Let $M$ be a point on the side $AB$ of triangle $ABC$. The length $AB = c$ and $\angle CMA=\phi$ are given. Find the distance between the orthocentres (intersection points of altitudes) of the triangles $AMC$ and $BMC$. (IF Sharygin)

Let $O$ be the circumcenter and $H$ the orthocenter of an acute-angled triangle $ABC$ such that $BC>CA$. Let $F$ be the foot of the altitude $CH$ of triangle $ABC$. The perpendicular to the line $OF$ at the point $F$ intersects the line $AC$ at $P$. Prove that $\measuredangle FHP=\measuredangle BAC$.

Let $ABC$ be an acute-angled triangle with $AB \ne AC$, and let $I$ and $O$ be its incenter and circumcenter, respectively. Let the incircle touch $BC, CA$ and $AB$ at $D, E$ and $F$, respectively. Assume that the line through $I$ parallel to $EF$, the line through $D$ parallel to$ AO$, and the altitude from $A$ are concurrent. Prove that the concurrency point is the orthocenter of the triangle $ABC$. Petar Nizic-Nikolac, Croatia

The inscribed circle of the non-isosceles triangle $ABC$ touches sides $AB, BC$ and $AC$ at points $C_1, A_1$ and $B_1$, respectively. The circumscribed circle of the triangle $A_1BC_1$ intersects the lines $B_1A_1$ and $B_1C_1$ at the points $A_0$ and $C_0$, respectively. Prove that the orthocenter of triangle $A_0BC_0$, the center of the inscribed circle of triangle $ABC$ and the midpoint of the $AC$ lie on one straight line.

A straight line passing through the center of the circumscribed circle and the intersection point of the heights of the non-equilateral triangle $ABC$ divides its perimeter and area in the same ratio.Find this ratio.

In triangle $ABC$, let $I$ be the incentre. Let $H$ be the orthocentre of triangle $BIC$. Show that $AH$ is parallel to $BC$ if and only if $H$ lies on the circle with diameter $AI$.

In triangle $ABC, \angle C= 60^o, \angle A= 45^o$. Let $M$ be the midpoint of $BC, H$ be the orthocenter of triangle $ABC$. Prove that line $MH$ passes through the midpoint of arc $AB$ of the circumcircle of triangle $ABC$.

At the altitude $AH_1$ of an acute non-isosceles triangle $ABC$ chose a point $X$ , from which draw the perpendiculars $XN$ and $XM$ on the sides $AB$ and $AC$ respectively. It turned out that $H_1A$ is the angle bisector $MH_1N$. Prove that $X$ is the point of intersection of the altitudes of the triangle $ABC$.

Let $BD$ and $CE$ be altitudes of an arbitrary scalene triangle $ABC$ with orthocenter $H$ and circumcenter $O$. Let $M$ and $N$ be the midpoints of sides $AB$, respectively $AC$, and $P$ the intersection point of lines $MN$ and $DE$. Prove that lines $AP$ and $OH$ are perpendicular. Liana Topan

Let $ABC$ be an acute triangle, with $\angle A > 60^\circ$, and let $H$ be it's orthocenter. Let $M$ and $N$ be points on $AB$ and $AC$, respectively, such that $\angle HMB = \angle HNC = 60^\circ$. Also, let $O$ be the circuncenter of $HMN$ and $D$ be a point on the semiplane determined by $BC$ that contains $A$ in such a way that $DBC$ is equilateral. Prove that $H$, $O$ and $D$ are collinear.

Given an acute triangle $ABC$. with $H$ as its orthocenter, lines $\ell_1$ and $\ell_2$ go through $H$ and are perpendicular to each other. Line $\ell_1$ cuts $BC$ and the extension of $AB$ on $D$ and $Z$ respectively. Whereas line $\ell_2$ cuts $BC$ and the extension of $AC$ on $E$ and $X$ respectively. If the line through $D$ and parallel to $AC$ and the line through $E$ parallel to $AB$ intersects at $Y$, prove that $X,Y,Z$ are collinear.

A circle passing through $A, B$ and the orthocenter of triangle $ABC$ meets sides $AC, BC$ at their inner points. Prove that $60^o < \angle C < 90^o$ . (A. Blinkov)

A triangle $ABC$ is given with barycentre $G$ and circumcentre $O$. The perpendicular bisectors of $GA, GB$ meet at $C_1$,of $GB,GC$ meet at $A _1$, and $GC,GA$ meet at $B_1$. Prove that $O$ is the barycenter of the triangle $A_1B_1C_1$.

Let $ABC$ a triangle inscribed in a circle $(O)$ with orthocenter $H$. Two lines $d_1$ and $d_2$ are mutually perpendicular at $H$. Let $d_1$ meet $BC,CA,AB$ at $X_1, Y_1,Z_1$ respectively. Let $A_1B_1C_1$ be a triangle formed by the line through $X_1$ perpendicular to $BC$, the line through $Y_1$ perpendicular to CA, the line through $Z_1$ perpendicular perpendicular to $AB$. Triangle $A_2B_2C_2$ is defined in the same manner. Prove that the circumcircles of triangles $A_1B_1C_1$ and $A_2B_2C_2$ touch each other at a point on $(O)$. Nguyễn Văn Linh

Let $ABC$ be an acute, non isosceles triangle with $O,H$ are circumcenter and orthocenter, respectively. Prove that the nine-point circles of $AHO,BHO,CHO$ has two common points.

Let $ABC$ be an acute triangle. Suppose the points $A',B',C'$ lie on its sides $BC,AC,AB$ respectively and the segments $AA',BB',CC'$ intersect in a common point $P$ inside the triangle. For each of those segments let us consider the circle such that the segment is its diameter, and the chord of this circle that contains the point $P$ and is perpendicular to this diameter. All three these chords occurred to have the same length. Prove that $P$ is the orthocenter of the triangle $ABC$. (Grigory Galperin)

Let $ABC$ be an acute triangle with incenter $O$, orthocenter $H$, altitude $AD. AO$ meets $BC$ at $E$. Line through $D$ parallel to $OH$ meet $AB,AC$ at $M,N$, respectively. Let $I$ be the midpoint of $AE$, and $DI$ intersect $AB,AC$ at $P,Q$ respectively. $MQ$ meets $NP$ at $T$. Prove that $D,O, T$ are collinear. Trần Quang Hùng

In triangle $ABC$, angle $B$ is not right. The circle inscribed in $ABC$ touches $AB$ and $BC$ at points $C_1$ and $A_1$, and the feet of the altitudes drawn to the sides $AB$ and $BC$ are points $C_2$ and $A_2$. Prove that the intersection point of the altitudes of triangle $A_1BC_1$ is the center of the circle inscribed in triangle $A_2BC_2$.

for acute triangle $\triangle ABC$ with orthocenter $H$, and $E,F$ the feet of altitudes for $B,C$, we have $P$ on $EF$ such as that $HO \perp HP$. $Q$ is on segment $AH$ so $HM \perp PQ$. prove $QA=3QH$

$H$ is the intersection point of the heights $AA'$ and $BB'$ of the acute-angled triangle $ABC$. A straight line, perpendicular to $AB$, intersects these heights at points $D$ and $E$, and side $AB$ at point $P$. Prove that the orthocenter of the triangle $DEH$ lies on segment $CP$.

Let $ Q $ be a point, $ H,O $ be the orthocenter and circumcenter, respectively, of a triangle $ ABC, $ and $ D,E,F, $ be the symmetric points of $ Q $ with respect to $ A,B,C, $ respectively. Also, $ M,N,P $ are the middle of the segments $ AE,BF,CD, $ and $ G,G',G'' $ are the centroids of $ ABC,MNP,DEF, $ respectively. Prove the following propositions: [b]a)[/b] $ \frac{1}{2}\overrightarrow{OG} =\frac{1}{3}\overrightarrow{OG'}=\frac{1}{4}\overrightarrow{OG''} $ [b]b)[/b] $ Q=O\implies \overrightarrow{OG'} =\overrightarrow{G'H} $ [b]c)[/b] $ Q=H\implies G'=O $ [i]Cătălin Zîrnă[/i]

In a quadrilateral $ABCD$ the intersection of the diagonals is called $P$. Point $X$ is the orthocentre of triangle $PAB$. (The orthocentre of a triangle is the point where the three altitudes of the triangle intersect.) Point $Y$ is the orthocentre of triangle $PCD$. Suppose that $X$ lies inside triangle $PAB$ and $Y$ lies inside triangle $PCD$. Moreover, suppose that $P$ is the midpoint of line segment $XY$ . Prove that $ABCD$ is a parallelogram. [asy] import geometry; unitsize (1.5 cm); pair A, B, C, D, P, X, Y; A = (0,0); B = (2,-0.5); C = (3.5,2.2); D = A + C - B; P = (A + C)/2; X = orthocentercenter(A,B,P); Y = orthocentercenter(C,D,P); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); draw(A--extension(A,X,B,P), dotted); draw(B--extension(B,X,A,P), dotted); draw(P--extension(P,X,A,B), dotted); draw(C--extension(C,Y,D,P), dotted); draw(D--extension(D,Y,C,P), dotted); draw(P--extension(P,Y,C,D), dotted); dot("$A$", A, W); dot("$B$", B, S); dot("$C$", C, E); dot("$D$", D, N); dot("$P$", P, E); dot("$X$", X, NW); dot("$Y$", Y, SE); [/asy]

Let $\Gamma$ be the circle of an acute triangle $ABC$ and let $H$ be its orthocenter. The circle $\omega$ with diameter $AH$ cuts $\Gamma$ at point $D$ ($D\ne A$). Let $M$ be the midpoint of the smaller arc $BC$ of $\Gamma$ . Let $N$ be the midpoint of the largest arc $BC$ of the circumcircle of the triangle $BHC$. Prove that there is a circle that passes through the points $D, H, M$ and $N$.

In triangle $ABC$, the point $H$ is the orthocenter. A circle centered at point $H$ and with radius $AH$ intersects the lines $AB$ and $AC$ at points $E$ and $D$, respectively. The point $X$ is the symmetric of the point $A$ with respect to the line $BC$ . Prove that $XH$ is the bisector of the angle $DXE$. (Matthew of Kursk)

An acute-angled triangle $ABC$ with an altitude $AD$ and orthocenter $H$ are given. $AD$ intersects the circumcircle of $ABC$ $\omega$ at $P$. $K$ is a point on segment $BC$ such that $KC=BD$. The circumcircle of $KPH$ intersects $\omega$ at $Q$ and $BC$ at $N$. A line perpendicular to $PQ$ and passing through $N$ intersects $AD$ at $T$. Prove that the center of $\omega$ lies on line $TK$. [i]U. Maksimenkau[/i]