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

(→‎Solution 3)
(Improved formatting, and made a common reference diagram.)
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\textbf{(E)}\ 12</math>
 
\textbf{(E)}\ 12</math>
  
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==Solution 1 (No trig)==
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==Diagram==
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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 = 2</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>
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<asy>
 +
/* Made by MRENTHUSIASM */
 +
 
 +
size(200);
 +
pair A, B, C, D, E, F, G, H;
 +
A = dir(120);
 +
B = dir(60);
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C = dir(0);
 +
D = dir(-60);
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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));
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dot("$F$",F,1.5*F,linewidth(4));
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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 <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 = 2</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
 
~mathboy100
Line 19: Line 48:
  
 
By the [[Law of Cosines]], we have:  
 
By the [[Law of Cosines]], we have:  
−
 
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
FG^2 &= AG^2 + AF^2 - 2 \cdot AG \cdot AF \cdot \cos \angle GAF \\
 
FG^2 &= AG^2 + AF^2 - 2 \cdot AG \cdot AF \cdot \cos \angle GAF \\
Line 27: Line 55:
 
FG &= \sqrt 7.
 
FG &= \sqrt 7.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
 
 
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, the perimeter of <math>GCHF</math> 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, the perimeter of <math>GCHF</math> is <math>4 \cdot FG = \boxed{\textbf{(D)} \ 4 \sqrt 7}</math>.
  
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== Solution 3 ==
 
== Solution 3 ==
  
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[[Image:2022 AMC 12B-10.PNG|thumb|none|509px]]
 
 
We use a coordinates approach. Letting the origin be the center of the hexagon, we can let <math>A = (-1, \sqrt{3}), B = (1, \sqrt{3}), C = (2, 0), D = (1, -\sqrt{3}), E = (-1, -\sqrt{3}), F = (-2, 0).</math> Then, <math>G = (0, \sqrt{3})</math> and <math>H = (0, -\sqrt{3}).</math>
 
We use a coordinates approach. Letting the origin be the center of the hexagon, we can let <math>A = (-1, \sqrt{3}), B = (1, \sqrt{3}), C = (2, 0), D = (1, -\sqrt{3}), E = (-1, -\sqrt{3}), F = (-2, 0).</math> Then, <math>G = (0, \sqrt{3})</math> and <math>H = (0, -\sqrt{3}).</math>
  
 
We use the distance formula four times to get <math>CH, HF, FG, \text{ and } GC.</math>
 
We use the distance formula four times to get <math>CH, HF, FG, \text{ and } GC.</math>
−
 
+
<cmath>\begin{alignat*}{8}
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<cmath>CH^2 = (2-0)^2 + (0-(-\sqrt{3}))^2 = 7 \rightarrow CH = \sqrt{7}</cmath>
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CH^2 &= (2-0)^2 + (0-(-\sqrt{3}))^2 &&= 7 &&\rightarrow CH &&= \sqrt{7}, \\
−
<cmath>HF^2 = (0-(-2))^2 + (-\sqrt{3}-0)^2 = 7 \rightarrow HF = \sqrt{7}</cmath>
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HF^2 &= (0-(-2))^2 + (-\sqrt{3}-0)^2 &&= 7 &&\rightarrow HF &&= \sqrt{7}, \\
−
<cmath>FG^2 = (-2-0)^2 + (0-\sqrt{3})^2 = 7 \rightarrow FG = \sqrt{7}</cmath>
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FG^2 &= (-2-0)^2 + (0-\sqrt{3})^2 &&= 7 &&\rightarrow FG &&= \sqrt{7}, \\
−
<cmath>GC^2 = (0-2)^2 + (\sqrt{3}-0)^2 = 7 \rightarrow GC = \sqrt{7}</cmath>
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GC^2 &= (0-2)^2 + (\sqrt{3}-0)^2 &&= 7 &&\rightarrow GC &&= \sqrt{7}.
−
 
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\end{alignat*}</cmath>
 
Thus, the perimeter of <math>GCHF = CH + HF + FG + GC = \sqrt{7} + \sqrt{7} + \sqrt{7} + \sqrt{7} = \boxed{\textbf{(D)} \ 4\sqrt{7}}</math>.
 
Thus, the perimeter of <math>GCHF = CH + HF + FG + GC = \sqrt{7} + \sqrt{7} + \sqrt{7} + \sqrt{7} = \boxed{\textbf{(D)} \ 4\sqrt{7}}</math>.
  
Line 50: Line 76:
 
Note: the last part of this solution could have been simplified by noting that <math>CH = HF = FG = GC.</math>
 
Note: the last part of this solution could have been simplified by noting that <math>CH = HF = FG = GC.</math>
  
−
 
+
== Solution 4 ==
−
== Solution 4 (Fast) ==
 
 
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>  
  

Revision as of 17:43, 9 January 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 $\sqrt{7} + \sqrt{7} + \sqrt{7} + \sqrt{7} = \boxed{\textbf{(D)} \ 4 \sqrt 7}$.

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

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 $GCHF = CH + HF + FG + GC = \sqrt{7} + \sqrt{7} + \sqrt{7} + \sqrt{7} = \boxed{\textbf{(D)} \ 4\sqrt{7}}$.

~sirswagger21

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

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, $FG+GC+CH+HF=4\cdot{FG}=\boxed{\textbf{(D)} \ 4\sqrt{7}}.$

-Benedict T (countmath1)

Video Solution 1

https://youtu.be/pIdM5l3CyUY

~Education, the Study of Everything

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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