2022 AMC 12B Problems/Problem 10

Problem

Regular hexagon $ABCDEF$ has side length $2$ . Let $G$ be the midpoint of $\overline{AB}$ , and let $H$ be the midpoint of $\overline{DE}$ . What is the perimeter of $GCHF$ ?

$\textbf{(A)}\ 4\sqrt3 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 4\sqrt5 \qquad \textbf{(D)}\ 4\sqrt7 \qquad \textbf{(E)}\ 12$

Solution

Consider triangle $AFG$ . Note that $AF = 2$ , $AG = 1$ , and $\angle GAF = 120 ^{\circ}$ because it is an interior angle of a regular hexagon.^{See note for details.}

By the Law of Cosines, we have:

$\begin{align*} FG^2 &= AG^2 + AF^2 - 2 \cdot AG \cdot AF \cdot \cos \angle GAF \\ FG^2 &= 1^2 + 2^2 - 2 \cdot 1 \cdot 2 \cdot \cos 120 ^{\circ} \\ FG^2 &= 1^2 + 2^2 - 4 \cdot \left( \frac 12 \right) \\ FG^2 &= 7 \\ FG &= \sqrt 7. \end{align*}$

By SAS Congruence, triangles $AFG$ , $BCG$ , $CDH$ , and $EFH$ are congruent, and by CPCTC, quadrilateral $GCHF$ is a rhombus. Therefore, its perimeter is $4 \cdot FG = \boxed{\textbf{(D)} \ 4 \sqrt 7}$ .

Note: The sum of the interior angles of any polygon with $n$ sides is given by $180 ^{\circ} (n - 2)$ . Therefore, the sum of the interior angles of a hexagon is $720 ^{\circ}$ , and each interior angle of a regular hexagon measures $\frac{720 ^{\circ}}{6} = 120 ^{\circ}$ .

2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AMC 12 Problems and Solutions

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2022 AMC 12B Problems/Problem 10

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