IMO-2008-6

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Tahmid Hasan
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IMO-2008-6

Unread post by Tahmid Hasan » Thu Mar 21, 2013 9:48 pm

Let $ABCD$ be a convex quadrilateral with $AB \neq BC$. Denote by $\omega_1$ and $\omega_2$ the incircles
of triangles $ABC$ and $ADC$. Suppose that there exists a circle $\omega$ inscribed in angle $ABC$, tangent to the extensions of line segments $AD$ and $CD$. Prove that the common external tangents of $\omega_1$ and $\omega_2$ intersect on $\omega$.
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Tahmid Hasan
Posts:665
Joined:Thu Dec 09, 2010 5:34 pm
Location:Khulna,Bangladesh.

Re: IMO-2008-6

Unread post by Tahmid Hasan » Thu Mar 21, 2013 10:19 pm

Solution (with proper incentives!):
Lemma: Let the incircle of $\triangle ABC$ touch $BC$ at $D$. Let the diametrically opposite point of $D$ be $D'$. $AD' \cap BC=E$. Then $BD=CE$.$E$ is the touch point of the $A$-excircle with $BC$. The converse is also true.
Let the touch points of $\omega$ with $AB,CD,AD,BC$ be $P,Q,R,S$ respectively. Let the touch points of $\omega_1,\omega_2$ with $AC$ be $E,F$ respectively. Let $E',F'$ be the diametrically opposite points of $E,F$ wrt $\omega_1,\omega_2$ respectively.
Immediately I noticed $B,E',F$ are collinear after drawing the figure. For that to be true $AE=CF$ according to the lemma.
So I claimed $AE=CF$.[Proof will be shown later.] Hence $B,E',F$ are collinear.
Which also implies $D,F',E$ are collinear. Hence $E$ is the touch point of the $D$- excircle of $\triangle ACD$ and $F$ is the touch point of the $B$-excircle of $\triangle ABC$.
Let $H$ be a point on $\omega$ such that the tangent at $H$ is parallel to $AC$. So $\omega$ is the $B$-excircle of the triangle formed by the tangent at $H$;$AB,BC$ and the $D$-excircle of the triangle formed by the tangent and $AD,CD$.
So from homothety we conclude $B,F,H$ are collinear and $D,H,E$ are collinear.
From the previous collinearity arguements we get $E',F,H$ and $E,F',H$ are collinear.
Since $EE',FF'$ are dimeters of $\omega_1,\omega_2$ and are parallel [both perpendicular to $AC$], so $EF' \cap FE'=H$ is the centre of homothety that sends $\omega_1$ to $\omega_2$
and since $E,F'$ are on the same side of the line joining the centres of $\omega_1,\omega_2$ the ratio of homothety is positive. Hence $H$ is the intersection of the external common tangents of $\omega_1,\omega_2$.
Proof of the claim: $AE=CF \Leftrightarrow \frac{AB+BC+CA}{2}-BC=\frac{CD+DA+AC}{2}-AD$
$\Leftrightarrow AB+AD=BC+CD$.
Now $AB+AD=AB+AR-DR=AB+AP-DQ=BP-DQ=BS-DQ$
$=BC+CS-DQ=BC+CQ-DQ=BC+CD$. So our claim was true.
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