Library / /53 of 91

, 2006

Geometry Difficulty 6.2 National Olympiad Prove it India

Let $ABCD$ be a parallelogram. A variable line $l$ through the point $A$ intersects the rays $BC$ and $DC$ at $X$ and $Y$ respectively. Let $K$ and $L$ be the centres of the excircles of triangles $ABX$ and $ADY$ , touching the sides $BX$ and $DY$ respectively. Prove that the size of $\angle KCL$ does not depend on the choice of the line $l$ . (Short-list, IMO-2005)

Solution

Let $\angle DAX = 2\gamma$ and $\angle YAP = 2\alpha$ . Now $K$ being the ex-centre of triangle $ABX$ , it lies on the bisectors of $\angle XAB$ and $\angle XBP$ . Note that $\angle ABX = 180^\circ - 2\alpha - 2\gamma$ and hence $\angle XBK = (2\alpha + 2\gamma)/2 = \alpha + \gamma$ . Thus the angles of triangle $ABK$ are $\alpha$ , $\gamma$ , $180^\circ - \alpha - \gamma$ .

Similarly, the angles of $ADL$ are also $x$ , $y$ , $180^\circ - x - y$ . Thus $ABK$ is similar to $LDA$ and hence $AB/LD = BK/AD$ . However $AB = DC$ and $AD = BC$ . Thus we obtain $DC/LD = BK/BC$ . Note also that $\angle CBK = \angle LDC$ . We hence get the similarity of triangles $CBK$ and $LDC$ . If we take $\angle ALC = \theta$ , then $\angle DCL = 180^\circ - 2x - \theta$ . But by similarity of $CBK$ and $LDC$ , we have $\angle DCL = \angle BKC$ , giving $\angle BKC = 180^\circ - 2x - \theta$ . This gives $\angle KCX = x + \theta$ .

This gives
$\angle KCL = 360^\circ - (x + \theta) - (2x + 2y) - (180^\circ - 2x - y - \theta) \\ = 180^\circ + x + y = 180^\circ + (\angle A)/2$
which depends only on the parallelogram but not on the line $l$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.