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Harvard-MIT Mathematics Tournament

Geometry Difficulty 4.8 AIME Prove it United States

Problem:

Convex quadrilateral $A B C D$ has sides $A B = B C = 7$ , $C D = 5$ , and $A D = 3$ . Given additionally that $m \angle A B C = 60^{\circ}$ , find $B D$ .

Solution

Solution:

Answer: $8$ . Triangle $A B C$ is equilateral, so $A C = 7$ as well. Now the law of cosines shows that $m \angle C D A = 120^{\circ}$ ; i.e., $A B C D$ is cyclic. Ptolemy's theorem now gives $A C \cdot B D = A B \cdot C D + A D \cdot B C$ , or simply $B D = C D + A D = 8$ .

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