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2012 AMC 12B Problems/Problem 21

Revision as of 22:15, 4 December 2012 by Lightest (talk | contribs)


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?

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

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


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$ (Error compiling LaTeX. ! Missing $ inserted.)a_1=EY=40$,$a_2=AF$, and$a_3=EF$.

Drop a perpendicular line from$ (Error compiling LaTeX. ! Missing $ inserted.)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 YLZ$is complementary to$\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}$

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