Olympiad problems

2019 Adygea Teachers' Geometry Olympiad, 4

From which two statements about the trapezoid follows the third: 1) the trapezoid is tangential, 2) the trapezoid is right, 3) its area is equal to the product of the bases?

2015 Thailand TSTST, 1

Tags: circumcircle , bisects segment , right angle , geometry

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 $EF$.

2005 Cuba MO, 5

Tags: right angle , perpendicular , geometry

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 $\angle POC = 90^o$ .

2016 Sharygin Geometry Olympiad, P15

Tags: geometry , nagel point , right angle , circumcircle

Let $O, M, N$ be the circumcenter, the centroid and the Nagel point of a triangle. Prove that angle $MON$ is right if and only if one of the triangle’s angles is equal to $60^o$.

2008 Postal Coaching, 1

Tags: incenter , circumcircle , right angle , arc midpoint , angle chasing , angle , geometry

In triangle $ABC,\angle B > \angle C, T$ is the midpoint of arc $BAC$ of the circumcicle of $ABC$, and $I$ is the incentre of $ABC$. Let $E$ be point such that $\angle AEI = 90^0$ and $AE$ is parallel to $BC$. If $TE$ intersects the circumcircle of $ABC$ at $P(\ne T)$ and $\angle B = \angle IPB$, determine $\angle A$.

2014 India PRMO, 12

Tags: geometry , inradius , sum , right angle

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 $BDC$ ?

2010 Contests, 1a

Tags: geometry , right angle , equal segments , 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 1984-93 - geometry, 1992.7.2

Tags: geometry , equilateral , right angle

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).

Kharkiv City MO Seniors - geometry, 2014.11.5

Tags: right angle , geometry , collinear , perpendicular

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 $\angle BCD = 90^o$.

2024 Polish MO Finals, 6

Tags: right angle , parallelogram , circumcircle , geometry

Let $ABCD$ be a parallelogram. Let $X \notin AC $ lie inside $ABCD$ so that $\angle AXB = \angle CXD = 90^ {\circ}$ and $\Omega$ denote the circumcircle of $AXC$. Consider a diameter $EF$ of $\Omega$ and assume neither $E, \ X, \ B$ nor $F, \ X, \ D$ are collinear. Let $K \neq X$ be an intersection point of circumcircles of $BXE$ and $DXF$ and $L \neq X$ be such point on $\Omega$ so that $\angle KXL = 90^{\circ}$. Prove that $AB = KL$.

Estonia Open Senior - geometry, 2004.2.4

Tags: circumcircle , right angle , geometry

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 $\angle POC = 90^o$.

2011 Saudi Arabia IMO TST, 2

Tags: geometry , incenter , right angle , circumcircle

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.

2009 Belarus Team Selection Test, 3

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

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

2016 Irish Math Olympiad, 6

Tags: centroid , right angle , geometry

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 $\angle KGH = 90^o$.

2021 Yasinsky Geometry Olympiad, 6

Tags: geometry , incenter , arc midpoint , right angle

In an acute-angled triangle $ABC$, point $I$ is the center of the inscribed circle, point $T$ is the midpoint of the arc $ABC$ of the circumcircle of triangle $ABC$. It turned out that $\angle AIT = 90^o$ . Prove that $AB + AC = 3BC$. (Matthew of Kursk)

2019 Polish Junior MO First Round, 5

Tags: geometry , right angle , parallelogram

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] .

2018 NZMOC Camp Selection Problems, 4

Tags: geometry , right angle , equal angles

Let $P$ be a point inside triangle $ABC$ such that $\angle CPA = 90^o$ and $\angle CBP = \angle CAP$. Prove that $\angle P XY = 90^o$, where $X$ and $Y$ are the midpoints of $AB$ and $AC$ respectively.

Kyiv City MO Seniors Round2 2010+ geometry, 2020.10.2

Tags: right angle , angle , geometry

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)

Ukraine Correspondence MO - geometry, 2005.7

Tags: geometry , trapezoid , right angle , 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 $\angle AOE$.

2017 Latvia Baltic Way TST, 10

Tags: geometry , right angle , incircle

In an obtuse triangle $ABC$, for which $AC < AB$, the radius of the inscribed circle is $R$, the midpoint of its arc $BC$ (which does not contain $A$) is $S$. A point $T$ is placed on the extension of altitude $AD$ such that $D$ is between $ A$ and $T$ and $AT = 2R$. Prove that $\angle AST = 90^o$.

2018 Saudi Arabia IMO TST, 2

Tags: geometry , circumcircle , right angle , angle bisector

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 $\angle EGF = 90^o$.

Kyiv City MO Juniors 2003+ geometry, 2004.7.3

Tags: geometry , right triangle , right angle

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 $\angle DOE = 90^o$.

Kyiv City MO Juniors Round2 2010+ geometry, 2011.8.3

Tags: geometry , right angle , square , equal segments

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)

Kyiv City MO Juniors 2003+ geometry, 2018.7.4

Tags: geometry , right angle , equal segments , angle

Inside the triangle $ABC $, the point $P $ is selected so that $BC = AP $ and $\angle APC = 180 {} ^ \circ - \angle ABC $. On the side $AB $ there is a point $K $, for which $AK = KB + PC $. Prove that $\angle AKC = 90 {} ^ \circ $. (Danilo Hilko)