id large_stringlengths 24 24 | competition large_stringclasses 22
values | year large_stringdate 1959-01-01 00:00:00 2024-01-01 00:00:00 | problem large_stringlengths 36 1.6k | trace large_stringlengths 2.75k 118k | solution large_stringlengths 1.09k 44.3k |
|---|---|---|---|---|---|
6733856f52d7e82fe11d4828 | Kosovo National Mathematical Olympiad | 2011 | Let $ a$, $ b$, $ c$ be the sides of a triangle, and $ S$ its area. Prove:
$$ a^{2} + b^{2} + c^{2}\geq 4S \sqrt {3}
$$
In what case does equality hold? | Okay, let's try to prove that for any triangle with sides \(a\), \(b\), \(c\) and area \(S\), the inequality \(a^2 + b^2 + c^2 \geq 4S\sqrt{3}\) holds, and find when equality occurs. Hmm, first, I remember that for triangles, there are several formulas related to the area, like Heron's formula, or using \(S = \frac{1}{... | To prove the inequality
$$
a^2 + b^2 + c^2 \geq 4S\sqrt{3}
$$
for any triangle with sides $a$, $b$, $c$ and area $S$, we proceed through a combination of trigonometric identities, geometric reasoning, and algebraic manipulation.
---
### **Step 1: Expressing the Area Using Trigonometry**
Let’s consider the angle ... |
6733857b52d7e82fe11d60b5 | Austrian-Polish | 1980 | Find the greatest natural number $n$ such there exist natural numbers $x_{1}, x_{2}, \ldots, x_{n}$ and natural $a_{1} \lt a_{2} \lt \ldots \lt a_{n-1}$ satisfying the following equations for $i =1,2,\ldots,n-1$: $$x_{1}x_{2}\ldots x_{n}= 1980 \quad \text{and}\quad x_{i}+\frac{1980}{x_{i}}= a_{i}.$$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so the problem is to find the greatest natural number n where we have natural numbers x₁ to xₙ, and natural numbers a₁ < a₂ < ... < aₙ₋₁ such that two conditions hold: first, the produ... | To solve the problem, we are given a natural number $ N = 1980 $ and asked to find the **greatest natural number** $ n $ such that there exist natural numbers $ x_1, x_2, \ldots, x_n $ and strictly increasing natural numbers $ a_1 < a_2 < \cdots < a_{n-1} $ satisfying:
$$
x_1 x_2 \cdots x_n = 1980 \quad \text{and} \qu... |
6733857b52d7e82fe11d6115 | Azerbaijan Team Selection Test | 2020 | You are given a set of $n$ blocks, each weighing at least $1$; their total weight is $2n$. Prove that for every real number $r$ with $0 \leq r \leq 2n-2$ you can choose a subset of the blocks whose total weight is at least $r$ but at most $r + 2$. | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so the problem says we have n blocks, each at least weight 1, total weight 2n. We need to show that for any r between 0 and 2n-2 inclusive, there's a subset with weight between r and r... | To solve the problem, we are given $ n $ blocks, each with weight at least 1, and their total weight is $ 2n $. We are to prove that for every real number $ r $ satisfying $ 0 \leq r \leq 2n - 2 $, there exists a subset of these blocks whose total weight lies in the interval $ [r, r + 2] $.
---
### **Key Observations... |
6733857b52d7e82fe11d6114 | Azerbaijan Team Selection Test | 2020 | Let $u_1, u_2, \dots, u_{2019}$ be real numbers satisfying $$u_{1}+u_{2}+\cdots+u_{2019}=0 \quad \text { and } \quad u_{1}^{2}+u_{2}^{2}+\cdots+u_{2019}^{2}=1.$$Let $a=\min \left(u_{1}, u_{2}, \ldots, u_{2019}\right)$ and $b=\max \left(u_{1}, u_{2}, \ldots, u_{2019}\right)$. Prove that
$$
a b \leqslant-\frac{1}{2019}.
... | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so we have 2019 real numbers that add up to zero, their squares add up to 1, and we need to show that the product of the minimum and maximum is at most -1/2019. Hmm, okay, let's start ... | We are given a sequence of real numbers $ u_1, u_2, \dots, u_{2019} $ satisfying:
$$
u_1 + u_2 + \cdots + u_{2019} = 0 \quad \text{and} \quad u_1^2 + u_2^2 + \cdots + u_{2019}^2 = 1,
$$
with $ a = \min(u_1, u_2, \ldots, u_{2019}) $ and $ b = \max(u_1, u_2, \ldots, u_{2019}) $. We are to prove that:
$$
ab \leq -\frac... |
6733857b52d7e82fe11d61b1 | Belarus Team Selection Test | 2014 | Let $\mathbb{Z} _{ \gt 0}$ be the set of positive integers. Find all functions $f: \mathbb{Z} _{ \gt 0}\rightarrow \mathbb{Z} _{ \gt 0}$ such that
$$ m^2 + f(n) \mid mf(m) +n $$
for all positive integers $m$ and $n$. | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so I need to find all functions f from positive integers to positive integers satisfying that for every m, n in positive integers, m squared plus f(n) divides m times f(m) plus n. Let'... | To solve the functional equation:
$$
m^2 + f(n) \mid mf(m) + n \quad \text{for all } m, n \in \mathbb{Z}_{>0},
$$
we aim to find all functions $ f: \mathbb{Z}_{>0} \to \mathbb{Z}_{>0} $ satisfying this condition.
---
### Step 1: Use Specific Values to Derive Constraints
Let us begin by setting $ m = 2 $ and $ n = ... |
6733858352d7e82fe11d6a23 | Germany Team Selection Test | 2022 | Find all positive integers $n\geq1$ such that there exists a pair $(a,b)$ of positive integers, such that $a^2+b+3$ is not divisible by the cube of any prime, and $$n=\frac{ab+3b+8}{a^2+b+3}.$$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so I need to find all positive integers n ≥ 1 where there's some positive integer pair (a,b) making n equal to (ab + 3b + 8)/(a² + b + 3), and also the denominator a² + b + 3 isn't div... | To solve the problem, we are to find all positive integers $ n \geq 1 $ such that there exists a pair $ (a, b) $ of **positive integers** for which:
$$
n = \frac{ab + 3b + 8}{a^2 + b + 3}
$$
and the denominator $ a^2 + b + 3 $ is **not divisible by the cube of any prime** (i.e., it is *cube-free*).
---
### Step 1: ... |
6733858752d7e82fe11d6ada | Germany Team Selection Test | 2007 | Prove the inequality:
$$\sum_{i \lt j}{\frac {a_{i}a_{j}}{a_{i} + a_{j}}}\leq \frac {n}{2(a_{1} + a_{2} +\cdots + a_{n})}\cdot \sum_{i \lt j}{a_{i}a_{j}}$$
for positive reals $ a_{1},a_{2},\ldots,a_{n}$.
Proposed by Dusan Dukic, Serbia | Okay, let's try to tackle this inequality. The problem is to prove that for positive real numbers \(a_1, a_2, \ldots, a_n\), the sum over all \(i < j\) of \(\frac{a_i a_j}{a_i + a_j}\) is less than or equal to \(\frac{n}{2S} \cdot \sum_{i < j} a_i a_j\), where \(S = a_1 + a_2 + \cdots + a_n\).
First, let's recall some... | To prove the inequality
$$
\sum_{i < j} \frac{a_i a_j}{a_i + a_j} \leq \frac{n}{2(a_1 + a_2 + \cdots + a_n)} \cdot \sum_{i < j} a_i a_j,
$$
for positive real numbers $ a_1, a_2, \ldots, a_n $, we proceed with a systematic and rigorous approach.
---
### **Step 1: Reformulate the Inequality**
Let us denote:
- $ S =... |
6733858752d7e82fe11d6b18 | Germany Team Selection Test | 2004 | Let $a_{ij}$ $i=1,2,3$; $j=1,2,3$ be real numbers such that $a_{ij}$ is positive for $i=j$ and negative for $i\neq j$.
Prove the existence of positive real numbers $c_{1}$, $c_{2}$, $c_{3}$ such that the numbers $$a_{11}c_{1}+a_{12}c_{2}+a_{13}c_{3},\qquad a_{21}c_{1}+a_{22}c_{2}+a_{23}c_{3},\qquad a_{31}c_{1}+a_{32}c... | Okay, let's try to tackle this problem. So, we have a 3x3 matrix with positive diagonal entries and negative off-diagonal entries. We need to find positive constants \( c_1, c_2, c_3 \) such that when we multiply the matrix by the vector \( (c_1, c_2, c_3)^T \), the resulting vector has all components non-negative, non... | To prove the existence of positive real numbers $ c_1, c_2, c_3 $ such that the vector
$$
\begin{pmatrix}
a_{11}c_1 + a_{12}c_2 + a_{13}c_3 \\
a_{21}c_1 + a_{22}c_2 + a_{23}c_3 \\
a_{31}c_1 + a_{32}c_2 + a_{33}c_3
\end{pmatrix}
$$
is either all positive, all negative, or all zero, we begin by analyzing the structure ... |
6733858752d7e82fe11d6b3d | Germany Team Selection Test | 1978 | Let $B$ be a set of $k$ sequences each having $n$ terms equal to $1$ or $-1$. The product of two such sequences $(a_1, a_2, \ldots , a_n)$ and $(b_1, b_2, \ldots , b_n)$ is defined as $(a_1b_1, a_2b_2, \ldots , a_nb_n)$. Prove that there exists a sequence $(c_1, c_2, \ldots , c_n)$ such that the intersection of $B$ a... | Okay, let's try to tackle this problem step by step. First, let's make sure I understand the problem correctly. We have a set B of k sequences, each of length n with entries ±1. The product of two sequences is component-wise multiplication, so if we have a sequence c, then multiplying B by c gives the set {b * c | b ∈ ... | To solve the problem, we consider a set $ B $ of $ k $ sequences, each of length $ n $ with entries in $ \{1, -1\} $. The operation defined on these sequences is component-wise multiplication, which makes the set of all such sequences into a group under this operation. This group is isomorphic to $ (\mathbb{Z}/2\mathbb... |
6733858752d7e82fe11d6b3e | Germany Team Selection Test | 1977 | We consider two sequences of real numbers $x_{1} \geq x_{2} \geq \ldots \geq x_{n}$ and $\ y_{1} \geq y_{2} \geq \ldots \geq y_{n}.$ Let $z_{1}, z_{2}, .\ldots, z_{n}$ be a permutation of the numbers $y_{1}, y_{2}, \ldots, y_{n}.$ Prove that $\sum \limits_{i=1}^{n} ( x_{i} -\ y_{i} )^{2} \leq \sum \limits_{i=1}^{n}$ $(... | Okay, let's try to tackle this problem. So, we have two non-increasing sequences, x₁ ≥ x₂ ≥ ... ≥ xₙ and y₁ ≥ y₂ ≥ ... ≥ yₙ. Then z is a permutation of the y's, and we need to show that the sum of (xᵢ - yᵢ)² is less than or equal to the sum of (xᵢ - zᵢ)² for any permutation z. Hmm, so basically, the sum of squared diff... | To prove the inequality
$$
\sum_{i=1}^{n} (x_i - y_i)^2 \leq \sum_{i=1}^{n} (x_i - z_i)^2,
$$
where $ x_1 \geq x_2 \geq \cdots \geq x_n $ and $ y_1 \geq y_2 \geq \cdots \geq y_n $ are non-increasing sequences of real numbers, and $ z_1, z_2, \ldots, z_n $ is any permutation of $ y_1, y_2, \ldots, y_n $, we proceed ... |
6733858752d7e82fe11d6dc2 | Indonesia TST | 2010 | Let $\Gamma_1$, $\Gamma_2$, $\Gamma_3$, $\Gamma_4$ be distinct circles such that $\Gamma_1$, $\Gamma_3$ are externally tangent at $P$, and $\Gamma_2$, $\Gamma_4$ are externally tangent at the same point $P$. Suppose that $\Gamma_1$ and $\Gamma_2$; $\Gamma_2$ and $\Gamma_3$; $\Gamma_3$ and $\Gamma_4$; $\Gamma_4$ and $\G... | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so we have four distinct circles, all tangent to each other at the same point P? Wait, no: Γ₁ and Γ₃ are externally tangent at P, Γ₂ and Γ₄ are externally tangent at P. So two pairs of... | To prove the identity
$$
\frac{AB \cdot BC}{AD \cdot DC} = \frac{PB^2}{PD^2},
$$
we consider a geometric configuration involving four distinct circles $\Gamma_1, \Gamma_2, \Gamma_3, \Gamma_4$, all passing through a common point $P$. The circles are arranged such that:
- $\Gamma_1$ and $\Gamma_3$ are externally tange... |
6733858a52d7e82fe11d71d6 | Morocco TST | 2010 | Find all triangles whose side lengths are consecutive integers, and one of whose angles is twice another. | Okay, let's try to tackle this problem: finding all triangles with side lengths as consecutive integers where one angle is twice another. First, let's recall that in a triangle, the sides are related to the angles via the Law of Sines and the Law of Cosines. Let's denote the sides as \( a \), \( a+1 \), \( a+2 \) where... | To solve the problem of finding all triangles with side lengths that are **consecutive integers** and where **one angle is twice another**, we proceed systematically using geometric and trigonometric principles.
---
### Step 1: Understanding the Triangle Structure
Let the side lengths be $ a, a+1, a+2 $, where $ a \... |
6733858a52d7e82fe11d71a4 | Moldova Team Selection Test | 1993 | A convex quadrilateral has equal diagonals. An equilateral triangle is constructed on the outside of each side of the quadrilateral. The centers of the triangles on opposite sides are joined. Show that the two joining lines are perpendicular.
Alternative formulation. Given a convex quadrilateral $ ABCD$ with congruent... | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so we have a convex quadrilateral ABCD with equal diagonals AC = BD. We need to construct equilateral triangles on each side externally, find their centers (which for equilateral trian... | To prove that the segments joining the centroids of equilateral triangles constructed externally on opposite sides of a convex quadrilateral with equal diagonals are perpendicular, we proceed using complex numbers to model the geometry and leverage rotational symmetry inherent in equilateral triangles.
---
### **Step... |
6733858d52d7e82fe11d74dc | Romania Team Selection Test | 1998 | An infinite arithmetic progression whose terms are positive integers contains the square of an integer and the cube of an integer. Show that it contains the sixth power of an integer. | Okay, let's try to tackle this problem. So, we have an infinite arithmetic progression (AP) with positive integer terms. It contains a square of an integer, say \( a^2 \), and a cube of an integer, say \( b^3 \). We need to show that it must contain a sixth power, like \( c^6 \).
First, let's recall that an arithmetic... | To solve the problem, we are given an infinite arithmetic progression (AP) with positive integer terms that contains both a square of an integer and a cube of an integer. We are to show that this AP must also contain the sixth power of some integer.
---
### **Step 1: Understanding the Structure of the AP**
An infini... |
6733858d52d7e82fe11d755e | Romania Team Selection Test | 1979 | Find all positive integer solutions $ x, y, z$ of the equation $ 3^x + 4^y = 5^z.$ | Okay, let's try to solve the equation \(3^x + 4^y = 5^z\) where \(x, y, z\) are positive integers. First, I need to consider small values of \(x, y, z\) because exponential equations often have small solutions, and maybe we can prove there are no larger ones.
Let's start with \(z = 1\). Then the right-hand side is \(5... | To solve the equation
$$
3^x + 4^y = 5^z
$$
for **positive integers** $ x, y, z $, we proceed with a systematic and rigorous analysis.
---
### Step 1: Try Small Values
We begin by testing small values of $ z $ to see if any solutions exist.
- **For $ z = 1 $:**
$ 5^1 = 5 $. The left-hand side becomes $ 3^x ... |
6733859452d7e82fe11d7d7c | IMO | 2005 | Determine all positive integers relatively prime to all the terms of the infinite sequence $$ a_n=2^n+3^n+6^n -1,\ n\geq 1. $$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so the problem is to find all positive integers that are relatively prime to every term in the infinite sequence aₙ = 2ⁿ + 3ⁿ + 6ⁿ - 1 for n ≥ 1. That means we need to find all positiv... | To determine all positive integers relatively prime to all terms of the infinite sequence
$$
a_n = 2^n + 3^n + 6^n - 1, \quad n \geq 1,
$$
we must find all positive integers $ m $ such that
$$
\gcd(m, a_n) = 1 \quad \text{for all } n \geq 1.
$$
---
### **Step 1: Understanding the Problem**
We are to find all pos... |
6733859452d7e82fe11d7d6d | IMO | 2007 | Real numbers $ a_{1}$, $ a_{2}$, $ \ldots$, $ a_{n}$ are given. For each $ i$, $ (1 \leq i \leq n )$, define
$$ d_{i} = \max \{ a_{j}\mid 1 \leq j \leq i \} - \min \{ a_{j}\mid i \leq j \leq n \}
$$
and let $ d = \max \{d_{i}\mid 1 \leq i \leq n \}$.
(a) Prove that, for any real numbers $ x_{1}\leq x_{2}\leq \cdots \l... | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so we have this problem about sequences, maxima, minima, and then comparing to a non-decreasing sequence with minimal maximum deviation. Part (a) says for any non-decreasing sequence x... | We are given a sequence of real numbers $ a_1, a_2, \ldots, a_n $, and for each index $ i $, we define:
$$
d_i = \max\{a_j \mid 1 \leq j \leq i\} - \min\{a_j \mid i \leq j \leq n\}
$$
Let $ d = \max\{d_i \mid 1 \leq i \leq n\} $. The problem asks us to prove two things:
---
### **(a) Prove that for any non-decreasi... |
6733859452d7e82fe11d7d68 | IMO | 2008 | (a) Prove that
$$\frac {x^{2}}{\left(x - 1\right)^{2}} + \frac {y^{2}}{\left(y - 1\right)^{2}} + \frac {z^{2}}{\left(z - 1\right)^{2}} \geq 1$$ for all real numbers $x$, $y$, $z$, each different from $1$, and satisfying $xyz=1$.
(b) Prove that equality holds above for infinitely many triples of rational numbers $x$, $... | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so part (a) is to prove this inequality for all real numbers x, y, z ≠ 1 with xyz=1. Let's start by recalling that when dealing with symmetric inequalities with a constraint like xyz=1... | To solve the given inequality and prove the equality condition, we proceed as follows.
---
### **Part (a): Prove that**
$$
\frac{x^2}{(x - 1)^2} + \frac{y^2}{(y - 1)^2} + \frac{z^2}{(z - 1)^2} \geq 1
$$
for all real numbers $ x, y, z \neq 1 $ such that $ xyz = 1 $.
---
#### **Step 1: Substitution to simplify the ... |
6733859452d7e82fe11d7d94 | IMO | 2001 | Let $n$ be an odd integer greater than 1 and let $c_1, c_2, \ldots, c_n$ be integers. For each permutation $a = (a_1, a_2, \ldots, a_n)$ of $\{1,2,\ldots,n\}$, define $S(a) = \sum_{i=1}^n c_i a_i$. Prove that there exist permutations $a \neq b$ of $\{1,2,\ldots,n\}$ such that $n!$ is a divisor of $S(a)-S(b)$. | Okay, let's try to tackle this problem. So, we need to prove that for any odd integer n > 1 and any integers c₁, c₂,..., cₙ, there exist two distinct permutations a and b of {1, 2,..., n} such that n! divides S(a) - S(b), where S(a) = Σcᵢaᵢ.
First, let's recall that the number of permutations of {1, 2,..., n} is n!. S... | To prove that for any odd integer $ n > 1 $ and integers $ c_1, c_2, \ldots, c_n $, there exist distinct permutations $ a \neq b $ of $ \{1, 2, \ldots, n\} $ such that $ n! \mid S(a) - S(b) $, we proceed as follows.
---
### **Step 1: Understanding the Problem**
We are given:
- An odd integer $ n > 1 $
- A sequence o... |
6733859452d7e82fe11d7db6 | IMO | 1995 | Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Prove that
$$ \frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}\geq \frac {3}{2}.
$$ | Okay, let's try to tackle this inequality problem. We need to prove that for positive real numbers \(a, b, c\) with \(abc = 1\), the following holds:
\[
\frac{1}{a^3(b + c)} + \frac{1}{b^3(c + a)} + \frac{1}{c^3(a + b)} \geq \frac{3}{2}.
\]
First, since \(abc = 1\), maybe we can use substitution to simplify the var... | To prove the inequality
$$
\frac{1}{a^3(b + c)} + \frac{1}{b^3(c + a)} + \frac{1}{c^3(a + b)} \geq \frac{3}{2}
$$
for all positive real numbers $a, b, c$ such that $abc = 1$, we proceed by transforming the expression using a standard substitution that simplifies the constraint.
---
### **Step 1: Substitution to Sim... |
6733859452d7e82fe11d7db5 | IMO | 1995 | Let $ A,B,C,D$ be four distinct points on a line, in that order. The circles with diameters $ AC$ and $ BD$ intersect at $ X$ and $ Y$. The line $ XY$ meets $ BC$ at $ Z$. Let $ P$ be a point on the line $ XY$ other than $ Z$. The line $ CP$ intersects the circle with diameter $ AC$ at $ C$ and $ M$, and the line $ BP... | Okay, let's try to tackle this geometry problem step by step. First, let's set up a coordinate system to make things concrete. Since all points are on a line, let's place them on the x-axis for simplicity. Let's denote the coordinates: let \( A = (a, 0) \), \( B = (b, 0) \), \( C = (c, 0) \), \( D = (d, 0) \), with \( ... | To prove that the lines $ AM $, $ DN $, and $ XY $ are concurrent, we proceed with a coordinate-based geometric approach. The key idea is to leverage the symmetry and algebraic structure of the problem by placing all points on the x-axis and using coordinate geometry to derive the necessary conditions for concurrency.
... |
6733859452d7e82fe11d7da4 | IMO | 1998 | In a contest, there are $m$ candidates and $n$ judges, where $n\geq 3$ is an odd integer. Each candidate is evaluated by each judge as either pass or fail. Suppose that each pair of judges agrees on at most $k$ candidates. Prove that $${\frac{k}{m}} \geq {\frac{n-1}{2n}}. $$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so we have a contest with m candidates and n judges, n is odd and at least 3. Each judge gives pass or fail to each candidate. Each pair of judges agrees on at most k candidates. Need ... | To prove the inequality
$$
\frac{k}{m} \geq \frac{n - 1}{2n}
$$
under the given conditions—where $ n \geq 3 $ is an odd integer, there are $ m $ candidates and $ n $ judges, each judge evaluates every candidate as either pass or fail, and each pair of judges agrees on at most $ k $ candidates—we proceed with a syst... |
6733859452d7e82fe11d7dbc | IMO | 1994 | Let $ ABC$ be an isosceles triangle with $ AB = AC$. $ M$ is the midpoint of $ BC$ and $ O$ is the point on the line $ AM$ such that $ OB$ is perpendicular to $ AB$. $ Q$ is an arbitrary point on $ BC$ different from $ B$ and $ C$. $ E$ lies on the line $ AB$ and $ F$ lies on the line $ AC$ such that $ E, Q, F$ are dis... | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so we have an isosceles triangle ABC with AB = AC, so it's symmetric about the altitude from A to BC, which is AM since M is the midpoint of BC. That makes sense, so AM is the axis of ... | To prove that $ OQ $ is perpendicular to $ EF $ if and only if $ QE = QF $, we proceed with a coordinate-based geometric analysis of the given isosceles triangle $ ABC $ with $ AB = AC $, and use the symmetry inherent in such triangles.
---
### **Coordinate Setup**
We place the triangle in the coordinate plane to ex... |
6733859452d7e82fe11d7ddd | IMO | 1989 | Prove that for each positive integer $ n$ there exist $ n$ consecutive positive integers none of which is an integral power of a prime number. | Okay, let's try to tackle this problem: Prove that for each positive integer \( n \), there exist \( n \) consecutive positive integers none of which is an integral power of a prime number. Hmm, so we need to find a sequence of \( n \) numbers in a row where each number is not a prime power. Prime powers are numbers li... | To prove that for each positive integer $ n $, there exist $ n $ consecutive positive integers none of which is an integral power of a prime number, we proceed as follows.
---
### **Understanding the Problem**
A *prime power* is a number of the form $ p^k $, where $ p $ is a prime and $ k \geq 1 $. This includes num... |
6733859452d7e82fe11d7dec | IMO | 1986 | Given a point $P_0$ in the plane of the triangle $A_1A_2A_3$. Define $A_s=A_{s-3}$ for all $s\ge4$. Construct a set of points $P_1,P_2,P_3,\ldots$ such that $P_{k+1}$ is the image of $P_k$ under a rotation center $A_{k+1}$ through an angle $120^o$ clockwise for $k=0,1,2,\ldots$. Prove that if $P_{1986}=P_0$, then the ... | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so we have this problem about rotating a point around a triangle's vertices repeatedly, and we need to show that if after 1986 rotations we get back to the start, the triangle must be ... | To solve the problem, we model the plane as the complex plane and represent points by complex numbers. Let $ A_1, A_2, A_3 $ correspond to complex numbers $ a_1, a_2, a_3 $, and let $ P_0 $ be a point represented by the complex number $ p_0 $. The sequence of points $ P_k $ is defined such that each $ P_{k+1} $ is obta... |
6733859452d7e82fe11d7e07 | IMO | 1982 | The diagonals $AC$ and $CE$ of the regular hexagon $ABCDEF$ are divided by inner points $M$ and $N$ respectively, so that $$ {AM\over AC}={CN\over CE}=r. $$ Determine $r$ if $B,M$ and $N$ are collinear. | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so we have a regular hexagon ABCDEF, and we need to find this ratio r where the diagonals AC and CE are divided by points M and N such that AM/AC = CN/CE = r, and then B, M, N are coll... | To determine the value of $ r $ such that points $ B $, $ M $, and $ N $ are collinear in a regular hexagon $ ABCDEF $, we proceed with a coordinate geometry approach.
---
### **Step 1: Assign Coordinates to Hexagon Vertices**
We place the regular hexagon on the unit circle centered at the origin. The vertices are l... |
6733859452d7e82fe11d7df8 | IMO | 1984 | Find one pair of positive integers $a,b$ such that $ab(a+b)$ is not divisible by $7$, but $(a+b)^7-a^7-b^7$ is divisible by $7^7$. | Okay, let's try to tackle this problem. We need to find positive integers \(a, b\) such that \(ab(a + b)\) is not divisible by 7, but \((a + b)^7 - a^7 - b^7\) is divisible by \(7^7\). First, let's recall some number theory related to divisibility by powers of primes, especially using the Lifting The Exponent (LTE) lem... | To solve the problem, we are tasked with finding **one pair of positive integers** $ a, b $ such that:
- $ ab(a + b) $ is **not divisible by 7**, and
- $ (a + b)^7 - a^7 - b^7 $ is **divisible by $ 7^7 $**.
---
### Step 1: Factorization Insight
We begin by recognizing a key algebraic identity:
$$
(a + b)^7 - a^7 -... |
6733859452d7e82fe11d7e1a | IMO | 1980 | Given a sequence $\{a_n\}$ of real numbers such that $|a_{k+m} - a_k - a_m| \leq 1$ for all positive integers $k$ and $m$, prove that, for all positive integers $p$ and $q$, $$|\frac{a_p}{p} - \frac{a_q}{q}| \lt \frac{1}{p} + \frac{1}{q}.$$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so I need to prove that for this sequence where the difference between a_{k+m} and a_k + a_m is at most 1 in absolute value for all positive integers k, m, then for any positive intege... | To prove the inequality
$$
\left| \frac{a_p}{p} - \frac{a_q}{q} \right| < \frac{1}{p} + \frac{1}{q}
$$
for all positive integers $ p $ and $ q $, given that the sequence $ \{a_n\} $ satisfies
$$
|a_{k+m} - a_k - a_m| \leq 1 \quad \text{for all } k, m \in \mathbb{Z}^+,
$$
we proceed with a systematic analysis.
... |
6733859b52d7e82fe11d7e38 | IMO | 1979 | Determine all real numbers a for which there exists positive reals $x_{1}, \ldots, x_{5}$ which satisfy the relations $ \sum_{k=1}^{5} kx_{k}=a,$ $ \sum_{k=1}^{5} k^{3}x_{k}=a^{2},$ $ \sum_{k=1}^{5} k^{5}x_{k}=a^{3}.$ | Okay, let's try to tackle this problem. We need to find all real numbers \( a \) such that there exist positive real numbers \( x_1, x_2, x_3, x_4, x_5 \) satisfying the three equations:
\[
\sum_{k=1}^{5} k x_k = a, \quad \sum_{k=1}^{5} k^3 x_k = a^2, \quad \sum_{k=1}^{5} k^5 x_k = a^3.
\]
First, let's denote the s... | To determine all real numbers $ a $ for which there exist **positive** real numbers $ x_1, x_2, x_3, x_4, x_5 $ satisfying the system:
$$
\sum_{k=1}^{5} kx_k = a, \quad \sum_{k=1}^{5} k^3 x_k = a^2, \quad \sum_{k=1}^{5} k^5 x_k = a^3,
$$
we begin by rephrasing the problem in terms of moments. Define $ y_k = kx_k $, s... |
6733859b52d7e82fe11d7e34 | IMO | 1979 | If $p$ and $q$ are natural numbers so that $$ \frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \ldots -\frac{1}{1318}+\frac{1}{1319}, $$ prove that $p$ is divisible with $1979$. | Okay, let's try to tackle this problem. The problem states that we have a sum of the form \( \frac{p}{q} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots - \frac{1}{1318} + \frac{1}{1319} \), where \( p \) and \( q \) are natural numbers, and we need to prove that \( p \) is divisible by 1979. First, let's recall... | To prove that the numerator $ p $ of the fraction
$$
\frac{p}{q} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots - \frac{1}{1318} + \frac{1}{1319}
$$
is divisible by 1979, we begin by recognizing this as an alternating harmonic series with 1319 terms. Since 1319 is odd, the last term is positive, and the sum c... |
6733859b52d7e82fe11d80e0 | IMO Shortlist | 2001 | Let $ABC$ be a triangle with centroid $G$. Determine, with proof, the position of the point $P$ in the plane of $ABC$ such that $AP{\cdot}AG + BP{\cdot}BG + CP{\cdot}CG$ is a minimum, and express this minimum value in terms of the side lengths of $ABC$. | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so we have triangle ABC with centroid G, and we need to find a point P that minimizes the sum AP·AG + BP·BG + CP·CG. First, let's recall some properties of centroids. The centroid divi... | To determine the point $ P $ in the plane of triangle $ ABC $ that minimizes the expression
$$
AP \cdot AG + BP \cdot BG + CP \cdot CG,
$$
we proceed with a combination of geometric insight and vector analysis.
---
### **Geometric Setup and Coordinate Choice**
Let us place the triangle in the coordinate plane such... |
6733859b52d7e82fe11d80ed | IMO Shortlist | 2001 | Let $a_0, a_1, a_2, \ldots$ be an arbitrary infinite sequence of positive numbers. Show that the inequality $1 + a_n \gt a_{n-1} \sqrt[n]{2}$ holds for infinitely many positive integers $n$. | Okay, let's try to tackle this problem. The problem says: given any infinite sequence of positive numbers \( a_0, a_1, a_2, \ldots \), we need to show that the inequality \( 1 + a_n > a_{n-1} \sqrt[n]{2} \) holds for infinitely many positive integers \( n \).
First, let's parse the inequality. For each \( n \geq 1 \) ... | To prove that the inequality
$$
1 + a_n > a_{n-1} \sqrt[n]{2}
$$
holds for **infinitely many positive integers** $ n $, given an arbitrary infinite sequence $ a_0, a_1, a_2, \ldots $ of **positive numbers**, we proceed by contradiction.
---
### Step 1: Reformulating the Inequality
We are given that all $ a_n > 0... |
6733859b52d7e82fe11d8109 | IMO Shortlist | 2000 | Determine all positive integers $ n\geq 2$ that satisfy the following condition: for all $ a$ and $ b$ relatively prime to $ n$ we have $$a \equiv b \pmod n\qquad\text{if and only if}\qquad ab\equiv 1 \pmod n.$$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so the problem is to find all positive integers n ≥ 2 such that for every a and b coprime to n, a ≡ b mod n if and only if ab ≡ 1 mod n. Hmm, let's parse that condition first. The "if ... | To determine all positive integers $ n \geq 2 $ such that for **all** $ a, b $ relatively prime to $ n $, the condition
$$
a \equiv b \pmod{n} \quad \text{if and only if} \quad ab \equiv 1 \pmod{n}
$$
holds, we begin by analyzing the logical structure of the biconditional.
---
### **Key Insight: Simplifying the Con... |
6733859b52d7e82fe11d8119 | IMO Shortlist | 1999 | Let $ABC$ be a triangle, $\Omega$ its incircle and $\Omega_{a}, \Omega_{b}, \Omega_{c}$ three circles orthogonal to $\Omega$ passing through $(B,C),(A,C)$ and $(A,B)$ respectively. The circles $\Omega_{a}$ and $\Omega_{b}$ meet again in $C'$; in the same way we obtain the points $B'$ and $A'$. Prove that the radius of ... | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so we have triangle ABC with incircle Ω. Then three circles Ω_a, Ω_b, Ω_c orthogonal to Ω, each passing through a pair of vertices: Ω_a through B,C; Ω_b through A,C; Ω_c through A,B. T... | To solve the problem, we consider triangle $ ABC $ with incircle $ \Omega $ centered at the incenter $ I $ and radius $ r $. We are given three circles $ \Omega_a $, $ \Omega_b $, and $ \Omega_c $ that are **orthogonal** to $ \Omega $ and pass through pairs of vertices: $ \Omega_a $ through $ B,C $, $ \Omega_b $ throug... |
6733859b52d7e82fe11d812c | IMO Shortlist | 1999 | Let $A$ be a set of $N$ residues $\pmod{N^{2}}$. Prove that there exists a set $B$ of of $N$ residues $\pmod{N^{2}}$ such that $A + B = \{a+b|a \in A, b \in B\}$ contains at least half of all the residues $\pmod{N^{2}}$. | Okay, let's try to tackle this problem. The question is: Given a set A of N residues modulo N², prove there exists a set B of N residues modulo N² such that A + B contains at least half of all residues modulo N².
First, let's recall some basics. The total number of residues modulo N² is N². So we need A + B to have si... | To prove that for any set $ A $ of $ N $ residues modulo $ N^2 $, there exists a set $ B $ of $ N $ residues modulo $ N^2 $ such that the sumset $ A + B = \{a + b \mid a \in A, b \in B\} $ contains at least half of all residues modulo $ N^2 $, we proceed using the **probabilistic method**.
---
### **Step 1: Understan... |
6733859b52d7e82fe11d8141 | IMO Shortlist | 1998 | Let $r_{1},r_{2},\ldots ,r_{n}$ be real numbers greater than or equal to 1. Prove that
$$ \frac{1}{r_{1} + 1} + \frac{1}{r_{2} + 1} + \cdots +\frac{1}{r_{n}+1} \geq \frac{n}{ \sqrt[n]{r_{1}r_{2} \cdots r_{n}}+1}. $$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so I need to prove this inequality for real numbers r₁ to rₙ all ≥1. The left side is the sum of 1/(rᵢ + 1), and the right side is n divided by (the nth root of the product of the rᵢs ... | To prove the inequality
$$
\frac{1}{r_1 + 1} + \frac{1}{r_2 + 1} + \cdots + \frac{1}{r_n + 1} \geq \frac{n}{\sqrt[n]{r_1 r_2 \cdots r_n} + 1}
$$
for all real numbers $ r_1, r_2, \ldots, r_n \geq 1 $, we proceed by transforming the problem into a form suitable for applying **Jensen's inequality**, a powerful tool in c... |
6733859b52d7e82fe11d818f | IMO Shortlist | 1995 | Let $ A_1A_2A_3A_4$ be a tetrahedron, $ G$ its centroid, and $ A'_1, A'_2, A'_3,$ and $ A'_4$ the points where the circumsphere of $ A_1A_2A_3A_4$ intersects $ GA_1,GA_2,GA_3,$ and $ GA_4,$ respectively. Prove that
$$ GA_1 \cdot GA_2 \cdot GA_3 \cdot GA_ \cdot4 \leq GA'_1 \cdot GA'_2 \cdot GA'_3 \cdot GA'_4$$
and
$$... | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so we have a tetrahedron A₁A₂A₃A₄ with centroid G. The centroid of a tetrahedron is the average of the vertices' coordinates, right? So maybe coordinate geometry would help here. Let's... | To prove the given inequalities for a tetrahedron $ A_1A_2A_3A_4 $ with centroid $ G $, and points $ A'_1, A'_2, A'_3, A'_4 $ where the circumsphere intersects the lines $ GA_1, GA_2, GA_3, GA_4 $ respectively, we proceed as follows.
---
### **Step 1: Coordinate Setup and Key Observations**
We place the centroid $ G... |
6733859b52d7e82fe11d8193 | IMO Shortlist | 1995 | Let $ \mathbb{Z}$ denote the set of all integers. Prove that for any integers $ A$ and $ B,$ one can find an integer $ C$ for which $ M_1 = \{x^2 + Ax + B : x \in \mathbb{Z}\}$ and $ M_2 = {2x^2 + 2x + C : x \in \mathbb{Z}}$ do not intersect. | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so the problem says: for any integers A and B, find an integer C such that the sets M₁ = {x² + A x + B | x ∈ ℤ} and M₂ = {2x² + 2x + C | x ∈ ℤ} have no elements in common. Hmm, okay. L... | To prove that for any integers $ A $ and $ B $, there exists an integer $ C $ such that the sets
$$
M_1 = \{x^2 + Ax + B : x \in \mathbb{Z}\}, \quad M_2 = \{2x^2 + 2x + C : x \in \mathbb{Z}\}
$$
do not intersect, we analyze the problem by considering two distinct cases based on the parity of $ A $. The key idea is ... |
6733859b52d7e82fe11d818b | IMO Shortlist | 1995 | Let $ A, B$ and $ C$ be non-collinear points. Prove that there is a unique point $ X$ in the plane of $ ABC$ such that $$ XA^2 + XB^2 + AB^2 = XB^2 + XC^2 + BC^2 = XC^2 + XA^2 + CA^2.$$ | Okay, let's try to tackle this problem. So, we have three non-collinear points A, B, C, and we need to prove there's a unique point X such that the three expressions XA² + XB² + AB², XB² + XC² + BC², and XC² + XA² + CA² are all equal. Hmm, let's start by writing down what the problem is saying. Let's denote the equal v... | To prove that there is a **unique point** $ X $ in the plane of triangle $ ABC $ (with $ A, B, C $ non-collinear) such that
$$
XA^2 + XB^2 + AB^2 = XB^2 + XC^2 + BC^2 = XC^2 + XA^2 + CA^2,
$$
we proceed by analyzing the given equalities and reducing them to a system of linear equations.
---
### Step 1: Simplify the... |
6733859b52d7e82fe11d819d | IMO Shortlist | 1995 | Suppose that $ x_1, x_2, x_3, \ldots$ are positive real numbers for which $$ x^n_n = \sum^{n-1}_{j=0} x^j_n$$ for $ n = 1, 2, 3, \ldots$ Prove that $ \forall n,$ $$ 2 - \frac{1}{2^{n-1}} \leq x_n \lt 2 - \frac{1}{2^n}.$$ | Okay, let's try to tackle this problem step by step. The problem states that for each positive integer n, we have a positive real number \( x_n \) satisfying \( x_n^n = \sum_{j=0}^{n-1} x_n^j \). We need to prove that for all n, \( 2 - \frac{1}{2^{n-1}} \leq x_n < 2 - \frac{1}{2^n} \).
First, let's recall that the sum... | To solve the problem, we are given a sequence of positive real numbers $ x_1, x_2, x_3, \ldots $ satisfying the equation:
$$
x_n^n = \sum_{j=0}^{n-1} x_n^j \quad \text{for all } n \geq 1.
$$
Our goal is to prove that for all $ n \geq 1 $, the following inequality holds:
$$
2 - \frac{1}{2^{n-1}} \leq x_n < 2 - \frac{... |
6733859b52d7e82fe11d81a9 | IMO Shortlist | 1994 | A circle $ C$ with center $ O.$ and a line $ L$ which does not touch circle $ C.$ $ OQ$ is perpendicular to $ L,$ $ Q$ is on $ L.$ $ P$ is on $ L,$ draw two tangents $ L_1, L_2$ to circle $ C.$ $ QA, QB$ are perpendicular to $ L_1, L_2$ respectively. ($ A$ on $ L_1,$ $ B$ on $ L_2$). Prove that, line $ AB$ intersect $ ... | Okay, let's try to solve this geometry problem. First, I need to make sure I understand the problem correctly. Let's parse the original formulation to set up coordinates, which might help. The problem says: there's a circle ω with center O, a line l not meeting the circle. E is the foot of the perpendicular from O to l... | To prove that the point $ F $, where line $ CD $ intersects $ OE $, is independent of the position of point $ M $ on the line $ l $, we proceed with a coordinate-based geometric analysis.
---
### **Coordinate Setup**
Let us place the circle $ \omega $ centered at $ O = (0, 0) $ with radius $ r $. The line $ l $ does... |
6733859b52d7e82fe11d816c | IMO Shortlist | 1996 | Let $ f$ be a function from the set of real numbers $ \mathbb{R}$ into itself such for all $ x \in \mathbb{R},$ we have $ |f(x)| \leq 1$ and
$$ f \left( x + \frac{13}{42} \right) + f(x) = f \left( x + \frac{1}{6} \right) + f \left( x + \frac{1}{7} \right).$$
Prove that $ f$ is a periodic function (that is, there exis... | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so we have this functional equation: for all real x, |f(x)| ≤ 1, and f(x + 13/42) + f(x) = f(x + 1/6) + f(x + 1/7). First, maybe I should simplify the constants to make them easier to ... | We are given a function $ f: \mathbb{R} \to \mathbb{R} $ satisfying the inequality $ |f(x)| \leq 1 $ for all $ x \in \mathbb{R} $, and the functional equation:
$$
f\left(x + \frac{13}{42}\right) + f(x) = f\left(x + \frac{1}{6}\right) + f\left(x + \frac{1}{7}\right), \quad \text{for all } x \in \mathbb{R}.
$$
---
###... |
6733859b52d7e82fe11d81da | IMO Shortlist | 1992 | Let $ f(x)$ be a polynomial with rational coefficients and $ \alpha$ be a real number such that $$ \alpha^3 - \alpha = [f(\alpha)]^3 - f(\alpha) = 33^{1992}.$$ Prove that for each $ n \geq 1,$ $$ \left [ f^{n}(\alpha) \right]^3 - f^{n}(\alpha) = 33^{1992},$$ where $ f^{n}(x) = f(f(\cdots f(x))),$ and $ n$ is a positive... | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so the problem says we have a polynomial f with rational coefficients, and a real number α such that α³ - α equals [f(α)]³ - f(α), and both equal 33¹⁹⁹². Then we need to show that no m... | We are given a polynomial $ f(x) $ with rational coefficients and a real number $ \alpha $ such that:
$$
\alpha^3 - \alpha = [f(\alpha)]^3 - f(\alpha) = 33^{1992}.
$$
Our goal is to prove that for all integers $ n \geq 1 $, the following holds:
$$
\left[ f^n(\alpha) \right]^3 - f^n(\alpha) = 33^{1992},
$$
where $ f... |
6733859b52d7e82fe11d81ce | IMO Shortlist | 1993 | A natural number $n$ is said to have the property $P,$ if, for all $a, n^2$ divides $a^n - 1$ whenever $n$ divides $a^n - 1.$
a.) Show that every prime number $n$ has property $P.$
b.) Show that there are infinitely many composite numbers $n$ that possess property $P.$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, let's tackle part (a) first: showing every prime number n has property P. So property P says that for all a, if n divides a^n - 1, then n² divides a^n - 1. Wait, no, wait: "for all a, ... | ### Part (a): Every prime number $ n $ has property $ P $
Let $ p $ be a prime number. We are to show that for all natural numbers $ a $, if $ p \mid a^p - 1 $, then $ p^2 \mid a^p - 1 $.
---
#### Step 1: Use Fermat's Little Theorem
By Fermat’s Little Theorem (FLT), for any integer $ a $ not divisible by $ p $, we ... |
6733859b52d7e82fe11d81fa | IMO Shortlist | 1991 | Let $ f(x)$ be a monic polynomial of degree $ 1991$ with integer coefficients. Define $ g(x) = f^2(x) - 9.$ Show that the number of distinct integer solutions of $ g(x) = 0$ cannot exceed $ 1995.$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so the problem says f(x) is a monic polynomial of degree 1991 with integer coefficients. Then g(x) = f(x)² - 9, and we need to show that the number of distinct integer solutions to g(x... | To solve the problem, we are given a monic polynomial $ f(x) $ of degree 1991 with integer coefficients, and we define $ g(x) = f(x)^2 - 9 $. We are to show that the number of **distinct integer solutions** to $ g(x) = 0 $ cannot exceed 1995.
---
### Step 1: Understanding the Equation
We begin by analyzing the equat... |
6733859b52d7e82fe11d81f7 | IMO Shortlist | 1991 | Find the highest degree $ k$ of $ 1991$ for which $ 1991^k$ divides the number $$ 1990^{1991^{1992}} + 1992^{1991^{1990}}.$$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so I need to find the highest power of 1991 that divides this big number: 1990^(1991^1992) + 1992^(1991^1990). First, let's factorize 1991 to see what primes we're dealing with, becaus... | To determine the highest degree $ k $ such that $ 1991^k $ divides the number
$$
N = 1990^{1991^{1992}} + 1992^{1991^{1990}},
$$
we begin by factoring $ 1991 $. It is well-known that:
$$
1991 = 11 \times 181,
$$
and both 11 and 181 are prime numbers. Since $ 1991 $ is square-free, the highest power of $ 1991 $ divi... |
6733859b52d7e82fe11d8200 | IMO Shortlist | 1991 | Determine the maximum value of the sum
$$ \sum_{i \lt j} x_ix_j (x_i + x_j)
$$
over all $ n -$tuples $ (x_1, \ldots, x_n),$ satisfying $ x_i \geq 0$ and $ \sum^n_{i = 1} x_i = 1.$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so I need to find the maximum of this sum over all non-negative n-tuples that add up to 1. The sum is over all i < j of x_i x_j (x_i + x_j). Let me first make sure I understand what th... | To determine the maximum value of the sum
$$
\sum_{i < j} x_i x_j (x_i + x_j)
$$
over all $ n $-tuples $ (x_1, \ldots, x_n) $ satisfying $ x_i \geq 0 $ and $ \sum_{i=1}^n x_i = 1 $, we proceed by analyzing and simplifying the expression.
---
### **Step 1: Reformulate the Sum**
We begin by observing that:
$$
x_i x... |
673385a052d7e82fe11d8228 | IMO Shortlist | 1989 | Let $ g: \mathbb{C} \rightarrow \mathbb{C}$, $ \omega \in \mathbb{C}$, $ a \in \mathbb{C}$, $ \omega^3 = 1$, and $ \omega \ne 1$. Show that there is one and only one function $ f: \mathbb{C} \rightarrow \mathbb{C}$ such that
$$ f(z) + f(\omega z + a) = g(z),z\in \mathbb{C}
$$ | Okay, let's try to tackle this problem. The question is about showing that there's a unique function \( f: \mathbb{C} \rightarrow \mathbb{C} \) satisfying the functional equation \( f(z) + f(\omega z + a) = g(z) \) for all \( z \in \mathbb{C} \), where \( \omega \) is a primitive 3rd root of unity (so \( \omega^3 = 1 \... | To solve the problem, we are given a functional equation involving a complex function $ g: \mathbb{C} \to \mathbb{C} $, and a primitive cube root of unity $ \omega \in \mathbb{C} $ (i.e., $ \omega^3 = 1 $, $ \omega \neq 1 $), along with a constant $ a \in \mathbb{C} $. The goal is to show that there exists **exactly on... |
673385a052d7e82fe11d821f | IMO Shortlist | 1990 | Prove that on the coordinate plane it is impossible to draw a closed broken line such that
(i) the coordinates of each vertex are rational;
(ii) the length each of its edges is 1;
(iii) the line has an odd number of vertices. | Okay, let's try to tackle this problem. The goal is to prove that you can't have a closed polygon (a broken line that starts and ends at the same point) with all vertices having rational coordinates, each edge length 1, and an odd number of vertices. Hmm, let's start by recalling some basics about polygons with rationa... | To prove that it is **impossible** to draw a closed broken line on the coordinate plane satisfying the following conditions:
- (i) All vertices have rational coordinates,
- (ii) Each edge has length 1,
- (iii) The number of vertices is odd,
we proceed by analyzing the geometric and algebraic constraints imposed by th... |
673385a052d7e82fe11d8235 | IMO Shortlist | 1989 | Let $ a, b \in \mathbb{Z}$ which are not perfect squares. Prove that if $$ x^2 - ay^2 - bz^2 + abw^2 = 0$$ has a nontrivial solution in integers, then so does $$ x^2 - ay^2 - bz^2 = 0.$$ | Okay, let's try to tackle this problem. The question is: given integers a and b, which are not perfect squares, if the equation \( x^2 - a y^2 - b z^2 + a b w^2 = 0 \) has a nontrivial integer solution, then the equation \( x^2 - a y^2 - b z^2 = 0 \) must also have a nontrivial integer solution.
First, let's recall th... | To prove that if the equation
$$
x^2 - a y^2 - b z^2 + ab w^2 = 0
$$
has a nontrivial integer solution, then so does
$$
x^2 - a y^2 - b z^2 = 0,
$$
we proceed by analyzing the structure of the given equation and leveraging algebraic identities to construct a solution to the second equation from the first.
---
... |
673385a052d7e82fe11d822f | IMO Shortlist | 1989 | Given seven points in the plane, some of them are connected by segments such that:
(i) among any three of the given points, two are connected by a segment;
(ii) the number of segments is minimal.
How many segments does a figure satisfying (i) and (ii) have? Give an example of such a figure. | Okay, let's try to tackle this problem step by step. The problem says we have seven points in the plane, and we need to connect some with segments (edges) such that two conditions are met: first, among any three points, at least two are connected (so no three points are all pairwise non-connected, which is called an in... | To solve the problem, we are given seven points in the plane and must connect some of them with segments (edges) such that:
1. **Among any three points, at least two are connected by a segment** — this means there is **no independent set of size 3** in the graph.
2. The number of segments (edges) is **minimal**.
---
... |
673385a052d7e82fe11d823c | IMO Shortlist | 1989 | The vertex $ A$ of the acute triangle $ ABC$ is equidistant from the circumcenter $ O$ and the orthocenter $ H.$ Determine all possible values for the measure of angle $ A.$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so the problem says in an acute triangle ABC, vertex A is equidistant from the circumcenter O and the orthocenter H. We need to find all possible measures of angle A. Hmm, let's start ... | To determine all possible values for the measure of angle $ A $ in an **acute triangle** $ ABC $ such that vertex $ A $ is equidistant from the **circumcenter** $ O $ and the **orthocenter** $ H $, we proceed with a systematic and rigorous analysis.
---
### Step 1: Understanding the Problem
We are given:
- Triangle ... |
673385a052d7e82fe11d8253 | IMO Shortlist | 1988 | Let $ Q$ be the centre of the inscribed circle of a triangle $ ABC.$ Prove that for any point $ P,$
$$ a(PA)^2 + b(PB)^2 + c(PC)^2 = a(QA)^2 + b(QB)^2 + c(QC)^2 + (a + b + c)(QP)^2,
$$
where $ a = BC, b = CA$ and $ c = AB.$ | Okay, let's try to tackle this problem. So, we need to prove that for any point P, the weighted sum of the squares of the distances from P to the vertices of triangle ABC, with weights a, b, c (where a, b, c are the lengths of the sides opposite to A, B, C respectively), equals the same weighted sum but with Q (the inc... | To prove the identity for any point $ P $ in the plane of triangle $ ABC $, where $ a = BC $, $ b = CA $, and $ c = AB $, and $ Q $ is the incenter of triangle $ ABC $, we aim to show:
$$
a(PA)^2 + b(PB)^2 + c(PC)^2 = a(QA)^2 + b(QB)^2 + c(QC)^2 + (a + b + c)(QP)^2.
$$
---
### **Step 1: Understanding the Incenter as... |
673385a052d7e82fe11d8254 | IMO Shortlist | 1988 | Let $ \{a_k\}^{\infty}_1$ be a sequence of non-negative real numbers such that:
$$ a_k - 2 a_{k + 1} + a_{k + 2} \geq 0
$$
and $ \sum^k_{j = 1} a_j \leq 1$ for all $ k = 1,2, \ldots$. Prove that:
$$ 0 \leq a_{k} - a_{k + 1} \lt \frac {2}{k^2}
$$
for all $ k = 1,2, \ldots$. | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so we have this sequence of non-negative real numbers {a_k} starting from k=1, satisfying two conditions: first, the second difference is non-negative, which means a_k - 2a_{k+1} + a_{... | We are given a sequence $\{a_k\}_{k=1}^\infty$ of non-negative real numbers satisfying two conditions:
1. The second difference condition:
$$
a_k - 2a_{k+1} + a_{k+2} \geq 0 \quad \text{for all } k \geq 1,
$$
2. The partial sums are bounded:
$$
\sum_{j=1}^k a_j \leq 1 \quad \text{for all } k \geq 1.
... |
673385a052d7e82fe11d826a | IMO Shortlist | 1987 | Let $p_n(k)$ be the number of permutations of the set $\{1,2,3,\ldots,n\}$ which have exactly $k$ fixed points. Prove that $\sum_{k=0}^nk p_n(k)=n!$.(IMO Problem 1)
Original formulation
Let $S$ be a set of $n$ elements. We denote the number of all permutations of $S$ that have exactly $k$ fixed points by $p_n(k).$ P... | Okay, let's tackle part (a) first. The problem says that \( p_n(k) \) is the number of permutations of \( \{1, 2, \ldots, n\} \) with exactly \( k \) fixed points. We need to prove that the sum of \( k p_n(k) \) from \( k=0 \) to \( n \) is \( n! \).
First, let's recall what a fixed point is: a fixed point of a permut... | To solve the given problem, we consider the function $ p_n(k) $, which counts the number of permutations of the set $ \{1, 2, \ldots, n\} $ that have exactly $ k $ fixed points. A fixed point in a permutation is an element $ i $ such that $ \sigma(i) = i $. The goal is to prove two identities:
---
### **Part (a):** ... |
673385a052d7e82fe11d8274 | IMO Shortlist | 1986 | Let $A, B$, and $C$ be three points on the edge of a circular chord such that $B$ is due west of $C$ and $ABC$ is an equilateral triangle whose side is $86$ meters long. A boy swam from $A$ directly toward $B$. After covering a distance of $x$ meters, he turned and swam westward, reaching the shore after covering a dis... | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, let's start by understanding the problem. We have three points A, B, C on the edge of a circular chord? Wait, no, hold on—wait, "on the edge of a circular chord"? That must be a typo. ... | To solve the problem, we begin by interpreting the geometric setup. The key is that points $ A $, $ B $, and $ C $ lie on a **circle** (not a "circular chord," which is not a valid term), forming an equilateral triangle with side length 86 meters. Given that $ B $ is due west of $ C $, we place these points in a coordi... |
673385a052d7e82fe11d8267 | IMO Shortlist | 1987 | Is it possible to put $1987$ points in the Euclidean plane such that the distance between each pair of points is irrational and each three points determine a non-degenerate triangle with rational area? (IMO Problem 5)
Proposed by Germany, DR | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so the problem is: can we place 1987 points in the plane so that every pair has irrational distance, and every three points form a triangle with rational area? Hmm, let's start by reca... | To determine whether it is possible to place 1987 points in the Euclidean plane such that:
1. The distance between each pair of points is irrational, and
2. Each set of three points forms a non-degenerate triangle with rational area,
we construct a specific configuration of points that satisfies both conditions.
-... |
673385a052d7e82fe11d8260 | IMO Shortlist | 1987 | Show that if $a, b, c$ are the lengths of the sides of a triangle and if $2S = a + b + c$, then
$$\frac{a^n}{b+c} + \frac{b^n}{c+a} +\frac{c^n}{a+b} \geq \left(\dfrac 23 \right)^{n-2}S^{n-1} \quad \forall n \in \mathbb N $$
Proposed by Greece. | Okay, let's try to tackle this inequality problem. The problem states that for a triangle with sides \(a, b, c\) and semiperimeter \(S = \frac{a + b + c}{2}\), we need to show that for all natural numbers \(n\), the inequality
\[
\frac{a^n}{b + c} + \frac{b^n}{c + a} + \frac{c^n}{a + b} \geq \left( \frac{2}{3} \right... | To prove the inequality
$$
\frac{a^n}{b+c} + \frac{b^n}{c+a} + \frac{c^n}{a+b} \geq \left( \frac{2}{3} \right)^{n-2} S^{n-1}
$$
for all natural numbers $ n \in \mathbb{N} $, where $ a, b, c $ are the side lengths of a triangle and $ S = \frac{a + b + c}{2} $ is the semiperimeter, we proceed as follows.
---
### **St... |
673385a052d7e82fe11d8269 | IMO Shortlist | 1987 | Let $x_1,x_2,\ldots,x_n$ be real numbers satisfying $x_1^2+x_2^2+\ldots+x_n^2=1$. Prove that for every integer $k\ge2$ there are integers $a_1,a_2,\ldots,a_n$, not all zero, such that $|a_i|\le k-1$ for all $i$, and $|a_1x_1+a_2x_2+\ldots+a_nx_n|\le{(k-1)\sqrt n\over k^n-1}$. (IMO Problem 3)
Proposed by Germany, FR | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so I need to prove this inequality for the IMO problem. Let's start by understanding what's being asked. We have real numbers x₁ to xₙ with sum of squares 1, and for any k ≥ 2, we need... | To prove the inequality for the given problem, we proceed as follows:
---
### **Problem Restatement**
Let $ x_1, x_2, \ldots, x_n $ be real numbers satisfying
$$
x_1^2 + x_2^2 + \cdots + x_n^2 = 1.
$$
We are to show that for every integer $ k \geq 2 $, there exist integers $ a_1, a_2, \ldots, a_n $, not all zero, ... |
673385a052d7e82fe11d8271 | IMO Shortlist | 1987 | Prove that for every natural number $k$ ($k \geq 2$) there exists an irrational number $r$ such that for every natural number $m$,
$$[r^m] \equiv -1 \pmod k .$$
Remark. An easier variant: Find $r$ as a root of a polynomial of second degree with integer coefficients.
Proposed by Yugoslavia. | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so the problem is to prove that for every natural number k ≥ 2, there's an irrational r such that for every natural number m, the floor of r^m is congruent to -1 mod k, which is the sa... | To solve the problem, we aim to prove that for every natural number $ k \geq 2 $, there exists an irrational number $ r $ such that for all natural numbers $ m $,
$$
\lfloor r^m \rfloor \equiv -1 \pmod{k}.
$$
---
### **Key Insight and Strategy**
We will construct such a number $ r $ as the larger root of a quadrati... |
673385a052d7e82fe11d8277 | IMO Shortlist | 1986 | Find four positive integers each not exceeding $70000$ and each having more than $100$ divisors. | Okay, let's try to tackle this problem: find four positive integers each not exceeding 70,000, each with more than 100 divisors. First, I need to remember how the number of divisors works. The number of positive divisors of a number is determined by its prime factorization. If a number \( n \) has a prime factorization... | To solve the problem of finding **four positive integers each not exceeding 70,000 and each having more than 100 divisors**, we begin by recalling a key number-theoretic principle:
---
### **Divisor Count Formula**
For any positive integer $ n $ with prime factorization
$$
n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k},... |
673385a052d7e82fe11d826b | IMO Shortlist | 1987 | Prove that there exists a four-coloring of the set $M = \{1, 2, \cdots, 1987\}$ such that any arithmetic progression with $10$ terms in the set $M$ is not monochromatic.
Alternative formulation
Let $M = \{1, 2, \cdots, 1987\}$. Prove that there is a function $f : M \to \{1, 2, 3, 4\}$ that is not constant on every se... | Okay, let's try to tackle this problem. We need to prove that there's a 4-coloring of the set {1, 2,..., 1987} such that no 10-term arithmetic progression (AP) is monochromatic. First, let's recall some concepts about colorings and arithmetic progressions.
I remember that van der Waerden's theorem states that for any ... | To prove that there exists a four-coloring of the set $ M = \{1, 2, \ldots, 1987\} $ such that no arithmetic progression (AP) with 10 terms is monochromatic, we employ the **probabilistic method**, a standard and rigorous technique in combinatorics for establishing existence without constructing an explicit example.
-... |
673385a052d7e82fe11d828e | IMO Shortlist | 1985 | Find a method by which one can compute the coefficients of $P(x) = x^6 + a_1x^5 + \cdots+ a_6$ from the roots of $P(x) = 0$ by performing not more than $15$ additions and $15$ multiplications. | Okay, let's try to figure out how to compute the coefficients of a 6th-degree polynomial from its roots using at most 15 additions and 15 multiplications. First, I remember that for a monic polynomial, the coefficients are related to the elementary symmetric sums of the roots. Let's denote the roots as \( r_1, r_2, r_3... | To compute the coefficients of the monic polynomial
$$
P(x) = x^6 + a_1x^5 + a_2x^4 + a_3x^3 + a_4x^2 + a_5x + a_6,
$$
from its six roots $ r_1, r_2, \dots, r_6 $, we use the **elementary symmetric sums** of the roots. These symmetric sums are directly related to the coefficients of the polynomial via the identity:... |
673385a052d7e82fe11d82bf | IMO Shortlist | 1983 | Find the greatest integer less than or equal to $\sum_{k=1}^{2^{1983}} k^{\frac{1}{1983} -1}.$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so I need to find the greatest integer less than or equal to the sum from k=1 to 2^1983 of k^(1/1983 - 1). Let me first rewrite the exponent to make it clearer. 1/1983 - 1 is the same ... | To evaluate the greatest integer less than or equal to the sum
$$
\sum_{k=1}^{2^{1983}} k^{\frac{1}{1983} - 1},
$$
we begin by simplifying the exponent. The expression $ k^{\frac{1}{1983} - 1} $ can be rewritten as:
$$
k^{-1 + \frac{1}{1983}} = \frac{k^{1/1983}}{k}.
$$
Let us define $ n = 1983 $ for clarity. Then the... |
673385a052d7e82fe11d829a | IMO Shortlist | 1984 | Prove:
(a) There are infinitely many triples of positive integers $m, n, p$ such that $4mn - m- n = p^2 - 1.$
(b) There are no positive integers $m, n, p$ such that $4mn - m- n = p^2.$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so I need to prove two parts here: part (a) says there are infinitely many positive integer triples (m, n, p) satisfying 4mn - m - n = p² - 1, and part (b) says there are none for the ... | To address the given problem, we analyze both parts (a) and (b) using algebraic manipulation, number-theoretic reasoning, and explicit construction of solutions.
---
### **Part (a): There are infinitely many triples of positive integers $ m, n, p $ such that $ 4mn - m - n = p^2 - 1 $.**
We begin by rewriting the equ... |
673385a052d7e82fe11d82cd | IMO Shortlist | 1982 | A box contains $p$ white balls and $q$ black balls. Beside the box there is a pile of black balls. Two balls are taken out of the box. If they have the same color, a black ball from the pile is put into the box. If they have different colors, the white ball is put back into the box. This procedure is repeated until the... | Okay, let's try to tackle this problem step by step. First, let's make sure I understand the problem correctly. We have a box with p white balls and q black balls. The process is: take out two balls. If they're the same color, add a black ball from the pile (so total balls decrease by 1: 2 out, 1 in, net -1). If they'r... | To determine the probability that the last ball remaining in the box is white, we analyze the process step by step and identify a key invariant.
---
### **Understanding the Process**
We begin with a box containing $ p $ white balls and $ q $ black balls. The total number of balls is $ n = p + q $. At each step, two ... |
673385a052d7e82fe11d82d6 | IMO Shortlist | 1981 | Find the minimum value of
$$\max(a + b + c, b + c + d, c + d + e, d + e + f, e + f + g)$$
subject to the constraints
(i) $a, b, c, d, e, f, g \geq 0,$
(ii)$ a + b + c + d + e + f + g = 1.$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so I need to find the minimum value of the maximum of these five sums: a+b+c, b+c+d, c+d+e, d+e+f, e+f+g, where all variables are non-negative and add up to 1. Hmm, let's start by unde... | To find the **minimum value of**
$$
\max(a + b + c, b + c + d, c + d + e, d + e + f, e + f + g)
$$
subject to the constraints:
- $ a, b, c, d, e, f, g \geq 0 $,
- $ a + b + c + d + e + f + g = 1 $,
we approach this as a **minimax problem**, where we aim to minimize the largest among five overlapping sums of consecut... |
673385a052d7e82fe11d82e0 | IMO Shortlist | 1981 | Let $P$ be a polynomial of degree $n$ satisfying
$$P(k) = \binom{n+1}{k}^{-1} \qquad \text{ for } k = 0, 1, . . ., n.$$
Determine $P(n + 1).$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so we have a polynomial P of degree n, and it's defined at n+1 points: k=0 to k=n, where each P(k) is the reciprocal of the binomial coefficient C(n+1, k). We need to find P(n+1). Hmm,... | To solve the problem, we are given a polynomial $ P $ of degree $ n $ satisfying
$$
P(k) = \binom{n+1}{k}^{-1} \quad \text{for } k = 0, 1, \dots, n,
$$
and we are to determine the value of $ P(n+1) $.
---
### Step 1: Understanding the Problem
We are working with a polynomial $ P(x) $ of degree $ n $ that is define... |
673385a052d7e82fe11d82dc | IMO Shortlist | 1981 | A sequence $(a_n)$ is defined by means of the recursion
$$a_1 = 1, a_{n+1} = \frac{1 + 4a_n +\sqrt{1+ 24a_n}}{16}.$$
Find an explicit formula for $a_n.$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so I need to find an explicit formula for this sequence defined by a recursive relation. The recursion is a₁ = 1, and then aₙ₊₁ = [1 + 4aₙ + √(1 + 24aₙ)] / 16. Hmm, recursive sequences... | To find an explicit formula for the sequence $(a_n)$ defined by the recurrence
$$
a_1 = 1, \quad a_{n+1} = \frac{1 + 4a_n + \sqrt{1 + 24a_n}}{16},
$$
we begin by observing that the square root term suggests a substitution might simplify the recurrence. Let us define:
$$
b_n = \sqrt{1 + 24a_n}.
$$
This implies:
$$
... |
673385a052d7e82fe11d830e | IMO Shortlist | 1978 | The set $M = \{1, 2, . . . , 2n\}$ is partitioned into $k$ nonintersecting subsets $M_1,M_2, \dots, M_k,$ where $n \ge k^3 + k.$ Prove that there exist even numbers $2j_1, 2j_2, \dots, 2j_{k+1}$ in $M$ that are in one and the same subset $M_i$ $(1 \le i \le k)$ such that the numbers $2j_1 - 1, 2j_2 - 1, \dots, 2j_{k+1}... | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so we have the set M = {1, 2, ..., 2n}, partitioned into k subsets. The condition is n ≥ k³ + k, and we need to show that there's a subset M_i containing k+1 even numbers, say 2j₁ to 2... | To prove the statement, we begin by analyzing the structure of the set $ M = \{1, 2, \ldots, 2n\} $ and its partition into $ k $ nonintersecting subsets $ M_1, M_2, \ldots, M_k $. The key idea is to focus on the even numbers in $ M $, which form the set $ E = \{2, 4, \ldots, 2n\} $, and their corresponding odd predeces... |
673385a052d7e82fe11d8304 | IMO Shortlist | 1979 | Find all bases of logarithms in which a real positive number can be equal to its logarithm or prove that none exist. | Okay, let's try to tackle this problem: "Find all bases of logarithms in which a real positive number can be equal to its logarithm or prove that none exist." Hmm, so we need to find all bases \( b > 0 \), \( b \neq 1 \) (since logarithm bases must be positive and not 1), such that there exists a real positive number \... | To solve the problem of finding all bases $ b > 0 $, $ b \neq 1 $, such that there exists a real positive number $ x > 0 $ satisfying $ x = \log_b x $, we begin by translating the logarithmic equation into an exponential form.
---
### Step 1: Reformulate the Equation
The equation $ x = \log_b x $ is equivalent to:
$... |
673385a052d7e82fe11d8314 | IMO Shortlist | 1978 | Let $S$ be the set of all the odd positive integers that are not multiples of $5$ and that are less than $30m$, $m$ being an arbitrary positive integer. What is the smallest integer $k$ such that in any subset of $k$ integers from $S$ there must be two different integers, one of which divides the other? | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so the problem is about finding the smallest k such that any subset of k elements from set S must contain two numbers where one divides the other. S is all odd positive integers not mu... | To solve the problem, we consider the set $ S $ of all **odd positive integers less than $ 30m $** that are **not multiples of 5**, where $ m $ is an arbitrary positive integer. The goal is to find the **smallest integer $ k $** such that **any subset of $ k $ elements from $ S $ must contain two distinct elements $ a ... |
673385a052d7e82fe11d8339 | IMO Shortlist | 1975 | Let $a_1, a_2, \ldots , a_n, \ldots $ be a sequence of real numbers such that $0 \leq a_n \leq 1$ and $a_n - 2a_{n+1} + a_{n+2} \geq 0$ for $n = 1, 2, 3, \ldots$. Prove that
$$0 \leq (n + 1)(a_n - a_{n+1}) \leq 2 \qquad \text{ for } n = 1, 2, 3, \ldots$$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so we have this sequence of real numbers between 0 and 1 inclusive, and it satisfies this inequality: a_n - 2a_{n+1} + a_{n+2} ≥ 0 for all n ≥ 1. That looks like a second difference co... | We are given a sequence $ a_1, a_2, \ldots $ of real numbers such that $ 0 \leq a_n \leq 1 $ for all $ n \geq 1 $, and it satisfies the inequality:
$$
a_n - 2a_{n+1} + a_{n+2} \geq 0 \quad \text{for all } n \geq 1.
$$
Our goal is to prove that for all $ n \geq 1 $,
$$
0 \leq (n + 1)(a_n - a_{n+1}) \leq 2.
$$
---
#... |
673385a052d7e82fe11d8349 | IMO Shortlist | 1974 | Let $a_i, b_i$ be coprime positive integers for $i = 1, 2, \ldots , k$, and $m$ the least common multiple of $b_1, \ldots , b_k$. Prove that the greatest common divisor of $a_1 \frac{m}{b_1} , \ldots, a_k \frac{m}{b_k}$ equals the greatest common divisor of $a_1, \ldots , a_k.$ | Okay, let's try to tackle this problem. The problem says: Let \( a_i, b_i \) be coprime positive integers for \( i = 1, 2, \ldots, k \), and \( m \) the least common multiple of \( b_1, \ldots, b_k \). We need to prove that the greatest common divisor of \( a_1 \frac{m}{b_1}, \ldots, a_k \frac{m}{b_k} \) equals the gre... | To prove that the greatest common divisor (gcd) of the numbers $ a_1 \frac{m}{b_1}, a_2 \frac{m}{b_2}, \ldots, a_k \frac{m}{b_k} $ equals the gcd of $ a_1, a_2, \ldots, a_k $, where $ m = \text{lcm}(b_1, b_2, \ldots, b_k) $ and each pair $ (a_i, b_i) $ is coprime, we proceed as follows.
---
### **Step 1: Define Key Q... |
673385a052d7e82fe11d833a | IMO Shortlist | 1975 | Let $M$ be the set of all positive integers that do not contain the digit $9$ (base $10$). If $x_1, \ldots , x_n$ are arbitrary but distinct elements in $M$, prove that
$$\sum_{j=1}^n \frac{1}{x_j} \lt 80 .$$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so the problem is about the set M of all positive integers without any digit 9, and we need to show that the sum of reciprocals of any distinct elements from M is less than 80. Wait, n... | To prove that for any finite set of distinct positive integers $ x_1, x_2, \ldots, x_n $ from the set $ M $ (where $ M $ is the set of all positive integers that do **not** contain the digit 9 in base 10), the sum
$$
\sum_{j=1}^n \frac{1}{x_j} < 80,
$$
we proceed by analyzing the **total sum of reciprocals over all e... |
673385a252d7e82fe11d8733 | IMO Longlists | 1972 | $f$ and $g$ are real-valued functions defined on the real line. For all $x$ and $y, f(x+y)+f(x-y)=2f(x)g(y)$. $f$ is not identically zero and $|f(x)|\le1$ for all $x$. Prove that $|g(x)|\le1$ for all $x$. | Okay, let's try to tackle this problem. So, we have two real-valued functions f and g defined on the real line. The functional equation given is f(x + y) + f(x - y) = 2f(x)g(y) for all real numbers x and y. We know that f isn't identically zero, and |f(x)| ≤ 1 for all x. We need to prove that |g(x)| ≤ 1 for all x.
Fir... | To prove that $ |g(x)| \leq 1 $ for all real $ x $, we begin by analyzing the given functional equation:
$$
f(x + y) + f(x - y) = 2f(x)g(y), \quad \text{for all } x, y \in \mathbb{R}.
$$
We are told that $ f $ is not identically zero and satisfies $ |f(x)| \leq 1 $ for all $ x $. Our goal is to show that $ |g(x)| \le... |
673385a252d7e82fe11d8735 | IMO Longlists | 1972 | The least number is $m$ and the greatest number is $M$ among $ a_1 ,a_2 ,\ldots,a_n$ satisfying $ a_1 +a_2 +...+a_n =0$. Prove that
$$ a_1^2 +\cdots +a_n^2 \le-nmM$$ | Okay, let's try to tackle this problem. The problem states that we have a set of numbers \( a_1, a_2, \ldots, a_n \) with sum zero, where \( m \) is the least number and \( M \) is the greatest number. We need to prove that the sum of their squares is at most \( -n m M \).
First, let's recall some basic concepts. Sinc... | To prove the inequality
$$
a_1^2 + a_2^2 + \cdots + a_n^2 \leq -n m M,
$$
where $ m = \min\{a_1, a_2, \ldots, a_n\} $, $ M = \max\{a_1, a_2, \ldots, a_n\} $, and $ a_1 + a_2 + \cdots + a_n = 0 $, we proceed as follows.
---
### **Key Observation**
Given that $ m \leq a_i \leq M $ for all $ i $, it follows that:
$... |
673385a252d7e82fe11d8757 | IMO Longlists | 1971 | Knowing that the system
$$x + y + z = 3,$$$$x^3 + y^3 + z^3 = 15,$$$$x^4 + y^4 + z^4 = 35,$$
has a real solution $x, y, z$ for which $x^2 + y^2 + z^2 \lt 10$, find the value of $x^5 + y^5 + z^5$ for that solution. | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so I need to find x⁵ + y⁵ + z⁵ given these three equations: sum of variables, sum of cubes, sum of fourth powers, and a condition that sum of squares is less than 10. Hmm, symmetric eq... | To solve the problem, we are given the following symmetric equations involving three real variables $ x, y, z $:
$$
x + y + z = 3, \quad x^3 + y^3 + z^3 = 15, \quad x^4 + y^4 + z^4 = 35,
$$
and the condition that $ x^2 + y^2 + z^2 < 10 $. We are to find $ x^5 + y^5 + z^5 $ for this solution.
---
### Step 1: Use Pow... |
673385a252d7e82fe11d8776 | IMO Longlists | 1971 | Prove the inequality
$$ \frac{a_1+ a_3}{a_1 + a_2} + \frac{a_2 + a_4}{a_2 + a_3} + \frac{a_3 + a_1}{a_3 + a_4} + \frac{a_4 + a_2}{a_4 + a_1} \geq 4, $$
where $a_i \gt 0, i = 1, 2, 3, 4.$ | Okay, let's try to prove this inequality. The problem is to show that for positive real numbers \(a_1, a_2, a_3, a_4\), the sum
\[
\frac{a_1 + a_3}{a_1 + a_2} + \frac{a_2 + a_4}{a_2 + a_3} + \frac{a_3 + a_1}{a_3 + a_4} + \frac{a_4 + a_2}{a_4 + a_1} \geq 4.
\]
First, I need to recall some inequalities that might hel... | To prove the inequality
$$
\frac{a_1 + a_3}{a_1 + a_2} + \frac{a_2 + a_4}{a_2 + a_3} + \frac{a_3 + a_1}{a_3 + a_4} + \frac{a_4 + a_2}{a_4 + a_1} \geq 4,
$$
where $ a_i > 0 $ for all $ i = 1, 2, 3, 4 $, we proceed with a systematic application of the **harmonic mean inequality** (a consequence of the AM–GM inequality)... |
673385a252d7e82fe11d87e8 | IMO Longlists | 1967 | Prove that all numbers of the sequence $$ \frac{107811}{3}, \quad \frac{110778111}{3}, \frac{111077781111}{3}, \quad \ldots $$ are exact cubes. | Okay, let's try to tackle this problem. We need to prove that all numbers in the sequence given are exact cubes. The sequence starts with 107811/3, then 110778111/3, then 111077781111/3, and so on. First, let's compute the first few terms to see what they are numerically, maybe that will help spot a pattern.
First ter... | To prove that all numbers in the sequence
$$
\frac{107811}{3}, \quad \frac{110778111}{3}, \quad \frac{111077781111}{3}, \quad \ldots
$$
are exact cubes, we begin by analyzing the structure of the numerators and identifying a general pattern.
---
### Step 1: Understanding the Numerator Pattern
Let us denote the $... |
673385a252d7e82fe11d87fc | IMO Longlists | 1967 | Without using tables, find the exact value of the product:
$$P = \prod^7_{k=1} \cos \left(\frac{k \pi}{15} \right).$$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so I need to find the product of cosines from k=1 to 7 of (kπ)/15, without using tables. Hmm, products of cosines... I remember there are some trigonometric identities for products, ma... | To find the exact value of the product
$$
P = \prod_{k=1}^{7} \cos\left(\frac{k\pi}{15}\right),
$$
we proceed using a combination of trigonometric identities and known results about products of sines and cosines.
---
### Step 1: Convert Cosines to Sines via Complementary Angles
We use the identity:
$$
\cos\theta... |
673385a252d7e82fe11d880b | IMO Longlists | 1967 | Prove that for arbitrary positive numbers the following inequality holds
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq \frac{a^8 + b^8 + c^8}{a^3b^3c^3}.$$ | Okay, let's try to tackle this inequality: for arbitrary positive numbers \(a, b, c\), we need to prove that \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq \frac{a^8 + b^8 + c^8}{a^3b^3c^3}\). First, let's rewrite the right-hand side (RHS) to make it clearer. The RHS is \(\frac{a^8}{a^3b^3c^3} + \frac{b^8}{a^3b^3c^3} +... | To prove the inequality for arbitrary positive real numbers $a, b, c$:
$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq \frac{a^8 + b^8 + c^8}{a^3b^3c^3},
$$
we begin by simplifying both sides. Multiplying both sides of the inequality by $a^3b^3c^3$ (which is positive, so the inequality direction remains unchanged), w... |
673385a252d7e82fe11d8818 | IMO Longlists | 1967 | The function $\varphi(x,y,z)$ defined for all triples $(x,y,z)$ of real numbers, is such that there are two functions $f$ and $g$ defined for all pairs of real numbers, such that
$$\varphi(x,y,z) = f(x+y,z) = g(x,y+z)$$
for all real numbers $x,y$ and $z.$ Show that there is a function $h$ of one real variable, such tha... | Okay, let's try to tackle this problem. So, we have a function φ(x, y, z) that can be written in two different ways: as f(x + y, z) where f is a function of two variables, and also as g(x, y + z) where g is another function of two variables. We need to show that φ must actually depend only on the sum x + y + z, meaning... | We are given that the function $\varphi(x, y, z)$ satisfies the identity
$$
\varphi(x, y, z) = f(x + y, z) = g(x, y + z)
$$
for all real numbers $x$, $y$, and $z$, where $f$ and $g$ are functions of two real variables. Our goal is to show that there exists a single-variable function $h$ such that
$$
\varphi(x, y, z)... |
673385a252d7e82fe11d87cd | IMO Longlists | 1969 | $(HUN 4)$IMO2 If $a_1, a_2, . . . , a_n$ are real constants, and if $y = \cos(a_1 + x) +2\cos(a_2+x)+ \cdots+ n \cos(a_n + x)$ has two zeros $x_1$ and $x_2$ whose difference is not a multiple of $\pi$, prove that $y = 0.$ | Okay, let's try to tackle this problem. The problem says: If \( a_1, a_2, \ldots, a_n \) are real constants, and \( y = \cos(a_1 + x) + 2\cos(a_2 + x) + \cdots + n\cos(a_n + x) \) has two zeros \( x_1 \) and \( x_2 \) whose difference is not a multiple of \( \pi \), prove that \( y = 0 \).
First, let's recall that \( ... | To solve the problem, we begin by analyzing the structure of the function:
$$
y(x) = \cos(a_1 + x) + 2\cos(a_2 + x) + \cdots + n\cos(a_n + x)
$$
We aim to prove that if this function has two zeros $ x_1 $ and $ x_2 $ such that their difference is **not** a multiple of $ \pi $, then $ y(x) = 0 $ for all $ x $.
---
#... |
673385a252d7e82fe11d87dd | IMO Longlists | 1969 | $(SWE 1)$ Six points $P_1, . . . , P_6$ are given in $3-$dimensional space such that no four of them lie in the same plane. Each of the line segments $P_jP_k$ is colored black or white. Prove that there exists one triangle $P_jP_kP_l$ whose edges are of the same color. | Okay, let's try to tackle this problem. So, we have six points in 3D space, no four coplanar, and every edge (line segment between two points) is colored either black or white. We need to prove there's a monochromatic triangle, meaning all three edges of the triangle are the same color.
First, I remember that this is ... | We are given six points $ P_1, P_2, \ldots, P_6 $ in 3-dimensional space such that no four lie in the same plane. Each line segment $ P_jP_k $ (for $ j < k $) is colored either black or white. We are to prove that there exists a triangle $ P_jP_kP_l $ whose edges are all of the same color.
---
### Step-by-Step Reason... |
673385a252d7e82fe11d87de | IMO Longlists | 1969 | $(SWE 2)$ For each $\lambda (0 \lt \lambda \lt 1$ and $\lambda = \frac{1}{n}$ for all $n = 1, 2, 3, \cdots)$, construct a continuous function $f$ such that there do not exist $x, y$ with $0 \lt \lambda \lt y = x + \lambda \le 1$ for which $f(x) = f(y).$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so the problem is: for each λ between 0 and 1, where λ is 1/n for some positive integer n (wait, no, actually it says λ = 1/n for all n=1,2,3,... but also 0 < λ < 1, so λ=1/2,1/3,... b... | The problem asks us to construct a continuous function $ f $ on the interval $[0,1]$ such that for each $ \lambda = \frac{1}{n} $ with $ n \in \mathbb{Z}_{\geq 2} $ (i.e., $ \lambda \in (0,1) $), there do **not** exist points $ x, y $ satisfying $ 0 < \lambda < y = x + \lambda \leq 1 $ and $ f(x) = f(y) $. That is, we ... |
673385a252d7e82fe11d87b6 | IMO Longlists | 1969 | $(BEL 6)$ Evaluate $\left(\cos\frac{\pi}{4} + i \sin\frac{\pi}{4}\right)^{10}$ in two different ways and prove that $\dbinom{10}{1}-\dbinom{10}{3}+\frac{1}{2}\dbinom{10}{5}=2^4$ | Okay, let's tackle this problem step by step. First, the problem says to evaluate \(\left(\cos\frac{\pi}{4} + i \sin\frac{\pi}{4}\right)^{10}\) in two different ways and then prove that \(\dbinom{10}{1} - \dbinom{10}{3} + \frac{1}{2}\dbinom{10}{5} = 2^4\). Let's start with the first part: evaluating the complex number ... | To evaluate the expression $\left(\cos\frac{\pi}{4} + i \sin\frac{\pi}{4}\right)^{10}$ in two different ways and prove the identity:
$$
\binom{10}{1} - \binom{10}{3} + \frac{1}{2}\binom{10}{5} = 2^4,
$$
we proceed as follows.
---
### **First Method: Using De Moivre’s Theorem**
De Moivre's Theorem states that for a... |
673385a252d7e82fe11d87e0 | IMO Longlists | 1969 | Which natural numbers can be expressed as the difference of squares of two integers? | Okay, let's try to figure out which natural numbers can be expressed as the difference of squares of two integers. First, I need to recall that the difference of squares formula is a² - b² = (a - b)(a + b). So, if a number n can be written as a difference of squares, then n must be expressible as a product of two integ... | To determine which natural numbers can be expressed as the difference of squares of two integers, we begin by recalling the algebraic identity:
$$
a^2 - b^2 = (a - b)(a + b)
$$
Let $ n = a^2 - b^2 $. Then $ n = x \cdot y $, where $ x = a - b $ and $ y = a + b $, with $ x \leq y $ and both $ x $ and $ y $ integers. Fo... |
673385a252d7e82fe11d8816 | IMO Longlists | 1967 | Determine all positive roots of the equation $ x^x = \frac{1}{\sqrt{2}}.$ | Okay, let's try to figure out how to solve the equation \( x^x = \frac{1}{\sqrt{2}} \) for positive roots. First, I need to recall that \( x^x \) is defined for positive real numbers \( x \), so we're only looking at \( x > 0 \).
First, maybe I should take the natural logarithm of both sides to simplify the equation. ... | To determine all positive roots of the equation
$$
x^x = \frac{1}{\sqrt{2}},
$$
we begin by analyzing the function $ f(x) = x^x $ for $ x > 0 $. The goal is to find all $ x > 0 $ such that $ f(x) = \frac{1}{\sqrt{2}} $.
---
### Step 1: Transform the Equation
Take the natural logarithm of both sides:
$$
\ln(x^x)... |
673385a252d7e82fe11d882a | IMO Longlists | 1966 | Find $x$ such that trigonometric
$$\frac{\sin 3x \cos (60^\circ -x)+1}{\sin(60^\circ - 7x) - \cos(30^\circ + x) + m}=0$$
where $m$ is a fixed real number. | Okay, let's try to solve this problem. The question is to find \( x \) such that the given trigonometric expression equals zero. The expression is:
\[
\frac{\sin 3x \cos (60^\circ - x) + 1}{\sin(60^\circ - 7x) - \cos(30^\circ + x) + m} = 0
\]
First, I remember that a fraction is zero if and only if the numerator is... | To solve the given trigonometric equation:
$$
\frac{\sin 3x \cos (60^\circ - x) + 1}{\sin(60^\circ - 7x) - \cos(30^\circ + x) + m} = 0,
$$
we begin by recognizing that a fraction equals zero **if and only if** its **numerator is zero** and its **denominator is not zero**. Therefore, we focus first on solving the **nu... |
673385a252d7e82fe11d8811 | IMO Longlists | 1967 | The equation
$$x^5 + 5 \lambda x^4 - x^3 + (\lambda \alpha - 4)x^2 - (8 \lambda + 3)x + \lambda \alpha - 2 = 0$$
is given. Determine $\alpha$ so that the given equation has exactly (i) one root or (ii) two roots, respectively, independent from $\lambda.$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so we have this quintic equation with parameters λ and α, and we need to find α such that the equation has exactly one root or exactly two roots, both independent of λ. Wait, "independ... | To determine the value of $\alpha$ such that the given quintic equation has exactly (i) one root or (ii) two roots, respectively, **independent from $\lambda$**, we begin by analyzing the structure of the polynomial:
$$
P(x, \lambda) = x^5 + 5\lambda x^4 - x^3 + (\lambda \alpha - 4)x^2 - (8\lambda + 3)x + \lambda \alp... |
673385a252d7e82fe11d87c8 | IMO Longlists | 1969 | $(GDR 2)^{IMO1}$ Prove that there exist infinitely many natural numbers $a$ with the following property: The number $z = n^4 + a$ is not prime for any natural number $n.$ | Okay, let's try to tackle this problem: Prove that there exist infinitely many natural numbers \( a \) such that \( z = n^4 + a \) is not prime for any natural number \( n \).
First, I need to recall that a natural number \( z \) is not prime if it's either 1, composite, or negative, but since \( n \) and \( a \) are ... | To prove that there exist infinitely many natural numbers $ a $ such that the number $ z = n^4 + a $ is not prime for any natural number $ n $, we analyze the structure of the expression $ n^4 + a $ and seek to construct a family of values for $ a $ that guarantees $ z $ is always composite.
---
### **Key Insight: Fa... |
673385a252d7e82fe11d880c | IMO Longlists | 1967 | Does there exist an integer such that its cube is equal to $3n^2 + 3n + 7,$ where $n$ is an integer. | Okay, let's try to figure out if there's an integer k such that k³ = 3n² + 3n + 7 for some integer n. First, I need to analyze the equation k³ = 3n² + 3n + 7. Maybe I can rearrange or complete the square on the right-hand side to see if that helps. Let's look at the quadratic in n: 3n² + 3n + 7. Let's factor out the 3 ... | To determine whether there exists an integer $ k $ such that its cube equals $ 3n^2 + 3n + 7 $ for some integer $ n $, we analyze the equation:
$$
k^3 = 3n^2 + 3n + 7
$$
---
### Step 1: Analyze the Right-Hand Side Modulo 9
We begin by examining the right-hand side modulo 9. The expression $ 3n^2 + 3n + 7 $ can be s... |
673385a252d7e82fe11d8837 | IMO Longlists | 1966 | There are $n\geq 2$ people at a meeting. Show that there exist two people at the meeting who have the same number of friends among the persons at the meeting. (It is assumed that if $A$ is a friend of $B,$ then $B$ is a friend of $A;$ moreover, nobody is his own friend.) | Okay, let's try to figure out this problem. So, we have n people at a meeting, where n is at least 2. We need to show that there must be two people who have the same number of friends. The rules are that friendship is mutual (if A is friends with B, then B is friends with A) and no one is friends with themselves.
Firs... | To solve the problem, we analyze the possible number of friends each person can have at a meeting with $ n \geq 2 $ people. The key lies in understanding the constraints on the number of friends and applying the **pigeonhole principle**.
---
### Step 1: Understanding the Constraints
Each person can be friends with a... |
673385a252d7e82fe11d87b7 | IMO Longlists | 1969 | $(BUL 1)$ Prove that the equation $\sqrt{x^3 + y^3 + z^3}=1969$ has no integral solutions. | Okay, let's try to tackle this problem: proving that the equation √(x³ + y³ + z³) = 1969 has no integral solutions. First, I need to understand what this equation means. If we square both sides, it's equivalent to x³ + y³ + z³ = 1969². So, we need to show that 1969 squared cannot be expressed as the sum of three intege... | To prove that the equation
$$
\sqrt{x^3 + y^3 + z^3} = 1969
$$
has no integral solutions, we begin by squaring both sides to eliminate the square root. This gives the equivalent equation:
$$
x^3 + y^3 + z^3 = 1969^2.
$$
---
### Step 1: Compute $ 1969^2 \mod 9 $
We first determine the value of $ 1969^2 \mod 9 $... |
673385a252d7e82fe11d87ed | IMO Longlists | 1967 | Solve the system of equations:
$
\begin{matrix}
|x+y| + |1-x| = 6 \\ |x+y+1| + |1-y| = 4.
\end{matrix}
$ | Okay, let's try to solve this system of equations involving absolute values. The system is:
\[
\begin{cases}
|x + y| + |1 - x| = 6 \\
|x + y + 1| + |1 - y| = 4
\end{cases}
\]
First, absolute value equations can be tricky because they depend on the sign of the expressions inside. So, I need to consider different ... | To solve the system of equations:
$$
\begin{cases}
|x + y| + |1 - x| = 6 \\
|x + y + 1| + |1 - y| = 4,
\end{cases}
$$
we begin by introducing a substitution to simplify the expressions. Let $ u = x + y $. Then the system becomes:
$$
\begin{cases}
|u| + |1 - x| = 6 \\
|u + 1| + |1 - y| = 4.
\end{cases}
$$
Since $ u ... |
673385a252d7e82fe11d882e | IMO Longlists | 1966 | Let $a_1, a_2, \ldots, a_n$ be positive real numbers. Prove the inequality
$$\binom n2 \sum_{i \lt j} \frac{1}{a_ia_j} \geq 4 \left( \sum_{i \lt j} \frac{1}{a_i+a_j} \right)^2$$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so I need to prove this inequality for positive real numbers a₁ to aₙ: the binomial coefficient n choose 2 times the sum over i<j of 1/(aᵢaⱼ) is at least 4 times the square of the sum ... | To prove the inequality
$$
\binom{n}{2} \sum_{i < j} \frac{1}{a_i a_j} \geq 4 \left( \sum_{i < j} \frac{1}{a_i + a_j} \right)^2,
$$
for all positive real numbers $ a_1, a_2, \ldots, a_n $, we proceed by combining two standard inequalities: the AM-GM inequality and the Cauchy-Schwarz inequality.
---
### **Step 1: Appl... |
673385a252d7e82fe11d8835 | IMO Longlists | 1966 | Prove that the volume $V$ and the lateral area $S$ of a right circular cone satisfy the inequality
$$\left( \frac{6V}{\pi}\right)^2 \leq \left( \frac{2S}{\pi \sqrt 3}\right)^3$$
When does equality occur? | Okay, let's try to tackle this problem. We need to prove that for a right circular cone, the volume V and lateral area S satisfy the inequality (6V/π)² ≤ (2S/(π√3))³, and find when equality holds. First, let's recall the formulas for volume and lateral (curved) surface area of a right circular cone.
A right circular c... | To prove the inequality
$$
\left( \frac{6V}{\pi} \right)^2 \leq \left( \frac{2S}{\pi \sqrt{3}} \right)^3,
$$
we begin by expressing the volume $ V $ and lateral surface area $ S $ of a right circular cone in terms of its base radius $ r $ and height $ h $. The slant height $ l $ is given by $ l = \sqrt{r^2 + h^2} $... |
673385a252d7e82fe11d87a0 | IMO Longlists | 1970 | We have $0\le x_i \lt b$ for $i=0,1,\ldots,n$ and $x_n \gt 0,x_{n-1} \gt 0$. If $a \gt b$, and $x_nx_{n-1}\ldots x_0$ represents the number $A$ base $a$ and $B$ base $b$, whilst $x_{n-1}x_{n-2}\ldots x_0$ represents the number $A'$ base $a$ and $B'$ base $b$, prove that $A'B \lt AB'$. | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, let's tackle this problem step by step. First, let's make sure I understand what's being asked. We have digits \( x_0, x_1, \ldots, x_n \) where each \( x_i \) is between 0 (inclusive)... | We are given a set of digits $ x_0, x_1, \ldots, x_n $ such that $ 0 \leq x_i < b $ for all $ i $, with $ x_n > 0 $ and $ x_{n-1} > 0 $. The number represented by the digit string $ x_n x_{n-1} \ldots x_0 $ in base $ a $ is denoted $ A $, and in base $ b $ it is denoted $ B $. Similarly, the digit string $ x_{n-1} x_{n... |
673385a252d7e82fe11d8809 | IMO Longlists | 1967 | Prove the identity $$\sum\limits_{k=0}^n\binom{n}{k}\left(\tan\frac{x}{2}\right)^{2k}\left(1+\frac{2^k}{\left(1-\tan^2\frac{x}{2}\right)^k}\right)=\sec^{2n}\frac{x}{2}+\sec^n x$$for any natural number $n$ and any angle $x.$ | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so I need to prove this identity: the sum from k=0 to n of binomial(n,k) times (tan(x/2))^(2k) times [1 + 2^k / (1 - tan²(x/2))^k] equals sec^(2n)(x/2) + sec^n(x). Hmm, let's start by ... | To prove the identity
$$
\sum_{k=0}^n \binom{n}{k} \left( \tan \frac{x}{2} \right)^{2k} \left( 1 + \frac{2^k}{\left(1 - \tan^2 \frac{x}{2} \right)^k} \right) = \sec^{2n} \frac{x}{2} + \sec^n x,
$$
we proceed by simplifying both sides using a substitution and applying the binomial theorem.
---
### Step 1: Substituti... |
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