Olympiad problems

2018 Hanoi Open Mathematics Competitions, 12

Tags: altitude , geometry , circumcircle , tangent , parallel

Let $ABC$ be an acute triangle with $AB < AC$, and let $BE$ and $CF$ be the altitudes. Let the median $AM$ intersect $BE$ at point $P$, and let line $CP$ intersect $AB$ at point $D$ (see Figure 2). Prove that $DE \parallel BC$, and $AC$ is tangent to the circumcircle of $\vartriangle DEF$. [img]https://cdn.artofproblemsolving.com/attachments/f/7/bbad9f6019a77c6aa46c3a821857f06233cb93.png[/img]

2023 Ukraine National Mathematical Olympiad, 9.3

Tags: altitude , geometry

You are given an acute triangle $ABC$ with circumcircle $\omega$. Points $F$ on $AC$, $E$ on $AB$ and $P, Q$ on $\omega$ are chosen so that $\angle AFB = \angle AEC = \angle APE = \angle AQF = 90^\circ$. Show that lines $BC, EF, PQ$ are concurrent or parallel. [i]Proposed by Fedir Yudin[/i]

Ukrainian TYM Qualifying - geometry, 2019.10

Tags: geometry , orthocenter , altitude

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 $ABC$.

2019 AIME Problems, 15

Tags: geometry , circumcircle , altitude , power of a point

In acute triangle $ABC$ points $P$ and $Q$ are the feet of the perpendiculars from $C$ to $\overline{AB}$ and from $B$ to $\overline{AC}$, respectively. Line $PQ$ intersects the circumcircle of $\triangle ABC$ in two distinct points, $X$ and $Y$. Suppose $XP=10$, $PQ=25$, and $QY=15$. The value of $AB\cdot AC$ can be written in the form $m\sqrt n$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$.

Cono Sur Shortlist - geometry, 2009.G4

Tags: altitude , geometry , perpendicular , incenter

Let $AA _1$ and $CC_1$ be altitudes of an acute triangle $ABC$. Let $I$ and $J$ be the incenters of the triangles $AA_1C$ and $AC_1C$ respectively. The $C_1J$ and $A_1 I$ lines cut into $T$. Prove that lines $AT$ and $TC$ are perpendicular.

2010 Swedish Mathematical Competition, 1

Tags: geometry , altitude

Exists a triangle whose three altitudes have lengths $1, 2$ and $3$ respectively?

2019 Saudi Arabia JBMO TST, 2

Tags: geometry , equal angles , equal segments , midpoint , altitude

In triangle $ABC$ point $M$ is the midpoint of side $AB$, and point $D$ is the foot of altitude $CD$. Prove that $\angle A = 2\angle B$ if and only if $AC = 2MD$

Estonia Open Senior - geometry, 2009.2.4

Tags: geometry , circumcircle , altitude , tangent

a) An altitude of a triangle is also a tangent to its circumcircle. Prove that some angle of the triangle is larger than $90^o$ but smaller than $135^o$. b) Some two altitudes of the triangle are both tangents to its circumcircle. Find the angles of the triangle.

1984 Tournament Of Towns, (061) O2

Tags: geometry , 3d geometry , tetrahedron , parallel , plane , altitude

Six altitudes are constructed from the three vertices of the base of a tetrahedron to the opposite sides of the three lateral faces. Prove that all three straight lines joining two base points of the altitudes in each lateral face are parallel to a certain plane. (IF Sharygin, Moscow)

2013 Singapore Senior Math Olympiad, 1

Tags: geometry , circumcircle , altitude , isosceles triangle , right triangle , length , triangle

In the Triangle ABC AB>AC, the extension of the altitude AD with D lying inside BC intersects the circum-circle of the Triangle ABC at P. The circle through P and tangent to BC at D intersects the circum-circle of Triangle ABC at Q distinct from P with PQ=DQ. Prove that AD=BD-DC

1987 Tournament Of Towns, (133) 2

Tags: altitude , parallel , geometry

In an acute angled triangle the feet of the altitudes are joined to form a new triangle. In this new triangle it is known that two sides are parallel to sides of the original triangle . Prove that the third side is also parallel to one of the sides of the original triangle .

2013 Junior Balkan Team Selection Tests - Romania, 4

Tags: altitude , geometry , angle bisector , circumcircle , equal segments

In the acute-angled triangle $ABC$, with $AB \ne AC$, $D$ is the foot of the angle bisector of angle $A$, and $E, F$ are the feet of the altitudes from $B$ and $C$, respectively. The circumcircles of triangles $DBF$ and $DCE$ intersect for the second time at $M$. Prove that $ME = MF$. Leonard Giugiuc

2006 Estonia Team Selection Test, 4

Tags: altitude , concyclic , cyclic , circles , geometry , cyclic quadrilateral

The side $AC$ of an acute triangle $ABC$ is the diameter of the circle $c_1$ and side $BC$ is the diameter of the circle $c_2$. Let $E$ be the foot of the altitude drawn from the vertex $B$ of the triangle and $F$ the foot of the altitude drawn from the vertex $A$. In addition, let $L$ and $N$ be the points of intersection of the line $BE$ with the circle $c_1$ (the point $L$ lies on the segment $BE$) and the points of intersection of $K$ and $M$ of line $AF$ with circle $c_2$ (point $K$ is in section $AF$). Prove that $K LM N$ is a cyclic quadrilateral.

2016 Kazakhstan National Olympiad, 4

Tags: altitude , geometry

In isosceles triangle $ABC$($CA=CB$),$CH$ is altitude and $M$ is midpoint of $BH$.Let $K$ be the foot of the perpendicular from $H$ to $AC$ and $L=BK \cap CM$ .Let the perpendicular drawn from $B$ to $BC$ intersects with $HL$ at $N$.Prove that $\angle ACB=2 \angle BCN$.(M. Kunhozhyn)

2012 Oral Moscow Geometry Olympiad, 3

Tags: altitude , orthocenter , geometry , concurrency

$H$ 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 $CP$.

2018 Estonia Team Selection Test, 7

Tags: altitude , equal segments , circumcircle , concurrency , geometry

Let $AD$ be the altitude $ABC$ of an acute triangle. On the line $AD$ are chosen different points $E$ and $F$ so that $|DE |= |DF|$ and point $E$ is in the interior of triangle $ABC$. The circumcircle of triangle $BEF$ intersects $BC$ and $BA$ for second time at points $K$ and $M$ respectively. The circumcircle of the triangle $CEF$ intersects the $CB$ and $CA$ for the second time at points $L$ and $N$ respectively. Prove that the lines $AD, KM$ and $LN$ intersect at one point.

2007 Estonia Team Selection Test, 2

Tags: circumradius , inradius , altitude , geometry , geometric inequalities , circumcircle , incircle

Let $D$ be the foot of the altitude of triangle $ABC$ drawn from vertex $A$. Let $E$ and $F$ be points symmetric to $D$ w.r.t. lines $AB$ and $AC$, respectively. Let $R_1$ and $R_2$ be the circumradii of triangles $BDE$ and $CDF$, respectively, and let $r_1$ and $r_2$ be the inradii of the same triangles. Prove that $|S_{ABD} - S_{ACD}| > |R_1r_1 - R_2r_2|$

2025 Junior Macedonian Mathematical Olympiad, 2

Tags: altitude , geometry , circumcircle

Let $B_1$ be the foot of the altitude from the vertex $B$ in the acute-angled $\triangle ABC$. Let $D$ be the midpoint of side $AB$, and $O$ be the circumcentre of $\triangle ABC$. Line $B_1D$ meets line $CO$ at $E$. Prove that the points $B, C, B_1$, and $E$ lie on a circle.

2006 Dutch Mathematical Olympiad, 2

Tags: altitude , geometry

Given is a acute angled triangle $ABC$. The lengths of the altitudes from $A, B$ and $C$ are successively $h_A, h_B$ and $h_C$. Inside the triangle is a point $P$. The distance from $P$ to $BC$ is $1/3 h_A$ and the distance from $P$ to $AC$ is $1/4 h_B$. Express the distance from $P$ to $AB$ in terms of $h_C$.

2011 Sharygin Geometry Olympiad, 6

Tags: geometry , circumcircle , altitude , collinear , collinearity , parallel lines

In triangle $ABC$ $AA_0$ and $BB_0$ are medians, $AA_1$ and $BB_1$ are altitudes. The circumcircles of triangles $CA_0B_0$ and $CA_1B_1$ meet again in point $M_c$. Points $M_a, M_b$ are defined similarly. Prove that points $M_a, M_b, M_c$ are collinear and lines $AM_a, BM_b, CM_c$ are parallel.

2018 Yasinsky Geometry Olympiad, 3

Tags: geometry , angle bisector , construction , altitude

Construct triangle $ABC$, given the altitude and the angle bisector both from $A$, if it is known for the sides of the triangle $ABC$ that $2BC = AB + AC$. (Alexey Karlyuchenko)

1935 Moscow Mathematical Olympiad, 008

Tags: arithmetic progression , inradius , altitude , geometry , arithmetic sequence

Prove that if the lengths of the sides of a triangle form an arithmetic progression, then the radius of the inscribed circle is one third of one of the heights of the triangle.

2014 Switzerland - Final Round, 8

Tags: altitude , concyclic , geometry

In the acute-angled triangle $ABC$, let $M$ be the midpoint of the atlitude $h_b$ through $B$ and $N$ be the midpoint of the height $h_c$ through $C$. Further let $P$ be the intersection of $AM$ and $h_c$ and $Q$ be the intersection of $AN$ and $h_b$. Show that $M, N, P$ and $Q$ lie on a circle.

2018 Ukraine Team Selection Test, 10

Tags: midpoint , geometry , circumcircle , altitude , equal angles

Let $ABC$ be a triangle with $AH$ altitude. The point $K$ is chosen on the segment $AH$ as follows such that $AH =3KH$. Let $O$ be the center of the circle circumscribed around by triangle $ABC, M$ and $N$ be the midpoints of $AC$ and AB respectively. Lines $KO$ and $MN$ intersect at the point $Z$, a perpendicular to $OK$ passing through point $Z$ intersects lines $AC$ and $AB$ at points $X$ and $Y$ respectively. Prove that $\angle XKY =\angle CKB$.

2014 Estonia Team Selection Test, 4

Tags: altitude , geometry , circumcircle , equal segments

In an acute triangle the feet of altitudes drawn from vertices $A$ and $B$ are $D$ and $E$, respectively. Let $M$ be the midpoint of side $AB$. Line $CM$ intersects the circumcircle of $CDE$ again in point $P$ and the circumcircle of $CAB$ again in point $Q$. Prove that $|MP| = |MQ|$.