Olympiad problems

2015 Caucasus Mathematical Olympiad, 5

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 $C_1 K = A_1L$.

2019 Sharygin Geometry Olympiad, 5

Tags: geometry , circumcircle , concurrency , altitude

Let $AA_1, BB_1, CC_1$ be the altitudes of triangle $ABC$, and $A0, C0$ be the common points of the circumcircle of triangle $A_1BC_1$ with the lines $A_1B_1$ and $C_1B_1$ respectively. Prove that $AA_0$ and $CC_0$ meet on the median of ABC or are parallel to it

2025 NEPALTST, 3

Tags: geometry , cyclic quadrilateral , altitude

Consider an acute triangle $\Delta ABC$. Let $D$ and $E$ be the feet of the altitudes from $A$ to $BC$ and from $B$ to $AC$ respectively. Define $D_1$ and $D_2$ as the reflections of $D$ across lines $AB$ and $AC$, respectively. Let $\Gamma$ be the circumcircle of $\Delta AD_1D_2$. Denote by $P$ the second intersection of line $D_1B$ with $\Gamma$, and by $Q$ the intersection of ray $EB$ with $\Gamma$. If $O$ is the circumcenter of $\Delta ABC$, prove that $O$, $D$, and $Q$ are collinear if and only if quadrilateral $BCQP$ can be inscribed within a circle. $\textbf{Proposed by Kritesh Dhakal, Nepal.}$

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.

2011 Oral Moscow Geometry Olympiad, 4

Tags: geometry , trapezoid , isosceles , midpoint , altitude , perpendicular

In the trapezoid $ABCD, AB = BC = CD, CH$ is the altitude. Prove that the perpendicular from $H$ on $AC$ passes through the midpoint of $BD$.

Mexican Quarantine Mathematical Olympiad, #4

Tags: geometry , altitude , orthocenter , circumcenter

Let $ABC$ be an acute triangle with orthocenter $H$. Let $A_1$, $B_1$ and $C_1$ be the feet of the altitudes of triangle $ABC$ opposite to vertices $A$, $B$, and $C$ respectively. Let $B_2$ and $C_2$ be the midpoints of $BB_1$ and $CC_1$, respectively. Let $O$ be the intersection of lines $BC_2$ and $CB_2$. Prove that $O$ is the circumcenter of triangle $ABC$ if and only if $H$ is the midpoint of $AA_1$. [i]Proposed by Dorlir Ahmeti[/i]

1995 Romania Team Selection Test, 3

Tags: geometry , inradius , sidelenghts , altitude

The altitudes of a triangle have integer length and its inradius is a prime number. Find all possible values of the sides of the triangle.

2003 Estonia National Olympiad, 3

Tags: altitude , geometry , equal segments , parallelogram

In the acute-angled triangle $ABC$ all angles are greater than $45^o$. Let $AM$ and $BN$ be the heights of this triangle and let $X$ and $Y$ be the points on $MA$ and $NB$, respecively, such that $|MX| =|MB|$ and $|NY| =|NA|$. Prove that $MN$ and $XY$ are parallel.

2006 Estonia Team Selection Test, 4

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

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.

1989 Chile National Olympiad, 5

Tags: geometry , rational , altitude

The lengths of the three sides of a $ \triangle ABC $ are rational. The altitude $ CD $ determines on the side $AB$ two segments $ AD $ and $ DB $. Prove that $ AD, DB $ are rational.

2017 Sharygin Geometry Olympiad, 5

Tags: geometry , parallel , reflection , altitude

Let $BH_b, CH_c$ be altitudes of an acute-angled triangle $ABC$. The line $H_bH_c$ meets the circumcircle of $ABC$ at points $X$ and $Y$. Points $P,Q$ are the reflections of $X,Y$ about $AB,AC$ respectively. Prove that $PQ \parallel BC$. [i]Proposed by Pavel Kozhevnikov[/i]

1988 Tournament Of Towns, (185) 2

Tags: geometry , altitude , angle

In a triangle two altitudes are not smaller than the sides on to which they are dropped. Find the angles of the triangle.

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?

2010 Bundeswettbewerb Mathematik, 3

Tags: concyclic , collinear , altitude , perpendicular , geometry , projection

Given an acute-angled triangle $ABC$. Let $CB$ be the altitude and $E$ a random point on the line $CD$. Finally, let $P, Q, R$ and $S$ are the projections of $D$ on the straight lines $AC, AE, BE$ and $BC$. Prove that the points $P, Q, R$ and $S$ lie either on a circle or on one straight line.

2009 Ukraine Team Selection Test, 12

Tags: geometric inequality , geometry , altitude

Denote an acute-angle $\vartriangle ABC $ with sides $a, b, c $ respectively by ${{H}_{a}}, {{H}_{b}}, {{H}_{c}} $ the feet of altitudes ${{h}_{a}}, {{h}_{b}}, {{h}_{c}} $. Prove the inequality: $$\frac {h_ {a} ^{2}} {{{a} ^{2}} - CH_ {a} ^{2}} + \frac{h_{b} ^{2}} {{{ b}^{2}} - AH_{b} ^{2}} + \frac{h_{c}^{2}}{{{c}^{2}} - BH_{c}^{2}} \ge 3 $$ (Dmitry Petrovsky)

2006 Denmark MO - Mohr Contest, 5

Tags: geometry , similar triangles , equal segments , altitude

We consider an acute triangle $ABC$. The altitude from $A$ is $AD$, the altitude from $D$ in triangle $ABD$ is $DE,$ and the altitude from $D$ in triangle $ACD$ is $DF$. a) Prove that the triangles $ABC$ and $AF E$ are similar. b) Prove that the segment $EF$ and the corresponding segments constructed from the vertices $B$ and $C$ all have the same length.

Kyiv City MO Juniors Round2 2010+ geometry, 2020.9.2

Tags: circumcircle , bisects segment , altitude , geometry

In the acute-angled triangle $ABC$ is drawn the altitude $CH$. A ray beginning at point $C$ that lies inside the $\angle BCA$ and intersects for second time the circles circumscribed circles of $\vartriangle BCH$ and $\vartriangle ABC$ at points $X$ and $Y$ respectively. It turned out that $2CX = CY$. Prove that the line $HX$ bisects the segment $AC$. (Hilko Danilo)

1997 Israel Grosman Mathematical Olympiad, 4

Tags: tetrahedron , 3d geometry , geometry , altitude

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

1996 Nordic, 3

Tags: geometry , altitude

The circle whose diameter is the altitude dropped from the vertex $A$ of the triangle $ABC$ intersects the sides $AB$ and $AC$ at $D$ and $E$, respectively $(A\ne D, A \ne E)$. Show that the circumcenter of $ABC$ lies on the altitude drawn from the vertex $A$ of the triangle $ADE$, or on its extension.

1990 All Soviet Union Mathematical Olympiad, 532

Tags: inequalities , geometric inequality , 3d geometry , tetrahedron , distance , altitude

If every altitude of a tetrahedron is at least $1$, show that the shortest distance between each pair of opposite edges is more than $2$.

2006 Tournament of Towns, 2

Tags: altitude , geometry

Suppose $ABC$ is an acute triangle. Points $A_1, B_1$ and $C_1$ are chosen on sides $BC, AC$ and $AB$ respectively so that the rays $A_1A, B_1B$ and $C_1C$ are bisectors of triangle $A_1B_1C_1$. Prove that $AA_1, BB_1$ and $CC_1$ are altitudes of triangle $ABC$. (6)

1994 Mexico National Olympiad, 5

Tags: geometry , convex quadrilateral , altitude

$ABCD$ is a convex quadrilateral. Take the $12$ points which are the feet of the altitudes in the triangles $ABC, BCD, CDA, DAB$. Show that at least one of these points must lie on the sides of $ABCD$.

2006 Sharygin Geometry Olympiad, 12

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

In the triangle $ABC$, the bisector of angle $A$ is equal to the half-sum of the height and median drawn from vertex $A$. Prove that if $\angle A$ is obtuse, then $AB = AC$.

2025 Bangladesh Mathematical Olympiad, P9

Tags: geometry , similarity , parallel lines , congruent triangles , altitude

Let $ABC$ be an acute triangle and $D$ be the foot of the altitude from $A$ onto $BC$. A semicircle with diameter $BC$ intersects segments $AB, AC$ and $AD$ in the points $F, E$ and $X$, respectively. The circumcircles of the triangles $DEX$ and $DXF$ intersect $BC$ in $L$ and $N$, respectively, other than $D$. Prove that $BN = LC$.

2015 Sharygin Geometry Olympiad, P6

Tags: geometry , symmetry , altitude

Let $AA', BB'$ and $CC'$ be the altitudes of an acute-angled triangle $ABC$. Points $C_a, C_b$ are symmetric to $C' $ wrt $AA'$ and $BB'$. Points $A_b, A_c, B_c, B_a$ are defined similarly. Prove that lines $A_bB_a, B_cC_b$ and $C_aA_c$ are parallel.