Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Find the answer

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 $BC/AB$ .

Your answer

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

Solution

Let $\alpha=\angle ADC$ and $\beta=\angle ABE$ . By 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, $AC=3AD=\frac{3}{2}AB$ . Using Pythagorean Theorem, we find that $BC=\frac{\sqrt{13}}{2}AB$ . So the answer is $\frac{\sqrt{13}}{2}$ .

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