# 2006 IMO Shortlist Problems/G2

## Problem

(Ukraine) Let $\displaystyle ABCD$ be a trapezoid with parallel sides ${} \displaystyle AB >CD$. Points $\displaystyle K$ and $\displaystyle L$ lie on the line segments $\displaystyle AB$ and $\displaystyle CD$, respectively, so that ${} \displaystyle AK/KB = DL/LC$. Suppose that there are points $\displaystyle P$ and $\displaystyle Q$ on the line segment $\displaystyle KL$ satisfying $\angle APB = \angle BCD$ and $\angle CQD = \angle ABC$.

Prove that the points $\displaystyle P$, $\displaystyle Q$, $\displaystyle B$, and ${} \displaystyle C$ are concyclic.

## Solution

Since ${} \displaystyle A,B,K$ and $\displaystyle D,C,L$ are collinear, the condition ${} \displaystyle AK/KB = DL/LC$ is equivalent to the condition that lines $\displaystyle AD$, $\displaystyle KL$, and $\displaystyle BC$ are concurrent. Let $\displaystyle X$ be the point of concurrence.

Let $\displaystyle \omega_1, \omega_2$ be the circumcircles of $\displaystyle APB, DQC$, respectively. Since $\angle XBA = \angle BCD = \angle APB$, the line $\displaystyle XB$ is tangent to $\displaystyle \omega_1$ at $\displaystyle B$. Similarly, $\displaystyle \omega_2$ is tangent to $\displaystyle XB$ at ${} \displaystyle C$. It follows there is a dilation $\displaystyle \gamma$ centered at $\displaystyle X$ which takes $\displaystyle \omega_2$ to $\displaystyle \omega_1$. Let $\displaystyle Q'$ denote the image of $\displaystyle Q$ under $\displaystyle \gamma$. Evidently, $\displaystyle A, B$ are the respective images of ${} \displaystyle D,C$ under $\displaystyle \gamma$.

Now, since $\displaystyle XB$ is tangent to $\displaystyle \omega_1$ at $\displaystyle B$, it follows that $\angle XQ'B = \angle PQ'B = \angle XBP = \angle CBP$.

But $\displaystyle Q'B$ is the image of ${} \displaystyle QC$ under the dilation $\displaystyle \gamma$, so these two lines are parallel. Hence $\angle PQC = \pi - \angle XQC = \pi - \angle XQ'B = \pi - \angle CBP$.

Therefore $\displaystyle P, Q, B, C$ are concyclic, as desired.