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Difference between revisions of "2012 AMC 12B Problems/Problem 21"
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\qquad\textbf{(E)}\ 21\sqrt{6}</math>
 
\qquad\textbf{(E)}\ 21\sqrt{6}</math>
  
−
==Solution==
+
==Solution (Long)==
  
−
Extend <math>AF</math> and <math>YE</math> so that they meet at <math>G</math>. Then <math>\angle FEG=\angle GFE=60^{\circ}</math>, so <math>\angle FGE=60^{\circ}</math> and therefore <math>AB</math> is parallel to <math>YE</math>. Also, since <math>AX</math> is parallel and equal to <math>YZ</math>, we get <math>\angle BAX = \angle ZYE</math>, hence <math>\triangle ABX</math> is congruent to <math>\triangle </math>YEZ<math>. We now get </math>YE=AB=40<math>.
+
Extend <math>AF</math> and <math>YE</math> so that they meet at <math>G</math>. Then <math>\angle FEG=\angle GFE=60^{\circ}</math>, so <math>\angle FGE=60^{\circ}</math> and therefore <math>AB</math> is parallel to <math>YE</math>. Also, since <math>AX</math> is parallel and equal to <math>YZ</math>, we get <math>\angle BAX = \angle ZYE</math>, hence <math>\triangle ABX</math> is congruent to <math>\triangle YEZ</math>. We now get <math>YE=AB=40</math>.
  
−
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 YLZ</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>
Line 26: Line 26:
 
<cmath>a_2=82-40=42.</cmath>
 
<cmath>a_2=82-40=42.</cmath>
  
−
So, we can now use the law of cosines in </math>\triangle AGY<math>:
+
So, we can now use the law of cosines in <math>\triangle AGY</math>:
  
−
<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> 2AZ^2 = AY^2 = AG^2 + YG^2 - 2AG\cdot YG \cdot \cos 60^{\circ}</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> = (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)</cmath>
 +
<cmath> = 6 \cdot 41^2 + 2 - 3 \cdot 41^2 + 1 = 3 (41^2 + 1) = 3\cdot 1682</cmath>
 
<cmath> AZ^2 = 3 \cdot 841 = 3 \cdot 29^2</cmath>
 
<cmath> AZ^2 = 3 \cdot 841 = 3 \cdot 29^2</cmath>
  
−
Therefore </math>AZ = 29\sqrt{3} ... \framebox{A}$
+
Therefore <math>AZ = 29\sqrt{3} ... \framebox{A}</math>

Revision as of 22:17, 4 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 $\angle YLZ$ 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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