Library / /371 of 740

, 2013

Geometry Difficulty 5.0 AIME Prove it United States

Problem:

Let $ABC$ be an isosceles triangle with $AB = AC$ . Let $D$ and $E$ be the midpoints of segments $AB$ and $AC$ , respectively. Suppose that there exists a point $F$ on ray $\overrightarrow{DE}$ outside of $ABC$ such that triangle $BFA$ is similar to triangle $ABC$ . Compute $\frac{AB}{BC}$ .

Solution

Solution:

$\boxed{\sqrt{2}}$

Let $\alpha = \angle ABC = \angle ACB$ , $AB = 2x$ , and $BC = 2y$ , so $AD = DB = AE = EC = x$ and $DE = y$ . Since $\triangle BFA \sim \triangle ABC$ and $BA = AC$ , we in fact have $\triangle BFA \cong \triangle ABC$ , so $BF = BA = 2x$ , $FA = 2y$ , and $\angle DAF = \alpha$ . But $DE \parallel BC$ yields $\angle ADF = \angle ABC = \alpha$ as well, whence $\triangle FAD \sim \triangle ABC$ gives
$\frac{2y}{x} = \frac{FA}{AD} = \frac{AB}{BC} = \frac{2x}{2y} \Longrightarrow \frac{AB}{BC} = \frac{x}{y} = \sqrt{2}.$

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: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.