Let us consider $a,b$ two integers. Prove that there exists and it is unique a pair of integers $(x,y)$ such that: \[(x+2y-a)^{2}+(2x-y-b)^{2}\leq 1.\]

Let $ABCD$ be a trapezium $(AB \parallel CD)$ and $M,N$ be the intersection points of the circles of diameters $AD$ and $BC$. Prove that $O \in MN$, where $O \in AC \cap BD$.

Click for solution I will also sketch a simple solution: $ O \in MN \Longleftrightarrow OP^{2} - \frac {AD^{2}}{4} = OQ^{2} - \frac {BC^{2}}{4}$, because $ MN$ is the radical axe of the circles, and where $ P,Q$ are respetively the midpoints of $ AD$, $ BC$. But $ OP^{2} = \frac {2(OA^{2} + OD^{2}) - AD^{2}}{4}$, and the same for $ OQ$, and then we optain : $ O \in MN \Longleftrightarrow OA^{2} + OD^{2} - AD^{2} = OB^{2} + OC^{2} - BC^{2}$, but $ AD^{2} = OA^{2} + OB^{2} - 2 OA \cdot OB \cos{\angle{AOB}}$, and then it is obvious.

A rectangularly paper is divided in polygons areas in the following way: at every step one of the existing surfaces is cut by a straight line, obtaining two new areas. Which is the minimum number of cuts needed such that between the obtained polygons there exists $251$ polygons with $11$ sides?

Find the positive integers $n$ with $n \geq 4$ such that $[\sqrt{n}]+1$ divides $n-1$ and $[\sqrt{n}]-1$ divides $n+1$. RemarkThis problem can be solved in a similar way with the one given at Cono Sur Olympiad 2006, problem 5.

Consider a convex quadrilateral $ABCD$. Denote $M, \ N$ the points of tangency of the circle inscribed in $\triangle ABD$ with $AB, \ AD$, respectively and $P, \ Q$ the points of tangency of the circle inscribed in $\triangle CBD$ with the sides $CD, \ CB$, respectively. Assume that the circles inscribed in $\triangle ABD, \ \triangle CBD$ are tangent. Prove that: a) $ABCD$ is circumscriptible. b) $MNPQ$ is cyclic. c) The circles inscribed in $\triangle ABC, \ \triangle ADC$ are tangent.

Let $ABC$ an isosceles triangle, $P$ a point belonging to its interior. Denote $M$, $N$ the intersection points of the circle $\mathcal{C}(A, AP)$ with the sides $AB$ and $AC$, respectively. Find the position of $P$ if $MN+BP+CP$ is minimum.

Let $ABC$ a triangle and $M,N,P$ points on $AB,BC$, respective $CA$, such that the quadrilateral $CPMN$ is a paralelogram. Denote $R \in AN \cap MP$, $S \in BP \cap MN$, and $Q \in AN \cap BP$. Prove that $[MRQS]=[NQP]$.

Click for solution First $[MRQS]=[NQP] \iff [MRN]=[NPS]$. But $\frac{[MRN]}{[MNP]}=\frac{MR}{MP}=\frac{BN}{BC}$. And $\frac{[NPS]}{[MNP]}=\frac{NS}{MN}=\frac{CP}{AC}$. Thus $[MRN]=[NPS] \iff \frac{BN}{BC}=\frac{CP}{AC}=\frac{MN}{AC}$, obvious by Thales. So, the problem is solved.

Solve in positive integers: $(x^{2}+2)(y^{2}+3)(z^{2}+4)=60xyz$.

Consider a $n$x$n$ table such that the unit squares are colored arbitrary in black and white, such that exactly three of the squares placed in the corners of the table are white, and the other one is black. Prove that there exists a $2$x$2$ square which contains an odd number of unit squares white colored.

Let $a, b, c$ three positive reals such that \[\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\geq 1. \] Show that \[a+b+c\geq ab+bc+ca. \]

Find all nonzero subsets $A$ of the set $\left\{2,3,4,5,\cdots\right\}$ such that $\forall n\in A$, we have that $n^{2}+4$ and $\left\lfloor{\sqrt{n}\right\rfloor}+1$ are both in $A$.

Click for solution pohoatza wrote: Find all nonzero subsets $A$ of the set $\left\{2,3,4,5,\cdots\right\}$ such that $\forall n\in A$, we have that $n^{2}+4$ and $\left\lfloor{\sqrt{n}\right\rfloor}+1$ are both in $A$. $n\in A$ $\Rightarrow$ $m=n^{2}+4\ \in A$ $\Rightarrow$ $\left\lfloor{\sqrt{m}\right\rfloor}+1\ \in A$. But, for $n>1$, $\left\lfloor{\sqrt{n^{2}+4}\right\rfloor}+1=n+1$. So, $n\in A$ $\Rightarrow$ $n+1\in A$ $\Rightarrow$ $A=\{a,a+1,a+2, ...\}$ and we must have $\left\lfloor{\sqrt{a}\right\rfloor}+1\geq a\geq 2$, which means $a=2$ Then the only nonempty subset is $\left\{2,3,4,5,\cdots\right\}$ itself. -- Patrick

Let $w_{1}$ and $w_{2}$ be two circles which intersect at points $A$ and $B$. Consider $w_{3}$ another circle which cuts $w_{1}$ in $D,E$, and it is tangent to $w_{2}$ in the point $C$, and also tangent to $AB$ in $F$. Consider $G \in DE \cap AB$, and $H$ the symetric point of $F$ w.r.t $G$. Find $\angle{HCF}$.

Consider the numbers from $1$ to $16$. The "solitar" game consists in the arbitrary grouping of the numbers in pairs and replacing each pair with the great prime divisor of the sum of the two numbers (i.e from $(1,2); (3,4); (5,6);...;(15,16)$ the numbers which result are $3,7,11,5,19,23,3,31$). The next step follows from the same procedure and the games continues untill we obtain only one number. Which is the maximum numbers with which the game ends.

Find all integer positive numbers $n$ such that: $n=[a,b]+[b,c]+[c,a]$, where $a,b,c$ are integer positive numbers and $[p,q]$ represents the least common multiple of numbers $p,q$.

Consider $ \rho$ a semicircle of diameter $ AB$. A parallel to $ AB$ cuts the semicircle at $ C, D$ such that $ AD$ separates $ B, C$. The parallel at $ AD$ through $ C$ intersects the semicircle the second time at $ E$. Let $ F$ be the intersection point of the lines $ BE$ and $ CD$. The parallel through $ F$ at $ AD$ cuts $ AB$ in $ P$. Prove that $ PC$ is tangent to $ \rho$. Author: Cosmin Pohoata

Click for solution Very nice solution, Jan! Although the problem is easy, I will post my solution. Because $ AD \| PF$, we have that $ \angle{PFC} = \angle{ADC} = \angle{BCD}$, thus the quadrilateral $ PBFC$ is an issoscel trapezoid. Therefore $ \angle{PCB} = \angle{PFB}\iff 90 + \angle{PCA} = 90 + \angle{DBF}$ $ \iff \angle{PCA} = \angle{DBF}$. But $ \angle{DBF} = \angle{DAE}$, because the quadrilateral $ ABDE$ is cyclic, and $ \angle{DAE} = \angle{ADC}$. Therefore $ \angle{PCA} = \angle{ADC}$, so $ PC$ is tangent to $ \rho$.

Let $x, y, z \ge 0$ be real numbers. Prove that: \[\frac{x^{3}+y^{3}+z^{3}}{3}\ge xyz+\frac{3}{4}|(x-y)(y-z)(z-x)| .\] Additional taskFind the maximal real constant $\alpha$ that can replace $\frac{3}{4}$ such that the inequality is still true for any non-negative $x,y,z$.

At a party there are eight guests, and each participant can't talk with at most three persons. Prove that we can group the persons in four pairs such that in every pair a conversation can take place.

We call a set of points free if there is no equilateral triangle with the vertices among the points of the set. Prove that every set of $n$ points in the plane contains a free subset with at least $\sqrt{n}$ elements.

Click for solution Another solution: Let $F$ be maximal free subset of set $A$ with $n$ elements. Let $|F|=k$, then $|F'| = n-k$, $F' = A\setminus F$. Connect point $X\in F'$ with unordered pair $\{Y,Z\}\subset F$ iff $XY = XZ=YZ$. Then each point $X$ has degree at least 1 since it would contradict maximality of set $F$ and each pair $\{Y,Z\}$ has degree at most 2 (by contradiction, if pair is connected with three points, say $A,B,C$ then all triangles $AYZ$, $BYZ$ and $CYZ$ are equilateral, so two points from set $\{A,B,C\}$ are equal). Now we have \[\;\;\;\;\;2\cdot{k\choose 2}\geq n-k\] and problem is solved.

Consider an 8x8 board divided in 64 unit squares. We call diagonal in this board a set of 8 squares with the property that on each of the rows and the columns of the board there is exactly one square of the diagonal. Some of the squares of this board are coloured such that in every diagonal there are exactly two coloured squares. Prove that there exist two rows or two columns whose squares are all coloured.

There are given the integers $1 \le m < n$. Consider the set $M = \{ (x,y);x,y \in \mathbb{Z_{+}}, 1 \le x,y \le n \}$. Determine the least value $v(m,n)$ with the property that for every subset $P \subseteq M$ with $|P| = v(m,n)$ there exist $m+1$ elements $A_{i}= (x_{i},y_{i}) \in P, i = 1,2,...,m+1$, for which the $x_{i}$ are all distinct, and $y_{i}$ are also all distinct.

Let $ABC$ be a right triangle with $A = 90^{\circ}$ and $D \in (AC)$. Denote by $E$ the reflection of $A$ in the line $BD$ and $F$ the intersection point of $CE$ with the perpendicular in $D$ to $BC$. Prove that $AF, DE$ and $BC$ are concurrent.

Click for solution Here is another solution to this nice problem: Let $DF\cap BC= O$, $AE\cap BD=M$, $AE\cap BC=N$ and assume $AD\geq DC$. Let $\measuredangle{CBD}=x$, $\measuredangle{DBA}=y$, and $\measuredangle{BCA}=c$. It`s clear that $x+y+c=90^{\circ}$ Because triangle $BAM$ is right-angled, we get $\measuredangle{BAM}=c+x$. Because the quadrilateral $ABDO$ is cyclic, we have $\measuredangle{OAD}=\measuredangle{OBD}=x$, thus $\measuredangle{OAE}=y-x$. In isosceles triangle $ABE$, we have $\measuredangle{BAE}=c+x$, therefore $\measuredangle{ABE}=2y$. But $\measuredangle{ABN}=x+y$, therefore $\measuredangle{EBO}=y-x=\measuredangle{OAE}$. This means that quadrilaterel $OABE$ is cyclic, and thus the pentagon $ADOEB$ is cyclic. This means that $\measuredangle{EDB}=\measuredangle{EAB}=c+x\Longrightarrow \measuredangle{ADE}=2(c+x)\Longrightarrow \measuredangle{EDF}=y-x$. In triangle $ABE$, $\ds N\in (AE)\Longrightarrow \frac{AN}{NE}=\frac{AB}{BE}\cdot\frac{\sin \measuredangle{ABN}}{\sin\measuredangle{EBN}}\Longrightarrow \frac{AN}{NE}=\frac{\sin(x+y)}{\sin(y-x)}$ In triangle $EDC$, $\ds F\in (EC)\Longrightarrow \frac{EF}{FC}=\frac{ED}{DC}\cdot\frac{\sin\measuredangle{EDF}}{\sin\measuredangle{FDC}}\Longrightarrow \frac{EF}{FC}=\frac{ED}{DC}\cdot\frac{\sin(y-x)}{\sin(x+y)}$ By multiplying the last two relations we obtain $\ds \frac{AN}{NE}\cdot\frac{EF}{FC}\cdot\frac{CD}{DA}=\frac{ED}{DA}=1$, therefore, by the converse of Ceva`s theorem, the lines $AF, CN, ED$ are concurent, QED.

We call a real number $x$ with $0 < x < 1$ interesting if $x$ is irrational and if in its decimal writing the first four decimals are equal. Determine the least positive integer $n$ with the property that every real number $t$ with $0 < t < 1$ can be written as the sum of $n$ pairwise distinct interesting numbers.