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Difference between revisions of "1985 IMO Problems/Problem 5"
(→‎Solution 3 (No Miquel's point))
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Thus <math>\angle AMP = \angle CMQ</math>, this implies <math>\angle PMQ = \angle AMC = \angle ABC = \angle PBQ</math>. Obviously BMPQ is cyclic, and so is BPOQ. Finally, we have <math>OM \perp MB</math>. ('''by gougutheorem''')
 
Thus <math>\angle AMP = \angle CMQ</math>, this implies <math>\angle PMQ = \angle AMC = \angle ABC = \angle PBQ</math>. Obviously BMPQ is cyclic, and so is BPOQ. Finally, we have <math>OM \perp MB</math>. ('''by gougutheorem''')
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==Solution 4 (Sparrow solution)==
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[[File:IMO 1985 5 Sparrow.png|300px|right]]
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Let <math>\Omega,Q,</math> and <math>R</math> be the circumcircle of <math>\triangle ABC,</math> circumcenter and radius of <math>\Omega.</math>
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Let <math>\omega,O',</math> and <math>r</math> be the circumcircle of <math>\triangle KBN,</math> circumcenter and radius of <math>\omega.</math>
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<cmath>BO' = MO' = r, BQ = MQ = R \implies</cmath>
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<cmath>QO' \perp MB, \angle BQO'  = \angle MQO'.</cmath>
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<math>AKNC</math> is cyclic, so <math>KN</math> is antiparallel <math>AC, O'O \perp KN.</math>
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We use [[Sparrow’s lemmas | Sparrow’s Lemma 3A]] for circle <math>\omega</math> and get that point <math>O'</math> lies on altitude of <math>\triangle ABC \implies BO' \perp AC.</math>
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Let <math>D</math> be the point on <math>\omega</math> opposite <math>B.</math>
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<math>BQ</math> is isogonal to <math>BO' \implies OO' || BQ.</math>
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<math>OQ</math> lies on bisector <math>AC \implies BO' || QO \implies BO'OQ</math> is parallelogram <math>\implies OO' = BQ = R, BO' = QO = r = O'D \implies DO'QO</math> is parallelogram.
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Let <math>\angle BO'M = 2 \varphi \implies \angle O'MD = \angle O'DM = \varphi.</math>
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<math>\angle DO'Q = 180^\circ - \varphi - (180^\circ - 2 \varphi) = \varphi \implies MD || O'Q \perp MB \blacksquare</math>
  
 
== See Also ==
 
== See Also ==
 
*[[Miquel's point]]
 
*[[Miquel's point]]
 
  {{IMO box|year=1985|num-b=4|num-a=6}}
 
  {{IMO box|year=1985|num-b=4|num-a=6}}

Revision as of 12:36, 21 September 2025

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

1985 IMO.png

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)

IMO 1985 5 Sparrow.png

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

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

See Also

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