Midpoints of arcs reflected on sides

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Midpoints of arcs reflected on sides

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Source: IMOTC 2015 Practice Test 2 Problem 1

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#1 h Jul 11, 2015, 10:28 AM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle in which $CA>BC>AB$ . Let $H$ be its orthocentre and $O$ its circumcentre. Let $D$ and $E$ be respectively the midpoints of the arc $AB$ not containing $C$ and arc $AC$ not containing $B$ . Let $D'$ and $E'$ be respectively the reflections of $D$ in $AB$ and $E$ in $AC$ . Prove that $O, H, D', E'$ lie on a circle if and only if $A, D', E'$ are collinear.

This post has been edited 1 time. Last edited by hajimbrak, Jul 11, 2015, 10:29 AM
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#2 h Jul 11, 2015, 9:43 PM • 1 Y

Y by Adventure10

Since $\angle BAD'=\angle BAD=\tfrac{1}{2}\angle ACB$ and $\angle CAE'=\angle CAE=\tfrac{1}{2}\angle ABC,$ we get $A,D',E'$ are collinear $\Longleftrightarrow$ $\angle BAD'+\angle CAE'=\angle BAC$ $\Longleftrightarrow$ $\angle ACB+\angle ABC=2\angle BAC$ $\Longleftrightarrow$ $\angle BAC=60^{\circ}.$

Let $I,N,Na$ denote the incenter, 9-point center and Nagel point of $\triangle ABC.$ By Furhmann theorem, $D'$ and $E'$ lie on the circle with diameter $\overline{HNa}$ for any $\triangle ABC,$ hence $O,H,D',E'$ are concyclic $\Longleftrightarrow$ $\angle HONa=90^{\circ}.$ As $I,N$ are the Nagel point and circumcenter of the medial triangle, then $ONa \parallel IN,$ so $\angle HONa=90^{\circ}$ $\Longleftrightarrow$ $IN \perp OH$ $\Longleftrightarrow$ $IH=IO$ $\Longleftrightarrow$ one of the angles of $\triangle ABC$ equals $60^{\circ}$ (well-known).

From the previous discussion we get: If $D',E'$ are collinear, then $\angle BAC=60^{\circ}$ $\Longrightarrow$ $AH=AO$ $\Longrightarrow$ $IN \perp OH$ $\Longrightarrow$ $\angle HONa=90^{\circ}$ $\Longrightarrow$ $O,H,D',E'$ are concyclic. But the converse is not necessarily true; for instance the concyclity yields $IH=IO$ $\Longrightarrow$ one of the angles of $\triangle ABC$ equals $60^{\circ},$ say $\angle ACB=60^{\circ}$ $\Longrightarrow$ $D' \equiv O$ $\Longrightarrow$ $D'E'$ is perpendicular bisector of $AC$ $\Longrightarrow$ $A \notin D'E'.$

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#3 h Jul 12, 2015, 12:32 AM • 2 Y

Y by Adventure10, Mango247

See http://www.artofproblemsolving.com/community/c6h567389p3324618

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