Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2012 AMC 12B Problems/Problem 21"
Line 7: Line 7:
 
<math>\textbf{(D)}\ 20\sqrt{2}+13\sqrt{3}  
 
<math>\textbf{(D)}\ 20\sqrt{2}+13\sqrt{3}  
 
\qquad\textbf{(E)}\ 21\sqrt{6}</math>
 
\qquad\textbf{(E)}\ 21\sqrt{6}</math>
 +
 +
==Solution==
 +
 +
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>.
 +
 +
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:
 +
 +
<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 </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> = (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 </math>AZ = 29\sqrt{3} ... \framebox{A}$

Revision as of 22:15, 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

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. Unknown error_msg)a_1=EY=40$,$a_2=AF$, and$a_3=EF$.

Drop a perpendicular line from$ (Error compiling LaTeX. Unknown error_msg)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. Unknown error_msg)\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. Unknown error_msg)AZ = 29\sqrt{3} ... \framebox{A}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_21&oldid=49787"