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Berkeley Math Circle

Geometry Difficulty 4.6 AIME Prove it United States

Problem:

In convex hexagon $A X B Y C Z$ , sides $A X$ , $B Y$ and $C Z$ are parallel to diagonals $B C$ , $X C$ and $X Y$ , respectively. Prove that $\triangle A B C$ and $\triangle X Y Z$ have the same area.

Solution

Solution:

Let $[\mathcal{P}]$ denote the area of a polygon $\mathcal{P}$ .

The important claim is that if $\overline{K L} \parallel \overline{M N}$ , then $[K L M] = [K L N]$ . This is a simple consequence of the formula $A = \frac{1}{2} b h$ .

Then, we find that
$\begin{aligned} & [A B C] = [X B C] \quad (\text{since } \overline{A X} \parallel \overline{B C}) \\ & = [X Y C] \quad (\text{since } \overline{B Y} \parallel \overline{X C}) \\ & = [X Y Z] \quad (\text{since } \overline{C Z} \parallel \overline{X Y}) \end{aligned}$
as desired.

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