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 $\ell_2, m_2$ .

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

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