Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.4 AIME, harder Prove it Taiwan

Let $O$ be the circumcenter of triangle $ABC$ , and $\omega$ be the circumcircle of triangle $BOC$ . Line $AO$ intersects with circle $\omega$ again at the point $G$ . Let $M$ be the midpoint of side $BC$ , and the perpendicular bisector of $BC$ meets circle $\omega$ at the points $O$ and $N$ .

Prove that the midpoint of the segment $AN$ lies on the radical axis of the circumcircle of triangle $OMG$ , and the circle whose diameter is $AO$ .

Solution

Let $H$ be the foot of the perpendicular from $A$ to $BC$ . It is easy to see that $AN$ , the circle with diameter $AO$ , and circle $BOC$ are concurrent at a point $S$ . Let $AM$ intersect circle $OMG$ again at the point $V$ , let $GV$ intersect circle $BOC$ again at the point $U$ , and let it intersect $AN$ at the point $T$ . Since $\angle NUG = \angle MOG = \angle MVU$ , that is, $UN \parallel AV$ , we have
$\angle TGA = \angle AMO = \angle HAM = \angle SAO = \angle TAG$
(because $AN$ is the symmedian, hence $AM$ , $AN$ are isogonal conjugate lines, and $AH$ , $AO$ are isogonal conjugate lines). Also $\angle AGN = 90^\circ$ , so $T$ is the midpoint of $AN$ . Combining the above, we obtain that $AVNU$ is a parallelogram, and $T$ is the midpoint of $VU$ . Therefore $TS \cdot TA = TS \cdot TN = TU \cdot TG = TV \cdot TG$ , so $T$ lies on the radical axis of the two circles. Q.E.D.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.