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Problem 929

AMC 12 late, AIME early

Geometry Difficulty 4.7 Find the answer HMMT November

In convex quadrilateral $ABCD$ with $AB=11$ and $CD=13$ , there is a point $P$ for which $\triangle ADP$ and $\triangle BCP$ are congruent equilateral triangles. Compute the side length of these triangles.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Evidently $ABCD$ is an isosceles trapezoid with $P$ as its circumcenter. Now, construct isosceles trapezoid $ABB'C$ (that is, $BB'$ is parallel to $AC$ .) Then $AB'PD$ is a rhombus, so $\angle B'CD=\frac{1}{2} \angle B'PD=60^{\circ}$ by the inscribed angle theorem. Also, $B'C=11$ because the quadrilateral $B'APC$ is a $60^{\circ}$ rotation of $ADPB$ about $P$ . Since $CD=13$ , we use the law of cosines to get that $B'D=7\sqrt{3}$ . Hence $AP=7$ .

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.