Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it Ukraine

The bisector of an angle $BAC$ of acute angled triangle $ABC$ ( $AC \neq AB$ ) intersects its circumcircle second time at the point $W$ . Let $O$ be the circumcenter of $\triangle ABC$ . Line $AW$ intersects second time the circumcircles of triangles $OWB$ and $OWC$ at points $N$ and $M$ respectively. Prove that $BN + MC = AW$ .

(V.Mitrofanov, D.Khilko)

Solution

Without loss of generality we can consider that $AC < AB$ . From the conditions (fig. 18) $2\angle CAW = \angle COW = \angle CMW$ , hence $\angle ACM = \angle CAM$ . Thus $AM = CM$ . Similarly $AN = NB$ . Then it is enough to prove that $AM + AN = AW$ , thus $AM = NW$ . Obviously, $CO = OB$ , $CW = WB$ . Thus $\angle OBW = \angle OCW$ . Hence $\angle OBW = \angle OCW$ . Therefore,

$\angle ONM = \angle OBW = \angle OCW = \angle OMN$ , hence $\triangle ONM$ is isosceles. It is also obvious that $\triangle OAW$ is isosceles, hence it is clear that $\triangle ONW = \triangle OMA$ , thus $AM = NW$ .

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