Olympiad problems

2019 Polish Junior MO First Round, 5

A parallelogram $ABCD$ is given. On the diagonal BD, a point $P$ is selected such that $AP = BD$ is satisfied. Point $Q$ is the midpoint of segment $CP$. Prove that $\angle BQD = 90^o$. [img]https://cdn.artofproblemsolving.com/attachments/2/0/4bc69ec0330e2afa6b560c56da5dd783b16efb.png[/img] .

2021 Bosnia and Herzegovina Junior BMO TST, 3

Tags: right angle , geometry , equal angles

In the convex quadrilateral $ABCD$, $AD = BD$ and $\angle ACD = 3 \angle BAC$. Let $M$ be the midpoint of side $AD$. If the lines $CM$ and $AB$ are parallel, prove that the angle $\angle ACB$ is right.

2016 Irish Math Olympiad, 6

Tags: geometry , centroid , right angle

Triangle $ABC$ has sides $a = |BC| > b = |AC|$. The points $K$ and $H$ on the segment $BC$ satisfy $|CH| = (a + b)/3$ and $|CK| = (a - b)/3$. If $G$ is the centroid of triangle $ABC$, prove that $\angle KGH = 90^o$.

Estonia Open Junior - geometry, 2007.2.2

Tags: angle bisector , right angle , square , 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 $BP$.

2010 Contests, 1a

Tags: equal segments , right angle , geometry , angle

The point $P$ lies on the edge $AB$ of a quadrilateral $ABCD$. The angles $BAD, ABC$ and $CPD$ are right, and $AB = BC + AD$. Show that $BC = BP$ or $AD = BP$.

Kyiv City MO Juniors Round2 2010+ geometry, 2021.7.41

Tags: geometric inequality , right angle , perimeter , geometry

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

Kharkiv City MO Seniors - geometry, 2014.11.5

Tags: right angle , perpendicular , geometry , collinear

In the convex quadrilateral of the $ABCD$, the diagonals of $AC$ and $BD$ are mutually perpendicular and intersect at point $E$. On the side of $AD$, a point $P$ is chosen such that $PE = EC$. The circumscribed circle of the triangle $BCD$ intersects the segment $AD$ at the point $Q$. The circle passing through point $A$ and tangent to the line $EP$ at point $P$ intersects the segment $AC$ at point $R$. It turns out that points $B, Q, R$ are collinear. Prove that $\angle BCD = 90^o$.

Ukraine Correspondence MO - geometry, 2005.7

Tags: trapezoid , right angle , geometry , angle

Let $O$ be the point of intersection of the diagonals of the trapezoid $ABCD$ with the bases $AB$ and $CD$. It is known that $\angle AOB = \angle DAB = 90^o$. On the sides $AD$ and $BC$ take the points $E$ and $F$ so that $EF\parallel AB$ and $EF = AD$. Find the angle $\angle AOE$.

Kyiv City MO Juniors Round2 2010+ geometry, 2020.8.2

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

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

2021 Saudi Arabia IMO TST, 5

Tags: geometry , right angle , circumradius , incenter

Let $ABC$ be a non isosceles triangle with incenter $I$ . The circumcircle of the triangle $ABC$ has radius $R$. Let $AL$ be the external angle bisector of $\angle BAC $with $L \in BC$. Let $K$ be the point on perpendicular bisector of $BC$ such that $IL \perp IK$.Prove that $OK=3R$.

Estonia Open Senior - geometry, 2004.2.4

Tags: geometry , right angle , circumcircle

On the circumcircle of triangle $ABC$, point $P$ is chosen, such that the perpendicular drawn from point $P$ to line $AC$ intersects the circle again at a point $Q$, the perpendicular drawn from point $Q$ to line $AB$ intersects the circle again at a point $R$ and the perpendicular drawn from point $R$ to line $BC$ intersects the circle again at the initial point $P$. Let $O$ be the centre of this circle. Prove that $\angle POC = 90^o$.

Cono Sur Shortlist - geometry, 2003.G3

Tags: square , right angle , geometry

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.

2013 Tournament of Towns, 5

Tags: right angle , angle , geometry , equal segments

In a quadrilateral $ABCD$, angle $B$ is equal to $150^o$, angle $C$ is right, and sides $AB$ and $CD$ are equal. Determine the angle between $BC$ and the line connecting the midpoints of sides $BC$ and $AD$.

Kyiv City MO 1984-93 - geometry, 1992.7.2

Tags: equilateral , right angle , geometry

Inside a right angle is given a point $A$. Construct an equilateral triangle, one of the vertices of which is point $A$, and two others lie on the sides of the angle (one on each side).

2000 Junior Balkan Team Selection Tests - Moldova, 7

Tags: concurrency , geometry , right angle

Let a triangle $ABC, A_1$ be the midpoint of the segment $[BC], B_1 \in (AC)$ ¸and $C_1 \in (AB)$ such that $[A_1B_1$ is the bisector of the angle $AA_1C$ and $A_1C_1$ is perpendicular to $AB$. Show that the lines $AA_1, BB_1$ and $CC_1$ are concurrent if and only if $ \angle BAC = 90^o$

2007 Estonia Math Open Junior Contests, 7

Tags: geometry , right angle , square , angle bisector

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 $BP$.

2009 Regional Olympiad of Mexico Center Zone, 1

Tags: geometry , right angle

Let $\Gamma$ be a circle with the center $O$ and let $A$, $A ^ \prime $ be two diametrically opposite points in $\Gamma$. Let $P$ be the midpoint of $OA ^ \prime$ and $\ell$ a line that passes through $P$, different from the line $AA ^ \prime$ and different from the line perpendicular on $AA ^ \prime$. Let $B$ and $C$ be the intersection points of $\ell$ with $\Gamma$, let $H$ be the foot of the altitude from $A$ on $BC$, let $M$ be the midpoint of $BC$, and let $D$ be the intersection of the line $A ^ \prime M$ with $AH$. Show that the angle $\angle ADO = 90 ^ \circ $.

1970 Czech and Slovak Olympiad III A, 2

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

2010 Abels Math Contest (Norwegian MO) Final, 1a

Tags: geometry , angle , right angle , equal segments

The point $P$ lies on the edge $AB$ of a quadrilateral $ABCD$. The angles $BAD, ABC$ and $CPD$ are right, and $AB = BC + AD$. Show that $BC = BP$ or $AD = BP$.

2012 Grand Duchy of Lithuania, 2

Tags: right angle , geometry , trapezoid

The base $AB$ of a trapezium $ABCD$ is longer than the base $CD$, and $\angle ADC$ is a right angle. The diagonals $AC$ and $BD$ are perpendicular. Let $E$ be the foot of the altitude from $D$ to the line $BC$. Prove that $$\frac{AE}{BE} =\frac{ AC \cdot CD}{AC^2 - CD^2}$$ .

2017 Saudi Arabia IMO TST, 1

Tags: geometry , orthocenter , right angle , circumcircle , reflection

Let $ABC$ be a triangle inscribed in circle $(O),$ with its altitudes $BE, CF$ intersect at orthocenter $H$ ($E \in AC, F \in AB$). Let $M$ be the midpoint of $BC, K$ be the orthogonal projection of $H$ on $AM$. $EF$ intersects $BC$ at $P$. Let $Q$ be the intersection of tangent of $(O)$ which passes through $A$ with $BC, T$ be the reflection of $Q$ through $P$. Prove that $\angle OKT = 90^o$.

1996 Singapore Team Selection Test, 1

Tags: incenter , right angle , equal segments , geometry

Let $C, B, E$ be three points on a straight line $\ell$ in that order. Suppose that $A$ and $D$ are two points on the same side of $\ell$ such that (i) $\angle ACE = \angle CDE = 90^o$ and (ii) $CA = CB = CD$. Let $F$ be the point of intersection of the segment $AB$ and the circumcircle of $\vartriangle ADC$. Prove that $F$ is the incentre of $\vartriangle CDE$.

Kyiv City MO Seniors Round2 2010+ geometry, 2018.11.2

Tags: right angle , geometry , circumcircle

In the quadrilateral $ABCD $, $AB = BC $, the point $K $ is the midpoint of the side $CD $, the rays $BK $ and $AD $ intersect at the point $M $ , the circumscribed circle $ \Delta ABM $ intersects the line $AC $ for the second time at the point $P $. Prove that $\angle BKP = 90 {} ^ \circ $. (Anton Trygub)

2014 India PRMO, 12

Tags: sum , right angle , inradius , geometry

Let $ABCD$ be a convex quadrilateral with $\angle DAB =\angle B DC = 90^o$. Let the incircles of triangles $ABD$ and $BCD$ touch $BD$ at $P$ and $Q$, respectively, with $P$ lying in between $B$ and $Q$. If $AD = 999$ and $PQ = 200$ then what is the sum of the radii of the incircles of triangles $ABD$ and $BDC$ ?

2016 Denmark MO - Mohr Contest, 3

Tags: geometry , area , right angle , equal segments

Prove that all quadrilaterals $ABCD$ where $\angle B = \angle D = 90^o$, $|AB| = |BC|$ and $|AD| + |DC| = 1$, have the same area. [img]https://1.bp.blogspot.com/-55lHuAKYEtI/XzRzDdRGDPI/AAAAAAAAMUk/n8lYt3fzFaAB410PQI4nMEz7cSSrfHEgQCLcBGAsYHQ/s0/2016%2Bmohr%2Bp3.png[/img]