Olympiad problems

1991 Chile National Olympiad, 6

Given a triangle with $ \triangle ABC $, with: $ \angle C = 36^o$ and $ \angle A = \angle B $. Consider the points $ D $ on $ BC $, $ E $ on $ AD $, $ F $ on $ BE $, $ G $ on $ DF $ and $ H $ on $ EG $, so that the rays $ AD, BE, DF, EG, FH $ bisect the angles $ A, B, D, E, F $ respectively. It is known that $ FH = 1 $. Calculate $ AC$.

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 $\angle BAP = \angle CBP$.

Brazil L2 Finals (OBM) - geometry, 2005.6

Tags: angle bisector , angle chasing , angle , geometry

The angle $B$ of a triangle $ABC$ is $120^o$. Let $M$ be a point on the side $AC$ and $K$ a point on the extension of the side $AB$, such that $BM$ is the internal bisector of the angle $\angle ABC$ and $CK$ is the external bisector corresponding to the angle $\angle ACB$ . The segment $MK$ intersects $BC$ at point $P$. Prove that $\angle APM = 30^o$.

2014 India IMO Training Camp, 3

Tags: circumcircle , isosceles triangle , angle chasing , inversion , geometry

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 Bulgaria JBMO TST, 2

Tags: geometry problem , angle chasing , cyclic quadrilateral , geometry , circumcircle

Let $ABC$ ($AB < AC$) be a triangle with circumcircle $k$. The tangent to $k$ at $A$ intersects the line $BC$ at $D$ and the point $E\neq A$ on $k$ is such that $DE$ is tangent to $k$. The point $X$ on line $BE$ is such that $B$ is between $E$ and $X$ and $DX = DA$ and the point $Y$ on the line $CX$ is such that $Y$ is between $C$ and $X$ and $DY = DA$. Prove that the lines $BC$ and $YE$ are perpendicular.

2018 India PRMO, 29

Tags: geometry , incenter , angle chasing

Let $D$ be an interior point of the side $BC$ of a triangle $ABC$. Let $I_1$ and $I_2$ be the incentres of triangles $ABD$ and $ACD$ respectively. Let $AI_1$ and $AI_2$ meet $BC$ in $E$ and $F$ respectively. If $\angle BI_1E = 60^o$, what is the measure of $\angle CI_2F$ in degrees?

2019 Korea Junior Math Olympiad., 7

Tags: plane geometry , geometry , 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 $\frac{\overline{AY}}{\overline{BY}}= \frac{\overline{EX}}{\overline{EO}}$.

2017 Polish MO Finals, 1

Tags: circumcircle , miquel point , spiral similarity , angle chasing , cyclic quadrilateral , geometry

Points $P$ and $Q$ lie respectively on sides $AB$ and $AC$ of a triangle $ABC$ and $BP=CQ$. Segments $BQ$ and $CP$ cross at $R$. Circumscribed circles of triangles $BPR$ and $CQR$ cross again at point $S$ different from $R$. Prove that point $S$ lies on the bisector of angle $BAC$.

2019 Yasinsky Geometry Olympiad, p5

Tags: geometry , angle chasing , angle

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 $\angle BED$.

India EGMO 2022 TST, 5

Tags: angle chasing , incenter , excenter , inversion , geometry , tangent circle

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]

2022 Germany Team Selection Test, 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.

1997 Canada National Olympiad, 4

Tags: parallelogram , angle chasing , geometry

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

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 $\angle NML = 90^o$.

2020 Balkan MO Shortlist, G2

Tags: right angle , euler line , angle chasing , geometry

Let $G, H$ be the centroid and orthocentre of $\vartriangle ABC$ which has an obtuse angle at $\angle B$. Let $\omega$ be the circle with diameter $AG$. $\omega$ intersects $\odot(ABC)$ again at $L \ne A$. The tangent to $\omega$ at $L$ intersects $\odot(ABC)$ at $K \ne L$. Given that $AG = GH$, prove $\angle HKG = 90^o$ . [i]Sam Bealing, United Kingdom[/i]

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 $ABC$.

2018 Dutch BxMO TST, 4

Tags: geometry , circumcircle , angle chasing , angle

In a non-isosceles triangle $\vartriangle ABC$ we have $\angle BAC = 60^o$. Let $D$ be the intersection of the angular bisector of $\angle BAC$ with side $BC, O$ the centre of the circumcircle of $\vartriangle ABC$ and $E$ the intersection of $AO$ and $BC$. Prove that $\angle AED + \angle ADO = 90^o$.

2019 Yasinsky Geometry Olympiad, p3

Tags: geometry , cyclic quadrilateral , angle chasing , angle

In the quadrilateral $ABCD$, the angles $B$ and $D$ are right . The diagonal $AC$ forms with the side $AB$ the angle of $40^o$, as well with side $AD$ an angle of $30^o$. Find the acute angle between the diagonals $AC$ and $BD$.

2019 Brazil Team Selection Test, 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.

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 $\angle DEP = \angle DFP$.

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.

Kharkiv City MO Seniors - geometry, 2014.10.4

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

Let $ABCD$ be a square. The points $N$ and $P$ are chosen on the sides $AB$ and $AD$ respectively, such that $NP=NC$. The point $Q$ on the segment $AN$ is such that that $\angle QPN=\angle NCB$. Prove that $\angle BCQ=\frac{1}{2}\angle AQP$.

2000 Saint Petersburg Mathematical Olympiad, 9.2

Tags: geometry , angle chasing

Let $AA_1$ and $CC_1$ be altitudes of acute angled triangle $ABC$. A point $D$ is chosen on $AA_1$ such that $A_1D=C_1D$. Let $E$ be the midpoint of $AC$. Prove that points $A$, $C_1$, $D$, $E$ are concylic. [I]Proposed by S. Berlov[/i]

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.

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

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 $\angle DEP = \angle DFP$.