Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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...")
 
(→‎Solution)
 
Line 3: Line 3:
  
 
==Solution==
 
==Solution==
−
{{Solution}}
+
 
 +
Since <math>A B C D</math> is a cyclic quadrilateral, opposite angles sum to <math>180^{\circ}</math> :
 +
<cmath>
 +
\angle A+\angle C=180^{\circ} \text { and } \angle B+\angle D=180^{\circ}
 +
</cmath>
 +
 
 +
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>.
 +
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.
 +
 
 +
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> :
 +
<cmath>
 +
\boxed{\angle P Q P=\angle P R P=\angle A C B=\angle B D A}
 +
</cmath>

Latest revision as of 17:03, 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 $B D$ : \[\boxed{\angle P Q P=\angle P R P=\angle A C B=\angle B D A}\]

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_Canadian_MO_Problems/Problem_4&oldid=219641"