Olympiad problems

2024 Oral Moscow Geometry Olympiad, 5

An acute-angled unequal triangle $ABC$ is drawn with its circumcircle and circumcenter $O$. The incenter $I$ is also marked. Using only a ruler (without divisions), construct the symedian (a line symmetrical to the median relative to the corresponding bisector) of the triangle, drawing no more than four lines.

2018 Junior Balkan MO, 4

Tags: geometry , circumcircle

Let $\triangle ABC$ and $A'$,$B'$,$C'$ the symmetrics of vertex over opposite sides.The intersection of the circumcircles of $\triangle ABB'$ and $\triangle ACC'$ is $A_1$.$B_1$ and $C_1$ are defined similarly.Prove that lines $AA_1$,$BB_1$ and $CC_1$ are concurent.

2007 China National Olympiad, 1

Tags: circumcircle , incenter , trigonometry , power of a point , radical axis , geometry

Let $O, I$ be the circumcenter and incenter of triangle $ABC$. The incircle of $\triangle ABC$ touches $BC, CA, AB$ at points $D, E, F$ repsectively. $FD$ meets $CA$ at $P$, $ED$ meets $AB$ at $Q$. $M$ and $N$ are midpoints of $PE$ and $QF$ respectively. Show that $OI \perp MN$.

2014 Argentina Cono Sur TST, 5

Tags: circumcircle , angle bisector , geometry

In an acute triangle $ABC$, let $D$ be a point in $BC$ such that $AD$ is the angle bisector of $\angle{BAC}$. Let $E \neq B$ be the point of intersection of the circumcircle of triangle $ABD$ with the line perpendicular to $AD$ drawn through $B$. Let $O$ be the circumcenter of triangle $ABC$. Prove that $E$, $O$, and $A$ are collinear.

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 $QB = QR$.

2023 Turkey EGMO TST, 6

Tags: circumcircle , geometry

Let $ABC$ be a scalene triangle and $l_0$ be a line that is tangent to the circumcircle of $ABC$ at point $A$. Let $l$ be a variable line which is parallel to line $l_0$. Let $l$ intersect segment $AB$ and $AC$ at the point $X$, $Y$ respectively. $BY$ and $CX$ intersects at point $T$ and the line $AT$ intersects the circumcirle of $ABC$ at $Z$. Prove that as $l$ varies, circumcircle of $XYZ$ passes through a fixed point.

2022 Sharygin Geometry Olympiad, 8.1

Tags: geometry , geometric transformation , reflection , circumcircle

Let $ABCD$ be a convex quadrilateral with $\angle{BAD} = 2\angle{BCD}$ and $AB = AD$. Let $P$ be a point such that $ABCP$ is a parallelogram. Prove that $CP = DP$.

2012 China Team Selection Test, 1

Tags: circumcircle , inversion , geometry

Given two circles ${\omega _1},{\omega _2}$, $S$ denotes all $\Delta ABC$ satisfies that ${\omega _1}$ is the circumcircle of $\Delta ABC$, ${\omega _2}$ is the $A$- excircle of $\Delta ABC$ , ${\omega _2}$ touches $BC,CA,AB$ at $D,E,F$. $S$ is not empty, prove that the centroid of $\Delta DEF$ is a fixed point.

2007 Germany Team Selection Test, 2

Tags: geometry , circumcircle , pentagon

Let $ ABCDE$ be a convex pentagon such that \[ \angle BAC \equal{} \angle CAD \equal{} \angle DAE\qquad \text{and}\qquad \angle ABC \equal{} \angle ACD \equal{} \angle ADE. \]The diagonals $BD$ and $CE$ meet at $P$. Prove that the line $AP$ bisects the side $CD$. [i]Proposed by Zuming Feng, USA[/i]

2007 Baltic Way, 14

Tags: circumcircle , geometry

In a convex quadrilateral $ABCD$ we have $ADC = 90^{\circ}$. Let $E$ and $F$ be the projections of $B$ onto the lines $AD$ and $AC$, respectively. Assume that $F$ lies between $A$ and $C$, that $A$ lies between $D$ and $E$, and that the line $EF$ passes through the midpoint of the segment $BD$. Prove that the quadrilateral $ABCD$ is cyclic.

2012 Brazil National Olympiad, 2

Tags: incenter , circumcircle , geometric transformation , homothety , ratio , geometry

$ABC$ is a non-isosceles triangle. $T_A$ is the tangency point of incircle of $ABC$ in the side $BC$ (define $T_B$,$T_C$ analogously). $I_A$ is the ex-center relative to the side BC (define $I_B$,$I_C$ analogously). $X_A$ is the mid-point of $I_BI_C$ (define $X_B$,$X_C$ analogously). Show that $X_AT_A$,$X_BT_B$,$X_CT_C$ meet in a common point, colinear with the incenter and circumcenter of $ABC$.

2006 All-Russian Olympiad, 6

Tags: circumcircle , geometry

Let $P$, $Q$, $R$ be points on the sides $AB$, $BC$, $CA$ of a triangle $ABC$ such that $AP=CQ$ and the quadrilateral $RPBQ$ is cyclic. The tangents to the circumcircle of triangle $ABC$ at the points $C$ and $A$ intersect the lines $RQ$ and $RP$ at the points $X$ and $Y$, respectively. Prove that $RX=RY$.

2014 ELMO Shortlist, 8

Tags: incenter , circumcircle , geometric transformation , homothety , inversion , geometry

In triangle $ABC$ with incenter $I$ and circumcenter $O$, let $A',B',C'$ be the points of tangency of its circumcircle with its $A,B,C$-mixtilinear circles, respectively. Let $\omega_A$ be the circle through $A'$ that is tangent to $AI$ at $I$, and define $\omega_B, \omega_C$ similarly. Prove that $\omega_A,\omega_B,\omega_C$ have a common point $X$ other than $I$, and that $\angle AXO = \angle OXA'$. [i]Proposed by Sammy Luo[/i]

2013 Online Math Open Problems, 38

Tags: geometry , circumcircle , geometric transformation , rotation , parallelogram

Triangle $ABC$ has sides $AB = 25$, $BC = 30$, and $CA=20$. Let $P,Q$ be the points on segments $AB,AC$, respectively, such that $AP=5$ and $AQ=4$. Suppose lines $BQ$ and $CP$ intersect at $R$ and the circumcircles of $\triangle{BPR}$ and $\triangle{CQR}$ intersect at a second point $S\ne R$. If the length of segment $SA$ can be expressed in the form $\frac{m}{\sqrt{n}}$ for positive integers $m,n$, where $n$ is not divisible by the square of any prime, find $m+n$. [i]Victor Wang[/i]

2016 Brazil National Olympiad, 6

Tags: circumcircle , angle bisector , geometry

Lei it $ABCD$ be a non-cyclical, convex quadrilateral, with no parallel sides. The lines $AB$ and $CD$ meet in $E$. Let it $M \not= E$ be the intersection of circumcircles of $ADE$ and $BCE$. The internal angle bisectors of $ABCD$ form an convex, cyclical quadrilateral with circumcenter $I$. The external angle bisectors of $ABCD$ form an convex, cyclical quadrilateral with circumcenter $J$. Show that $I,J,M$ are colinear.

2008 Kazakhstan National Olympiad, 2

Tags: circumcircle , inradius , trigonometry , incenter , geometry

Let $ \triangle ABC$ be a triangle and let $ K$ be some point on the side $ AB$, so that the tangent line from $ K$ to the incircle of $ \triangle ABC$ intersects the ray $ AC$ at $ L$. Assume that $ \omega$ is tangent to sides $ AB$ and $ AC$, and to the circumcircle of $ \triangle AKL$. Prove that $ \omega$ is tangent to the circumcircle of $ \triangle ABC$ as well.

2013 India IMO Training Camp, 2

Tags: geometry , geometric transformation , reflection , circumcircle , trigonometry , parallelogram , homothety

In a triangle $ABC$, with $\widehat{A} > 90^\circ$, let $O$ and $H$ denote its circumcenter and orthocenter, respectively. Let $K$ be the reflection of $H$ with respect to $A$. Prove that $K, O$ and $C$ are collinear if and only if $\widehat{A} - \widehat{B} = 90^\circ$.

2017 Sharygin Geometry Olympiad, 2

Tags: circumcircle , perpendicular bisector , angle bisector , geometry

Let $H$ and $O$ be the orthocenter and circumcenter of an acute-angled triangle $ABC$, respectively. The perpendicular bisector of $BH$ meets $AB$ and $BC$ at points $A_1$ and $C_1$, respectively. Prove that $OB$ bisects the angle $A_1OC_1$.

2001 Bundeswettbewerb Mathematik, 3

Tags: circumcircle , parallelogram , trigonometry , geometry

Let $ ABC$ an acute triangle with circumcircle center $ O.$ The line $ (BO)$ intersects the circumcircle again in $ D,$ and the extension of the altitude from $ A$ intersects the circle in $ E.$ Prove that the quadrilateral $ BECD$ and the triangle $ ABC$ have the same area.

2019 Iran MO (3rd Round), 1

Tags: geometry , incenter , circumcircle , geometric transformation , reflection , law of sines , menelaus

Consider a triangle $ABC$ with incenter $I$. Let $D$ be the intersection of $BI,AC$ and $CI$ intersects the circumcircle of $ABC$ at $M$. Point $K$ lies on the line $MD$ and $\angle KIA=90^\circ$. Let $F$ be the reflection of $B$ about $C$. Prove that $BIKF$ is cyclic.

2011 Costa Rica - Final Round, 1

Tags: geometry , incenter , geometric transformation , reflection , circumcircle , dilation , symmetry

Let $ABC$ be a triangle with orthocenter $H$. Let $P,Q,R$ be the reflections of $H$ with respect to sides $BC,AC,AB$, respectively. Show that $H$ is incenter of $PQR$.

2010 JBMO Shortlist, 1

Tags: geometry , circumcircle , reflection

$\textbf{Problem G1}$ Consider a triangle $ABC$ with $\angle ACB=90^{\circ}$. Let $F$ be the foot of the altitude from $C$. Circle $\omega$ touches the line segment $FB$ at point $P$, the altitude $CF$ at point $Q$ and the circumcircle of $ABC$ at point $R$. Prove that points $A, Q, R$ are collinear and $AP = AC$.

2010 APMO, 1

Tags: circumcircle , perpendicular bisector , geometry

Let $ABC$ be a triangle with $\angle BAC \neq 90^{\circ}.$ Let $O$ be the circumcenter of the triangle $ABC$ and $\Gamma$ be the circumcircle of the triangle $BOC.$ Suppose that $\Gamma$ intersects the line segment $AB$ at $P$ different from $B$, and the line segment $AC$ at $Q$ different from $C.$ Let $ON$ be the diameter of the circle $\Gamma.$ Prove that the quadrilateral $APNQ$ is a parallelogram.

2014 ITAMO, 2

Tags: symmetry , circumcircle , geometry

Let $ABC$ be a triangle. Let $H$ be the foot of the altitude from $C$ on $AB$. Suppose that $AH = 3HB$. Suppose in addition we are given that (a) $M$ is the midpoint of $AB$; (b) $N$ is the midpoint of $AC$; (c) $P$ is a point on the opposite side of $B$ with respect to the line $AC$ such that $NP = NC$ and $PC = CB$. Prove that $\angle APM = \angle PBA$.

2009 Iran MO (2nd Round), 3

Tags: circumcircle , trigonometry , angle bisector , geometric inequalities , geometry , inequalities

Let $ ABC $ be a triangle and the point $ D $ is on the segment $ BC $ such that $ AD $ is the interior bisector of $ \angle A $. We stretch $ AD $ such that it meets the circumcircle of $ \Delta ABC $ at $ M $. We draw a line from $ D $ such that it meets the lines $ MB,MC $ at $ P,Q $, respectively ($ M $ is not between $ B,P $ and also is not between $ C,Q $). Prove that $ \angle PAQ\geq\angle BAC $.