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</math> = 1 and <math>BC</math> = 2. <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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<cmath> 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.</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 17:45, 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 $BCG$. $BG$ = 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 $4 \cdot CG = \boxed{\textbf{(D)} \ 4 \sqrt 7}$.

See Also

2022 AMC 12B (ProblemsAnswer KeyResources)
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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