# Difference between revisions of "1985 IMO Problems/Problem 5"

## Problem

A circle with center $O$ passes through the vertices $A$ and $C$ of the triangle $ABC$ and intersects the segments $AB$ and $BC$ again at distinct points $K$ and $N$ respectively. Let $M$ be the point of intersection of the circumcircles of triangles $ABC$ and $KBN$ (apart from $B$). Prove that $\angle OMB = 90^{\circ}$.

## Solution

$M$ is the Miquel Point of quadrilateral $ACNK$, so there is a spiral similarity centered at $M$ that takes $KN$ to $AC$. Let $M_1$ be the midpoint of $KA$ and $M_2$ be the midpoint of $NC$. Thus the spiral similarity must also send $M_1$ to $M_2$ and so $BMM_1 M_2$ is cyclic. $OM_1 B M_2$ is also cyclic with diameter $BO$ and thus $M$ must lie on the same circumcircle as $B$, $M_1$, and $M_2$ so $\angle OMB = 90^{\circ}$.

## Solution 2

Let $\Omega, \Omega', \omega$ and $O,O',O''$ be the circumcircles and circumcenters of $AKNC, ABC, BNKM,$ respectively.

Let $\angle ACB = \gamma, AKNC$ is cyclic $\implies \angle BKN = \gamma.$

The radius of $\omega$ is $MO'' = BO'' = \frac {BN}{2 \sin \gamma}.$

Let $D$ and $E$ be midpoints of $BC$ and $NC$ respectively.

$OE \perp BC, OD \perp BC, OO' \perp AC, DE = \frac {BC}{2} - \frac {NC}{2} = \frac {BN}{2}$ $\implies OO' = \frac {DE}{\sin \gamma} = \frac {BN}{2 \sin \gamma} = MO''.$

$M$ is the Miquel Point of quadrilateral $ACNK,$ so $MO''O'O$ is cyclic. $MO''O'O$ is trapezium $\implies O''O' || MO.$ $O''O' \perp BM \implies MO\perp BM$ as desired.