Difference between revisions of "2021 Fall AMC 12A Problems/Problem 14"
MRENTHUSIASM (talk | contribs) (→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. | ||
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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>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 11:53, 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 \[\begin{align*}[ACE]&=\frac{\sqrt3}{4}\left(2\cdotCG\right)^2\\ &=\sqrt3(2+sqrt3)\end{align*}\]
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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