Olympiad problems

2015 Bangladesh Mathematical Olympiad, 7

In triangle $\triangle ABC$, the points $A', B', C'$ are on sides $BC, AC, AB$ respectively. Also, $AA', BB', CC'$ intersect at the point $O$(they are concurrent at $O$). Also, $\frac {AO}{OA'}+\frac {BO}{OB'}+\frac {CO}{OC'} = 92$. Find the value of $\frac {AO}{OA'}\times \frac {BO}{OB'}\times \frac {CO}{OC'}$.

2023 Bulgaria EGMO TST, 6

Tags: mixtilinear , brianchon , ceva theorem , radical axis , concurrency , incircle , geometry

Let $ABC$ be a triangle with incircle $\gamma$. The circle through $A$ and $B$ tangent to $\gamma$ touches it at $C_2$ and the common tangent at $C_2$ intersects $AB$ at $C_1$. Define the points $A_1$, $B_1$, $A_2$, $B_2$ analogously. Prove that: a) the points $A_1$, $B_1$, $C_1$ are collinear; b) the lines $AA_2$, $BB_2$, $CC_2$ are concurrent.

2017 Kürschák Competition, 1

Tags: probability , algebra , plane geometry , ceva theorem

Let $ABC$ be a triangle. Choose points $A'$, $B'$ and $C'$ independently on side segments $BC$, $CA$ and $AB$ respectively with a uniform distribution. For a point $Z$ in the plane, let $p(Z)$ denote the probability that $Z$ is contained in the triangle enclosed by lines $AA'$, $BB'$ and $CC'$. For which interior point $Z$ in triangle $ABC$ is $p(Z)$ maximised?

2021 Harvard-MIT Mathematics Tournament., 7

Tags: geometry , tangency , ceva theorem

In triangle $ABC$, let $M$ be the midpoint of $BC$ and $D$ be a point on segment $AM$. Distinct points $Y$ and $Z$ are chosen on rays $\overrightarrow{CA}$ and $\overrightarrow{BA}$ , respectively, such that $\angle DYC=\angle DCB$ and $\angle DBC=\angle DZB$. Prove that the circumcircle of $\Delta DYZ$ is tangent to the circumcircle of $\Delta DBC$.