Olympiad problems

2014 Sharygin Geometry Olympiad, 3

Points $M$ and $N$ are the midpoints of sides $AC$ and $BC$ of a triangle $ABC$. It is known that $\angle MAN = 15^o$ and $\angle BAN = 45^o$. Find the value of angle $\angle ABM$. (A. Blinkov)

2011 Dutch IMO TST, 3

Tags: circles , angle chasing , equal angles , geometry

The circles $\Gamma_1$ and $\Gamma_2$ intersect at $D$ and $P$. The common tangent line of the two circles closest to point $D$ touches $\Gamma_1$ in A and $\Gamma_2$ in $B$. The line $AD$ intersects $\Gamma_2$ for the second time in $C$. Let $M$ be the midpoint of line segment $BC$. Prove that $\angle DPM = \angle BDC$.

2023 pOMA, 2

Tags: cyclic quadrilateral , circumcircle , area of a triangle , angle chasing , geometry

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.

2004 IMO, 1

Tags: incenter , angle chasing , angle bisector , 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 $BC$.

Brazil L2 Finals (OBM) - geometry, 2012.4

Tags: pentagon , angle chasing , region , equilateral , angle

The figure below shows a regular $ABCDE$ pentagon inscribed in an equilateral triangle $MNP$ . Determine the measure of the angle $CMD$. [img]http://4.bp.blogspot.com/-LLT7hB7QwiA/Xp9fXOsihLI/AAAAAAAAL14/5lPsjXeKfYwIr5DyRAKRy0TbrX_zx1xHQCK4BGAYYCw/s200/2012%2Bobm%2Bl2.png[/img]

2004 IMO Shortlist, 1

Tags: incenter , angle chasing , angle bisector , 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 $BC$.

2021 Macedonian Mathematical Olympiad, Problem 3

Tags: circles , geometry , trapezoid , parallel lines , angle chasing , tangent , cyclic quadrilateral

Let $ABCD$ be a trapezoid with $AD \parallel BC$ and $\angle BCD < \angle ABC < 90^\circ$. Let $E$ be the intersection point of the diagonals $AC$ and $BD$. The circumcircle $\omega$ of $\triangle BEC$ intersects the segment $CD$ at $X$. The lines $AX$ and $BC$ intersect at $Y$, while the lines $BX$ and $AD$ intersect at $Z$. Prove that the line $EZ$ is tangent to $\omega$ iff the line $BE$ is tangent to the circumcircle of $\triangle BXY$.

2018 Yasinsky Geometry Olympiad, 6

Tags: angle chasing , altitude , projection , equal angles , geometry

$AH$ is the altitude of the acute triangle $ABC$, $K$ and $L$ are the feet of the perpendiculars, from point $H$ on sides $AB$ and $AC$ respectively. Prove that the angles $BKC$ and $BLC$ are equal.

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 $ABC$.

2021 Argentina National Olympiad, 3

Tags: right triangle , angle chasing , geometry

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 $\angle MAN.$

2022 USA TSTST, 7

Tags: parallelogram , angle chasing , trapezoid , geometry

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]

Kyiv City MO Juniors Round2 2010+ geometry, 2013.7.3

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

In the square $ABCD$ on the sides $AD$ and $DC$, the points $M$ and $N$ are selected so that $\angle BMA = \angle NMD = 60 { } ^ \circ $. Find the value of the angle $MBN$.

2024 Middle European Mathematical Olympiad, 3

Tags: circumcircle , geometry , fixed point , angle chasing , circles

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 $T$.

2022 Brazil Team Selection Test, 2

Tags: concurrency , angle chasing , geometry , cyclic quadrilateral

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.

2004 IMO Shortlist, 4

Tags: geometry , isogonal conjugate , angle chasing , easier p5

In a convex quadrilateral $ABCD$, the diagonal $BD$ bisects neither the angle $ABC$ nor the angle $CDA$. The point $P$ lies inside $ABCD$ and satisfies \[\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.\] Prove that $ABCD$ is a cyclic quadrilateral if and only if $AP=CP$.

2015 ITAMO, 3

Tags: angle bisector , angle chasing , geometry

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.

2017 Iran MO (3rd round), 2

Tags: angle chasing , trapezoid , geometry

Let $ABCD$ be a trapezoid ($AB<CD,AB\parallel CD$) and $P\equiv AD\cap BC$. Suppose that $Q$ be a point inside $ABCD$ such that $\angle QAB=\angle QDC=90-\angle BQC$. Prove that $\angle PQA=2\angle QCD$.

2012 Sharygin Geometry Olympiad, 1

Tags: angle chasing , geometry

Let $M$ be the midpoint of the base $AC$ of an acute-angled isosceles triangle $ABC$. Let $N$ be the reflection of $M$ in $BC$. The line parallel to $AC$ and passing through $N$ meets $AB$ at point $K$. Determine the value of $\angle AKC$. (A.Blinkov)

2018 Yasinsky Geometry Olympiad, 1

Tags: angle chasing , equilateral triangle , square , geometry

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 $BF$.

2011 Dutch IMO TST, 3

Tags: angle chasing , circles , equal angles , geometry

The circles $\Gamma_1$ and $\Gamma_2$ intersect at $D$ and $P$. The common tangent line of the two circles closest to point $D$ touches $\Gamma_1$ in A and $\Gamma_2$ in $B$. The line $AD$ intersects $\Gamma_2$ for the second time in $C$. Let $M$ be the midpoint of line segment $BC$. Prove that $\angle DPM = \angle BDC$.

2012 Sharygin Geometry Olympiad, 4

Tags: angle chasing , geometry , isosceles

Let $ABC$ be an isosceles triangle with $\angle B = 120^o$ . Points $P$ and $Q$ are chosen on the prolongations of segments $AB$ and $CB$ beyond point $B$ so that the rays $AQ$ and $CP$ intersect and are perpendicular to each other. Prove that $\angle PQB = 2\angle PCQ$. (A.Akopyan, D.Shvetsov)

2019 Germany Team Selection Test, 2

Tags: symmetry , angle chasing , miquel point , geometry

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.

Russian TST 2018, P2

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

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 $ABC$.

2019 Taiwan TST Round 1, 1

Tags: symmetry , angle chasing , miquel point , geometry

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.

2019 Yasinsky Geometry Olympiad, p5

Tags: angle , angle chasing , geometry

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 $\angle BED$.