Let $ABC$ be a triangle with $\angle{C} = 90^{\circ}$, and let $H$ be the foot of the altitude from $C$. A point $D$ is chosen inside the triangle $CBH$ so that $CH$ bisects $AD$. Let $P$ be the intersection point of the lines $BD$ and $CH$. Let $\omega$ be the semicircle with diameter $BD$ that meets the segment $CB$ at an interior point. A line through $P$ is tangent to $\omega$ at $Q$. Prove that the lines $CQ$ and $AD$ meet on $\omega$.

In triangle $ABC$ $(\angle A\neq 90^\circ)$, let $O$, $H$ be the circumcenter and the foot of the altitude from $A$ respectively. Suppose $M$, $N$ are the midpoints of $BC$, $AH$ respectively. Let $D$ be the intersection of $AO$ and $BC$ and let $H'$ be the reflection of $H$ about $M$. Suppose that the circumcircle of $OH'D$ intersects the circumcircle of $BOC$ at $E$. Prove that $NO$ and $AE$ are concurrent on the circumcircle of $BOC$. [i]Proposed by Mehran Talaei[/i]

$BB_1$ is the angle bisector of $\triangle ABC$, and $I$ is its incenter. The perpendicular bisector of segment $AC$ intersects the circumcircle of $\triangle AIC$ at $D$ and $E$. Point $F$ is on the segment $B_1C$ such that $AB_1=CF$.Prove that the four points $B, D, E$ and $F$ are concyclic.

Let $ABC$ be a scalene triangle with circumcircle $\Gamma$. Let $M$ be the midpoint of $BC$. A variable point $P$ is selected in the line segment $AM$. The circumcircles of triangles $BPM$ and $CPM$ intersect $\Gamma$ again at points $D$ and $E$, respectively. The lines $DP$ and $EP$ intersect (a second time) the circumcircles to triangles $CPM$ and $BPM$ at $X$ and $Y$, respectively. Prove that as $P$ varies, the circumcircle of $\triangle AXY$ passes through a fixed point $T$ distinct from $A$.

The tangents at $B$ and $A$ to the circumcircle of an acute angled triangle $ABC$ meet the tangent at $C$ at $T$ and $U$ respectively. $AT$ meets $BC$ at $P$, and $Q$ is the midpoint of $AP$; $BU$ meets $CA$ at $R$, and $S$ is the midpoint of $BR$. Prove that $\angle ABQ=\angle BAS$. Determine, in terms of ratios of side lengths, the triangles for which this angle is a maximum.

Let $\Omega$ be the circumcircle of cyclic quadrilateral $ABCD$. Consider such pairs of points $P$, $Q$ of diagonal $AC$ that the rays $BP$ and $BQ$ are symmetric with respect the bisector of angle $B$. Find the locus of circumcenters of triangles $PDQ$.

Let $ ABC $ be an acute triangle. Denote by $ D $ the foot of the perpendicular line drawn from the point $ A $ to the side $ BC $, by $M$ the midpoint of $ BC $, and by $ H $ the orthocenter of $ ABC $. Let $ E $ be the point of intersection of the circumcircle $ \Gamma $ of the triangle $ ABC $ and the half line $ MH $, and $ F $ be the point of intersection (other than $E$) of the line $ ED $ and the circle $ \Gamma $. Prove that $ \tfrac{BF}{CF} = \tfrac{AB}{AC} $ must hold. (Here we denote $XY$ the length of the line segment $XY$.)

Let $ABC$ be a triangle inscribed in a circle with center $O$. Let $A_1$ be a point on arc $BC$ that does not contain $ A$ such that the line perpendicular to $OA$ at $A_1$ intersects the lines $AB$ and $AC$ at two points and the line segment joining those two points has as midpoint $A_1$. Points $B_1$, $C_1$ are determined similarly. Prove that the lines $AA_1$, $BB_1$, $CC_1$ are concurrent.

Consider a convex pentagon circumscribed about a circle. We name the lines that connect vertices of the pentagon with the opposite points of tangency with the circle [i]gergonnians[/i]. (a) Prove that if four gergonnians are conncurrent, the all five of them are concurrent. (b) Prove that if there is a triple of gergonnians that are concurrent, then there is another triple of gergonnians that are concurrent.

Let $ABC$ be a triangle with $\angle{C} = 90^{\circ}$, and let $H$ be the foot of the altitude from $C$. A point $D$ is chosen inside the triangle $CBH$ so that $CH$ bisects $AD$. Let $P$ be the intersection point of the lines $BD$ and $CH$. Let $\omega$ be the semicircle with diameter $BD$ that meets the segment $CB$ at an interior point. A line through $P$ is tangent to $\omega$ at $Q$. Prove that the lines $CQ$ and $AD$ meet on $\omega$.

Given an acute non-isosceles triangle $ABC$ with circumcircle $\Gamma$. $M$ is the midpoint of segment $BC$ and $N$ is the midpoint of arc $BC$ of $\Gamma$ (the one that doesn't contain $A$). $X$ and $Y$ are points on $\Gamma$ such that $BX\parallel CY\parallel AM$. Assume there exists point $Z$ on segment $BC$ such that circumcircle of triangle $XYZ$ is tangent to $BC$. Let $\omega$ be the circumcircle of triangle $ZMN$. Line $AM$ meets $\omega$ for the second time at $P$. Let $K$ be a point on $\omega$ such that $KN\parallel AM$, $\omega_b$ be a circle that passes through $B$, $X$ and tangents to $BC$ and $\omega_c$ be a circle that passes through $C$, $Y$ and tangents to $BC$. Prove that circle with center $K$ and radius $KP$ is tangent to 3 circles $\omega_b$, $\omega_c$ and $\Gamma$. [i]Proposed by Tran Quan - Vietnam[/i]

Denote by $ M$ midpoint of side $ BC$ in an isosceles triangle $ \triangle ABC$ with $ AC = AB$. Take a point $ X$ on a smaller arc $ \overarc{MA}$ of circumcircle of triangle $ \triangle ABM$. Denote by $ T$ point inside of angle $ BMA$ such that $ \angle TMX = 90$ and $ TX = BX$. Prove that $ \angle MTB - \angle CTM$ does not depend on choice of $ X$. [i]Author: Farzan Barekat, Canada[/i]

Let $ABC$ be triangle with circumcircle $(O)$ of fixed $BC$, $AB \ne AC$ and $BC$ not a diameter. Let $I$ be the incenter of the triangle $ABC$ and $D = AI \cap BC, E = BI \cap CA, F = CI \cap AB$. The circle passing through $D$ and tangent to $OA$ cuts for second time $(O)$ at $G$ ($G \ne A$). $GE, GF$ cut $(O)$ also at $M, N$ respectively. a) Let $H = BM \cap CN$. Prove that $AH$ goes through a fixed point. b) Suppose $BE, CF$ cut $(O)$ also at $L, K$ respectively and $AH \cap KL = P$. On $EF$ take $Q$ for $QP = QI$. Let $J$ be a point of the circimcircle of triangle $IBC$ so that $IJ \perp IQ$. Prove that the midpoint of $IJ$ belongs to a fixed circle.

Let $(O_1), (O_2)$ be given two circles intersecting at $A$ and $B$. The tangent lines of $(O_1)$ at $A, B$ intersect at $O$. Let $I$ be a point on the circle $(O_1)$ but outside the circle $(O_2)$. The lines $IA, IB$ intersect circle $(O_2)$ at $C, D$. Denote by $M$ the midpoint of $C D$. Prove that $I, M, O$ are collinear.