Olympiad problems

2017 Junior Regional Olympiad - FBH, 5

Points $K$ and $L$ are on side $AB$ of triangle $ABC$ such that $KL=BC$ and $AK=LB$. Let $M$ be a midpoint of $AC$. Prove that $\angle KML = 90^{\circ}$

2005 Cuba MO, 5

Tags: geometry , right angle , perpendicular

On the circumcircle of triangle $ABC$, point $P$ is taken in such a way that the perpendicular drawn by the point $P$ to the line $AC$ cuts the circle also at the point $Q$, the perpendicular drawn by the point $Q$ to the line $AB$ cuts the circle also at point R and the perpendicular drawn by point $R$ to the line BC cuts the circle also at the point $P$. Let $O$ be the center of this circle. Prove that $\angle POC = 90^o$ .

Indonesia MO Shortlist - geometry, g9

Tags: circles , right angle , arc midpoint , geometry

Given two circles $\Gamma_1$ and $\Gamma_2$ which intersect at points $A$ and $B$. A line through $A$ intersects $\Gamma_1$ and $\Gamma_2$ at points $C$ and $D$, respectively. Let $M$ be the midpoint of arc $BC$ in $\Gamma_1$ ,which does not contains $A$, and $N$ is the midpoint of the arc $BD$ in $\Gamma_2$, which does not contain $A$. If $K$ is the midpoint of $CD$, prove that $\angle MKN = 90^o.$

2018 Oral Moscow Geometry Olympiad, 2

Tags: right angle , diagonal , perpendicular , trapezoid , geometry

The diagonals of the trapezoid $ABCD$ are perpendicular ($AD//BC, AD>BC$) . Point $M$ is the midpoint of the side of $AB$, the point $N$ is symmetric of the center of the circumscribed circle of the triangle $ABD$ wrt $AD$. Prove that $\angle CMN = 90^o$. (A. Mudgal, India)

2015 BMT Spring, 3

Tags: geometry , right angle

A quadrilateral $ABCD$ has a right angle at $\angle ABC$ and satisfies $AB = 12$, $BC = 9$, $CD = 20$, and $DA = 25$. Determine $BD^2$. .

1971 Czech and Slovak Olympiad III A, 2

Tags: geometry , right angle

Let $ABC$ be a triangle. Four distinct points $D,A,B,E$ lie on the line $AB$ in this order such that $DA=AB=BE.$ Find necessary and sufficient condition for lengths $a=BC,b=AC$ such that the angle $\angle DCE$ is right.

2021 Abels Math Contest (Norwegian MO) Final, 4a

Tags: tetrahedron , 3d geometry , right angle , geometry

A tetrahedron $ABCD$ satisfies $\angle BAC=\angle CAD=\angle DAB=90^o$. Show that the areas of its faces satisfy the equation $area(BAC)^2 + area(CAD)^2 + area(DAB)^2 = area(BCD)^2$. .

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 $\angle I_1DI_2 = 90^o$.

Kyiv City MO Seniors Round2 2010+ geometry, 2020.10.2

Tags: geometry , angle , right angle

Let $M$ be the midpoint of the side $AC$ of triangle $ABC$. Inside $\vartriangle BMC$ was found a point $P$ such that $\angle BMP = 90^o$, $\angle ABC+ \angle APC =180^o$. Prove that $\angle PBM + \angle CBM = \angle PCA$. (Anton Trygub)

2021 Dutch IMO TST, 2

Tags: right triangle , geometry , right angle , centroid

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 $ABC$.

2014 Costa Rica - Final Round, 5

Tags: geometry , incircle , excircle , right angle

Let $ABC$ be a triangle, with $A'$, $B'$, and $C'$ the points of tangency of the incircle with $BC$, $CA$, and $AB$ respectively. Let $X$ be the intersection of the excircle with respect to $A$ with $AB$, and $M$ the midpoint of $BC$. Let $D$ be the intersection of $XM$ with $B'C'$. Show that $\angle C'A'D' = 90^o$.

May Olympiad L2 - geometry, 2001.2

Tags: trapezoid , right angle , geometry

On the trapezoid $ABCD$ , side $DA$ is perpendicular to the bases $AB$ and $CD$ . The base $AB$ measures $45$, the base $CD$ measures $20$ and the $BC$ side measures $65$. Let $P$ on the $BC$ side such that $BP$ measures $45$ and $M$ is the midpoint of $DA$. Calculate the measure of the $PM$ segment.

2014 Bosnia and Herzegovina Junior BMO TST, 2

Tags: geometry , triangle , right angle

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 $\angle BAC= 90^{\circ}$

2007 Austria Beginners' Competition, 4

Tags: right triangle , right angle , parallelogram , geometry

Consider a parallelogram $ABCD$ such that the midpoint $M$ of the side $CD$ lies on the angle bisector of $\angle BAD$. Show that $\angle AMB$ is a right angle.

2001 May Olympiad, 2

Tags: trapezoid , right angle , geometry

On the trapezoid $ABCD$ , side $DA$ is perpendicular to the bases $AB$ and $CD$ . The base $AB$ measures $45$, the base $CD$ measures $20$ and the $BC$ side measures $65$. Let $P$ on the $BC$ side such that $BP$ measures $45$ and $M$ is the midpoint of $DA$. Calculate the measure of the $PM$ segment.

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 $ADB$.

2019 Saudi Arabia Pre-TST + Training Tests, 3.2

Tags: geometry , circles , right angle

Let $ABC$ be a triangle, the circle having $BC$ as diameter cuts $AB,AC$ at $F,E$ respectively. Let $P$ a point on this circle. Let $C',B$' be the projections of $P$ upon the sides $AB,AC$ respectively. Let $H$ be the orthocenter of the triangle $AB'C'$. Show that $\angle EHF = 90^o$.

Kharkiv City MO Seniors - geometry, 2015.11.3

Tags: geometry , midpoint , rectangle , right angle

In the rectangle $ABCD$, point $M$ is the midpoint of the side $BC$. The points $P$ and $Q$ lie on the diagonal $AC$ such that $\angle DPC = \angle DQM = 90^o$. Prove that $Q$ is the midpoint of the segment $AP$.

2011 Oral Moscow Geometry Olympiad, 2

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

In an isosceles triangle $ABC$ ($AB=AC$) on the side $BC$, point $M$ is marked so that the segment $CM$ is equal to the altitude of the triangle drawn on this side, and on the side $AB$, point $K$ is marked so that the angle $\angle KMC$ is right. Find the angle $\angle ACK$.

2012 Dutch IMO TST, 4

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

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 $\angle NML = 90^o$.

2009 Bosnia And Herzegovina - Regional Olympiad, 1

Tags: geometry , incircle , right angle

In triangle $ABC$ such that $\angle ACB=90^{\circ}$, let point $H$ be foot of perpendicular from point $C$ to side $AB$. Show that sum of radiuses of incircles of $ABC$, $BCH$ and $ACH$ is $CH$

2017 Singapore Senior Math Olympiad, 2

Tags: perpendicular , right angle , cyclic , cyclic quadrilateral , geometry

In the cyclic quadrilateral $ABCD$, the sides $AB, DC$ meet at $Q$, the sides $AD,BC$ meet at $P, M$ is the midpoint of $BD$, If $\angle APQ=90^o$, prove that $PM$ is perpendicular to $AB$.

Kyiv City MO Juniors 2003+ geometry, 2004.7.3

Tags: right triangle , right angle , geometry

Given a right triangle $ABC$ ($\angle A <45^o$,$ \angle C = 90^o$), on the sides $AC$ and $AB$ which are selected points $D,E$ respectively, such that $BD = AD$ and $CB = CE$. Let the segments $BD$ and $CE$ intersect at the point $O$. Prove that $\angle DOE = 90^o$.

2020 Tournament Of Towns, 5

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

Let $ABCD$ be an inscribed trapezoid. The base $AB$ is $3$ times longer than $CD$. Tangents to the circumscribed circle at the points $A$ and $C$ intersect at the point $K$. Prove that the angle $KDA$ is a right angle. Alexandr Yuran

2009 China Northern MO, 2

Tags: trigonometry , angle bisector , geometry , right angle

In an acute triangle $ABC$ , $AB>AC$ , $ \cos B+ \cos C=1$ , $E,F$ are on the extend line of $AB,AC$ such that $\angle ABF = \angle ACE = 90$ . (1) Prove :$BE+CF=EF$ ; (2) Assume the bisector of $\angle EBC$ meet $EF$ at $P$ , prove that $CP$ is the bisector of $\angle BCF$. [img]https://cdn.artofproblemsolving.com/attachments/a/2/c554c2bc0b4e044c45f88138568f5234d544a8.png[/img]