Let $ABC$ be a triangle, and let $D$ be a point on side $AB$. Circle $\omega_1$ passes through $A$ and $D$ and is tangent to line $AC$ at $A$. Circle $\omega_2$ passes through $B$ and $D$ and is tangent to line $BC$ at $B$. Circles $\omega_1$ and $\omega_2$ meet at $D$ and $E$. Point $F$ is the reflection of $C$ across the perpendicular bisector of $AB$. Prove that points $D$, $E$, and $F$ are collinear.

In how many distinguishable ways can $10$ distinct pool balls be formed into a pyramid ($6$ on the bottom, $3$ in the middle, one on top), assuming that all rotations of the pyramid are indistinguishable?

Let a positive integer $n$.Consider square table $3*3$.One use $n$ colors to color all cell of table such that each cell is colored by exactly one color. Two colored table is same if we can receive them from other by a rotation through center of $3*3$ table How many way to color this square table satifies above conditions.

The circles $W_1, W_2, W_3$ in the plane are pairwise externally tangent to each other. Let $P_1$ be the point of tangency between circles $W_1$ and $W_3$, and let $P_2$ be the point of tangency between circles $W_2$ and $W_3$. $A$ and $B$, both different from $P_1$ and $P_2$, are points on $W_3$ such that $AB$ is a diameter of $W_3$. Line $AP_1$ intersects $W_1$ again at $X$, line $BP_2$ intersects $W_2$ again at $Y$, and lines $AP_2$ and $BP_1$ intersect at $Z$. Prove that $X, Y$, and $Z$ are collinear.

Let $ABC$ be an acute triangle with $AB<AC$, and let $D$ and $E$ be points on side $BC$ such that $BD=CE$ and $D$ lies between $B$ and $E$. Suppose there exists a point $P$ inside $ABC$ such that $PD\parallel AE$ and $\angle PAB=\angle EAC$. Prove that $\angle PBA=\angle PCA$. [i]Calvin Deng.[/i]

Let $ABC$ be an acute triangle with $AB<AC$, and let $D$ and $E$ be points on side $BC$ such that $BD=CE$ and $D$ lies between $B$ and $E$. Suppose there exists a point $P$ inside $ABC$ such that $PD\parallel AE$ and $\angle PAB=\angle EAC$. Prove that $\angle PBA=\angle PCA$. [i]Calvin Deng.[/i]

In acute triangle $ ABC,$ points $ P,Q$ lie on its sidelines $ AB,AC,$ respectively. The circumcircle of triangle $ ABC$ intersects of triangle $ APQ$ at $ X$ (different from $ A$). Let $ Y$ be the reflection of $ X$ in line $ PQ.$ Given $ PX>PB.$ Prove that $ S_{\bigtriangleup XPQ}>S_{\bigtriangleup YBC}.$ Where $ S_{\bigtriangleup XYZ}$ denotes the area of triangle $ XYZ.$

A triangle $ABC$ is inscribed in a circle $\omega$. A variable line $\ell$ chosen parallel to $BC$ meets segments $AB$, $AC$ at points $D$, $E$ respectively, and meets $\omega$ at points $K$, $L$ (where $D$ lies between $K$ and $E$). Circle $\gamma_1$ is tangent to the segments $KD$ and $BD$ and also tangent to $\omega$, while circle $\gamma_2$ is tangent to the segments $LE$ and $CE$ and also tangent to $\omega$. Determine the locus, as $\ell$ varies, of the meeting point of the common inner tangents to $\gamma_1$ and $\gamma_2$. [i](Russia) Vasily Mokin and Fedor Ivlev[/i]

Let $ m,n$ be two natural nonzero numbers and sets $ A \equal{} \{ 1,2,...,n\}, B \equal{} \{1,2,...,m\}$. We say that subset $ S$ of Cartesian product $ A \times B$ has property $ (j)$ if $ (a \minus{} x)(b \minus{} y)\le 0$ for each pairs $ (a,b),(x,y) \in S$. Prove that every set $ S$ with propery $ (j)$ has at most $ m \plus{} n \minus{} 1$ elements. [color=#FF0000]The statement was edited, in order to reflect the actual problem asked. The sign of the inequality was inadvertently reversed into $ (a \minus{} x)(b \minus{} y)\ge 0$, and that accounts for the following two posts.[/color]

The point $ P$ is a fixed point on a circle and $ Q$ is a fixed point on a line. The point $ R$ is a variable point on the circle such that $ P,Q,$ and $ R$ are not collinear. The circle through $ P,Q,$ and $ R$ meets the line again at $ V$. Show that the line $ VR$ passes through a fixed point.

Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$. Let $P$ be a point in the plane such that $AP \perp BC$. Let $Q$ and $R$ be the reflections of $P$ in the lines $CA$ and $AB$, respectively. Let $Y$ be the orthogonal projection of $R$ onto $CA$. Let $Z$ be the orthogonal projection of $Q$ onto $AB$. Assume that $H \neq O$ and $Y \neq Z$. Prove that $YZ \perp HO$. [asy] import olympiad; unitsize(30); pair A,B,C,H,O,P,Q,R,Y,Z,Q2,R2,P2; A = (-14.8, -6.6); B = (-10.9, 0.3); C = (-3.1, -7.1); O = circumcenter(A,B,C); H = orthocenter(A,B,C); P = 1.2 * H - 0.2 * A; Q = reflect(A, C) * P; R = reflect(A, B) * P; Y = foot(R, C, A); Z = foot(Q, A, B); P2 = foot(A, B, C); Q2 = foot(P, C, A); R2 = foot(P, A, B); draw(B--(1.6*A-0.6*B)); draw(B--C--A); draw(P--R, blue); draw(R--Y, red); draw(P--Q, blue); draw(Q--Z, red); draw(A--P2, blue); draw(O--H, darkgreen+linewidth(1.2)); draw((1.4*Z-0.4*Y)--(4.6*Y-3.6*Z), red+linewidth(1.2)); draw(rightanglemark(R,Y,A,10), red); draw(rightanglemark(Q,Z,B,10), red); draw(rightanglemark(C,Q2,P,10), blue); draw(rightanglemark(A,R2,P,10), blue); draw(rightanglemark(B,P2,H,10), blue); label("$\textcolor{blue}{H}$",H,NW); label("$\textcolor{blue}{P}$",P,N); label("$A$",A,W); label("$B$",B,N); label("$C$",C,S); label("$O$",O,S); label("$\textcolor{blue}{Q}$",Q,E); label("$\textcolor{blue}{R}$",R,W); label("$\textcolor{red}{Y}$",Y,S); label("$\textcolor{red}{Z}$",Z,NW); dot(A, filltype=FillDraw(black)); dot(B, filltype=FillDraw(black)); dot(C, filltype=FillDraw(black)); dot(H, filltype=FillDraw(blue)); dot(P, filltype=FillDraw(blue)); dot(Q, filltype=FillDraw(blue)); dot(R, filltype=FillDraw(blue)); dot(Y, filltype=FillDraw(red)); dot(Z, filltype=FillDraw(red)); dot(O, filltype=FillDraw(black)); [/asy]

Let $ABC$ be a triangle which it not right-angled. Define a sequence of triangles $A_iB_iC_i$, with $i \ge 0$, as follows: $A_0B_0C_0$ is the triangle $ABC$ and, for $i \ge 0$, $A_{i+1},B_{i+1},C_{i+1}$ are the reflections of the orthocentre of triangle $A_iB_iC_i$ in the sides $B_iC_i$,$C_iA_i$,$A_iB_i$, respectively. Assume that $\angle A_m = \angle A_n$ for some distinct natural numbers $m,n$. Prove that $\angle A = 60^{\circ}$.

Let points $A_1$, $A_2$ and $A_3$ lie on the circle $\Gamma$ in a counter-clockwise order, and let $P$ be a point in the same plane. For $i \in \{1,2,3\}$, let $\tau_i$ denote the counter-clockwise rotation of the plane centred at $A_i$, where the angle of rotation is equial to the angle at vertex $A_i$ in $\triangle A_1A_2A_3$. Further, define $P_i$ to be the point $\tau_{i+2}(\tau_{i}(\tau_{i+1}(P)))$, where the indices are taken modulo $3$ (i.e., $\tau_4 = \tau_1$ and $\tau_5 = \tau_2$). Prove that the radius of the circumcircle of $\triangle P_1P_2P_3$ is at most the radius of $\Gamma$. [i]Proposed by Anant Mudgal[/i]