Olympiad problems

2017 Junior Balkan Team Selection Tests - Romania, 1

Let $P$ be a point in the interior of the acute-angled triangle $ABC$. Prove that if the reflections of $P$ with respect to the sides of the triangle lie on the circumcircle of the triangle, then $P$ is the orthocenter of $ABC$.

2018 Saudi Arabia BMO TST, 4

Tags: geometry , circles , orthocenter , perpendicular , incenter

Let $ABC$ be an acute, non isosceles with $I$ is its incenter. Denote $D, E$ as tangent points of $(I)$ on $AB,AC$, respectively. The median segments respect to vertex $A$ of triangles $ABE$ and $ACD$ meet$ (I)$ at$ P,Q,$ respectively. Take points $M, N$ on the line $DE$ such that $AM \parallel BE$ and $AN \parallel C D$ respectively. a) Prove that $A$ lies on the radical axis of $(MIP)$ and $(NIQ)$. b) Suppose that the orthocenter $H$ of triangle $ABC$ lies on $(I)$. Prove that there exists a line which is tangent to three circles of center $A, B, C$ and all pass through $H$.

Ukraine Correspondence MO - geometry, 2010.11

Tags: geometry , orthocenter , incenter , angle

Let $ABC$ be an acute-angled triangle in which $\angle BAC = 60^o$ and $AB> AC$. Let $H$ and $I$ denote the points of intersection of the altitudes and angle bisectors of this triangle, respectively. Find the ratio $\angle ABC: \angle AHI$.

2018 Ukraine Team Selection Test, 9

Tags: altitude , geometry , circumcircle , concyclic , orthocenter

Let $AA_1, BB_1, CC_1$ be the heights of triangle $ABC$ and $H$ be its orthocenter. Liune $\ell$ parallel to $AC$, intersects straight lines $AA_1$ and $CC_1$ at points $A_2$ and $C_2$, respectively. Suppose that point $B_1$ lies outside the circumscribed circle of triangle $A_2 HC_2$. Let $B_1P$ and $B_1T$ be tangent to of this circle. Prove that points $A_1, C_1, P$, and $T$ are cyclic.

2022-IMOC, G1

Tags: geometry , circumcenter , orthocenter , concurrency

The circumcenter and orthocenter of $ABC$ are $O$ and $H$, respectively. Let $XACH$ be a parallelogram. Show that if $OH$ is parallel to $BC$, then $OX$ and $AB$ intersect at some point on the perpendicular bisector of $AH$. [i]proposed by USJL[/i]

Ukrainian TYM Qualifying - geometry, 2016.1

Tags: geometry , orthocenter , right triangle

The points $K$ and $N$ lie on the hypotenuse $AB$ of a right triangle $ABC$. Prove that orthocenters the triangles $BCK$ and $ACN$ coincide if and only if $\frac{BN}{AK}=\tan^2 A.$

Mathley 2014-15, 4

Tags: circumcenter , equal segments , orthocenter , isosceles , trigonometry

Points $E, F$ are in the plane of triangle $ABC$ so that triangles $ABE$ and $ACF$ are the opposite directed, and the two triangles are isosceles in that $BE = AE, AF = CF$. Let $H, K$ be the orthocenter of triangle $ABE, ACF$ respectively. Points $M, N$ are the intersections of $BE$ and $CF, CK$ and $CH$. Prove that $MN$ passes through the center of the circumcircle of triangle $ABC$. Nguyen Minh Ha, High School for Education, Hanoi Pedagogical University

2011 Belarus Team Selection Test, 2

Tags: geometry , circumcircle , angle bisector , orthocenter

The external angle bisector of the angle $A$ of an acute-angled triangle $ABC$ meets the circumcircle of $\vartriangle ABC$ at point $T$. The perpendicular from the orthocenter $H$ of $\vartriangle ABC$ to the line $TA$ meets the line $BC$ at point $P$. The line $TP$ meets the circumcircce of $\vartriangle ABC$ at point $D$. Prove that $AB^2+DC^2=AC^2+BD^2$ A. Voidelevich

2015 Sharygin Geometry Olympiad, 7

Tags: angle , orthocenter , centroid , geometry

Let $ABC$ be an acute-angled, nonisosceles triangle. Altitudes $AA'$ and $BB' $meet at point $H$, and the medians of triangle $AHB$ meet at point $M$. Line $CM$ bisects segment $A'B'$. Find angle $C$. (D. Krekov)

2001 Saint Petersburg Mathematical Olympiad, 11.5

Tags: geometry , orthocenter , incenter , circumcircle

Let $I$ and $H$ be the incenter and orthocenter of an acute triangle $ABC$. $M$ is the midpoint of arc $AC$ of circumcircle of triangle $ABC$ which does not contain point $B$. If $MI=MH$, find the measure of angle $\angle ABC$. [I]Proposed by F. Bakharev[/i]

IV Soros Olympiad 1997 - 98 (Russia), 9.3

Tags: ratio , geometry , orthocenter , angle

What is angle $B$ of triangle$ ABC$, if it is known that the altitudes drawn from $A$ and $C$ intersect inside the triangle and one of them is divided by of intersection point into equal parts, and the other one in the ratio of $2: 1$, counting from the vertex?

2009 Brazil Team Selection Test, 1

Tags: right triangle , orthocenter , geometry , cyclic , pentagon

Let $A, B, C, D, E$ points in circle of radius r, in that order, such that $AC = BD = CE = r$. The points $H_1, H_2, H_3$ are the orthocenters of the triangles $ACD$, $BCD$ and $BCE$, respectively. Prove that $H_1H_2H_3$ is a right triangle .

1996 Iran MO (3rd Round), 5

Tags: geometry , circumcircle , symmetry , trigonometry , orthocenter

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 $\measuredangle FHP=\measuredangle BAC$.

2018 Bosnia and Herzegovina Team Selection Test, 1

Tags: geometry , perpendicular , orthocenter

In acute triangle $ABC$ $(AB < AC)$ let $D$, $E$ and $F$ be foots of perpedicular from $A$, $B$ and $C$ to $BC$, $CA$ and $AB$, respectively. Let $P$ and $Q$ be points on line $EF$ such that $DP \perp EF$ and $BQ=CQ$. Prove that $\angle ADP = \angle PBQ$

Estonia Open Junior - geometry, 2015.2.5

Tags: geometry , parallelogram , circumcircle , orthocenter

Let $ABC$ be an acute-angled triangle, $H$ the intersection point of its altitudes , and $AA'$ the diameter of the circumcircle of triangle $ABC$. Prove that the quadrilateral $HB A'C$ is a parallelogram.

2015 Sharygin Geometry Olympiad, P2

Tags: geometry , circumcenter , orthocenter , incenter , angle

Let $O$ and $H$ be the circumcenter and the orthocenter of a triangle $ABC$. The line passing through the midpoint of $OH$ and parallel to $BC$ meets $AB$ and $AC$ at points $D$ and $E$. It is known that $O$ is the incenter of triangle $ADE$. Find the angles of $ABC$.

2019 Ukraine Team Selection Test, 3

Tags: geometry , circumcircle , orthocenter , altitude , tangent circle

Given an acute triangle $ABC$ . It's altitudes $AA_1 , BB_1$ and $CC_1$ intersect at a point $H$ , the orthocenter of $\vartriangle ABC$. Let the lines $B_1C_1$ and $AA_1$ intersect at a point $K$, point $M$ be the midpoint of the segment $AH$. Prove that the circumscribed circle of $\vartriangle MKB_1$ touches the circumscribed circle of $\vartriangle ABC$ if and only if $BA1 = 3A1C$. (Bondarenko Mykhailo)

2019 Tournament Of Towns, 2

Tags: geometry , fixed point , fixed , orthocenter , circles

Let $\omega$ be a circle with the center $O$ and $A$ and $C$ be two different points on $\omega$. For any third point $P$ of the circle let $X$ and $Y$ be the midpoints of the segments $AP$ and $CP$. Finally, let $H$ be the orthocenter (the point of intersection of the altitudes) of the triangle $OXY$ . Prove that the position of the point H does not depend on the choice of $P$. (Artemiy Sokolov)

2013 BAMO, 3

Tags: geometry , congruent triangles , circumcenter , orthocenter

Let $H$ be the orthocenter of an acute triangle $ABC$. (The orthocenter is the point at the intersection of the three altitudes. An acute triangle has all angles less than $90^o$.) Draw three circles: one passing through $A, B$, and $H$, another passing through $B, C$, and $H$, and finally, one passing through $C, A$, and $H$. Prove that the triangle whose vertices are the centers of those three circles is congruent to triangle $ABC$.

Russian TST 2015, P3

Tags: geometry , incenter , orthocenter

The triangle $ABC$ is given. Let $A'$ be the midpoint of the side $BC$, $B_c{}$ be the projection of $B{}$ onto the bisector of the angle $ACB{}$ and $C_b$ be the projection of the point $C{}$ onto the bisector of the angle $ABC$. Let $A_0$ be the center of the circle passing through $A', B_c, C_b$. The points $B_0$ and $C_0$ are defined similarly. Prove that the incenter of the triangle $ABC$ coincides with the orthocenter of the triangle $A_0B_0C_0$.

2016 Romania National Olympiad, 1

Tags: geometry , circumcircle , orthocenter , pure geometry

The orthocenter $ H $ of a triangle $ ABC $ is distinct from its vertices and its circumcenter $ O. $ $ M,N,P $ are the circumcenters of the triangles $ HBC,HCA, $ respectively, $ HAB. $ Prove that $ AM,BN,CP $ and $ OH $ are concurrent.

2005 Sharygin Geometry Olympiad, 5

Tags: geometry , locus , orthocenter , circumcenter , centroid , circumcircle

There are two parallel lines $p_1$ and $p_2$. Points $A$ and $B$ lie on $p_1$, and $C$ on $p_2$. We will move the segment $BC$ parallel to itself and consider all the triangles $AB'C '$ thus obtained. Find the locus of the points in these triangles: a) points of intersection of heights, b) the intersection points of the medians, c) the centers of the circumscribed circles.

2005 Sharygin Geometry Olympiad, 9.4

Tags: geometry , orthocenter , midpoint

Let $P$ be the intersection point of the diagonals of the quadrangle $ABCD$, $M$ the intersection point of the lines connecting the midpoints of its opposite sides, $O$ the intersection point of the perpendicular bisectors of the diagonals, $H$ the intersection point of the lines connecting the orthocenters of the triangles $APD$ and $BCP$, $APB$ and $CPD$. Prove that $M$ is the midpoint of $OH$.

1979 Brazil National Olympiad, 3

Tags: orthocenter , geometry

The vertex C of the triangle ABC is allowed to vary along a line parallel to AB. Find the locus of the orthocenter.

2022 Iran MO (3rd Round), 2

Tags: geometry , dilation , homothety , orthocenter

Constant points $B$ and $C$ lie on the circle $\omega$. The point middle of $BC$ is named $M$ by us. Assume that $A$ is a variable point on the $\omega$ and $H$ is the orthocenter of the triangle $ABC$. From the point $H$ we drop a perpendicular line to $MH$ to intersect the lines $AB$ and $AC$ at $X$ and $Y$ respectively. Prove that with the movement of $A$ on the $\omega$, the orthocenter of the triangle $AXY$ also moves on a circle.