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Team Selection Tests for BMO, 2018

Geometry Difficulty 8.2 Shortlist Prove it Saudi Arabia

Let $ABC$ be a triangle with $M, N, P$ as midpoints of the segments $BC, CA, AB$ respectively. Suppose that $I$ is the intersection of angle bisectors of $\angle BPM$ , $\angle MNP$ and $J$ is the intersection of angle bisectors of $\angle CNM$ , $\angle MPN$ . Denote ( $\omega_{1}$ ) as the circle of center $I$ and tangent to $MP$ at $D$ , $\left(\omega_{2}\right)$ as the circle of center $J$ and tangent to $MN$ at $E$ .

1. Prove that $DE$ is parallel to $BC$ .

2. Prove that the radical axis of two circles ( $\omega_{1}$ ), ( $\omega_{2}$ ) bisects the segment $DE$ .

Solution

1) Note that $\angle MNC = \angle MPB = \angle A$ then by angle chasing, we have $IP \parallel JN$ . Denote $K = PJ \cap IN$ then $K$ is the incenter of triangle $MNP$ . Hence, $MK$ is the angle bisector of $\angle NMP$ , thus $MK \parallel IP$ . Denote $X = IN \cap MP$ then
$\frac{IP}{MK} = \frac{XP}{XM} = \frac{NP}{NM}.$
Similarly, $\frac{JN}{MK} = \frac{NP}{MP}$ .
Thus $\frac{IP}{JN} = \frac{MP}{MN}$ . Since $\triangle IPD \sim \triangle JNE$ then $\frac{IP}{JN} = \frac{PD}{NE}$ . Therefore, $\frac{MP}{MN} = \frac{PD}{NE}$ which implies that $DE \parallel NP \parallel BC$ .

2) Suppose that $DE$ cuts ( $\omega_{1}$ ), ( $\omega_{2}$ ) at $R, S$ respectively. We have
$\frac{DS}{ER} = \frac{ID \cdot \cos \angle IDR}{JE \cdot \cos \angle JES} = \frac{PD \cdot \sin \angle PDR}{NE \cdot \sin \angle NES} = \frac{AC \cdot \sin C}{AB \cdot \sin B} = 1.$

Hence $DS = ER$ . Denote $T$ as midpoint of $DE$ then
$\mathscr{P}_{T/(\omega_{1})} = TD \cdot TR = TE \cdot TS = \mathscr{P}_{T/(\omega_{2})},$
which implies that $T$ lies on the radical axis of $\left(\omega_{1}\right), \left(\omega_{2}\right)$ . $\square$

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