Olympiad problems

2017 239 Open Mathematical Olympiad, 2

Inside the circle $\omega$ through points $A, B$ point $C$ is chosen. An arbitrary point $X$ is selected on the segment $BC$. The ray $AX$ cuts the circle in $Y$. Prove that all circles $CXY$ pass through a two fixed points that is they intersect and are coaxial, independent of the position of $X$.

2015 İberoAmerican, 2

Tags: geometry , angle bisector , inversion

A line $r$ contains the points $A$, $B$, $C$, $D$ in that order. Let $P$ be a point not in $r$ such that $\angle{APB} = \angle{CPD}$. Prove that the angle bisector of $\angle{APD}$ intersects the line $r$ at a point $G$ such that: $\frac{1}{GA} + \frac{1}{GC} = \frac{1}{GB} + \frac{1}{GD}$

2008 Peru Iberoamerican Team Selection Test, P2

Tags: geometry , circumcircle , homothety , trigonometry , quadrilateral , inversion

Let $ ABCD$ be a convex quadrilateral and let $ P$ and $ Q$ be points in $ ABCD$ such that $ PQDA$ and $ QPBC$ are cyclic quadrilaterals. Suppose that there exists a point $ E$ on the line segment $ PQ$ such that $ \angle PAE \equal{} \angle QDE$ and $ \angle PBE \equal{} \angle QCE$. Show that the quadrilateral $ ABCD$ is cyclic. [i]Proposed by John Cuya, Peru[/i]

2012 ELMO Shortlist, 7

Tags: circumcircle , trigonometry , inversion , geometry

Let $\triangle ABC$ be an acute triangle with circumcenter $O$ such that $AB<AC$, let $Q$ be the intersection of the external bisector of $\angle A$ with $BC$, and let $P$ be a point in the interior of $\triangle ABC$ such that $\triangle BPA$ is similar to $\triangle APC$. Show that $\angle QPA + \angle OQB = 90^{\circ}$. [i]Alex Zhu.[/i]

2017 India IMO Training Camp, 2

Tags: cyclic quadrilateral , geometry , incenter , reflection , inversion

Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.

2004 Iran MO (2nd round), 5

Tags: circumcircle , incenter , power of a point , inversion , geometry

The interior bisector of $\angle A$ from $\triangle ABC$ intersects the side $BC$ and the circumcircle of $\Delta ABC$ at $D,M$, respectively. Let $\omega$ be a circle with center $M$ and radius $MB$. A line passing through $D$, intersects $\omega$ at $X,Y$. Prove that $AD$ bisects $\angle XAY$.

2014 Contests, 2

Tags: incenter , circumcircle , reflection , homothety , inversion

Let $D$ and $E$ be points in the interiors of sides $AB$ and $AC$, respectively, of a triangle $ABC$, such that $DB = BC = CE$. Let the lines $CD$ and $BE$ meet at $F$. Prove that the incentre $I$ of triangle $ABC$, the orthocentre $H$ of triangle $DEF$ and the midpoint $M$ of the arc $BAC$ of the circumcircle of triangle $ABC$ 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 $QB = QR$.

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.

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]

2010 Contests, 2

Tags: length , parallelogram , homothety , inversion , geometry

Let $\Gamma_1$, $\Gamma_2$, $\Gamma_3$, $\Gamma_4$ be distinct circles such that $\Gamma_1$, $\Gamma_3$ are externally tangent at $P$, and $\Gamma_2$, $\Gamma_4$ are externally tangent at the same point $P$. Suppose that $\Gamma_1$ and $\Gamma_2$; $\Gamma_2$ and $\Gamma_3$; $\Gamma_3$ and $\Gamma_4$; $\Gamma_4$ and $\Gamma_1$ meet at $A$, $B$, $C$, $D$, respectively, and that all these points are different from $P$. Prove that \[ \frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}. \]

2015 Iberoamerican Math Olympiad, 2

Tags: geometry , angle bisector , inversion

A line $r$ contains the points $A$, $B$, $C$, $D$ in that order. Let $P$ be a point not in $r$ such that $\angle{APB} = \angle{CPD}$. Prove that the angle bisector of $\angle{APD}$ intersects the line $r$ at a point $G$ such that: $\frac{1}{GA} + \frac{1}{GC} = \frac{1}{GB} + \frac{1}{GD}$

2017 Taiwan TST Round 1, 2

Tags: cyclic quadrilateral , geometry , incenter , reflection , inversion

Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.

2018 Serbia Team Selection Test, 5

Tags: inversion , geometry

Let $H $ be the orthocenter of $ABC $ ,$AB\neq AC $ ,and let $F $ be a point on circumcircle of $ABC $ such that $\angle AFH=90^{\circ} $.$K $ is the symmetric point of $H $ wrt $B $.Let $P $ be a point such that $\angle PHB=\angle PBC=90^{\circ} $,and $Q $ is the foot of $B $ to $CP $.Prove that $HQ $ is tangent to tge circumcircle of $FHK $.

2013 IMO Shortlist, G4

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 $QB = QR$.

2017 USA Team Selection Test, 2

Tags: inversion , circumcircle , euler line , geometry

Let $ABC$ be an acute scalene triangle with circumcenter $O$, and let $T$ be on line $BC$ such that $\angle TAO = 90^{\circ}$. The circle with diameter $\overline{AT}$ intersects the circumcircle of $\triangle BOC$ at two points $A_1$ and $A_2$, where $OA_1 < OA_2$. Points $B_1$, $B_2$, $C_1$, $C_2$ are defined analogously. [list=a][*] Prove that $\overline{AA_1}$, $\overline{BB_1}$, $\overline{CC_1}$ are concurrent. [*] Prove that $\overline{AA_2}$, $\overline{BB_2}$, $\overline{CC_2}$ are concurrent on the Euler line of triangle $ABC$. [/list][i]Evan Chen[/i]

2019 Tuymaada Olympiad, 8

Tags: geometry , geometric transformation , reflection , arc midpoint , circumcircle , moving points , inversion

In $\triangle ABC$ $\angle B$ is obtuse and $AB \ne BC$. Let $O$ is the circumcenter and $\omega$ is the circumcircle of this triangle. $N$ is the midpoint of arc $ABC$. The circumcircle of $\triangle BON$ intersects $AC$ on points $X$ and $Y$. Let $BX \cap \omega = P \ne B$ and $BY \cap \omega = Q \ne B$. Prove that $P, Q$ and reflection of $N$ with respect to line $AC$ are collinear.

Istek Lyceum Math Olympiad 2016, 2

Tags: inversion , circles , geometry

Let $\omega$ be the semicircle with diameter $PQ$. A circle $k$ is tangent internally to $\omega$ and to segment $PQ$ at $C$. Let $AB$ be the tangent to $K$ perpendicular to $PQ$, with $A$ on $\omega$ and $B$ on the segment $CQ$. Show that $AC$ bisects angle $\angle PAB$

2015 Peru IMO TST, 8

Tags: inversion , circles , tangent , geometry

Let $I$ be the incenter of the $ABC$ triangle. The circumference that passes through $I$ and has center in $A$ intersects the circumscribed circumference of the $ABC$ triangle at points $M$ and $N$. Prove that the line $MN$ is tangent to the inscribed circle of the $ABC$ triangle.

2012 ELMO Shortlist, 7

Tags: circumcircle , trigonometry , inversion , geometry

Let $\triangle ABC$ be an acute triangle with circumcenter $O$ such that $AB<AC$, let $Q$ be the intersection of the external bisector of $\angle A$ with $BC$, and let $P$ be a point in the interior of $\triangle ABC$ such that $\triangle BPA$ is similar to $\triangle APC$. Show that $\angle QPA + \angle OQB = 90^{\circ}$. [i]Alex Zhu.[/i]

2016 IMO Shortlist, G4

Tags: cyclic quadrilateral , geometry , incenter , reflection , inversion

Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.

2010 ELMO Problems, 3

Tags: circumcircle , incenter , homothety , inversion , spiral similarity , geometry

Let $ABC$ be a triangle with circumcircle $\omega$, incenter $I$, and $A$-excenter $I_A$. Let the incircle and the $A$-excircle hit $BC$ at $D$ and $E$, respectively, and let $M$ be the midpoint of arc $BC$ without $A$. Consider the circle tangent to $BC$ at $D$ and arc $BAC$ at $T$. If $TI$ intersects $\omega$ again at $S$, prove that $SI_A$ and $ME$ meet on $\omega$. [i]Amol Aggarwal.[/i]

1971 Spain Mathematical Olympiad, 7

Tags: geometry , inversion , concentric circles , equal circles

Transform by inversion two concentric and coplanar circles into two equal.

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.

2016 Singapore MO Open, 1

Tags: inversion , similarity , geometry , triangle

Let $D$ be a point in the interior of $\triangle{ABC}$ such that $AB=ab$, $AC=ac$, $BC=bc$, $AD=ad$, $BD=bd$, $CD=cd$. Show that $\angle{ABD}+\angle{ACD}=60^{\circ}$. Source: 2016 Singapore Mathematical Olympiad (Open) Round 2, Problem 1