Track / Stage 5 / 56 of 400 #656 of 2000

Problem 656

AIME late

Geometry Difficulty 5.2 Prove it HMMT February · United States

Problem:

Convex quadrilateral $ABCD$ satisfies $\angle CAB = \angle ADB = 30^{\circ}$ , $\angle ABD = 77^{\circ}$ , $BC = CD$ , and $\angle BCD = n^{\circ}$ for some positive integer $n$ . Compute $n$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Solution:

Let $O$ be the circumcenter of $\triangle ABD$ . From $\angle ADB = 30^{\circ}$ , we get that $\triangle AOB$ is equilateral. Moreover, since $\angle BAC = 30^{\circ}$ , we have that $AC$ bisects $\angle BAO$ , and thus must be the perpendicular bisector of $BO$ . Therefore, we have $CB = CD = CO$ , so $C$ is actually the circumcenter of $\triangle BDO$ . Hence,
$\begin{aligned} \angle BCD & = 2\left(180^{\circ} - \angle BOD\right) \\ & = 2\left(180^{\circ} - 2 \angle BAD\right) \\ & = 2\left(180^{\circ} - 146^{\circ}\right) = 68^{\circ} \end{aligned}$

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