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Difference between revisions of "2012 AMC 12B Problems/Problem 21"
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Let <math>a_1=EY=40</math>, <math>a_2=AF</math>, and <math>a_3=EF</math>.
 
Let <math>a_1=EY=40</math>, <math>a_2=AF</math>, and <math>a_3=EF</math>.
  
−
Drop a perpendicular line from <math>A</math> to the line of <math>EF</math> that meets line <math>EF</math> at <math>K</math>, and a perpendicular line from <math>Y</math> to the line of <math>EF</math> that meets <math>EF</math> at <math>L</math>, then <math>\triangle AKZ</math> is congruent to <math>\triangle ZLY</math> since <math>\angle YLZ</math> is complementary to <math>\angle KZA</math>. Then we have the following equations:
+
Drop a perpendicular line from <math>A</math> to the line of <math>EF</math> that meets line <math>EF</math> at <math>K</math>, and a perpendicular line from <math>Y</math> to the line of <math>EF</math> that meets <math>EF</math> at <math>L</math>, then <math>\triangle AKZ</math> is congruent to <math>\triangle ZLY</math> since <math>\angle YZL</math> is complementary to <math>\angle KZA</math>. 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_2 = AK=ZL = ZE+\frac{1}{2} a_1</cmath>

Revision as of 11:56, 28 December 2012

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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