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

(→‎Solution 2: Removed repetitive solution. Prof. Chen agreed to this through PM ...)
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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 33: Line 33:
 
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 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);
 +
draw(D--G);
 +
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
 +
<cmath>6\sqrt3=[ACE]-3\cdot[DCE].</cmath>
 +
Letting <math>x=CD,</math> the perimeter will be <math>6x.</math>
 +
 +
We know that <math>\angle CDG=75^{\circ}</math> and using such, we have
 +
<cmath>CG=x\sin(75^{\circ})=\frac{\sqrt6+\sqrt2}{4}x</cmath>
 +
and
 +
<cmath>DG=x\cos(75^{\circ})=\frac{\sqrt6-\sqrt2}{4}x.</cmath>
 +
Thus, we have
 +
\[\begin{align*}[ACE]&=\frac{\sqrt3}{4}\left(2\cdotCG\right)^2\\
 +
&=\sqrt3(2+sqrt3)\end{align*}\]
 
== 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}}

Revision as of 12:53, 6 January 2022

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 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); draw(D--G); 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 \[6\sqrt3=[ACE]-3\cdot[DCE].\] Letting $x=CD,$ the perimeter will be $6x.$

We know that $\angle CDG=75^{\circ}$ and using such, we have \[CG=x\sin(75^{\circ})=\frac{\sqrt6+\sqrt2}{4}x\] and \[DG=x\cos(75^{\circ})=\frac{\sqrt6-\sqrt2}{4}x.\] Thus, we have \[\begin{align*}[ACE]&=\frac{\sqrt3}{4}\left(2\cdotCG\right)^2\\ &=\sqrt3(2+sqrt3)\end{align*}\]

See Also

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