2012 AMC 12B Problems/Problem 21
Problem
Square is inscribed in equiangular hexagon
with
on
,
on
, and
on
. Suppose that
, and
. What is the side-length of the square?
Solution
Extend and
so that they meet at
. Then
, so
and therefore
is parallel to
. Also, since
is parallel and equal to
, we get
, hence
is congruent to
YEZ
YE=AB=40$.
Let$ (Error compiling LaTeX. Unknown error_msg)a_1=EY=40a_2=AF
a_3=EF$.
Drop a perpendicular line from$ (Error compiling LaTeX. Unknown error_msg)AEF
EF
K
Y
EF
EF
L
\triangle AKZ
\triangle ZLY
\angle YLZ
\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}$