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Problem 1. 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.
|
Solution. 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 $p \geqslant 7$. Workin... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to find all prime numbers p where there exist positive integers x, y, z such that the expression x^p + y^p + z^p - x - y - z is a product of exactly three distinct prime numbers. Hmm, let me break this down step by step.
First, since p is a prime number, possible values for p are 2, 3, 5, 7, et... |
Problem 1. 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.
|
Solution. 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 $p \geqslant 7$. Workin... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to find all prime numbers p such that there exist positive integers x, y, z where the expression x^p + y^p + z^p - x - y - z is a product of exactly three distinct prime numbers. Hmm, let's break this down step by step.
First, since p is a prime number, possible values are 2, 3, 5, 7, etc. The ... |
Problem 2. 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 $-\sqrt{c}<a b<0$.
|
Solution. 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
$$
\begin{aligned}
2 c & =a^{4}-2019 a+b^{4}-2019 b \\
& =a^{4}+b^{4}-2019(a+b) \\
& =a^{4}+b^{4}-(a+b)^{2}\left(a^{2}+b^{2}\right) \\
& ... | proof | Algebra | proof | Yes | Incomplete | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to prove that if two distinct real numbers a and b, and a positive real number c satisfy the equations a⁴ - 2019a = b⁴ - 2019b = c, then the product ab is between -√c and 0. Hmm, let me start by understanding what the problem is saying.
First, both a and b are distinct real numbers, and c is po... |
Problem 2. Let $a, b$ be two distinct real numbers and let $c$ be a positive real number such that
$$
a^{4}-2019 a=b^{4}-2019 b=c .
$$
Prove that $-\sqrt{c}<a b<0$.
|
Solution. Firstly, we see that
$$
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) \\
& ... | proof | Algebra | proof | Yes | Incomplete | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I have this problem here where there are two distinct real numbers, a and b, and a positive real number c. The equations given are a⁴ - 2019a = c and b⁴ - 2019b = c. I need to prove that -√c is less than ab, which is less than 0. Hmm, interesting. Let me try to unpack this step by step.
First, since a... |
Problem 3. 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 $... |
Solution. 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 $M$ a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to prove that lines HM and AN meet on the circumcircle of triangle ABC. Let me start by understanding the problem step by step.
First, we have triangle ABC with AB < AC. The perpendicular bisector of BC intersects AB at P and AC at Q. H is the orthocenter of ABC. M is the midpoint of BC, which ... |
Problem 3. 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 $... |
Solution. 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 $M$ a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Alright, so I have this geometry problem here. Let me read it again to make sure I understand all the parts.
We have triangle ABC where AB is less than AC. That's important because it tells me that triangle ABC isn't isoceles at A; AC is longer. Then, the perpendicular bisector of side BC intersects lines AB ... |
Problem 4. A $5 \times 100$ table is divided into 500 unit square cells, where $n$ of them are coloured black and the rest are coloured white. Two unit square cells are called adjacent if they share a common side. Each of the unit square cells has at most two adjacent black unit square cells. Find the largest possible... |
Solution. If we colour all the cells along all edges of the board together with the entire middle row except the second and the last-but-one cell, the condition is satisfied and there are 302 black cells. The figure below exhibits this colouring for the $5 \times 8$ case.
 in a 5x100 grid where each black cell has at most two adjacent black cells. Let me think through this step by step.
First, the problem states that each cell, whether black or white, has at most two adjacent black cells. Wait, actually, let ... |
Problem 1. 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})
$$
the quadrilateral $I R N C$ is ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, let's see. I need to prove that the incenter of the trapezoid ABCD lies on the line MN, where M and N are the points where the incircle of triangle ABC touches AB and AC, respectively. Hmm, trapezoid is circumscribed, which means it's tangential, right? So, it has an incircle tangent to all four sides. Wa... |
Problem 1. A trapezoid $A B C D(A B \| C D, A B>C D)$ is circumscribed. The incircle of the triangle $A B C$ touches the lines $A B$ and $A C$ at the points $M$ and $N$, respectively. Prove that the incenter of the trapezoid $A B C D$ lies on the line $M N$.
=90^{\circ}-\frac{1}{2} m(\widehat{M A N}) \quad \text { and } \quad m(\widehat{B I C})=90^{\circ}+\frac{1}{2} m(\widehat{M A N})
$$
the quadrilateral $I R N C$ is ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Alright, so I have this problem about a trapezoid ABCD where AB is parallel to CD and AB is longer than CD. The trapezoid is circumscribed, which I think means it has an incircle tangent to all four sides. Then, there's the incircle of triangle ABC that touches AB and AC at points M and N, respectively. I need ... |
Problem 2. Let $a, b$ and $c$ be positive real numbers. Prove that
$$
\frac{8}{(a+b)^{2}+4 a b c}+\frac{8}{(b+c)^{2}+4 a b c}+\frac{8}{(c+a)^{2}+4 a b c}+a^{2}+b^{2}+c^{2} \geq \frac{8}{a+3}+\frac{8}{b+3}+\frac{8}{c+3}
$$
|
Solution. Since $2 a b \leq a^{2}+b^{2}$, it follows that $(a+b)^{2} \leq 2\left(a^{2}+b^{2}\right)$
and $4 a b c \leq 2 c\left(a^{2}+b^{2}\right)$, for any positive reals $a, b, c$.
Adding these inequalities, we find
$$
(a+b)^{2}+4 a b c \leq 2\left(a^{2}+b^{2}\right)(c+1)
$$
so that
$$
\frac{8}{(a+b)^{2}+4 a b ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, let me try to tackle this inequality problem. Hmm, the problem states that for positive real numbers a, b, c, the following inequality holds:
$$
\frac{8}{(a+b)^{2}+4 a b c} + \frac{8}{(b+c)^{2}+4 a b c} + \frac{8}{(c+a)^{2}+4 a b c} + a^{2} + b^{2} + c^{2} \geq \frac{8}{a+3} + \frac{8}{b+3} + \frac{8}{c+... |
Problem 2. Let $a, b$ and $c$ be positive real numbers. Prove that
$$
\frac{8}{(a+b)^{2}+4 a b c}+\frac{8}{(b+c)^{2}+4 a b c}+\frac{8}{(c+a)^{2}+4 a b c}+a^{2}+b^{2}+c^{2} \geq \frac{8}{a+3}+\frac{8}{b+3}+\frac{8}{c+3}
$$
|
Solution. Since $2 a b \leq a^{2}+b^{2}$, it follows that $(a+b)^{2} \leq 2\left(a^{2}+b^{2}\right)$
and $4 a b c \leq 2 c\left(a^{2}+b^{2}\right)$, for any positive reals $a, b, c$.
Adding these inequalities, we find
$$
(a+b)^{2}+4 a b c \leq 2\left(a^{2}+b^{2}\right)(c+1)
$$
so that
$$
\frac{8}{(a+b)^{2}+4 a b ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, let's see. I need to prove this inequality involving three positive real numbers a, b, and c. The left side has three fractions with denominators like (a+b)^2 + 4abc, and then the sum of a² + b² + c². The right side has three terms that are 8 divided by (a+3), (b+3), and (c+3).
First, I should probably ... |
Problem 3. Find all the triples of integers $(a, b, c)$ such that the number
$$
N=\frac{(a-b)(b-c)(c-a)}{2}+2
$$
is a power of 2016 .
(A power of 2016 is an integer of the form $2016^{n}$, where $n$ is a non-negative integer.)
|
Solution. Let $a, b, c$ be integers and $n$ be a positive integer such that
$$
(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)... | (,b,)=(k+2,k+1,k),k\in\mathbb{Z} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to find all integer triples (a, b, c) such that N, defined as [(a - b)(b - c)(c - a)/2] + 2, is a power of 2016. A power of 2016 means it can be written as 2016^n where n is a non-negative integer. Let me break this down step by step.
First, let me understand the structure of N. It's given by [... |
Problem 3. 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.)
|
Solution. 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)... | (,b,)=(k+2,k+1,k),k\in\mathbb{Z} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, let's see. The problem is asking for all integer triples (a, b, c) such that the number N, defined as [(a - b)(b - c)(c - a)]/2 + 2, is a power of 2016. A power of 2016 means it's like 2016^n where n is a non-negative integer. So, we need to find all triples (a, b, c) where this expression equals 2016^n f... |
Problem 2. 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 ... |
Solution. 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 (w... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2 | [
true,
true
] | 2 | <think>
Okay, so I need to find the largest possible number of three-digit numbers, let's call this number n, that satisfy all the given conditions. Let me read the problem again and make sure I understand each part.
First, the numbers are three-digit, and they have to meet five conditions. Let me list them again:
1.... |
Problem 2. 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 ... |
Solution. 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 (w... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2 | [
true,
true
] | 2 | <think>
Okay, let's see. I need to find the largest possible value of n, which is the number of three-digit numbers that satisfy all these conditions. Let's go through the conditions again to make sure I understand them correctly.
First, each number is a three-digit number without any zeros. So each digit—hundreds, te... |
Problem 4. 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}$ ar... |
Solution. 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 intersectio... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1 | [
false
] | 0 | <think>
Alright, so I need to prove that the lines AA₁, BB₁, and CC₁ are concurrent in an acute triangle ABC. Let me start by understanding the problem step by step.
First, the problem mentions reflecting the vertices over the opposite sides. So, for example, A' is the reflection of A over BC. Similarly, B' is the ref... |
# Problem 1
Let $a, b$ and $c$ be positive real numbers such that $a+b+c=1$. Prove that
$$
\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}+6 \geq 2 \sqrt{2}\left(\sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}\right)
$$
When does equality hold?
| ## Solution
Replacing $1-a, 1-b, 1-c$ with $b+c, c+a, a+b$ respectively on the right hand side, the given inequality becomes
and equivalently
$$
\left(\frac{b+c}{a}-2 \sqrt{2} \sqrt{\frac{... | =b==\frac{1}{3} | Inequalities | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Alright, let me try to tackle this problem. So, we have three positive real numbers a, b, c such that their sum is 1. We need to prove this inequality involving fractions and square roots. The equality condition is also required. Hmm, okay.
First, let me write down the given condition and the inequality again ... |
# Problem 1
Let $a, b$ and $c$ be positive real numbers such that $a+b+c=1$. Prove that
$$
\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}+6 \geq 2 \sqrt{2}\left(\sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}\right)
$$
When does equality hold?
| ## Solution
Replacing $1-a, 1-b, 1-c$ with $b+c, c+a, a+b$ respectively on the right hand side, the given inequality becomes
and equivalently
$$
\left(\frac{b+c}{a}-2 \sqrt{2} \sqrt{\frac{... | =b==\frac{1}{3} | Inequalities | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to prove this inequality:
$$
\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}+6 \geq 2 \sqrt{2}\left(\sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}\right)
$$
given that \( a, b, c \) are positive real numbers with \( a + b + c = 1 \). Also, I have to find... |
## Problem 2
Let the circles $k_{1}$ and $k_{2}$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_{1}$ and $k_{2}$, that touches $k_{1}$ and $k_{2}$ at $M$ and $N$, respectively. If $t \perp A M$ and $M N=2 A M$, evaluate $\angle N M B$.
| ## Solution 1
Let $P$ be the symmetric of $A$ with respect to $M$ (Figure 1). Then $A M=M P$ and $t \perp A P$, hence the triangle $A P N$ is isosceles with $A P$ 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 we have
$$
180^{\circ}... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, let me try to figure out this geometry problem. So, we have two circles, k₁ and k₂, intersecting at points A and B. There's a common tangent t that touches k₁ at M and k₂ at N. We know that this tangent t is perpendicular to AM, and the length of MN is twice the length of AM. We need to find the measure o... |
## Problem 2
Let the circles $k_{1}$ and $k_{2}$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_{1}$ and $k_{2}$, that touches $k_{1}$ and $k_{2}$ at $M$ and $N$, respectively. If $t \perp A M$ and $M N=2 A M$, evaluate $\angle N M B$.
| ## Solution 1
Let $P$ be the symmetric of $A$ with respect to $M$ (Figure 1). Then $A M=M P$ and $t \perp A P$, hence the triangle $A P N$ is isosceles with $A P$ as its base, so $\angle N A P=\angle N P A$. We have $\angle B A P=\angle B A M=\angle B M N$ and $\angle B A N=\angle B N M$.
Thus we have
$$
180^{\circ}... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, so I need to find the measure of angle NMB given the conditions in the problem. Let me start by visualizing the problem and then break it down step by step.
First, there are two circles, k₁ and k₂, intersecting at points A and B. There's a common tangent t that touches k₁ at M and k₂ at N. The tangent li... |
## Problem 2
Let the circles $k_{1}$ and $k_{2}$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_{1}$ and $k_{2}$, that touches $k_{1}$ and $k_{2}$ at $M$ and $N$, respectively. If $t \perp A M$ and $M N=2 A M$, evaluate $\angle N M B$.
| ## Solution 1
Let $P$ be the symmetric of $A$ with respect to $M$ (Figure 1). Then $A M=M P$ and $t \perp A P$, hence the triangle $A P N$ is isosceles with $A P$ 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 we have
$$
180^{\circ}... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, so I need to solve this geometry problem. Let me read it again and make sure I understand all the parts.
We have two circles, k₁ and k₂, intersecting at points A and B. There's a common tangent t that touches k₁ at M and k₂ at N. The tangent t is perpendicular to AM, and the length of MN is twice the len... |
## Problem 2
Let the circles $k_{1}$ and $k_{2}$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_{1}$ and $k_{2}$, that touches $k_{1}$ and $k_{2}$ at $M$ and $N$, respectively. If $t \perp A M$ and $M N=2 A M$, evaluate $\angle N M B$.
| ## Solution 1
Let $P$ be the symmetric of $A$ with respect to $M$ (Figure 1). Then $A M=M P$ and $t \perp A P$, hence the triangle $A P N$ is isosceles with $A P$ as its base, so $\angle N A P=\angle N P A$. We have $\angle B A P=\angle B A M=\angle B M N$ and $\angle B A N=\angle B N M$.
Thus we have
$$
180^{\circ}... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, let's tackle this geometry problem step by step. First, let me understand the problem statement again to make sure I have all the details right.
We have two circles, k₁ and k₂, intersecting at two distinct points A and B. There's a common tangent t to both circles. This tangent touches k₁ at point M and ... |
## Problem 2
Let the circles $k_{1}$ and $k_{2}$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_{1}$ and $k_{2}$, that touches $k_{1}$ and $k_{2}$ at $M$ and $N$, respectively. If $t \perp A M$ and $M N=2 A M$, evaluate $\angle N M B$.
| ## Solution 1
Let $P$ be the symmetric of $A$ with respect to $M$ (Figure 1). Then $A M=M P$ and $t \perp A P$, hence the triangle $A P N$ is isosceles with $A P$ 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 we have
$$
180^{\circ}... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, so I have this geometry problem here. Let me read it again and try to visualize. Two circles, k₁ and k₂, intersect at points A and B. There's a common tangent t that touches k₁ at M and k₂ at N. The tangent t is perpendicular to AM, and the length of MN is twice the length of AM. I need to find the measur... |
## Problem 2
Let the circles $k_{1}$ and $k_{2}$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_{1}$ and $k_{2}$, that touches $k_{1}$ and $k_{2}$ at $M$ and $N$, respectively. If $t \perp A M$ and $M N=2 A M$, evaluate $\angle N M B$.
| ## Solution 1
Let $P$ be the symmetric of $A$ with respect to $M$ (Figure 1). Then $A M=M P$ and $t \perp A P$, hence the triangle $A P N$ is isosceles with $A P$ as its base, so $\angle N A P=\angle N P A$. We have $\angle B A P=\angle B A M=\angle B M N$ and $\angle B A N=\angle B N M$.
Thus we have
$$
180^{\circ}... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, let me try to solve this geometry problem step by step. So, we have two intersecting circles, k₁ and k₂, meeting at points A and B. There's a common tangent t that touches k₁ at M and k₂ at N. The tangent t is perpendicular to AM, and the length of MN is twice the length of AM. We need to find the measure... |
## Problem 3
On a board there are $n$ nails each two connected by a string. Each string is colored in one of $n$ given distinct colors. For each three distinct colors, there exist three nails connected with strings in these three colors. Can $n$ be
a) 6 ?
b) 7 ?
|
Solution. (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 colo... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, let's try to figure out this problem. So, we have a board with n nails, and each pair of nails is connected by a string of one of n distinct colors. The condition is that for any three distinct colors, there exist three nails connected with strings in exactly those three colors. The question is asking if ... |
## Problem 3
On a board there are $n$ nails each two connected by a string. Each string is colored in one of $n$ given distinct colors. For each three distinct colors, there exist three nails connected with strings in these three colors. Can $n$ be
a) 6 ?
b) 7 ?
|
Solution. (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 colo... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to figure out whether n can be 6 or 7 for this problem. Let me try to understand the problem first.
We have a board with n nails. Each pair of nails is connected by a string, and each string is colored with one of n distinct colors. The key condition here is that for every three distinct color... |
## Problem 4
Find all positive integers $x, y, z$ and $t$ such that
$$
2^{x} \cdot 3^{y}+5^{z}=7^{t}
$$
| ## Solution
Reducing modulo 3 we get $5^{z} \equiv 1$, therefore $z$ is even, $z=2 c, c \in \mathbb{N}$.
Next we prove that $t$ is even:
Obviously, $t \geq 2$. Let us suppose that $t$ is odd, say $t=2 d+1, d \in \mathbb{N}$. The equation becomes $2^{x} \cdot 3^{y}+25^{c}=7 \cdot 49^{d}$. If $x \geq 2$, reducing modu... | 3,1,=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3 | [
false,
false,
false
] | 0 | <think>
Okay, let's see. I need to find all positive integers x, y, z, t such that 2^x * 3^y + 5^z = 7^t. Hmm, this looks like an exponential Diophantine equation. These can be tricky, but maybe I can find some patterns or use modular arithmetic to limit the possibilities.
First, since all variables are positive integ... |
## Problem 4
Find all positive integers $x, y, z$ and $t$ such that
$$
2^{x} \cdot 3^{y}+5^{z}=7^{t}
$$
| ## Solution
Reducing modulo 3 we get $5^{z} \equiv 1$, therefore $z$ is even, $z=2 c, c \in \mathbb{N}$.
Next we prove that $t$ is even:
Obviously, $t \geq 2$. Let us suppose that $t$ is odd, say $t=2 d+1, d \in \mathbb{N}$. The equation becomes $2^{x} \cdot 3^{y}+25^{c}=7 \cdot 49^{d}$. If $x \geq 2$, reducing modu... | 3,1,=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3 | [
false,
false,
false
] | 0 | <think>
Okay, let's try to solve this problem where we need to find all positive integers x, y, z, and t such that 2^x * 3^y + 5^z = 7^t. Hmm, so we have an equation involving exponents with different bases. I remember that equations like this often require checking small cases or using modular arithmetic to find contr... |
## Problem 4
Find all positive integers $x, y, z$ and $t$ such that
$$
2^{x} \cdot 3^{y}+5^{z}=7^{t}
$$
| ## Solution
Reducing modulo 3 we get $5^{z} \equiv 1$, therefore $z$ is even, $z=2 c, c \in \mathbb{N}$.
Next we prove that $t$ is even:
Obviously, $t \geq 2$. Let us suppose that $t$ is odd, say $t=2 d+1, d \in \mathbb{N}$. The equation becomes $2^{x} \cdot 3^{y}+25^{c}=7 \cdot 49^{d}$. If $x \geq 2$, reducing modu... | 3,1,=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3 | [
false,
false,
false
] | 0 | <think>
Okay, so I need to find all positive integers x, y, z, and t such that 2^x * 3^y + 5^z = 7^t. Hmm, let's start by understanding what this equation is saying. On the left side, we have the sum of two numbers: one is a product of powers of 2 and 3, and the other is a power of 5. The right side is a power of 7. So... |
Problem 1. The real numbers $a, b, c, d$ satisfy simultaneously the equations
$$
a b c-d=1, b c d-a=2, c d a-b=3, d a b-c=-6
$$
Prove that $a+b+c+d \neq 0$.
|
Solution. Suppose that $a+b+c+d=0$. Then
$$
a b c+b c d+c d a+d a b=0
$$
If $a b c d=0$, then one of numbers, say $d$, must be 0 . In this case $a b c=0$, and so at least two of the numbers $a, b, c, d$ will be equal to 0 , making one of the given equations impossible. Hence $a b c d \neq 0$ and, from (1),
$$
\frac... | proof | Algebra | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to prove that a + b + c + d is not equal to zero given these four equations. Let me start by writing down the equations again to get a clear picture:
1. \( abc - d = 1 \)
2. \( bcd - a = 2 \)
3. \( cda - b = 3 \)
4. \( dab - c = -6 \)
Hmm, all four variables a, b, c, d are real numbers... |
Problem 1. The real numbers $a, b, c, d$ satisfy simultaneously the equations
$$
a b c-d=1, b c d-a=2, c d a-b=3, d a b-c=-6
$$
Prove that $a+b+c+d \neq 0$.
|
Solution. Suppose that $a+b+c+d=0$. Then
$$
a b c+b c d+c d a+d a b=0
$$
If $a b c d=0$, then one of numbers, say $d$, must be 0 . In this case $a b c=0$, and so at least two of the numbers $a, b, c, d$ will be equal to 0 , making one of the given equations impossible. Hence $a b c d \neq 0$ and, from (1),
$$
\frac... | proof | Algebra | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, let's see. I have this problem where four real numbers a, b, c, d satisfy these four equations:
1. abc - d = 1
2. bcd - a = 2
3. cda - b = 3
4. dab - c = -6
And I need to prove that a + b + c + d ≠ 0. Hmm. So, the problem is to show that the sum of these variables can't be zero given those equations. Al... |
Problem 3. Let $A L$ and $B K$ be angle bisectors in the non-isosceles triangle $A B C$ ( $L$ lies on the side $B C, K$ lies on the side $A C$ ). The perpendicular bisector of $B K$ intersects the line $A L$ at point $M$. Point $N$ lies on the line $B K$ such that $L N$ is parallel to $M K$. Prove that $L N=N A$.
|
Solution. The point $M$ lies on the circumcircle of $\triangle A B K$ (since both $A L$ and the perpendicular bisector of $B K$ bisect the arc $B K$ of this circle). Then $\angle C B K=$ $\angle A B K=\angle A M K=\angle N L A$. Thus $A B L N$ is cyclic, whence $\angle N A L=\angle N B L=$ $\angle C B K=\angle N L A$.... | LN=NA | Geometry | proof | Yes | Yes | olympiads | false | 2 | [
true,
true
] | 2 | <think>
Okay, so I need to prove that in triangle ABC, which is not isosceles, the points M and N are constructed in a certain way, and then LN is equal to NA. Let me try to visualize the problem first.
We have triangle ABC with angle bisectors AL and BK. L is on BC, K is on AC. Then, the perpendicular bisector of BK ... |
Problem 3. Let $A L$ and $B K$ be angle bisectors in the non-isosceles triangle $A B C$ ( $L$ lies on the side $B C, K$ lies on the side $A C$ ). The perpendicular bisector of $B K$ intersects the line $A L$ at point $M$. Point $N$ lies on the line $B K$ such that $L N$ is parallel to $M K$. Prove that $L N=N A$.
|
Solution. The point $M$ lies on the circumcircle of $\triangle A B K$ (since both $A L$ and the perpendicular bisector of $B K$ bisect the arc $B K$ of this circle). Then $\angle C B K=$ $\angle A B K=\angle A M K=\angle N L A$. Thus $A B L N$ is cyclic, whence $\angle N A L=\angle N B L=$ $\angle C B K=\angle N L A$.... | LN=NA | Geometry | proof | Yes | Yes | olympiads | false | 2 | [
true,
true
] | 2 | <think>
Okay, so I need to prove that in triangle ABC, which isn't isosceles, with angle bisectors AL and BK, where L is on BC and K is on AC, the perpendicular bisector of BK intersects AL at point M. Then, there's a point N on BK such that LN is parallel to MK, and I have to show that LN equals NA. Hmm, that seems a ... |
Problem 4. A $9 \times 7$ rectangle is tiled with tiles of the two types shown in the picture below (the tiles are composed by three, respectively four unit squares and the L-shaped tiles can be rotated repeatedly with $90^{\circ}$ ).
 that can be used in tiling a 9x7 rectangle. The other tiles are L-shaped, made of three unit squares, and they can be rotated. The problem says n is the number of 2x2 tiles, and ... |
Problem 1. 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)
$$
where $x$ is the unknown real variable.
(a) Solve the equation for $n=8$.
(b) Prove that there exists an integer $n$ for which the equation has at least 2021 solutions. (For any re... |
Solution. Let $k=\left[\frac{1}{2 x}\right], k \in \mathbb{Z}$.
(a) For $n=8$, the equation becomes
$$
k=\left[\frac{1}{2 x}\right]=8-36 x \Rightarrow x \neq 0 \text { and } x=\frac{8-k}{36}
$$
Since $x \neq 0$, we have $k \neq 8$, and the last relation implies $k=\left[\frac{1}{2 x}\right]=\left[\frac{18}{8-k}\rig... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, let's tackle part (a) first. The problem is to solve the equation for n=8:
2 · floor(1/(2x)) - 8 + 1 = (8 + 1)(1 - 8x)
Simplify both sides first. Left side: 2·floor(1/(2x)) -7. Right side: 9·(1 -8x) = 9 -72x.
So the equation becomes:
2·floor(1/(2x)) -7 = 9 -72x
Let's rearrange terms:
2·floor(1/(2x))... | |
Problem 1. Let $n(n \geq 1)$ be an integer. Consider the equation
$$
2 \cdot\left\lfloor\frac{1}{2 x}\right\rfloor-n+1=(n+1)(1-n x)
$$
where $x$ is the unknown real variable.
(a) Solve the equation for $n=8$.
(b) Prove that there exists an integer $n$ for which the equation has at least 2021 solutions. (For any re... |
Solution. Let $k=\left[\frac{1}{2 x}\right], k \in \mathbb{Z}$.
(a) For $n=8$, the equation becomes
$$
k=\left[\frac{1}{2 x}\right]=8-36 x \Rightarrow x \neq 0 \text { and } x=\frac{8-k}{36}
$$
Since $x \neq 0$, we have $k \neq 8$, and the last relation implies $k=\left[\frac{1}{2 x}\right]=\left[\frac{18}{8-k}\rig... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to solve part (a) where n=8. The equation is given as:
2 · floor(1/(2x)) - 8 + 1 = (8 + 1)(1 - 8x)
First, let me simplify both sides. On the left-hand side (LHS), 2·floor(1/(2x)) minus 8 plus 1. That simplifies to 2·floor(1/(2x)) - 7. On the right-hand side (RHS), (8+1) is 9, so it becomes 9(1... | |
Problem 3. Let $A B C$ be an acute scalene triangle with circumcenter $O$. Let $D$ be the foot of the altitude from $A$ to the side $B C$. The lines $B C$ and $A O$ intersect at $E$. Let $s$ be the line through $E$ perpendicular to $A O$. The line $s$ intersects $A B$ and $A C$ at $K$ and $L$, respectively. Denote by ... | ## Solution.
Let us denote angles of triangle $A B C$ with $\alpha, \beta, \gamma$ in a standard way. By basic anglechasing we have
$$
\angle B A D=90^{\circ}-\beta=\angle O A C \text { and ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1 | [
false
] | 0 | <think>
Alright, let's tackle this geometry problem step by step. First, I need to make sure I understand all the components involved. The problem involves an acute scalene triangle ABC with circumcenter O. So, O is the intersection point of the perpendicular bisectors of the sides of triangle ABC.
Then, D is the foo... |
Problem 4. 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 $M$.
|
Solution. 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|... | 1006 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5 | [
true,
true,
true,
true,
true
] | 5 | <think>
Okay, so I need to find the largest possible subset M of the integers from 1 to 2021 such that for any three elements a, b, c in M, the absolute value |a + b - c| is greater than 10. Hmm, interesting. Let me try to break this down step by step.
First, let's make sure I understand the problem correctly. We have... |
Problem 4. 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 $M$.
|
Solution. 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|... | 1006 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5 | [
true,
true,
true,
true,
true
] | 5 | <think>
Okay, let me try to figure out this problem. So, we need to find the largest subset M of the integers from 1 to 2021 such that for any three elements a, b, c in M, the absolute value |a + b - c| is greater than 10. Hmm, interesting.
First, let me understand the condition. For any three elements in M, when you... |
Problem 4. 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 $M$.
|
Solution. 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|... | 1006 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5 | [
true,
true,
true,
true,
true
] | 5 | <think>
Okay, so I need to find the largest subset M of {1,2,3,...,2021} such that for any three elements a, b, c in M, the absolute value |a + b - c| is greater than 10. Hmm, interesting. Let me try to break this down.
First, the problem says that for any three elements in M, the expression |a + b - c| must be greate... |
Problem 4. 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 $M$.
|
Solution. 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|... | 1006 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5 | [
true,
true,
true,
true,
true
] | 5 | <think>
Okay, so I have this problem here where I need to find the largest possible subset M of the integers from 1 to 2021. The condition is that for any three elements a, b, c in M, the absolute value |a + b - c| must be greater than 10. Hmm, interesting. Let me try to break this down step by step.
First, let me mak... |
Problem 4. 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 $M$.
|
Solution. 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|... | 1006 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5 | [
true,
true,
true,
true,
true
] | 5 | <think>
Okay, let me try to work through this problem step by step. The problem is about finding the largest possible subset M of the integers from 1 to 2021 such that for any three elements a, b, c in M, the absolute value |a + b - c| is greater than 10. So, I need to figure out how to choose as many numbers as possib... |
Problem 1. Find all distinct prime numbers $p, q$ and $r$ such that
$$
3 p^{4}-5 q^{4}-4 r^{2}=26
$$
|
Solution. First notice that if both primes $q$ and $r$ differ from 3 , then $q^{2} \equiv r^{2} \equiv 1(\bmod 3)$, hence the left hand side of the given equation is congruent to zero modulo 3 , which is impossible since 26 is not divisible by 3 . Thus, $q=3$ or $r=3$. We consider two cases.
Case 1. $q=3$.
The equat... | p=5,q=3,r=19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
true,
false,
true,
false,
false,
false,
false
] | 2 | <think>
Okay, so I have this problem where I need to find all distinct prime numbers p, q, and r such that 3p⁴ - 5q⁴ - 4r² = 26. Hmm, primes... they have to be distinct, so p, q, r can't be the same. Let me start by thinking about possible small primes since exponents are high, which can make numbers big quickly. The p... |
Problem 1. Find all distinct prime numbers $p, q$ and $r$ such that
$$
3 p^{4}-5 q^{4}-4 r^{2}=26
$$
|
Solution. First notice that if both primes $q$ and $r$ differ from 3 , then $q^{2} \equiv r^{2} \equiv 1(\bmod 3)$, hence the left hand side of the given equation is congruent to zero modulo 3 , which is impossible since 26 is not divisible by 3 . Thus, $q=3$ or $r=3$. We consider two cases.
Case 1. $q=3$.
The equat... | p=5,q=3,r=19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
true,
false,
true,
false,
false,
false,
false
] | 2 | <think>
Okay, let's see. I need to find distinct prime numbers p, q, and r such that 3p⁴ - 5q⁴ - 4r² = 26. Hmm, primes are 2, 3, 5, 7, 11, etc. Since they have to be distinct, p, q, r can't be the same. Let me think about how to approach this.
First, maybe I can try plugging in small primes and see if the equation hol... |
Problem 1. Find all distinct prime numbers $p, q$ and $r$ such that
$$
3 p^{4}-5 q^{4}-4 r^{2}=26
$$
|
Solution. First notice that if both primes $q$ and $r$ differ from 3 , then $q^{2} \equiv r^{2} \equiv 1(\bmod 3)$, hence the left hand side of the given equation is congruent to zero modulo 3 , which is impossible since 26 is not divisible by 3 . Thus, $q=3$ or $r=3$. We consider two cases.
Case 1. $q=3$.
The equat... | p=5,q=3,r=19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
true,
false,
true,
false,
false,
false,
false
] | 2 | <think>
Okay, so I need to find all distinct prime numbers p, q, and r such that 3p⁴ - 5q⁴ - 4r² = 26. Alright, primes are numbers like 2, 3, 5, 7, 11, etc. Since they have to be distinct, p, q, and r can't be the same. Let me think about how to approach this.
First, maybe I can try plugging in small prime numbers and... |
Problem 1. Find all distinct prime numbers $p, q$ and $r$ such that
$$
3 p^{4}-5 q^{4}-4 r^{2}=26
$$
|
Solution. First notice that if both primes $q$ and $r$ differ from 3 , then $q^{2} \equiv r^{2} \equiv 1(\bmod 3)$, hence the left hand side of the given equation is congruent to zero modulo 3 , which is impossible since 26 is not divisible by 3 . Thus, $q=3$ or $r=3$. We consider two cases.
Case 1. $q=3$.
The equat... | p=5,q=3,r=19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
true,
false,
true,
false,
false,
false,
false
] | 2 | <think>
Okay, let's see. I have to find all distinct prime numbers p, q, and r such that 3p⁴ - 5q⁴ - 4r² = 26. Hmm, primes are 2, 3, 5, 7, 11, etc. Since they have to be distinct, each of p, q, r has to be different primes.
First, maybe I can start by considering small primes because the equation involves fourth powe... |
Problem 1. Find all distinct prime numbers $p, q$ and $r$ such that
$$
3 p^{4}-5 q^{4}-4 r^{2}=26
$$
|
Solution. First notice that if both primes $q$ and $r$ differ from 3 , then $q^{2} \equiv r^{2} \equiv 1(\bmod 3)$, hence the left hand side of the given equation is congruent to zero modulo 3 , which is impossible since 26 is not divisible by 3 . Thus, $q=3$ or $r=3$. We consider two cases.
Case 1. $q=3$.
The equat... | p=5,q=3,r=19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
true,
false,
true,
false,
false,
false,
false
] | 2 | <think>
Okay, let's try to tackle this problem. So, we need to find all distinct prime numbers p, q, and r such that 3p⁴ - 5q⁴ - 4r² = 26. Hmm, primes are numbers like 2, 3, 5, 7, 11, etc. They have to be distinct, so p, q, and r can't be the same as each other. Let me think step by step.
First, maybe I should try plu... |
Problem 1. Find all distinct prime numbers $p, q$ and $r$ such that
$$
3 p^{4}-5 q^{4}-4 r^{2}=26
$$
|
Solution. First notice that if both primes $q$ and $r$ differ from 3 , then $q^{2} \equiv r^{2} \equiv 1(\bmod 3)$, hence the left hand side of the given equation is congruent to zero modulo 3 , which is impossible since 26 is not divisible by 3 . Thus, $q=3$ or $r=3$. We consider two cases.
Case 1. $q=3$.
The equat... | p=5,q=3,r=19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
true,
false,
true,
false,
false,
false,
false
] | 2 | <think>
Okay, let me try to solve this problem. So, we need to find all distinct prime numbers p, q, and r such that 3p⁴ - 5q⁴ - 4r² = 26. Hmm, primes are numbers like 2, 3, 5, 7, 11, etc. And they have to be distinct, so p, q, r can't be the same. Alright, let's start thinking.
First, maybe I should rearrange the equ... |
Problem 1. Find all distinct prime numbers $p, q$ and $r$ such that
$$
3 p^{4}-5 q^{4}-4 r^{2}=26
$$
|
Solution. First notice that if both primes $q$ and $r$ differ from 3 , then $q^{2} \equiv r^{2} \equiv 1(\bmod 3)$, hence the left hand side of the given equation is congruent to zero modulo 3 , which is impossible since 26 is not divisible by 3 . Thus, $q=3$ or $r=3$. We consider two cases.
Case 1. $q=3$.
The equat... | p=5,q=3,r=19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
true,
false,
true,
false,
false,
false,
false
] | 2 | <think>
Okay, let's see. I need to find distinct prime numbers p, q, and r such that 3p^4 - 5q^4 -4r^2 = 26. Hmm, primes are numbers like 2, 3, 5, 7, 11, etc. They have to be distinct, so p, q, r can't be the same. Let's start by thinking about possible primes and see if the equation holds.
First, since primes can be ... |
Problem 1. Find all distinct prime numbers $p, q$ and $r$ such that
$$
3 p^{4}-5 q^{4}-4 r^{2}=26
$$
|
Solution. First notice that if both primes $q$ and $r$ differ from 3 , then $q^{2} \equiv r^{2} \equiv 1(\bmod 3)$, hence the left hand side of the given equation is congruent to zero modulo 3 , which is impossible since 26 is not divisible by 3 . Thus, $q=3$ or $r=3$. We consider two cases.
Case 1. $q=3$.
The equat... | p=5,q=3,r=19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
true,
false,
true,
false,
false,
false,
false
] | 2 | <think>
Okay, let's see. I need to find all distinct prime numbers p, q, and r such that 3p⁴ - 5q⁴ - 4r² = 26. Hmm, primes are 2, 3, 5, 7, 11, etc. Since they have to be distinct, each of p, q, r has to be different primes. The equation involves p to the fourth power, q to the fourth power, and r squared. The coefficie... |
Problem 2. 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}... |
Solution 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 $\overline{T R}=\frac{1}{2} \overline{M C}$ ... | S | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4 | [
true,
true,
true,
true
] | 4 | <think>
Okay, so I need to find the area of quadrilateral AH₁BH₂ in terms of the area S of the original acute triangle ABC. Let me try to break this down step by step.
First, let's visualize the problem. We have triangle ABC, which is acute, so all altitudes are inside the triangle. CD is the altitude from C to AB, me... |
Problem 2. 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}... |
Solution 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 $\overline{T R}=\frac{1}{2} \overline{M C}$ ... | S | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4 | [
true,
true,
true,
true
] | 4 | <think>
Okay, so I have this problem about an acute triangle ABC with area S. The problem involves constructing several perpendiculars and then finding the area of a quadrilateral formed by the orthocenters of two smaller triangles. Let me try to parse this step by step.
First, let me visualize triangle ABC. Since it'... |
Problem 2. 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}... |
Solution 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 $\overline{T R}=\frac{1}{2} \overline{M C}$ ... | S | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4 | [
true,
true,
true,
true
] | 4 | <think>
Okay, so I have this problem here about an acute triangle ABC with area S. There are several perpendiculars drawn: CD is perpendicular to AB, with D on AB. Then from D, they drop two more perpendiculars: DM perpendicular to AC, meeting AC at M, and DN perpendicular to BC, meeting BC at N. Then they mention H1 a... |
Problem 2. 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}... |
Solution 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 $\overline{T R}=\frac{1}{2} \overline{M C}$ ... | S | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4 | [
true,
true,
true,
true
] | 4 | <think>
Okay, so I have this geometry problem here. Let me try to parse it step by step. We have an acute triangle ABC with area S. Then, CD is perpendicular to AB, with D lying on AB. Then from D, we draw DM perpendicular to AC, meeting AC at M, and DN perpendicular to BC, meeting BC at N. Then H1 is the orthocenter o... |
Problem 3. 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 1. By using AM-GM $\left(x^{2}+y^{2}+z^{2} \geq x y+y z+z x\right)$ we have
$$
\begin{aligned}
\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} & \geq\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)+\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)+\left(c+\... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to prove that for positive real numbers a, b, c with abc = 1, the expression (a + 1/b)^2 + (b + 1/c)^2 + (c + 1/a)^2 is greater than or equal to 3 times (a + b + c + 1). Hmm, let's start by understanding what the problem is asking. We have three variables constrained by their product being 1, an... |
Problem 3. Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that
$$
\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)
$$
When does equality hold?
|
Solution 1. By using AM-GM $\left(x^{2}+y^{2}+z^{2} \geq x y+y z+z x\right)$ we have
$$
\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+\... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to prove this inequality:
$$
\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} \geq 3(a+b+c+1)
$$
given that $a, b, c$ are positive real numbers with $abc = 1$. Hmm, let's start by understanding what's given and what's required.
First, the conditio... |
Problem 4. 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 ... |
Solution. 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 tw... | n-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3 | [
true,
true,
true
] | 3 | <think>
Alright, let's try to tackle this problem step by step. So, we have two players, A and B, playing a game with a pile of s stones. They take turns, starting with A. On each turn, a player can take either 1 stone, a prime number of stones, or a multiple of n stones. The person who takes the last stone wins. We ne... |
Problem 4. 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 ... |
Solution. 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 tw... | n-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3 | [
true,
true,
true
] | 3 | <think>
Okay, let me try to figure out this problem. So, we have a game where two players, A and B, take turns picking stones from a pile. The rules are that on each turn, a player can take either 1 stone, a prime number of stones, or a multiple of n stones. The person who takes the last stone wins. We need to determin... |
Problem 4. 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 ... |
Solution. 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 tw... | n-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3 | [
true,
true,
true
] | 3 | <think>
Okay, let's try to figure out this problem. So, we have two players, A and B, who take turns picking stones from a pile. The pile starts with s stones, and they can take either 1 stone, a prime number of stones, or a multiple of n stones on each turn. The person who takes the last stone wins. We need to determi... |
Problem 1. Find all pairs $(a, b)$ of positive integers such that
$$
11 a b \leq a^{3}-b^{3} \leq 12 a b
$$
|
Solution 1. Let $a-b=t$. Due to $a^{3}-b^{3} \geq 11 a b$ we conclude that $a>b$ so $t$ is a positive integer and the condition can be written as
$$
11 b(b+t) \leq t\left[b^{2}+b(b+t)+(b+t)^{2}\right] \leq 12 b(b+t)
$$
Since
$$
t\left[b^{2}+b(b+t)+(b+t)^{2}\right]=t\left(b^{2}+b^{2}+b t+b^{2}+2 b t+t^{2}\right)=3 t... | (5,2) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4 | [
true,
true,
true,
true
] | 4 | <think>
Okay, so I need to find all pairs of positive integers (a, b) that satisfy the inequalities 11ab ≤ a³ - b³ ≤ 12ab. Hmm, let's start by understanding what this means. Both a and b are positive integers, so they must be at least 1. The inequalities involve cubic terms and a product term. Maybe I can manipulate th... |
Problem 1. Find all pairs $(a, b)$ of positive integers such that
$$
11 a b \leq a^{3}-b^{3} \leq 12 a b
$$
|
Solution 1. Let $a-b=t$. Due to $a^{3}-b^{3} \geq 11 a b$ we conclude that $a>b$ so $t$ is a positive integer and the condition can be written as
$$
11 b(b+t) \leq t\left[b^{2}+b(b+t)+(b+t)^{2}\right] \leq 12 b(b+t)
$$
Since
$$
t\left[b^{2}+b(b+t)+(b+t)^{2}\right]=t\left(b^{2}+b^{2}+b t+b^{2}+2 b t+t^{2}\right)=3 t... | (5,2) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4 | [
true,
true,
true,
true
] | 4 | <think>
Okay, let's try to tackle this problem. We need to find all pairs of positive integers (a, b) such that 11ab ≤ a³ - b³ ≤ 12ab. Hmm, so both a and b are positive integers, which means they are at least 1. Let me start by understanding the inequalities given.
First, let's write down the inequalities:
1. a³ - b³... |
Problem 1. Find all pairs $(a, b)$ of positive integers such that
$$
11 a b \leq a^{3}-b^{3} \leq 12 a b
$$
|
Solution 1. Let $a-b=t$. Due to $a^{3}-b^{3} \geq 11 a b$ we conclude that $a>b$ so $t$ is a positive integer and the condition can be written as
$$
11 b(b+t) \leq t\left[b^{2}+b(b+t)+(b+t)^{2}\right] \leq 12 b(b+t)
$$
Since
$$
t\left[b^{2}+b(b+t)+(b+t)^{2}\right]=t\left(b^{2}+b^{2}+b t+b^{2}+2 b t+t^{2}\right)=3 t... | (5,2) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4 | [
true,
true,
true,
true
] | 4 | <think>
Okay, let's see. I have this problem where I need to find all pairs of positive integers (a, b) such that 11ab is less than or equal to a³ - b³, which in turn is less than or equal to 12ab. Hmm, okay. So, the inequalities are 11ab ≤ a³ - b³ ≤ 12ab. Both a and b are positive integers.
First, I need to understa... |
Problem 1. Find all pairs $(a, b)$ of positive integers such that
$$
11 a b \leq a^{3}-b^{3} \leq 12 a b
$$
|
Solution 1. Let $a-b=t$. Due to $a^{3}-b^{3} \geq 11 a b$ we conclude that $a>b$ so $t$ is a positive integer and the condition can be written as
$$
11 b(b+t) \leq t\left[b^{2}+b(b+t)+(b+t)^{2}\right] \leq 12 b(b+t)
$$
Since
$$
t\left[b^{2}+b(b+t)+(b+t)^{2}\right]=t\left(b^{2}+b^{2}+b t+b^{2}+2 b t+t^{2}\right)=3 t... | (5,2) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4 | [
true,
true,
true,
true
] | 4 | <think>
Okay, so I need to find all pairs of positive integers (a, b) such that 11ab is less than or equal to a³ minus b³, which in turn is less than or equal to 12ab. Hmm, let me start by understanding the inequalities given here.
First, the problem states that 11ab ≤ a³ - b³ ≤ 12ab. Both a and b are positive integer... |
Problem 2. 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 $\... |
Solution 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... | proof | Geometry | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to prove that lines SM and TN are parallel in this given geometric configuration. Let me start by understanding all the components of the problem.
First, we have an acute triangle ABC. The orthocenter is H, and D is the foot of the altitude from A to BC. The condition given is AH = HD. That mea... |
Problem 2. 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 $\... |
Solution 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... | proof | Geometry | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I have this geometry problem here. Let me try to parse it step by step.
We have an acute triangle ABC. The orthocenter is H, and D is the foot of the altitude from A to BC. The condition given is that AH = HD. Then, there's a line ℓ through H that's tangent to the circumcircle of triangle BHC. The in... |
Problem 4. 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\},\... |
Solution. 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)... | 2^{2022}-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 1 | [
false
] | 0 | <think>
Okay, so I need to figure out how many nice even positive integers are smaller than \(3^{2022}\). A number \(n\) is nice if the set \(\{1, 2, \ldots, n\}\) can be partitioned into \(\frac{n}{2}\) pairs where each pair sums to a power of 3. The example given is 6, which works because we can pair \(\{1,2\}\), \(\... |
## Problem 1.
Find all pairs $(a, b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of 5 .
|
Solution. The condition is symmetric so we can assume that $b \leq a$.
The first case is when $a=b$. In this case, $a!+a=5^{m}$ for some positive integer $m$. We can rewrite this as $a \cdot((a-1)!+1)=5^{m}$. This means that $a=5^{k}$ for some integer $k \geq 0$. It is clear that $k$ cannot be 0 . If $k \geq 2$, then... | (1,4),(4,1),(5,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
false,
false,
true,
false,
false,
true,
true
] | 3 | <think>
Okay, so I have this problem where I need to find all pairs of positive integers (a, b) such that both a! + b and b! + a are powers of 5. Hmm, that sounds a bit tricky, but let me try to break it down step by step.
First, let's recall what a power of 5 is. A power of 5 is any number of the form 5^k where k is ... |
## Problem 1.
Find all pairs $(a, b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of 5 .
|
Solution. The condition is symmetric so we can assume that $b \leq a$.
The first case is when $a=b$. In this case, $a!+a=5^{m}$ for some positive integer $m$. We can rewrite this as $a \cdot((a-1)!+1)=5^{m}$. This means that $a=5^{k}$ for some integer $k \geq 0$. It is clear that $k$ cannot be 0 . If $k \geq 2$, then... | (1,4),(4,1),(5,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
false,
false,
true,
false,
false,
true,
true
] | 3 | <think>
Alright, let's try to tackle this problem. So, we need to find all pairs of positive integers (a, b) such that both a! + b and b! + a are powers of 5. Hmm, okay. Powers of 5 are numbers like 5, 25, 125, 625, etc., which can be written as 5^k where k is a positive integer. So, both a! + b and b! + a have to be o... |
## Problem 1.
Find all pairs $(a, b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of 5 .
|
Solution. The condition is symmetric so we can assume that $b \leq a$.
The first case is when $a=b$. In this case, $a!+a=5^{m}$ for some positive integer $m$. We can rewrite this as $a \cdot((a-1)!+1)=5^{m}$. This means that $a=5^{k}$ for some integer $k \geq 0$. It is clear that $k$ cannot be 0 . If $k \geq 2$, then... | (1,4),(4,1),(5,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
false,
false,
true,
false,
false,
true,
true
] | 3 | <think>
Okay, so I need to find all pairs of positive integers (a, b) such that both a! + b and b! + a are powers of 5. Hmm, powers of 5 are numbers like 1, 5, 25, 125, 625, etc. So both expressions have to equal some 5^k and 5^m where k and m are non-negative integers. Let me start by considering small values of a and... |
## Problem 1.
Find all pairs $(a, b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of 5 .
|
Solution. The condition is symmetric so we can assume that $b \leq a$.
The first case is when $a=b$. In this case, $a!+a=5^{m}$ for some positive integer $m$. We can rewrite this as $a \cdot((a-1)!+1)=5^{m}$. This means that $a=5^{k}$ for some integer $k \geq 0$. It is clear that $k$ cannot be 0 . If $k \geq 2$, then... | (1,4),(4,1),(5,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
false,
false,
true,
false,
false,
true,
true
] | 3 | <think>
Okay, so I need to find all pairs of positive integers (a, b) such that both a! + b and b! + a are powers of 5. Hmm, powers of 5 are numbers like 5, 25, 125, 625, etc. So both expressions have to equal 5^k for some integers k. Let me start by considering small values for a and b because factorials grow really f... |
## Problem 1.
Find all pairs $(a, b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of 5 .
|
Solution. The condition is symmetric so we can assume that $b \leq a$.
The first case is when $a=b$. In this case, $a!+a=5^{m}$ for some positive integer $m$. We can rewrite this as $a \cdot((a-1)!+1)=5^{m}$. This means that $a=5^{k}$ for some integer $k \geq 0$. It is clear that $k$ cannot be 0 . If $k \geq 2$, then... | (1,4),(4,1),(5,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
false,
false,
true,
false,
false,
true,
true
] | 3 | <think>
Okay, so I need to find all pairs of positive integers (a, b) such that both a! + b and b! + a are powers of 5. Hmm, powers of 5 are numbers like 5, 25, 125, 625, etc. So, each of these expressions has to equal 5^k for some integer k ≥ 1.
First, let's understand what a factorial is. For a positive integer n, ... |
## Problem 1.
Find all pairs $(a, b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of 5 .
|
Solution. The condition is symmetric so we can assume that $b \leq a$.
The first case is when $a=b$. In this case, $a!+a=5^{m}$ for some positive integer $m$. We can rewrite this as $a \cdot((a-1)!+1)=5^{m}$. This means that $a=5^{k}$ for some integer $k \geq 0$. It is clear that $k$ cannot be 0 . If $k \geq 2$, then... | (1,4),(4,1),(5,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
false,
false,
true,
false,
false,
true,
true
] | 3 | <think>
Okay, let's try to solve this problem where we need to find all pairs of positive integers (a, b) such that both a! + b and b! + a are powers of 5. Hmm, powers of 5 are numbers like 5, 25, 125, 625, etc. So, we need a! + b = 5^k and b! + a = 5^m for some non-negative integers k and m. Since a and b are positive... |
## Problem 1.
Find all pairs $(a, b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of 5 .
|
Solution. The condition is symmetric so we can assume that $b \leq a$.
The first case is when $a=b$. In this case, $a!+a=5^{m}$ for some positive integer $m$. We can rewrite this as $a \cdot((a-1)!+1)=5^{m}$. This means that $a=5^{k}$ for some integer $k \geq 0$. It is clear that $k$ cannot be 0 . If $k \geq 2$, then... | (1,4),(4,1),(5,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
false,
false,
true,
false,
false,
true,
true
] | 3 | <think>
Okay, let's tackle this problem where we need to find all pairs of positive integers (a, b) such that both a! + b and b! + a are powers of 5. Hmm, powers of 5 are numbers like 5, 25, 125, 625, etc. So both expressions have to be equal to some 5^k and 5^m where k and m are positive integers.
First, since a and... |
## Problem 1.
Find all pairs $(a, b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of 5 .
|
Solution. The condition is symmetric so we can assume that $b \leq a$.
The first case is when $a=b$. In this case, $a!+a=5^{m}$ for some positive integer $m$. We can rewrite this as $a \cdot((a-1)!+1)=5^{m}$. This means that $a=5^{k}$ for some integer $k \geq 0$. It is clear that $k$ cannot be 0 . If $k \geq 2$, then... | (1,4),(4,1),(5,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8 | [
false,
false,
false,
true,
false,
false,
true,
true
] | 3 | <think>
Okay, let's try to tackle this problem where we need to find all pairs of positive integers (a, b) such that both a! + b and b! + a are powers of 5. Hmm, powers of 5, so numbers like 1, 5, 25, 125, 625, etc. Since we're dealing with factorials and additions, maybe starting with small values of a and b will help... |
## Problem 2.
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.
|
Solution. 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)... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Alright, let's try to tackle this inequality problem. The problem states that for all non-negative real numbers \( x, y, z \), not all zero, 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... |
## Problem 2.
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.
|
Solution. 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)... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to prove that for all non-negative real numbers x, y, z, not all zero, this inequality holds:
\[
\frac{2 x^{2} - x + y + z}{x + y^{2} + z^{2}} + \frac{2 y^{2} + x - y + z}{x^{2} + y + z^{2}} + \frac{2 z^{2} + x + y - z}{x^{2} + y^{2} + z} \geqslant 3
\]
And also find all triples (x, y, z) wher... |
## Problem 3.
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 the... |
Solution. 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... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so Alice and Bob are playing this game on a 100x100 grid. Alice goes first, they take turns placing numbers from 1 to 100² in the cells. Once all cells are filled, Alice's score is the maximum row sum, and Bob's score is the maximum column sum. Whoever has the higher score wins. The question is, does eith... |
## Problem 3.
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 the... |
Solution. 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... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so Alice and Bob are playing this game on a 100x100 grid. They take turns putting numbers from 1 to 100² in the cells, with Alice going first. When the grid is full, Alice's score is the maximum row sum, and Bob's score is the maximum column sum. The question is whether either player has a winning strateg... |
## Problem 4.
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... | ## Solution.
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 gene... | proof | Geometry | proof | Yes | Incomplete | olympiads | false | 1 | [
false
] | 0 | <think>
Alright, let's tackle this geometry problem step by step. I need to prove that points A, Ob, M, and Oc are concyclic given the conditions of the problem. First, let me parse the problem and note down all the given information and what needs to be proven.
We have an acute triangle ABC with circumcenter O. D is ... |
Problem 1. Find all triples $(a, b, c)$ of real numbers such that the following system holds:
$$
\left\{\begin{array}{l}
a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \\
a^{2}+b^{2}+c^{2}=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}
\end{array}\right.
$$
|
Solution. 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
$$
(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\rig... | (,b,)=(,\frac{1}{},1),(,\frac{1}{},-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3 | [
false,
false,
false
] | 0 | <think>
Okay, so I need to solve this system of equations where we have triples (a, b, c) of real numbers. The equations are:
First: a + b + c equals the sum of their reciprocals, 1/a + 1/b + 1/c.
Second: The sum of their squares, a² + b² + c², equals the sum of the reciprocals of their squares, 1/a² + 1/b² + 1/c².
... |
Problem 1. Find all triples $(a, b, c)$ of real numbers such that the following system holds:
$$
\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.
$$
|
Solution. First of all if $(a, b, c)$ is a solution of the system then also $(-a,-b,-c)$ is a solution. Hence we can suppose that $a b c>0$. From the first condition we have
$$
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}\rig... | (,b,)=(,\frac{1}{},1),(,\frac{1}{},-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3 | [
false,
false,
false
] | 0 | <think>
Okay, let's see. I have to find all triples (a, b, c) of real numbers that satisfy both equations:
1. a + b + c = 1/a + 1/b + 1/c
2. a² + b² + c² = 1/a² + 1/b² + 1/c²
Hmm, interesting. So both the sums and the sums of squares of the variables equal the sums of their reciprocals. Let me think about how to appr... |
Problem 1. Find all triples $(a, b, c)$ of real numbers such that the following system holds:
$$
\left\{\begin{array}{l}
a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \\
a^{2}+b^{2}+c^{2}=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}
\end{array}\right.
$$
|
Solution. 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}\rig... | (,b,)=(,\frac{1}{},1),(,\frac{1}{},-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3 | [
false,
false,
false
] | 0 | <think>
Okay, so I need to find all triples (a, b, c) of real numbers that satisfy these two equations:
First equation: a + b + c = 1/a + 1/b + 1/c
Second equation: a² + b² + c² = 1/a² + 1/b² + 1/c²
Hmm, let me think. Maybe I can start by simplifying the first equation. Let me write it out again:
a + b + c = (1/a) ... |
Problem 2. 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 intersecti... |
Solution. 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... | proof | Geometry | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I have this geometry problem here, and I need to prove that the points T, E, B, Z are concyclic. Let me start by understanding the problem setup and then break it down step by step.
First, we have a right-angled triangle ABC with a right angle at A. So, angle BAC is 90 degrees. Then, E is the foot of ... |
Problem 2. 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 intersecti... |
Solution. 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... | proof | Geometry | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, let me try to work through this problem step by step. It's a geometry problem involving circles, perpendiculars, and various points. I need to prove that four points are concyclic, which usually means showing that they lie on a circle. The points in question are T, E, B, Z.
First, let me parse the probl... |
Problem 3. 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 chos... |
Solution. 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 ins... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 1 | [
false
] | 0 | <think>
Okay, so Alice and Bob are playing this game where they take turns picking numbers from the set {1, 2, ..., n}. The rules are that Bob goes first and can pick any number he wants. Then each subsequent pick has to be a number that hasn't been chosen yet and must differ by 1 from an already chosen number. The gam... |
Problem 4. Find all pairs $(p, q)$ of prime numbers such that
$$
1+\frac{p^{q}-q^{p}}{p+q}
$$
is a prime number.
|
Solution. It is clear that $p \neq q$. We set
$$
1+\frac{p^{q}-q^{p}}{p+q}=r
$$
and we have that
$$
p^{q}-q^{p}=(r-1)(p+q)
$$
From Fermat's Little Theorem we have
$$
p^{q}-q^{p} \equiv-q \quad(\bmod p)
$$
Since we also have that
$$
(r-1)(p+q) \equiv-r q-q \quad(\bmod p)
$$
from (3) we get that
$$
r q \equiv 0... | (2,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, so I need to find all pairs of prime numbers (p, q) such that the expression 1 + (p^q - q^p)/(p + q) is a prime number. Hmm, primes, exponents, and fractions... Let me start by understanding the problem step by step.
First, primes are numbers greater than 1 that have no divisors other than 1 and themselv... |
Problem 4. Find all pairs $(p, q)$ of prime numbers such that
$$
1+\frac{p^{q}-q^{p}}{p+q}
$$
is a prime number.
|
Solution. It is clear that $p \neq q$. We set
$$
1+\frac{p^{q}-q^{p}}{p+q}=r
$$
and we have that
$$
p^{q}-q^{p}=(r-1)(p+q)
$$
From Fermat's Little Theorem we have
$$
p^{q}-q^{p} \equiv-q \quad(\bmod p)
$$
Since we also have that
$$
(r-1)(p+q) \equiv-r q-q \quad(\bmod p)
$$
from (3) we get that
$$
r q \equiv 0... | (2,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, so I have this problem here: Find all pairs (p, q) of prime numbers such that 1 + (p^q - q^p)/(p + q) is a prime number. Hmm, primes p and q... I need to find all such pairs where this expression results in a prime. Let me start by understanding the problem better.
First, the expression is 1 plus (p^q - ... |
Problem 4. Find all pairs $(p, q)$ of prime numbers such that
$$
1+\frac{p^{q}-q^{p}}{p+q}
$$
is a prime number.
|
Solution. It is clear that $p \neq q$. We set
$$
1+\frac{p^{q}-q^{p}}{p+q}=r
$$
and we have that
$$
p^{q}-q^{p}=(r-1)(p+q)
$$
From Fermat's Little Theorem we have
$$
p^{q}-q^{p} \equiv-q \quad(\bmod p)
$$
Since we also have that
$$
(r-1)(p+q) \equiv-r q-q \quad(\bmod p)
$$
from (3) we get that
$$
r q \equiv 0... | (2,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, so I need to find all pairs of prime numbers (p, q) such that the expression 1 + (p^q - q^p)/(p + q) is a prime number. Hmm, primes p and q, so both are primes. Let me start by understanding the expression step by step.
First, let's write the expression again:
1 + (p^q - q^p)/(p + q)
This needs to be a... |
Problem 4. Find all pairs $(p, q)$ of prime numbers such that
$$
1+\frac{p^{q}-q^{p}}{p+q}
$$
is a prime number.
|
Solution. It is clear that $p \neq q$. We set
$$
1+\frac{p^{q}-q^{p}}{p+q}=r
$$
and we have that
$$
p^{q}-q^{p}=(r-1)(p+q)
$$
From Fermat's Little Theorem we have
$$
p^{q}-q^{p} \equiv-q \quad(\bmod p)
$$
Since we also have that
$$
(r-1)(p+q) \equiv-r q-q \quad(\bmod p)
$$
from (3) we get that
$$
r q \equiv 0... | (2,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Alright, so I need to find all pairs of prime numbers (p, q) such that the expression 1 + (p^q - q^p)/(p + q) is a prime number. Hmm, okay. Let me start by understanding what this expression actually is. Let's rewrite it for clarity:
The expression is 1 + (p^q - q^p)/(p + q). So, first, compute p^q and q^p, su... |
Problem 4. Find all pairs $(p, q)$ of prime numbers such that
$$
1+\frac{p^{q}-q^{p}}{p+q}
$$
is a prime number.
|
Solution. It is clear that $p \neq q$. We set
$$
1+\frac{p^{q}-q^{p}}{p+q}=r
$$
and we have that
$$
p^{q}-q^{p}=(r-1)(p+q)
$$
From Fermat's Little Theorem we have
$$
p^{q}-q^{p} \equiv-q \quad(\bmod p)
$$
Since we also have that
$$
(r-1)(p+q) \equiv-r q-q \quad(\bmod p)
$$
from (3) we get that
$$
r q \equiv 0... | (2,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, so I need to find all pairs of prime numbers (p, q) such that the expression 1 + (p^q - q^p)/(p + q) is a prime number. Hmm, primes, exponents, and fractions. Let me start by breaking down the problem step by step.
First, since p and q are primes, the possible values they can take are 2, 3, 5, 7, 11, etc... |
Problem 4. Find all pairs $(p, q)$ of prime numbers such that
$$
1+\frac{p^{q}-q^{p}}{p+q}
$$
is a prime number.
|
Solution. It is clear that $p \neq q$. We set
$$
1+\frac{p^{q}-q^{p}}{p+q}=r
$$
and we have that
$$
p^{q}-q^{p}=(r-1)(p+q)
$$
From Fermat's Little Theorem we have
$$
p^{q}-q^{p} \equiv-q \quad(\bmod p)
$$
Since we also have that
$$
(r-1)(p+q) \equiv-r q-q \quad(\bmod p)
$$
from (3) we get that
$$
r q \equiv 0... | (2,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6 | [
true,
true,
true,
true,
true,
true
] | 6 | <think>
Okay, let's see. I need to find all pairs of prime numbers (p, q) such that the expression 1 + (p^q - q^p)/(p + q) is a prime number. Hmm, primes... Let me start by understanding the problem step by step.
First, primes are numbers greater than 1 that have no divisors other than 1 and themselves. So p and q can... |
Problem 2. Let $x, y, z$ be positive integers such that $x \neq y \neq z \neq x$. Prove that
$$
(x+y+z)(x y+y z+z x-2) \geq 9 x y z
$$
When does the equality hold?
|
Solution. Since $x, y, z$ are distinct positive integers, the required inequality is symmetric and WLOG we can suppose that $x \geq y+1 \geq z+2$. We consider 2 possible cases:
Case 1. $y \geq z+2$. Since $x \geq y+1 \geq z+3$ it follows that
$$
(x-y)^{2} \geq 1, \quad(y-z)^{2} \geq 4, \quad(x-z)^{2} \geq 9
$$
whic... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to prove that for positive integers x, y, z, all distinct (since x ≠ y ≠ z ≠ x), the inequality (x + y + z)(xy + yz + zx - 2) ≥ 9xyz holds. And also find when equality occurs. Alright, let's start by understanding what this inequality is about.
First, the left side is the product of two terms: ... |
Problem 2. Let $x, y, z$ be positive integers such that $x \neq y \neq z \neq x$. Prove that
$$
(x+y+z)(x y+y z+z x-2) \geq 9 x y z
$$
When does the equality hold?
|
Solution. Since $x, y, z$ are distinct positive integers, the required inequality is symmetric and WLOG we can suppose that $x \geq y+1 \geq z+2$. We consider 2 possible cases:
Case 1. $y \geq z+2$. Since $x \geq y+1 \geq z+3$ it follows that
$$
(x-y)^{2} \geq 1, \quad(y-z)^{2} \geq 4, \quad(x-z)^{2} \geq 9
$$
whic... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to prove that for positive integers x, y, z, all distinct (since x ≠ y ≠ z ≠ x), the inequality (x + y + z)(xy + yz + zx - 2) ≥ 9xyz holds. And also find when equality occurs. Hmm, let me think step by step.
First, since x, y, z are positive integers and distinct, maybe I can start by testing s... |
Problem 3. Let $A B C$ be an acute triangle such that $A B \neq A C$, with circumcircle $\Gamma$ and circumcenter $O$. Let $M$ be the midpoint of $B C$ and $D$ be a point on $\Gamma$ such that $A D \perp B C$. Let $T$ be a point such that $B D C T$ is a parallelogram and $Q$ a point on the same side of $B C$ as $A$ su... |
Solution. Let $X^{\prime}$ be symmetric point to $Q$ in line $B C$. Now since $\angle C B A=\angle C Q M=\angle C X^{\prime} M$, $\angle B C A=\angle B Q M=\angle B X^{\prime} M$, we have
$$
\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 hav... | proof | Geometry | proof | Yes | Yes | olympiads | false | 1 | [
false
] | 0 | <think>
Okay, let's try to tackle this problem step by step. First, I need to make sure I understand all the given information and what we need to prove. So, we have an acute triangle ABC with AB ≠ AC. The circumcircle is Γ, and O is the circumcenter. M is the midpoint of BC. Then there's a point D on Γ such that AD is... |
Problem 4. 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 ... |
Solution 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... | 6n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, let's try to tackle this problem. So, we have a regular 2n-gon, and we need to color its sides with three different colors. The conditions are that every side is colored, each color is used at least once, and from any external point, you can only see points of at most two different colors. Hmm, that seems... |
Problem 4. 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 ... |
Solution 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... | 6n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2 | [
false,
false
] | 0 | <think>
Okay, so I need to solve this problem about coloring the sides of a regular 2n-gon with three colors. The conditions are that each color is used at least once, and from any external point, you can only see points of at most two different colors. Hmm, that's interesting. Let me try to break this down.
First, th... |
Problem 1. 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
$$
| # Solution:
The relation $9 \cdot k^{2}+1 \equiv 1(\bmod 3)$ implies
$$
a^{2}+b^{2}+16 \cdot c^{2} \equiv 1(\bmod 3) \Leftrightarrow a^{2}+b^{2}+c^{2} \equiv 1(\bmod 3)
$$
Since $a^{2} \equiv 0,1(\bmod 3), \quad b^{2} \equiv 0,1(\bmod 3), c^{2} \equiv 0,1(\bmod 3)$, we have:
| $a^{2}$ | 0 | 0 | 0 | 0 | 1 | 1 | 1 |... | (37,3,3,13),(17,3,3,7),(3,37,3,13),(3,17,3,7),(3,3,2,3) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5 | [
false,
true,
false,
false,
false
] | 1 | <think>
Okay, so I need to find all prime numbers a, b, c and positive integers k that satisfy the equation a² + b² + 16c² = 9k² + 1. Hmm, primes are numbers greater than 1 that have no divisors other than 1 and themselves. The primes start with 2, 3, 5, 7, 11, etc. Since the equation involves squares of a, b, c, and k... |
Problem 1. Find all prime numbers $a, b, c$ and positive integers $k$ which satisfy the equation
$$
a^{2}+b^{2}+16 \cdot c^{2}=9 \cdot k^{2}+1
$$
| # Solution:
The relation $9 \cdot k^{2}+1 \equiv 1(\bmod 3)$ implies
$$
a^{2}+b^{2}+16 \cdot c^{2} \equiv 1(\bmod 3) \Leftrightarrow a^{2}+b^{2}+c^{2} \equiv 1(\bmod 3)
$$
Since $a^{2} \equiv 0,1(\bmod 3), \quad b^{2} \equiv 0,1(\bmod 3), c^{2} \equiv 0,1(\bmod 3)$, we have:
| $a^{2}$ | 0 | 0 | 0 | 0 | 1 | 1 | 1 |... | (37,3,3,13),(17,3,3,7),(3,37,3,13),(3,17,3,7),(3,3,2,3) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5 | [
false,
true,
false,
false,
false
] | 1 | <think>
Okay, so I need to find all prime numbers a, b, c and positive integers k that satisfy the equation a² + b² + 16c² = 9k² + 1. Let me start by understanding the problem and thinking about possible approaches.
First, all a, b, c are primes. Primes are numbers greater than 1 that have no divisors other than 1 and... |
Problem 1. Find all prime numbers $a, b, c$ and positive integers $k$ which satisfy the equation
$$
a^{2}+b^{2}+16 \cdot c^{2}=9 \cdot k^{2}+1
$$
| # Solution:
The relation $9 \cdot k^{2}+1 \equiv 1(\bmod 3)$ implies
$$
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 |... | (37,3,3,13),(17,3,3,7),(3,37,3,13),(3,17,3,7),(3,3,2,3) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5 | [
false,
true,
false,
false,
false
] | 1 | <think>
Okay, so I need to find all prime numbers a, b, c and positive integers k that satisfy the equation a² + b² + 16c² = 9k² + 1. Hmm, primes and squares... Let me start by breaking this down.
First, primes are numbers greater than 1 that have no divisors other than 1 and themselves. So a, b, c can be 2, 3, 5, 7, ... |
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