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

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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>  
 
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>  
  
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By the Law of Cosines,  
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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>
  
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<cmath>FG=\sqrt{1^2+2^2-2\cdot{1}\cdot{-\frac{1}{2}}}=\sqrt{7}.</cmath>
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-Benedict T (countmath1)
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== Solution 5 (Answer Choices + Pythagorean Theorem Extension) ==
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Like the previous solutions, note that <math>\triangle{GAF}, \triangle{FEH}, \triangle{HDC},</math> and <math>\triangle{CBG}</math> are all congruent by SAS. It follows that quadrilateral <math>GCHF</math> is a rhombus.
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Recall the Pythagorean Theorem, which states <math>a^2+b^2=c^2</math> for all right triangles, where <math>c</math> is the hypotenuse of the triangle. However, by drawing a quick diagram of an obtuse triangle, we can see that <math>a^2+b^2<c^2</math>, in any given obtuse triangle.
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Since <math>ABCDEF</math> is a regular hexagon, all of its angles are obtuse. It follows that <math>\triangle{GAF}</math> is an obtuse triangle. Using the extended Pythagorean Theorem for obtuse triangles, we have: <cmath>AG^2+AF^2<GF^2 \implies 1^2+2^2<GF^2 \implies \sqrt{5}<GF</cmath>
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Since <math>GCHF</math> is a rhombus, the perimeter is <math>4GF</math>. This eliminates all answer choices but <math>D</math> and <math>E</math>, since in all of those options <math>GF<\sqrt{5}</math>. Lastly, <math>E</math> is eliminated due to the triangle inequality, as <math>1+2</math> is not greater than <math>12/4=3</math>.
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Hence, the answer is <math>\boxed{\textbf{(D)}\ 4\sqrt7}</math>.
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~SwordOfJustice
  
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Therefore, <math>FG+GC+CH+HF=4\cdot{FG}=\boxed{\textbf{(D)}\ 4\sqrt7}.</math>
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==Video Solution 1 by mop 2024==
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https://youtu.be/ezGvZgBLe8k&t=0s
  
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-Benedict T (countmath1)
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~r00tsOfUnity
  
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==Video Solution 1==
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==Video Solution 2 (Under 1 min!)==
−
https://youtu.be/pIdM5l3CyUY
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https://youtu.be/XUIm4yOwVd0
  
 
~Education, the Study of Everything
 
~Education, the Study of Everything
  
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==Video Solution 2 by SpreadTheMathLove==
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==Video Solution 3 by SpreadTheMathLove==
 
https://www.youtube.com/watch?v=6I3ZNpI7qwE
 
https://www.youtube.com/watch?v=6I3ZNpI7qwE
  

Latest revision as of 21:16, 27 October 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)

Solution 5 (Answer Choices + Pythagorean Theorem Extension)

Like the previous solutions, note that $\triangle{GAF}, \triangle{FEH}, \triangle{HDC},$ and $\triangle{CBG}$ are all congruent by SAS. It follows that quadrilateral $GCHF$ is a rhombus.

Recall the Pythagorean Theorem, which states $a^2+b^2=c^2$ for all right triangles, where $c$ is the hypotenuse of the triangle. However, by drawing a quick diagram of an obtuse triangle, we can see that $a^2+b^2<c^2$, in any given obtuse triangle.

Since $ABCDEF$ is a regular hexagon, all of its angles are obtuse. It follows that $\triangle{GAF}$ is an obtuse triangle. Using the extended Pythagorean Theorem for obtuse triangles, we have: \[AG^2+AF^2<GF^2 \implies 1^2+2^2<GF^2 \implies \sqrt{5}<GF\]

Since $GCHF$ is a rhombus, the perimeter is $4GF$. This eliminates all answer choices but $D$ and $E$, since in all of those options $GF<\sqrt{5}$. Lastly, $E$ is eliminated due to the triangle inequality, as $1+2$ is not greater than $12/4=3$.

Hence, the answer is $\boxed{\textbf{(D)}\ 4\sqrt7}$.

~SwordOfJustice

Video Solution 1 by mop 2024

https://youtu.be/ezGvZgBLe8k&t=0s

~r00tsOfUnity

Video Solution 2 (Under 1 min!)

https://youtu.be/XUIm4yOwVd0

~Education, the Study of Everything

Video Solution 3 by SpreadTheMathLove

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

Video Solution(1-16)

https://youtu.be/SCwQ9jUfr0g

~~Hayabusa1

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

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