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Difference between revisions of "2014 Canadian MO Problems/Problem 4"

(Created page with "== Problem== The quadrilateral <math>ABCD</math> is inscribed in a circle. The point <math>P</math> lies in the interior of <math>ABCD</math>, and <math>\angle P AB = \angle P...")
 
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==Solution==
 
==Solution==
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{{Solution}}
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Since <math>A B C D</math> is a cyclic quadrilateral, opposite angles sum to <math>180^{\circ}</math> :
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<cmath>
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\angle A+\angle C=180^{\circ} \text { and } \angle B+\angle D=180^{\circ}
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</cmath>
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Point <math>P</math> ensures that <math>\angle P A B=\angle P B C=\angle P C D=\angle P D A=\theta</math>.
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Triangles <math>P A D</math> and <math>P B C</math> are perspective from <math>Q</math>, and triangles <math>P A B</math> and <math>P C D</math> are perspective from <math>R</math>. This implies the points <math>P, Q, R</math> are collinear.
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Hence, the lines <math>P Q</math> and <math>P R</math> form the same angle as the diagonals <math>A C</math> and <math>B D</math> :
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<cmath>
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\boxed{\angle P Q P=\angle P R P=\angle A C B=\angle B D A}
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</cmath>

Latest revision as of 16:03, 24 May 2024

Problem

The quadrilateral $ABCD$ is inscribed in a circle. The point $P$ lies in the interior of $ABCD$, and $\angle P AB = \angle P BC = \angle P CD = \angle P DA$. The lines $AD$ and $BC$ meet at $Q$, and the lines $AB$ and $CD$ meet at $R$. Prove that the lines $PQ$ and $PR$ form the same angle as the diagonals of $ABCD$.

Solution

Since $A B C D$ is a cyclic quadrilateral, opposite angles sum to $180^{\circ}$ : \[\angle A+\angle C=180^{\circ} \text { and } \angle B+\angle D=180^{\circ}\]

Point $P$ ensures that $\angle P A B=\angle P B C=\angle P C D=\angle P D A=\theta$. Triangles $P A D$ and $P B C$ are perspective from $Q$, and triangles $P A B$ and $P C D$ are perspective from $R$. This implies the points $P, Q, R$ are collinear.

Hence, the lines $P Q$ and $P R$ form the same angle as the diagonals $A C$ and $B D$ : \[\boxed{\angle P Q P=\angle P R P=\angle A C B=\angle B D A}\]

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