Olympiad problems

2023 pOMA, 2

Let $\triangle ABC$ be an acute triangle, and let $D,E,F$ respectively be three points on sides $BC,CA,AB$ such that $AEDF$ is a cyclic quadrilateral. Let $O_B$ and $O_C$ be the circumcenters of $\triangle BDF$ and $\triangle CDE$, respectively. Finally, let $D'$ be a point on segment $BC$ such that $BD'=CD$. Prove that $\triangle BD'O_B$ and $\triangle CD'O_C$ have the same surface.

2021 Argentina National Olympiad, 3

Tags: geometry , Angle Chasing , Argentina

Let $ABCD$ be a quadrilateral inscribed in a circle such that $\angle ABC=60^{\circ}.$ a) Prove that if $BC=CD$ then $AB= CD+DA.$ b) Is it true that if $AB= CD+DA$ then $BC=CD$?

2012 Dutch IMO TST, 4

Tags: geometry , right angle , equal angles , Angle Chasing

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 $\angle NML = 90^o$.

2019 Korea National Olympiad, 6

Tags: geometry , Korea , geometry solved , Angle Chasing , similar triangles , incenter , arc midpoint

In acute triangle $ABC$, $AB>AC$. Let $I$ the incenter, $\Omega$ the circumcircle of triangle $ABC$, and $D$ the foot of perpendicular from $A$ to $BC$. $AI$ intersects $\Omega$ at point $M(\neq A)$, and the line which passes $M$ and perpendicular to $AM$ intersects $AD$ at point $E$. Now let $F$ the foot of perpendicular from $I$ to $AD$. Prove that $ID\cdot AM=IE\cdot AF$.

2012 India PRMO, 14

Tags: geometry , incenter , Angle Chasing

$O$ and $I$ are the circumcentre and incentre of $\vartriangle ABC$ respectively. Suppose $O$ lies in the interior of $\vartriangle ABC$ and $I$ lies on the circle passing through $B, O$, and $C$. What is the magnitude of $\angle B AC$ in degrees?

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

2024 Bangladesh Mathematical Olympiad, P5

Tags: geometry , incenter , parallelogram , Angle Chasing

Consider $\triangle XPQ$ and $\triangle YPQ$ such that $X$ and $Y$ are on the opposite sides of $PQ$ and the circumradius of $\triangle XPQ$ and the circumradius of $\triangle YPQ$ are the same. $I$ and $J$ are the incenters of $\triangle XPQ$ and $\triangle YPQ$ respectively. Let $M$ be the midpoint of $PQ$. Suppose $I, M, J$ are collinear. Prove that $XPYQ$ is a parallelogram.

2023 Romanian Master of Mathematics Shortlist, G1

Tags: trigonometry , Angle Chasing , concurrency , geometry , RMM Shortlist , circumcircle

Let $ABC$ be a triangle with incentre $I$ and circumcircle $\omega$. The incircle of the triangle $ABC$ touches the sides $BC$, $CA$ and $AB$ at $D$, $E$ and $F$, respectively. The circumcircle of triangle $ADI$ crosses $\omega$ again at $P$, and the lines $PE$ and $PF$ cross $\omega$ again at $X$and $Y$, respectively. Prove that the lines $AI$, $BX$ and $CY$ are concurrent.

2022 Belarusian National Olympiad, 10.6

Tags: geometry , Angle Chasing

Circles $\omega_1$ and $\omega_2$ intersect at $X$ and $Y$. Through point $Y$ two lines pass, one of which intersects $\omega_1$ and $\omega_2$ for the second time at $A$ and $B$, and the other at $C$ and $D$. Line $AD$ intersects for the second time circles $\omega_1$ and $\omega_2$ at $P$ and $Q$. It turned out that $YP=YQ$ Prove that the circumcircles of triangles $BCY$ and $PQY$ are tangent to each other.

2004 Greece JBMO TST, 1

Tags: geometry , Angle Chasing

Let $ABCD$ be a convex quadrilateral with $\angle A=60^o$. Let $E$ and $Z$ be the symmetric points of $A$ wrt $BC$ and $CD$ respectively. If the points $B,D,E$ and $Z$ are collinear, then calculate the angle $\angle BCD$.

2024 Middle European Mathematical Olympiad, 3

Tags: geometry , Fixed point , circle , circumcircle , Angle Chasing

Let $ABC$ be an acute scalene triangle. Choose a circle $\omega$ passing through $B$ and $C$ which intersects the segments $AB$ and $AC$ at the interior points $D$ and $E$, respectively. The lines $BE$ and $CD$ intersects at $F$. Let $G$ be a point on the circumcircle of $ABF$ such that $GB$ is tangent to $\omega$ and let $H$ be a point on the circumcircle of $ACF$ such that $HC$ is tangent to $\omega$. Prove that there exists a point $T\neq A$, independent of the choice of $\omega$, such that the circumcircle of triangle $AGH$ passes through $T$.

2013 IMO Shortlist, G4

Tags: geometry , circumcircle , IMO Shortlist , geometry solved , Isosceles Triangle , Angle Chasing , Inversion

Let $ABC$ be a triangle with $\angle B > \angle C$. Let $P$ and $Q$ be two different points on line $AC$ such that $\angle PBA = \angle QBA = \angle ACB $ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $BQ$ for which $PD=PB$. Let the ray $AD$ intersect the circle $ABC$ at $R \neq A$. Prove that $QB = QR$.

1997 Bosnia and Herzegovina Team Selection Test, 2

Tags: geometry , circumcircle , incenter , Angle Chasing , cyclic quadrilateral

In isosceles triangle $ABC$ with base side $AB$, on side $BC$ it is given point $M$. Let $O$ be a circumcenter and $S$ incenter of triangle $ABC$. Prove that $$ SM \mid \mid AC \Leftrightarrow OM \perp BS$$

2002 Korea - Final Round, 2

Tags: geometry unsolved , geometry , Angle Chasing

Let $ABC$ be an acute triangle and let $\omega$ be its circumcircle. Let the perpendicular line from $A$ to $BC$ meet $\omega$ at $D$. Let $P$ be a point on $\omega$, and let $Q$ be the foot of the perpendicular line from $P$ to the line $AB$. Prove that if $Q$ is on the outside of $\omega$ and $2\angle QPB = \angle PBC$, then $D,P,Q$ are collinear.

2010 Sharygin Geometry Olympiad, 5

Tags: geometry , incircle , Angle Chasing , excircle

The incircle of a right-angled triangle $ABC$ ($\angle ABC =90^o$) touches $AB, BC, AC$ in points $C_1, A_1, B_1$, respectively. One of the excircles touches the side $BC$ in point $A_2$. Point $A_0$ is the circumcenter or triangle $A_1A_2B_1$, point $C_0$ is defined similarly. Find angle $A_0BC_0$.

2019 Silk Road, 1

Tags: geometry , Angle Chasing , equal angles , altitudes

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 $.

2018 Bosnia and Herzegovina Team Selection Test, 6

Tags: geometry , IMO Shortlist , Angle Chasing , projective geometry , similar triangles , ISL 2017 , G3

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 $ABC$.

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 $CK\parallel AN \parallel TS$.

2018 Yasinsky Geometry Olympiad, 1

Tags: geometry , Angle Chasing , midpoint , equal angles , altitude

In the triangle $ABC$, $AD$ is altitude, $M$ is the midpoint of $BC$. It is known that $\angle BAD = \angle DAM = \angle MAC$. Find the values of the angles of the triangle $ABC$

2013 India Regional Mathematical Olympiad, 5

Tags: geometry , Angle Chasing , similar triangles

Let $ABC$ be a triangle with $\angle A=90^{\circ}$ and $AB=AC$. Let $D$ and $E$ be points on the segment $BC$ such that $BD:DE:EC = 1:2:\sqrt{3}$. Prove that $\angle DAE= 45^{\circ}$

2014 Flanders Math Olympiad, 3

Tags: geometry , equal segments , angles , Angle Chasing

Let $PQRS$ be a quadrilateral with $| P Q | = | QR | = | RS |$, $\angle Q= 110^o$ and $\angle R = 130^o$ . Determine $\angle P$ and $\angle S$ .

2015 ITAMO, 3

Tags: geometry , Angle Chasing , angle bisector

Let ABC a triangle, let K be the foot of the bisector relative to BC and J be the foot of the trisectrix relative to BC closer to the side AC (3* m(JAC)=m(CAB) ). Let C' and B' be two point on the line AJ on the side of J with respect to A, such that AC'=AC and AB=AB'. Prove that ABB'C is cyclic if and only if lines C'K and BB' are parallel.

2010 Middle European Mathematical Olympiad, 3

Tags: geometry , circumcircle , cyclic quadrilateral , radical axis , geometry proposed , Angle Chasing

We are given a cyclic quadrilateral $ABCD$ with a point $E$ on the diagonal $AC$ such that $AD=AE$ and $CB=CE$. Let $M$ be the center of the circumcircle $k$ of the triangle $BDE$. The circle $k$ intersects the line $AC$ in the points $E$ and $F$. Prove that the lines $FM$, $AD$ and $BC$ meet at one point. [i](4th Middle European Mathematical Olympiad, Individual Competition, Problem 3)[/i]

2022 Taiwan TST Round 3, 1

Tags: ISL 2021 , IMO Shortlist , geometry , cyclic quadrilateral , concurrency , Angle Chasing , imo shortlist g4

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.

2023 pOMA, 6

Tags: geometry , Angle Chasing

Let $\Omega$ be a circle, and let $A$, $B$, $C$, $D$ and $K$ be distinct points on it, in that order, and such that lines $BC$ and $AD$ are parallel. Let $A'\neq A$ be a point on line $AK$ such that $BA=BA'$. Similarly, let $C'\neq C$ be a point on line $CK$ such that $DC=DC'$. Prove that segments $AC$ and $A'C'$ have the same length.