Competition problem — Geometry — Maths Olympiad Prep

Library / /364 of 841

Geometry Difficulty 5.1 AIME, harder Find the answer

Let $ABCD$ be a trapezoid with $AB \parallel CD$ and $\angle D=90^{\circ}$ . Suppose that there is a point $E$ on $CD$ such that $AE=BE$ and that triangles $AED$ and $CEB$ are similar, but not congruent. Given that $\frac{CD}{AB}=2014$ , find $\frac{BC}{AD}$ .

Your answer

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

Solution

Let $M$ be the midpoint of $AB$ . Let $AM=MB=ED=a, ME=AD=b$ , and $AE=BE=c$ . Since $\triangle BEC \sim \triangle DEA$ , but $\triangle BEC$ is not congruent to $\triangle DAE$ , we must have $\triangle BEC \sim \triangle DEA$ . Thus, $BC / BE=AD / DE=b / a$ , so $BC=bc / a$ , and $CE / EB=AE / ED=c / a$ , so $EC=c^{2} / a$ . We are given that $CD / AB=\frac{c^{2}/a+a}{2a}=\frac{c^{2}}{2a^{2}}+\frac{1}{2}=2014 \Rightarrow \frac{c^{2}}{a^{2}}=4027$ . Thus, $BC / AD=\frac{bc / a}{b}=c / a=\sqrt{4027}$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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