Let $ABC$ be an acute-angled triangle and $D$ a point on the altitude through $C$. Let $E$, $F$, $G$ and $H$ be the midpoints of the segments $AD$, $BD$, $BC$ and $AC$. Show that $E$, $F$, $G$, and $H$ form a rectangle. (G. Anegg, Innsbruck)

In triangle $ABC$ $CL$ is a bisector($L$ lies $AB$) $I$ is center incircle of $ABC$. $G$ is intersection medians. If $a=BC, b=AC, c=AB$ and $CL\perp GI$ then prove that $\frac{a+b+c}{3}=\frac{2ab}{a+b}$

The orthocenter of an acute triangle $ABC$ is $H$. Let $K$ and $P$ be the midpoints of lines $BC$ and $AH$, respectively. The angle bisector drawn from the vertex $A$ of the triangle $ABC$ intersects with line $KP$ at $D$. Prove that $HD\perp AD$.

The equation $ x^3\plus{}6x^2\plus{}11x\plus{}6\equal{}0$ has: $ \textbf{(A)}\ \text{no negative real roots} \qquad \textbf{(B)}\ \text{no positive real roots} \qquad \textbf{(C)}\ \text{no real roots} \\ \textbf{(D)}\ \text{1 positive and 2 negative roots} \qquad \textbf{(E)}\ \text{2 positive and 1 negative root}$

Let $ABC$ be an acute triangle. Let $M_A, M_B$ and $M_C$ be the midpoints of sides $BC,CA$, respectively $AB$. Let $M'_A , M'_B$ and $M'_C$ be the the midpoints of the arcs $BC, CA$ and $AB$ respectively of the circumscriberd circle of triangle $ABC$. Let $P_A$ be the intersection of the straight line $M_BM_C$ and the perpendicular to $M'_BM'_C$ through $A$. Define $P_B$ and $P_C$ similarly. Show that the straight line $M_AP_A, M_BP_B$ and $M_CP_C$ intersect at one point.

Find all pairs of integers $(m,n)$ such that $m^3-n^3=2mn +8$

In acute triangle $ABC$, $\angle{A}<\angle{B}$ and $\angle{A}<\angle{C}$. Let $P$ be a variable point on side $BC$. Points $D$ and $E$ lie on sides $AB$ and $AC$, respectively, such that $BP=PD$ and $CP=PE$. Prove that as $P$ moves along side $BC$, the circumcircle of triangle $ADE$ passes through a fixed point other than $A$.

Two children take turns breaking chocolate bar that is 5*10 squares. They can only break the bar using the divisions between squares and can only do 1 break at a time.. The first player that when breaking the chocolate bar breaks off only a single square wins. Is there a winning strategy for any player?

The incircle of $ \triangle ABC$ touches $ BC$, $ AC$, and $ AB$ at $ A_1$, $ B_1$, and $ C_1$, respectively. The line $ AA_1$ intersects the incircle at $ Q$, again. $ A_1C_1$ and $ A_1B_1$ intersect the line, passing through $ A$ and parallel to $ BC$, at $ P$ and $ R$, respectively. If $ \angle PQC_1 \equal{} 45^\circ$ and $ \angle RQB_1 \equal{} 65^\circ$, then $ \angle PQR$ will be ? $\textbf{(A)}\ 110^\circ \qquad\textbf{(B)}\ 115^\circ \qquad\textbf{(C)}\ 120^\circ \qquad\textbf{(D)}\ 125^\circ \qquad\textbf{(E)}\ 130^\circ$

Let $ABC$ be a triangle. Suppose that the perpendicular bisector of $BC$ meets the circle of diameter $AB$ at a point $D$ at the opposite side of $BC$ with respect to $A$, and meets the circle through $A, C, D$ again at $E$. Prove that $\angle ACE=\angle BCD$. [i]Proposed by José Manuel Guerra and Victor Domínguez[/i]

Suppose $a,b,c>0$ and $abc=64$, show that $$\sum_{cyc}\frac{a^2}{\sqrt{a^3+8}\sqrt{b^3+8}}\ge\frac{2}{3}$$

In space there are given four distinct points $ A $, $ B $, $ C $, $ D $. Prove that the three segments connecting the midpoints of the segments $ AB $ and $ CD $, $ AC $ and $ BD $, $ AD $ and $ BC $ have a common midpoint.

A segment $AB$ of length $100$ is divided into $100$ little segments of length $1$ by $99$ intermediate points. Endpoint $A$ is assigned $0$ and endpoint $B$ is assigned $1$. Gustavo assigns each of the $99$ intermediate points a $0$ or a $1$, at his choice, and then color each segment of length $1$ blue or red, respecting the following rule: [i]The segments that have the same number at their ends are red, and the segments that have different numbers at their ends are blue. [/i] Determine if Gustavo can assign the $0$'s and $1$'s so as to get exactly $30$ blue segments. And $35$ blue segments? (In each case, if the answer is yes, show a distribution of $0$'s and $1$'s, and if the answer is no, explain why).

Let $\Omega$ be a family of subsets of $M$ such that: $(\text{i})$ If $|A\cap B|\ge 2$ for $A,B\in\Omega$, then $A=B$; $(\text{ii})$ There exist different subsets $A,B,C\in\Omega$ with $|A\cap B\cap C|=1$; $(\text{iii})$ For every $A\in\Omega$ and $a\in M \ A$, there is a unique $B\in\Omega$ such that $a\in B$ and $A\cap B=\emptyset$. Prove that there are numbers $p$ and $s$ such that: $(1)$ Each $a\in M$ is contained in exactly $p$ sets in $\Omega$; $(2)$ $|A|=s$ for all $A\in\Omega$; $(3)$ $s+1\ge p$.

A right circular cone has base radius 1 cm and slant height 3 cm is given. $P$ is a point on the circumference of the base and the shortest path from $P$ around the cone and back to $P$ is drawn (see diagram). What is the minimum distance from the vertex $V$ to this path? [asy] import graph; unitsize(1 cm); filldraw(shift(-0.15,0.37)*rotate(17)*yscale(0.3)*xscale(1.41)*(Circle((0,0),1)),gray(0.9),nullpen); draw(yscale(0.3)*(arc((0,0),1.5,0,180)),dashed); draw(yscale(0.3)*(arc((0,0),1.5,180,360))); draw((1.5,0)--(0,4)--(-1.5,0)); draw((0,0)--(1.5,0),Arrows); draw(((1.5,0) + (0.3,0.1))--((0,4) + (0.3,0.1)),Arrows); draw(shift(-0.15,0.37)*rotate(17)*yscale(0.3)*xscale(1.41)*(arc((0,0),1,0,180)),dashed); draw(shift(-0.15,0.37)*rotate(17)*yscale(0.3)*xscale(1.41)*(arc((0,0),1,180,360))); label("$V$", (0,4), N); label("1 cm", (0.75,-0.5), N); label("$P$", (-1.5,0), SW); label("3 cm", (1.7,2)); [/asy]

Let $ABCD$ be a rectangle, $AB = a, BC = b$. Consider the family of parallel and equidistant straight lines (the distance between two consecutive lines being $d$) that are at an the angle $\phi, 0 \leq \phi \leq 90^{\circ},$ with respect to $AB$. Let $L$ be the sum of the lengths of all the segments intersecting the rectangle. Find: [i](a)[/i] how $L $ varies, [i](b)[/i] a necessary and sufficient condition for $L$ to be a constant, and [i](c)[/i] the value of this constant.

Prove that the number $\binom np-\left\lfloor\frac np\right\rfloor$ is divisible by $p$ for every prime number and integer $n\ge p$.

Any positive integer $n$ can be written in the form $n = 2^b(2c+1)$. We call $2c+1$ the[i] odd part[/i] of $n$. Given an odd integer $n > 0$, define the sequence $ a_0, a_1, a_2, ...$ as follows: $a_0 = 2^n-1, a_{k+1} $ is the [i]odd part[/i] of $3a_k+1$. Find $a_n$.

Determine all real numbers $a,b,c,d$ that satisfy the following equations \[\begin{cases} abc + ab + bc + ca + a + b + c = 1\\ bcd + bc + cd + db + b + c + d = 9\\ cda + cd + da + ac + c + d + a = 9\\ dab + da + ab + bd + d + a + b = 9\end{cases}\]

In a group of people, every two of them have exactly one mutual friend in the group. Prove that there is one person who is friends with all the other people in the group. Note: the friendship is mutual, that is, if $X$ is friends with $Y$, then $Y$ is friends with $X$.

Let $a_1,a_2,...,a_9$ be nine real numbers, not necessarily distinct, with average $m$. Let $A$ denote the number of triples $1 \le i < j < k \le 9$ for which $a_i + a_j + a_k \ge 3m$. What is the minimum possible value of $A$? [i]Proposed by Ray Li[/i]

Solve in prime numbers the equation $x^y - y^x = xy^2 - 19$.

$n$ cockroaches are sitting on the plane at the vertices of the regular $ n $ -gon. They simultaneously begin to move at a speed of $ v $ on the sides of the polygon in the direction of the clockwise adjacent cockroach, then they continue moving in the initial direction with the initial speed. Vasya a entomologist moves on a straight line in the plane at a speed of $u$. After some time, it turned out that Vasya has crushed three cockroaches. Prove that $ v = u $.

Two circles $\mathcal{C}_1$ and $\mathcal{C}_2$ intersect each other at two distinct points $M$ and $N$. A common tangent lines touches $\mathcal{C}_1$ at $P$ and $\mathcal{C}_2$ at $Q$, the line being closer to $N$ than to $M$. The line $PN$ meets the circle $\mathcal{C}_2$ again at the point $R$. Prove that the line $MQ$ is a bisector of the angle $\angle{PMR}$.

We color some points in the plane with red, in such way that if $P,Q$ are red and $X$ is a point such that triangle $\triangle PQX$ has angles $30º, 60º, 90º$ in some order, then $X$ is also red. If we have vertices $A, B, C$ all red, prove that the barycenter of triangle $\triangle ABC$ is also red.