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.

MrOreoJuice

06.05.2022 20:08

Let

$E'$ ,

$F'$ be the intersections of

$\overline{BG}$ ,

$\overline{CG}$ with

$(ABC)$ . By homothety

$GE' = 2GY$ and

$GF' = 2GZ$ and we have

$GA=2GD$ . Immediate power of point at

$G$ gives that

$BYDX$ and

$CZDX$ are cyclic, hence

$\angle YXZ = \angle YXD + \angle DXZ = \angle YBD + \angle DCZ = \pi - \angle YGZ.$

samrocksnature

06.05.2022 21:01

Oops I found the hidden cyclic quads through geogebra

ALM_04

06.05.2022 21:58

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, $\measuredangle{GYX}=\measuredangle{BYX}=\measuredangle{BDX}=\measuredangle{CDX}=\measuredangle{CZX}=\measuredangle{GZX}$ .

Ejaifeobuks

18.05.2022 03:36

Here is a link to a video solution https://youtu.be/MWfTog7WvRg