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Saudi Arabian Mathematical Competitions

Geometry Difficulty 5.0 AIME, harder Prove it Saudi Arabia

A circle with center $O$ passes through points $A$ and $C$ and intersects the sides $AB$ and $BC$ of triangle $ABC$ at points $K$ and $N$ , respectively. The circumcircles of triangles $ABC$ and $KBN$ meet at distinct points $B$ and $M$ . Prove that $\angle OMB = 90^{\circ}$ .

Solution

Consider an inversion with center $B$ and radius $r > 0$ . The points $A'$ , $B'$ , $C'$ are collinear and the points $K'$ , $N'$ , $M'$ are, too.
The circumcircle of quadrilateral $ACNK$ becomes the circumcircle $\omega$ of quadrilateral $A' C' N' K'$ . In case $B$ is outside $\omega$ , let $X$ , $Y$ be the tangent points of 2 tangents from $B$ to $\omega$ , then $O'$ is the midpoint of $XY$ .
Since $M'$ , $X$ , $Y$ lies on the polar of $B$ relative to $\omega$ then $\angle M' O' B = 90^{\circ}$ , this implies that $\angle OMB = 90^{\circ}$ .
In the other cases, we also have $O'$ lies on the polar of $B$ relative to $\omega$ , and the argument is similar.

Remark. This problem can be solve by using the spiral similarity of center $M$ or by using the Brocard's theorem.

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