Six points are chosen on the sides of an equilateral triangle $ABC$: $A_1$, $A_2$ on $BC$, $B_1$, $B_2$ on $CA$ and $C_1$, $C_2$ on $AB$, such that they are the vertices of a convex hexagon $A_1A_2B_1B_2C_1C_2$ with equal side lengths. Prove that the lines $A_1B_2$, $B_1C_2$ and $C_1A_2$ are concurrent. Bogdan Enescu, Romania

Let $a_1,a_2,\ldots$ be a sequence of integers with infinitely many positive and negative terms. Suppose that for every positive integer $n$ the numbers $a_1,a_2,\ldots,a_n$ leave $n$ different remainders upon division by $n$. Prove that every integer occurs exactly once in the sequence $a_1,a_2,\ldots$.

Let $x,y,z$ be three positive reals such that $xyz\geq 1$. Prove that \[ \frac { x^5-x^2 }{x^5+y^2+z^2} + \frac {y^5-y^2}{x^2+y^5+z^2} + \frac {z^5-z^2}{x^2+y^2+z^5} \geq 0 . \] Hojoo Lee, Korea

Click for solution We begin by using the standard trick to reduce the problem to the case $xyz=1$. We make the replacements $x=\ell x'$, $y=\ell y'$, $z=\ell z'$, with $x'y'z'=1$, and $\ell \geq 1$. The inequality becomes \[ 3 - \left( \sum { \ell^2 x'^2 } \right) \cdot \left ( \sum \frac 1 { \ell^5 x'^5 + \ell^2 ( y'^2+z'^2 ) } \right) \geq 0 . \] We denote the LHS from the above inequality with $f(\ell)$, $f:[1,+\infty) \to \mathbb{R}$. We work a little bit on the algebra to get a nicer expression for the function $f$: \begin{eqnarray*} f(\ell ) &=& 3 - \ell^2 \left( \sum x'^2 \right) \cdot \frac 1{\ell^2 } \left ( \sum \frac 1 { \ell^3 x'^5 + y'^2 + z'^2 } \right) \\ \ &=& 3 - C \cdot \left ( \frac 1 { \ell^3 x'^5 + y'^2 + z'^2 } + \frac 1 { \ell^3 y'^5 + z'^2 + x'^2 } + \frac 1 { \ell^3 z'^5 + x'^2 + y'^2 } \right) \\ \end{eqnarray*} where $C=\sum x'^2$ is a positive constant. It is easy to see that each of the functions $\ell \mapsto \dfrac 1 { \ell^3 x'^5 + y'^2 + z'^2 }$ is decreasing, so $f$ is an increasing function. So it is enough to prove that $f(1) \geq 0$. We again use a standard trick to get rid of the constraint $x'y'z'=1$ namely we replace them respectively by $\dfrac {bc}{a^2}$, $\dfrac {ca}{b^2}$, $\dfrac{ab}{c^2}$. The inequality then becomes \[ 3 \geq \left( \sum \frac { b^2c^2 }{a^4 } \right) \cdot \left( \sum \frac 1 { \displaystyle \frac {b^5c^5}{a^{10}} + \frac {c^2a^2}{b^4} + \frac {a^2b^2}{c^4} } \right) \Leftrightarrow \] \[ 3 \geq \frac 1{a^4b^4c^4} \left( \sum {b^6c^6} \right) \cdot \left( \sum \frac { a^{10}b^4c^4 }{ b^9c^9 + c^6a^{12} + b^6 a^{12} } \right) \ \Leftrightarrow \] \[ 3 \geq \left( \sum b^6 c^6 \right) \cdot \left( \sum \frac { a^6 }{ b^9c^9 + c^6a^{12} + b^6 a^{12} } \right) \ \Leftrightarrow \] \[ 3 \geq ( n^2p^2 + p^2 m^2 + m^2n^2 ) \cdot \left( \sum \frac {m^2}{n^3p^3 + p^2m^4 + n^2 m^4 } \right) , \] where we denoted $a^3,b^3,c^3$ with $m,n,p$ respectively. At this point we have a sum of squares in the RHS, which is very hard to majorate. So we must turn our attention to the sum of the fractions, namely to try to minorate the denominator. After a few trials one finds that \[ (n^3p^3 + p^2m^4 + n^2 m^4 )( np + p^2 + n^2 ) \geq ( n^2p^2 + p^2 m^2 + m^2n^2 )^2 \] from Cauchy. Using this minoration we have to prove that \[ 3 \geq \frac { \sum ( m^2np + m^2p^2 + m^2n^2 ) } { \sum n^2p^2 } \] which is obviously true as \[\sum n^2p^2 \geq \sum (mn)(mp) \Leftrightarrow \frac 12 \left ( \sum (mn-mp)^2 \right) \geq 0 . \]

Determine all positive integers relatively prime to all the terms of the infinite sequence \[ a_n=2^n+3^n+6^n -1,\ n\geq 1. \]

Let $ABCD$ be a fixed convex quadrilateral with $BC=DA$ and $BC$ not parallel with $DA$. Let two variable points $E$ and $F$ lie of the sides $BC$ and $DA$, respectively and satisfy $BE=DF$. The lines $AC$ and $BD$ meet at $P$, the lines $BD$ and $EF$ meet at $Q$, the lines $EF$ and $AC$ meet at $R$. Prove that the circumcircles of the triangles $PQR$, as $E$ and $F$ vary, have a common point other than $P$.

In a mathematical competition, in which $6$ problems were posed to the participants, every two of these problems were solved by more than $\frac 25$ of the contestants. Moreover, no contestant solved all the $6$ problems. Show that there are at least $2$ contestants who solved exactly $5$ problems each. Radu Gologan and Dan Schwartz