Olympiad problems

2022 Harvard-MIT Mathematics Tournament, 5

Let $ABC$ be a triangle with centroid $G$, and let $E$ and $F$ be points on side $BC$ such that $BE = EF = F C$. Points $X$ and $Y$ lie on lines $AB$ and $AC$, respectively, so that $X$, $Y$ , and $G$ are not collinear. If the line through $E$ parallel to $XG$ and the line through $F$ parallel to $Y G$ intersect at $P\ne G$, prove that $GP$ passes through the midpoint of $XY$.

2008 Balkan MO Shortlist, G4

Tags: perpendicular bisector , centroid , orthocenter , barycenter , geometry

A triangle $ABC$ is given with barycentre $G$ and circumcentre $O$. The perpendicular bisectors of $GA, GB$ meet at $C_1$,of $GB,GC$ meet at $A _1$, and $GC,GA$ meet at $B_1$. Prove that $O$ is the barycenter of the triangle $A_1B_1C_1$.

1978 Romania Team Selection Test, 7

Tags: analytic geometry , barycenter , linear algebra , geometry

[b]a)[/b] Prove that for any natural number $ n\ge 1, $ there is a set $ \mathcal{M} $ of $ n $ points from the Cartesian plane such that the barycenter of every subset of $ \mathcal{M} $ has integral coordinates (both coordinates are integer numbers). [b]b)[/b] Show that if a set $ \mathcal{N} $ formed by an infinite number of points from the Cartesian plane is given such that no three of them are collinear, then there exists a finite subset of $ \mathcal{N} , $ the barycenter of which has non-integral coordinates.

2024 Czech-Polish-Slovak Junior Match, 1

Tags: parallelogram , barycenter , geometry

Let $G$ be the barycenter of triangle $ABC$. Let $D$ be a point such that $AGDB$ is a parallelogram. Show that $BG \parallel CD$.

2024 Austrian MO National Competition, 4

Tags: orthocenter , geometry , centroid , barycenter

Let $ABC$ be an obtuse triangle with orthocenter $H$ and centroid $S$. Let $D$, $E$ and $F$ be the midpoints of segments $BC$, $AC$, $AB$, respectively. Show that the circumcircle of triangle $ABC$, the circumcircle of triangle $DEF$ and the circle with diameter $HS$ have two distinct points in common. [i](Josef Greilhuber)[/i]