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

Revision as of 14:29, 24 April 2023

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

https://youtu.be/pIdM5l3CyUY

~Education, the Study of Everything

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=6I3ZNpI7qwE

Video Solution(1-16)

https://youtu.be/SCwQ9jUfr0g

~~Hayabusa1

@@ Line 79: / Line 79: @@
 Note that triangles <math>\triangle{GAF}, \triangle{FEH}, \triangle{HDC},</math> and <math>\triangle{CBG}</math> are all congruent, since they have side lengths of <math>1</math> and <math>2</math> and an included angle of <math>120^{\circ}.</math>
-By the Law of Cosines,
+By the Law of Cosines, <cmath>FG=\sqrt{1^2+2^2-2\cdot{1}\cdot{-\frac{1}{2}}}=\sqrt{7}.</cmath> Therefore, <math>FG+GC+CH+HF=4\cdot{FG}=\boxed{\textbf{(D)}\ 4\sqrt7}.</math>
-<cmath>FG=\sqrt{1^2+2^2-2\cdot{1}\cdot{-\frac{1}{2}}}=\sqrt{7}.</cmath>
-Therefore, <math>FG+GC+CH+HF=4\cdot{FG}=\boxed{\textbf{(D)}\ 4\sqrt7}.</math>
 -Benedict T (countmath1)

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