O, H, D', E' concylic if and only if A,D',E' are collinear

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O, H, D', E' concylic if and only if A,D',E' are collinear

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Source: Hong Kong National Olympiad 2013 Problem 3

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WakeUp

#1 h Dec 16, 2013, 4:32 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle with $CA>BC>AB$ . Let $O$ and $H$ be the circumcentre and orthocentre of triangle $ABC$ respectively. Denote by $D$ and $E$ the midpoints of the arcs $AB$ and $AC$ of the circumcircle of triangle $ABC$ not containing the opposite vertices. Let $D'$ be the reflection of $D$ about $AB$ and $E'$ the reflection of $E$ about $AC$ . Prove that $O,H,D',E'$ are concylic if and only if $A,D',E'$ are collinear.

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#2 h Dec 16, 2013, 5:20 PM • 1 Y

Y by Adventure10

We use complex numbers. WLOG, assume that the circumcircle of $\triangle ABC$ is the unit circle and that $a=1$ . Then note that since $DM \perp AB$ where $m=\frac{1+b}{2}$ , we must have $\frac{2d-b-1}{2(b-1)}=-\frac{2/d-1/b-1}{2(1/b-1)} \implies d=\sqrt{b}$ Now it is well known that the reflection of a point $z$ over a chord $AB$ on the unit circle is $a+b-ab\bar z$ . Thus $d'=1+b-b\sqrt{b}$ . Similarly, $e'=1+c-c\sqrt{c}$ . To make computations easier, we can let $b=b_1^2$ and $c=c_1^2$ so then $d'=1+b_1^2-b_1^3$ and $e'=1+c_1^2-c_1^3$ . Now all that remains is to bash out the conditions for collinearity and concyclity which get slightly ugly, but are still very manageable in the time limit.

I will edit in the full solution when I find time.

This post has been edited 2 times. Last edited by sicilianfan, Dec 22, 2013, 8:54 PM

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MMEEvN

#3 h Dec 16, 2013, 6:05 PM • 2 Y

Y by Adventure10, Mango247

First by simple angle chasing we can observe that both $AD'HB$ and $AHE'C$ are con-cyclic
$\angle D'HE'=\angle AHE'-\angle AHD'=(180-\angle ACE')-\angle ABD' =180 -\frac{(B+C)}{2}$
.If $A,D',E'$ are collinear,
$\angle BAD'+\angle CAD'=A \Rightarrow \frac{(B+C)}{2}=A$ .From easy angle chase $D'OE= 180-A \Rightarrow D'HE'O$ are concyclic .
If $H,D',E' ,O$ are cocncyclic. $\angle D'HE=\angle D'OE \Rightarrow \frac{(B+C)}{2}=A \Rightarrow \angle BAD'+\angle CAE'=A$
Done!

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SayanRoychowdhuri

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SayanRoychowdhuri

#4 h Dec 16, 2013, 6:14 PM • 1 Y

Y by Adventure10

As $AC>AB$ we arrange the following lines up to down $BN,BE,CD,CM$ w.r.t $O$ the circumcenter of $\triangle ABC$ ,where $M,N$ are intersections of circumcircle with $CH,BH$ respectievly, due to this order clearly we have $BE=CD\Leftrightarrow\angle BNE=\angle DMC$ and clearly we have $HMDD'$ and $HE'EN$ are isoscleses trapizoids, hence we have $A,D',D$ are coliner $\Leftrightarrow CE\parallel BD\Leftrightarrow BE=CD\Leftrightarrow \angle BNE=\angle DMC$ or $\angle HE'O=\angle DD'H\Leftrightarrow$ $D',H,E',O$ are concyclic, done!

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#5 h Dec 17, 2013, 4:45 PM • 2 Y

Y by Adventure10, Mango247

NOTE: $A=\dfrac{B+C}{2}\Leftrightarrow 2.A=B+C\Leftrightarrow 3.A=A+B+C=180^0\Leftrightarrow \boxed{A=60^0}.$

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