$
\left\{\begin{array}{l}
a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \\
a^{2}+b^{2}+c^{2}=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}
\end{array}\right.
$$ | First of all if $(a, b, c)$ is a solution of the system then also $(-a,-b,-c)$ is a solution. Hence we can suppose that $a b c>0$. From the first condition we have
$$
a+b+c=\frac{a b+b c+c a}{a b c}
$$
Now, from the first condition and the second condition we get
$$
(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)=\left(\... | {
"problem_match": "\nProblem 1.",
"resource_path": "JBMO/segmented/en-official/en-jbmo_2020_solutions.jsonl",
"solution_match": "\nSolution."
} | 106 | 860 |
2020 | T3 | 2 | null | JBMO | Let $\triangle A B C$ be a right-angled triangle with $\angle B A C=90^{\circ}$ and let $E$ be the foot of the perpendicular from $A$ on $B C$. Let $Z \neq A$ be a point on the line $A B$ with $A B=B Z$. Let (c) be the circumcircle of the triangle $\triangle A E Z$. Let $D$ be the second point of intersection of $(c)$ ... | We will first show that $P A$ is tangent to $(c)$ at $A$.
Since $E, D, Z, A$ are concyclic, then $\angle E D C=\angle E A Z=\angle E A B$. Since also the triangles $\triangle A B C$ and $\triangle E B A$ are similar, then $\angle E A B=\angle B C A$, therefore $\angle E D C=\angle B C A$.
Since $\angle F E D=90^{\cir... | {
"problem_match": "\nProblem 2.",
"resource_path": "JBMO/segmented/en-official/en-jbmo_2020_solutions.jsonl",
"solution_match": "\nSolution."
} | 186 | 535 |
2020 | T3 | 3 | null | JBMO | Alice and Bob play the following game: Alice picks a set $A=\{1,2, \ldots, n\}$ for some natural number $n \geqslant 2$. Then starting with Bob, they alternatively choose one number from the set $A$, according to the following conditions: initially Bob chooses any number he wants, afterwards the number chosen at each s... | To say that Alice has a winning strategy means that she can find a number $n$ to form the set A, so that she can respond appropriately to all choices of Bob and always get at the end a composite number for the sum of her choices. If such $n$ does not exist, this would mean that Bob has a winning strategy instead.
Alic... | {
"problem_match": "\nProblem 3.",
"resource_path": "JBMO/segmented/en-official/en-jbmo_2020_solutions.jsonl",
"solution_match": "\nSolution."
} | 144 | 1,003 |
2023 | T3 | 2 | null | JBMO | Prove that for all non-negative real numbers $x, y, z$, not all equal to 0 , the following inequality holds
$$
\frac{2 x^{2}-x+y+z}{x+y^{2}+z^{2}}+\frac{2 y^{2}+x-y+z}{x^{2}+y+z^{2}}+\frac{2 z^{2}+x+y-z}{x^{2}+y^{2}+z} \geqslant 3
$$
Determine all the triples $(x, y, z)$ for which the equality holds. | Let us first write the expression $L$ on the left hand side in the following way
$$
\begin{aligned}
L & =\left(\frac{2 x^{2}-x+y+z}{x+y^{2}+z^{2}}+2\right)+\left(\frac{2 y^{2}+x-y+z}{x^{2}+y+z^{2}}+2\right)+\left(\frac{2 z^{2}+x+y-z}{x^{2}+y^{2}+z}+2\right)-6 \\
& =\left(2 x^{2}+2 y^{2}+2 z^{2}+x+y+z\right)\left(\frac... | {
"problem_match": "## Problem 2.",
"resource_path": "JBMO/segmented/en-official/en-jbmo_2023_final_paper_-_with_solutions_1.jsonl",
"solution_match": "\nSolution."
} | 128 | 717 |
2023 | T3 | 3 | null | JBMO | Alice and Bob play the following game on a $100 \times 100$ grid, taking turns, with Alice starting first. Initially the grid is empty. At their turn, they choose an integer from 1 to $100^{2}$ that is not written yet in any of the cells and choose an empty cell, and place it in the chosen cell. When there is no empty ... | We denote by $(i, j)$ the cell in the $i$-th line and in the $j$-th column for every $1 \leq i, j \leq n$. Bob associates the following pair of cells : $(i, 2 k+1),(i, 2 k+2)$ for $1 \leq i \leq 100$ and $0 \leq k \leq 49$ except for $(i, k)=(100,0)$ and $(100,1)$, and the pairs $(100,1),(100,3)$ and $(100,2),(100,4)$.... | {
"problem_match": "## Problem 3.",
"resource_path": "JBMO/segmented/en-official/en-jbmo_2023_final_paper_-_with_solutions_1.jsonl",
"solution_match": "\nSolution."
} | 187 | 675 |
2023 | T3 | 4 | null | JBMO | Let $A B C$ be an acute triangle with circumcenter $O$. Let $D$ be the foot of the altitude from $A$ to $B C$ and let $M$ be the midpoint of $O D$. The points $O_{b}$ and $O_{c}$ are the circumcenters of triangles $A O C$ and $A O B$, respectively. If $A O=A D$, prove that the points $A, O_{b}, M$ and $O_{c}$ are concy... | Note that $A B=A C$ cannot hold since $A O=A D$ would imply that $O$ is the midpoint of $B C$, which is not possible for an acute triangle. So we may assume without loss of generality that $A B<A C$.
Let $M_{b}$ and $M_{c}$ be the midpoints of $A C$ and $A B$, respectively. Since $\angle A M_{b} O=\angle A M_{c} O=$ $... | {
"problem_match": "## Problem 4.",
"resource_path": "JBMO/segmented/en-official/en-jbmo_2023_final_paper_-_with_solutions_1.jsonl",
"solution_match": "## Solution."
} | 110 | 707 |
2019 | T3 | 1 | null | JBMO | Find all prime numbers $p$ for which there exist positive integers $x, y$ and $z$ such that the number
$$
x^{p}+y^{p}+z^{p}-x-y-z
$$
is a product of exactly three distinct prime numbers. | Let $A=x^{p}+y^{p}+z^{p}-x-y-z$. For $p=2$, we take $x=y=4$ and $z=3$. Then $A=30=2 \cdot 3 \cdot 5$. For $p=3$ we can take $x=3$ and $y=2$ and $z=1$. Then again $A=30=2 \cdot 3 \cdot 5$. For $p=5$ we can take $x=2$ and $y=1$ and $z=1$. Again $A=30=2 \cdot 3 \cdot 5$.
Assume now that $p \geqslant 7$. Working modulo 2 ... | {
"problem_match": "\nProblem 1.",
"resource_path": "JBMO/segmented/en-official/en-optimizedjbmo_2019_problems_and_solutions_english.jsonl",
"solution_match": "\nSolution."
} | 59 | 661 |
2019 | T3 | 2 | null | JBMO | Let $a, b$ be two distinct real numbers and let $c$ be a positive real number such that
$$
a^{4}-2019 a=b^{4}-2019 b=c .
$$
Prove that $-\sqrt{c}<a b<0$. | Firstly, we see that
$$
2019(a-b)=a^{4}-b^{4}=(a-b)(a+b)\left(a^{2}+b^{2}\right)
$$
Since $a \neq b$, we get $(a+b)\left(a^{2}+b^{2}\right)=2019$, so $a+b \neq 0$. Thus
$$
\begin{aligned}
2 c & =a^{4}-2019 a+b^{4}-2019 b \\
& =a^{4}+b^{4}-2019(a+b) \\
& =a^{4}+b^{4}-(a+b)^{2}\left(a^{2}+b^{2}\right) \\
& =-2 a b\lef... | {
"problem_match": "\nProblem 2.",
"resource_path": "JBMO/segmented/en-official/en-optimizedjbmo_2019_problems_and_solutions_english.jsonl",
"solution_match": "\nSolution."
} | 62 | 1,022 |
2019 | T3 | 3 | null | JBMO | Triangle $A B C$ is such that $A B<A C$. The perpendicular bisector of side $B C$ intersects lines $A B$ and $A C$ at points $P$ and $Q$, respectively. Let $H$ be the orthocentre of triangle $A B C$, and let $M$ and $N$ be the midpoints of segments $B C$ and $P Q$, respectively. Prove that lines $H M$ and $A N$ meet on... | We have
$$
\angle A P Q=\angle B P M=90^{\circ}-\angle M B P=90^{\circ}-\angle C B A=\angle H C B
$$
and
$$
\angle A Q P=\angle M Q C=90^{\circ}-\angle Q C M=90^{\circ}-\angle A C B=\angle C B H
$$
From these two equalities, we see that the triangles $A P Q$ and $H C B$ are similar. Moreover, since $M$ and $N$ are ... | {
"problem_match": "\nProblem 3.",
"resource_path": "JBMO/segmented/en-official/en-optimizedjbmo_2019_problems_and_solutions_english.jsonl",
"solution_match": "\nSolution."
} | 113 | 574 |
2015 | T3 | 1 | null | JBMO | Find all prime numbers $a, b, c$ and positive integers $k$ which satisfy the equation
$$
a^{2}+b^{2}+16 \cdot c^{2}=9 \cdot k^{2}+1
$$ | The relation $9 \cdot k^{2}+1 \equiv 1(\bmod 3)$ implies
$$
a^{2}+b^{2}+16 \cdot c^{2} \equiv 1(\bmod 3) \Leftrightarrow a^{2}+b^{2}+c^{2} \equiv 1(\bmod 3)
$$
Since $a^{2} \equiv 0,1(\bmod 3), \quad b^{2} \equiv 0,1(\bmod 3), c^{2} \equiv 0,1(\bmod 3)$, we have:
| $a^{2}$ | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| :---: |... | {
"problem_match": "\nProblem 1.",
"resource_path": "JBMO/segmented/en-official/en-solutions-english-jbmo2015.jsonl",
"solution_match": "# Solution:"
} | 53 | 928 |
2015 | T3 | 2 | null | JBMO | Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Find the minimum value of the expression
$$
A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}
$$
$19^{\text {th }}$ Junior B... | We can rewrite $A$ as follows:
$$
\begin{aligned}
& A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-a^{2}-b^{2}-c^{2}= \\
& 2\left(\frac{a b+b c+c a}{a b c}\right)-\left(a^{2}+b^{2}+c^{2}\right)=2\left(\frac{a b+b c+c a}{a b c}\right)-\left((a+b+c)^{2}-2(a b+b ... | {
"problem_match": "\nProblem 2.",
"resource_path": "JBMO/segmented/en-official/en-solutions-english-jbmo2015.jsonl",
"solution_match": "## Solution:"
} | 171 | 535 |
2015 | T3 | 4 | null | JBMO | An $L$-figure is one of the following four pieces, each consisting of three unit squares:
A $5 \times 5$ board, consisting of 25 unit squares, a positive integer $k \leq 25$ and an unlimited... | We will show that player $\boldsymbol{A}$ wins if $k=1,2,3$, but player $\boldsymbol{B}$ wins if $k=4$. Thus the smallest $k$ for which $\boldsymbol{B}$ has a winning strategy exists and is equal to 4 .
If $k=1$, player $\boldsymbol{A}$ marks the upper left corner of the square and then fills it as follows.
## $19^{\... | {
"problem_match": "## Problem 4.",
"resource_path": "JBMO/segmented/en-official/en-solutions-english-jbmo2015.jsonl",
"solution_match": "## Solution:"
} | 286 | 629 |
2014 | T3 | 2 | null | JBMO | Consider an acute triangle $A B C$ with area S. Let $C D \perp A B \quad(D \in A B)$, $D M \perp A C \quad(M \in A C)$ and $\quad D N \perp B C \quad(N \in B C)$. Denote by $H_{1}$ and $H_{2}$ the orthocentres of the triangles $M N C$ and $M N D$ respectively. Find the area of the quadrilateral $\mathrm{AH}_{1} \mathrm... | 1. Let $O, P, K, R$ and $T$ be the mid-points of the segments $C D, M N$, $C N, C H_{1}$ and $M H_{1}$, respectively. From $\triangle M N C$ we have that $\overline{P K}=\frac{1}{2} \overline{M C}$ and $P K \| M C$. Analogously, from $\Delta M H_{1} C$ we have that $\overline{T R}=\frac{1}{2} \overline{M C}$ and $T R \... | {
"problem_match": "\nProblem 2.",
"resource_path": "JBMO/segmented/en-official/en-solutions_en-jbmo2014.jsonl",
"solution_match": "\nSolution "
} | 128 | 768 |
2014 | T3 | 3 | null | JBMO | Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that
$$
\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} \geq 3(a+b+c+1)
$$
When does equality hold? | ## Solution 3.
By using $x^{2}+y^{2}+z^{2} \geq x y+y z+z x$
$$
\begin{aligned}
& \left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2}=a^{2}+b^{2}+c^{2}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{a^{2}}+\frac{2 a}{b}+\frac{2 b}{c}+\frac{2 c}{a} \geq \\
& \geq a b+a c+b c+\frac{1}... | {
"problem_match": "\nProblem 3.",
"resource_path": "JBMO/segmented/en-official/en-solutions_en-jbmo2014.jsonl",
"solution_match": "\nSolution "
} | 90 | 569 |
2014 | T3 | 3 | null | JBMO | Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that
$$
\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} \geq 3(a+b+c+1)
$$
When does equality hold? | Solution 4. $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$
$$
\begin{aligned}
& \left(\frac{x}{y}+\frac{z}{y}\right)^{2}+\left(\frac{y}{z}+\frac{x}{z}\right)^{2}+\left(\frac{z}{x}+\frac{y}{x}\right)^{2} \geq 3\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+1\right) \\
& (x+z)^{2} x^{2} z^{2}+(y+x)^{2} y^{2} x^{2}+(z+y)^{2} z... | {
"problem_match": "\nProblem 3.",
"resource_path": "JBMO/segmented/en-official/en-solutions_en-jbmo2014.jsonl",
"solution_match": "\nSolution "
} | 90 | 568 |
2014 | T3 | 4 | null | JBMO | For a positive integer $n$, two players A and B play the following game: Given a pile of $s$ stones, the players take turn alternatively with A going first. On each turn the player is allowed to take either one stone, or a prime number of stones, or a multiple of $n$ stones. The winner is the one who takes the last sto... | Denote by $k$ the sought number and let $\left\{s_{1}, \mathrm{~s}_{2}, \ldots, \mathrm{s}_{k}\right\}$ be the corresponding values for $s$. We call each $s_{i}$ a losing number and every other nonnegative integer a winning numbers.
## Clearly every multiple of $n$ is a winning number.
Suppose there are two different... | {
"problem_match": "\nProblem 4.",
"resource_path": "JBMO/segmented/en-official/en-solutions_en-jbmo2014.jsonl",
"solution_match": "\nSolution."
} | 106 | 891 |
2011 | T3 | A6 | Algebra | JBMO-SL | Let $x_{i}>1$, for all $i \in\{1,2,3, \ldots, 2011\}$. Prove the inequality $\sum_{i=1}^{2011} \frac{x_{i}^{2}}{x_{i+1}-1} \geq 8044$ where $x_{2012}=x_{1}$. When does equality hold? | Realize that $\left(x_{i}-2\right)^{2} \geq 0 \Leftrightarrow x_{i}^{2} \geq 4\left(x_{i}-1\right)$. So we get:
$\frac{x_{1}^{2}}{x_{2}-1}+\frac{x_{2}^{2}}{x_{3}-1}+\ldots+\frac{x_{2011}^{2}}{x_{1}-1} \geq 4\left(\frac{x_{1}-1}{x_{2}-1}+\frac{x_{2}-1}{x_{3}-1}+\ldots+\frac{x_{2011}-1}{x_{1}-1}\right)$. By $A M-G M$ :
... | {
"problem_match": "\nA6 ",
"resource_path": "JBMO/segmented/en-shortlist/en-alg-20111.jsonl",
"solution_match": "## Solution 1"
} | 97 | 1,015 |
2011 | T3 | A7 | Algebra | JBMO-SL | Let $a, b, c$ be positive real numbers with $a b c=1$. Prove the inequality:
$$
\frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1}+\frac{2 b^{2}+\frac{1}{b}}{c+\frac{1}{b}+1}+\frac{2 c^{2}+\frac{1}{c}}{a+\frac{1}{c}+1} \geq 3
$$ | By $A M-G M$ we have $2 x^{2}+\frac{1}{x}=x^{2}+x^{2}+\frac{1}{x} \geq 3 \sqrt[3]{\frac{x^{4}}{x}}=3 x$ for all $x>0$, so we have:
$\sum_{\text {cyc }} \frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1} \geq \sum_{c y c} \frac{3 a}{1+b+b c}=3\left(\sum_{c y c} \frac{a^{2}}{1+a+a b}\right) \geq \frac{3(a+b+c)^{2}}{3+a+b+c+a b... | {
"problem_match": "\nA7 ",
"resource_path": "JBMO/segmented/en-shortlist/en-alg-20111.jsonl",
"solution_match": "## Solution 1"
} | 115 | 729 |
2011 | T3 | A9 | Algebra | JBMO-SL | Let $x_{1}, x_{2}, \ldots, x_{n}$ be real numbers satisfying $\sum_{k=1}^{n-1} \min \left(x_{k} ; x_{k+1}\right)=\min \left(x_{1}, x_{n}\right)$.
Prove that $\sum_{k=2}^{n-1} x_{k} \geq 0$. | Case I. If $\min \left(x_{1}, x_{n}\right)=x_{1}$, we know that $x_{k} \geq \min \left(x_{k} ; x_{k+1}\right)$ for all $k \in\{1,2,3, \ldots, n-1\}$. So $x_{1}+x_{2}+\ldots+x_{n-1} \geq \sum_{k=1}^{n-1} \min \left(x_{k} ; x_{k+1}\right)=\min \left(x_{1}, x_{n}\right)=x_{1}$, hence $\sum_{k=2}^{n-1} x_{k} \geq 0$.
Case... | {
"problem_match": "\nA9 ",
"resource_path": "JBMO/segmented/en-shortlist/en-alg-20111.jsonl",
"solution_match": "## Solution 1"
} | 91 | 651 |
2011 | T3 | C4 | Combinatorics | JBMO-SL | In a group of $n$ people, each one had a different ball. They performed a sequence of swaps; in each swap, two people swapped the ball they had at that moment. Each pair of people performed at least one swap. In the end each person had the ball he/she had at the start. Find the least possible number of swaps, if: $a$ )... | We will denote the people by $A, B, C, \ldots$ and their initial balls by the corresponding small letters. Thus the initial state is $A a, B b, C c, D d, E e(, F f)$. A swap is denoted by the (capital) letters of the people involved.
a) Five people form 10 pairs, so at least 10 swaps are necessary.
In fact, 10 swaps ... | {
"problem_match": "\nC4 ",
"resource_path": "JBMO/segmented/en-shortlist/en-combi-2011.jsonl",
"solution_match": "## Solution"
} | 90 | 597 |
2011 | T3 | C7 | Combinatorics | JBMO-SL | Consider a rectangle whose lengths of sides are natural numbers. If someone places as many squares as possible, each with area 3 , inside of the given rectangle, such that
the sides of the squares are parallel to the rectangle sides, then the maximal number of these squares fill exactly half of the area of the rectangl... | Let $A B C D$ be a rectangle with $A B=m$ and $A D=n$ where $m, n$ are natural numbers such that $m \geq n \geq 2$. Suppose that inside of the rectangle $A B C D$ is placed a rectangular lattice consisting of some identical squares whose areas are equals to 3 , where $k$ of them are placed along the side $A B$ and $l$ ... | {
"problem_match": "\nC7 ",
"resource_path": "JBMO/segmented/en-shortlist/en-combi-2011.jsonl",
"solution_match": "## Solution"
} | 74 | 1,326 |
2011 | T3 | G4 | Geometry | JBMO-SL | Point $D$ lies on the side $[B C]$ of $\triangle A B C$. The circumcenters of $\triangle A D C$ and $\triangle B A D$ are $O_{1}$ and $O_{2}$, respectively and $O_{1} O_{2} \| A B$. The orthocenter of $\triangle A D C$ is $H$ and $A H=O_{1} O_{2}$. Find the angles of $\triangle A B C$ if $2 m(<C)=3 m(<B)$. | As $A D$ is the radical axis of the circumcircles of $\triangle A D C$ and $\triangle B A D$, we have that $O_{1} O_{2} \perp A D$, therefore $\widehat{D A B}=90^{\circ}$. Let $F$ be the midpoint of $[C D]$ and $[C E]$ be a diameter of the circumcircle of $\triangle A D C$. Then $E D \perp C D$ and $E A \perp C A$, so ... | {
"problem_match": "\nG4 ",
"resource_path": "JBMO/segmented/en-shortlist/en-geome-2011.jsonl",
"solution_match": "## Solution 1"
} | 117 | 727 |
2003 | T3 | ALG 7 | Algebra | JBMO-SL | .
Let $x, y, z$ be real numbers greater than -1 . Prove that
$$
\frac{1+x^{2}}{1+y+z^{2}}+\frac{1+y^{2}}{1+z+x^{2}}+\frac{1+z^{2}}{1+x+y^{2}} \geq 2
$$ | We have $y \leq \frac{1+y^{2}}{2}$, hence $\quad$
$$
\frac{1+x^{2}}{1+y+z^{2}} \geq \frac{1+x^{2}}{1+z^{2}+\frac{1+\dot{y}^{2}}{2}}
$$
and the similar inequalities.
Setting $a=1+x^{2}, b=1+y^{2}, c=1+z^{2}$, it sufices to prove that
$$
\frac{a}{2 c+b}+\frac{b}{2 a+c}+\frac{c}{2 b+a} \geq 1
$$
for all $a, b, c \geq... | {
"problem_match": "\nALG 7 ",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl",
"solution_match": "\nSolution."
} | 75 | 619 |
2003 | T3 | ALG 8 | Algebra | JBMO-SL | Prove that there exist two sets $A=\{x, y, z\}$ and $B=\{m, n, p\}$ of positive integers greater than 2003 such that the sets have no common elements and the equalities $x+y+z=m+n+p$ and $x^{2}+y^{2}+z^{2}=m^{2}+n^{2}+p^{2}$ hold. | Let $A B C$ be a triangle with $B C=a, A C=b, A B=c$ and $a<b<c$. Denote by $m_{a}, m_{b}, m_{c}$ the lengths of medianes drawing from the vertices $A, B, C$ respectively. Using the formulas
$$
4 m_{a}^{2}=2\left(b^{2}+c^{2}\right)-a^{2}, 4 m_{b}^{2}=2\left(a^{2}+c^{2}\right)-b^{2}, 4 m_{c}^{2}=2\left(a^{2}+b^{2}\righ... | {
"problem_match": "\nALG 8.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl",
"solution_match": "\nSolution."
} | 92 | 833 |
2003 | T3 | COM 5 | Combinatorics | JBMO-SL | If $m$ is a number from the set $\{1,2,3,4\}$ and each point of the plane is painted in red or blue, prove that in the plane there exists at least an equilateral triangle with the vertices of the same colour and with length side $m$. | Suppose that in the plane there no exists an equilateral triangle with the vertices of the same colour and length side $m=1,2,3,4$.
First assertion: we shall prove that in the plane there no exists a segment with the length 2 such that the ends and the midpint of this segment have the same colour. Suppose that the seg... | {
"problem_match": "\nCOM 5.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl",
"solution_match": "\nSolution."
} | 61 | 723 |
2003 | T3 | GEO 5 | Geometry | JBMO-SL | Let three congruent circles intersect in one point $M$ and $A_{1}, A_{2}$ and $A_{3}$ be the other intersection points for those circles. Prove that $M$ is a.orthocenter for a triangle $A_{1} A_{2} A_{3}$. | The quadrilaterals $\mathrm{O}_{3} M O_{2} A_{1}, \mathrm{O}_{3} M O_{1} A_{2}$ and $O_{1} M O_{2} A_{3}$ are rombes. Therefore, $O_{2} A_{1} \| M O_{3}$ and $M O_{3} \| O_{1} A_{2}$, which imply $O_{2} A_{1} \| O_{1} A_{2}$. Because $O_{2} A_{1}=O_{3}{ }^{*} M=O_{1} A_{2}$ the quadrilateral $O_{2} A_{1} A_{2} O_{1}$ i... | {
"problem_match": "\nGEO 5.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2003_shl.jsonl",
"solution_match": "\nSolution:"
} | 66 | 709 |
2004 | T3 | NT4 | Number Theory | JBMO-SL | If the positive integers $x$ and $y$ are such that both $3 x+4 y$ and $4 x+3 y$ are perfect squares, prove that both $x$ and $y$ are multiples of 7 . | Let
$$
3 x+4 y=m^{2}, \quad 4 x+3 y=n^{2}
$$
Then
$$
7(x+y)=m^{2}+n^{2} \Rightarrow 7 \mid m^{2}+n^{2}
$$
Considering $m=7 k+r, \quad r \in\{0,1,2,3,4,5,6\}$ we find that $m^{2} \equiv u(\bmod 7), \quad u \in$ $\{0,1,2,4\}$ and similarly $n^{2} \equiv v(\bmod 7), \quad v \in\{0,1,2,4\}$. Therefore we have either $m... | {
"problem_match": "\nNT4.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl",
"solution_match": "\nSolution."
} | 50 | 540 |
2004 | T3 | G1 | Geometry | JBMO-SL | Two circles $k_{1}$ and $k_{2}$ intersect a.t points $A$ and $B$. A circle $k_{3}$ centered at $A$ meet $k_{1}$ at $M$ and $P$ and $k_{2}$ at $N$ and $Q$, such that $N$ and $Q$ are on different sides of $M P$ and $A B>A M$.
Prove rhat the angles $\angle M B Q$ and $\angle N B P$ are equal. | As $A M=A P$, we have
$$
\angle M B A=\frac{1}{2} \operatorname{arcAM}=\frac{1}{2} \operatorname{arc} A P=\angle A B P
$$
and likewise
$$
\angle Q B A=\frac{1}{2} \operatorname{arc} A Q=\frac{1}{2} \operatorname{arc} c A N=\angle A B N
$$
Summing these equalities yields $\angle M B Q=\angle N B P$ as needed.
Q2. L... | {
"problem_match": "\nG1.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl",
"solution_match": "\nSolution "
} | 112 | 1,707 |
2004 | T3 | G3 | Geometry | JBMO-SL | Let $A B C$ be scalene triangle inscribed in the circle $k$. Circles $\alpha, \beta, \gamma$ are internally tangent to $k$ at points $A_{1}, B_{1}, C_{1}$ respectively, and tangent to the sides $B C, C A, A B$ at points $A_{2}, B_{2}, C_{2}$ respectively, so that $A$ and $A_{1}$ are on opposite sides of $B C^{\prime}, ... | The idea is to observe that $A^{\prime}, B^{\prime}, C^{\prime \prime}$ are the midpoints of the arcs $B C$, $C A$ and $A B$ of the circle $k$ which do not contain the points $A, B, C$ respectively. To prove this, consider the dilatation with the center $A_{1}$ taking' $\alpha$ to $k$. The line $B C$, which touches $\a... | {
"problem_match": "\nG3.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl",
"solution_match": "\nSolution."
} | 254 | 701 |
2004 | T3 | C2 | Combinatorics | JBMO-SL | Given $m \times n$ table, each cell signed with "-". The following operations are
(i) to change all the signs in entire row to the opposite, i. e. every "-" to "+", and every "+" to "-";
(ii) to change all the signs in entire column to the opposite, i. e. every "-" to "+" and every "+" to " -".
(a) Prove that if $m=... | If we apply (i) to $l$ rows and (ii) to $k$ columns we obtain $(m-k) l+(n-l) k$
(a) We have equation $(100-k) l+(100-l) k=2004$, or $100 l+100 k-2 l k=2004$, le
$$
50 l+50 k-1 k=1002
$$
Rewrite the lasc equation as
$$
(50-l)(50-h)=2.500-100.2=1498
$$
Since $1498=2 \cdot 7 \cdot 107$, this equation has no solitions... | {
"problem_match": "\nC2.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2004_shl.jsonl",
"solution_match": "\nSolution."
} | 145 | 644 |
2005 | T3 | A3 | Algebra | JBMO-SL | Find the maximum value of the area of a triangle having side lengths $a, b, c$ with
$$
a^{2}+b^{2}+c^{2}=a^{3}+b^{3}+c^{3}
$$ | Without any loss of generality, we may assume that $a \leq b \leq c$.
On the one hand, Tchebyshev's inequality gives
$$
(a+b+c)\left(a^{2}+b^{2}+c^{2}\right) \leq 3\left(a^{3}+b^{3}+c^{3}\right)
$$
Therefore using the given equation we get
$$
a+b+c \leq 3 \text { or } p \leq \frac{3}{2}
$$
where $p$ denotes the se... | {
"problem_match": "\nA3.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl",
"solution_match": "## Solution"
} | 53 | 631 |
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