$AH$ is the altitude of triangle $ABC$ and $H^\prime$ is the reflection of $H$ trough the midpoint of $BC$. If the tangent lines to the circumcircle of $ABC$ at $B$ and $C$, intersect each other at $X$ and the perpendicular line to $XH^\prime$ at $H^\prime$, intersects $AB$ and $AC$ at $Y$ and $Z$ respectively, prove that $\angle ZXC=\angle YXB$.

Let $ ABC$ be an acute angled triangle; let $ D,F$ be the midpoints of $ BC,AB$ respectively. Let the perpendicular from $ F$ to $ AC$ and the perpendicular from $ B$ ti $ BC$ meet in $ N$: Prove that $ ND$ is the circumradius of $ ABC$. [15 points out of 100 for the 6 problems]

Let the angles of a triangle be $\alpha, \beta$, and $\gamma$, the perimeter $2p$ and the radius of the circumcircle $R$. Prove the inequality $\cot^2 \alpha + \cot^2 \beta + \cot^2 \gamma \ge 3 \left(\frac{9R^2}{p^2}-1\right)$. When is the equality achieved?

Semicircles $POQ$ and $ROS$ pass through the center of circle $O$. What is the ratio of the combined areas of the two semicircles to the area of circle $O$? [asy] import graph; size(7.5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-6.27,xmax=10.01,ymin=-5.65,ymax=10.98; draw(circle((0,0),2)); draw((-3,0)--(3,0),EndArrow(6)); draw((0,-3)--(0,3),EndArrow(6)); draw(shift((0.01,1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,179.76,359.76)); draw(shift((-0.01,-1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,-0.38,179.62)); draw((-1.4,1.43)--(1.41,1.41)); draw((-1.42,-1.41)--(1.4,-1.42)); label("$ P(-1,1) $",(-2.57,2.17),SE*lsf); label("$ Q(1,1) $",(1.55,2.21),SE*lsf); label("$ R(-1,-1) $",(-2.72,-1.45),SE*lsf); label("$S(1,-1)$",(1.59,-1.49),SE*lsf); dot((0,0),ds); label("$O$",(-0.24,-0.35),NE*lsf); dot((1.41,1.41),ds); dot((-1.4,1.43),ds); dot((1.4,-1.42),ds); dot((-1.42,-1.41),ds); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy] $ \textbf{(A)}\ \frac{\sqrt 2}4 \qquad\textbf{(B)}\ \frac 12 \qquad\textbf{(C)}\ \frac{2}{\pi} \qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{\sqrt 2}{2} $

Two players by turn paint the circles on the given picture each with his colour. At the end, the rest of the area of each of small triangles is painted by the colour of the majority of vertices of this triangle. The winner is one who gets larger area of his colour (the area of circles is taken into account). Does any of them have winning strategy? If yes, then who wins? \[ \begin{picture}(60,60) \put(5,3){\put(3,0){\line(6,0){8}} \put(17,0){\line(6,0){8}} \put(31,0){\line(6,0){8}} \put(45,0){\line(6,0){8}} \put(10,14){\line(6,0){8}} \put(24,14){\line(6,0){8}} \put(38,14){\line(6,0){8}} \put(17,28){\line(6,0){8}} \put(31,28){\line(6,0){8}} \put(24,42){\line(6,0){8}} \put(1,2){\line(1,2){5}} \put(15,2){\line(1,2){5}} \put(29,2){\line(1,2){5}} \put(43,2){\line(1,2){5}} \put(8,16){\line(1,2){5}} \put(22,16){\line(1,2){5}} \put(36,16){\line(1,2){5}} \put(15,30){\line(1,2){5}} \put(29,30){\line(1,2){5}} \put(22,44){\line(1,2){5}} \put(13,2){\line( \minus{} 1,2){5}} \put(27,2){\line( \minus{} 1,2){5}} \put(41,2){\line( \minus{} 1,2){5}} \put(55,2){\line( \minus{} 1,2){5}} \put(20,16){\line( \minus{} 1,2){5}} \put(34,16){\line( \minus{} 1,2){5}} \put(48,16){\line( \minus{} 1,2){5}} \put(27,30){\line( \minus{} 1,2){5}} \put(41,30){\line( \minus{} 1,2){5}} \put(34,44){\line( \minus{} 1,2){5}} \put(0,0){\circle{6}} \put(14,0){\circle{6}} \put(28,0){\circle{6}} \put(42,0){\circle{6}} \put(56,0){\circle{6}} \put(7,14){\circle{6}} \put(21,14){\circle{6}} \put(35,14){\circle{6}} \put(49,14){\circle{6}} \put(14,28){\circle{6}} \put(28,28){\circle{6}} \put(42,28){\circle{6}} \put(21,42){\circle{6}} \put(35,42){\circle{6}} \put(28,56){\circle{6}}} \end{picture}\]

The incircle of triangle $ABC$ is tangent to the sides $AB,AC$ and $BC$ at the points $M,N$ and $K$ respectively. The median $AD$ of the triangle $ABC$ intersects $MN$ at the point $L$. Prove that $K,I$ and $L$ are collinear, where $I$ is the incenter of the triangle $ABC$.

A parallelogram of paper with sides $25$ and $10$ is given. The distance between the longer sides is $6$. The paper should be cut into exactly two parts in such a way that one can stick both the pieces together and fold it in a suitable manner to form a cube of suitable edge length without any further cuts and overlaps. Show that it is really possible and describe such a fragmentation.

Let $ ABCD$ be a trapezoid with $ AB\parallel{}CD$. Let $ a \equal{} AB$ and $ b \equal{} CD$. For $ MN\parallel{}AB$ such that $ M$ lies on $ AD,$ $ N$ lies on $ BC$, and the trapezoids $ ABNM$ and $ MNCD$ have the same area, the length of $ MN$ equals [img]http://i250.photobucket.com/albums/gg265/geometry101/NielsHenrikAbel1997Number6.jpg[/img] A. $ \sqrt{ab}$ B. $ \frac{a\plus{}b}{2}$ C. $ \frac{a^2 \plus{} b^2}{a\plus{}b}$ D. $ \sqrt{\frac{a^2 \plus{} b^2}{2}}$ E. $ \frac{a^2 \plus{} (2 \sqrt{2} \minus{} 2)ab \plus{} b^2}{\sqrt{2} (a\plus{}b)}$

A triangle $ABC$ with circumcenter $U$ is given, so that $\angle CBA = 60^o$ and $\angle CBU = 45^o$ apply. The straight lines $BU$ and $AC$ intersect at point $D$. Prove that $AD = DU$. (Karl Czakler)

Let $ABC$ be an acute-angled triangle with $AC>AB$, and $\Omega$ is your circumcircle. Let $P$ be the midpoint of the arc $BC$ of $\Omega$ (not containing $A$) and $Q$ be the midpoint of the arc $BC$ of $\Omega$(containing the point $A$). Let $M$ be the foot of perpendicular of $Q$ on the line $AC$. Prove that the circumcircle of $\triangle AMB$ cut the segment $AP$ in your midpoint.

Let $ABCD$ be a convex quadrilateral which admits an incircle. Let $AB$ produced beyond $B$ meet $DC$ produced towards $C$, at $E$. Let $BC$ produced beyond $C$ meet $AD$ produced towards $D$, at $F$. Let $G$ be the point on line $AB$ so that $FG \parallel CD$, and let $H$ be the point on line $BC$ so that $EH \parallel AD$. Prove that the (concave) quadrilateral $EGFH$ admits an excircle tangent to $\overline{EG}, \overline{EH}, \overrightarrow{FG}, \overrightarrow{FH}$. [i]Proposed by Rijul Saini[/i]

Two non-intersecting circles, not lying inside each other, are drawn in the plane. Two lines pass through a point P which lies outside each circle. The first line intersects the first circle at A and A′ and the second circle at B and B′; here A and B are closer to P than A′ and B′, respectively, and P lies on segment AB. Analogously, the second line intersects the first circle at C and C′ and the second circle at D and D′. Prove that the points A, B, C, D are concyclic if and only if the points A′, B′, C′, D′ are concyclic.

Given a circle centered at the origin. Prove that there is a circle of smaller radius that has no less points with integer coordinates.

Let $Q(x)$ be a polynomial with integer coefficients. Prove that there exists a polynomial $P(x)$ with integer coefficients such that for every integer $n\ge\deg{Q}$, \[\sum_{i=0}^{n}\frac{!i P(i)}{i!(n-i)!} = Q(n),\]where $!i$ denotes the number of derangements (permutations with no fixed points) of $1,2,\ldots,i$. [i]Calvin Deng.[/i]

A circle center $O$ meets a circle center $O'$ at $A$ and $B.$ The line $TT'$ touches the first circle at $T$ and the second at $T'$. The perpendiculars from $T$ and $T'$ meet the line $OO'$ at $S$ and $S'$. The ray $AS$ meets the first circle again at $R$, and the ray $AS'$ meets the second circle again at $R'$. Show that $R, B$ and $R'$ are collinear.

$ABC$ is a triangle with $AB = 33$, $AC = 21$ and $BC = m$, an integer. There are points $D$, $E$ on the sides $AB$, $AC$ respectively such that $AD = DE = EC = n$, an integer. Find $m$.

In isosceles triangle $ABC$ with base side $AB$, on side $BC$ it is given point $M$. Let $O$ be a circumcenter and $S$ incenter of triangle $ABC$. Prove that $$ SM \mid \mid AC \Leftrightarrow OM \perp BS$$

We are given an isosceles triangle $ABC$ such that $BC=a$ and $AB=BC=b$. The variable points $M\in (AC)$ and $N\in (AB)$ satisfy $a^2\cdot AM \cdot AN = b^2 \cdot BN \cdot CM$. The straight lines $BM$ and $CN$ intersect in $P$. Find the locus of the variable point $P$. [i]Dan Branzei[/i]

Let $ABC$ be a triangle with $\angle BCA=90^{\circ}$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK=BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL=AC$. Let $M$ be the point of intersection of $AL$ and $BK$. Show that $MK=ML$. [i]Proposed by Josef Tkadlec, Czech Republic[/i]

A [i]cuboctahedron[/i], shown below, is a polyhedron with 8 equilateral triangle faces and 6 square faces. Each edge has the same length and each of the 24 vertices borders 2 squares and 2 triangles. An \textit{octahedron} is a regular polyhedron with 6 vertices and 8 equilateral triangle faces. Compute the sum of the volumes of an octahedron with side length 5 and a cuboctahedron with side length 5. [img]http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvMi82LzBmNjM1OTM2M2ExYTQzOTFhODEwODkwM2FiYmM1MTljOGQzNmJhLmpwZw==&rn=Q3Vib2N0YWhlZHJvbi5qcGc=[/img] [i]2016 CCA Math Bonanza Team #7[/i]

In a triangle $ ABC$ the points $ A'$, $ B'$ and $ C'$ lie on the sides $ BC$, $ CA$ and $ AB$, respectively. It is known that $ \angle AC'B' \equal{} \angle B'A'C$, $ \angle CB'A' \equal{} \angle A'C'B$ and $ \angle BA'C' \equal{} \angle C'B'A$. Prove that $ A'$, $ B'$ and $ C'$ are the midpoints of the corresponding sides.

Let $ABCD$ be a convex quadrilateral with $AC = BD = AD$; $E$ and $F$ the midpoints of $AB$ and $CD$ respectively; $O$ the common point of the diagonals.Prove that $EF$ passes through the touching points of the incircle of triangle $AOD$ with $AO$ and $OD$ [i]Proposed by N.Moskvitin[/i]

The quadrilateral $ABCD$, in which $\angle BAC < \angle DCB$ , is inscribed in a circle $c$, with center $O$. If $\angle BOD = \angle ADC = \alpha$. Find out which values of $\alpha$ the inequality $AB <AD + CD$ occurs.

In a quadrilateral ABCD, it is given that AB = AD = 13, BC = CD = 20, BD = 24. If r is the radius of the circle inscribable in the quadrilateral, then what is the integer closest to r?

Given $n$ points $A_1, A_2, \ldots, A_n,$ no three collinear, show that the $n$- gon $A_1 A_2 \ldots A_n,$ is inscribed in a circle if and only if $A_1 A_2 \cdot A_3 A_n \cdot \ldots \cdot A_{n-1} A_n + A_2 A_3 \cdot A_4 A_n \cdot \ldots A_{n-1} A_n \cdot A_1 A_n + \ldots$ $+ A_{n-1} A_{n-2} \cdot A_1 A_n \cdot \ldots \cdot A_{n-3} A_n$ $= A_1 A_{n-1} \cdot A_2 A_n \cdot \ldots \cdot A_{n-2} A_n$, where $XY$ denotes the length of the segment $XY.$