Let $O$ be the center of circle $\Gamma$, and $A$, $B$ be two points on $\Gamma$ so that $O, A$ and $B$ are not collinear. Let $M$ be the midpoint of $AB$. Let $P$ and $Q$ be points on $OA$ and $OB$, respectively, so that $P \neq A$ and $P, M, Q$ are collinear. Let $X$ be the intersection of the line passing through $P$ and parallel to $AB$ and the line passing through $Q$ and parallel to $OM$. Let $Y$ be the intersection of the line passing through $X$ and parallel to $OA$ and the line passing through $B$ and orthogonal to $OX$. Prove that: if $X$ is on $\Gamma$, then $Y$ is also on $\Gamma$. [i] Proposed by usjl[/i]

Let $ ABCDE$ be a convex pentagon such that \[ \angle BAC \equal{} \angle CAD \equal{} \angle DAE\qquad \text{and}\qquad \angle ABC \equal{} \angle ACD \equal{} \angle ADE. \]The diagonals $BD$ and $CE$ meet at $P$. Prove that the line $AP$ bisects the side $CD$. [i]Proposed by Zuming Feng, USA[/i]

Given a triangle $ABC$, let $D$ be the midpoint of the side $AC$ and let $M$ be the point that divides the segment $BD$ in the ratio $1/2$; that is, $MB/MD=1/2$. The rays $AM$ and $CM$ meet the sides $BC$ and $AB$ at points $E$ and $F$, respectively. Assume the two rays perpendicular: $AM\perp CM$. Show that the quadrangle $AFED$ is cyclic if and only if the median from $A$ in triangle $ABC$ meets the line $EF$ at a point situated on the circle $ABC$.

In triangle $ABC$ let $F$ denote the midpoint of side $BC$. Let the circle passing through point $A$ and tangent to side $BC$ at point $F$ intersect sides $AB$ and $AC$ at points $M$ and $N$, respectively. Let the line segments $CM$ and $BN$ intersect in point $X$. Let $P$ be the second point of intersection of the circumcircles of triangles $BMX$ and $CNX$. Prove that points $A, F$ and $P$ are collinear. Proposed by Imolay András, Budapest

Let $I$ be the center of the incircle of non-isosceles triangle $ABC,A_{1}=AI\cap BC$ and $B_{1}=BI\cap AC.$ Let $l_{a}$ be the line through $A_{1}$ which is parallel to $AC$ and $l_{b}$ be the line through $B_{1}$ parallel to $BC.$ Let $l_{a}\cap CI=A_{2}$ and $l_{b}\cap CI=B_{2}.$ Also $N=AA_{2}\cap BB_{2}$ and $M$ is the midpoint of $AB.$ If $CN\parallel IM$ find $\frac{CN}{IM}$.

We have an acute-angled triangle $ABC$, and $AA',BB'$ are its altitudes. A point $D$ is chosen on the arc $ACB$ of the circumcircle of $ABC$. If $P=AA'\cap BD,Q=BB'\cap AD$, show that the midpoint of $PQ$ lies on $A'B'$.

Let $ABCD$ be a parallelogram and a circle $k$ passes through $A, C$ and meets rays $AB, AD$ at $E, F$. If $BD, EF$ and the tangent at $C$ concur, show that $AC$ is diameter of $k$.

Let $ ABC$ be a triangle and $ K$ and $ L$ be two points on $ (AB)$, $ (AC)$ such that $ BK \equal{} CL$ and let $ P \equal{} CK\cap BL$. Let the parallel through $ P$ to the interior angle bisector of $ \angle BAC$ intersect $ AC$ in $ M$. Prove that $ CM \equal{} AB$.

Given $\vartriangle ABC$. Choose two points $P, Q$ on $AB, AC$ such that $BP = CQ$. Let $M, T$ be the midpoints of $BC, PQ$. Show that $MT$ is parallel to the angle bisevtor of $\angle BAC$ [img]http://4.bp.blogspot.com/-MgMtdnPtq1c/XnSHHFl1LDI/AAAAAAAALdY/8g8541DnyGo_Gqd19-7bMBpVRFhbXeYPACK4BGAYYCw/s1600/imoc2017%2Bg1.png[/img]

Let $ABCD$ be a convex quadrilateral. $I=AC\cap BD$, and $E$, $H$, $F$ and $G$ are points on $AB$, $BC$, $CD$ and $DA$ respectively, such that $EF \cap GH= I$. If $M=EG \cap AC$, $N=HF \cap AC$, show that \[\frac{AM}{IM}\cdot \frac{IN}{CN}=\frac{IA}{IC}.\]

Quadrilateral $ABCD$ is inscribed in circle $O$ and has sides $AB = 3$, $BC = 2$, $CD = 6$, and $DA = 8$. Let $X$ and $Y$ be points on $\overline{BD}$ such that \[\frac{DX}{BD} = \frac{1}{4} \quad \text{and} \quad \frac{BY}{BD} = \frac{11}{36}.\] Let $E$ be the intersection of intersection of line $AX$ and the line through $Y$ parallel to $\overline{AD}$. Let $F$ be the intersection of line $CX$ and the line through $E$ parallel to $\overline{AC}$. Let $G$ be the point on circle $O$ other than $C$ that lies on line $CX$. What is $XF \cdot XG$? $\textbf{(A) }17\qquad\textbf{(B) }\frac{59 - 5\sqrt{2}}{3}\qquad\textbf{(C) }\frac{91 - 12\sqrt{3}}{4}\qquad\textbf{(D) }\frac{67 - 10\sqrt{2}}{3}\qquad\textbf{(E) }18$

In a parallelogram $\Box ABCD$ $(AB < BC)$ The incircle of $\triangle ABC$ meets $\overline {BC}$ and $\overline {CA}$ at $P, Q$. The incircle of $\triangle ACD$ and $\overline {CD}$ meets at $R$. Let $S$ = $PQ$ $\cap$ $AD$ $U$ = $AR$ $\cap$ $CS$ $T$, a point on $\overline {BC}$ such that $\overline {AB} = \overline {BT}$ Prove that $AT, BU, PQ$ are concurrent

Let $A,B,C$ be three points in that order on a line $\ell$ in the plane, and suppose $AB>BC$. Draw semicircles $\Gamma_1$ and $\Gamma_2$ respectively with $AB$ and $BC$ as diameters, both on the same side of $\ell$. Let the common tangent to $\Gamma_1$ and $\Gamma_2$ touch them respectively at $P$ and $Q$, $P\ne Q$. Let $D$ and $E$ be points on the segment $PQ$ such that the semicircle $\Gamma_3$ with $DE$ as diameter touches $\Gamma_2$ in $S$ and $\Gamma_1$ in $T$. [list=1][*]Prove that $A,C,S,T$ are concyclic. [*]Prove that $A,C,D,E$ are concyclic.[/list]

We have an acute-angled triangle $ABC$, and $AA',BB'$ are its altitudes. A point $D$ is chosen on the arc $ACB$ of the circumcircle of $ABC$. If $P=AA'\cap BD,Q=BB'\cap AD$, show that the midpoint of $PQ$ lies on $A'B'$.