Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 8.4 Shortlist Prove it Turkey

Let $ABC$ be an acute triangle with circumcircle $\omega$ centered at $O$ . A point $P$ is chosen on the extension of the segment $BC$ . The line $AP$ meets $\omega$ again at $L$ . Let $K$ be the reflection of $L$ over the line $OP$ and $M$ be the point of intersection of lines $AK$ and $OP$ . Prove that $\overline{OMB} + \overline{OMC} = 180^\circ$ .

Solution

Note that $K$ lies on $\omega$ . Therefore,
$\widehat{OKM} = \widehat{OKA} = \widehat{OAK} = \widehat{OAM}.$
On the other hand, since $OP$ is the perpendicular bisector of $[KL]$ we get $\widehat{OLM} = \widehat{OKM}$ . Therefore, $\widehat{OAM} = \widehat{OLM}$ , which in turn implies that $A, L, M, O$ are concyclic. By writing powers of the point $P$ with respect to the circles $\omega$ and $(ALMO)$ , we get that
$|PM||PO| = |PL||PA| = |PC||PB|.$
Hence, $B, O, M, C$ are concyclic. Therefore,
$\widehat{OMB} + \widehat{OMC} = (90^\circ - \hat{A}) + (90^\circ + \hat{A}) = 180^\circ$

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