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.

vladimir.shelomovskii@gmail.com, vvsss

Solution 3 (No Miquel's point)

Consider $\triangle MKA$ and $\triangle MNC$ , they are similar because $\angle MAK$ = $\angle MCN$ , and also $\angle MKA = \angle MNC$ .

Now draw $OP \perp AB$ , and intersecting $AB$ at $P$ ; $OQ \perp BC$ , at $Q$ . Naturally $OP$ bisects $AK$ , and $OQ$ bisects $CN$ . We claim $\triangle MAP \sim \triangle MCQ$ , because $\frac {AP}{CQ} = \frac {AK}{CN} = \frac {AM}{CB}.$

Thus $\angle AMP = \angle CMQ$ , this implies $\angle PMQ = \angle AMC = \angle ABC = \angle PBQ$ . Obviously BMPQ is cyclic, and so is BPOQ. Finally, we have $OM \perp MB$ . (by gougutheorem)

Solution 4 (Sparrow solution)

Let $\Omega,Q,$ and $R$ be the circumcircle of $\triangle ABC,$ circumcenter and radius of $\Omega.$

Let $\omega,O',$ and $r$ be the circumcircle of $\triangle KBN,$ circumcenter and radius of $\omega.$ $BO' = MO' = r, BQ = MQ = R \implies$ $QO' \perp MB, \angle BQO' = \angle MQO'.$

$AKNC$ is cyclic, so $KN$ is antiparallel $AC, O'O \perp KN.$

We use Sparrow’s Lemma 3A for circle $\omega$ and get that point $O'$ lies on altitude of $\triangle ABC \implies BO' \perp AC.$

Let $D$ be the point on $\omega$ opposite $B.$

$BQ$ is isogonal to $BO' \implies OO' || BQ.$

$OQ$ lies on bisector $AC \implies BO' || QO \implies BO'OQ$ is parallelogram $\implies OO' = BQ = R, BO' = QO = r = O'D \implies DO'QO$ is parallelogram.

Let $\angle BO'M = 2 \varphi \implies \angle O'MD = \angle O'DM = \varphi.$

$\angle DO'Q = 180^\circ - \varphi - (180^\circ - 2 \varphi) = \varphi \implies MD || O'Q \perp MB \blacksquare$

1985 IMO (Problems) • Resources
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6
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