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2019 IMO Problems/Problem 2: Difference between revisions

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Prove that points <math>P,Q,P_1</math>, and <math>Q_1</math> are concyclic.
Prove that points <math>P,Q,P_1</math>, and <math>Q_1</math> are concyclic.
==Solution==
[[File:2019 IMO 2.png|500px|right]]
The essence of the proof is to build a circle through the points <math>P, Q,</math> and two additional points <math>A_0</math> and <math>B_0,</math> then we prove that the points <math>P_1</math> and <math>Q_1</math> lie on the same circle.
Let the circumcircle of <math>\triangle ABC</math> be <math>\Omega</math>. Let <math>A_0</math> and <math>B_0</math> be the points of intersection of <math>AP</math> and <math>BQ</math> with <math>\Omega</math>. Let <math>\angle BAP = \delta.</math>
<cmath>PQ||AB \implies \angle QPA_0 = \delta.</cmath>
<math>\angle BAP = \angle BB_0A_0 = \delta</math> since they intersept the arc <math>BA_0</math> of the circle <math>\Omega</math>.
<math>\angle QPA_0 = \angle QB_0A_0 \implies QPB_0A_0</math> is cyclic (in circle <math>\omega.</math>)

Revision as of 10:53, 13 August 2022

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.

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 $\omega.$)

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