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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 14"

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==Solution 1==
 
==Solution 1==
  
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Isosceles triangles <math>ABF</math>, <math>CBD</math>, and <math>EDF</math> are congruent by [[Congruent_(geometry)#SAS_Congruence|SAS congruence]]. By [[CPCTC]], <math>BF=BD=DF</math>, so triangle <math>BDF</math> is equilateral.  
+
Divide the equilateral hexagon <math>ABCDEF</math> into isosceles triangles <math>ABF</math>, <math>CBD</math>, and <math>EDF</math> and triangle <math>BDF</math>. The three isosceles triangles are congruent by SAS congruence. By CPCTC, <math>BF=BD=DF</math>, so triangle <math>BDF</math> is equilateral.  
  
−
Let the side length of the hexagon be <math>s</math>. The area of each isosceles triangle is <cmath>\frac{1}{2}s \cdot s \cdot \sin{30}=\frac{1}{4}s^2</cmath> by the fourth formula [[Area#Area_of_Triangle|here]].
+
Let the side length of the hexagon be <math>s</math>. The area of each isosceles triangle is <cmath>\frac{1}{2} a b \sin\angle C = \frac{1}{2} \cdot s \cdot s \cdot \sin{30^{\circ}} = \frac{1}{4}s^2.</cmath>
  
−
By the [[Law of Cosines]] on triangle <math>ABF</math>, <cmath>BF^2=s^2+s^2-2s^2\cos{30^\circ}=2s^2-\sqrt{3}s^2.</cmath> Hence, the [[Area_of_an_equilateral_triangle|area of the equilateral triangle]] <math>BDF</math> is <cmath>\frac{\sqrt{3}}{4}\left(2s^2-\sqrt{3}s^2\right)=\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2.</cmath>
+
By the [[Law of Cosines]] on triangle <math>ABF</math>, <cmath>BF^2=s^2+s^2-2s^2\cos{30^{\circ}}=2s^2-\sqrt{3}s^2.</cmath>  
−
The total area of the hexagon is thrice the area of each isosceles triangle plus the area of the equilateral triangle, or <cmath>3\left(\frac{1}{4}s^2\right)+\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2=\frac{\sqrt{3}}{2}s^2=6\sqrt{3}.</cmath> Hence, <math>s=2\sqrt{3}</math> and the perimeter is <math>6s=\boxed{\textbf{(E)} \: 12\sqrt3}</math>.
+
 
 +
Hence, the [[Area_of_an_equilateral_triangle|area of the equilateral triangle]] <math>BDF</math> is <cmath>\frac{\sqrt{3}}{4} BF^2 = \frac{\sqrt{3}}{4}\left(2s^2-\sqrt{3}s^2\right)=\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2.</cmath>
 +
 
 +
The total area of the hexagon is thrice the area of each isosceles triangle plus the area of the equilateral triangle, or <cmath>3\left(\frac{1}{4}s^2\right)+ \left( \frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2 \right)=\frac{\sqrt{3}}{2}s^2=6\sqrt{3}.</cmath> Hence, <math>s=2\sqrt{3}</math>, and the perimeter of the hexagon is <math>6s=\boxed{\textbf{(E)} \: 12\sqrt3}</math>.
  
 
==Solution 2 ==
 
==Solution 2 ==
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pair C=(0,0),D=(cos(pi/12),sin(pi/12)),E=rotate(150,D)*C,F=rotate(-30,E)*D,A=rotate(150,F)*E,B=rotate(-30,A)*F,G=(1/2)*(C+E);
 
pair C=(0,0),D=(cos(pi/12),sin(pi/12)),E=rotate(150,D)*C,F=rotate(-30,E)*D,A=rotate(150,F)*E,B=rotate(-30,A)*F,G=(1/2)*(C+E);
 
draw(C--D--E--F--A--B--cycle,p);
 
draw(C--D--E--F--A--B--cycle,p);
−
draw(C--E--A--C);
+
draw(C--E--A--C,p+dashed);
−
draw(D--G);
+
draw(D--G,p+dashed);
 
dot(A,q);
 
dot(A,q);
 
dot(B,q);
 
dot(B,q);
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~ASAB
 
~ASAB
 +
==Solution 3==
 +
We will be using this diagram:
 +
 +
==Video Solution by TheBeautyofMath==
 +
Cleverly done with Trig
 +
https://youtu.be/AgSE7HPCVR0
 +
 +
==Video Solution (Logic and Trigonometry)==
 +
https://youtu.be/4j8e47S1cr0
 +
 +
~Education, the Study of Everything
 +
 +
~IceMatrix
  
 +
==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=g6Dk6An2ALY&t=208s
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2021 Fall|ab=A|num-b=13|num-a=15}}
 
{{AMC12 box|year=2021 Fall|ab=A|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:16, 1 November 2023

Problem

In the figure, equilateral hexagon $ABCDEF$ has three nonadjacent acute interior angles that each measure $30^\circ$. The enclosed area of the hexagon is $6\sqrt{3}$. What is the perimeter of the hexagon? [asy] size(10cm); pen p=black+linewidth(1),q=black+linewidth(5); pair C=(0,0),D=(cos(pi/12),sin(pi/12)),E=rotate(150,D)*C,F=rotate(-30,E)*D,A=rotate(150,F)*E,B=rotate(-30,A)*F; draw(C--D--E--F--A--B--cycle,p); dot(A,q); dot(B,q); dot(C,q); dot(D,q); dot(E,q); dot(F,q); label("$C$",C,2*S); label("$D$",D,2*S); label("$E$",E,2*S); label("$F$",F,2*dir(0)); label("$A$",A,2*N); label("$B$",B,2*W); [/asy] $\textbf{(A)} \: 4 \qquad \textbf{(B)} \: 4\sqrt3 \qquad \textbf{(C)} \: 12 \qquad \textbf{(D)} \: 18 \qquad \textbf{(E)} \: 12\sqrt3$

Solution 1

Divide the equilateral hexagon $ABCDEF$ into isosceles triangles $ABF$, $CBD$, and $EDF$ and triangle $BDF$. The three isosceles triangles are congruent by SAS congruence. By CPCTC, $BF=BD=DF$, so triangle $BDF$ is equilateral.

Let the side length of the hexagon be $s$. The area of each isosceles triangle is \[\frac{1}{2} a b \sin\angle C = \frac{1}{2} \cdot s \cdot s \cdot \sin{30^{\circ}} = \frac{1}{4}s^2.\]

By the Law of Cosines on triangle $ABF$, \[BF^2=s^2+s^2-2s^2\cos{30^{\circ}}=2s^2-\sqrt{3}s^2.\]

Hence, the area of the equilateral triangle $BDF$ is \[\frac{\sqrt{3}}{4} BF^2 = \frac{\sqrt{3}}{4}\left(2s^2-\sqrt{3}s^2\right)=\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2.\]

The total area of the hexagon is thrice the area of each isosceles triangle plus the area of the equilateral triangle, or \[3\left(\frac{1}{4}s^2\right)+ \left( \frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2 \right)=\frac{\sqrt{3}}{2}s^2=6\sqrt{3}.\] Hence, $s=2\sqrt{3}$, and the perimeter of the hexagon is $6s=\boxed{\textbf{(E)} \: 12\sqrt3}$.

Solution 2

We will be referring to the following diagram:

[asy] size(10cm); pen p=black+linewidth(1),q=black+linewidth(5); pair C=(0,0),D=(cos(pi/12),sin(pi/12)),E=rotate(150,D)*C,F=rotate(-30,E)*D,A=rotate(150,F)*E,B=rotate(-30,A)*F,G=(1/2)*(C+E); draw(C--D--E--F--A--B--cycle,p); draw(C--E--A--C,p+dashed); draw(D--G,p+dashed); dot(A,q); dot(B,q); dot(C,q); dot(D,q); dot(E,q); dot(F,q); dot(G,q); label("$C$",C,2*S); label("$D$",D,2*N); label("$E$",E,2*S); label("$F$",F,2*dir(0)); label("$A$",A,2*N); label("$G$",G,2*S); label("$B$",B,2*W); [/asy]

Observe that \begin{align}6\sqrt3=[ACE]-3\cdot[DCE].\end{align} Letting $x=CD,$ the perimeter will be $6x.$

We know that $\angle CDG=75^{\circ}$ and using such, we have \begin{alignat*}{8} CG &= x\sin(75^{\circ}) &&= \frac{\sqrt6+\sqrt2}{4}x, \\ DG &= x\cos(75^{\circ}) &&= \frac{\sqrt6-\sqrt2}{4}x. \end{alignat*} Thus, we have \begin{align*}[ACE]&=\frac{\sqrt3}{4}\left(2\cdot CG\right)^2\\ &=\frac{\sqrt3}{4}(2+\sqrt3)x^2 \\ &=\frac{3+2\sqrt3}{4} x^2.\end{align*} Computing the area of $DCE,$ we have \begin{align*}[DCE]&=\frac12 \cdot 2\cdot CG\cdot DG \\ &=CG\cdot DG\\ &=\frac{x^2}{4}.\end{align*} Plugging back into $(1),$ we have \[6\sqrt3=\frac{3+2\sqrt3}{4} x^2 -\frac{3x^2}{4}=\frac{\sqrt3}{2}x^2,\] which means $x=2\sqrt3$ and $6x=\boxed{\textbf{(E)} \: 12\sqrt3}.$

~ASAB

Solution 3

We will be using this diagram:

Video Solution by TheBeautyofMath

Cleverly done with Trig https://youtu.be/AgSE7HPCVR0

Video Solution (Logic and Trigonometry)

https://youtu.be/4j8e47S1cr0

~Education, the Study of Everything

~IceMatrix

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=g6Dk6An2ALY&t=208s

See Also

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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All AMC 12 Problems and Solutions

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