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 $BCG$ . $BG = 1$ and $BC = 2$ . $\angle CBG = 120 ^{\circ}$ because it is an interior angle of a regular hexagon. By the Law of Cosines, we have:

$CG^2 = BG^2 + BC^2 - 2 \cdot BG \cdot BC \cdot \cos \angle CBG$

$CG^2 = 1^2 + 2^2 - 2 \cdot 1 \cdot 2 \cdot \cos 120 ^{\circ}$

$CG = \sqrt 7.$

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

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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