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

Revision as of 00:59, 26 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 (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 $6s=\boxed{\textbf{(E)} \: 12\sqrt{3}}$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ 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>.
-== Solution 2 ==
+== See Also ==
-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}}
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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 14"

Revision as of 00:59, 26 November 2021

Problem

Solution (Law of Cosines and Equilateral Triangle Area)

See Also