2021 Fall AMC 12A Problems/Problem 14

Problem

In the figure, equilateral hexagon $ABCDEF$ has three nonadjacent acute interior angles that each measure $30^\circ$ . The enclosed area of the hexagon is $6\sqrt{3}$ . What is the perimeter of the hexagon? $[asy] size(10cm); pen p=black+linewidth(1),q=black+linewidth(5); pair C=(0,0),D=(cos(pi/12),sin(pi/12)),E=rotate(150,D)*C,F=rotate(-30,E)*D,A=rotate(150,F)*E,B=rotate(-30,A)*F; draw(C--D--E--F--A--B--cycle,p); dot(A,q); dot(B,q); dot(C,q); dot(D,q); dot(E,q); dot(F,q); label("$C$",C,2*S); label("$D$",D,2*S); label("$E$",E,2*S); label("$F$",F,2*dir(0)); label("$A$",A,2*N); label("$B$",B,2*W); [/asy]$ $\textbf{(A)} \: 4 \qquad \textbf{(B)} \: 4\sqrt3 \qquad \textbf{(C)} \: 12 \qquad \textbf{(D)} \: 18 \qquad \textbf{(E)} \: 12\sqrt3$

Solution (Law of Cosines and Equilateral Triangle Area)

Isosceles triangles $ABF$ , $CBD$ , and $EDF$ are congruent by SAS congruence. By CPCTC, $BF=BD=DF$ , so triangle $BDF$ is equilateral.

Let the side length of the hexagon be $s$ .

The area of each isosceles triangle is $\frac{1}{2}s \cdot s \cdot \sin{30}=\frac{1}{4}s^2$ by the fourth formula here.

By the Law of Cosines on triangle $ABF$ , $BF^2=s^2+s^2-2s^2\cos{30^\circ}=2s^2-\sqrt{3}s^2$ . Hence, the area of the equilateral triangle $BDF$ is $\frac{\sqrt{3}}{4}\left(2s^2-\sqrt{3}s^2\right)=\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2$ .

The total area of the hexagon is thrice the area of each isosceles triangle plus the area of the equilateral triangle, or $3\left(\frac{1}{4}s^2\right)+\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2=\frac{\sqrt{3}}{2}s^2=6\sqrt{3}$ . Hence, $s=2\sqrt{3}$ and the perimeter is $6s=\boxed{\textbf{(E)} \: 12\sqrt{3}}$ .

Solution 2

We denote by $x$ the side length.

Following from SAS, $\triangle ABF \cong \triangle CDB \cong \triangle EFD$ .

Hence, $BF =BD = DF$ . Hence, $\triangle BDF$ is equilateral.

In $\triangle ABF$ , by applying the law of cosines, we have $\begin{align*} BF^2 & = AB^2 + AF^2 - 2 AB \cdot AF \cdot \cos A \\ & = x^2 + x^2 - 2x^2 \cos 30^\circ \\ & = x^2 \left( 2 - \sqrt{3} \right) . \end{align*}$

Therefore, $\begin{align*} {\rm Area} \ ABCDEF & = {\rm Area} \ \triangle ABF + {\rm Area} \ \triangle CDB + {\rm Area} \ \triangle EFD + {\rm Area} \ \triangle BDF \\ & = 3 {\rm Area} \ \triangle ABF + {\rm Area} \ \triangle BDF \\ & = 3 \cdot \frac{1}{2} AB \cdot AF \cdot \sin A + \frac{\sqrt{3}}{4} BF^2 \\ & = 3 \cdot \frac{1}{2} x^2 \sin 30^\circ + \frac{\sqrt{3}}{4} x^2 \left( 2 - \sqrt{3} \right) \\ & = \frac{\sqrt{3}}{2} x^2 . \end{align*}$

Because ${\rm Area} \ ABCDEF = 6 \sqrt{3}$ , $x = 2 \sqrt{3}$ . Therefore, the perimeter of $ABCDEF$ is $6 x = 12 \sqrt{3}$ .

Therefore, the answer is $\boxed{\textbf{(E) }12 \sqrt{3}}$ .

~Steven Chen (www.professorchenedu.com)

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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All AMC 12 Problems and Solutions

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2021 Fall AMC 12A Problems/Problem 14

Problem

Solution (Law of Cosines and Equilateral Triangle Area)

Solution 2