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Difference between revisions of "2019 IMO Problems/Problem 2"
 
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==Problem==
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In triangle <math>ABC</math>, point <math>A_1</math> lies on side <math>BC</math> and point <math>B_1</math> lies on side <math>AC</math>. Let <math>P</math> and <math>Q</math> be points on segments <math>AA_1</math> and <math>BB_1</math>, respectively, such that <math>PQ</math> is parallel to <math>AB</math>. Let <math>P_1</math> be a point on line <math>PB_1</math>, such that <math>B_1</math> lies strictly between <math>P</math> and <math>P_1</math>, and <math>\angle PP_1C=\angle BAC</math>. Similarly, let <math>Q_1</math> be the point on line <math>QA_1</math>, such that <math>A_1</math> lies strictly between <math>Q</math> and <math>Q_1</math>, and <math>\angle CQ_1Q=\angle CBA</math>.
 
In triangle <math>ABC</math>, point <math>A_1</math> lies on side <math>BC</math> and point <math>B_1</math> lies on side <math>AC</math>. Let <math>P</math> and <math>Q</math> be points on segments <math>AA_1</math> and <math>BB_1</math>, respectively, such that <math>PQ</math> is parallel to <math>AB</math>. Let <math>P_1</math> be a point on line <math>PB_1</math>, such that <math>B_1</math> lies strictly between <math>P</math> and <math>P_1</math>, and <math>\angle PP_1C=\angle BAC</math>. Similarly, let <math>Q_1</math> be the point on line <math>QA_1</math>, such that <math>A_1</math> lies strictly between <math>Q</math> and <math>Q_1</math>, and <math>\angle CQ_1Q=\angle CBA</math>.
  
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==Solution==
 
==Solution==
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[[File:2019 IMO 2.png|500px|right]]
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[[File:2019 IMO 2.png|420px|right]]
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[[File:2019 IMO 2a.png|420px|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.
 
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.
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We assume that the intersection point of <math>AP</math> and <math>BQ</math> lies on the segment <math>PA_1.</math> If it lies on segment <math>AP,</math> then the proof is the same, but some angles will be replaced with additional ones up to <math>180^\circ</math>.
  
 
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>
 
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>
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<math>\angle QPA_0 = \angle QB_0A_0 \implies QPB_0A_0</math> is cyclic (in circle <math>\omega.</math>)
 
<math>\angle QPA_0 = \angle QB_0A_0 \implies QPB_0A_0</math> is cyclic (in circle <math>\omega.</math>)
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Let <math>\angle BAC = \alpha, \angle AA_0B_0 = \varphi.</math>
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<math>\angle PP_1C = \alpha, \angle BB_0C = \alpha</math> since they intersept the arc <math>BC</math> of the circle <math>\Omega.</math>
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So <math>B_0P_1CB_1</math> is cyclic.
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<math>\angle ACB_0 = \angle AA_0B_0 =  \varphi</math> (they intersept the arc <math>A_0B_0</math> of the circle  <math>\Omega).</math>
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<math>\angle B_1CB_0 = \varphi.</math>
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<math>\angle B_1P_1B_0 = \angle B_1CB_0 =  \varphi</math> (since they intersept the arc <math>B_1B_0</math> of the circle <math>B_0P_1CB_1).</math>
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Hence <math>\angle PA_0B_0 = \angle PP_1B_0 = \varphi,</math>  the point <math>P_1</math> lies on <math>\omega.</math>
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Similarly, point <math>Q_1</math> lies on  <math>\omega.</math>
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'''vladimir.shelomovskii@gmail.com, vvsss'''
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==See Also==
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{{IMO box|year=2019|num-b=1|num-a=3}}

Latest revision as of 00:49, 19 November 2023

Problem

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

2019 IMO 2.png
2019 IMO 2a.png

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 $\omega.$

vladimir.shelomovskii@gmail.com, vvsss

See Also

2019 IMO (Problems) • Resources
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 3
All IMO Problems and Solutions
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