Acute triangle $ABC$ is inscribed in circle $\omega$. Let $H$ and $O$ denote its orthocenter and circumcenter, respectively. Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively. Rays $MH$ and $NH$ meet $\omega$ at $P$ and $Q$, respectively. Lines $MN$ and $PQ$ meet at $R$. Prove that $OA\perp RA$.

Let $I$ be the incenter of an acute triangle $ABC$. Assume that $K_1$ is the point such that $AK_1 \perp BC$ and the circle with center $K_1$ of radius $K_1A$ is internally tangent to the incircle of triangle $ABC$ at $A_1$. The points $B_1, C_1$ are defined similarly. a) Prove that $AA_1, BB_1, CC_1$ are concurrent at a point $P$. b) Let $\omega_1,\omega_2,\omega_3$ be the excircles of triangle $ABC$ with respect to $A, B, C$, respectively. The circles $\gamma_1,\gamma_2\gamma_3$ are the reflections of $\omega_1,\omega_2,\omega_3$ with respect to the midpoints of $BC, CA, AB$, respectively. Prove that P is the radical center of $\gamma_1,\gamma_2,\gamma_3$.

Let $P$ be a point and $\omega$ be a circle with center $O$ and radius $r$ . We define the power of the point $P$ with respect to the circle $\omega$ to be $OP^2 - r^2$ , and we denote this by pow $(P, \omega)$. We define the radical axis of two circles $\omega_1$ and $\omega_2$ to be the locus of all points P such that pow $(P,\omega_1) =$ pow $(P,\omega_2)$. It turns out that the pairwise radical axes of three circles are either concurrent or pairwise parallel. The concurrence point is referred to as the radical center of the three circles. In $\vartriangle ABC$, let $I$ be the incenter, $\Gamma$ be the circumcircle, and $O$ be the circumcenter. Let $A_1,B_1,C_1$ be the point of tangency of the incircle of $\vartriangle ABC$ with side $BC,CA, AB$, respectively. Let $X_1,X_2 \in \Gamma$ such that $X_1,B_1,C_1,X_2$ are collinear in this order. Let $M_A$ be the midpoint of $BC$, and define $\omega_A$ as the circumcircle of $\vartriangle X_1X_2M_A$. Define $\omega_B$ ,$\omega_C$ analogously. The goal of this problem is to show that the radical center of $\omega_A$, $\omega_B$, $\omega_C$ lies on line $OI$. (a) Let$ A'_1$ denote the intersection of $B_1C_1$ and $BC$. Show that $\frac{A_1B}{A_1C}=\frac{A'_1B}{A'_1C}$. (b) Prove that $A_1$ lies on $\omega_A$. (c) Prove that $A_1$ lies on the radical axis of $\omega_B$ and $\omega_C$ . (d) Prove that the radical axis of $\omega_B$ and $\omega_C$ is perpendicular to $B_1C_1$. (e) Prove that the radical center of $\omega_A$, $\omega_B$, $\omega_C$ is the orthocenter of $\vartriangle A_1B_1C_1$. (f ) Conclude that the radical center of $\omega_A$, $\omega_B$, $\omega_C$ , $O$, and $I$ are collinear. PS. You had better use hide for answers.

Let $P$ be a point and $\omega$ be a circle with center $O$ and radius $r$ . We define the power of the point $P$ with respect to the circle $\omega$ to be $OP^2 - r^2$ , and we denote this by pow $(P, \omega)$. We define the radical axis of two circles $\omega_1$ and $\omega_2$ to be the locus of all points P such that pow $(P,\omega_1) =$ pow $(P,\omega_2)$. It turns out that the pairwise radical axes of three circles are either concurrent or pairwise parallel. The concurrence point is referred to as the radical center of the three circles. In $\vartriangle ABC$, let $I$ be the incenter, $\Gamma$ be the circumcircle, and $O$ be the circumcenter. Let $A_1,B_1,C_1$ be the point of tangency of the incircle of $\vartriangle ABC$ with side $BC,CA, AB$, respectively. Let $X_1,X_2 \in \Gamma$ such that $X_1,B_1,C_1,X_2$ are collinear in this order. Let $M_A$ be the midpoint of $BC$, and define $\omega_A$ as the circumcircle of $\vartriangle X_1X_2M_A$. Define $\omega_B$ ,$\omega_C$ analogously. The goal of this problem is to show that the radical center of $\omega_A$, $\omega_B$, $\omega_C$ lies on line $OI$. (a) Let$ A'_1$ denote the intersection of $B_1C_1$ and $BC$. Show that $\frac{A_1B}{A_1C}=\frac{A'_1B}{A'_1C}$. (b) Prove that $A_1$ lies on $\omega_A$. (c) Prove that $A_1$ lies on the radical axis of $\omega_B$ and $\omega_C$ . (d) Prove that the radical axis of $\omega_B$ and $\omega_C$ is perpendicular to $B_1C_1$. (e) Prove that the radical center of $\omega_A$, $\omega_B$, $\omega_C$ is the orthocenter of $\vartriangle A_1B_1C_1$. (f ) Conclude that the radical center of $\omega_A$, $\omega_B$, $\omega_C$ , $O$, and $I$ are collinear. PS. You had better use hide for answers.

Let $ABC$ be an acute, scalene triangle with altitudes $AD, BE, CF$ with $D \in BC, E \in CA$ and $F \in AB$. Let $H, O, I$ be the orthocenter, circumcenter, incenter of triangle $ABC$ respectively and let $M, N, P$ be the midpoint of segments $BC, CA, AB$ respectively. Let $X, Y, Z$ be the intersection of pairs of lines $(AI, NP), (BI, PM)$ and $(CI, MN)$ respectively. a) Prove that the circumcircle of triangles $AXD, BYE, CZF$ have two common points that lie on line $OH$. b) Lines $XP, YM, ZN$ meet the circumcircle of triangles $AXD, BYE, CZF$ again at $X', Y', Z'$ ($X' \neq X, Y' \neq Y, Z' \neq Z$). Let $J$ be the reflection of $I$ across $O$. Prove that $X', Y', Z'$ lie on a line perpendicular to $HJ$.

Two circles in the plane do not intersect and do not lie inside each other. We choose diameters $A_1B_1$ and $A_2B_2$ of these circles such that the segments $A_1A_2$ and $B_1B_2'$ intersect. Let $A$ and $B$ be the midpoints of the segments $A_1A_2$ and $B_1B_2$, and $C$ be the intersection point of these segments. Prove that the orthocenter of the triangle $ABC$ belongs to a fixed line that does not depend on the choice of diameters.