Let $ABCD$ be a cyclic quadrilateral. The lines $BC$ and $AD$ meet at a point $P$. Let $Q$ be the point on the line $BP$, different from $B$, such that $PQ=BP$. Consider the parallelograms $CAQR$ and $DBCS$. Prove that the points $C,Q,R,S$ lie on a circle.

Let $ABC$ be a triangle. An interior point $P$ of $ABC$ is said to be [i]good [/i]if we can find exactly $27$ rays emanating from $P$ intersecting the sides of the triangle $ABC$ such that the triangle is divided by these rays into $27$ [i]smaller triangles of equal area.[/i] Determine the number of good points for a given triangle $ABC$.

Sam colors each tile in a 4 by 4 grid white or black. A coloring is called [i]rotationally symmetric[/i] if the grid can be rotated 90, 180, or 270 degrees to achieve the same pattern. Two colorings are called [i]rotationally distinct[/i] if neither can be rotated to match the other. How many rotationally distinct ways are there for Sam to color the grid such that the colorings are [i]not[/i] rotationally symmetric? [i]Proposed by Gabriel Wu[/i]

[b]5.[/b] Prove that neither the closed nor the open interval can be decomposed into finitely many mutually disjoint proper subsets which are all congruent by translation. [b](St. 2)[/b]

Let $ABC$ be an acute triangle, $H$ its orthocentre, $D$ a point on the side $[BC]$, and $P$ a point such that $ADPH$ is a parallelogram. Show that $\angle BPC > \angle BAC$.

There are $n$ fleas on an infinite sheet of triangulated paper. Initially the fleas are in different small triangles, all of which are inside some equilateral triangle consisting of $n^2$ small triangles. Once a second each flea jumps from its original triangle to one of the three small triangles having a common vertex but no common side with it. For which natural numbers $n$ does there exist an initial configuration such that after a finite number of jumps all the $n$ fleas can meet in a single small triangle?

Let $H$ be a rectangle with angle between two diagonal $\leq 45^{0}$. Rotation $H$ around the its center with angle $0^{0}\leq x\leq 360^{0}$ we have rectangle $H_{x}$. Find $x$ such that $[H\cap H_{x}]$ minimum, where $[S]$ is area of $S$.

Let $ABCD$ be a quadrilateral with $AB\parallel CD$ and $AC \perp BD$. Let $O$ be the intersection of $AC$ and $BD$. On the rays $(OA$ and $(OB$ we consider the points $M$ and $N$ respectively such that $\angle ANC = \angle BMD = 90^\circ$. We denote with $E$ the midpoint of the segment $MN$. Prove that a) $\triangle OMN \sim \triangle OBA$; b) $OE \perp AB$. [i]Claudiu-Stefan Popa[/i]

In hexagon $ABCDEF$, which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy $\angle A=3\angle D$, $\angle C=3\angle F$, and $\angle E=3\angle B$. Furthermore $AB=DE$, $BC=EF$, and $CD=FA$. Prove that diagonals $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$ are concurrent.

We define a [i]chessboard polygon[/i] to be a polygon whose sides are situated along lines of the form $ x \equal{} a$ or $ y \equal{} b$, where $ a$ and $ b$ are integers. These lines divide the interior into unit squares, which are shaded alternately grey and white so that adjacent squares have different colors. To tile a chessboard polygon by dominoes is to exactly cover the polygon by non-overlapping $ 1 \times 2$ rectangles. Finally, a [i]tasteful tiling[/i] is one which avoids the two configurations of dominoes shown on the left below. Two tilings of a $ 3 \times 4$ rectangle are shown; the first one is tasteful, while the second is not, due to the vertical dominoes in the upper right corner. [asy]size(300); pathpen = linewidth(2.5); void chessboard(int a, int b, pair P){ for(int i = 0; i < a; ++i) for(int j = 0; j < b; ++j) if((i+j) % 2 == 1) fill(shift(P.x+i,P.y+j)*unitsquare,rgb(0.6,0.6,0.6)); D(P--P+(a,0)--P+(a,b)--P+(0,b)--cycle); } chessboard(2,2,(2.5,0));fill(unitsquare,rgb(0.6,0.6,0.6));fill(shift(1,1)*unitsquare,rgb(0.6,0.6,0.6)); chessboard(4,3,(6,0)); chessboard(4,3,(11,0)); MP("\mathrm{Distasteful\ tilings}",(2.25,3),fontsize(12)); /* draw lines */ D((0,0)--(2,0)--(2,2)--(0,2)--cycle); D((1,0)--(1,2)); D((2.5,1)--(4.5,1)); D((7,0)--(7,2)--(6,2)--(10,2)--(9,2)--(9,0)--(9,1)--(7,1)); D((8,2)--(8,3)); D((12,0)--(12,2)--(11,2)--(13,2)); D((13,1)--(15,1)--(14,1)--(14,3)); D((13,0)--(13,3));[/asy] a) Prove that if a chessboard polygon can be tiled by dominoes, then it can be done so tastefully. b) Prove that such a tasteful tiling is unique.

Let $\triangle ABC$ be a triangle with $|AC|=2|AB|$ and let $O$ be its circumcenter. Let $D$ be the intersection of the bisector of $\angle A$ with $BC$. Let $E$ be the orthogonal projection of $O$ to $AD$ and let $F\ne D$ be the point on $AD$ satisfying $|CD|=|CF|$. Prove that $\angle EBF=\angle ECF$.

Points $ A’,B,C’ $ are arbitrarily taken on edges $ SA,SB, $ respectively, $ SC $ of a tetrahedron $ SABC. $ Plane forrmed by $ ABC $ intersects the plane $ \rho , $ formed by $ A’B’C’, $ in a line $ d. $ Prove that, meanwhile the plane $ \rho $ rotates around $ d, $ the lines $ AA’,BB’ $ and $ CC’ $ are, and remain concurrent. Find de locus of the respective intersections.

Let $p \geq 5$ be a prime. (a) Show that exists a prime $q \neq p$ such that $q| (p-1)^{p}+1$ (b) Factoring in prime numbers $(p-1)^{p}+1 = \prod_{i=1}^{n}p_{i}^{a_{i}}$ show that: \[\sum_{i=1}^{n}p_{i}a_{i}\geq \frac{p^{2}}2 \]

In $\triangle ABC$, $H$ is the orthocenter, and $AD,BE$ are arbitrary cevians. Let $\omega_1, \omega_2$ denote the circles with diameters $AD$ and $BE$, respectively. $HD,HE$ meet $\omega_1,\omega_2$ again at $F,G$. $DE$ meets $\omega_1,\omega_2$ again at $P_1,P_2$ respectively. $FG$ meets $\omega_1,\omega_2$ again $Q_1,Q_2$ respectively. $P_1H,Q_1H$ meet $\omega_1$ at $R_1,S_1$ respectively. $P_2H,Q_2H$ meet $\omega_2$ at $R_2,S_2$ respectively. Let $P_1Q_1\cap P_2Q_2 = X$, and $R_1S_1\cap R_2S_2=Y$. Prove that $X,Y,H$ are collinear. [i]Ray Li.[/i]