Olympiad problems

2018 Junior Regional Olympiad - FBH, 5

In triangle $ABC$ length of altitude $CH$, with $H \in AB$, is equal to half of side $AB$. If $\angle BAC = 45^{\circ}$ find $\angle ABC$

Estonia Open Senior - geometry, 2015.2.5

Tags: geometry , similar triangles , altitude

The triangle $K_2$ has as its vertices the feet of the altitudes of a non-right triangle $K_1$. Find all possibilities for the sizes of the angles of $K_1$ for which the triangles $K_1$ and $K_2$ are similar.

2014 Estonia Team Selection Test, 4

Tags: geometry , circumcircle , equal segments , altitude

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 $|MP| = |MQ|$.

2019 BAMO, E/3

Tags: geometry , angle bisector , median , altitude , equilateral

In triangle $\vartriangle ABC$, we have marked points $A_1$ on side $BC, B_1$ on side $AC$, and $C_1$ on side $AB$ so that $AA_1$ is an altitude, $BB_1$ is a median, and $CC_1$ is an angle bisector. It is known that $\vartriangle A_1B_1C_1$ is equilateral. Prove that $\vartriangle ABC$ is equilateral too. (Note: A median connects a vertex of a triangle with the midpoint of the opposite side. Thus, for median $BB_1$ we know that $B_1$ is the midpoint of side $AC$ in $\vartriangle ABC$.)

Durer Math Competition CD 1st Round - geometry, 2015.D4

Tags: geometry , circumcircle , altitude

The altitude of the acute triangle $ABC$ drawn from $A$ , intersects the side $BC$ at $A_1$ and the circumscribed circle at $A_2$ (different from $A$). Similarly, we get the points $B_1$, $B_2$, $C_1$, $C_2$. Prove that $$\frac{AA_2}{AA_1}+\frac{BB_2}{BB_1}+\frac{CC_2}{CC_1}= 4.$$

2011 Belarus Team Selection Test, 2

Tags: geometry , angle bisector , altitude , ratio

Points $L$ and $H$ are marked on the sides $AB$ of an acute-angled triangle ABC so that $CL$ is a bisector and $CH$ is an altitude. Let $P,Q$ be the feet of the perpendiculars from $L$ to $AC$ and $BC$ respectively. Prove that $AP \cdot BH = BQ \cdot AH$. I. Gorodnin

2012 Greece JBMO TST, 3

Tags: geometry , perpendicular , symmetry , circumcircle , circles , altitude

Let $ABC$ be an acute triangle with $AB<AC<BC$, inscribed in circle $c(O,R)$ (with center $O$ and radius $R$). Let $O_1$ be the symmetric point of $O$ wrt $AC$. Circle $c_1(O_1,R)$ intersects $BC$ at $Z$. If the extension of the altitude $AD$ intersects the cicrumscribed circle $c(O,R)$ at point $E$, prove that $EC$ is perpendicular on $AZ$.

Geometry Mathley 2011-12, 1.2

Tags: orthocenter , geometry , altitude

Let $ABC$ be an acute triangle with its altitudes $BE,CF$. $M$ is the midpoint of $BC$. $N$ is the intersection of $AM$ and $EF. X$ is the projection of $N$ on $BC$. $Y,Z$ are respectively the projections of $X$ onto $AB,AC$. Prove that $N$ is the orthocenter of triangle $AYZ$. Nguyễn Minh Hà

2019 Estonia Team Selection Test, 7

Tags: geometry , equal angles , altitude

An acute-angled triangle $ABC$ has two altitudes $BE$ and $CF$. The circle with diameter $AC$ intersects the segment $BE$ at point $P$. A circle with diameter $AB$ intersects the segment $CF$ at point $Q$ and the extension of this altitude at point $Q'$. Prove that $\angle PQ'Q = \angle PQB$.

2023 Sharygin Geometry Olympiad, 16

Tags: geometry , reflection , altitude

Let $AH_A$ and $BH_B$ be the altitudes of a triangle $ABC$. The line $H_AH_B$ meets the circumcircle of $ABC$ at points $P$ and $Q$. Let $A'$ be the reﬂection of $A$ about $BC$, and $B'$ be the reﬂection of $B$ about $CA$. Prove that $A',B', P,Q$ are concyclic.

2010 Swedish Mathematical Competition, 1

Tags: geometry , altitude

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

1985 Bundeswettbewerb Mathematik, 2

Tags: collinear , altitude , projection , geometry

Prove that in every triangle for each of its altitudes: If you project the foof of one altitude on the other two altitudes and on the other two sides of the triangle, those four projections lie on the same line.

2007 Sharygin Geometry Olympiad, 1

Tags: geometry , construction , angle bisector , circumcircle , altitude

In an acute triangle $ABC$, altitudes at vertices $A$ and $B$ and bisector line at angle $C$ intersect the circumcircle again at points $A_1, B_1$ and $C_0$. Using the straightedge and compass, reconstruct the triangle by points $A_1, B_1$ and $C_0$.

2014 Belarus Team Selection Test, 1

Tags: geometry , parallel , altitude

Let $AA_1, BB_1$ be the altitudes of an acute non-isosceles triangle $ABC$. Circumference of the triangles $ABC$ meets that of the triangle $A_1B_1C$ at point $N$ (different from $C$). Let $M$ be the midpoint of $AB$ and $K$ be the intersection point of $CN$ and $AB$. Prove that the line of centers the circumferences of the triangles $ABC$ and $KMC$ is parallel to the line $AB$. (I. Kachan)

2010 Ukraine Team Selection Test, 7

Tags: geometry , geometric inequality , trigonometry , altitude

Denote in the triangle $ABC$ by $h$ the length of the height drawn from vertex $A$, and by $\alpha = \angle BAC$. Prove that the inequality $AB + AC \ge BC \cdot \cos \alpha + 2h \cdot \sin \alpha$ . Are there triangles for which this inequality turns into equality?

1956 Moscow Mathematical Olympiad, 333

Tags: geometry , altitude , circumcenter , circumcircle , concurrency

Let $O$ be the center of the circle circumscribed around $\vartriangle ABC$, let $A_1, B_1, C_1$ be symmetric to $O$ through respective sides of $\vartriangle ABC$. Prove that all altitudes of $\vartriangle A_1B_1C_1$ pass through $O$, and all altitudes of $\vartriangle ABC$ pass through the center of the circle circumscribed around $\vartriangle A_1B_1C_1$.

2015 Hanoi Open Mathematics Competitions, 12

Tags: integer , geometry , altitude

Give a triangle $ABC$ with heights $h_a = 3$ cm, $h_b = 7$ cm and $h_c = d$ cm, where $d$ is an integer. Determine $d$.

2016 Sharygin Geometry Olympiad, 1

Tags: geometry , right triangle , altitude , median

An altitude $AH$ of triangle $ABC$ bisects a median $BM$. Prove that the medians of triangle $ABM$ are sidelengths of a right-angled triangle. by Yu.Blinkov

2015 Irish Math Olympiad, 1

Tags: altitude , midpoint , geometry

In the triangle $ABC$, the length of the altitude from $A$ to $BC$ is equal to $1$. $D$ is the midpoint of $AC$. What are the possible lengths of $BD$?

2007 Balkan MO Shortlist, G4

Tags: geometric inequality , altitude , geometry

Points $M,N$ and $P$ on the sides $BC, CA$ and $AB$ of $\vartriangle ABC$ are such that $\vartriangle MNP$ is acute. Denote by $h$ and $H$ the lengths of the shortest altitude of $\vartriangle ABC$ and the longest altitude of $\vartriangle MNP$. Prove that $h \le 2H$.

2009 Puerto Rico Team Selection Test, 3

Tags: geometry , altitude

On an arbitrary triangle $ ABC$ let $ E$ be a point on the height from $ A$. Prove that $ (AC)^2 - (CE)^2 = (AB)^2 - (EB)^2$.

2019 Estonia Team Selection Test, 7

Tags: geometry , equal angles , altitude

An acute-angled triangle $ABC$ has two altitudes $BE$ and $CF$. The circle with diameter $AC$ intersects the segment $BE$ at point $P$. A circle with diameter $AB$ intersects the segment $CF$ at point $Q$ and the extension of this altitude at point $Q'$. Prove that $\angle PQ'Q = \angle PQB$.

Kharkiv City MO Seniors - geometry, 2012.10.4

Tags: geometry , altitude , midpoint , concyclic

In the acute-angled triangle $ABC$ on the sides $AC$ and $BC$, points $D$ and $E$ are chosen such that points $A, B, E$, and $D$ lie on one circle. The circumcircle of triangle $DEC$ intersects side $AB$ at points $X$ and $Y$. Prove that the midpoint of segment $XY$ is the foot of the altitude of the triangle, drawn from point $C$.

2017 Puerto Rico Team Selection Test, 2

Tags: perpendicular bisector , geometry , altitude

For an acute triangle $ ABC $ let $ H $ be the point of intersection of the altitudes $ AA_1 $, $ BB_1 $, $ CC_1 $. Let $ M $ and $ N $ be the midpoints of the $ BC $ and $ AH $ segments, respectively. Show that $ MN $ is the perpendicular bisector of segment $ B_1C_1 $.

1952 Moscow Mathematical Olympiad, 212

Tags: geometry , ratio , orthocenter , equilateral , altitude

Prove that if the orthocenter divides all heights of a triangle in the same proportion, the triangle is equilateral.