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?

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

$\textbf{(D)}\ 20\sqrt{2}+13\sqrt{3} \qquad\textbf{(E)}\ 21\sqrt{6}$

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

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2012 AMC 12B Problems/Problem 21

Problem

Solution (Long)