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...") |
Anyu tsuruko (talk | contribs) (→Solution) |
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==Solution== | ==Solution== | ||
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+ | 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 16:03, 24 May 2024
Problem
The quadrilateral is inscribed in a circle. The point
lies in the interior of
, and
. The lines
and
meet at
, and the lines
and
meet at
. Prove that the lines
and
form the same angle as the diagonals of
.
Solution
Since is a cyclic quadrilateral, opposite angles sum to
:
Point ensures that
.
Triangles
and
are perspective from
, and triangles
and
are perspective from
. This implies the points
are collinear.
Hence, the lines and
form the same angle as the diagonals
and
: