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, 2024

Geometry Difficulty 5.2 AIME, harder Prove it United States

Problem:

Let $ABC$ be a triangle with $\angle BAC = 90^\circ$ . Let $D$ , $E$ , and $F$ be the feet of the altitude, angle bisector, and median from $A$ to $BC$ , respectively. If $DE = 3$ and $EF = 5$ , compute the length of $BC$ .

Solutions — 2

Solution 1

Since $F$ is the circumcenter of $\triangle ABC$ , we have that $AE$ bisects $\angle DAF$ . So by the angle bisector theorem, we can set $AD = 3x$ and $AF = 5x$ . Applying Pythagorean theorem to $\triangle ADE$ then gives
$(3x)^2 + (5+3)^2 = (5x)^2 \Longrightarrow x = 2$
So $AF = 5x = 10$ and $BC = 2AF = 20$ .

Solution 2

Let $BF = FC = x$ . We know that $\triangle BAD \sim \triangle ACD$ so $\frac{BA}{AC} = \frac{BD}{DA} = \frac{DA}{DC}$ and thus $\frac{BA}{AC} = \sqrt{\frac{BD}{DC}} = \sqrt{\frac{x-8}{x+8}}$ . By Angle Bisector Theorem, we also have $\frac{AB}{AC} = \frac{BE}{EC} = \frac{x-5}{x+5}$ , which means that
$\sqrt{\frac{x-8}{x+8}} = \frac{x-5}{x+5} \Longrightarrow (x-8)(x+5)^2 = (x+8)(x-5)^2$
which expands to
$x^3 + 2x^2 - 55x - 200 = x^3 - 2x^2 - 55x + 200 \Longrightarrow 4x^2 = 400$
This solves to $x = 10$ , and so $BC = 2x = 20$ .

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