$\Delta ABC$ is isosceles with a circumscribed circle $\omega (O)$. Let $H$ be the foot of the altitude from $C$ to $AB$ and let $M$ be the middle point of $AB$. We define a point $X$ as the second intersection point of the circle with diameter $CM$ and $\omega$ and let $XH$ intersect $\omega$ for a second time in $Y$. If $CO\cap AB=D$, then prove that the circumscribed circle of $\Delta YHD$ is tangent to $\omega$.

Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.

Let $\Gamma_1$, $\Gamma_2$, $\Gamma_3$, $\Gamma_4$ be distinct circles such that $\Gamma_1$, $\Gamma_3$ are externally tangent at $P$, and $\Gamma_2$, $\Gamma_4$ are externally tangent at the same point $P$. Suppose that $\Gamma_1$ and $\Gamma_2$; $\Gamma_2$ and $\Gamma_3$; $\Gamma_3$ and $\Gamma_4$; $\Gamma_4$ and $\Gamma_1$ meet at $A$, $B$, $C$, $D$, respectively, and that all these points are different from $P$. Prove that \[ \frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}. \]

Let $ ABCD$ be a convex quadrilateral and let $ P$ and $ Q$ be points in $ ABCD$ such that $ PQDA$ and $ QPBC$ are cyclic quadrilaterals. Suppose that there exists a point $ E$ on the line segment $ PQ$ such that $ \angle PAE \equal{} \angle QDE$ and $ \angle PBE \equal{} \angle QCE$. Show that the quadrilateral $ ABCD$ is cyclic. [i]Proposed by John Cuya, Peru[/i]

We are given a $\Delta ABC$. Point $D$ on the circumscribed circle k is such that $CD$ is a symmedian in $\Delta ABC$. Let $X$ and $Y$ be on the rays $\overrightarrow{CB}$ and $\overrightarrow{CA}$, so that $CX=2CA$ and $CY=2CB$. Prove that the circle, tangent externally to $k$ and to the lines $CA$ and $CB$, is tangent to the circumscribed circle of $\Delta XDY$.

In convex cyclic quadrilateral $ABCD$, we know that lines $AC$ and $BD$ intersect at $E$, lines $AB$ and $CD$ intersect at $F$, and lines $BC$ and $DA$ intersect at $G$. Suppose that the circumcircle of $\triangle ABE$ intersects line $CB$ at $B$ and $P$, and the circumcircle of $\triangle ADE$ intersects line $CD$ at $D$ and $Q$, where $C,B,P,G$ and $C,Q,D,F$ are collinear in that order. Prove that if lines $FP$ and $GQ$ intersect at $M$, then $\angle MAC = 90^\circ$. [i]Proposed by Kada Williams[/i]

Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.

Given a triangle $ABC$, with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$. Let $E$ be a point on the arc $BDC$, and $F$ a point on the segment $BC$, such that $\angle BAF=\angle CAE < \dfrac12\angle BAC$. If $G$ is the midpoint of $IF$, prove that the meeting point of the lines $EI$ and $DG$ lies on $\Gamma$. [i]Proposed by Tai Wai Ming and Wang Chongli, Hong Kong[/i]

Let $\triangle ABC$ be an acute triangle with circumcenter $O$ such that $AB<AC$, let $Q$ be the intersection of the external bisector of $\angle A$ with $BC$, and let $P$ be a point in the interior of $\triangle ABC$ such that $\triangle BPA$ is similar to $\triangle APC$. Show that $\angle QPA + \angle OQB = 90^{\circ}$. [i]Alex Zhu.[/i]

Let $ABC$ be a triangle such that $\angle A=90^{\circ }$ and $\angle B<\angle C$. The tangent at $A$ to the circumcircle $\omega$ of triangle $ABC$ meets the line $BC$ at $D$. Let $E$ be the reflection of $A$ in the line $BC$, let $X$ be the foot of the perpendicular from $A$ to $BE$, and let $Y$ be the midpoint of the segment $AX$. Let the line $BY$ intersect the circle $\omega$ again at $Z$. Prove that the line $BD$ is tangent to the circumcircle of triangle $ADZ$. [hide="comment"] [i]Edited by Orl.[/i] [/hide]

Let $ ABCD$ be a convex quadrilateral and let $ P$ and $ Q$ be points in $ ABCD$ such that $ PQDA$ and $ QPBC$ are cyclic quadrilaterals. Suppose that there exists a point $ E$ on the line segment $ PQ$ such that $ \angle PAE \equal{} \angle QDE$ and $ \angle PBE \equal{} \angle QCE$. Show that the quadrilateral $ ABCD$ is cyclic. [i]Proposed by John Cuya, Peru[/i]

Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.

Given two circles ${\omega _1},{\omega _2}$, $S$ denotes all $\Delta ABC$ satisfies that ${\omega _1}$ is the circumcircle of $\Delta ABC$, ${\omega _2}$ is the $A$- excircle of $\Delta ABC$ , ${\omega _2}$ touches $BC,CA,AB$ at $D,E,F$. $S$ is not empty, prove that the centroid of $\Delta DEF$ is a fixed point.

Let $I$ be the incenter of the $ABC$ triangle. The circumference that passes through $I$ and has center in $A$ intersects the circumscribed circumference of the $ABC$ triangle at points $M$ and $N$. Prove that the line $MN$ is tangent to the inscribed circle of the $ABC$ triangle.

Determine the poles of the inversions that transform four collienar points $A,B, C, D$, aligned in this order, at four points $A' $, $B' $, $C'$ , $D'$ that are vertices of a rectangle, and such that $A'$ and $C'$ are opposite vertices.

Given a triangle $ABC$, with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$. Let $E$ be a point on the arc $BDC$, and $F$ a point on the segment $BC$, such that $\angle BAF=\angle CAE < \dfrac12\angle BAC$. If $G$ is the midpoint of $IF$, prove that the meeting point of the lines $EI$ and $DG$ lies on $\Gamma$. [i]Proposed by Tai Wai Ming and Wang Chongli, Hong Kong[/i]

Let $ABCD$ be an isosceles trapezoid with $AB\parallel CD$. Let $E$ be the midpoint of $AC$. Denote by $\omega$ and $\Omega$ the circumcircles of the triangles $ABE$ and $CDE$, respectively. Let $P$ be the crossing point of the tangent to $\omega$ at $A$ with the tangent to $\Omega$ at $D$. Prove that $PE$ is tangent to $\Omega$. [i]Jakob Jurij Snoj, Slovenia[/i]

Given convex hexagon $ABCDEF$, inscribed in the circle. Prove that $AC*BD*DE*CE*EA*FB \geq 27 AB * BC * CD * DE * EF * FA$

A collection of circles in the upper half-plane, all tangent to the $x$-axis, is constructed in layers as follows. Layer $L_0$ consists of two circles of radii $70^2$ and $73^2$ that are externally tangent. For $k\geq 1$, the circles in $\textstyle\bigcup_{j=0}^{k-1} L_j$ are ordered according to their points of tangency with the $x$-axis. For every pair of consecutive circles in this order, a new circle is constructed externally tangent to each of the two circles in the pair. Layer $L_k$ consists of the $2^{k-1}$ circles constructed in this way. Let $S=\textstyle\bigcup_{j=0}^6 L_j$, and for every circle $C$ denote by $r(C)$ its radius. What is \[\sum_{C\in S}\dfrac1{\sqrt{r(C)}}?\] [asy] import olympiad; size(350); defaultpen(linewidth(0.7)); // define a bunch of arrays and starting points pair[] coord = new pair[65]; int[] trav = {32,16,8,4,2,1}; coord[0] = (0,73^2); coord[64] = (2*73*70,70^2); // draw the big circles and the bottom line path arc1 = arc(coord[0],coord[0].y,260,360); path arc2 = arc(coord[64],coord[64].y,175,280); fill((coord[0].x-910,coord[0].y)--arc1--cycle,gray(0.78)); fill((coord[64].x+870,coord[64].y+425)--arc2--cycle,gray(0.78)); draw(arc1^^arc2); draw((-930,0)--(70^2+73^2+850,0)); // We now apply the findCenter function 63 times to get // the location of the centers of all 63 constructed circles. // The complicated array setup ensures that all the circles // will be taken in the right order for(int i = 0;i<=5;i=i+1) { int skip = trav[i]; for(int k=skip;k<=64 - skip; k = k + 2*skip) { pair cent1 = coord[k-skip], cent2 = coord[k+skip]; real r1 = cent1.y, r2 = cent2.y, rn=r1*r2/((sqrt(r1)+sqrt(r2))^2); real shiftx = cent1.x + sqrt(4*r1*rn); coord[k] = (shiftx,rn); } // Draw the remaining 63 circles } for(int i=1;i<=63;i=i+1) { filldraw(circle(coord[i],coord[i].y),gray(0.78)); }[/asy] $\textbf{(A) }\dfrac{286}{35}\qquad\textbf{(B) }\dfrac{583}{70}\qquad\textbf{(C) }\dfrac{715}{73}\qquad\textbf{(D) }\dfrac{143}{14}\qquad\textbf{(E) }\dfrac{1573}{146}$

Given a triangle $ABC$, with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$. Let $E$ be a point on the arc $BDC$, and $F$ a point on the segment $BC$, such that $\angle BAF=\angle CAE < \dfrac12\angle BAC$. If $G$ is the midpoint of $IF$, prove that the meeting point of the lines $EI$ and $DG$ lies on $\Gamma$. [i]Proposed by Tai Wai Ming and Wang Chongli, Hong Kong[/i]

Let $ ABCD$ be a convex quadrilateral and let $ P$ and $ Q$ be points in $ ABCD$ such that $ PQDA$ and $ QPBC$ are cyclic quadrilaterals. Suppose that there exists a point $ E$ on the line segment $ PQ$ such that $ \angle PAE \equal{} \angle QDE$ and $ \angle PBE \equal{} \angle QCE$. Show that the quadrilateral $ ABCD$ is cyclic. [i]Proposed by John Cuya, Peru[/i]

Let $ABC$ be a triangle with circumcircle $\omega$, incenter $I$, and $A$-excenter $I_A$. Let the incircle and the $A$-excircle hit $BC$ at $D$ and $E$, respectively, and let $M$ be the midpoint of arc $BC$ without $A$. Consider the circle tangent to $BC$ at $D$ and arc $BAC$ at $T$. If $TI$ intersects $\omega$ again at $S$, prove that $SI_A$ and $ME$ meet on $\omega$. [i]Amol Aggarwal.[/i]

Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.

Transform by inversion two concentric and coplanar circles into two equal.

Let $\triangle ABC$ be an acute triangle with circumcenter $O$ such that $AB<AC$, let $Q$ be the intersection of the external bisector of $\angle A$ with $BC$, and let $P$ be a point in the interior of $\triangle ABC$ such that $\triangle BPA$ is similar to $\triangle APC$. Show that $\angle QPA + \angle OQB = 90^{\circ}$. [i]Alex Zhu.[/i]