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. | ||
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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>\begin{align}6\sqrt3=[ACE]-3\cdot[DCE]\end{align}.</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 | ||
+ | <cmath>\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*}</cmath> | ||
+ | Computing the area of <math>DCE,</math> we have | ||
+ | <cmath>\begin{align*}[DCE]&=\frac12 \cdot 2\cdot CG\cdot DG \\ | ||
+ | &=CG\cdot DG\\ | ||
+ | &=\frac{x^2}{4}.\end{align*}</cmath> | ||
+ | Plugging back into <math>(1),</math> we have | ||
+ | <cmath>\begin{align*}6\sqrt3&=\frac{3+2\sqrt3}{4} x^2 -\frac{3x^2}{4} \\ | ||
+ | &=\frac{\sqrt3}{2}x^2\end{align*}</cmath> | ||
+ | which means <math>x=2\sqrt3</math> and | ||
+ | <cmath>6x=\boxed{\textbf{(E)} \: 12\sqrt3}.</cmath> | ||
+ | |||
+ | ~ASAB | ||
== 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 12:15, 6 January 2022
Contents
Problem
In the figure, equilateral hexagon has three nonadjacent acute interior angles that each measure . The enclosed area of the hexagon is . What is the perimeter of the hexagon?
Solution 1 (Law of Cosines and Equilateral Triangle Area)
Isosceles triangles , , and are congruent by SAS congruence. By CPCTC, , so triangle is equilateral.
Let the side length of the hexagon be .
The area of each isosceles triangle is by the fourth formula here.
By the Law of Cosines on triangle , . Hence, the area of the equilateral triangle is .
The total area of the hexagon is thrice the area of each isosceles triangle plus the area of the equilateral triangle, or . Hence, and the perimeter is .
Solution 2
We will be referring to the following diagram.
Observe that Letting the perimeter will be
We know that and using such, we have and Thus, we have Computing the area of we have Plugging back into we have which means and
~ASAB
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.