This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

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Found problems: 7

2004 Pan African, 2

Tags: PAMO , 2004
Each of the digits $1$, $3$, $7$ and $9$ occurs at least once in the decimal representation of some positive integers. Prove that one can permute the digits of this integer such that the resulting integer is divisible by $7$.

2004 Germany Team Selection Test, 1

A function $f$ satisfies the equation \[f\left(x\right)+f\left(1-\frac{1}{x}\right)=1+x\] for every real number $x$ except for $x = 0$ and $x = 1$. Find a closed formula for $f$.

2004 Germany Team Selection Test, 2

In a triangle $ABC$, let $D$ be the midpoint of the side $BC$, and let $E$ be a point on the side $AC$. The lines $BE$ and $AD$ meet at a point $F$. Prove: If $\frac{BF}{FE}=\frac{BC}{AB}+1$, then the line $BE$ bisects the angle $ABC$.

2004 All-Russian Olympiad, 4

Let $n > 3$ be a natural number, and let $x_1$, $x_2$, ..., $x_n$ be $n$ positive real numbers whose product is $1$. Prove the inequality \[ \frac {1}{1 + x_1 + x_1\cdot x_2} + \frac {1}{1 + x_2 + x_2\cdot x_3} + ... + \frac {1}{1 + x_n + x_n\cdot x_1} > 1. \]

2004 Germany Team Selection Test, 1

A function $f$ satisfies the equation \[f\left(x\right)+f\left(1-\frac{1}{x}\right)=1+x\] for every real number $x$ except for $x = 0$ and $x = 1$. Find a closed formula for $f$.

2004 Germany Team Selection Test, 2

In a triangle $ABC$, let $D$ be the midpoint of the side $BC$, and let $E$ be a point on the side $AC$. The lines $BE$ and $AD$ meet at a point $F$. Prove: If $\frac{BF}{FE}=\frac{BC}{AB}+1$, then the line $BE$ bisects the angle $ABC$.

2004 Romania Team Selection Test, 12

Let $n\geq 2$ be an integer and let $a_1,a_2,\ldots,a_n$ be real numbers. Prove that for any non-empty subset $S\subset \{1,2,3,\ldots, n\}$ we have \[ \left( \sum_{i \in S} a_i \right)^2 \leq \sum_{1\leq i \leq j \leq n } (a_i + \cdots + a_j ) ^2 . \] [i]Gabriel Dospinescu[/i]