Olympiad problems

2019 Romanian Masters In Mathematics, 2

Let $ABCD$ be an isosceles trapezoid with $AB\parallel CD$. Let $E$ be the midpoint of $AC$. Denote by $\omega$ and $\Omega$ the circumcircles of the triangles $ABE$ and $CDE$, respectively. Let $P$ be the crossing point of the tangent to $\omega$ at $A$ with the tangent to $\Omega$ at $D$. Prove that $PE$ is tangent to $\Omega$. [i]Jakob Jurij Snoj, Slovenia[/i]

2009 Germany Team Selection Test, 2

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]

2019 Romanian Master of Mathematics, 2

Tags: geometry , complex geometry , inversion

Let $ABCD$ be an isosceles trapezoid with $AB\parallel CD$. Let $E$ be the midpoint of $AC$. Denote by $\omega$ and $\Omega$ the circumcircles of the triangles $ABE$ and $CDE$, respectively. Let $P$ be the crossing point of the tangent to $\omega$ at $A$ with the tangent to $\Omega$ at $D$. Prove that $PE$ is tangent to $\Omega$. [i]Jakob Jurij Snoj, Slovenia[/i]

2010 IMO, 2

Tags: geometry , inversion , p2

Given a triangle $ABC$, with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$. Let $E$ be a point on the arc $BDC$, and $F$ a point on the segment $BC$, such that $\angle BAF=\angle CAE < \dfrac12\angle BAC$. If $G$ is the midpoint of $IF$, prove that the meeting point of the lines $EI$ and $DG$ lies on $\Gamma$. [i]Proposed by Tai Wai Ming and Wang Chongli, Hong Kong[/i]

2016 ELMO Problems, 6

Tags: circumcircle , angle bisector , incircle , inversion , tangent circle , geometry

Elmo is now learning olympiad geometry. In triangle $ABC$ with $AB\neq AC$, let its incircle be tangent to sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. The internal angle bisector of $\angle BAC$ intersects lines $DE$ and $DF$ at $X$ and $Y$, respectively. Let $S$ and $T$ be distinct points on side $BC$ such that $\angle XSY=\angle XTY=90^\circ$. Finally, let $\gamma$ be the circumcircle of $\triangle AST$. (a) Help Elmo show that $\gamma$ is tangent to the circumcircle of $\triangle ABC$. (b) Help Elmo show that $\gamma$ is tangent to the incircle of $\triangle ABC$. [i]James Lin[/i]

2009 USAMO, 5

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

Trapezoid $ ABCD$, with $ \overline{AB}\parallel{}\overline{CD}$, is inscribed in circle $ \omega$ and point $ G$ lies inside triangle $ BCD$. Rays $ AG$ and $ BG$ meet $ \omega$ again at points $ P$ and $ Q$, respectively. Let the line through $ G$ parallel to $ \overline{AB}$ intersects $ \overline{BD}$ and $ \overline{BC}$ at points $ R$ and $ S$, respectively. Prove that quadrilateral $ PQRS$ is cyclic if and only if $ \overline{BG}$ bisects $ \angle CBD$.

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 $QB = QR$.

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]

2017 Brazil Team Selection Test, 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 $BDE$.

2015 Romania Team Selection Tests, 1

Tags: tangent , circles , inversion , geometry

Two circles $\gamma $ and $\gamma'$ cross one another at points $A$ and $B$ . The tangent to $\gamma'$ at $A$ meets $\gamma$ again at $C$ , the tangent to $\gamma$ at $A$ meets $\gamma'$ again at $C'$ , and the line $CC'$ separates the points $A$ and $B$ . Let $\Gamma$ be the circle externally tangent to $\gamma$ , externally tangent to $\gamma'$ , tangent to the line $CC'$, and lying on the same side of $CC'$ as $B$ . Show that the circles $\gamma$ and $\gamma'$ intercept equal segments on one of the tangents to $\Gamma$ through $A$ .

1969 German National Olympiad, 2

Tags: geometry , inversion , geometric transformation

There is a circle $k$ in a plane with center $M$ and radius $r$. The following illustration, through which every point $P \ne M$., is called a “reflection on the circle $k$” from $\varepsilon$ a point $P'$ from $\varepsilon$ is assigned: (1) $P'$ lies on the ray emanating from$ M$ and passing through $P$. (2) It is $MP \cdot MP' = r^2$. a) Construct the mirror point $ P'$ for any given point $P \ne M$ inside $k$. b) Let another circle $k_1$ be given arbitrarily, but such that $M$ lies outside $k_1$.Construct $k'_1$ , i.e. the set of all mirror points $P'$ of the points $P$ of $k_1$.

2010 Contests, 2

Tags: geometry , inversion , p2

Given a triangle $ABC$, with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$. Let $E$ be a point on the arc $BDC$, and $F$ a point on the segment $BC$, such that $\angle BAF=\angle CAE < \dfrac12\angle BAC$. If $G$ is the midpoint of $IF$, prove that the meeting point of the lines $EI$ and $DG$ lies on $\Gamma$. [i]Proposed by Tai Wai Ming and Wang Chongli, Hong Kong[/i]

2023 Euler Olympiad, Round 2, 3

Tags: inversion , euler , geometry

Let $ABCD$ be a convex quadrilateral with side lengths satisfying the equality: $$ AB \cdot CD = AD \cdot BC = AC \cdot BD.$$ Determine the sum of the acute angles of quadrilateral $ABCD$. [i]Proposed by Zaza Meliqidze, Georgia[/i]

2019 USMCA, 3

Tags: inversion , projective , angle chasing , length chase , trigonometry , geometry

Let $ABC$ be a scalene triangle. The incircle of $ABC$ touches $\overline{BC}$ at $D$. Let $P$ be a point on $\overline{BC}$ satisfying $\angle BAP = \angle CAP$, and $M$ be the midpoint of $\overline{BC}$. Define $Q$ to be on $\overline{AM}$ such that $\overline{PQ} \perp \overline{AM}$. Prove that the circumcircle of $\triangle AQD$ is tangent to $\overline{BC}$.

2020 International Zhautykov Olympiad, 3

Tags: geometry , hexagon , inequality , inversion

Given convex hexagon $ABCDEF$, inscribed in the circle. Prove that $AC*BD*DE*CE*EA*FB \geq 27 AB * BC * CD * DE * EF * FA$

2024 Thailand October Camp, 3

Tags: geometry , spiral similarity , cartesian coordinates , complex bash , inversion

Let triangle $ ABC $ be an acute-angled triangle. Square $ AEFB $ and $ ADGC $ lie outside triangle $ ABC $. $ BD $ intersects $ CE $ at point $ H $, and $ BG $ intersects $ CF $ at point $ I $. The circumcircle of triangle $ BFI $ intersects the circumcircle of triangle $ CGI $ again at point $ K $. Prove that line segment $ HK $ bisects $ BC $.

2017 Azerbaijan Team Selection Test, 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 $BDE$.

2010 ELMO Shortlist, 4

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]

2014 EGMO, 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 Brazil Team Selection Test, 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 $QB = QR$.

Russian TST 2017, P1

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 $BDE$.

1998 IMO Shortlist, 8

Tags: geometry , harmonic division , inversion , right triangle

Let $ABC$ be a triangle such that $\angle A=90^{\circ }$ and $\angle B<\angle C$. The tangent at $A$ to the circumcircle $\omega$ of triangle $ABC$ meets the line $BC$ at $D$. Let $E$ be the reflection of $A$ in the line $BC$, let $X$ be the foot of the perpendicular from $A$ to $BE$, and let $Y$ be the midpoint of the segment $AX$. Let the line $BY$ intersect the circle $\omega$ again at $Z$. Prove that the line $BD$ is tangent to the circumcircle of triangle $ADZ$. [hide="comment"] [i]Edited by Orl.[/i] [/hide]

2009 Ukraine Team Selection Test, 10

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]

2020 Iran MO (3rd Round), 3

Tags: geometry , excircle , inversion

The circle $\Omega$ with center $I_A$, is the $A$-excircle of triangle $ABC$. Which is tangent to $AB,AC$ at $F,E$ respectivly. Point $D$ is the reflection of $A$ through $I_AB$. Lines $DI_A$ and $EF$ meet at $K$. Prove that ,circumcenter of $DKE$ , midpoint of $BC$ and $I_A$ are collinear.

2023 Bangladesh Mathematical Olympiad, P9

Tags: geometry , parallel , cyclic quadrilateral , radical axis , angle chasing , inversion

Let $\Delta ABC$ be an acute angled triangle. $D$ is a point on side $BC$ such that $AD$ bisects angle $\angle BAC$. A line $l$ is tangent to the circumcircles of triangles $ADB$ and $ADC$ at point $K$ and $L$, respectively. Let $M$, $N$ and $P$ be its midpoints of $BD$, $DC$ and $KL$, respectively. Prove that $l$ is tangent to the circumcircle of $\Delta MNP$.