Library / /829 of 1394

, 2023

Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:

Suppose $ABCD$ is a convex quadrilateral with $\angle ABD = 105^{\circ}$ , $\angle ADB = 15^{\circ}$ , $AC = 7$ , and $BC = CD = 5$ . Compute the sum of all possible values of $BD$ .

Solution

Solution:

Let $O$ be the circumcenter of triangle $ABD$ . By the inscribed angle theorem, $\angle AOC = 90^{\circ}$ and $\angle BOC = 60^{\circ}$ . Let $AO = BO = CO = x$ and $CO = y$ . By the Pythagorean theorem on triangle $AOC$ ,
$x^{2} + y^{2} = 49$
and by the Law of Cosines on triangle $BOC$ ,
$x^{2} - x y + y^{2} = 25$
It suffices to find the sum of all possible values of $BD = \sqrt{3} x$ .
Since the two conditions on $x$ and $y$ are both symmetric, the answer is equal to
$\sqrt{3}(x + y) = \sqrt{9(x^{2} + y^{2}) - 6(x^{2} - x y + y^{2})} = \sqrt{291}.$
It is easy to check that both solutions generate valid configurations.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.