Olympiad problems

2018 Peru Iberoamerican Team Selection Test, P2

Let $ABC$ be a triangle with $AB = AC$ and let $D$ be the foot of the height drawn from $A$ to $BC$. Let $P$ be a point inside the triangle $ADC$ such that $\angle APB> 90^o$ and $\angle PAD + \angle PBD = \angle PCD$. The $CP$ and $AD$ lines are cut at $Q$ and the $BP$ and $AD$ lines cut into $R$. Let $T$ be a point in segment $AB$ such that $\angle TRB = \angle DQC$ and let S be a point in the extension of the segment $AP$ (on the $P$ side) such that $\angle PSR = 2 \angle PAR$. Prove that $RS = RT$.

2018 Moldova Team Selection Test, 3

Tags: geometry , angle chasing , projective geometry , similar triangles

Let $O$ be the circumcenter of an acute triangle $ABC$. Line $OA$ intersects the altitudes of $ABC$ through $B$ and $C$ at $P$ and $Q$, respectively. The altitudes meet at $H$. Prove that the circumcenter of triangle $PQH$ lies on a median of triangle $ABC$.

2004 IMO Shortlist, 1

Tags: incenter , angle bisector , angle chasing , geometry

1. Let $ABC$ be an acute-angled triangle with $AB\neq AC$. The circle with diameter $BC$ intersects the sides $AB$ and $AC$ at $M$ and $N$ respectively. Denote by $O$ the midpoint of the side $BC$. The bisectors of the angles $\angle BAC$ and $\angle MON$ intersect at $R$. Prove that the circumcircles of the triangles $BMR$ and $CNR$ have a common point lying on the side $BC$.

2024 Auckland Mathematical Olympiad, 4

Tags: angle chasing , geometry

The altitude $AH$ and the bisector $CL$ of triangle $ABC$ intersect at point $O$. Find the angle $BAC$, if it is known that the difference between angle $COH$ and half of angle $ABC$ is $46$.

1986 China Team Selection Test, 1

Tags: incenter , rectangle , cyclic quadrilateral , angle chasing , geometry

If $ABCD$ is a cyclic quadrilateral, then prove that the incenters of the triangles $ABC$, $BCD$, $CDA$, $DAB$ are the vertices of a rectangle.

2009 May Olympiad, 2

Tags: angle chasing , isosceles , equilateral triangle , convex quadrilateral , geometry

Let $ABCD$ be a convex quadrilateral such that the triangle $ABD$ is equilateral and the triangle $BCD$ is isosceles, with $\angle C = 90^o$. If $E$ is the midpoint of the side $AD$, determine the measure of the angle $\angle CED$.

2021 Science ON Seniors, 4

Tags: geometry , angle chasing , circles , projective

$ABCD$ is a cyclic convex quadrilateral whose diagonals meet at $X$. The circle $(AXD)$ cuts $CD$ again at $V$ and the circle $(BXC)$ cuts $AB$ again at $U$, such that $D$ lies strictly between $C$ and $V$ and $B$ lies strictly between $A$ and $U$. Let $P\in AB\cap CD$.\\ \\ If $M$ is the intersection point of the tangents to $U$ and $V$ at $(UPV)$ and $T$ is the second intersection of circles $(UPV)$ and $(PAC)$, prove that $\angle PTM=90^o$.\\ \\ [i](Vlad Robu)[/i]

2016 Iranian Geometry Olympiad, 2

Tags: geometry , equal segments , circumcircle , angle chasing

Let $\omega$ be the circumcircle of triangle $ABC$ with $AC > AB$. Let $X$ be a point on $AC$ and $Y$ be a point on the circle $\omega$, such that $CX = CY = AB$. (The points $A$ and $Y$ lie on different sides of the line $BC$). The line $XY$ intersects $\omega$ for the second time in point $P$. Show that $PB = PC$. by Iman Maghsoudi

2007 District Olympiad, 1

Tags: geometry , equal segments , angle chasing , circumcenter , circumcircle , angle

Point $O$ is the intersection of the perpendicular bisectors of the sides of the triangle $\vartriangle ABC$ . Let $D$ be the intersection of the line $AO$ with the segment $[BC]$. Knowing that $OD = BD = \frac 13 BC$, find the measures of the angles of the triangle $\vartriangle ABC$.

2018 Estonia Team Selection Test, 5

Tags: geometry , angle chasing , projective geometry , similar triangles

Let $O$ be the circumcenter of an acute triangle $ABC$. Line $OA$ intersects the altitudes of $ABC$ through $B$ and $C$ at $P$ and $Q$, respectively. The altitudes meet at $H$. Prove that the circumcenter of triangle $PQH$ lies on a median of triangle $ABC$.

2022 Saudi Arabia IMO TST, 2

Tags: geometry , cyclic quadrilateral , concurrency , angle chasing

Let $ABCD$ be a quadrilateral inscribed in a circle $\Omega.$ Let the tangent to $\Omega$ at $D$ meet rays $BA$ and $BC$ at $E$ and $F,$ respectively. A point $T$ is chosen inside $\triangle ABC$ so that $\overline{TE}\parallel\overline{CD}$ and $\overline{TF}\parallel\overline{AD}.$ Let $K\ne D$ be a point on segment $DF$ satisfying $TD=TK.$ Prove that lines $AC,DT,$ and $BK$ are concurrent.

2002 IMO Shortlist, 3

Tags: geometry , incenter , circumcircle , perpendicular bisector , angle chasing

The circle $S$ has centre $O$, and $BC$ is a diameter of $S$. Let $A$ be a point of $S$ such that $\angle AOB<120{{}^\circ}$. Let $D$ be the midpoint of the arc $AB$ which does not contain $C$. The line through $O$ parallel to $DA$ meets the line $AC$ at $I$. The perpendicular bisector of $OA$ meets $S$ at $E$ and at $F$. Prove that $I$ is the incentre of the triangle $CEF.$

2018 Romania Team Selection Tests, 1

Tags: geometry , angle chasing , projective geometry , similar triangles

Let $O$ be the circumcenter of an acute triangle $ABC$. Line $OA$ intersects the altitudes of $ABC$ through $B$ and $C$ at $P$ and $Q$, respectively. The altitudes meet at $H$. Prove that the circumcenter of triangle $PQH$ lies on a median of triangle $ABC$.

2015 Dutch BxMO/EGMO TST, 4

Tags: geometry , circumcircle , equal angles , angle chasing , angle bisector

In a triangle $ABC$ the point $D$ is the intersection of the interior angle bisector of $\angle BAC$ and side $BC$. Let $P$ be the second intersection point of the exterior angle bisector of $\angle BAC$ with the circumcircle of $\angle ABC$. A circle through $A$ and $P$ intersects line segment $BP$ internally in $E$ and line segment $CP$ internally in $F$. Prove that $\angle DEP = \angle DFP$.

2023 Bangladesh Mathematical Olympiad, P6

Tags: geometry , cyclic quadrilateral , angle chasing

Let $\triangle ABC$ be an acute angle triangle and $\omega$ be its circumcircle. Let $N$ be a point on arc $AC$ not containing $B$ and $S$ be a point on line $AB$. The line tangent to $\omega$ at $N$ intersects $BC$ at $T$, $NS$ intersects $\omega$ at $K$. Assume that $\angle NTC = \angle KSB$. Prove that $CK\parallel AN \parallel TS$.

2022 Thailand TST, 2

Tags: geometry , cyclic quadrilateral , concurrency , angle chasing

Let $ABCD$ be a quadrilateral inscribed in a circle $\Omega.$ Let the tangent to $\Omega$ at $D$ meet rays $BA$ and $BC$ at $E$ and $F,$ respectively. A point $T$ is chosen inside $\triangle ABC$ so that $\overline{TE}\parallel\overline{CD}$ and $\overline{TF}\parallel\overline{AD}.$ Let $K\ne D$ be a point on segment $DF$ satisfying $TD=TK.$ Prove that lines $AC,DT,$ and $BK$ are concurrent.

2018 Pan African, 4

Tags: geometry , cyclic quadrilateral , angle chasing

Given a triangle $ABC$, let $D$ be the intersection of the line through $A$ perpendicular to $AB$, and the line through $B$ perpendicular to $BC$. Let $P$ be a point inside the triangle. Show that $DAPB$ is cyclic if and only if $\angle BAP = \angle CBP$.

2019 Silk Road, 1

Tags: geometry , angle chasing , equal angles , altitude

The altitudes of the acute-angled non-isosceles triangle $ ABC $ intersect at the point $ H $. On the segment $ C_1H $, where $ CC_1 $ is the altitude of the triangle, the point $ K $ is marked. Points $ L $ and $ M $ are the feet of perpendiculars from point $ K $ on straight lines $ AC $ and $ BC $, respectively. The lines $ AM $ and $ BL $ intersect at $ N $. Prove that $ \angle ANK = \angle HNL $.

2012 Indonesia TST, 3

Tags: angle chasing , complex number , cyclic quadrilateral , geometry

The incircle of a triangle $ABC$ is tangent to the sides $AB,AC$ at $M,N$ respectively. Suppose $P$ is the intersection between $MN$ and the bisector of $\angle ABC$. Prove that $BP$ and $CP$ are perpendicular.

2019 USMCA, 3

Tags: inversion , projective , angle chasing , length chase , geometry , trigonometry

Let $ABC$ be a scalene triangle. The incircle of $ABC$ touches $\overline{BC}$ at $D$. Let $P$ be a point on $\overline{BC}$ satisfying $\angle BAP = \angle CAP$, and $M$ be the midpoint of $\overline{BC}$. Define $Q$ to be on $\overline{AM}$ such that $\overline{PQ} \perp \overline{AM}$. Prove that the circumcircle of $\triangle AQD$ is tangent to $\overline{BC}$.

STEMS 2021 Math Cat A, Q3

Tags: geometry , angle chasing

An acute scalene triangle $\triangle{ABC}$ with altitudes $\overline{AD}, \overline{BE},$ and $\overline{CF}$ is inscribed in circle $\Gamma$. Medians from $B$ and $C$ meet $\Gamma$ again at $K$ and $L$ respectively. Prove that the circumcircles of $\triangle{BFK}, \triangle{CEL}$ and $\triangle{DEF}$ concur.

2023 Rioplatense Mathematical Olympiad, 2

Tags: excenter , angle chasing , geometry

Let $ABCD$ be a convex quadrilateral, such that $AB = CD$, $\angle BCD = 2 \angle BAD$, $\angle ABC = 2 \angle ADC$ and $\angle BAD \neq \angle ADC$. Determine the measure of the angle between the diagonals $AC$ and $BD$.

2017 All-Russian Olympiad, 8

Tags: incenter , angle chasing , incircle , geometry

Given a convex quadrilateral $ABCD$. We denote $I_A,I_B, I_C$ and $I_D$ centers of $\omega_A, \omega_B,\omega_C $and $\omega_D$,inscribed In the triangles $DAB, ABC, BCD$ and $CDA$, respectively.It turned out that $\angle BI_AA + \angle I_CI_AI_D = 180^\circ$. Prove that $\angle BI_BA + \angle I_CI_BI_D = 180^{\circ}$. (A. Kuznetsov)

2016 Bangladesh Mathematical Olympiad, 3

Tags: geometry , angle chasing

$\triangle ABC$ is isosceles $AB = AC$. $P$ is a point inside $\triangle ABC$ such that $\angle BCP = 30$ and $\angle APB = 150$ and $\angle CAP = 39$. Find $\angle BAP$.

2022 Iran MO (3rd Round), 1

Tags: angle chasing , spiral similarity , circumcircle , geometric transformation , reflection , angle bisector , geometry

Triangle $ABC$ is assumed. The point $T$ is the second intersection of the symmedian of vertex $A$ with the circumcircle of the triangle $ABC$ and the point $D \neq A$ lies on the line $AC$ such that $BA=BD$. The line that at $D$ tangents to the circumcircle of the triangle $ADT$, intersects the circumcircle of the triangle $DCT$ for the second time at $K$. Prove that $\angle BKC = 90^{\circ}$(The symmedian of the vertex $A$, is the reflection of the median of the vertex $A$ through the angle bisector of this vertex).