Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2022 AMC 12B Problems/Problem 10"
m (→‎Solution)
Line 23: Line 23:
 
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>.
 
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>.
  
−
Note: The sum of the internal angles of any polygon with <math>n</math> sides is given by <math>180 ^{\circ} (n - 2)</math>. Therefore, the sum of the internal angles of a hexagon is <math>720 ^{\circ}</math>, and each internal angle of a regular hexagon measures <math>\frac{720 ^{\circ}}{6} = 120 ^{\circ}</math>.
+
Note: The sum of the interior angles of any polygon with <math>n</math> sides is given by <math>180 ^{\circ} (n - 2)</math>. Therefore, the sum of the interior angles of a hexagon is <math>720 ^{\circ}</math>, and each interior angle of a regular hexagon measures <math>\frac{720 ^{\circ}}{6} = 120 ^{\circ}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 18:28, 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. 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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12B_Problems/Problem_10&oldid=182034"