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Harvard-MIT Mathematics Tournament

Geometry Difficulty 4.9 AIME Prove it United States

Problem:
Let $ABC$ be a right triangle with $\angle A = 90^{\circ}$ . Let $D$ be the midpoint of $AB$ and let $E$ be a point on segment $AC$ such that $AD = AE$ . Let $BE$ meet $CD$ at $F$ . If $\angle BFC = 135^{\circ}$ , determine $BC / AB$ .

Solution

Solution:
Answer: $\frac{\sqrt{13}}{2}$

Let $\alpha = \angle ADC$ and $\beta = \angle ABE$ . By the exterior angle theorem, $\alpha = \angle BFD + \beta = 45^{\circ} + \beta$ . Also, note that $\tan \beta = AE / AB = AD / AB = 1/2$ . Thus,
$1 = \tan 45^{\circ} = \tan (\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} = \frac{\tan \alpha - \frac{1}{2}}{1 + \frac{1}{2} \tan \alpha}$
Solving for $\tan \alpha$ gives $\tan \alpha = 3$ . Therefore, $AC = 3 AD = \frac{3}{2} AB$ . Using the Pythagorean Theorem, we find that $BC = \frac{\sqrt{13}}{2} AB$ . So the answer is $\frac{\sqrt{13}}{2}$ .

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