Difference between revisions of "2021 Fall AMC 12A Problems/Problem 14"
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− | == | + | ==Problem== |
+ | In the figure, equilateral hexagon <math>ABCDEF</math> has three nonadjacent acute interior angles that each measure <math>30^\circ</math>. The enclosed area of the hexagon is <math>6\sqrt{3}</math>. 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> | ||
+ | <math>\textbf{(A)} \: 4 \qquad \textbf{(B)} \: 4\sqrt3 \qquad \textbf{(C)} \: 12 \qquad \textbf{(D)} \: 18 \qquad \textbf{(E)} \: 12\sqrt3</math> | ||
− | + | ==Solution 1== | |
− | + | Divide the equilateral hexagon <math>ABCDEF</math> into isosceles triangles <math>ABF</math>, <math>CBD</math>, and <math>EDF</math> and triangle <math>BDF</math>. The three isosceles triangles are congruent by SAS congruence. By CPCTC, <math>BF=BD=DF</math>, so triangle <math>BDF</math> is equilateral. | |
− | The area of each isosceles triangle is < | + | Let the side length of the hexagon be <math>s</math>. The area of each isosceles triangle is <cmath>\frac{1}{2} a b \sin\angle C = \frac{1}{2} \cdot s \cdot s \cdot \sin{30^{\circ}} = \frac{1}{4}s^2.</cmath> |
− | By the [[Law of Cosines]] | + | By the [[Law of Cosines]] on triangle <math>ABF</math>, <cmath>BF^2=s^2+s^2-2s^2\cos{30^{\circ}}=2s^2-\sqrt{3}s^2.</cmath> |
− | + | Hence, the [[Area_of_an_equilateral_triangle|area of the equilateral triangle]] <math>BDF</math> is <cmath>\frac{\sqrt{3}}{4} BF^2 = \frac{\sqrt{3}}{4}\left(2s^2-\sqrt{3}s^2\right)=\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2.</cmath> | |
+ | |||
+ | The total area of the hexagon is thrice the area of each isosceles triangle plus the area of the equilateral triangle, or <cmath>3\left(\frac{1}{4}s^2\right)+ \left( \frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2 \right)=\frac{\sqrt{3}}{2}s^2=6\sqrt{3}.</cmath> Hence, <math>s=2\sqrt{3}</math>, and the perimeter of the hexagon is <math>6s=\boxed{\textbf{(E)} \: 12\sqrt3}</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,p+dashed); | ||
+ | draw(D--G,p+dashed); | ||
+ | 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>\begin{alignat*}{8} | ||
+ | CG &= x\sin(75^{\circ}) &&= \frac{\sqrt6+\sqrt2}{4}x, \\ | ||
+ | DG &= x\cos(75^{\circ}) &&= \frac{\sqrt6-\sqrt2}{4}x. | ||
+ | \end{alignat*}</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>6\sqrt3=\frac{3+2\sqrt3}{4} x^2 -\frac{3x^2}{4}=\frac{\sqrt3}{2}x^2,</cmath> | ||
+ | which means <math>x=2\sqrt3</math> and | ||
+ | <math>6x=\boxed{\textbf{(E)} \: 12\sqrt3}.</math> | ||
+ | |||
+ | ~ASAB | ||
+ | ==Solution 3== | ||
+ | We will be using this diagram: | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | Cleverly done with Trig | ||
+ | https://youtu.be/AgSE7HPCVR0 | ||
+ | |||
+ | ==Video Solution (Logic and Trigonometry)== | ||
+ | https://youtu.be/4j8e47S1cr0 | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==Video Solution by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=g6Dk6An2ALY&t=208s | ||
+ | == See Also == | ||
+ | {{AMC12 box|year=2021 Fall|ab=A|num-b=13|num-a=15}} | ||
+ | {{MAA Notice}} |
Latest revision as of 23:16, 1 November 2023
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
Divide the equilateral hexagon into isosceles triangles
,
, and
and triangle
. The three isosceles triangles 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 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 of the hexagon 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
Thus, we have
Computing the area of
we have
Plugging back into
we have
which means
and
~ASAB
Solution 3
We will be using this diagram:
Video Solution by TheBeautyofMath
Cleverly done with Trig https://youtu.be/AgSE7HPCVR0
Video Solution (Logic and Trigonometry)
~Education, the Study of Everything
~IceMatrix
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=g6Dk6An2ALY&t=208s
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.