AoPS - Problem

Source: 2022 Nigerian Senior MO Round 2/Problem 2

Tags: geometry, Concyclic, cyclic quadrilateral

Rukevwe

06.05.2022 18:37

Let $G$ be the centroid of $\triangle ABC $ and let $D, E $ and $F$ be the midpoints of the line segments $BC, CA $ and $AB$ respectively. Suppose the circumcircle of $\triangle ABC $ meets $AD $ again at $X$, the circumcircle of $\triangle DEF $ meets $BE$ again at $Y$ and the circumcircle of $\triangle DEF $ meets $CF$ again at $Z$. Show that $G, X, Y $ and $Z$ are concyclic.

Let $\overline{BE}\cap (ABC)=M$ and $\overline{CF}\cap (ABC)=N$. Homothety centered at $G$ with a ratio $-2$ maps $(DEF)$ to $(ABC)$ and this gives $2\overline{GY}=\overline{GM}$ and $2\overline{GZ}=\overline{GN}$. We also have $2\overline{GD}=\overline{GA}$. Combining the relations with PoP, we get $\overline{GB}\cdot \overline{GY}=\overline{GD}\cdot \overline{GX}$ and $\overline{GC}\cdot \overline{GZ}=\overline{GD}\cdot \overline{GX}$ and thus, $BYDX$ and $CZDX$ are cyclic. So, $$\measuredangle{GYX}=\measuredangle{BYX}=\measuredangle{BDX}=\measuredangle{CDX}=\measuredangle{CZX}=\measuredangle{GZX}$$.