# 2019 IMO Problems/Problem 2

In triangle $ABC$, point $A_1$ lies on side $BC$ and point $B_1$ lies on side $AC$. Let $P$ and $Q$ be points on segments $AA_1$ and $BB_1$, respectively, such that $PQ$ is parallel to $AB$. Let $P_1$ be a point on line $PB_1$, such that $B_1$ lies strictly between $P$ and $P_1$, and $\angle PP_1C=\angle BAC$. Similarly, let $Q_1$ be the point on line $QA_1$, such that $A_1$ lies strictly between $Q$ and $Q_1$, and $\angle CQ_1Q=\angle CBA$.

Prove that points $P,Q,P_1$, and $Q_1$ are concyclic.

## Solution

The essence of the proof is to build a circle through the points $P, Q,$ and two additional points $A_0$ and $B_0,$ then we prove that the points $P_1$ and $Q_1$ lie on the same circle.

We assume that the intersection point of $AP$ and $BQ$ lies on the segment $PA_1.$ If it lies on segment $AP,$ then the proof is the same, but some angles will be replaced with additional ones up to $180^\circ$.

Let the circumcircle of $\triangle ABC$ be $\Omega$. Let $A_0$ and $B_0$ be the points of intersection of $AP$ and $BQ$ with $\Omega$. Let $\angle BAP = \delta.$

$$PQ||AB \implies \angle QPA_0 = \delta.$$

$\angle BAP = \angle BB_0A_0 = \delta$ since they intersept the arc $BA_0$ of the circle $\Omega$.

$\angle QPA_0 = \angle QB_0A_0 \implies QPB_0A_0$ is cyclic (in circle $\omega.$)

Let $\angle BAC = \alpha, \angle AA_0B_0 = \varphi.$

$\angle PP_1C = \alpha, \angle BB_0C = \alpha$ since they intersept the arc $BC$ of the circle $\Omega.$ So $B_0P_1CB_1$ is cyclic.

$\angle ACB_0 = \angle AA_0B_0 = \varphi$ (they intersept the arc $A_0B_0$ of the circle $\Omega).$

$\angle B_1CB_0 = \varphi.$ $\angle B_1P_1B_0 = \angle B_1CB_0 = \varphi$ (since they intersept the arc $B_1B_0$ of the circle $B_0P_1CB_1).$

Hence $\angle PA_0B_0 = \angle PP_1B_0 = \varphi,$ the point $P_1$ lies on $\omega.$

Similarly, point $Q_1$ lies on $\omega.$