Let $BC$ be a fixed segment in the plane, and let $A$ be a variable point in the plane not on the line $BC$. Distinct points $X$ and $Y$ are chosen on the rays $CA^\to$ and $BA^\to$, respectively, such that $\angle CBX = \angle YCB = \angle BAC$. Assume that the tangents to the circumcircle of $ABC$ at $B$ and $C$ meet line $XY$ at $P$ and $Q$, respectively, such that the points $X$, $P$, $Y$ and $Q$ are pairwise distinct and lie on the same side of $BC$. Let $\Omega_1$ be the circle through $X$ and $P$ centred on $BC$. Similarly, let $\Omega_2$ be the circle through $Y$ and $Q$ centred on $BC$. Prove that $\Omega_1$ and $\Omega_2$ intersect at two fixed points as $A$ varies. [i]Daniel Pham Nguyen, Denmark[/i]

Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ meet at $E$. The extensions of the sides $AD$ and $BC$ beyond $A$ and $B$ meet at $F$. Let $G$ be the point such that $ECGD$ is a parallelogram, and let $H$ be the image of $E$ under reflection in $AD$. Prove that $D,H,F,G$ are concyclic.

Let $\triangle ABC$ be a triangle with incenter $I$. Points $P$ and $Q$ are chosen on segmets $BI$ and $CI$ such that $2\angle PAQ=\angle BAC$. If $D$ is the touch point of incircle and side $BC$ prove that $\angle PDQ=90$.

$AH$ is the altitude of triangle $ABC$ and $H^\prime$ is the reflection of $H$ trough the midpoint of $BC$. If the tangent lines to the circumcircle of $ABC$ at $B$ and $C$, intersect each other at $X$ and the perpendicular line to $XH^\prime$ at $H^\prime$, intersects $AB$ and $AC$ at $Y$ and $Z$ respectively, prove that $\angle ZXC=\angle YXB$.

Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ meet at $E$. The extensions of the sides $AD$ and $BC$ beyond $A$ and $B$ meet at $F$. Let $G$ be the point such that $ECGD$ is a parallelogram, and let $H$ be the image of $E$ under reflection in $AD$. Prove that $D,H,F,G$ are concyclic.

Let $ ABC$ be a triangle, and $ H$ its orthocenter. Let $ P$ be a point on the circumcircle of triangle $ ABC$ (distinct from the vertices $ A$, $ B$, $ C$), and let $ E$ be the foot of the altitude of triangle $ ABC$ from the vertex $ B$. Let the parallel to the line $ BP$ through the point $ A$ meet the parallel to the line $ AP$ through the point $ B$ at a point $ Q$. Let the parallel to the line $ CP$ through the point $ A$ meet the parallel to the line $ AP$ through the point $ C$ at a point $ R$. The lines $ HR$ and $ AQ$ intersect at some point $ X$. Prove that the lines $ EX$ and $ AP$ are parallel.

Let $D$ be an interior point of the acute triangle $ABC$ with $AB > AC$ so that $\angle DAB = \angle CAD.$ The point $E$ on the segment $AC$ satisfies $\angle ADE =\angle BCD,$ the point $F$ on the segment $AB$ satisfies $\angle FDA =\angle DBC,$ and the point $X$ on the line $AC$ satisfies $CX = BX.$ Let $O_1$ and $O_2$ be the circumcenters of the triangles $ADC$ and $EXD,$ respectively. Prove that the lines $BC, EF,$ and $O_1O_2$ are concurrent.

Let $ABCD$ be a trapezium, $AD\parallel BC$, and let $E,F$ be points on the sides$AB$ and $CD$, respectively. The circumcircle of $AEF$ meets $AD$ again at $A_1$, and the circumcircle of $CEF$ meets $BC$ again at $C_1$. Prove that $A_1C_1,BD,EF$ are concurrent.

Let $ABCD$ be a cyclic quadrilateral such that $AD=BD=AC$. A point $P$ moves along the circumcircle $\omega$ of triangle $ABCD$. The lined $AP$ and $DP$ meet the lines $CD$ and $AB$ at points $E$ and $F$ respectively. The lines $BE$ and $CF$ meet point $Q$. Find the locus of $Q$.

Let $O$ and $H$ be the circumcenter and the orthocenter, respectively, of an acute triangle $ABC$. Points $D$ and $E$ are chosen from sides $AB$ and $AC$, respectively, such that $A$, $D$, $O$, $E$ are concyclic. Let $P$ be a point on the circumcircle of triangle $ABC$. The line passing $P$ and parallel to $OD$ intersects $AB$ at point $X$, while the line passing $P$ and parallel to $OE$ intersects $AC$ at $Y$. Suppose that the perpendicular bisector of $\overline{HP}$ does not coincide with $XY$, but intersect $XY$ at $Q$, and that points $A$, $Q$ lies on the different sides of $DE$. Prove that $\angle EQD = \angle BAC$. [i]Proposed by Shuang-Yen Lee[/i]