Olympiad problems

1994 Poland - Second Round, 5

The incircle $\omega$ of a triangle $ABC$ is tangent to the sides $AB$ and $BC$ at $P$ and $Q$ respectively. The angle bisector at $A$ meets $PQ$ at point $S$. Prove $\angle ASC = 90^o$ .

Kyiv City MO Juniors Round2 2010+ geometry, 2021.7.41

Tags: geometry , perimeter , geometric inequality , right angle

Point $C$ lies inside the right angle $AOB$. Prove that the perimeter of triangle $ABC$ is greater than $2 OC$.

Indonesia MO Shortlist - geometry, g5

Tags: geometry , circles , right angle

Two circles intersect at points $A$ and $B$. The line $\ell$ through A intersects the circles at $C$ and $D$, respectively. Let $M, N$ be the midpoints of arc $BC$ and arc $BD$. which does not contain $A$, and suppose that $K$ is the midpoint of the segment $CD$ . Prove that $\angle MKN=90^o$.

2013 Sharygin Geometry Olympiad, 1

Tags: geometry , pentagon , right angle , equal segments

Let $ABCDE$ be a pentagon with right angles at vertices $B$ and $E$ and such that $AB = AE$ and $BC = CD = DE$. The diagonals $BD$ and $CE$ meet at point $F$. Prove that $FA = AB$.

2023 Euler Olympiad, Round 2, 4

Tags: trapezoid , euler , right angle , geometry , harmonic bundles

Let $ABCD$ be a trapezoid, with $AD \parallel BC$, let $M$ be the midpoint of $AD$, and let $C_1$ be symmetric point to $C$ with respect to line $BD$. Segment $BM$ meets diagonal $AC$ at point $K$, and ray $C_1K$ meets line $BD$ at point $H$. Prove that $\angle{AHD}$ is a right angle. [i]Proposed by Giorgi Arabidze, Georgia[/i]

Kyiv City MO Juniors Round2 2010+ geometry, 2021.9.2

Tags: geometry , circumcircle , right angle

In an acute triangle $AB$ the heights $ BE$ and $CF$ intersect at the orthocenter $H$, and $M$ is the midpoint of $BC$. The line $EF$ intersects the lines $MH$ and $BC$ at the points $P$ and $T$ , respectively. $AP$ intersects the cirumcscribed circle of $\vartriangle ABC$ for second time at the point $Q$ . Prove that $\angle AQT= 90^o$. (Fedir Yudin)

2006 Junior Balkan Team Selection Tests - Romania, 2

Tags: geometry , rectangle , circles , circumcenter , right angle , tangent circle

Let $C (O)$ be a circle (with center $O$ ) and $A, B$ points on the circle with $\angle AOB = 90^o$. Circles $C_1 (O_1)$ and $C_2 (O_2)$ are tangent internally with circle $C$ at $A$ and $B$, respectively, and, also, are tangent to each other. Consider another circle $C_3 (O_3)$ tangent externally to the circles $C_1, C_2$ and tangent internally to circle $C$, located inside angle $\angle AOB$. Show that the points $O, O_1, O_2, O_3$ are the vertices of a rectangle.

1970 Czech and Slovak Olympiad III A, 2

Tags: geometry , 3d geometry , tetrahedron , right triangle , right angle , similar , similar triangles

Determine whether there is a tetrahedron $ABCD$ with the longest edge of length 1 such that all its faces are similar right triangles with right angles at vertices $B,C.$ If so, determine which edge is the longest, which is the shortest and what is its length.

2011 Sharygin Geometry Olympiad, 7

Tags: geometry , locus , locus problems , right angle , axes

Let a point $M$ not lying on coordinates axes be given. Points $Q$ and $P$ move along $Y$ - and $X$-axis respectively so that angle $P M Q$ is always right. Find the locus of points symmetric to $M$ wrt $P Q$.

Estonia Open Junior - geometry, 2007.2.2

Tags: angle bisector , square , right angle , geometry

The center of square $ABCD$ is $K$. The point $P$ is chosen such that $P \ne K$ and the angle $\angle APB$ is right . Prove that the line $PK$ bisects the angle between the lines $AP$ and $BP$.

Kyiv City MO Juniors Round2 2010+ geometry, 2020.8.2

Tags: geometry , equal segments , equal angles , right angle

Given a convex quadrilateral $ABCD$, in which $\angle CBD = 90^o$, $\angle BCD =\angle CAD$ and $AD= 2BC$. Prove that $CA =CD$. (Anton Trygub)

2020 OMpD, 4

Tags: geometry , incenter , right triangle , right angle

Let $ABC$ be a triangle and $P$ be any point on the side $BC$. Let $I_1$,$I_2$ be the incenters of triangles $ABP$ and $ACP$, respectively. If $D$ is the point of tangency of the incircle of $ABC$ with the side $BC$, prove that $\angle I_1DI_2 = 90^o$.

2012 Dutch IMO TST, 4

Tags: geometry , right angle , equal angles , angle chasing

Let $\vartriangle ABC$ be a triangle. The angle bisector of $\angle CAB$ intersects$ BC$ at $L$. On the interior of line segments $AC$ and $AB$, two points, $M$ and $N$, respectively, are chosen in such a way that the lines $AL, BM$ and $CN$ are concurrent, and such that $\angle AMN = \angle ALB$. Prove that $\angle NML = 90^o$.

2020 Balkan MO Shortlist, G2

Tags: right angle , euler line , angle chasing , geometry

Let $G, H$ be the centroid and orthocentre of $\vartriangle ABC$ which has an obtuse angle at $\angle B$. Let $\omega$ be the circle with diameter $AG$. $\omega$ intersects $\odot(ABC)$ again at $L \ne A$. The tangent to $\omega$ at $L$ intersects $\odot(ABC)$ at $K \ne L$. Given that $AG = GH$, prove $\angle HKG = 90^o$ . [i]Sam Bealing, United Kingdom[/i]

Cono Sur Shortlist - geometry, 2003.G3

Tags: square , geometry , right angle

An interior $P$ point to a square $ABCD$ is such that $PA = a, PB = b$ and $PC = b + c$, where the numbers $a, b$ and $c$ satisfy the relationship $a^2 = b^2 + c^2$. Prove that the angle $BPC$ is right.

Ukrainian From Tasks to Tasks - geometry, 2010.13

Tags: geometry , pentagon , right angle , tangential

You can inscribe a circle in the pentagon $ABCDE$. It is also known that $\angle ABC = \angle BAE = \angle CDE = 90^o$. Find the measure of the angle $ADB$.

1992 Swedish Mathematical Competition, 5

Tags: right triangle , right angle , geometry , circumradius

A triangle has sides $a, b, c$ with longest side $c$, and circumradius $R$. Show that if $a^2 + b^2 = 2cR$, then the triangle is right-angled.

2021 Polish Junior MO Second Round, 2

Tags: geometry , right angle , square

Given is the square $ABCD$. Point $E$ lies on the diagonal $AC$, where $AE> EC$. On the side $AB$, a different point from $B$ has been selected for which $EF = DE$. Prove that $\angle DEF = 90^o$.

Champions Tournament Seniors - geometry, 2011.2

Tags: right angle , isosceles , geometry

Let $ABC$ be an isosceles triangle in which $AB = AC$. On its sides $BC$ and $AC$ respectively are marked points $P$ and $Q$ so that $PQ\parallel AB$. Let $F$ be the center of the circle circumscribed about the triangle $PQC$, and $E$ the midpoint of the segment $BQ$. Prove that $\angle AEF = 90^o $.

2004 District Olympiad, 4

Tags: trigonometry , 3d geometry , geometry , palnes , trapezoid , right angle , angle

In the right trapezoid $ABCD$ with $AB \parallel CD, \angle B = 90^o$ and $AB = 2DC$. At points $A$ and $D$ there is therefore a part of the plane $(ABC)$ perpendicular to the plane of the trapezoid, on which the points $N$ and $P$ are taken, ($AP$ and $PD$ are perpendicular to the plane) such that $DN = a$ and $AP = \frac{a}{2}$ . Knowing that $M$ is the midpoint of the side $BC$ and the triangle $MNP$ is equilateral, determine: a) the cosine of the angle between the planes $MNP$ and $ABC$. b) the distance from $D$ to the plane $MNP$

2018 Peru MO (ONEM), 3

Tags: geometry , isosceles , angle chasing , right angle , orthocenter

Let $ABC$ be an acute triangle such that $BA = BC$. On the sides $BA$ and $BC$ points $D$ and $E$ are chosen respectively, such that $DE$ and $AC$ are parallel. Let $H$ be the orthocenter of the triangle $DBE$ and $M$ be the midpoint of $AE$. If $\angle HMC = 90^o$, determine the measure of angle $\angle ABC$.

2015 India Regional MathematicaI Olympiad, 1

Tags: geometry , circumcircle , incenter , right angle , cyclic quadrilateral

In a cyclic quadrilateral $ABCD$, let the diagonals $AC$ and $BD$ intersect at $X$. Let the circumcircles of triangles $AXD$ and $BXC$ intersect again at $Y$ . If $X$ is the incentre of triangle $ABY$ , show that $\angle CAD = 90^o$.

2020 China Team Selection Test, 2

Tags: geometry , right angle , equal segments , tangent circle

Given an isosceles triangle $\triangle ABC$, $AB=AC$. A line passes through $M$, the midpoint of $BC$, and intersects segment $AB$ and ray $CA$ at $D$ and $E$, respectively. Let $F$ be a point of $ME$ such that $EF=DM$, and $K$ be a point on $MD$. Let $\Gamma_1$ be the circle passes through $B,D,K$ and $\Gamma_2$ be the circle passes through $C,E,K$. $\Gamma_1$ and $\Gamma_2$ intersect again at $L \neq K$. Let $\omega_1$ and $\omega_2$ be the circumcircle of $\triangle LDE$ and $\triangle LKM$. Prove that, if $\omega_1$ and $\omega_2$ are symmetric wrt $L$, then $BF$ is perpendicular to $BC$.

2016 Saudi Arabia BMO TST, 2

Tags: geometry , circles , right angle

A circle with center $O$ passes through points $A$ and $C$ and intersects the sides $AB$ and $BC$ of triangle $ABC$ at points $K$ and $N$, respectively. The circumcircles of triangles $ABC$ and $KBN$ meet at distinct points $B$ and $M$. Prove that $\angle OMB = 90^o$.

2014 Bosnia and Herzegovina Junior BMO TST, 2

Tags: right angle , geometry , triangle

In triangle $ABC$, on line $CA$ it is given point $D$ such that $CD = 3 \cdot CA$ (point $A$ is between points $C$ and $D$), and on line $BC$ it is given point $E$ ($E \neq B$) such that $CE=BC$. If $BD=AE$, prove that $\angle BAC= 90^{\circ}$