The diagonals of a cyclic quadrilateral meet at point $M$. A circle $\omega$ touches segments $MA$ and $MD$ at points $P,Q$ respectively and touches the circumcircle of $ABCD$ at point $X$. Prove that $X$ lies on the radical axis of circles $ACQ$ and $BDP$. [i](Proposed by Ivan Frolov)[/i]

From the point $P$ outside a circle $\omega$ with center $O$ draw the tangents $PA$ and $PB$ where $A$ and $B$ belong to $\omega$.In a random point $M$ in the chord $AB$ we draw the perpendicular to $OM$, which intersects $PA$ and $PB$ in $C$ and $D$. Prove that $M$ is the midpoint $CD$.

Let $ABC$ be an isosceles triangle with $AC=BC$, whose incentre is $I$. Let $P$ be a point on the circumcircle of the triangle $AIB$ lying inside the triangle $ABC$. The lines through $P$ parallel to $CA$ and $CB$ meet $AB$ at $D$ and $E$, respectively. The line through $P$ parallel to $AB$ meets $CA$ and $CB$ at $F$ and $G$, respectively. Prove that the lines $DF$ and $EG$ intersect on the circumcircle of the triangle $ABC$. [i]Proposed by Hojoo Lee, Korea[/i]

Inside of convex quadrilateral $ABCD$ found a point $M$ such that $\angle AMB=\angle ADM+\angle BCM$ and $\angle AMD=\angle ABM+\angle DCM$.Prove that $$AM\cdot CM+BM\cdot DM\ge \sqrt{AB\cdot BC\cdot CD\cdot DA}.$$

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.

In an inscribed quadrilateral $ABCD$, we have $BC=CD$ but $AB\neq AD$. Points $I$ and $J$ are the incenters of triangles $ABC$ and $ACD$ respectively. Point $K$ was chosen on segment $AC$ so that $IK=JK$. Points $M$ and $N$ are the incenters of triangles $AIK$ and $AJK$. Prove that the perpendicular to $CD$ at $D$ and the perpendicular to $KI$ at $I$ intersect on the circumcircle of $MAN$.

Let $ABCD$ be a quadrilateral inscribed in circle $\omega$. Define $E = AA \cap CD$, $F = AA \cap BC$, $G = BE \cap \omega$, $H = BE \cap AD$, $I = DF \cap \omega$, and $J = DF \cap AB$. Prove that $GI$, $HJ$, and the $B$-symmedian are concurrent. [i]Proposed by Robin Park[/i]

Let $ABC$ be an acute-angled triangle with $AC > AB$, let $O$ be its circumcentre, and let $D$ be a point on the segment $BC$. The line through $D$ perpendicular to $BC$ intersects the lines $AO, AC,$ and $AB$ at $W, X,$ and $Y,$ respectively. The circumcircles of triangles $AXY$ and $ABC$ intersect again at $Z \ne A$. Prove that if $W \ne D$ and $OW = OD,$ then $DZ$ is tangent to the circle $AXY.$

Let $ABC$ be a nondegenerate acute triangle with circumcircle $\omega$ and let its incircle $\gamma$ touch $AB, AC, BC$ at $X, Y, Z$ respectively. Let $XY$ hit arcs $AB, AC$ of $\omega$ at $M, N$ respectively, and let $P \neq X, Q \neq Y$ be the points on $\gamma$ such that $MP=MX, NQ=NY$. If $I$ is the center of $\gamma$, prove that $P, I, Q$ are collinear if and only if $\angle BAC=90^\circ$. [i]Proposed by David Stoner[/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 $BDE$.

The quadrilateral $ABCD$ is inscribed in circle $\Omega$ with center $O$, not lying on either of the diagonals. Suppose that the circumcircle of triangle $AOC$ passes through the midpoint of the diagonal $BD$. Prove that the circumcircle of triangle $BOD$ passes through the midpoint of diagonal $AC$. [i](A. Zaslavsky)[/i] (Translated from [url=http://sasja.shap.homedns.org/Turniry/TG/index.html]here.[/url])

Let $ABC$ be an acute-angled triangle with circumcircle $\omega$. A circle $\Gamma$ is internally tangent to $\omega$ at $A$ and also tangent to $BC$ at $D$. Let $AB$ and $AC$ intersect $\Gamma$ at $P$ and $Q$ respectively. Let $M$ and $N$ be points on line $BC$ such that $B$ is the midpoint of $DM$ and $C$ is the midpoint of $DN$. Lines $MP$ and $NQ$ meet at $K$ and intersect $\Gamma$ again at $I$ and $J$ respectively. The ray $KA$ meets the circumcircle of triangle $IJK$ again at $X\neq K$. Prove that $\angle BXP = \angle CXQ$. [i]Kian Moshiri, United Kingdom[/i]

Let be given a semicircle with diameter $AB$ and center $O$, and a line intersecting the semicircle at $C$ and $D$ and the line $AB$ at $M$ ($MB < MA$, $MD < MC$). The circumcircles of the triangles $AOC$ and $DOB$ meet again at $L$. Prove that $\angle MKO$ is right. [i]L. Kuptsov[/i]

Let $P$ be a point in the interior of an acute triangle $ABC$, and let $Q$ be its isogonal conjugate. Denote by $\omega_P$ and $\omega_Q$ the circumcircles of triangles $BPC$ and $BQC$, respectively. Suppose the circle with diameter $\overline{AP}$ intersects $\omega_P$ again at $M$, and line $AM$ intersects $\omega_P$ again at $X$. Similarly, suppose the circle with diameter $\overline{AQ}$ intersects $\omega_Q$ again at $N$, and line $AN$ intersects $\omega_Q$ again at $Y$. Prove that lines $MN$ and $XY$ are parallel. (Here, the points $P$ and $Q$ are [i]isogonal conjugates[/i] with respect to $\triangle ABC$ if the internal angle bisectors of $\angle BAC$, $\angle CBA$, and $\angle ACB$ also bisect the angles $\angle PAQ$, $\angle PBQ$, and $\angle PCQ$, respectively. For example, the orthocenter is the isogonal conjugate of the circumcenter.) [i]Proposed by Sammy Luo[/i]