Olympiad problems

2013 IMO, 3

Let the excircle of triangle $ABC$ opposite the vertex $A$ be tangent to the side $BC$ at the point $A_1$. Define the points $B_1$ on $CA$ and $C_1$ on $AB$ analogously, using the excircles opposite $B$ and $C$, respectively. Suppose that the circumcentre of triangle $A_1B_1C_1$ lies on the circumcircle of triangle $ABC$. Prove that triangle $ABC$ is right-angled. [i]Proposed by Alexander A. Polyansky, Russia[/i]

Kvant 2024, M2809

Tags: conditional geometry , angle measure , construction , right angle , beautiful geometry

Given is a triangle $ABC$ and the points $M, P$ lie on the segments $AB, BC$, respectively, such that $AM=BC$ and $CP=BM$. If $AP$ and $CM$ meet at $O$ and $2\angle AOM=\angle ABC$, find the measure of $\angle ABC$.

2024 Bulgaria National Olympiad, 2

Tags: right angle , conditional geometry , angle measure , beautiful geometry , construction

Given is a triangle $ABC$ and the points $M, P$ lie on the segments $AB, BC$, respectively, such that $AM=BC$ and $CP=BM$. If $AP$ and $CM$ meet at $O$ and $2\angle AOM=\angle ABC$, find the measure of $\angle ABC$.

2024 All-Russian Olympiad, 4

Tags: conditional geometry , geometry , cyclic quadrilateral , ratio

In cyclic quadrilateral $ABCD$, $\angle A+ \angle D=\frac{\pi}{2}$. $AC$ intersects $BD$ at ${E}$. A line ${l}$ cuts segment $AB, CD, AE, DE$ at $X, Y, Z, T$ respectively. If $AZ=CE$ and $BE=DT$, prove that the diameter of the circumcircle of $\triangle EZT$ equals $XY$.

2023 Turkey Olympic Revenge, 2

Tags: conditional geometry , triangle geometry , geometry

Let $ABC$ be a triangle. A point $D$ lies on line $BC$ and points $E,F$ are taken on $AC,AB$ such that $DE \parallel AB$ and $DF\parallel AC$. Let $G = (AEF) \cap (ABC) \neq A$ and $I = (DEF) \cap BC\neq D$. Let $H$ and $O$ denote the orthocenter and the circumcenter of triangle $DEF$. Prove that $A,O,I$ are collinear if and only if $G,H,I$ are collinear. [i]Proposed by Kaan Bilge[/i]

2018 Brazil Team Selection Test, 4

Tags: tangent , geometry , isosceles triangle , conditional geometry , circumcenter , isosceles

Consider an isosceles triangle $ABC$ with $AB = AC$. Let $\omega(XYZ)$ be the circumcircle of the triangle $XY Z$. The tangents to $\omega(ABC)$ through $B$ and $C$ meet at the point $D$. The point $F$ is marked on the arc $AB$ of $\omega(ABC)$ that does not contain $C$. Let $K$ be the point of intersection of lines $AF$ and $BD$ and $L$ the point of intersection of the lines $AB$ and $CF$. Let $T$ and $S$ be the centers of the circles $\omega(BLC)$ and $\omega(BLK)$, respectively. Suppose that the circles $\omega(BTS)$ and $\omega(CFK)$ are tangent to each other at the point $P$. Prove that $P$ belongs to the line $AB$.

2013 IMO Shortlist, G6

Tags: geometry , conditional geometry

Let the excircle of triangle $ABC$ opposite the vertex $A$ be tangent to the side $BC$ at the point $A_1$. Define the points $B_1$ on $CA$ and $C_1$ on $AB$ analogously, using the excircles opposite $B$ and $C$, respectively. Suppose that the circumcentre of triangle $A_1B_1C_1$ lies on the circumcircle of triangle $ABC$. Prove that triangle $ABC$ is right-angled. [i]Proposed by Alexander A. Polyansky, Russia[/i]

2024 Bulgaria MO Regional Round, 12.1

Tags: geometry , conditional geometry , right angle , incentre , isosceles

Let $ABC$ be an acute triangle with midpoint $M$ of $AB$. The point $D$ lies on the segment $MB$ and $I_1, I_2$ denote the incenters of $\triangle ADC$ and $\triangle BDC$. Given that $\angle I_1MI_2=90^{\circ}$, show that $CA=CB$.