Olympiad problems

2000 Saint Petersburg Mathematical Olympiad, 10.2

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]

2020 Tournament Of Towns, 2

Tags: incenter , geometry , circumcircle , circumradius , altitude

At heights $AA_0, BB_0, CC_0$ of an acute-angled non-equilateral triangle $ABC$, points $A_1, B_1, C_1$ were marked, respectively, so that $AA_1 = BB_1 = CC_1 = R$, where $R$ is the radius of the circumscribed circle of triangle $ABC$. Prove that the center of the circumscribed circle of the triangle $A_1B_1C_1$ coincides with the center of the inscribed circle of triangle $ABC$. E. Bakaev

2011 Sharygin Geometry Olympiad, 19

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

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?

1935 Moscow Mathematical Olympiad, 008

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

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.

Ukrainian TYM Qualifying - geometry, 2019.10

Tags: orthocenter , altitude , geometry

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 $ABC$.

Denmark (Mohr) - geometry, 2006.5

Tags: equal segments , geometry , similar triangles , 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.

1972 IMO Longlists, 21

Tags: 3d geometry , tetrahedron , altitude , geometry

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

1993 Italy TST, 3

Tags: geometry , inradius , exradius , altitude

Let $ABC$ be an isosceles triangle with base $AB$ and $D$ be a point on side $AB$ such that the incircle of triangle $ACD$ is congruent to the excircle of triangle $DCB$ across $C$. Prove that the diameter of each of these circles equals half the altitude of $\vartriangle ABC$ from $A$

2017 Puerto Rico Team Selection Test, 2

Tags: geometry , perpendicular bisector , 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 $.

2010 Sharygin Geometry Olympiad, 5

Tags: angle bisector , altitude , median , geometry

Let $AH$, $BL$ and $CM$ be an altitude, a bisectrix and a median in triangle $ABC$. It is known that lines $AH$ and $BL$ are an altitude and a bisectrix of triangle $HLM$. Prove that line $CM$ is a median of this triangle.

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?

2013 Junior Balkan Team Selection Tests - Romania, 4

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

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

2019 Ukraine Team Selection Test, 3

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

Given an acute triangle $ABC$ . It's altitudes $AA_1 , BB_1$ and $CC_1$ intersect at a point $H$ , the orthocenter of $\vartriangle ABC$. Let the lines $B_1C_1$ and $AA_1$ intersect at a point $K$, point $M$ be the midpoint of the segment $AH$. Prove that the circumscribed circle of $\vartriangle MKB_1$ touches the circumscribed circle of $\vartriangle ABC$ if and only if $BA1 = 3A1C$. (Bondarenko Mykhailo)

2006 Sharygin Geometry Olympiad, 12

Tags: angle bisector , geometry , altitude , median , 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 $AB = AC$.

2006 Dutch Mathematical Olympiad, 2

Tags: geometry , altitude

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 $h_C$.

2009 Federal Competition For Advanced Students, P1, 4

Tags: geometry , concurrency , midpoint , altitude

Let $D, E$, and $F$ be respectively the midpoints of the sides $BC, CA$, and $AB$ of $\vartriangle ABC$. Let $H_a, H_b, H_c$ be the feet of perpendiculars from $A, B, C$ to the opposite sides, respectively. Let $P, Q, R$ be the midpoints of the $H_bH_c, H_cH_a$, and $H_aH_b$ respectively. Prove that $PD, QE$, and $RF$ are concurrent.

2018 Austria Beginners' Competition, 2

Tags: geometry , equal segments , midpoint , altitude

Let $ABC$ be an acute-angled triangle, $M$ the midpoint of the side $AC$ and $F$ the foot on $AB$ of the altitude through the vertex $C$. Prove that $AM = AF$ holds if and only if $\angle BAC = 60^o$. (Karl Czakler)

2011 Tournament of Towns, 3

Tags: geometry , bisects segment , equal segments , altitude , projection

In triangle $ABC$, points $A_1,B_1,C_1$ are bases of altitudes from vertices $A,B,C$, and points $C_A,C_B$ are the projections of $C_1$ to $AC$ and $BC$ respectively. Prove that line $C_AC_B$ bisects the segments $C_1A_1$ and $C_1B_1$.

1956 Moscow Mathematical Olympiad, 333

Tags: geometry , concurrency , altitude , circumcenter , circumcircle

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 $\vartriangle A_1B_1C_1$.

2000 Saint Petersburg Mathematical Olympiad, 11.5

Tags: circumscribed quadrilateral , altitude , geometry

Let $AA_1$, $BB_1$, $CC_1$ be the altitudes of an acute angled triangle $ABC$. On the side $BC$ point $K$ is taken such that $\angle BB_1K=\angle A$. On the side $AB$ a point $M$ is taken such that $\angle BB_1M\angle C$. Let $L$ be the intersection of $BB_1$ and $A_1C_1$. Prove that the quadrilateral $B_1KLM$ is circumscribed. [I]Proposed by A. Khrabrov, D. Rostovski[/i]

1992 Czech And Slovak Olympiad IIIA, 6

Tags: geometry , altitude , concyclic , circles

Let $ABC$ be an acute triangle. The altitude from $B$ meets the circle with diameter $AC$ at points $P,Q$, and the altitude from $C$ meets the circle with diameter $AB$ at $M,N$. Prove that the points $M,N,P,Q$ lie on a circle.

2016 Latvia Baltic Way TST, 15

Tags: geometry , altitude , concyclic , midpoint

Let $ABC$ be a triangle. Let its altitudes $AD$, $BE$ and $CF$ concur at $H$. Let $K, L$ and $M$ be the midpoints of $BC$, $CA$ and $AB$, respectively. Prove that, if $\angle BAC = 60^o$, then the midpoints of the segments $AH$, $DK$, $EL$, $FM$ are concyclic.

2003 Bosnia and Herzegovina Team Selection Test, 4

Tags: geometry , altitude , isosceles , orthogonal , angle chasing

In triangle $ABC$ $AD$ and $BE$ are altitudes. Let $L$ be a point on $ED$ such that $ED$ is orthogonal to $BL$. If $LB^2=LD\cdot LE$ prove that triangle $ABC$ is isosceles

2020 Ukrainian Geometry Olympiad - December, 5

Tags: geometry , circumcircle , altitude , angle

In an acute triangle $ABC$ with an angle $\angle ACB =75^o$, altitudes $AA_3,BB_3$ intersect the circumscribed circle at points $A_1,B_1$ respectively. On the lines $BC$ and $CA$ select points $A_2$ and $B_2$ respectively suchthat the line $B_1B_2$ is parallel to the line $BC$ and the line $A_1A_2$ is parallel to the line $AC$ . Let $M$ be the midpoint of the segment $A_2B_2$. Find in degrees the measure of the angle $\angle B_3MA_3$.

2006 Denmark MO - Mohr Contest, 5

Tags: geometry , altitude , similar triangles , equal segments

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.