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Harvard-MIT Math Tournament

Geometry Difficulty 4.9 AIME Prove it United States

Problem:

$ABC$ is a triangle with points $E, F$ on sides $AC, AB$ , respectively. Suppose that $BE, CF$ intersect at $X$ . It is given that $AF / FB = (AE / EC)^2$ and that $X$ is the midpoint of $BE$ . Find the ratio $CX / XF$ .

Solution

Solution:

Let $x = AE / EC$ . By Menelaus's theorem applied to triangle $ABE$ and line $CXF$ ,
$1 = \frac{AF}{FB} \cdot \frac{BX}{XE} \cdot \frac{EC}{CA} = \frac{x^2}{x+1}.$
Thus, $x^2 = x + 1$ , and $x$ must be positive, so $x = (1 + \sqrt{5}) / 2$ .

Now apply Menelaus to triangle $ACF$ and line $BXE$ , obtaining
$1 = \frac{AE}{EC} \cdot \frac{CX}{XF} \cdot \frac{FB}{BA} = \frac{CX}{XF} \cdot \frac{x}{x^2 + 1},$
so $CX / XF = (x^2 + 1) / x = (2x^2 - x) / x = 2x - 1 = \sqrt{5}$ .

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