A point $P$ is chosen inside a triangle $ABC$ with circumcircle $\Omega$. Let $\Gamma$ be the circle passing through the circumcenters of the triangles $APB$, $BPC$, and $CPA$. Let $\Omega$ and $\Gamma$ intersect at points $X$ and $Y$. Let $Q$ be the reflection of $P$ in the line $XY$ . Prove that $\angle BAP = \angle CAQ$.

Let $ M$ and $ N$ be two points inside triangle $ ABC$ such that \[ \angle MAB \equal{} \angle NAC\quad \mbox{and}\quad \angle MBA \equal{} \angle NBC. \] Prove that \[ \frac {AM \cdot AN}{AB \cdot AC} \plus{} \frac {BM \cdot BN}{BA \cdot BC} \plus{} \frac {CM \cdot CN}{CA \cdot CB} \equal{} 1. \]

Let $E$ be an interior point of the convex quadrilateral $ABCD$. Construct triangles $\triangle ABF,\triangle BCG,\triangle CDH$ and $\triangle DAI$ on the outside of the quadrilateral such that the similarities $\triangle ABF\sim\triangle DCE,\triangle BCG\sim \triangle ADE,\triangle CDH\sim\triangle BAE$ and $ \triangle DAI\sim\triangle CBE$ hold. Let $P,Q,R$ and $S$ be the projections of $E$ on the lines $AB,BC,CD$ and $DA$, respectively. Prove that if the quadrilateral $PQRS$ is cyclic, then \[EF\cdot CD=EG\cdot DA=EH\cdot AB=EI\cdot BC.\]

Let $ABCD$ be a quadrilateral circumscribed around a circle $\omega$ with center $I$. Assume $P$ and $Q$ are distinct points and are isogonal conjugates such that $P, Q$, and $I$ are collinear. Show that $ABCD$ is a kite, that is, it has two disjoint pairs of consecutive equal sides.

Let $P$ be a given point inside quadrilateral $ABCD$. Points $Q_1$ and $Q_2$ are located within $ABCD$ such that \[\angle Q_1BC=\angle ABP,\quad\angle Q_1CB=\angle DCP,\quad\angle Q_2AD=\angle BAP,\quad\angle Q_2DA=\angle CDP.\] Prove that $\overline{Q_1Q_2}\parallel\overline{AB}$ if and only if $\overline{Q_1Q_2}\parallel\overline{CD}$.

A convex quadrilateral $ABCD$ satisfies $AB\cdot CD = BC\cdot DA$. Point $X$ lies inside $ABCD$ so that \[\angle{XAB} = \angle{XCD}\quad\,\,\text{and}\quad\,\,\angle{XBC} = \angle{XDA}.\] Prove that $\angle{BXA} + \angle{DXC} = 180^\circ$. [i]Proposed by Tomasz Ciesla, Poland[/i]

In a convex quadrilateral $ABCD$, the diagonal $BD$ bisects neither the angle $ABC$ nor the angle $CDA$. The point $P$ lies inside $ABCD$ and satisfies \[\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.\] Prove that $ABCD$ is a cyclic quadrilateral if and only if $AP=CP$.

Inside the acute-angled triangle $ABC$ we take $P$ and $Q$ two isogonal conjugate points. The perpendicular lines on the interior angle-bisector of $\angle BAC$ passing through $P$ and $Q$ intersect the segments $AC$ and $AB$ at the points $B_p\in AC$, $B_q\in AC$, $C_p\in AB$ and $C_q\in AB$, respectively. Let $W$ be the midpoint of the arc $BAC$ of the circle $(ABC)$. The line $WP$ intersects the circle $(ABC)$ again at $P_1$ and the line $WQ$ intersects the circle $(ABC)$ again at $Q_1$. Prove that the points $P_1$, $Q_1$, $B_p$, $B_q$, $C_p$ and $C_q$ lie on a circle. [i]Proposed by P. Bibikov[/i]

Points $A_1$, $B_1$, $C_1$ are midpoints of sides $BC$, $AC$, $AB$ of triangle $ABC$. On midlines $C_1B_1$ and $A_1B_1$ points $E$ and $F$ are chosen such that $BE$ is the angle bisector of $AEB_1$ and $BF$ is the angle bisector of $CFB_1$. Prove that bisectors of angles $ABC$ and $FBE$ coincide. [I]Proposed by F. Baharev[/i]

Let $M_A, M_B, M_C$ be the midpoints of the sides $BC, CA, AB$ respectively of a non-isosceles triangle $ABC$. Points $H_A, H_B, H_C$ lie on the corresponding sides, different from $M_A, M_B, M_C$ such that $M_AH_B=M_AH_C, $ $M_BH_A=M_BH_C,$ and $M_CH_A=M_CH_B$. Prove that $H_A, H_B, H_C$ are the feet of the corresponding altitudes.

Let $ M$ and $ N$ be two points inside triangle $ ABC$ such that \[ \angle MAB \equal{} \angle NAC\quad \mbox{and}\quad \angle MBA \equal{} \angle NBC. \] Prove that \[ \frac {AM \cdot AN}{AB \cdot AC} \plus{} \frac {BM \cdot BN}{BA \cdot BC} \plus{} \frac {CM \cdot CN}{CA \cdot CB} \equal{} 1. \]

The insphere and the exsphere opposite to the vertex $D$ of a (not necessarily regular) tetrahedron $ABCD$ touch the face $ABC$ in the points $X$ and $Y$, respectively. Show that $\measuredangle XAB=\measuredangle CAY$.

Let $ABC$ be a triangle, let $O$ be its circumcenter, let $A'$ be the orthogonal projection of $A$ on the line $BC$, and let $X$ be a point on the open ray $AA'$ emanating from $A$. The internal bisectrix of the angle $BAC$ meets the circumcircle of $ABC$ again at $D$. Let $M$ be the midpoint of the segment $DX$. The line through $O$ and parallel to the line $AD$ meets the line $DX$ at $N$. Prove that the angles $BAM$ and $CAN$ are equal.

Canmoo is trying to do constructions, but doesn't have a ruler or compass. Instead, Canmoo has a device that, given four distinct points $A$, $B$, $C$, $P$ in the plane, will mark the isogonal conjugate of $P$ with respect to triangle $ABC$, if it exists. Show that if two points are marked on the plane, then Canmoo can construct their midpoint using this device, a pencil for marking additional points, and no other tools. (Recall that the [i]isogonal conjugate[/i] of $P$ with respect to triangle $ABC$ is the point $Q$ such that lines $AP$ and $AQ$ are reflections around the bisector of $\angle BAC$, lines $BP$ and $BQ$ are reflections around the bisector of $\angle CBA$, lines $CP$ and $CQ$ are reflections around the bisector of $\angle ACB$. Additional points marked by the pencil can be assumed to be in general position, meaning they don't lie on any line through two existing points or any circle through three existing points.) [i]Maxim Li[/i]

Let a point $D$ lie on the median $AM$ of a triangle $ABC$. The tangents to the circumcircle of triangle $BDC$ at points $B$ and $C$ meet at point $K$. Prove that $DD'$ is parallel to $AK$, where $D'$ is isogonally conjugated to $D$ with respect to $ABC$.

Inside the acute-angled triangle $ABC$ we take $P$ and $Q$ two isogonal conjugate points. The perpendicular lines on the interior angle-bisector of $\angle BAC$ passing through $P$ and $Q$ intersect the segments $AC$ and $AB$ at the points $B_p\in AC$, $B_q\in AC$, $C_p\in AB$ and $C_q\in AB$, respectively. Let $W$ be the midpoint of the arc $BAC$ of the circle $(ABC)$. The line $WP$ intersects the circle $(ABC)$ again at $P_1$ and the line $WQ$ intersects the circle $(ABC)$ again at $Q_1$. Prove that the points $P_1$, $Q_1$, $B_p$, $B_q$, $C_p$ and $C_q$ lie on a circle. [i]Proposed by P. Bibikov[/i]

Let $A_1A_2A_3A_4A_5A_6$ be a convex hexagon. Suppose that there exists 2 points $P,Q$ in its interior such that $\angle A_{i-1}A_iP=\angle QA_iA_{i+1}$ for $i=1,2,\ldots,6$ where $A_0\equiv A_6,A_1\equiv A_7$. Prove that \[\angle A_1PA_2+\angle A_3PA_4+\angle A_5PA_6=180^\circ.\]

Let $ABC$ be a triangle, let $O$ be its circumcenter, let $A'$ be the orthogonal projection of $A$ on the line $BC$, and let $X$ be a point on the open ray $AA'$ emanating from $A$. The internal bisectrix of the angle $BAC$ meets the circumcircle of $ABC$ again at $D$. Let $M$ be the midpoint of the segment $DX$. The line through $O$ and parallel to the line $AD$ meets the line $DX$ at $N$. Prove that the angles $BAM$ and $CAN$ are equal.

A convex quadrilateral $ABCD$ satisfies $AB\cdot CD = BC\cdot DA$. Point $X$ lies inside $ABCD$ so that \[\angle{XAB} = \angle{XCD}\quad\,\,\text{and}\quad\,\,\angle{XBC} = \angle{XDA}.\] Prove that $\angle{BXA} + \angle{DXC} = 180^\circ$. [i]Proposed by Tomasz Ciesla, Poland[/i]

Let $ABC$ be a triangle with circumscribed and inscribed circles $(O)$ and $(I)$ respectively. $AA'$,$BB'$,$CC'$ are the bisectors of triangle $ABC$. Prove that $OI$ passes through the the isogonal conjugate of point $I$ with respect to triangle $A'B'C'$.

Let $ABCD$ be a convex quadrilateral, and $P$ be a point in its interior. The projections of $P$ on the sides of the quadrilateral lie on a circle with center $O$. Show that $O$ lies on the line through the midpoints of $AC$ and $BD$.

In a convex quadrilateral $ABCD$, the diagonal $BD$ bisects neither the angle $ABC$ nor the angle $CDA$. The point $P$ lies inside $ABCD$ and satisfies \[\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.\] Prove that $ABCD$ is a cyclic quadrilateral if and only if $AP=CP$.

In a quadrilateral $ABCD$, $\angle ABC=\angle ADC <90^{\circ}$. The circle with diameter $AC$ intersects $BC$ and $CD$ again at $E,F$, respectively. $M$ is the midpoint of $BD$, and $AN \perp BD$ at $N$. Prove that $M,N,E,F$ is concyclic.