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# Difference between revisions of "2012 AMC 12B Problems/Problem 21"

## Problem

Square $AXYZ$ is inscribed in equiangular hexagon $ABCDEF$ with $X$ on $\overline{BC}$, $Y$ on $\overline{DE}$, and $Z$ on $\overline{EF}$. Suppose that $AB=40$, and $EF=41(\sqrt{3}-1)$. What is the side-length of the square?

$\textbf{(A)}\ 29\sqrt{3} \qquad\textbf{(B)}\ \frac{21}{2}\sqrt{2}+\frac{41}{2}\sqrt{3}\qquad\textbf{(C)}\ 20\sqrt{3}+16 \qquad\textbf{(D)}\ 20\sqrt{2}+13 \sqrt{3} \qquad\textbf{(E)}\ 21\sqrt{6}$

## Solution (Long)

Extend $AF$ and $YE$ so that they meet at $G$. Then $\angle FEG=\angle GFE=60^{\circ}$, so $\angle FGE=60^{\circ}$ and therefore $AB$ is parallel to $YE$. Also, since $AX$ is parallel and equal to $YZ$, we get $\angle BAX = \angle ZYE$, hence $\triangle ABX$ is congruent to $\triangle YEZ$. We now get $YE=AB=40$.

Let $a_1=EY=40$, $a_2=AF$, and $a_3=EF$.

Drop a perpendicular line from $A$ to the line of $EF$ that meets line $EF$ at $K$, and a perpendicular line from $Y$ to the line of $EF$ that meets $EF$ at $L$, then $\triangle AKZ$ is congruent to $\triangle ZLY$ since $\angle YZL$ is complementary to $\angle KZA$. Then we have the following equations:

$$\frac{\sqrt{3}}{2}a_2 = AK=ZL = ZE+\frac{1}{2} a_1$$ $$\frac{\sqrt{3}}{2}a_1 = YL =ZK = ZF+\frac{1}{2} a_2$$

The sum of these two yields that

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

So, we can now use the law of cosines in $\triangle AGY$:

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

Therefore $AZ = 29\sqrt{3} ... \framebox{A}$