Olympiad problems

2020 Sharygin Geometry Olympiad, 18

Tags: geometry , Sharygin 2020 , geometry solved , Angle Chasing , tangent circles , circumcircle , incenter

Bisectors $AA_1$, $BB_1$, and $CC_1$ of triangle $ABC$ meet at point $I$. The perpendicular bisector to $BB_1$ meets $AA_1,CC_1$ at points $A_0,C_0$ respectively. Prove that the circumcircles of triangles $A_0IC_0$ and $ABC$ touch.

2016 Postal Coaching, 1

Tags: geometry , angle , Angle Chasing

Let $ABCD$ be a convex quadrilateral in which $$\angle BAC = 48^{\circ}, \angle CAD = 66^{\circ}, \angle CBD = \angle DBA.$$Prove that $\angle BDC = 24^{\circ}$.

1997 German National Olympiad, 3

Tags: Angle Chasing , angles , geometry

In a convex quadrilateral $ABCD$ we are given that $\angle CBD = 10^o$, $\angle CAD = 20^o$, $\angle ABD = 40^o$, $\angle BAC = 50^o$. Determine the angles $\angle BCD$ and $\angle ADC$.

2018 Pan-African Shortlist, G3

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 Yasinsky Geometry Olympiad, p5

Tags: geometry , angles , Angle Chasing

In the triangle $ABC$, $\angle ABC = \angle ACB = 78^o$. On the sides $AB$ and $AC$, respectively, the points $D$ and $E$ are chosen such that $\angle BCD = 24^o$, $\angle CBE = 51^o$. Find the measure of angle $\angle BED$.

2020 Macedonia Additional BMO TST, 2

Tags: geometry , Angle Chasing , parallel

Let $ABCD$ be a convex quadrilateral. On the sides $AB$ and $CD$ there are interior points $K$ and $L$, respectively, such that $\angle BAL = \angle CDK$. Prove that the following statements are equivalent: $i) \angle BLA= \angle CKD$ $ii) AD \parallel BC $

2018 Yasinsky Geometry Olympiad, 1

Tags: geometry , Angle Chasing , Equilateral Triangle , square

Points $A, B$ and $C$ lie on the same line so that $CA = AB$. Square $ABDE$ and the equilateral triangle $CFA$, are constructed on the same side of line $CB$. Find the acute angle between straight lines $CE$ and $BF$.

2013 Czech-Polish-Slovak Junior Match, 4

Tags: geometry , circumcircle , equal angles , Angle Chasing

Let $ABCD$ be a convex quadrilateral with $\angle DAB =\angle ABC =\angle BCD > 90^o$. The circle circumscribed around the triangle $ABC$ intersects the sides $AD$ and $CD$ at points $K$ and $L$, respectively, different from any vertex of the quadrilateral $ABCD$ . Segments $AL$ and $CK$ intersect at point $P$. Prove that $\angle ADB =\angle PDC$.

2022 USA TSTST, 7

Tags: geometry , USA TSTST , parallelogram , geometry solved , USA , trapezoid , Angle Chasing

Let $ABCD$ be a parallelogram. Point $E$ lies on segment $CD$ such that \[2\angle AEB=\angle ADB+\angle ACB,\] and point $F$ lies on segment $BC$ such that \[2\angle DFA=\angle DCA+\angle DBA.\] Let $K$ be the circumcenter of triangle $ABD$. Prove that $KE=KF$. [i]Merlijn Staps[/i]

2024 Bundeswettbewerb Mathematik, 3

Tags: geometry , circumcircle , geometry proposed , Angle Chasing , parallelogram

Let $ABCD$ be a parallelogram whose diagonals intersect in $M$. Suppose that the circumcircle of $ABM$ intersects the segment $AD$ in a point $E \ne A$ and that the circumcircle of $EMD$ intersects the segment $BE$ in a point $F \ne E$. Show that $\angle ACB=\angle DCF$.

2019 Korea Junior Math Olympiad., 7

Tags: Plane Geometry , geometry , KJMO , circumcircle , Angle Chasing

Let $O$ be the circumcenter of an acute triangle $ABC$. Let $D$ be the intersection of the bisector of the angle $A$ and $BC$. Suppose that $\angle ODC = 2 \angle DAO$. The circumcircle of $ABD$ meets the line segment $OA$ and the line $OD$ at $E (\neq A,O)$, and $F(\neq D)$, respectively. Let $X$ be the intersection of the line $DE$ and the line segment $AC$. Let $Y$ be the intersection of the bisector of the angle $BAF$ and the segment $BE$. Prove that $\frac{\overline{AY}}{\overline{BY}}= \frac{\overline{EX}}{\overline{EO}}$.

2023 Sharygin Geometry Olympiad, 9.8

Tags: geometry , Sharygin Geometry Olympiad , Sharygin 2023 , Angle Chasing

Let $ABC$ be a triangle with $\angle A = 120^\circ$, $I$ be the incenter, and $M$ be the midpoint of $BC$. The line passing through $M$ and parallel to $AI$ meets the circle with diameter $BC$ at points $E$ and $F$ ($A$ and $E$ lie on the same semiplane with respect to $BC$). The line passing through $E$ and perpendicular to $FI$ meets $AB$ and $AC$ at points $P$ and $Q$ respectively. Find the value of $\angle PIQ$.

2019 Thailand TST, 1

Tags: IMO Shortlist , geometry , geometry solved , symmetry , Angle Chasing , Miquel point

Let $ABC$ be a triangle with $AB=AC$, and let $M$ be the midpoint of $BC$. Let $P$ be a point such that $PB<PC$ and $PA$ is parallel to $BC$. Let $X$ and $Y$ be points on the lines $PB$ and $PC$, respectively, so that $B$ lies on the segment $PX$, $C$ lies on the segment $PY$, and $\angle PXM=\angle PYM$. Prove that the quadrilateral $APXY$ is cyclic.

2016 Sharygin Geometry Olympiad, 1

Tags: geometry , Angle Chasing , parallelogram , circumcircle

The diagonals of a parallelogram $ABCD$ meet at point $O$. The tangent to the circumcircle of triangle $BOC$ at $O$ meets ray $CB$ at point $F$. The circumcircle of triangle $FOD$ meets $BC$ for the second time at point $G$. Prove that $AG=AB$.

2014 Taiwan TST Round 3, 4

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 Canada National Olympiad, 4

Tags: geometry , parallelogram , geometry solved , Angle Chasing , Canada , CMO

The point $O$ is situated inside the parallelogram $ABCD$ such that $\angle AOB+\angle COD=180^{\circ}$. Prove that $\angle OBC=\angle ODC$.

2021 Argentina National Olympiad, 3

Tags: geometry , right triangle , Angle Chasing , contests

Let $ABC$ be an isosceles right triangle at $A$ with $AB=AC$. Let $M$ and $N$ be on side $BC$, with $M$ between $B$ and $N,$ such that $$BM^2+ NC^2= MN^2.$$ Determine the measure of the angle $\angle MAN.$

India EGMO 2022 TST, 5

Tags: geometry , geometry solved , tangent circles , Angle Chasing , incenter , excenter , Inversion

Let $I$ and $I_A$ denote the incentre and excentre opposite to $A$ of scalene $\triangle ABC$ respectively. Let $A'$ be the antipode of $A$ in $\odot (ABC)$ and $L$ be the midpoint of arc $(BAC)$. Let $LB$ and $LC$ intersect $AI$ at points $Y$ and $Z$ respectively. Prove that $\odot (LYZ)$ is tangent to $\odot (A'II_A)$. [i]~Mahavir Gandhi[/i]

2023 Bulgaria EGMO TST, 1

Tags: geometry , symmedian , Angle Chasing , similar triangles , circumcircle

Let $ABC$ be a triangle with circumcircle $k$. The tangents at $A$ and $C$ intersect at $T$. The circumcircle of triangle $ABT$ intersects the line $CT$ at $X$ and $Y$ is the midpoint of $CX$. Prove that the lines $AX$ and $BY$ intersect on $k$.

2021 Science ON all problems, 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]

2014 India IMO Training Camp, 3

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 $QB = QR$.

2022 Abelkonkurransen Finale, 2a

Tags: geometry , Angle Chasing , harmonic division

A triangle $ABC$ with circumcircle $\omega$ satisfies $|AB| > |AC|$. Points $X$ and $Y$ on $\omega$ are different from $A$, such that the line $AX$ passes through the midpoint of $BC$, $AY$ is perpendicular to $BC$, and $XY$ is parallel to $BC$. Find $\angle BAC$.

2013 India Regional Mathematical Olympiad, 4

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 $\angle DAE= 45^{\circ}$

2019 Yasinsky Geometry Olympiad, p1

Tags: angles , geometry , inradius , Angle Chasing

It is known that in the triangle $ABC$ the distance from the intersection point of the angle bisector to each of the vertices of the triangle does not exceed the diameter of the circle inscribed in this triangle. Find the angles of the triangle $ABC$. (Grigory Filippovsky)

2018 Polish MO Finals, 1

Tags: geometry , Poland , geometry solved , radical axis , Angle Chasing

An acute triangle $ABC$ in which $AB<AC$ is given. The bisector of $\angle BAC$ crosses $BC$ at $D$. Point $M$ is the midpoint of $BC$. Prove that the line though centers of circles escribed on triangles $ABC$ and $ADM$ is parallel to $AD$.