AoPS - Problem

Source: Swiss Imo Selection 2006

Tags: geometry, circumcircle, power of a point, radical axis, geometry proposed

BogG

26.05.2006 00:44

Let $D$ be inside $\triangle ABC$ and $E$ on $AD$ different to $D$. Let $\omega_1$ and $\omega_2$ be the circumscribed circles of $\triangle BDE$ and $\triangle CDE$ respectively. $\omega_1$ and $\omega_2$ intersect $BC$ in the interior points $F$ and $G$ respectively. Let $X$ be the intersection between $DG$ and $AB$ and $Y$ the intersection between $DF$ and $AC$. Show that $XY$ is $\|$ to $BC$.

Let $P=DE\cap BC$. By Menelaus in $ABP,ACP$ with transversals $DG$ and $DF$ respectively, we see that $\frac{XB}{XA}=\frac{DP}{DA}\cdot\frac{GB}{GP}$, and $\frac{YC}{YA}=\frac{DP}{DA}\cdot\frac{FC}{FP}$. Proving that the two LHS's in the expressions above are equal is then equivalent to proving that $\frac{GB}{GP}=\frac{FC}{FP}$, which, in turn, is equivalent to $PF\cdot PB=PG\cdot PC$. This is just another way of saying that the powers of $P$ wrt $\omega_1,\omega_2$ are equal, which is clear because $P$ belongs to the radical axis $DE$ of the two circles. P.S. You said $\omega_1$ is the circumcircle of $BDC$. I'm pretty sure, however, that you meant $BDE$, right?