# 2022 AMC 12B Problems/Problem 10

## 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 $GCHF$? $\textbf{(A)}\ 4\sqrt3 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 4\sqrt5 \qquad \textbf{(D)}\ 4\sqrt7 \qquad \textbf{(E)}\ 12$

## Diagram $[asy] /* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, E, F, G, H; A = dir(120); B = dir(60); C = dir(0); D = dir(-60); E = dir(-120); F = dir(180); G = midpoint(A--B); H = midpoint(D--E); filldraw(G--C--H--F--cycle,yellow); draw(polygon(6)); dot("A",A,1.5*A,linewidth(4)); dot("B",B,1.5*B,linewidth(4)); dot("C",C,1.5*C,linewidth(4)); dot("D",D,1.5*D,linewidth(4)); dot("E",E,1.5*E,linewidth(4)); dot("F",F,1.5*F,linewidth(4)); dot("G",G,1.5*dir(90),linewidth(4)); dot("H",H,1.5*dir(-90),linewidth(4)); [/asy]$ ~MRENTHUSIASM

## Solution 1

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, $CO = 2$, 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 $\sqrt{7} + \sqrt{7} + \sqrt{7} + \sqrt{7} = \boxed{\textbf{(D)}\ 4\sqrt7}$.

~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 $4 \cdot FG = \boxed{\textbf{(D)}\ 4\sqrt7}$.

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 $\frac{720 ^{\circ}}{6} = 120 ^{\circ}$.

## Solution 3

We use a coordinates approach. Letting the origin be the center of the hexagon, we can let $A = (-1, \sqrt{3}), B = (1, \sqrt{3}), C = (2, 0), D = (1, -\sqrt{3}), E = (-1, -\sqrt{3}), F = (-2, 0).$ Then, $G = (0, \sqrt{3})$ and $H = (0, -\sqrt{3}).$

We use the distance formula four times to get $CH, HF, FG, \text{ and } GC.$ \begin{alignat*}{8} CH^2 &= (2-0)^2 + (0-(-\sqrt{3}))^2 &&= 7 &&\rightarrow CH &&= \sqrt{7}, \\ HF^2 &= (0-(-2))^2 + (-\sqrt{3}-0)^2 &&= 7 &&\rightarrow HF &&= \sqrt{7}, \\ FG^2 &= (-2-0)^2 + (0-\sqrt{3})^2 &&= 7 &&\rightarrow FG &&= \sqrt{7}, \\ GC^2 &= (0-2)^2 + (\sqrt{3}-0)^2 &&= 7 &&\rightarrow GC &&= \sqrt{7}. \end{alignat*} Thus, the perimeter of $GCHF = CH + HF + FG + GC = \sqrt{7} + \sqrt{7} + \sqrt{7} + \sqrt{7} = \boxed{\textbf{(D)}\ 4\sqrt7}$.

~sirswagger21

Note: the last part of this solution could have been simplified by noting that $CH = HF = FG = GC = \sqrt{7}.$

## Solution 4

Note that triangles $\triangle{GAF}, \triangle{FEH}, \triangle{HDC},$ and $\triangle{CBG}$ are all congruent, since they have side lengths of $1$ and $2$ and an included angle of $120^{\circ}.$

By the Law of Cosines, $$FG=\sqrt{1^2+2^2-2\cdot{1}\cdot{-\frac{1}{2}}}=\sqrt{7}.$$ Therefore, $FG+GC+CH+HF=4\cdot{FG}=\boxed{\textbf{(D)}\ 4\sqrt7}.$

-Benedict T (countmath1)

## Video Solution 1

~Education, the Study of Everything

## Video Solution(1-16)

~~Hayabusa1

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