2005 IMO Shortlist Problems/G3

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Problem

(Ukraine) Let $\displaystyle ABCD$ be a parallelogram. A variable line $\ell$ passing through the point $\displaystyle A$ intersects the rays $\displaystyle BC$ and $\displaystyle DC$ at points $\displaystyle X$ and $\displaystyle Y$, respectively. Let $\displaystyle K$ and $\displaystyle L$ be the centres of the excircles of triangles $\displaystyle ABX$ and $\displaystyle ADY$, touching the sides $\displaystyle BX$ and $\displaystyle DY$, respectively. Prove that the size of angle $\displaystyle KCL$ does not depend on the choice of $\ell$.

This was also Problem 3 of the 2006 3rd German TST, and a problem at the 2006 India IMO Training Camp. It also appeared in modified form as Problem 3, Day 3 of the 2006 Iran TST.

Solution

Let $\ell_1, \ell_2$ be the interior angle bisectors of $\displaystyle ABX, YAD$. Let $\displaystyle m_1, m_2$ be the exterior angle bisectors of $\displaystyle ABC, CDA$. Then $\displaystyle K$ is the intersection of $\ell_1, m_1$ and $\displaystyle L$ is the intersection of $\ell_2, m_2$.

ISL2005G3.png

Let us denote $\displaystyle x,y$ as the measures of $\angle BAK, \angle LAD$, and denote $\displaystyle \alpha = x+y$. Then $\angle BAD \equiv \angle DCB \equiv 2 \alpha$. Furthermore, since $\displaystyle BK$ is the exterior angle bisector of $\displaystyle ABC$, we know that the exterior angle at $\displaystyle ABK$ is $\displaystyle \alpha = x+y$, so $\angle AKB \equiv y$. Similarly, $\angle ALD \equiv x$. It follows that triangles $\displaystyle ABK, LDA$ are similar. Then since $\displaystyle ABCD$ is a parallelogram,

$\frac{BK}{DC} = \frac{BK}{AB} = \frac{DA}{LD} = \frac{CB}{LD}$.

Since we know $\angle KBC \equiv \angle CDL \equiv \alpha$, this implies that triangles $\displaystyle KBC, CDL$ are similar. This means that


$\angle KCL \equiv 2\pi - (\angle LCD + \angle DCB + \angle BCK) \equiv 2\pi - [(\angle CBK + \angle BCK) + \angle DCB]$ $\equiv 2\pi - [ (\pi- \alpha) + 2\alpha] \equiv \pi - \alpha$,


which is independent of $\ell$, since $\displaystyle \alpha$ is always half the measure of $\angle BAD$.


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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