Olympiad problems

1999 USAMO, 6

Tags: incenter , trapezoid , symmetry , isosceles triangle , geometry , reflection , geometric transformation

Let $ABCD$ be an isosceles trapezoid with $AB \parallel CD$. The inscribed circle $\omega$ of triangle $BCD$ meets $CD$ at $E$. Let $F$ be a point on the (internal) angle bisector of $\angle DAC$ such that $EF \perp CD$. Let the circumscribed circle of triangle $ACF$ meet line $CD$ at $C$ and $G$. Prove that the triangle $AFG$ is isosceles.

May Olympiad L1 - geometry, 2014.4

Tags: geometry , right triangle , isosceles triangle , equilateral triangle , angle

Let $ABC$ be a right triangle and isosceles, with $\angle C = 90^o$. Let $M$ be the midpoint of $AB$ and $N$ the midpoint of $AC$. Let $ P$ be such that $MNP$ is an equilateral triangle with $ P$ inside the quadrilateral $MBCN$. Calculate the measure of $\angle CAP$

2020 IMO Shortlist, G1

Tags: right triangle , geometry , isosceles triangle

Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$. Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.

2007 Korea Junior Math Olympiad, 7

Tags: circumcenter , isosceles , isosceles triangle , incenter , geometry

Let the incircle of $\triangle ABC$ meet $BC,CA,AB$ at $J,K,L$. Let $D(\ne B, J),E(\ne C,K), F(\ne A,L)$ be points on $BJ,CK,AL$. If the incenter of $\triangle ABC$ is the circumcenter of $\triangle DEF$ and $\angle BAC = \angle DEF$, prove that $\triangle ABC$ and $\triangle DEF$ are isosceles triangles.

2018 Yasinsky Geometry Olympiad, 5

Tags: geometry , incircle , equal angles , isosceles triangle , isosceles

The inscribed circle of the triangle $ABC$ touches its sides $AB, BC, CA$, at points $K,N, M$ respectively. It is known that $\angle ANM = \angle CKM$. Prove that the triangle $ABC$ is isosceles. (Vyacheslav Yasinsky)

2021 Estonia Team Selection Test, 2

Tags: geometry , right triangle , isosceles triangle

Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$. Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.

2013 IMO Shortlist, G4

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

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 $QB = QR$.

2014 India IMO Training Camp, 3

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

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 $QB = QR$.