Olympiad problems

VI Soros Olympiad 1999 - 2000 (Russia), 9.9

The center of a circle, the radius of which is $r$, lies on the bisector of the right angle $A$ at a distance $a$ from its sides ($a > r$). A tangent to the circle intersects the sides of the angle at points $B$ and $C$. Find the smallest possible value of the area of triangle $ABC$.

2018 Saudi Arabia IMO TST, 2

Tags: circumcircle , angle bisector , geometry , right angle

Let $ABC$ be an acute-angled triangle inscribed in circle $(O)$. Let $G$ be a point on the small arc $AC$ of $(O)$ and $(K)$ be a circle passing through $A$ and $G$. Bisector of $\angle BAC$ cuts $(K)$ again at $P$. The point $E$ is chosen on $(K)$ such that $AE$ is parallel to $BC$. The line $PK$ meets the perpendicular bisector of $BC$ at $F$. Prove that $\angle EGF = 90^o$.

2023 Euler Olympiad, Round 2, 4

Tags: geometry , trapezoid , euler , right angle , 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]

2004 District Olympiad, 4

Tags: trapezoid , geometry , trigonometry , 3d geometry , palnes , 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 $MNP$

2017 Ukrainian Geometry Olympiad, 4

Tags: geometry , circumcircle , parallelogram , circumcenter , right angle

Let $ABCD$ be a parallelogram and $P$ be an arbitrary point of the circumcircle of $\Delta ABD$, different from the vertices. Line $PA$ intersects the line $CD$ at point $Q$. Let $O$ be the center of the circumcircle $\Delta PCQ$. Prove that $\angle ADO = 90^o$.

2009 Belarus Team Selection Test, 3

Tags: tangential , trapezoid , geometry , right angle , parallel , tangential quadrilateral

Given trapezoid $ABCD$ ($AD\parallel BC$) with $AD \perp AB$ and $T=AC\cap BD$. A circle centered at point $O$ is inscribed in the trapezoid and touches the side $CD$ at point $Q$. Let $P$ be the intersection point (different from $Q$) of the side $CD$ and the circle passing through $T,Q$ and $O$. Prove that $TP \parallel AD$. I. Voronovich

2002 IMO Shortlist, 8

Tags: geometry , circles , right angle

Let two circles $S_{1}$ and $S_{2}$ meet at the points $A$ and $B$. A line through $A$ meets $S_{1}$ again at $C$ and $S_{2}$ again at $D$. Let $M$, $N$, $K$ be three points on the line segments $CD$, $BC$, $BD$ respectively, with $MN$ parallel to $BD$ and $MK$ parallel to $BC$. Let $E$ and $F$ be points on those arcs $BC$ of $S_{1}$ and $BD$ of $S_{2}$ respectively that do not contain $A$. Given that $EN$ is perpendicular to $BC$ and $FK$ is perpendicular to $BD$ prove that $\angle EMF=90^{\circ}$.

2020 China Team Selection Test, 2

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

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 $BC$.

1999 All-Russian Olympiad Regional Round, 8.6

Tags: geometry , right angle , perpendicular

Given triangle $ABC$. Point $A_1$ is symmetric to vertex $A$ wrt line $BC$, and point $C_1$ is symmetric to vertex $C$ wrt line $AB$. Prove that if points $A_1$, $B$ and $C_1$ lie on the same line and $C_1B = 2A_1B$, then angle $\angle CA_1B$ is right.

1994 Poland - Second Round, 5

Tags: geometry , angle bisector , right angle , incircle

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 $\angle ASC = 90^o$ .

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 $\angle OMB = 90^o$.

2021 Dutch IMO TST, 2

Tags: geometry , right angle , centroid , right triangle

Let $ABC $be a right triangle with $\angle C = 90^o$ and let $D$ be the foot of the altitude from $C$. Let $E$ be the centroid of triangle $ACD$ and let $F$ be the centroid of triangle $BCD$. The point $P$ satisfies $\angle CEP = 90^o$ and $|CP| = |AP|$, while point $Q$ satisfies $\angle CFQ = 90^o$ and $|CQ| = |BQ|$. Prove that $PQ$ passes through the centroid of triangle $ABC$.

2019 Saudi Arabia Pre-TST + Training Tests, 1.3

Tags: incircle , equal segments , trapezoid , geometry , right angle

Let $ABCD$ be a trapezoid with $\angle A = \angle B = 90^o$ and a point $E$ lies on the segment $CD$. Denote $(\omega)$ as incircle of triangle $ABE$ and it is tangent to $AB,AE,BE$ respectively at $P, F,K$. Suppose that $KF$ cuts $BC,AD$ at $M,N$ and $PM,PN$ cut $(\omega)$ at $H, T$. Prove that $PH = PT$.

2015 Saudi Arabia JBMO TST, 3

Tags: geometry , circumcircle , right triangle , right angle

A right triangle $ABC$ with $\angle C=90^o$ is inscribed in a circle. Suppose that $K$ is the midpoint of the arc $BC$ that does not contain $A$. Let $N$ be the midpoint of the segment $AC$, and $M$ be the intersection point of the ray $KN$ and the circle.The tangents to the circle drawn at $A$ and $C$ meet at $E$. prove that $\angle EMK = 90^o$

Kyiv City MO Juniors Round2 2010+ geometry, 2011.8.3

Tags: equal segments , geometry , right angle , square

On the sides $AD , BC$ of the square $ABCD$ the points $M, N$ are selected $N$, respectively, such that $AM = BN$. Point $X$ is the foot of the perpendicular from point $D$ on the line $AN$. Prove that the angle $MXC$ is right. (Mirchev Borislav)

2013 Tournament of Towns, 2

Tags: geometry , right angle , right triangle , square

Let $C$ be a right angle in triangle $ABC$. On legs $AC$ and$BC$ the squares $ACKL, BCMN$ are constructed outside of triangle. If $CE$ is an altitude of the triangle prove that $LEM$ is a right angle.

Champions Tournament Seniors - geometry, 2011.2

Tags: right angle , geometry , isosceles

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 $.

2024 Bulgaria MO Regional Round, 12.1

Tags: geometry , conditional geometry , right angle , incentre , isosceles

Let $ABC$ be an acute triangle with midpoint $M$ of $AB$. The point $D$ lies on the segment $MB$ and $I_1, I_2$ denote the incenters of $\triangle ADC$ and $\triangle BDC$. Given that $\angle I_1MI_2=90^{\circ}$, show that $CA=CB$.

2017 Auckland Mathematical Olympiad, 1

Tags: right angle , geometry

A $6$ meter ladder rests against a vertical wall. The midpoint of the ladder is twice as far from the ground as it is from the wall. At what height on the wall does the ladder reach?

2021 New Zealand MO, 1

Tags: right angle , geometry , equal segments

Let $ABCD$ be a convex quadrilateral such that $AB + BC = 2021$ and $AD = CD$. We are also given that $\angle ABC = \angle CDA = 90^o$. Determine the length of the diagonal $BD$.

2005 Oral Moscow Geometry Olympiad, 4

Tags: geometry , hexagon , perpendicular , equal segments , right angle

Given a hexagon $ABCDEF$, in which $AB = BC, CD = DE, EF = FA$, and angles $A$ and $C$ are right. Prove that lines $FD$ and $BE$ are perpendicular. (B. Kukushkin)

2011 Saudi Arabia IMO TST, 2

Tags: circumcircle , geometry , incenter , right angle

Let $ABC$ be a triangle with $AB\ne AC$. Its incircle has center $I$ and touches the side $BC$ at point $D$. Line $AI$ intersects the circumcircle $\omega$ of triangle $ABC$ at $M$ and $DM$ intersects again $\omega$ at $P$. Prove that $\angle API= 90^o$.

1950 Polish MO Finals, 5

Tags: trigonometry , geometry , right triangle , right angle

Prove that if for angles $A,B,C$ of a triangle holds $$\sin^2 A+\sin^2 B +\sin^2 C=2$$ iff the triangle $ABC$ is right.

2015 Thailand TSTST, 1

Tags: geometry , circumcircle , bisects segment , right angle

Let $O$ be the circumcenter of an acute $\vartriangle ABC$ which has altitude $AD$. Let $AO$ intersect the circumcircle of $\vartriangle BOC$ again at $X$. If $E$ and $F$ are points on lines $AB$ and $AC$ such that $\angle XEA = \angle XFA = 90^o$ , then prove that the line $DX$ bisects the segment $EF$.

Croatia MO (HMO) - geometry, 2020.3

Tags: geometry , right angle , perpendicular

Given a triangle $ABC$ such that $AB<AC$ . On sides $AB$ and $BC$, points $P$ and $Q$ are marked respectively such that the lines $AQ$ and $CP$ are perpendicular, and the circle inscribed in the triangle $ABC$ touches the length $PQ$. The line $CP$ intersects the circle circumscribed around the triangle $ABC$ at the points $C$ and $T$. If the lines $CA,PQ$ and $BT$ intersect at one point, prove that the angle $\angle CAB$ is right.