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, 2014

Geometry Difficulty 4.9 AIME Prove it United States

Problem:

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 $\frac{BC}{AD}$ .

Solution

Solution:

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

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