Let $ I(A)$ and $ I(B)$ be the centers of the excircles of a triangle $ ABC,$ which touches the sides $ BC$ and $ CA$ in its interior. Furthermore let $ P$ a point on the circumcircle $ \omega$ of the triangle $ ABC.$ Show that the center of the segment which connects the circumcenters of the triangles $ I(A)CP$ and $ I(B)CP$ coincides with the center of the circle $ \omega.$

In square $ABCD$ (labeled clockwise), let $P$ be any point on $BC$ and construct square $APRS$ (labeled clockwise). Prove that line $CR$ is tangent to the circumcircle of triangle $ABC$.

Let $ I $ be the incenter of triangle $ ABC $. The incircle touches $ BC, CA, AB$ at points $ P, Q, R $. A circle passing through $ B , C $ is tangent to the circle $I$ at point $ X $, a circle passing through $ C , A $ is tangent to the circle $I$ at point $ Y $, and a circle passing through $ A , B $ is tangent to the circle $I$ at point $ Z $, respectively. Prove that three lines $ PX, QY, RZ $ are concurrent.

Point $D$ is an arbitary point on side $BC$ of triangle $ABC$. $I$,$I_1$ and$I_2$ are the incenters of triangles $ABC$,$ABD$ and $ACD$ respectively. $M\not=A$ and $N\not=A$ are the intersections of circumcircle of triangle $ABC$ and circumcircles of triangles $IAI_1$ and $IAI_2$ respectively. Prove that regardless of point $D$, line $MN$ goes through a fixed point.

$ABC$ is an acute triangle with $AB> AC$. $\Gamma_B$ is a circle that passes through $A,B$ and is tangent to $AC$ on $A$. Define similar for $ \Gamma_C$. Let $D$ be the intersection $\Gamma_B$ and $\Gamma_C$ and $M$ be the midpoint of $BC$. $AM$ cuts $\Gamma_C$ at $E$. Let $O$ be the center of the circumscibed circle of the triangle $ABC$. Prove that the circumscibed circle of the triangle $ODE$ is tangent to $\Gamma_B$.

Let $ABCDE$ be a convex pentagon such that $BC \parallel AE,$ $AB = BC + AE,$ and $\angle ABC = \angle CDE.$ Let $M$ be the midpoint of $CE,$ and let $O$ be the circumcenter of triangle $BCD.$ Given that $\angle DMO = 90^{\circ},$ prove that $2 \angle BDA = \angle CDE.$ [i]Proposed by Nazar Serdyuk, Ukraine[/i]

Let $h_a, h_b, h_c$ be the lengths of the altitudes of a triangle $ABC$ from $A, B, C$ respectively. Let $P$ be any point inside the triangle. Show that \[\frac{PA}{h_b+h_c} + \frac{PB}{h_a+h_c} + \frac{PC}{h_a+h_b} \ge 1.\]

Triangle $\vartriangle ABC$ has $AB = 13$, $BC = 14$, and $CA = 15$. $\vartriangle ABC$ has incircle $\gamma$ and circumcircle $\omega$. $\gamma$ has center at $I$. Line $AI$ is extended to hit $\omega$ at $P$. What is the area of quadrilateral $ABPC$?

Quadrilateral $ABCD$ is given with $AD \nparallel BC$. The midpoints of $AD$ and $BC$ are denoted by $M$ and $N$, respectively. The line $MN$ intersects the diagonals $AC$ and $BD$ in points $K$ and $L$, respectively. Prove that the circumcircles of the triangles $AKM$ and $BNL$ have common point on the line $AB$.( Proposed by Emil Stoyanov ) [img]http://estoyanov.net/wp-content/uploads/2015/09/est.png[/img]

Let $ AB$ be the diameter of semicircle $O$ , $C, D $ be points on the arc $AB$, $P, Q$ be respectively the circumcenter of $\triangle OAC $ and $\triangle OBD $ . Prove that:$CP\cdot CQ=DP \cdot DQ$.[asy] import cse5; import olympiad; unitsize(3.5cm); dotfactor=4; pathpen=black; real h=sqrt(55/64); pair A=(-1,0), O=origin, B=(1,0),C=shift(-3/8,h)*O,D=shift(4/5,3/5)*O,P=circumcenter(O,A,C), Q=circumcenter(O,D,B); D(arc(O,1,0,180),darkgreen); D(MP("A",A,W)--MP("C",C,N)--MP("P",P,SE)--MP("D",D,E)--MP("Q",Q,E)--C--MP("O",O,S)--D--MP("B",B,E)--cycle,deepblue); D(O); [/asy]

A point $P$ is given inside an acute-angled triangle $ABC$. Let $A',B',C'$ be the orthogonal projections of $P$ on sides $BC, CA, AB$ respectively. Determine the locus of points $P$ for which $\angle BAC = \angle B'A'C'$ and $\angle CBA = \angle C'B'A'$

Given a nondegenerate triangle $ABC$, with circumcentre $O$, orthocentre $H$, and circumradius $R$, prove that $|OH| < 3R$.

A non-isosceles triangle $\triangle ABC$ has incenter $I$ and the incircle hits $BC, CA, AB$ at $D, E, F$. Let $EF$ hit the circumcircle of $CEI$ at $P \not= E$. Prove that $\triangle ABC = 2 \triangle ABP$.

Let $ABC$ be a scalene triangle and $\omega$ is your incircle. The sides $BC,CA$ and $AB$ are tangents to $\omega$ in $X,Y,Z$ respectively. Let $M$ be the midpoint of $BC$ and $D$ is the intersection point of $BC$ with the angle bisector of $\angle BAC$. Prove that $\angle BAX=\angle MAC$ if and only if $YZ$ passes by the midpoint of $AD$.

Let $ABCD$ be a cyclic quadrilateral whose sides have pairwise different lengths. Let $O$ be the circumcenter of $ABCD$. The internal angle bisectors of $\angle ABC$ and $\angle ADC$ meet $AC$ at $B_1$ and $D_1$, respectively. Let $O_B$ be the center of the circle which passes through $B$ and is tangent to $\overline{AC}$ at $D_1$. Similarly, let $O_D$ be the center of the circle which passes through $D$ and is tangent to $\overline{AC}$ at $B_1$. Assume that $\overline{BD_1} \parallel \overline{DB_1}$. Prove that $O$ lies on the line $\overline{O_BO_D}$.

The incircle of triangle $ABC$ is tangent to $BC,CA,AB$ at $D,E,F$ respectively. Consider the triangle formed by the line joining the midpoints of $AE,AF$, the line joining the midpoints of $BF,BD$, and the line joining the midpoints of $CD,CE$. Prove that the circumcenter of this triangle coincides with the circumcenter of triangle $ABC$.

Let $ABCD$ be a cyclic quadrilateral. Let $E$ and $F$ be variable points on the sides $AB$ and $CD$, respectively, such that $AE:EB=CF:FD$. Let $P$ be the point on the segment $EF$ such that $PE:PF=AB:CD$. Prove that the ratio between the areas of triangles $APD$ and $BPC$ does not depend on the choice of $E$ and $F$.

Let $ABC$ be an arbitrary triangle and $O$ is the circumcenter of $\triangle {ABC}$.Points $X,Y$ lie on $AB,AC$,respectively such that the reflection of $BC$ WRT $XY$ is tangent to circumcircle of $\triangle {AXY}$.Prove that the circumcircle of triangle $AXY$ is tangent to circumcircle of triangle $BOC$.

Cyclic quadrilateral $ABCD$ is inscribed in circle $k$ with center $O$. The angle bisector of $ABD$ intersects $AD$ and $k$ in $K,M$ respectively, and the angle bisector of $CBD$ intersects $CD$ and $k$ in $L,N$ respectively. If $KL\parallel MN$ prove that the circumcircle of triangle $MON$ bisects segment $BD$.

Given an acute $\vartriangle ABC$ with orthocenter $H$. Let $M_a$ be the midpoint of $BC. M_aH$ intersects the circumcircle of $\vartriangle ABC$ at $X_a$ and $AX_a$ intersects $BC$ at $Y_a$. Define $Y_b, Y_c$ in a similar way. Prove that $Y_a, Y_b,Y_c$ are collinear. [img]https://2.bp.blogspot.com/-yjISBHtRa0s/XnSKTrhhczI/AAAAAAAALds/e_rvs9glp60L1DastlvT0pRFyP7GnJnCwCK4BGAYYCw/s320/imoc2017%2Bg4.png[/img]

Given a parallelogram $ABCD$ with $AB<BC$, show that the circumcircles of the triangles $APQ$ share a second common point (apart from $A$) as $P,Q$ move on the sides $BC,CD$ respectively s.t. $CP=CQ$.

Given a scalene triangle $ \triangle ABC $. $ B', C' $ are points lie on the rays $ \overrightarrow{AB}, \overrightarrow{AC} $ such that $ \overline{AB'} = \overline{AC}, \overline{AC'} = \overline{AB} $. Now, for an arbitrary point $ P $ in the plane. Let $ Q $ be the reflection point of $ P $ w.r.t $ \overline{BC} $. The intersections of $ \odot{\left(BB'P\right)} $ and $ \odot{\left(CC'P\right)} $ is $ P' $ and the intersections of $ \odot{\left(BB'Q\right)} $ and $ \odot{\left(CC'Q\right)} $ is $ Q' $. Suppose that $ O, O' $ are circumcenters of $ \triangle{ABC}, \triangle{AB'C'} $ Show that 1. $ O', P', Q' $ are colinear 2. $ \overline{O'P'} \cdot \overline{O'Q'} = \overline{OA}^{2} $

Let $ABC$ be a triangle with incenter $I$. The line through $I$, perpendicular to $AI$, intersects the circumcircle of $ABC$ at points $P$ and $Q$. It turns out there exists a point $T$ on the side $BC$ such that $AB + BT = AC + CT$ and $AT^2 = AB \cdot AC$. Determine all possible values of the ratio $IP/IQ$.

Let $ABCD$ be a cyclic quadrilateral. Let $E$ be the intersection of lines parallel to $AC$ and $BD$ passing through points $B$ and $A$, respectively. The lines $EC$ and $ED$ intersect the circumcircle of $AEB$ again at $F$ and $G$, respectively. Prove that points $C$, $D$, $F$, and $G$ lie on a circle.

An isosceles trapezoid $ABCD$, inscribed in $\omega$, satisfies $AB=CD, AD<BC, AD<CD$. A circle with center $D$ and passing $A$ hits $BD, CD, \omega$ at $E, F, P(\not= A)$, respectively. Let $AP \cap EF = Q$, and $\omega$ meet $CQ$ and the circumcircle of $\triangle BEQ$ at $R(\not= C), S(\not= B)$, respectively. Prove that $\angle BER= \angle FSC$.