# 2012 AMC 12B Problems/Problem 21

## Problem

Square is inscribed in equiangular hexagon with on , on , and on . Suppose that , and . What is the side-length of the square?

## Solution

Extend and so that they meet at . Then , so and therefore is parallel to . Also, since is parallel and equal to , we get , hence is congruent to YEZYE=AB=40$.

Let$ (Error compiling LaTeX. ! Missing $ inserted.)a_1=EY=40a_2=AFa_3=EF$.

Drop a perpendicular line from$ (Error compiling LaTeX. ! Missing $ inserted.)AEFEFKYEFEFL\triangle AKZ\triangle ZLY\angle YLZ\angle KZA$. Then we have the following equations:

<cmath>\frac{\sqrt{3}}{2}a_2 = AK=ZL = ZE+\frac{1}{2} a_1</cmath> <cmath>\frac{\sqrt{3}}{2}a_1 = YL =ZK = ZF+\frac{1}{2} a_2</cmath>

The sum of these two yields that

<cmath>\frac{\sqrt{3}}{2}(a_1+a_2) = \frac{1}{2} (a_1+a_2) + ZE+ZF = \frac{1}{2} (a_1+a_2) + EF</cmath> <cmath>\frac{\sqrt{3}-1}{2}(a_1+a_2) = 41(\sqrt{3}-1)</cmath> <cmath>a_1+a_2=82</cmath> <cmath>a_2=82-40=42.</cmath>

So, we can now use the law of cosines in$ (Error compiling LaTeX. ! Missing $ inserted.)\triangle AGY$:

<cmath> 2AZ^2 = AY^2 = AG^2 + YG^2 - 2AG\cdot YG \cdot \cos 60^{\circ} = (a_2+a_3)^2 + (a_1+a_3)^2 - (a_2+a_3)(a_1+a_3)</cmath> <cmath> = (41\sqrt{3}+1)^2 + (41\sqrt{3}-1)^2 - (41\sqrt{3}+1)(41\sqrt{3}-1) = 6 \cdot 41^1 + 2 - 3 \cdot 41^2 + 1 = 3 (\cdot 41^2 + 1) = 3\cdot 1682</cmath> <cmath> AZ^2 = 3 \cdot 841 = 3 \cdot 29^2</cmath>

Therefore$ (Error compiling LaTeX. ! Missing $ inserted.)AZ = 29\sqrt{3} ... \framebox{A}$