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 |
|---|---|---|---|---|---|---|---|---|---|
2005 | T3 | A5 | Algebra | JBMO-SL | Solve the equation
$$
8 x^{3}+8 x^{2} y+8 x y^{2}+8 y^{3}=15\left(x^{2}+y^{2}+x y+1\right)
$$
in the set of integers. | We transform the equation to the following one
$$
\left(x^{2}+y^{2}\right)(8 x+8 y-15)=15(x y+1)
$$
Since the right side is divisible by 3 , then $3 /\left(x^{2}+y^{2}\right)(8 x+8 y-15)$. But if $3 /\left(x^{2}+y^{2}\right)$, then $3 / x$ and $3 / y, 009$ will wive $15(x y+1)$ and $3 /(x y+1)$, which is impossible. ... | {
"problem_match": "\nA5.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl",
"solution_match": "## Solution"
} | 61 | 513 |
2005 | T3 | G7 | Geometry | JBMO-SL | Let $A B C D$ be a parallelogram, $\mathrm{P}$ a point on $C D$, and $Q$ a point on $A B$. Let also $M=A P \cap D Q, \quad N=B P \cap C Q, K=M N \cap A D$, and $L=M N \cap B C$. Show that $B L=D K$. | Let $O$ be the intersection of the diagonals. Let $P_{1}$ be on $A B$ such that $P P_{1} / / A D$, and let $Q_{1}$ be on $C D$ such that $\mathrm{Q} Q_{1} / / A D$. Let $\sigma$ be the central symmetry with center $\mathrm{O}$. Let $\left.P^{\prime}=\sigma(P), Q^{\prime}=\sigma(Q), P_{1}^{\prime}=\sigma\left(P_{1}\righ... | {
"problem_match": "\nG7.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl",
"solution_match": "## Solution"
} | 83 | 1,081 |
2005 | T3 | NT5 | Number Theory | JBMO-SL | Let $p$ be a prime number and let $a$ be an integer. Show that if $n^{2}-5$ is not divisible by $p$ for any integer $n$, there exist infinitely many integers $m$ so that $p$ divides $m^{5}+a$. | We start with a simple fact:
Lemma: If $b$ is an integer not divisible by $p$ then there is an integer $s$ so that $s b$ has the remainder $l$ when divided by $p$.
For a proof, just note that numbers $b, 2 b, \ldots,(p-1) b$ have distinct non-zero remainders when divided by $p$, and hence one of them is equal to 1 .
... | {
"problem_match": "\nNT5.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2005_shl.jsonl",
"solution_match": "## Solution"
} | 62 | 673 |
2007 | T3 | C3 | Combinatorics | JBMO-SL | The nonnegative integer $n$ and $(2 n+1) \times(2 n+1)$ chessboard with squares colored alternatively black and white are given. For every natural number $m$ with $1<m<2 n+1$, an $m \times m$ square of the given chessboard that has more than half of its area colored in black, is called a $B$-square. If the given chessb... | Every square with even side length will have an equal number of black and white $1 \times 1$ squares, so it isn't a $B$-square. In a square with odd side length, there is one more $1 \times 1$ black square than white squares, if it has black corner squares. So, a square with odd side length is a $B$-square either if it... | {
"problem_match": "\nC3 ",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl",
"solution_match": "## Solution"
} | 120 | 893 |
2007 | T3 | NT1 | Number Theory | JBMO-SL | Find all the pairs positive integers $(x, y)$ such that
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{[x, y]}+\frac{1}{(x, y)}=\frac{1}{2}
$$
where $(x, y)$ is the greatest common divisor of $x, y$ and $[x, y]$ is the least common multiple of $x, y$. | We put $x=d u$ and $y=d v$ where $d=(x, y)$. So we have $(u, v)=1$. From the conclusion we obtain $2(u+1)(v+1)=d u v$. Because $(v, v+1)=1, v$ divides $2(u+1)$.
Case 1. $u=v$. Hence $x=y=[x, y]=(x, y)$, which leads to the solution $x=8$ and $y=8$.
Case 2. $u<v$. Then $u+1 \leq v \Leftrightarrow 2(u+1) \leq 2 v \Leftr... | {
"problem_match": "\nNT1 ",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl",
"solution_match": "## Solution"
} | 94 | 519 |
2007 | T3 | NT3 | Number Theory | JBMO-SL | Let $n>1$ be a positive integer and $p$ a prime number such that $n \mid(p-1)$ and $p \mid\left(n^{6}-1\right)$. Prove that at least one of the numbers $p-n$ and $p+n$ is a perfect square. | Since $n \mid p-1$, then $p=1+n a$, where $a \geq 1$ is an integer. From the condition $p \mid n^{6}-1$, it follows that $p|n-1, p| n+1, p \mid n^{2}+n+1$ or $p \mid n^{2}-n+1$.
- Let $p \mid n-1$. Then $n \geq p+1>n$ which is impossible.
- Let $p \mid n+1$. Then $n+1 \geq p=1+n a$ which is possible only when $a=1$ an... | {
"problem_match": "\nNT3 ",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2007_shl.jsonl",
"solution_match": "## Solution"
} | 66 | 515 |
2008 | T3 | A6 | Algebra | JBMO-SL | If the real numbers $a, b, c, d$ are such that $0<a, b, c, d<1$, show that
$$
1+a b+b c+c d+d a+a c+b d>a+b+c+d
$$ | If $1 \geq a+b+c$ then we write the given inequality equivalently as
$$
\begin{gathered}
1-(a+b+c)+d[(a+b+c)-1]+a b+b c+c a>0 \\
\Leftrightarrow[1-(a+b+c)](1-d)+a b+b c+c a>0
\end{gathered}
$$
which is of course true.
If instead $a+b+c>1$, then $d(a+b+c)>d$ i.e.
$$
d a+d b+d c>d
$$
We are going to prove that also
... | {
"problem_match": "\nA6 ",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl",
"solution_match": "## Solution"
} | 52 | 1,159 |
2008 | T3 | A7 | Algebra | JBMO-SL | Let $a, b$ and $c$ be a positive real numbers such that $a b c=1$. Prove the inequality
$$
\left(a b+b c+\frac{1}{c a}\right)\left(b c+c a+\frac{1}{a b}\right)\left(c a+a b+\frac{1}{b c}\right) \geq(1+2 a)(1+2 b)(1+2 c)
$$ | By Cauchy-Schwarz inequality and $a b c=1$ we get
$$
\begin{gathered}
\sqrt{\left(b c+c a+\frac{1}{a b}\right)\left(a b+b c+\frac{1}{c a}\right)}=\sqrt{\left(b c+c a+\frac{1}{a b}\right)\left(\frac{1}{c a}+a b+b c\right)} \geq \\
\left(\sqrt{a b} \cdot \sqrt{\frac{1}{a b}}+\sqrt{b c} \cdot \sqrt{b c}+\sqrt{\frac{1}{c ... | {
"problem_match": "\nA7 ",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl",
"solution_match": "## Solution 1"
} | 97 | 920 |
2008 | T3 | G6 | Geometry | JBMO-SL | Let $A B C$ be a triangle with $A<90^{\circ}$. Outside of a triangle we consider isosceles triangles $A B E$ and $A C Z$ with bases $A B$ and $A C$, respectively. If the midpoint $D$ of the side $B C$ is such that $D E \perp D Z$ and $E Z=2 \cdot E D$, prove that $\widehat{A E B}=2 \cdot \widehat{A Z C}$. | Since $D$ is the midpoint of the side $B C$, in the extension of the line segment $Z D$ we take a point $H$ such that $Z D=D H$. Then the quadrilateral $B H C Z$ is parallelogram and therefore we have
$$
B H=Z C=Z A
$$
Also from the isosceles triangle $A B E$ we get
$$
B E=A E
$$
Since $D E \perp D Z, E D$ is altit... | {
"problem_match": "\nG6 ",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl",
"solution_match": "## Solution"
} | 113 | 771 |
2008 | T3 | NT3 | Number Theory | JBMO-SL | Let $s(a)$ denote the sum of digits of a given positive integer $a$. The sequence $a_{1}, a_{2}, \ldots a_{n}, \ldots$ of positive integers is such that $a_{n+1}=a_{n}+s\left(a_{n}\right)$ for each positive integer $n$. Find the greatest possible $n$ for which it is possible to have $a_{n}=2008$. | Since $a_{n-1} \equiv s\left(a_{n-1}\right)$ (all congruences are modulo 9 ), we have $2 a_{n-1} \equiv a_{n} \equiv 2008 \equiv 10$, so $a_{n-1} \equiv 5$. But $a_{n-1}<2008$, so $s\left(a_{n-1}\right) \leq 28$ and thus $s\left(a_{n-1}\right)$ can equal 5,14 or 23 . We check $s(2008-5)=s(2003)=5, s(2008-14)=s(1994)=23... | {
"problem_match": "\nNT3 ",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl",
"solution_match": "## Solution"
} | 98 | 705 |
2008 | T3 | NT5 | Number Theory | JBMO-SL | Is it possible to arrange the numbers $1^{1}, 2^{2}, \ldots, 2008^{2008}$ one after the other, in such a way that the obtained number is a perfect square? (Explain your answer.) | We will use the following lemmas.
Lemma 1. If $x \in \mathbb{N}$, then $x^{2} \equiv 0$ or $1(\bmod 3)$.
Proof: Let $x \in \mathbb{N}$, then $x=3 k, x=3 k+1$ or $x=3 k+2$, hence
$$
\begin{aligned}
& x^{2}=9 k^{2} \equiv 0(\bmod 3) \\
& x^{2}=9 k^{2}+6 k+1 \equiv 1(\bmod 3), \\
& x^{2}=9 k^{2}+12 k+4 \equiv 1(\bmod 3... | {
"problem_match": "\nNT5 ",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl",
"solution_match": "## Solution"
} | 56 | 1,201 |
2008 | T3 | NT6 | Number Theory | JBMO-SL | Let $f: \mathbb{N} \rightarrow \mathbb{R}$ be a function, satisfying the following condition:
for every integer $n>1$, there exists a prime divisor $p$ of $n$ such that $f(n)=f\left(\frac{n}{p}\right)-f(p)$. If
$$
f\left(2^{2007}\right)+f\left(3^{2008}\right)+f\left(5^{2009}\right)=2006
$$
determine the value of
$$... | If $n=p$ is prime number, we have
$$
f(p)=f\left(\frac{p}{p}\right)-f(p)=f(1)-f(p)
$$
i.e.
$$
f(p)=\frac{f(1)}{2}
$$
If $n=p q$, where $p$ and $q$ are prime numbers, then
$$
f(n)=f\left(\frac{n}{p}\right)-f(p)=f(q)-f(p)=\frac{f(1)}{2}-\frac{f(1)}{2}=0
$$
If $n$ is a product of three prime numbers, we have
$$
f(n... | {
"problem_match": "\nNT6 ",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl",
"solution_match": "## Solution"
} | 166 | 813 |
2008 | T3 | NT8 | Number Theory | JBMO-SL | Let $a, b, c, d, e, f$ are nonzero digits such that the natural numbers $\overline{a b c}, \overline{d e f}$ and $\overline{a b c d e f}$ are squares.
a) Prove that $\overline{a b c d e f}$ can be represented in two different ways as a sum of three squares of natural numbers.
b) Give an example of such a number. | a) Let $\overline{a b c}=m^{2}, \overline{d e f}=n^{2}$ and $\overline{a b c d e f}=p^{2}$, where $11 \leq m \leq 31,11 \leq n \leq 31$ are natural numbers. So, $p^{2}=1000 \cdot m^{2}+n^{2}$. But $1000=30^{2}+10^{2}=18^{2}+26^{2}$. We obtain the following relations
$$
\begin{gathered}
p^{2}=\left(30^{2}+10^{2}\right)... | {
"problem_match": "\nNT8 ",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl",
"solution_match": "\nSolution"
} | 96 | 526 |
2008 | T3 | NT9 | Number Theory | JBMO-SL | Let $p$ be a prime number. Find all positive integers $a$ and $b$ such that:
$$
\frac{4 a+p}{b}+\frac{4 b+p}{a}
$$
and
$$
\frac{a^{2}}{b}+\frac{b^{2}}{a}
$$
are integers. | Since $a$ and $b$ are symmetric we can assume that $a \leq b$. Let $d=(a, b), a=d u, b=d v$ and $(u, v)=1$. Then we have:
$$
\frac{a^{2}}{b}+\frac{b^{2}}{a}=\frac{d\left(u^{3}+v^{3}\right)}{u v}
$$
Since,
$$
\left(u^{3}+v^{3}, u\right)=\left(u^{3}+v^{3}, v\right)=1
$$
we deduce that $u \mid d$ and $v \mid d$. But a... | {
"problem_match": "\nNT9 ",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2008_shl.jsonl",
"solution_match": "## Solution"
} | 74 | 1,049 |
2009 | T3 | A2 | Algebra | JBMO-SL | Find the maximum value of $z+x$, if $(x, y, z, t)$ satisfies the conditions:
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=4 \\
z^{2}+t^{2}=9 \\
x t+y z \geq 6
\end{array}\right.
$$ | I: From the conditions we have
$$
36=\left(x^{2}+y^{2}\right)\left(z^{2}+t^{2}\right)=(x t+y z)^{2}+(x z-y t)^{2} \geq 36+(x z-y t)^{2}
$$
and this implies $x z-y t=0$.
Now it is clear that
$$
x^{2}+z^{2}+y^{2}+t^{2}=(x+z)^{2}+(y-t)^{2}=13
$$
and the maximum value of $z+x$ is $\sqrt{13}$. It is achieved for $x=\fr... | {
"problem_match": "\nA2 ",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl",
"solution_match": "\nSolution "
} | 75 | 524 |
2009 | T3 | A4 | Algebra | JBMO-SL | Real numbers $x, y, z$ satisfy
$$
0<x, y, z<1
$$
and
$$
x y z=(1-x)(1-y)(1-z) .
$$
Show that
$$
\frac{1}{4} \leq \max \{(1-x) y,(1-y) z,(1-z) x\}
$$ | It is clear that $a(1-a) \leq \frac{1}{4}$ for any real numbers $a$ (equivalent to $0<(2 a-1)^{2}$ ). Thus,
$$
\begin{gathered}
x y z=(1-x)(1-y)(1-z) \\
(x y z)^{2}=[x(1-x)][y(1-y)][z(1-z)] \leq \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4}=\frac{1}{4^{3}} \\
x y z \leq \frac{1}{2^{3}}
\end{gathered}
$$
It implies ... | {
"problem_match": "\nA4 ",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl",
"solution_match": "\nSolution:"
} | 80 | 845 |
2009 | T3 | A5 | Algebra | JBMO-SL | Let $x, y, z$ be positive real numbers. Prove that:
$$
\left(x^{2}+y+1\right)\left(x^{2}+z+1\right)\left(y^{2}+z+1\right)\left(y^{2}+x+1\right)\left(z^{2}+x+1\right)\left(z^{2}+y+1\right) \geq(x+y+z)^{6}
$$ | I: Applying Cauchy-Schwarz's inequality:
$$
\left(x^{2}+y+1\right)\left(z^{2}+y+1\right)=\left(x^{2}+y+1\right)\left(1+y+z^{2}\right) \geq(x+y+z)^{2}
$$
Using the same reasoning we deduce:
$$
\left(x^{2}+z+1\right)\left(y^{2}+z+1\right) \geq(x+y+z)^{2}
$$
and
$$
\left(y^{2}+x+1\right)\left(z^{2}+x+1\right) \geq(x+... | {
"problem_match": "\nA5 ",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl",
"solution_match": "\nSolution "
} | 103 | 530 |
2009 | T3 | C2 | Combinatorics | JBMO-SL | Five players $(A, B, C, D, E)$ take part in a bridge tournament. Every two players must play (as partners) against every other two players. Any two given players can be partners not more than once per day. What is the least number of days needed for this tournament? | A given pair must play with three other pairs and these plays must be in different days, so at three days are needed. Suppose that three days suffice. Let the pair $A B$ play against $C D$ on day $x$. Then $A B-D E$ and $C D-B E$ cannot play on day $x$. Then one of the other two plays of $D E$ (with $A C$ and $B C$ ) m... | {
"problem_match": "\nC2 ",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl",
"solution_match": "\nSolution:"
} | 61 | 790 |
2009 | T3 | C4 | Combinatorics | JBMO-SL | Determine all pairs $(m, n)$ for which it is possible to tile the table $m \times n$ with "corners" as in the figure below, with the condition that in the tiling there is no rectangle (except for the $m \times n$ one) regularly covered with corners.
![](https://cdn.mathpix.com/cropped/2024_06_05_17e61ec1e039d21af3a0g-... | Every "corner" covers exactly 3 squares, so a necessary condition for the tiling to exists is $3 \mid m n$.
First, we shall prove that for a tiling with our condition to exist, it is necessary that both $m, n$ for $m, n>3$ to be even. Suppose the contrary, i.e. suppose that that $m>3$ is odd (without losing generality... | {
"problem_match": "\nC4 ",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl",
"solution_match": "\nSolution:"
} | 137 | 702 |
2009 | T3 | NT2 | Number Theory | JBMO-SL | A group of $n>1$ pirates of different age owned total of 2009 coins. Initially each pirate (except for the youngest one) had one coin more than the next younger.
a) Find all possible values of $n$.
b) Every day a pirate was chosen. The chosen pirate gave a coin to each of the other pirates. If $n=7$, find the largest... | a) If $n$ is odd, then it is a divisor of $2009=7 \times 7 \times 41$. If $n>49$, then $n$ is at least $7 \times 41$, while the average pirate has 7 coins, so the initial division is impossible. So, we can have $n=7, n=41$ or $n=49$. Each of these cases is possible (e.g. if $n=49$, the average pirate has 41 coins, so t... | {
"problem_match": "\nNT2 ",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl",
"solution_match": "## Solution:"
} | 95 | 606 |
2009 | T3 | NT4 | Number Theory | JBMO-SL | Determine all prime numbers $p_{1}, p_{2}, \ldots, p_{12}, p_{13}, p_{1} \leq p_{2} \leq \ldots \leq p_{12} \leq p_{13}$, such that
$$
p_{1}^{2}+p_{2}^{2}+\ldots+p_{12}^{2}=p_{13}^{2}
$$
and one of them is equal to $2 p_{1}+p_{9}$. | Obviously, $p_{13} \neq 2$, because sum of squares of 12 prime numbers is greater or equal to $12 \times 2^{2}=48$. Thus, $p_{13}$ is odd number and $p_{13} \geq 7$.
We have that $n^{2} \equiv 1(\bmod 8)$, when $n$ is odd. Let $k$ be the number of prime numbers equal to 2 . Looking at equation modulo 8 we get:
$$
4 k... | {
"problem_match": "\nNT4 ",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2009_shl.jsonl",
"solution_match": "\nSolution:"
} | 122 | 1,602 |
2010 | T3 | A5 | Algebra | JBMO-SL | The real positive numbers $x, y, z$ satisfy the relations $x \leq 2$, $y \leq 3, x+y+z=11$. Prove that $\sqrt{x y z} \leq 6$. | For $x=2, y=3$ and $z=6$ the equality holds.
After the substitutions $x=2-u, y=3-v$ with $u \in[0,2), v \in[0,3)$, we obtain that $z=6+u+v$ and the required inequality becomes
$$
(2-u)(3-v)(6+u+v) \leqslant 36
$$
We shall need the following lemma.
Lemma. If real numbers $a$ and $b$ satisfy the relations $0<b \leq a... | {
"problem_match": "\nProblem A5.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl",
"solution_match": "\nSolution."
} | 52 | 667 |
2010 | T3 | C1 | Combinatorics | JBMO-SL | There are two piles of coins, each containing 2010 pieces. Two players A and B play a game taking turns (A plays first). At each turn, the player on play has to take one or more coins from one pile or exactly one coin from each pile. Whoever takes the last coin is the winner. Which player will win if they both play in ... | B wins.
In fact, we will show that A will lose if the total number of coins is a multiple of 3 and the two piles differ by not more than one coin (call this a balanced position). To this end, firstly notice that it is not possible to move from one balanced position to another. The winning strategy for B consists in re... | {
"problem_match": "\nProblem C1.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo-2010_shl.jsonl",
"solution_match": "\nSolution."
} | 80 | 600 |
2013 | T3 | G1 | Geometry | JBMO-SL | Let $A B$ be a diameter of a circle $\omega$ with center $O$ and $O C$ be a radius of $\omega$ which is perpendicular to $A B$. Let $M$ be a point on the line segment $O C$. Let $N$ be the second point of intersection of the line $A M$ with $\omega$, and let $P$ be the point of intersection of the lines tangent to $\om... | Since the lines $P N$ and $B P$ are tangent to $\omega, N P=P B$ and $O P$ is the bisector of $\angle N O B$. Therefore the lines $O P$ and $N B$ are perpendicular. Since $\angle A N B=90^{\circ}$, it follows that the lines $A N$ and $O P$ are parallel. As $M O$ and $P B$ are also parallel and $A O=O B$, the triangles ... | {
"problem_match": "\nG1.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl",
"solution_match": "\nSolution."
} | 119 | 609 |
2013 | T3 | G5 | Geometry | JBMO-SL | A circle passing through the midpoint $M$ of the side $B C$ and the vertex $A$ of a triangle $A B C$ intersects the sides $A B$ and $A C$ for the second time at the points $P$ and $Q$, respectively. Show that if $\angle B A C=60^{\circ}$ then
$$
A P+A Q+P Q<A B+A C+\frac{1}{2} B C
$$ | Since the quadrilateral $A P M Q$ is cyclic, we have $\angle P M Q=180^{\circ}-\angle P A Q=$ $180^{\circ}-\angle B A C=120^{\circ}$. Therefore $\angle P M B+\angle Q M C=180^{\circ}-\angle P M Q=60^{\circ}$.
Let the point $B^{\prime}$ be the symmetric of the point $B$ with respect to the line $P M$ and the point $C^{... | {
"problem_match": "\nG5.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl",
"solution_match": "\nSolution."
} | 99 | 1,240 |
2013 | T3 | N3 | Number Theory | JBMO-SL | Find all ordered pairs $(a, b)$ of positive integers for which the numbers $\frac{a^{3} b-1}{a+1}$ and $\frac{b^{3} a+1}{b-1}$ are positive integers. | As $a^{3} b-1=b\left(a^{3}+1\right)-(b+1)$ and $a+1 \mid a^{3}+1$, we have $a+1 \mid b+1$.
As $b^{3} a+1=a\left(b^{3}-1\right)+(a+1)$ and $b-1 \mid b^{3}-1$, we have $b-1 \mid a+1$.
So $b-1 \mid b+1$ and hence $b-1 \mid 2$.
- If $b=2$, then $a+1 \mid b+1=3$ gives $a=2$. Hence $(a, b)=(2,2)$ is the only solution in t... | {
"problem_match": "\nN3.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2013.jsonl",
"solution_match": "\nSolution."
} | 51 | 724 |
2014 | T3 | A2 | Algebra | JBMO-SL | Let $a, b$ and $c$ be positive real numbers such that abc $=\frac{1}{8}$. Prove the inequality
$$
a^{2}+b^{2}+c^{2}+a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \geq \frac{15}{16}
$$
When does equality hold? | By using The Arithmetic-Geometric Mean Inequality for 15 positive numbers, we find that
$$
\begin{aligned}
& a^{2}+b^{2}+c^{2}+a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}= \\
& \quad=\frac{a^{2}}{4}+\frac{a^{2}}{4}+\frac{a^{2}}{4}+\frac{a^{2}}{4}+\frac{b^{2}}{4}+\frac{b^{2}}{4}+\frac{b^{2}}{4}+\frac{b^{2}}{4}+\frac{c^{2}}{4}+... | {
"problem_match": "## A2",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl",
"solution_match": "\nSolution1."
} | 93 | 563 |
2014 | T3 | A3 | Algebra | JBMO-SL | 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? | By using AM-GM $\left(x^{2}+y^{2}+z^{2} \geq x y+y z+z x\right)$ we have
$$
\begin{aligned}
\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} & \geq\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)+\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)+\left(c+\frac{1}{a}\ri... | {
"problem_match": "## A3",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl",
"solution_match": "\nSolution1."
} | 90 | 916 |
2014 | T3 | A4 | Algebra | JBMO-SL | Let $a, b, c$ be positive real numbers such that $a+b+c=1$. Prove that
$$
\frac{7+2 b}{1+a}+\frac{7+2 c}{1+b}+\frac{7+2 a}{1+c} \geq \frac{69}{4}
$$
When does equality hold? | Solution. Equality holds when $x=y=z=0$.
Apply AM-GM to $x+y+z=x y z$,
$$
\begin{aligned}
& x y z=x+y+z \geq 3 \sqrt[3]{x y z} \Rightarrow(x y z)^{3} \geq(3 \sqrt[3]{x y z})^{3} \\
& \Rightarrow x^{3} y^{3} z^{3} \geq 27 x y z \\
& \Rightarrow x^{2} y^{2} z^{2} \geq 27 \\
& \Rightarrow \sqrt[3]{x^{2} y^{2} z^{2}} \ge... | {
"problem_match": "## A4",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl",
"solution_match": "\nSolution1."
} | 78 | 720 |
2014 | T3 | A8 | Algebra | JBMO-SL | Let $x, y$ and $z$ be positive real numbers such that $x y z=1$. Prove the inequality
$$
\frac{1}{x(a y+b)}+\frac{1}{y(a z+b)}+\frac{1}{z(a x+b)} \geq 3 \text {, if: }
$$
a) $a=0$ and $b=1$;
b) $a=1$ and $b=0$;
c) $a+b=1$ for $a, b>0$
When does the equality hold true? | a) The inequality reduces to $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq 3$, which follows directly from the AM-GM inequality.
Equality holds only when $x=y=z=1$.
b) Here the inequality reduces to $\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x} \geq 3$, i.e. $x+y+z \geq 3$, which also follows from the AM-GM inequality.
Equ... | {
"problem_match": "## A8",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl",
"solution_match": "\nSolution."
} | 125 | 634 |
2014 | T3 | A9 | Algebra | JBMO-SL | Let $n$ be a positive integer, and let $x_{1}, \ldots, x_{n}, y_{1}, \ldots, y_{n}$ be positive real numbers such that $x_{1}+\ldots+x_{n}=y_{1}+\ldots+y_{n}=1$. Show that
$$
\left|x_{1}-y_{1}\right|+\ldots\left|x_{n}-y_{n}\right| \leq 2-\min _{1 \leq i \leq n} \frac{x_{i}}{y_{i}}-\min _{1 \leq i \leq n} \frac{y_{i}}{... | Up to reordering the real numbers $x_{i}$ and $y_{i}$, we may assume that $\frac{x_{1}}{y_{1}} \leq \ldots \leq \frac{x_{n}}{y_{n}}$. Let $A=\frac{x_{1}}{y_{1}}$ and $B=\frac{x_{n}}{y_{n}}$, and $\mathrm{S}=\left|x_{1}-y_{1}\right|+\ldots\left|x_{n}-y_{n}\right|$. Our aim is to prove that $S \leq 2-A-\frac{1}{B}$.
Fir... | {
"problem_match": "## A9",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl",
"solution_match": "\nSolution."
} | 156 | 682 |
2014 | T3 | C3 | Combinatorics | JBMO-SL | For a given positive integer n, two players $A$ and B play the following game: Given is pile of $\boldsymbol{\Omega}$ stones. The players take turn alternatively with A going first. On each turn the player is allowed to take one stone, a prime number of stones, or a multiple of $n$ stones. The winner is the one who tak... | Denote by $k$ the sought number and let $\left\{a_{1}, a_{2}, \ldots, a_{k}\right\}$ be the corresponding values for $a$. We will call each $a_{i}$ a losing number and every other positive integer a winning numbers. Clearly every multiple of $n$ is a winning number.
Suppose there are two different losing numbers $a_{i... | {
"problem_match": "## C3",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl",
"solution_match": "\nSolution."
} | 103 | 622 |
2014 | T3 | C4 | Combinatorics | JBMO-SL | Let $A=1 \cdot 4 \cdot 7 \cdot \ldots \cdot 2014$ be the product of the numbers less or equal to 2014 that give remainder 1 when divided by 3 . Find the last non-zero digit of $A$. | Grouping the elements of the product by ten we get:
$$
\begin{aligned}
& (30 k+1)(30 k+4)(30 k+7)(30 k+10)(30 k+13)(30 k+16) \\
& (30 k+19)(30 k+22)(30 k+25)(30 k+28)= \\
& =(30 k+1)(15 k+2)(30 k+7)(120 k+40)(30 k+13)(15 k+8) \\
& (30 k+19)(15 k+11)(120 k+100)(15 k+14)
\end{aligned}
$$
(We divide all even numbers not... | {
"problem_match": "## C4",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl",
"solution_match": "\nSolution."
} | 62 | 979 |
2014 | T3 | G3 | Geometry | JBMO-SL | Let $C D \perp A B(D \in A B), D M \perp A C(M \in A C)$ and $D N \perp B C(N \in B C)$ for an acute triangle ABC with area $S$. If $H_{1}$ and $H_{2}$ are the orthocentres of the triangles $M N C$ and MND respectively. Evaluate the area of the quadrilateral $\mathrm{AH}_{1} \mathrm{BH}_{2}$. | Let $O, P, K, R$ and $T$ be the midpoints of the segments $C D, M N, C N, C H_{1}$ and $M H_{1}$, respectively. From $\triangle M N C$ we have that $\overline{P K}=\frac{1}{2} \overline{M C}$ and $P K \| M C$. Analogously, from $\Delta M H_{1} C$ we have that $\overline{T R}=\frac{1}{2} \overline{M C}$ and $T R \| M C$... | {
"problem_match": "## G3",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl",
"solution_match": "\nSolution1."
} | 106 | 700 |
2014 | T3 | G4 | Geometry | JBMO-SL | Let $A B C$ be a triangle such that $\overline{A B} \neq \overline{A C}$. Let $M$ be a midpoint of $\overline{B C}, H$ the orthocenter of $A B C, O_{1}$ the midpoint of $\overline{A H}$ and $O_{2}$ the circumcenter of $B C H$. Prove that $O_{1} A M O_{2}$ is a parallelogram. | Let $O_{2}^{\prime}$ be the point such that $O_{1} A M O_{2}^{\prime}$ is a parallelogram. Note that $\overrightarrow{M O_{2}}=\overrightarrow{A O_{1}}=\overrightarrow{O_{1} H}$. Therefore, $O_{1} H O_{2}^{\prime} M$ is a parallelogram and $\overrightarrow{M O_{1}}=\overrightarrow{O_{2} H}$.
Since $M$ is the midpoint ... | {
"problem_match": "## G4",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl",
"solution_match": "\nSolution1."
} | 105 | 768 |
2014 | T3 | G5 | Geometry | JBMO-SL | Let $A B C$ be a triangle with $\overline{A B} \neq \overline{B C}$, and let $B D$ be the internal bisector of $\measuredangle A B C(D \in A C)$. Denote the midpoint of the arc $A C$ which contains point BbyM. The circumcircle of the triangle $B D M$ intersects the segment $A B$ at point $K \neq B$, and let $J$ be the ... | Let the circumcircle of the triangle $B D M$ intersect the line segment $B C$ at point $L \neq B$. From $\measuredangle C B D=\measuredangle D B A$ we have $\overline{D L}=\overline{D K}$. Since $\measuredangle L C M=\measuredangle B C M=\measuredangle B A M=\measuredangle K A M, \overline{M C}=\overline{M A}$ and
$$
... | {
"problem_match": "## G5",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl",
"solution_match": "## Solution1."
} | 150 | 877 |
2014 | T3 | G6 | Geometry | JBMO-SL | Let $A B C D$ be a quadrilateral whose sides $A B$ and $C D$ are not parallel, and let $O$ be the intersection of its diagonals. Denote with $H_{1}$ and $H_{2}$ the orthocenters of the triangles $O A B$ and OCD, respectively. If $M$ and $N$ are the midpoints of the segments $\overline{A B}$ and $\overline{C D}$, respec... | Let $A^{\prime}$ and $B^{\prime}$ be the feet of the altitudes drawn from $A$ and $B$ respectively in the triangle $A O B$, and $C^{\prime}$ and $D^{\prime}$ are the feet of the altitudes drawn from $C$ and $D$ in the triangle $C O D$. Obviously, $A^{\prime}$ and $D^{\prime}$ belong to the circle $c_{1}$ of diameter $\... | {
"problem_match": "## G6",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl",
"solution_match": "## Solution."
} | 148 | 556 |
2014 | T3 | N1 | Number Theory | JBMO-SL | Each letter of the word OHRID corresponds to a different digit belonging to the set $(1,2,3,4,5)$. Decipher the equality $(O+H+R+I+D)^{2}:(O-H-R+I+D)=O^{H^{R^{I_{D}^{D}}}}$. | Since $O, H, R, I$ and $D$ are distinct numbers from $\{1,2,3,4,5\}$, we have $O+H+R+I+D=15$ and $O-H-R+I+D=O+H+R+I+D-2(H+R)<15$. From this $O^{H^{R^{I^{D}}}}=\frac{(O+H+R+I+D)^{2}}{O-H-R+I+D}=\frac{225}{15-2(H+R)}$, hence $O^{H^{R^{R^{D}}}}>15$ and divides 225 , which is only possible for $O^{H^{R^{I^{D}}}}=25$ (must ... | {
"problem_match": "## N1",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl",
"solution_match": "\nSolution."
} | 71 | 649 |
2014 | T3 | N4 | Number Theory | JBMO-SL | Prove there are no integers $a$ and $b$ satisfying the following conditions:
i) $16 a-9 b$ is a prime number
ii) $\quad a b$ is a perfect square
iii) $a+b$ is a perfect square | Suppose $a$ and $b$ be integers satisfying the given conditions. Let $p$ be a prime number, $n$ and $m$ be integers. Then we can write the conditions as follows:
$$
\begin{aligned}
& 16 a-9 b=p \\
& a b=n^{2} \\
& a+b=m^{2}
\end{aligned}
$$
Moreover, let $d=g d c(a, b)$ and $a=d x, b=d y$ for some relatively prime in... | {
"problem_match": "\nN4",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl",
"solution_match": "\nSolution."
} | 55 | 1,492 |
2014 | T3 | N6 | Number Theory | JBMO-SL | Vukasin, Dimitrije, Dusan, Stefan and Filip asked their professor to guess a three consecutive positive integer numbers after they had told him these (true) sentences:
Vukasin: "Sum of the digits of one of them is a prime number. Sum of the digits of some of the other two is an even perfect number ( $n$ is perfect if ... | Let the middle number be $n$, so the numbers are $n-1, n$ and $n+1$. Since 4 does not divide any of them, $n \equiv 2(\bmod 4)$. Furthermore, neither 3,5 nor 7 divides $n$. Also $n+1+11 \equiv 2(\bmod 4)$ cannot be a square. Then 3 must divide $n-1$ or $n+1$. If $n-1+11$ is a square, then $3 \mid n+1$ which implies $3 ... | {
"problem_match": "## N6",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2014.jsonl",
"solution_match": "\nSolution."
} | 190 | 1,103 |
2017 | T3 | A1 | Algebra | JBMO-SL | Let $a, b, c$ be positive real numbers such that $a+b+c+a b+b c+c a+a b c=7$. Prove that
$$
\sqrt{a^{2}+b^{2}+2}+\sqrt{b^{2}+c^{2}+2}+\sqrt{c^{2}+a^{2}+2} \geq 6
$$ | First we see that $x^{2}+y^{2}+1 \geq x y+x+y$. Indeed, this is equivalent to
$$
(x-y)^{2}+(x-1)^{2}+(y-1)^{2} \geq 0
$$
Therefore
$$
\begin{aligned}
& \sqrt{a^{2}+b^{2}+2}+\sqrt{b^{2}+c^{2}+2}+\sqrt{c^{2}+a^{2}+2} \\
\geq & \sqrt{a b+a+b+1}+\sqrt{b c+b+c+1}+\sqrt{c a+c+a+1} \\
= & \sqrt{(a+1)(b+1)}+\sqrt{(b+1)(a+1)... | {
"problem_match": "\nA1.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl",
"solution_match": "\nSolution."
} | 88 | 555 |
2017 | T3 | A3 | Algebra | JBMO-SL | Let $a, b, c, d$ be real numbers such that $0 \leq a \leq b \leq c \leq d$. Prove the inequality
$$
a b^{3}+b c^{3}+c d^{3}+d a^{3} \geq a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}
$$ | The inequality is equivalent to
$$
\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\right)^{2} \geq\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}\right)^{2}
$$
By the Cauchy-Schwarz inequality,
$$
\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\right)\left(a^{3} b+b^{3} c+c^{3} d+d^{3} a\right) \geq\left(a^{2} b^{2}+b^{2} c^{2}+c... | {
"problem_match": "\nA3.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl",
"solution_match": "\nSolution."
} | 103 | 900 |
2017 | T3 | A4 | Algebra | JBMO-SL | Let $x, y, z$ be three distinct positive integers. 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": "\nA4.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl",
"solution_match": "\nSolution."
} | 50 | 625 |
2017 | T3 | C1 | Combinatorics | JBMO-SL | Consider a regular $2 n+1$-gon $P$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$, if the line segment $S E$ contains no other points that lie on the sides of $P$ except $S$. We want to color the sides of $P$ in 3 co... | Answer: $n=1$ is clearly a solution, we can just color each side of the equilateral triangle in a different color, and the conditions are satisfied. We prove there is no larger $n$ that fulfills the requirements.
Lemma 1. Given a regular $2 n+1$-gon in the plane, and a sequence of $n+1$ consecutive sides $s_{1}, s_{2}... | {
"problem_match": "\nC1.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl",
"solution_match": "\nSolution."
} | 174 | 700 |
2017 | T3 | C2 | Combinatorics | JBMO-SL | Consider a regular $2 n$-gon $P$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$, if the line segment $S E$ contains no other points that lie on the sides of $P$ except $S$. We want to color the sides of $P$ in 3 colo... | 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": "\nC2.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl",
"solution_match": "\nSolution."
} | 187 | 1,438 |
2017 | T3 | C3 | Combinatorics | JBMO-SL | We have two piles with 2000 and 2017 coins respectively. Ann and Bob take alternate turns making the following moves: The player whose turn is to move picks a pile with at least two coins, removes from that pile $t$ coins for some $2 \leqslant t \leqslant 4$, and adds to the other pile 1 coin. The players can choose a ... | Denote the number of coins in the two piles by $X$ and $Y$. We say that the pair $(X, Y)$ is losing if the player who begins the game loses and that the pair $(X, Y)$ is winning otherwise. We shall prove that $(X, Y)$ is loosing if $X-Y \equiv 0,1,7 \bmod 8$, and winning if $X-Y \equiv 2,3,4,5,6 \bmod 8$.
Lemma 1. If ... | {
"problem_match": "\nC3.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl",
"solution_match": "\nSolution."
} | 117 | 512 |
2017 | T3 | G1 | Geometry | JBMO-SL | Given a parallelogram $A B C D$. The line perpendicular to $A C$ passing through $C$ and the line perpendicular to $B D$ passing through $A$ intersect at point $P$. The circle centered at point $P$ and radius $P C$ intersects the line $B C$ at point $X,(X \neq C)$ and the line $D C$ at point $Y$, $(Y \neq C)$. Prove th... | Denote the feet of the perpendiculars from $P$ to the lines $B C$ and $D C$ by $M$ and $N$ respectively and let $O=A C \cap B D$. Since the points $O, M$ and $N$ are midpoints of $C A, C X$ and $C Y$ respectively it suffices to prove that $M, N$ and $O$ are collinear. According to Menelaus's theorem for $\triangle B C ... | {
"problem_match": "\nG1.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl",
"solution_match": "\nSolution."
} | 112 | 620 |
2017 | T3 | G2 | Geometry | JBMO-SL | Let $A B C$ be an acute triangle such that $A B$ is the shortest side of the triangle. Let $D$ be the midpoint of the side $A B$ and $P$ be an interior point of the triangle such that
$$
\Varangle C A P=\Varangle C B P=\Varangle A C B
$$
Denote by $M$ and $N$ the feet of the perpendiculars from $P$ to $B C$ and $A C$... | If $\gamma=\Varangle A C B$ then $\Varangle C A P=\Varangle C B P=\Varangle A C B=\gamma$. Let $E=K N \cap A P$ and $F=K M \cap B P$. We show that points $E$ and $F$ are midpoints of $A P$ and $B P$, respectively.
Indeed, consider the triangle $A E N$. Since $K N \| B C$, we have $\Varangle E N A=\Varangle B C A=\gamm... | {
"problem_match": "\nG2.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl",
"solution_match": "\nSolution."
} | 170 | 735 |
2017 | T3 | G3 | Geometry | JBMO-SL | Consider triangle $A B C$ such that $A B \leq A C$. Point $D$ on the arc $B C$ of the circumcirle of $A B C$ not containing point $A$ and point $E$ on side $B C$ are such that
$$
\Varangle B A D=\Varangle C A E<\frac{1}{2} \Varangle B A C .
$$
Let $S$ be the midpoint of segment $A D$. If $\Varangle A D E=\Varangle A ... | Let the tangent to the circumcircle of $\triangle A B C$ at point $A$ intersect line $B C$ at $T$. Since $A B \leq A C$ we get that $B$ lies between $T$ and $C$. Since $\Varangle B A T=\Varangle A C B$ and $\Varangle A B T=\Varangle 180^{\circ}-\Varangle A B C$ we get $\Varangle E T A=\Varangle B T A=\Varangle A B C-\V... | {
"problem_match": "\nProblem G3.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl",
"solution_match": "\nSolution."
} | 146 | 1,189 |
2017 | T3 | G4 | Geometry | JBMO-SL | Let $A B C$ be a scalene triangle with circumcircle $\Gamma$ and circumcenter $O$. Let $M$ be the midpoint of $B C$ and $D$ be a point on $\Gamma$ such that $A D \perp B C$. Let $T$ be a point such that $B D C T$ is a parallelogram and $Q$ a point on the same side of $B C$ as $A$ such that
$$
\Varangle B Q M=\Varangle... | Let $X^{\prime}$ be symmetric point to $Q$ in line $B C$. Now since $\Varangle C B A=\Varangle C Q M=$ $\Varangle C X^{\prime} M, \Varangle B C A=\Varangle B Q M=\Varangle B X^{\prime} M$, we have
$$
\Varangle B X^{\prime} C=\Varangle B X^{\prime} M+\Varangle C X^{\prime} M=\Varangle C B A+\Varangle B C A=180^{\circ}-... | {
"problem_match": "\nProblem G4.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl",
"solution_match": "\nSolution."
} | 193 | 959 |
2017 | T3 | G5 | Geometry | JBMO-SL | A point $P$ lies in the interior of the triangle $A B C$. The lines $A P, B P$, and $C P$ intersect $B C, C A$, and $A B$ at points $D, E$, and $F$, respectively. Prove that if two of the quadrilaterals $A B D E, B C E F, C A F D, A E P F, B F P D$, and $C D P E$ are concyclic, then all six are concyclic. | We first prove the following lemma:
Lemma 1. Let $A B C D$ be a convex quadrilateral and let $A B \cap C D=E$ and $B C \cap D A=F$. Then the circumcircles of triangles $A B F, C D F, B C E$ and $D A E$ all pass through a common point $P$. This point lies on line $E F$ if and only if $A B C D$ in concyclic.
Proof. Let... | {
"problem_match": "\nProblem G5.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl",
"solution_match": "\nSolution."
} | 111 | 1,105 |
2017 | T3 | NT5 | Number Theory | JBMO-SL | Find all positive integers $n$ such that there exists a prime number $p$, such that
$$
p^{n}-(p-1)^{n}
$$
is a power of 3 .
Note. A power of 3 is a number of the form $3^{a}$ where $a$ is a positive integer. | Suppose that the positive integer $n$ is such that
$$
p^{n}-(p-1)^{n}=3^{a}
$$
for some prime $p$ and positive integer $a$.
If $p=2$, then $2^{n}-1=3^{a}$ by $(1)$, whence $(-1)^{n}-1 \equiv 0(\bmod 3)$, so $n$ should be even. Setting $n=2 s$ we obtain $\left(2^{s}-1\right)\left(2^{s}+1\right)=3^{a}$. It follows tha... | {
"problem_match": "\nNT5.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_2017_short_list.jsonl",
"solution_match": "\nSolution."
} | 70 | 603 |
2019 | T3 | A3 | Algebra | JBMO-SL | Let $A$ and $B$ be two non-empty subsets of $X=\{1,2, \ldots, 11\}$ with $A \cup B=X$. Let $P_{A}$ be the product of all elements of $A$ and let $P_{B}$ be the product of all elements of $B$. Find the minimum and maximum possible value of $P_{A}+P_{B}$ and find all possible equality cases. | For the maximum, we use the fact that $\left(P_{A}-1\right)\left(P_{B}-1\right) \geqslant 0$, to get that $P_{A}+P_{B} \leqslant P_{A} P_{B}+1=11!+1$. Equality holds if and only if $A=\{1\}$ or $B=\{1\}$.
For the minimum observe, first that $P_{A} \cdot P_{B}=11!=c$. Without loss of generality let $P_{A} \leqslant P_{... | {
"problem_match": "\nA3.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2019-1.jsonl",
"solution_match": "\nSolution."
} | 97 | 1,234 |
2019 | T3 | A4 | Algebra | JBMO-SL | 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
$$
\begin{aligned}
2 c & =a^{4}-2019 a+b^{4}-2019 b \\
& =a^{4}+b^{4}-2019(a+b) \\
& =a^{4}+b^{4}-(a+b)^{2}\left(a^{2}+b^{2}\right) \\
& =-2 a b\lef... | {
"problem_match": "\nA4.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2019-1.jsonl",
"solution_match": "\nSolution."
} | 62 | 753 |
2019 | T3 | A5 | Algebra | JBMO-SL | Let $a, b, c, d$ be positive real numbers such that $a b c d=1$. Prove the inequality
$$
\frac{1}{a^{3}+b+c+d}+\frac{1}{a+b^{3}+c+d}+\frac{1}{a+b+c^{3}+d}+\frac{1}{a+b+c+d^{3}} \leqslant \frac{a+b+c+d}{4}
$$ | From the Cauchy-Schwarz Inequality, we obtain
$$
(a+b+c+d)^{2} \leqslant\left(a^{3}+b+c+d\right)\left(\frac{1}{a}+b+c+d\right)
$$
Using this, together with the other three analogous inequalities, we get
$$
\begin{aligned}
\frac{1}{a^{3}+b+c+d}+\frac{1}{a+b^{3}+c+d}+\frac{1}{a+b+c^{3}+d} & +\frac{1}{a+b+c+d^{3}} \\
&... | {
"problem_match": "\nA5.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2019-1.jsonl",
"solution_match": "\nSolution."
} | 101 | 745 |
2019 | T3 | A6 | Algebra | JBMO-SL | Let $a, b, c$ be positive real numbers. Prove the inequality
$$
\left(a^{2}+a c+c^{2}\right)\left(\frac{1}{a+b+c}+\frac{1}{a+c}\right)+b^{2}\left(\frac{1}{b+c}+\frac{1}{a+b}\right)>a+b+c
$$ | By the Cauchy-Schwarz Inequality, we have
$$
\frac{1}{a+b+c}+\frac{1}{a+c} \geqslant \frac{4}{2 a+b+2 c}
$$
and
$$
\frac{1}{b+c}+\frac{1}{a+b} \geqslant \frac{4}{a+2 b+c}
$$
Since
$$
a^{2}+a c+c^{2}=\frac{3}{4}(a+c)^{2}+\frac{1}{4}(a-c)^{2} \geqslant \frac{3}{4}(a+c)^{2}
$$
then, writing $L$ for the Left Hand Sid... | {
"problem_match": "\nA6.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2019-1.jsonl",
"solution_match": "\nSolution."
} | 83 | 718 |
2019 | T3 | A7 | Algebra | JBMO-SL | Show that for any positive real numbers $a, b, c$ such that $a+b+c=a b+b c+c a$, the following inequality holds
$$
3+\sqrt[3]{\frac{a^{3}+1}{2}}+\sqrt[3]{\frac{b^{3}+1}{2}}+\sqrt[3]{\frac{c^{3}+1}{2}} \leqslant 2(a+b+c)
$$ | Using the condition we have
$$
a^{2}-a+1=a^{2}-a+1+a b+b c+c a-a-b-c=(c+a-1)(a+b-1)
$$
Hence we have
$$
\sqrt[3]{\frac{a^{3}+1}{2}}=\sqrt[3]{\frac{(a+1)\left(a^{2}-a+1\right)}{2}}=\sqrt[3]{\left(\frac{a+1}{2}\right)(c+a-1)(a+b-1)}
$$
Using the last equality together with the AM-GM Inequality, we have
$$
\begin{ali... | {
"problem_match": "\nA7.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2019-1.jsonl",
"solution_match": "\nSolution."
} | 98 | 1,006 |
2019 | T3 | C2 | Combinatorics | JBMO-SL | In a certain city there are $n$ straight streets, such that every two streets intersect, and no three streets pass through the same intersection. The City Council wants to organize the city by designating the main and the side street on every intersection. Prove that this can be done in such way that if one goes along ... | Pick any street $s$ and organize the intersections along $s$ such that the intersections of the two types alternate, as in the statement of the problem.
On every other street $s_{1}$, exactly one intersection has been organized, namely the one where $s_{1}$ intersects $s$. Call this intersection $I_{1}$. We want to or... | {
"problem_match": "\nC2.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2019-1.jsonl",
"solution_match": "\nSolution."
} | 102 | 800 |
2019 | T3 | C4 | Combinatorics | JBMO-SL | We have a group of $n$ kids. For each pair of kids, at least one has sent a message to the other one. For each kid $A$, among the kids to whom $A$ has sent a message, exactly $25 \%$ have sent a message to $A$. How many possible two-digit values of $n$ are there? | If the number of pairs of kids with two-way communication is $k$, then by the given condition the total number of messages is $4 k+4 k=8 k$. Thus the number of pairs of kids is $\frac{n(n-1)}{2}=7 k$. This is possible only if $n \equiv 0,1 \bmod 7$.
- In order to obtain $n=7 m+1$, arrange the kids in a circle and let ... | {
"problem_match": "\nC4.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2019-1.jsonl",
"solution_match": "\nSolution."
} | 74 | 1,237 |
2019 | T3 | C5 | Combinatorics | JBMO-SL | An economist and a statistician play a game on a calculator which does only one operation. The calculator displays only positive integers and it is used in the following way: Denote by $n$ an integer that is shown on the calculator. A person types an integer, $m$, chosen from the set $\{1,2, \ldots, 99\}$ of the first ... | First of all, the game finishes because the number on the calculator always decreases. By picking $m \%$ of a positive integer $n$, players get the number
$$
\frac{m \cdot n}{100}=\frac{m \cdot n}{2^{2} 5^{2}}
$$
We see that at least one of the powers of 2 and 5 that divide $n$ decreases after one move, as $m$ is not... | {
"problem_match": "\nC5.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2019-1.jsonl",
"solution_match": "\nSolution."
} | 281 | 2,084 |
2019 | T3 | G1 | Geometry | JBMO-SL | Let $A B C$ be a right-angled triangle with $\hat{A}=90^{\circ}$ and $\hat{B}=30^{\circ}$. The perpendicular at the midpoint $M$ of $B C$ meets the bisector $B K$ of the angle $\hat{B}$ at the point $E$. The perpendicular bisector of $E K$ meets $A B$ at $D$. Prove that $K D$ is perpendicular to $D E$. | Alternative Solution by PSC. Let $P$ be the point of intersection of $E M$ with $A C$ and let $I$ be the midpoint of $E K$. Then the triangle $P B C$ is equilateral. We also have $\angle P E I=\angle B E M=75^{\circ}$ and $\angle P K E=75^{\circ}$, so $P E K$ is isosceles. We also have $P I \perp E K$ and $D I \perp E ... | {
"problem_match": "\nG1.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2019-1.jsonl",
"solution_match": "\nSolution."
} | 104 | 1,098 |
2019 | T3 | G2 | Geometry | JBMO-SL | Let $A B C$ be a triangle and let $\omega$ be its circumcircle. Let $\ell_{B}$ and $\ell_{C}$ be two parallel lines passing through $B$ and $C$ respectively. The lines $\ell_{B}$ and $\ell_{C}$ intersect with $\omega$ for the second time at the points $D$ and $E$ respectively, with $D$ belonging on the arc $A B$, and $... | ## Alternative Solution by PSC.
Let $L$ and $Z$ be the points of intesecrion of $O O_{1}$ with $\ell_{b}$ and $D A$ respectively. Since $L Z$ is perpendicular on $D A$, and since $\ell_{b}$ is parallel to $\ell_{c}$, then
$$
\angle D L O=90^{\circ}-\angle L D Z=90^{\circ}-\angle D F E=90^{\circ}-\angle A F E
$$
Sinc... | {
"problem_match": "\nG2.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2019-1.jsonl",
"solution_match": "\nSolution."
} | 218 | 539 |
2019 | T3 | G3 | Geometry | JBMO-SL | Let $A B C$ be a triangle with incenter $I$. The points $D$ and $E$ lie on the segments $C A$ and $B C$ respectively, such that $C D=C E$. Let $F$ be a point on the segment $C D$. Prove that the quadrilateral $A B E F$ is circumscribable if and only if the quadrilateral $D I E F$ is cyclic. | Since $C D=C E$ it means that $E$ is the reflection of $D$ on the bisector of $\angle A C B$, i.e. the line $C I$. Let $G$ be the reflection of $F$ on $C I$. Then $G$ lies on the segment $C E$, the segment $E G$ is the reflection of the segment $D F$ on the line $C I$. Also, the quadraliteral $D E G F$ is cyclic since ... | {
"problem_match": "\nG3.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2019-1.jsonl",
"solution_match": "\nSolution."
} | 94 | 667 |
2019 | T3 | G4 | Geometry | JBMO-SL | Let $A B C$ be a triangle such that $A B \neq A C$, and let the perpendicular bisector of the side $B C$ intersect lines $A B$ and $A C$ at points $P$ and $Q$, respectively. If $H$ is the orthocenter of the triangle $A B C$, and $M$ and $N$ are the midpoints of the segments $B C$ and $P Q$ respectively, prove that $H M... | We have
$$
\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": "\nG4.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2019-1.jsonl",
"solution_match": "\nSolution."
} | 120 | 550 |
2019 | T3 | G5 | Geometry | JBMO-SL | Let $P$ be a point in the interior of a triangle $A B C$. The lines $A P, B P$ and $C P$ intersect again the circumcircles of the triangles $P B C, P C A$, and $P A B$ at $D, E$ and $F$ respectively. Prove that $P$ is the orthocenter of the triangle $D E F$ if and only if $P$ is the incenter of the triangle $A B C$. | If $P$ is the incenter of $A B C$, then $\angle B P D=\angle A B P+\angle B A P=\frac{\hat{A}+\hat{B}}{2}$, and $\angle B D P=\angle B C P=\frac{\hat{C}}{2}$. From triangle $B D P$, it follows that $\angle P B D=90^{\circ}$, i.e. that $E B$ is one of the altitudes of the triangle $D E F$. Similarly, $A D$ and $C F$ are... | {
"problem_match": "\nG5.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2019-1.jsonl",
"solution_match": "\nSolution."
} | 106 | 689 |
2019 | T3 | G7 | Geometry | JBMO-SL | Let $A B C$ be a right-angled triangle with $\hat{A}=90^{\circ}$. Let $K$ be the midpoint of $B C$, and let $A K L M$ be a parallelogram with centre $C$. Let $T$ be the intersection of the line $A C$ and the perpendicular bisector of $B M$. Let $\omega_{1}$ be the circle with centre $C$ and radius $C A$ and let $\omega... | Let $M^{\prime}$ be the symmetric point of $M$ with respect to $T$. Observe that $T$ is equidistant from $B$ and $M$, therefore $M$ belongs on $\omega_{2}$ and $M^{\prime} M$ is a diameter of $\omega_{2}$. It suffices to prove that $M^{\prime} A$ is perpendicular to $L M$, or equivalently, to $A K$. To see this, let $S... | {
"problem_match": "\nG7.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2019-1.jsonl",
"solution_match": "\nSolution."
} | 149 | 806 |
2019 | T3 | N1 | Number Theory | JBMO-SL | Find all prime numbers $p$ for which there are non-negative integers $x, y$ and $z$ such that the number
$$
A=x^{p}+y^{p}+z^{p}-x-y-z
$$
is a product of exactly three distinct prime numbers. | For $p=2$, we take $x=y=4$ and $z=3$. Then $A=30=2 \cdot 3 \cdot 5$. For $p=3$ we can take $x=3$ and $y=2$ and $z=1$. Then again $A=30=2 \cdot 3 \cdot 5$. For $p=5$ we can take $x=2$ and $y=1$ and $z=1$. Again $A=30=2 \cdot 3 \cdot 5$.
Assume now that $p \geqslant 7$. Working modulo 2 and modulo 3 we see that $A$ is d... | {
"problem_match": "\nN1.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2019-1.jsonl",
"solution_match": "\nSolution."
} | 61 | 606 |
2019 | T3 | N2 | Number Theory | JBMO-SL | Find all triples $(p, q, r)$ of prime numbers such that all of the following numbers are integers
$$
\frac{p^{2}+2 q}{q+r}, \quad \frac{q^{2}+9 r}{r+p}, \quad \frac{r^{2}+3 p}{p+q}
$$ | We consider the following cases:
1st Case: If $r=2$, then $\frac{r^{2}+3 p}{p+q}=\frac{4+3 p}{p+q}$. If $p$ is odd, then $4+3 p$ is odd and therefore $p+q$ must be odd. From here, $q=2$ and $\frac{r^{2}+3 p}{p+q}=\frac{4+3 p}{p+2}=3-\frac{2}{p+2}$ which is not an integer. Thus $p=2$ and $\frac{r^{2}+3 p}{p+q}=\frac{10... | {
"problem_match": "\nN2.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2019-1.jsonl",
"solution_match": "\nSolution."
} | 73 | 500 |
2020 | T3 | A 1 | Algebra | JBMO-SL | Find all triples $(a, b, c)$ of real numbers such that the following system holds:
$$
\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": "\nA 1.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl",
"solution_match": "\nSolution."
} | 106 | 858 |
2020 | T3 | A 2 | Algebra | JBMO-SL | Consider the sequence $a_{1}, a_{2}, a_{3}, \ldots$ defined by $a_{1}=9$ and
$$
a_{n+1}=\frac{(n+5) a_{n}+22}{n+3}
$$
for $n \geqslant 1$.
Find all natural numbers $n$ for which $a_{n}$ is a perfect square of an integer. | Define $b_{n}=a_{n}+11$. Then
$$
22=(n+3) a_{n+1}-(n+5) a_{n}=(n+3) b_{n+1}-11 n-33-(n+5) b_{n}+11 n+55
$$
giving $(n+3) b_{n+1}=(n+5) b_{n}$. Then
$b_{n+1}=\frac{n+5}{n+3} b_{n}=\frac{(n+5)(n+4)}{(n+3)(n+2)} b_{n-1}=\frac{(n+5)(n+4)}{(n+2)(n+1)} b_{n-2}=\cdots=\frac{(n+5)(n+4)}{5 \cdot 4} b_{1}=(n+5)(n+4)$.
Theref... | {
"problem_match": "\nA 2.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl",
"solution_match": "\nSolution:"
} | 94 | 1,150 |
2020 | T3 | C 3 | Combinatorics | JBMO-SL | Alice and Bob play the following game: Alice begins by picking a natural number $n \geqslant 2$. Then, with Bob starting first, they alternately choose one number from the set $A=\{1,2, \ldots, n\}$ according to the following condition: The number chosen at each step should be distinct from all the already chosen numbe... | Alice has a winning strategy. She initially picks $n=8$. We will give a strategy so that she can end up with $S$ even, or $S=15$, or $S=21$, so she wins.
Case 1: If Bob chooses 1, then the game ends with Alice choosing 2,4,6,8 so $S$ is even (larger than 2) and Alice wins.
Case 2: If Bob chooses 2, then Alice chooses... | {
"problem_match": "\nC 3.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl",
"solution_match": "\nSolution."
} | 212 | 738 |
2020 | T3 | G 3 | Geometry | JBMO-SL | 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": "\nG 3.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl",
"solution_match": "\nSolution."
} | 187 | 535 |
2020 | T3 | NT 3 | Number Theory | JBMO-SL | Find the largest integer $k(k \geq 2)$, for which there exists an integer $n(n \geq k)$ such that from any collection of $n$ consecutive positive integers one can always choose $k$ numbers, which verify the following conditions:
1. each chosen number is not divisible by 6 , by 7 and by 8 ;
2. the positive difference o... | An integer is divisible by 6,7 and 8 if and only if it is divisible by their Least Common Multiple, which equals $6 \times 7 \times 4=168$.
Let $n$ be a positive integer and let $A$ be an arbitrary set of $n$ consecutive positive integers. Replace each number $a_{i}$ from $A$ with its remainder $r_{i}$ ( mod 168). The... | {
"problem_match": "\nNT 3.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl",
"solution_match": "\nSolution."
} | 104 | 947 |
2020 | T3 | NT 7 | Number Theory | JBMO-SL | Prove that there doesn't exist any prime $p$ such that every power of $p$ is a palindrome (palindrome is a number that is read the same from the left as it is from the right; in particular, number that ends in one or more zeros cannot be a palindrome). | Note that by criterion for divisibility by 11 and the definition of a palindrome we have that every palindrome that has even number of digits is divisible by 11 .
Since $11^{5}=161051$ is not a palindrome and since 11 cannot divide $p^{k}$ for any prime other than 11 we are now left to prove that no prime whose all po... | {
"problem_match": "\nNT 7.",
"resource_path": "JBMO/segmented/en-shortlist/en-jbmo_shortlist_2020.jsonl",
"solution_match": "\nSolution."
} | 60 | 1,737 |
2011 | T3 | NT3 | Number Theory | JBMO-SL | Find all positive integers $n$ such that the equation $y^{2}+x y+3 x=n\left(x^{2}+x y+3 y\right)$ has at least a solution $(x, y)$ in positive integers. | Clearly for $n=1$, each pair $(x, y)$ with $x=y$ is a solution. Now, suppose that $n>1$ which implies $x \neq y$. We have $0<n-1=\frac{y^{2}+x y+3 x}{x^{2}+x y+3 y}-1=\frac{(x+y-3)(y-x)}{x^{2}+x y+3 y}$.
Since $x+y \geq 3$, we conclude that $x+y>3$ and $y>x$. Take $d=\operatorname{gcd}(x+y-$ $\left.3 ; x^{2}+x y+3 y\r... | {
"problem_match": "\nNT3 ",
"resource_path": "JBMO/segmented/en-shortlist/en-nt20111.jsonl",
"solution_match": "## Solution"
} | 52 | 780 |
2012 | T3 | A4 | Algebra | JBMO-SL | Solve the following equation for $x, y, z \in \mathbb{N}$
$$
\left(1+\frac{x}{y+z}\right)^{2}+\left(1+\frac{y}{z+x}\right)^{2}+\left(1+\frac{z}{x+y}\right)^{2}=\frac{27}{4}
$$ | Call $a=1+\frac{x}{y+z}, b=1+\frac{y}{z+x}, c=1+\frac{z}{x+y}$ to get
$$
a^{2}+b^{2}+c^{2}=\frac{27}{4}
$$
Since it is also true that
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2
$$
the quadratic-harmonic means inequality implies
$$
\frac{3}{2}=\sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}} \geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\... | {
"problem_match": "\nA4.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl",
"solution_match": "\nSolution 1."
} | 82 | 838 |
2012 | T3 | A5 | Algebra | JBMO-SL | Find the largest positive integer $n$ for which the inequality
$$
\frac{a+b+c}{a b c+1}+\sqrt[n]{a b c} \leq \frac{5}{2}
$$
holds for all $a, b, c \in[0,1]$. Here $\sqrt[1]{a b c}=a b c$. | Let $n_{\max }$ be the sought largest value of $n$, and let $E_{a, b, c}(n)=\frac{a+b+c}{a b c+1}+\sqrt[n]{a b c}$. Then $E_{a, b, c}(m)-E_{a, b, c}(n)=\sqrt[m]{a b c}-\sqrt[n]{a b c}$ and since $a . b c \leq 1$ we clearly have $E_{a, b, c}(m) \geq$ $E_{a, b, c}(n)$ for $m \geq n$. So if $E_{a, b, c}(n) \geq \frac{5}{2... | {
"problem_match": "\nA5.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl",
"solution_match": "\nSolution."
} | 79 | 1,235 |
2012 | T3 | G1 | Geometry | JBMO-SL | Let $A B C$ be an equilateral triangle, and $P$ a point on the circumcircle of the triangle $A B C$ and distinct from $A, B$ and $C$. If the lines through $P$ and parallel to $B C, C A, A B$ intersect the lines $C A, A B, B C$ at $M, N$ and $Q$ respectively, prove that $M, N$ and $Q$ are collinear. | Solution 2. Let $D$ be a point on $c$ such that $A D<A C$, and let $E, Z$ be the second points of intersection of lines $A D$ and $B D$ with $c$ respectively. Let also $N$ be the second point of intersection of line $B E$ with the circle $c$. Figure $5 \mathrm{~b}$ shows $Z$ between $B, D$. The argument below can be tr... | {
"problem_match": "\nG1.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl",
"solution_match": "\nSolution."
} | 102 | 560 |
2012 | T3 | G4 | Geometry | JBMO-SL | Let $A B C$ be an acute-angled triangle with circumcircle $\Gamma$, and let $O, H$ be the triangle's circumcenter and orthocenter respectively. Let also $A^{\prime}$ be the point where the angle bisector of angle $B A C$ meets $\Gamma$. If $A^{\prime} H=A H$, find the measure of angle $B A C$.
. Then $A M=M P$ and $t \perp A P$, hence the triangle $A P N$ is isosceles with $\mathrm{AP}$ as its base, so $\angle N A P=\angle N P A$. We have $\angle B A P=\angle B A M=\angle B M N$ and $\angle B A N=\angle B N M$. Thus
$$
180^{\circ}-\angl... | {
"problem_match": "\nG4.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl",
"solution_match": "\nSolution."
} | 171 | 543 |
2012 | T3 | G4 | Geometry | JBMO-SL | Let $A B C$ be an acute-angled triangle with circumcircle $\Gamma$, and let $O, H$ be the triangle's circumcenter and orthocenter respectively. Let also $A^{\prime}$ be the point where the angle bisector of angle $B A C$ meets $\Gamma$. If $A^{\prime} H=A H$, find the measure of angle $B A C$.
. Call $Q, M$ the tangency points of $B A, B C$ with $c^{\prime}$ and $c$ respectively, and call $H$ the midpoint of segment $A C$. It is well known that
$$
A Q=\frac{1}{2}\left(A O_{1}+A B-B O_{1}\right) \... | {
"problem_match": "\nG4.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl",
"solution_match": "\nSolution."
} | 171 | 841 |
2012 | T3 | G4 | Geometry | JBMO-SL | Let $A B C$ be an acute-angled triangle with circumcircle $\Gamma$, and let $O, H$ be the triangle's circumcenter and orthocenter respectively. Let also $A^{\prime}$ be the point where the angle bisector of angle $B A C$ meets $\Gamma$. If $A^{\prime} H=A H$, find the measure of angle $B A C$.
. Then $A B+B C+C D+D A \in \mathbb{N}$. But $(A B+B C)+(C D+D A)>A C+A C \geq 2$, hence $A B+B C+C D+D A>2$.
On the other hand, $A B+B C+C D+D A<(A N+N B)+(B P+P C)+(C Q+Q D)+$ $(D M+M A)=4$, hence $A B+B C+C D+D A=3$.
Obviously on... | {
"problem_match": "\nG4.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl",
"solution_match": "\nSolution."
} | 171 | 919 |
2012 | T3 | C2 | Combinatorics | JBMO-SL | $n$ nails nailed on a board are connected by two via a string. Each string is colored in one of $n$ given colors. For any three colors there exist three nails conne.cted by two with strings in these three colors. Can $n$ be: (a) 6, (b) 7? | (a) The answer is no:
Suppose it is possible. Consider some color, say blue. Each blue string is the side of 4 triangles formed with vertices on the given points. As there exist $\binom{5}{2}=\frac{5 \cdot 4}{2}=10$ pairs of colors other than blue, and for any such pair of colors together with the blue color there exi... | {
"problem_match": "\nC2.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl",
"solution_match": "\nSolution."
} | 67 | 516 |
2012 | T3 | C3 | Combinatorics | JBMO-SL | In a circle of diameter 1 consider 65 points no three of which are collinear. Prove that there exist 3 among these points which form a triangle with area less then or equal to $\frac{1}{72}$. | Solution. Write the given equation in the form
$$
p^{2}(p-1)\left(p^{2}+p+1\right)=q(q+1)
$$
First observe that it must not be $p=q$, since in this case the left hand side of (9) is greater than its right hand side. Hence, since $p$ and $q$ are distinct prines, (9) immediately yields $p^{2} \mid q+1$, that is
$$
q=a... | {
"problem_match": "\nC3.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl",
"solution_match": "\nSolution."
} | 50 | 851 |
2012 | T3 | C3 | Combinatorics | JBMO-SL | In a circle of diameter 1 consider 65 points no three of which are collinear. Prove that there exist 3 among these points which form a triangle with area less then or equal to $\frac{1}{72}$. | Solution. Denote
$$
x=\overline{V E R}-\overline{I A}, y=\overline{G R E}+\overline{E C E}, z=G^{R^{E}}
$$
Then obviously, we have
$$
\begin{aligned}
& (201+131 \text { or } 231+101) \leq y \leq(879+969 \text { or } 869+979 \text { or. } 769+989) \\
\Rightarrow \quad & 332 \leq y \leq 1848 \Rightarrow 102-98 \leq x ... | {
"problem_match": "\nC3.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl",
"solution_match": "\nSolution."
} | 50 | 1,527 |
2012 | T3 | C3 | Combinatorics | JBMO-SL | In a circle of diameter 1 consider 65 points no three of which are collinear. Prove that there exist 3 among these points which form a triangle with area less then or equal to $\frac{1}{72}$. | Solution. By Fermat's theorem $n^{2 p} \equiv n^{2}(\bmod p)$, therefore $m^{2}+n^{2}+p+1 \equiv n^{2}(\bmod p) \Rightarrow$ $m^{2} \equiv-1(\bmod p)$.
Case 1: $p=4 k+3$. We have $\left(m^{2}\right)^{2 k+1} \equiv(-1)^{2 k+1}(\bmod p)$. Therefore,
$$
m^{p-1} \equiv-1(\bmod p)
$$
and $p$ does not divide $m$. On the o... | {
"problem_match": "\nC3.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-2012.jsonl",
"solution_match": "\nSolution."
} | 50 | 511 |
2015 | T3 | A1 | Algebra | JBMO-SL | MLD
Let $x, y, z$ be real numbers, satisfying the relations
$$
\left\{\begin{array}{l}
x \geq 20 \\
y \geq 40 \\
z \geq 1675 \\
x+y+z=2015
\end{array}\right.
$$
Find the greatest value of the product $P=x \cdot y \cdot z$. | By virtue of $z \geq 1675$ we have
$$
y+z<2015 \Leftrightarrow y<2015-z \leq 2015-1675<1675
$$
It follows that $(1675-y) \cdot(1675-z) \leq 0 \Leftrightarrow y \cdot z \leq 1675 \cdot(y+z-1675)$.
By using the inequality $u \cdot v \leq\left(\frac{u+v}{2}\right)^{2}$ for all real numbers $u, v$ we obtain
$$
\begin{g... | {
"problem_match": "## A1 ",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl",
"solution_match": "## Solution 1:"
} | 92 | 893 |
2015 | T3 | A3 | Algebra | JBMO-SL | MNE
Let $a, b, c$ be positive real numbers. Prove that
$$
\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}>2
$$ | Starting from the double expression on the left-hand side of given inequality, and applying twice the Arithmetic-Geometric mean inequality, we find that
$$
\begin{aligned}
2 \frac{a}{b}+2 \sqrt{\frac{b}{c}}+2 \sqrt[3]{\frac{c}{a}} & =\frac{a}{b}+\left(\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt{\frac{b}{c}}\right)+2 \sqrt[3]... | {
"problem_match": "## A3 ",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl",
"solution_match": "## Solution:"
} | 53 | 1,074 |
2015 | T3 | A5 | Algebra | JBMO-SL | MKCD
Let $x, y, z$ be positive real numbers that satisfy the equality $x^{2}+y^{2}+z^{2}=3$. Prove that
$$
\frac{x^{2}+y z}{x^{2}+y z+1}+\frac{y^{2}+z x}{y^{2}+z x+1}+\frac{z^{2}+x y}{z^{2}+x y+1} \leq 2
$$ | We have
$$
\begin{aligned}
& \frac{x^{2}+y z}{x^{2}+y z+1}+\frac{y^{2}+z x}{y^{2}+z x+1}+\frac{z^{2}+x y}{z^{2}+x y+1} \leq 2 \Leftrightarrow \\
& \frac{x^{2}+y z+1}{x^{2}+y z+1}+\frac{y^{2}+z x+1}{y^{2}+z x+1}+\frac{z^{2}+x y+1}{z^{2}+x y+1} \leq 2+\frac{1}{x^{2}+y z+1}+\frac{1}{y^{2}+z x+1}+\frac{1}{z^{2}+x y+1} \Le... | {
"problem_match": "## A5 ",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl",
"solution_match": "## Solution:"
} | 111 | 539 |
2015 | T3 | G3 | Geometry | JBMO-SL | GRE
Let $c \equiv c(O, K)$ be a circle with center $O$ and radius $R$ and $A, B$ be two points on it, not belonging to the same diameter. The bisector of the angle $A \hat{B} O$ intersects the circle $c$ at point $C$, the circumcircle of the triangle $A O B$, say ${ }^{c_{1}}$ at point $K$ and the circumcircle of the ... | The segments $O B, O C$ are equal, as radii of the circle ${ }^{c}$. Hence $O B C$ is an isosceles triangle and
$$
\hat{B}_{1}=\hat{C}_{1}=\hat{x}
$$
The chord $B C$ is the bisector of the angle $O \hat{B} A$, and hence
$$
\hat{B}_{1}=\hat{B}_{2}=\hat{x}
$$
The angles $\hat{B}_{2}$ and $\hat{O}_{1}$ are inscribed t... | {
"problem_match": "## G3 ",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl",
"solution_match": "## Solution:"
} | 159 | 644 |
2015 | T3 | G5 | Geometry | JBMO-SL | ROU
Let $A B C$ be an acute triangle with $A B \neq A C$. The incircle $\omega$ of the triangle touches the sides $B C, C A$ and $A B$ at $D, E$ and $F$, respectively. The perpendicular line erected at $C$ onto $B C$ meets $E F$ at $M$, and similarly, the perpendicular line erected at $B$ onto $B C$ meets $E F$ at $N$... | ## Proof 1.1.
Let $\{T\}=E F \cap B C$. Applying Menelaus' theorem to the triangle $A B C$ and the transversal line $E-F-T$ we obtain $\frac{T B}{T C} \cdot \frac{E C}{E A} \cdot \frac{F A}{F B}=1$, i.e. $\frac{T B}{T C} \cdot \frac{s-c}{s-a} \cdot \frac{s-a}{s-b}=1$, or $\frac{T B}{T C}=\frac{s-b}{s-c}$, where the no... | {
"problem_match": "\nG5 ",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl",
"solution_match": "## Solution:"
} | 221 | 526 |
2015 | T3 | NT3 | Number Theory | JBMO-SL | ALB
a) Show that the product of all differences of possible couples of six given positive integers is divisible by 960 (original from Albania).
b) Show that the product of all differences of possible couples of six given positive integers. is divisible by 34560 (modified by problem selecting committee). | a) Since we have six numbers then at least two of them have a same residue when divided by 3 , so at least one of the differences in our product is divisible by 3 .
Since we have six numbers then at least two of them have a same residuc when divided by 5 , so at least one of the differences in our product is divisible... | {
"problem_match": "## NT3 ",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl",
"solution_match": "## Solution:"
} | 66 | 731 |
2015 | T3 | NT4 | Number Theory | JBMO-SL | MLD
Find all prime numbers $a, b, c$ and integers $k$ which satisfy the equation $a^{2}+b^{2}+16 \cdot c^{2}=9 \cdot k^{2}+1$. | 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 $a^{2} \equiv 0,1(\bmod 3), b^{2} \equiv 0,1(\bmod 3), c^{2} \equiv 0,1(\bmod 3)$, we have:
| $a^{2}$ | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| :--- | :--- |... | {
"problem_match": "\nNT4 ",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl",
"solution_match": "## Solution:"
} | 126 | 1,183 |
2015 | T3 | C1 | Combinatorics | JBMO-SL | BUL
A board $n \times n(n \geq 3)$ is divided into $n^{2}$ unit squares. Integers from 0 to $n$ included are written down: one integer in each unit square, in such a way that the sums of integers in each $2 \times 2$ square of the board are different. Find all $n$ for which such boards exist. | The number of the $2 \times 2$ squares in a board $n \times n$ is equal to $(n-1)^{2}$. All possible sums of the numbers in such squares are $0,1, \ldots, 4 n$. A necessary condition for the existence of a board with the required property is $4 n+1 \geq(n-1)^{2}$ and consequently $n(n-6) \leq 0$. Thus $n \leq 6$. The e... | {
"problem_match": "## C1 ",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl",
"solution_match": "## Solution:"
} | 85 | 519 |
2015 | T3 | C3 | Combinatorics | JBMO-SL | ALB
Positive integers are put into the following table
| 1 | 3 | 6 | 10 | 15 | 21 | 28 | 36 | | |
| ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| 2 | 5 | 9 | 14 | 20 | 27 | 35 | 44 | | |
| 4 | 8 | 13 | 19 | 26 | 34 | 43 | 53 | | |
| 7 | 12 | 18 | 25 | 33 | 42 | | | | |
| 11 | 17 | ... | We shall observe straights lines as on the next picture. We can call these lines diagonals.
| 1 | $\sqrt{3}$ | 6 | 10 | 15 | 21 | 28 | 36 | |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 2 | 5 | 9 | 14 | 20 | 27 | 35 | 44 | |
| 4 | 8 | 13 | 19 | 26 | 34 | 43 | 53 | |
| | 12 | 18 | 25 | 3... | {
"problem_match": "## C3 ",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl",
"solution_match": "## Solution 1:"
} | 300 | 1,157 |
2015 | T3 | C4 | Combinatorics | JBMO-SL | GRE
Let $n \geq 1$ be a positive integer. A square of side length $n$ is divided by lines parallel to each side into $n^{2}$ squares of side length 1 . Find the number of parallelograms which have vertices among the vertices of the $n^{2}$ squares of side length 1 , with both sides smaller or equal to 2 , and which ha... | We can divide all these parallelograms into 7 classes (types I-VII), according to Figure.
Type 1: There are $n$ ways to choose the strip for the horizontal (shorter) side of the parallelogram, and $(n-1)$ ways to choose the strip (of the width 2 ) for the vertical (longer) side. So there are $n(n-1)$ parallelograms of... | {
"problem_match": "\nC4",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-2015.jsonl",
"solution_match": "\nSolution:"
} | 92 | 1,102 |
2018 | T3 | A 1 | Algebra | JBMO-SL | Let $x, y$ and $z$ be positive numbers. Prove that
$$
\frac{x}{\sqrt{\sqrt[4]{y}+\sqrt[4]{z}}}+\frac{y}{\sqrt{\sqrt[4]{z}+\sqrt[4]{x}}}+\frac{z}{\sqrt{\sqrt[4]{x}+\sqrt[4]{y}}} \geq \frac{\sqrt[4]{(\sqrt{x}+\sqrt{y}+\sqrt{z})^{7}}}{\sqrt{2 \sqrt{27}}}
$$ | Replacing $x=a^{2}, y=b^{2}, z=c^{2}$, where $a, b, c$ are positive numbers, our inequality is equivalent to
$$
\frac{a^{2}}{\sqrt{\sqrt{b}+\sqrt{c}}}+\frac{b^{2}}{\sqrt{\sqrt{c}+\sqrt{a}}}+\frac{c^{2}}{\sqrt{\sqrt{a}+\sqrt{b}}} \geq \frac{\sqrt[4]{(a+b+c)^{7}}}{\sqrt{2 \sqrt{27}}}
$$
Using the Cauchy-Schwarz inequal... | {
"problem_match": "\nA 1.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-jbmo2018.jsonl",
"solution_match": "\nSolution."
} | 122 | 877 |
2018 | T3 | A 2 | Algebra | JBMO-SL | Find the maximum positive integer $k$ such that for any positive integers $m, n$ such that $m^{3}+n^{3}>$ $(m+n)^{2}$, we have
$$
m^{3}+n^{3} \geq(m+n)^{2}+k
$$ | We see that for $m=3$ and $n=2$ we have $m^{3}+n^{3}>(m+n)^{2}$, thus
$$
3^{3}+2^{3} \geq(3+2)^{2}+k \Rightarrow k \leq 10
$$
We will show that $k=10$ is the desired maximum. In other words, we have to prove that
$$
m^{3}+n^{3} \geq(m+n)^{2}+10
$$
The last inequality is equivalent to
$$
(m+n)\left(m^{2}+n^{2}-m n-... | {
"problem_match": "\nA 2.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-jbmo2018.jsonl",
"solution_match": "\nSolution."
} | 67 | 879 |
2018 | T3 | A 3 | Algebra | JBMO-SL | Let $a, b, c$ be positive real numbers. Prove that
$$
\frac{1}{a b(b+1)(c+1)}+\frac{1}{b c(c+1)(a+1)}+\frac{1}{c a(a+1)(b+1)} \geq \frac{3}{(1+a b c)^{2}}
$$ | The required inequality is equivalent to
$$
\frac{c(a+1)+a(b+1)+b(c+1)}{a b c(a+1)(b+1)(c+1)} \geq \frac{3}{(1+a b c)^{2}}
$$
or equivalently to,
$$
(1+a b c)^{2}(a b+b c+c a+a+b+c) \geq 3 a b c(a b+b c+c a+a+b+c+a b c+1)
$$
Let $m=a+b+c, n=a b+b c+c a$ and $x^{3}=a b c$, then the above can be rewritten as
$$
(m+n... | {
"problem_match": "\nA 3.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-jbmo2018.jsonl",
"solution_match": "\nSolution."
} | 82 | 942 |
2018 | T3 | A 4 | Algebra | JBMO-SL | Let $k>1, n>2018$ be positive integers, and let $n$ be odd. The nonzero rational numbers $x_{1}$, $x_{2}, \ldots, x_{n}$ are not all equal and satisfy
$$
x_{1}+\frac{k}{x_{2}}=x_{2}+\frac{k}{x_{3}}=x_{3}+\frac{k}{x_{4}}=\cdots=x_{n-1}+\frac{k}{x_{n}}=x_{n}+\frac{k}{x_{1}}
$$
Find:
a) the product $x_{1} x_{2} \ldots ... | a) If $x_{i}=x_{i+1}$ for some $i$ (assuming $x_{n+1}=x_{1}$ ), then by the given identity all $x_{i}$ will be equal, a contradiction. Thus $x_{1} \neq x_{2}$ and
$$
x_{1}-x_{2}=k \frac{x_{2}-x_{3}}{x_{2} x_{3}}
$$
Analogously
$$
x_{1}-x_{2}=k \frac{x_{2}-x_{3}}{x_{2} x_{3}}=k^{2} \frac{x_{3}-x_{4}}{\left(x_{2} x_{3... | {
"problem_match": "\nA 4.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-jbmo2018.jsonl",
"solution_match": "\nSolution."
} | 197 | 873 |
2018 | T3 | A 5 | Algebra | JBMO-SL | Let $a, b, c, d$ and $x, y, z, t$ be real numbers such that
$$
0 \leq a, b, c, d \leq 1, \quad x, y, z, t \geq 1 \text { and } a+b+c+d+x+y+z+t=8
$$
Prove that
$$
a^{2}+b^{2}+c^{2}+d^{2}+x^{2}+y^{2}+z^{2}+t^{2} \leq 28
$$
When does the equality hold? | We observe that if $u \leq v$ then by replacing $(u, v)$ with $(u-\varepsilon, v+\varepsilon)$, where $\varepsilon>0$, the sum of squares increases. Indeed,
$$
(u-\varepsilon)^{2}+(v+\varepsilon)^{2}-u^{2}-v^{2}=2 \varepsilon(v-u)+2 \varepsilon^{2}>0
$$
Then, denoting
$$
E(a, b, c, d, x, y, z, t)=a^{2}+b^{2}+c^{2}+d... | {
"problem_match": "\nA 5.",
"resource_path": "JBMO/segmented/en-shortlist/en-shl-jbmo2018.jsonl",
"solution_match": "\nSolution."
} | 137 | 916 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.