Olympiad problems

1994 All-Russian Olympiad, 7

The altitudes $AA_1,BB_1,CC_1,DD_1$ of a tetrahedron $ABCD$ intersect in the center $H$ of the sphere inscribed in the tetrahedron $A_1B_1C_1D_1$. Prove that the tetrahedron $ABCD$ is regular. (D. Tereshin)

Indonesia MO Shortlist - geometry, g6

Tags: altitude , geometric inequality , geometry

Suppose the points $D, E, F$ lie on sides $BC, CA, AB$, respectively, so that $AD, BE, CF$ are the altitudes. Also, let $AD$ and $EF$ intersect at $P$. Prove that $$\frac{AP}{AD} \ge 1 - \frac{BC^2}{AB^2 + CA^2}$$

2011 Sharygin Geometry Olympiad, 1

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

Altitudes $AA_1$ and $BB_1$ of triangle ABC meet in point $H$. Line $CH$ meets the semicircle with diameter $AB$, passing through $A_1, B_1$, in point $D$. Segments $AD$ and $BB_1$ meet in point $M$, segments $BD$ and $AA_1$ meet in point $N$. Prove that the circumcircles of triangles $B_1DM$ and $A_1DN$ touch.

2009 Postal Coaching, 3

Tags: concyclic , circumcenter , altitude , geometry , incenter

Let $ABC$ be a triangle with circumcentre $O$ and incentre $I$ such that $O$ is different from $I$. Let $AK, BL, CM$ be the altitudes of $ABC$, let $U, V , W$ be the mid-points of $AK, BL, CM$ respectively. Let $D, E, F$ be the points at which the in-circle of $ABC$ respectively touches the sides $BC, CA, AB$. Prove that the lines $UD, VE, WF$ and $OI$ are concurrent.

2018 Junior Regional Olympiad - FBH, 5

Tags: perimeter , geometry , altitude , angle chasing

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

2014 Belarus Team Selection Test, 1

Tags: parallel , altitude , geometry

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)

2015 Oral Moscow Geometry Olympiad, 1

Tags: geometry , bisects , altitude

In triangle $ABC$, the altitude $AH$ passes through midpoint of the median $BM$. Prove that in the triangle $BMC$ also one of the altitudes passes through the midpoint of one of the medians.

2015 Caucasus Mathematical Olympiad, 5

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

Let $AA_1$ and $CC_1$ be the altitudes of the acute-angled triangle $ABC$. Let $K,L$ and $M$ be the midpoints of the sides $AB,BC$ and $CA$ respectively. Prove that if $\angle C_1MA_1 =\angle ABC$, then $C_1 K = A_1L$.

2009 Bosnia and Herzegovina Junior BMO TST, 1

Tags: altitude , number theory , area , geometry

Lengths of sides of triangle $ABC$ are positive integers, and smallest side is equal to $2$. Determine the area of triangle $P$ if $v_c = v_a + v_b$, where $v_a$, $v_b$ and $v_c$ are lengths of altitudes in triangle $ABC$ from vertices $A$, $B$ and $C$, respectively.

1999 Estonia National Olympiad, 4

Tags: altitude , vector , median , geometry

For the given triangle $ABC$, prove that a point $X$ on the side $AB$ satisfies the condition $\overrightarrow{XA} \cdot\overrightarrow{XB} +\overrightarrow{XC} \cdot \overrightarrow{XC} = \overrightarrow{CA} \cdot \overrightarrow{CB} $, iff $X$ is the basepoint of the altitude or median of the triangle $ABC$.

1970 All Soviet Union Mathematical Olympiad, 135

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

The angle bisector $[AD]$, the median $[BM]$ and the height $[CH]$ of the acute-angled triangle $ABC$ intersect in one point. Prove that the $\angle BAC> 45^o$.

2008 Balkan MO Shortlist, G6

Tags: geometry , median , altitude , circles , collinear , tangent

On triangle $ABC$ the $AM$ ($M\in BC$) is median and $BB_1$ and $CC_1$ ($B_1 \in AC,C_1 \in AB$) are altitudes. The stright line $d$ is perpendicular to $AM$ at the point $A$ and intersect the lines $BB_1$ and $CC_1$ at the points $E$ and $F$ respectively. Let denoted with $\omega$ the circle passing through the points $E, M$ and $F$ and with $\omega_1$ and with $\omega_2$ the circles that are tangent to segment $EF$ and with $\omega$ at the arc $EF$ which is not contain the point $M$. If the points $P$ and $Q$ are intersections points for $\omega_1$ and $\omega_2$ then prove that the points $P, Q$ and $M$ are collinear.

2014 Estonia Team Selection Test, 4

Tags: geometry , circumcircle , altitude , 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 $|MP| = |MQ|$.

Croatia MO (HMO) - geometry, 2016.3

Tags: geometry , altitude , concyclic , cyclic quadrilateral

Given a cyclic quadrilateral $ABCD$ such that the tangents at points $B$ and $D$ to its circumcircle $k$ intersect at the line $AC$. The points $E$ and $F$ lie on the circle $k$ so that the lines $AC, DE$ and $BF$ parallel. Let $M$ be the intersection of the lines $BE$ and $DF$. If $P, Q$ and $R$ are the feet of the altitides of the triangle $ABC$, prove that the points $P, Q, R$ and $M$ lie on the same circle

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]

1997 Israel Grosman Mathematical Olympiad, 4

Tags: tetrahedron , 3d geometry , altitude , geometry

Prove that if two altitudes of a tetrahedron intersect, then so do the other two altitudes.

2018 Estonia Team Selection Test, 7

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

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.

2016 Singapore Junior Math Olympiad, 3

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

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.

Kharkiv City MO Seniors - geometry, 2019.11.5

Tags: altitude , bisects segment , projection , geometry

In the acute-angled triangle $ABC$, let $CD, AE$ be the altitudes. Points $F$ and $G$ are the projections of $A$ and $C$ on the line $DE$, respectively, $H$ and $K$ are the projections of $D$ and $E$ on the line $AC$, respectively. The lines $HF$ and $KG$ intersect at point $P$. Prove that line $BP$ bisects the segment $DE$.

Geometry Mathley 2011-12, 1.2

Tags: geometry , orthocenter , 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

2023 Pan-American Girls’ Mathematical Olympiad, 3

Tags: circumcircle , geometry , altitude

Let $ABC$ an acute triangle and $D,E$ and $F$ be the feet of altitudes from $A,B$ and $C$, respectively. The line $EF$ and the circumcircle of $ABC$ intersect at $P$, such that $F$ it´s between $E$ and $P$. Lines $BP$ and $DF$ intersect at $Q$. Prove that if $ED=EP$, then $CQ$ and $DP$ are parallel.

1960 IMO, 4

Tags: geometry , construction , altitude

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

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)

2013 Singapore Senior Math Olympiad, 1

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

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

2011 Sharygin Geometry Olympiad, 6

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

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.