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

Revision as of 20:09, 19 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 1 (No trig)

Let the center of the hexagon be $O$ . $\triangle AOB$ , $\triangle BOC$ , $\triangle COD$ , $\triangle DOE$ , $\triangle EOF$ , and $\triangle FOA$ are all equilateral triangles with side length $2$ . Thus, $CO = 1$ , and $GO = \sqrt{AO^2 - AG^2} = \sqrt{3}$ . By symmetry, $\angle COG = 90^{\circ}$ . Thus, by the Pythagorean theorem, $CG = \sqrt{2^2 + \sqrt{3}^2} = \sqrt{7}$ . Because $CO = OF$ and $GO = OH$ , $CG = HC = FH = GF = \sqrt{7}$ . Thus, the solution to our problem is $\sqrt{7} + \sqrt{7} + \sqrt{7} + \sqrt{7} = \boxed{\textbf{(D)} \ 4 \sqrt 7}$

~mathboy100

Solution 2

Consider triangle $AFG$ . Note that $AF = 2$ , $AG = 1$ , and $\angle GAF = 120 ^{\circ}$ because it is an interior angle of a regular hexagon. (See note for details.)

By the Law of Cosines, we have:

$\begin{align*} FG^2 &= AG^2 + AF^2 - 2 \cdot AG \cdot AF \cdot \cos \angle GAF \\ FG^2 &= 1^2 + 2^2 - 2 \cdot 1 \cdot 2 \cdot \cos 120 ^{\circ} \\ FG^2 &= 5 + 4 \cdot \left( \frac 12 \right) \\ FG^2 &= 7 \\ FG &= \sqrt 7. \end{align*}$

By SAS Congruence, triangles $AFG$ , $BCG$ , $CDH$ , and $EFH$ are congruent, and by CPCTC, quadrilateral $GCHF$ is a rhombus. Therefore, the perimeter of $GCHF$ 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 $\frac{720 ^{\circ}}{6} = 120 ^{\circ}$ .

Video Solution 1

https://youtu.be/pIdM5l3CyUY

~Education, the Study of Everything

@@ Line 9: / Line 9: @@
 \textbf{(E)}\ 12</math>
-==Solution==
+==Solution 1 (No trig)==
+Let the center of the hexagon be <math>O</math>. <math>\triangle AOB</math>, <math>\triangle BOC</math>, <math>\triangle COD</math>, <math>\triangle DOE</math>, <math>\triangle EOF</math>, and <math>\triangle FOA</math> are all equilateral triangles with side length <math>2</math>. Thus, <math>CO = 1</math>, and <math>GO = \sqrt{AO^2 - AG^2} = \sqrt{3}</math>. By symmetry, <math>\angle COG = 90^{\circ}</math>. Thus, by the Pythagorean theorem, <math>CG = \sqrt{2^2 + \sqrt{3}^2} = \sqrt{7}</math>. Because <math>CO = OF</math> and <math>GO = OH</math>, <math>CG = HC = FH = GF = \sqrt{7}</math>. Thus, the solution to our problem is <math>\sqrt{7} + \sqrt{7} + \sqrt{7} + \sqrt{7} = \boxed{\textbf{(D)} \ 4 \sqrt 7}</math>
+~mathboy100
+==Solution 2==
 Consider triangle <math>AFG</math>. Note that <math>AF = 2</math>, <math>AG = 1</math>, and <math>\angle GAF = 120 ^{\circ}</math> because it is an interior angle of a regular hexagon. (See note for details.)

2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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