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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 14"
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<math>\textbf{(A)} \: 4 \qquad \textbf{(B)} \: 4\sqrt3 \qquad \textbf{(C)} \: 12 \qquad \textbf{(D)} \: 18 \qquad \textbf{(E)} \: 12\sqrt3</math>
 
<math>\textbf{(A)} \: 4 \qquad \textbf{(B)} \: 4\sqrt3 \qquad \textbf{(C)} \: 12 \qquad \textbf{(D)} \: 18 \qquad \textbf{(E)} \: 12\sqrt3</math>
  
−
==Solution (Law of Cosines and Equilateral Triangle Area)==
+
==Solution 1 (Law of Cosines and Equilateral Triangle Area)==
  
 
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.  
 
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.  
Line 32: Line 32:
  
 
The total area of the hexagon is thrice the area of each isosceles triangle plus the area of the equilateral triangle, or <math>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}</math>. Hence, <math>s=2\sqrt{3}</math> and the perimeter is <math>6s=\boxed{\textbf{(E)} \: 12\sqrt{3}}</math>.
 
The total area of the hexagon is thrice the area of each isosceles triangle plus the area of the equilateral triangle, or <math>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}</math>. Hence, <math>s=2\sqrt{3}</math> and the perimeter is <math>6s=\boxed{\textbf{(E)} \: 12\sqrt{3}}</math>.
 +
 +
== Solution 2 ==
 +
We denote by <math>x</math> the side length.
 +
 +
Following from SAS, <math>\triangle ABF \cong \triangle CDB \cong \triangle EFD</math>.
 +
 +
Hence, <math>BF =BD = DF</math>. Hence, <math>\triangle BDF</math> is equilateral.
 +
 +
In <math>\triangle ABF</math>, by applying the law of cosines, we have
 +
<cmath>
 +
\begin{align*}
 +
BF^2 & = AB^2 + AF^2 - 2 AB \cdot AF \cdot \cos A \\
 +
& = x^2 + x^2 - 2x^2 \cos 30^\circ \\
 +
& = x^2 \left( 2 - \sqrt{3} \right) .
 +
\end{align*}
 +
</cmath>
 +
 +
Therefore,
 +
<cmath>
 +
\begin{align*}
 +
{\rm Area} \ ABCDEF
 +
& = {\rm Area} \ \triangle ABF + {\rm Area} \ \triangle CDB + {\rm Area} \ \triangle EFD
 +
+ {\rm Area} \ \triangle BDF \\
 +
& = 3  {\rm Area} \ \triangle ABF + {\rm Area} \ \triangle BDF \\
 +
& = 3 \cdot \frac{1}{2} AB \cdot AF \cdot \sin A
 +
+ \frac{\sqrt{3}}{4} BF^2 \\
 +
& = 3 \cdot \frac{1}{2} x^2 \sin 30^\circ
 +
+ \frac{\sqrt{3}}{4} x^2 \left( 2 - \sqrt{3} \right) \\
 +
& = \frac{\sqrt{3}}{2} x^2 .
 +
\end{align*}
 +
</cmath>
 +
 +
Because <math>{\rm Area} \ ABCDEF = 6 \sqrt{3}</math>, <math>x = 2 \sqrt{3}</math>.
 +
Therefore, the perimeter of <math>ABCDEF</math> is <math>6 x = 12 \sqrt{3}</math>.
 +
 +
Therefore, the answer is <math>\boxed{\textbf{(E) }12 \sqrt{3}}</math>.
 +
 +
~Steven Chen (www.professorchenedu.com)
 +
  
 
{{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}}

Revision as of 20:46, 25 November 2021

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 (Law of Cosines and Equilateral Triangle Area)

Isosceles triangles $ABF$, $CBD$, and $EDF$ 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 $\frac{1}{2}s \cdot s \cdot \sin{30}=\frac{1}{4}s^2$ by the fourth formula here.

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 $\frac{\sqrt{3}}{4}\left(2s^2-\sqrt{3}s^2\right)=\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2$.

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

Solution 2

We denote by $x$ the side length.

Following from SAS, $\triangle ABF \cong \triangle CDB \cong \triangle EFD$.

Hence, $BF =BD = DF$. Hence, $\triangle BDF$ is equilateral.

In $\triangle ABF$, by applying the law of cosines, we have \begin{align*} BF^2 & = AB^2 + AF^2 - 2 AB \cdot AF \cdot \cos A \\ & = x^2 + x^2 - 2x^2 \cos 30^\circ \\ & = x^2 \left( 2 - \sqrt{3} \right) . \end{align*}

Therefore, \begin{align*} {\rm Area} \ ABCDEF & = {\rm Area} \ \triangle ABF + {\rm Area} \ \triangle CDB + {\rm Area} \ \triangle EFD + {\rm Area} \ \triangle BDF \\ & = 3 {\rm Area} \ \triangle ABF + {\rm Area} \ \triangle BDF \\ & = 3 \cdot \frac{1}{2} AB \cdot AF \cdot \sin A + \frac{\sqrt{3}}{4} BF^2 \\ & = 3 \cdot \frac{1}{2} x^2 \sin 30^\circ + \frac{\sqrt{3}}{4} x^2 \left( 2 - \sqrt{3} \right) \\ & = \frac{\sqrt{3}}{2} x^2 . \end{align*}

Because ${\rm Area} \ ABCDEF = 6 \sqrt{3}$, $x = 2 \sqrt{3}$. Therefore, the perimeter of $ABCDEF$ is $6 x = 12 \sqrt{3}$.

Therefore, the answer is $\boxed{\textbf{(E) }12 \sqrt{3}}$.

~Steven Chen (www.professorchenedu.com)


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