Olympiad problems

1960 IMO Shortlist, 4

Construct triangle $ABC$, given $h_a$, $h_b$ (the altitudes from $A$ and $B$), and $m_a$, the median from vertex $A$.

Kyiv City MO 1984-93 - geometry, 1991.10.2

Tags: orthocenter , altitude , geometry

In an acute-angled triangle $ABC$ on the sides $AB$, $BC$, $AC$, the points $C_1$, $A_1$, and $B_1$ are marked such that the segments $AA_1$, $BB_1$, $CC_1$ intersect at some point $O$ and the angles $AA_1C$, $BB_1A$, $CC_1B$ are equal. Prove that $AA_1$, $BB_1$, and $CC_1$ are the altitudes of the triangle.

2018 Hanoi Open Mathematics Competitions, 5

Tags: geometry , median , angle bisector , altitude

Let $ABC$ be an acute triangle with $AB = 3$ and $AC = 4$. Suppose that $AH,AO$ and $AM$ are the altitude, the bisector and the median derived from $A$, respectively. If $HO = 3 MO$, then the length of $BC$ is [img]https://cdn.artofproblemsolving.com/attachments/e/c/26cc00629f4c0ab27096b8bdc562c56ff01ce5.png[/img] A. $3$ B. $\frac72$ C. $4$ D. $\frac92$ E. $5$

1942 Eotvos Mathematical Competition, 1

Tags: geometry , altitude , geometric inequalities

Prove that in any triangle, at most one side can be shorter than the altitude from the opposite vertex.

2019 Estonia Team Selection Test, 7

Tags: altitude , geometry , equal angles

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 $\angle PQ'Q = \angle PQB$.

2016 Singapore Junior Math Olympiad, 3

Tags: geometry , angle bisector , collinear , right triangle , altitude , perpendicular

In the triangle $ABC$, $\angle A=90^\circ$, the bisector of $\angle B$ meets the altitude $AD$ at the point $E$, and the bisector of $\angle CAD$ meets the side $CD$ at $F$. The line through $F$ perpendicular to $BC$ intersects $AC$ at $G$. Prove that $B,E,G$ are collinear.

2019 Silk Road, 1

Tags: altitude , geometry , angle chasing , equal angles

The altitudes of the acute-angled non-isosceles triangle $ ABC $ intersect at the point $ H $. On the segment $ C_1H $, where $ CC_1 $ is the altitude of the triangle, the point $ K $ is marked. Points $ L $ and $ M $ are the feet of perpendiculars from point $ K $ on straight lines $ AC $ and $ BC $, respectively. The lines $ AM $ and $ BL $ intersect at $ N $. Prove that $ \angle ANK = \angle HNL $.

2009 Belarus Team Selection Test, 3

Tags: angle bisector , altitude , trigonometry , geometry

Points $T,P,H$ lie on the side $BC,AC,AB$ respectively of triangle $ABC$, so that $BP$ and $AT$ are angle bisectors and $CH$ is an altitude of $ABC$. Given that the midpoint of $CH$ belongs to the segment $PT,$ find the value of $\cos A + \cos B$ I. Voronovich

1972 IMO Shortlist, 5

Tags: altitude , geometry , 3d geometry , tetrahedron

Prove the following assertion: The four altitudes of a tetrahedron $ABCD$ intersect in a point if and only if \[AB^2 + CD^2 = BC^2 + AD^2 = CA^2 + BD^2.\]

2017 Puerto Rico Team Selection Test, 2

Tags: altitude , perpendicular bisector , geometry

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 $.

2011 Sharygin Geometry Olympiad, 6

Tags: incenter , altitude , orthocenter , contact triangle , incircle , geometry

Let $BB_1$ and $CC_1$ be the altitudes of acute-angled triangle $ABC$, and $A_0$ is the midpoint of $BC$. Lines $A_0B_1$ and $A_0C_1$ meet the line passing through $A$ and parallel to $BC$ in points $P$ and $Q$. Prove that the incenter of triangle $PA_0Q$ lies on the altitude of triangle $ABC$.

2020 Malaysia IMONST 1, 18

Tags: altitude , triangle , geometry

In a triangle, the ratio of the interior angles is $1 : 5 : 6$, and the longest side has length $12$. What is the length of the altitude (height) of the triangle that is perpendicular to the longest side?

2011 Sharygin Geometry Olympiad, 19

Tags: geometry , median , altitude , bisector , equal segments

Does there exist a nonisosceles triangle such that the altitude from one vertex, the bisectrix from the second one and the median from the third one are equal?

Indonesia Regional MO OSP SMA - geometry, 2007.4

Tags: altitude , geometric inequality , geometry

In acute triangles $ABC$, $AD, BE ,CF$ are altitudes, with $D, E, F$ on the sides $BC, CA, AB$, respectively. Prove that $$DE + DF \le BC$$

2010 Sharygin Geometry Olympiad, 5

Tags: geometry , equal segments , right triangle , altitude , circumcircle

Let $BH$ be an altitude of a right-angled triangle $ABC$ ($\angle B = 90^o$). The incircle of triangle $ABH$ touches $AB,AH$ in points $H_1, B_1$, the incircle of triangle $CBH$ touches $CB,CH$ in points $H_2, B_2$, point $O$ is the circumcenter of triangle $H_1BH_2$. Prove that $OB_1 = OB_2$.

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?

1985 Tournament Of Towns, (080) T1

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

A median , a bisector and an altitude of a certain triangle intersect at an inner point $O$ . The segment of the bisector from the vertex to $O$ is equal to the segment of the altitude from the vertex to $O$ . Prove that the triangle is equilateral .

2005 Sharygin Geometry Olympiad, 9.5

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

It is given that for no side of the triangle from the height drawn to it, the bisector and the median it is impossible to make a triangle. Prove that one of the angles of the triangle is greater than $135^o$

1984 Austrian-Polish Competition, 1

Tags: geometry , 3d geometry , tetrahedron , incenter , altitude

Prove that if the feet of the altitudes of a tetrahedron are the incenters of the corresponding faces, then the tetrahedron is regular.

2012 Sharygin Geometry Olympiad, 7

Tags: altitude , geometry

The altitudes $AA_1$ and $CC_1$ of an acute-angled triangle $ABC$ meet at point $H$. Point $Q$ is the reflection of the midpoint of $AC$ in line $AA_1$, point $P$ is the midpoint of segment $A_1C_1$. Prove that $\angle QPH = 90^o$. (D.Shvetsov)

1997 Romania National Olympiad, 3

Tags: geometry , altitude

The triangle $ABC$ has $\angle ACB = 30^o$, $BC = 4$ cm and $AB = 3$ cm . Compute the altitudes of the triangle.

2022 Canadian Junior Mathematical Olympiad, 1

Tags: altitude , geometry

Let $\triangle{ABC}$ has circumcircle $\Gamma$, drop the perpendicular line from $A$ to $BC$ and meet $\Gamma$ at point $D$, similarly, altitude from $B$ to $AC$ meets $\Gamma$ at $E$. Prove that if $AB=DE, \angle{ACB}=60^{\circ}$ (sorry it is from my memory I can't remember the exact problem, but it means the same)

2000 Saint Petersburg Mathematical Olympiad, 10.2

Tags: midpoint , altitude , geometry , cyclic quadrilateral

Let $AA_1$ and $BB_1$ be the altitudes of acute angled triangle $ABC$. Points $K$ and $M$ are midpoints of $AB$ and $A_1B_1$ respectively. Segments $AA_1$ and $KM$ intersect at point $L$. Prove that points $A$, $K$, $L$ and $B_1$ are noncyclic. [I]Proposed by S. Berlov[/i]

2011 Sharygin Geometry Olympiad, 7

Tags: altitude , geometry , cyclic quadrilateral , circumcenter , circles , circumcircle , concurrency

Point $O$ is the circumcenter of acute-angled triangle $ABC$, points $A_1,B_1, C_1$ are the bases of its altitudes. Points $A', B', C'$ lying on lines $OA_1, OB_1, OC_1$ respectively are such that quadrilaterals $AOBC', BOCA', COAB'$ are cyclic. Prove that the circumcircles of triangles $AA_1A', BB_1B', CC_1C'$ have a common point.

2020 Yasinsky Geometry Olympiad, 3

Tags: altitude , geometry , circumcircle , incenter , inradius

The segments $BF$ and $CN$ are the altitudes in the acute-angled triangle $ABC$. The line $OI$, which connects the centers of the circumscribed and inscribed circles of triangle $ABC$, is parallel to the line $FN$. Find the length of the altitude $AK$ in the triangle $ABC$ if the radii of its circumscribed and inscribed circles are $R$ and $r$, respectively. (Grigory Filippovsky)