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

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==Solution==
 
==Solution==
  
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Consider triangle <math>BCG</math>. <math>BG = 1</math> and <math>BC = 2</math>. <math>\angle CBG = 120 ^{\circ}</math> because it is an interior angle of a regular hexagon. 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. By the [[Law of Cosines]], we have:  
  
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<cmath> CG^2 = BG^2 + BC^2 - 2 \cdot BG \cdot BC \cdot \cos \angle CBG</cmath>
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<cmath>\begin{align*}
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FG^2 &= AG^2 + AF^2 - 2 \cdot AG \cdot AF \cdot \cos \angle GAF \\
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FG^2 &= 1^2 + 2^2 - 2 \cdot 1 \cdot 2 \cdot \cos 120 ^{\circ} \\
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FG^2 &= 1^2 + 2^2 - 4 \cdot \left( \frac 12 \right) \\
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FG^2 &= 7 \\
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FG &= \sqrt 7.
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\end{align*}</cmath>
  
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<cmath>CG^2 = 1^2 + 2^2 - 2 \cdot 1 \cdot 2 \cdot \cos 120 ^{\circ}</cmath>
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By [[SAS Congruence]], triangles <math>AFG</math>, <math>BCG</math>, <math>CDH</math>, and <math>EFH</math> are congruent, and by [[CPCTC]], quadrilateral <math>GCHF</math> is a rhombus. Therefore, its perimeter is <math>4 \cdot FG = \boxed{\textbf{(D)} \ 4 \sqrt 7}</math>.
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<cmath>CG = \sqrt 7.</cmath>
 
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By [[SAS Congruence]], triangles <math>BCG</math>, <math>CDH</math>, <math>HEF</math>, and <math>AFG</math> are congruent, and by [[CPCTC]], quadrilateral <math>GCHF</math> is a rhombus. Therefore, its perimeter is <math>4 \cdot CG = \boxed{\textbf{(D)} \ 4 \sqrt 7}</math>.
 
  
 
== See Also ==
 
== See Also ==

Revision as of 19:21, 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. 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 $4 \cdot FG = \boxed{\textbf{(D)} \ 4 \sqrt 7}$.

See Also

2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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