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Difference between revisions of "2022 AMC 12B Problems/Problem 10"

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Consider triangle <math>AFG</math>. <math>AG = 1</math> and <math>AF = 2</math>. <math>\angle GAF = 120 ^{\circ}</math> because it is an interior angle of a regular hexagon.<sup>See note.</sup> By the [[Law of Cosines]], we have:  
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Consider triangle <math>AFG</math>. <math>AG = 1</math> and <math>AF = 2</math>. <math>\angle GAF = 120 ^{\circ}</math> because it is an interior angle of a regular hexagon.<sup>See note for details.</sup> By the [[Law of Cosines]], we have:  
  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}

Revision as of 18:29, 17 November 2022

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 $AFG$. $AG = 1$ and $AF = 2$. $\angle GAF = 120 ^{\circ}$ 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 $\frac{720 ^{\circ}}{6} = 120 ^{\circ}$.

See Also

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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