MMO Round 4

Geometry Difficulty 6.4 National Olympiad Prove it Mongolia

Let $I$ be the incenter of triangle $ABC$ , which is inscribed in the circle $\omega$ centered at $O$ . Suppose the line $BI$ intersects $\omega$ again at point $M$ . Let $I_B$ be the reflection of $I$ over the line $AC$ . Suppose that the line $MI_B$ intersects $\omega$ again at point $D \neq M$ , and the line $DO$ intersects $\omega$ again at point $E$ . Prove that the lines $OI$ and $BE$ are parallel.
(Batzorig Undrakh)

Solution

Note that since $OM \parallel IIB$ , it follows that $\angle IMO = \angle I_BIM$ . Also, since $BO = OM$ , we have
$\angle IMO = \angle IBO = \angle I_BIM.$
Let us compute the power of point $I$ with respect to the circle $\omega$ . Since $BI \cdot IM = R^2 - OI^2 = 2Rr$ , we get
$BI \cdot IM = 2Rr = 2 \cdot BO \cdot \frac{IIB}{2} = BO \cdot IIB.$

$\angle BOI = \angle IMI_B = \angle BMD = \angle BED = \angle OBE$ ,

so we conclude that $OI \parallel BE$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.