Let $H$ be the orthocenter of acute-angled triangle $ABC$ and $M$ be the midpoint of $AH$. Line $BH$ cuts $AC$ at $D$. Consider point $E$ such that $BC$ is the perpendicular bisector of $DE$. Segments $CM$ and $AE$ intersect at $F$. Show that $BF$ is perpendicular to $CM$. [i]Proposed by Germán Puga[/i]

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.

Let $\vartriangle ABC$ be an acute triangle with altitudes $AA_1, BB_1, CC_1$ and orthocenter $H$. Let $K, L$ be the midpoints of $BC_1, CB_1$. Let $\ell_A$ be the external angle bisector of $\angle BAC$. Let $\ell_B, \ell_C$ be the lines through $B, C$ perpendicular to $\ell_A$. Let $\ell_H$ be the line through $H$ parallel to $\ell_A$. Prove that the centers of the circumcircles of $\vartriangle A_1B_1C_1, \vartriangle AKL$ and the rectangle formed by $\ell_A, \ell_B, \ell_C, \ell_H$ lie on the same line.

Given an acute $\vartriangle ABC$ with orthocenter $H$. Let $M_a$ be the midpoint of $BC. M_aH$ intersects the circumcircle of $\vartriangle ABC$ at $X_a$ and $AX_a$ intersects $BC$ at $Y_a$. Define $Y_b, Y_c$ in a similar way. Prove that $Y_a, Y_b,Y_c$ are collinear. [img]https://2.bp.blogspot.com/-yjISBHtRa0s/XnSKTrhhczI/AAAAAAAALds/e_rvs9glp60L1DastlvT0pRFyP7GnJnCwCK4BGAYYCw/s320/imoc2017%2Bg4.png[/img]

The diagonals of a convex quadrilateral $ABCD$ intersect at $E$. The triangles $ABE, BCE, CDE$ and $DAE$ have centroids $K,L,M$ and $N$, and orthocentres $Q,R,S$ and $T$. Show that the quadrilaterals $QRST$ and $LMNK$ are similar.

Let $H$ be the orthocenter of acute-angled $\vartriangle ABC$, and $X, Y$ points on the ray $AB, AC$. ($B$ lies between $X, A$, and $C$ lies between $Y, A$.) Lines $HX, HY$ intersect $BC$ at $D, E$ respectively. Let the line through $D$ parallel to $AC$ intersect $XY$ at $Z$. Prove that $\angle XHY = 90^o$ if and only if $ZE \parallel AB$.

Let $ABC$ be an acute-angled triangle with $AB> AC$ and orthocenter $H$. Let $D$ the projection of $A$ on $BC$. Let $E$ be the reflection of $C$ wrt $D$. The lines $AE$ and $BH$ intersect at point $S$. Let $N$ be the midpoint of $AE$ and let $M$ be the midpoint of $BH$. Prove that $MN$ is perpendicular to $DS$.

Triangle $ABC$ is given. An arbitrary circle with center $J$, passing through $B$ and $C$, intersects the sides $AC$ and $AB$ at $E$ and $F$, respectively. Let $X$ be a point such that triangle $FXB$ is similar to triangle $EJC$ (with the same order) and the points $X$ and $C$ lie on the same side of the line $AB$. Similarly, let $Y$ be a point such that triangle $EYC$ is similar to triangle $FJB$ (with the same order) and the points $Y$ and $B$ lie on the same side of the line $AC$. Prove that the line $XY$ passes through the orthocenter of the triangle $ABC$. [i]Proposed by Nguyen Van Linh - Vietnam[/i]

Let $ABC$ be a triangle inscribed in the circle $(O)$. The bisector of $\angle BAC$ cuts the circle $(O)$ again at $D$. Let $DE$ be the diameter of $(O)$. Let $G$ be a point on arc $AB$ which does not contain $C$. The lines $GD$ and $BC$ intersect at $F$. Let $H$ be a point on the line $AG$ such that $FH \parallel AE$. Prove that the circumcircle of triangle $HAB$ passes through the orthocenter of triangle $HAC$.

Let $ABC$ be an acute-angled triangle with height intersection $H$. Let $G$ be the intersection of parallel of $AB$ through $H$ with the parallel of $AH$ through $B$. Let $I$ be the point on the line $GH$, so that $AC$ bisects segment $HI$. Let $J$ be the second intersection of $AC$ and the circumcircle of the triangle $CGI$. Show that $IJ = AH$

Let scalene triangle $ABC$ have orthocentre $H$ and circumcircle $\Gamma$. $AH$ meets $\Gamma$ at $D$ distinct from $A$. $BH$ and $CH$ meet $CA$ and $AB$ at $E$ and $F$ respectively, and $EF$ meets $BC$ at $P$. The tangents to $\Gamma$ at $B$ and $C$ meet at $T$. Show that $AP$ and $DT$ are concurrent on the circumcircle of $AFE$.

Let $ABC$ be an acute-angled triangle. $M ,N$ are any two points on the sides $AB , AC$ respectively. The circles with the diameters $BN$ and $CM$ intersect at points $P$ and $Q$. Show that the points $P, Q$ and the orthocenter of the triangle $ABC$ lie on a straight line.

Let $I, O$ be the incenter, circumcenter of triangle $ABC$ and $A_1, B_1, C_1 $be arbitrary points on the segments $AI, BI, CI$ respectively. The perpendicular bisectors of $AA_1, BB_1, CC_1$ intersect each other at $X, Y$ and $Z$. Prove that the circumcenter of triangle $XYZ$ coincides with $O$ if and only if $I$ is the orthocenter of triangle $A_1B_1C_1$

Let \( ABC \) be an acute scalene triangle with orthocenter \( H \), and consider \( M \) to be the midpoint of side \( BC \). Define \( P \neq A \) as the intersection point of the circle with diameter \( AH \) and the circumcircle of triangle \( ABC \), and let \( Q \) be the intersection of \( AP \) with \( BC \). Let \( G \neq M \) be the intersection of the circumcircle of triangle \( MPQ \) with the circumcircle of triangle \( AHM \). Show that \( G \) lies on the circle that passes through the feet of the altitudes of triangle \( ABC \).

Let $ABC$ be a triangle inscribed in circle $(O),$ with its altitudes $BE, CF$ intersect at orthocenter $H$ ($E \in AC, F \in AB$). Let $M$ be the midpoint of $BC, K$ be the orthogonal projection of $H$ on $AM$. $EF$ intersects $BC$ at $P$. Let $Q$ be the intersection of tangent of $(O)$ which passes through $A$ with $BC, T$ be the reflection of $Q$ through $P$. Prove that $\angle OKT = 90^o$.

Let \( H \) be the orthocenter of an acute triangle \( ABC \), and let \( AT \) be the diameter of the circumcircle of this triangle. Points \( X \) and \( Y \) are chosen on sides \( AC \) and \( AB \), respectively, such that \( TX = TY \) and \( \angle XTY + \angle XAY = 90^\circ \). Prove that \( \angle XHY = 90^\circ \). [i] Proposed by Matthew Kurskyi[/i]

Consider acute triangles $ABC$ and $BCD$, with $\angle BAC = \angle BDC$, such that $A$ and $D$ are on opposite sides of line $BC$. Denote by $E$ the foot of the perpendicular line to $AC$ through $B$ and by $F$ the foot of the perpendicular line to $BD$ through $C$. Let $H_1$ be the orthocenter of triangle $ABC$ and $H_2$ be the orthocenter of $BCD$. Show that lines $AD, EF$ and $H_1H_2$ are concurrent.

Let $ABC$ be a triangle, $O{}$ be its circumcenter, $I{}$ its incenter and $I_A,I_B,I_C$ the excenters. Let $M$ be the midpoint of $BC$ and $H_1$ and $H_2$ be the orthocenters of the triangles $MII_A$ and $MI_BI_C$. Prove that the parallel to $BC$ through $O$ passes through the midpoint of the segment $H_1H_2$. [i]Proposed by David Anghel[/i]

Let $ABC$ be a triangle with orthocenter $H$, $\Gamma$ its circumcircle, and $A' \neq A$, $B' \neq B$, $C' \neq C$ points on $\Gamma$. Define $l_a$ as the line that passes through the projections of $A'$ over $AB$ and $AC$. Define $l_b$ and $l_c$ similarly. Let $O$ be the circumcenter of the triangle determined by $l_a$, $l_b$ and $l_c$ and $H'$ the orthocenter of $A'B'C'$. Show that $O$ is midpoint of $HH'$.

An acute-angle non-isosceles triangle $ ABC $ is given. The point $ H $ is its orthocenter, the points $ O $ and $ I $ are the centers of its circumscribed and inscribed circles, respectively. The circumcircle of the triangle $ OIH $ passes through the vertex $ A $. Prove that one of the angles of the triangle is $ 60^\circ $.

Let $BE$ and $CF$ be the altitudes of acute triangle $ABC$. Let $H$ be the orthocenter of $ABC$ and $M$ be the midpoint of side $BC$. The points of intersection of the midperpendicular line to $BC$ with segments $BE$ and $CF$ are denoted by $K$ and $L$ respectively. The point $Q$ is the orthocenter of triangle $KLH$. Prove that $Q$ belongs to the median $AM$. (Bohdan Zheliabovskyi)

In the triangle $ABC$ the altitude $BD$ and $CT$ are drawn, they intersect at the point $H$. The point $Q$ is the foot of the perpendicular drawn from the point $H$ on the bisector of the angle $A$. Prove that the bisector of the external angle $A$ of the triangle $ABC$, the bisector of the angle $BHC$ and the line $QM$, where $M$ is the midpoint of the segment $DT$, intersect at one point. (Matvsh Kursky)

$ABCD$ is a convex quadrangle, $G$ is its center of gravity as a homogeneous plate (i.e., the intersection point of two lines, each of which connects the centroids of triangles having a common diagonal). a) Suppose that around $ABCD$ we can circumscribe a circle centered on $O$. We define $H$ similarly to $G$, taking orthocenters instead of centroids. Then the points of $H, G, O$ lie on the same line and $HG: GO = 2: 1$. b) Suppose that in $ABCD$ we can inscribe a circle centered on $I$. The Nagel point N of the circumscribed quadrangle is the intersection point of two lines, each of which passes through points on opposite sides of the quadrangle that are symmetric to the tangent points of the inscribed circle relative to the midpoints of the sides. (These lines divide the perimeter of the quadrangle in half). Then $N, G, I$ lie on one straight line, with $NG: GI = 2: 1$.

The circle circumscribed about an acute triangle $ABC$ and the vertex $C$ are fixed. Orthocenter $H$ moves in a circle with center at point $C$. Find the locus of the midpoints of the segments connecting the feet of altitudes drawn from vertices $A$ and $B$.

Let $ABC$ be acute-angled triangle with circumcircle $\Gamma$. Points $H$ and $M$ are the orthocenter and the midpoint of $BC$ respectively. The line $HM$ meets the circumcircle $\omega$ of triangle $BHC$ at point $N\not= H$. Point $P$ lies on the arc $BC$ of $\omega$ not containing $H$ in such a way that $\angle HMP = 90^\circ$. The segment $PM$ meets $\Gamma$ at point $Q$. Points $B'$ and $C'$ are the reflections of $A$ about $B$ and $C$ respectively. Prove that the circumcircles of triangles $AB'C'$ and $PQN$ are tangent.