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

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(→‎Solution (Law of Cosines and Equilateral Triangle Area))
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Isosceles triangles <math>ABE</math>, <math>CBD</math>, and <math>EDF</math> are identical by SAS similarity. <math>BF=BD=DF</math> by CPCTC, and triangle <math>BDF</math> is equilateral.  
 
Isosceles triangles <math>ABE</math>, <math>CBD</math>, and <math>EDF</math> are identical by SAS similarity. <math>BF=BD=DF</math> by CPCTC, and triangle <math>BDF</math> is equilateral.  
  
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Let the side length of the hexagon be <math>s</math>. The area of each equilateral triangle is <math>\frac{1}{2}s^2\sin{30}=\frac{3}{4}s^2</math>. By the [[Law of Cosines]], the square of the side length of equilateral triangle BDF is <math>2s^2-2x^2\cos{30}=(2-\sqrt{3})s^2</math>. Hence, the area of the triangle is <math>\frac{\sqrt{3}}{4}(2-\sqrt{3})s^2=\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2</math>. So the total area of the hexagon is <math>\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2+3(\frac{3}{4}s^2)=\frac{\sqrt{3}}{2}s^2=6\sqrt{3}</math> and <math>6s=\boxed{12\sqrt{3}}</math>.
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Let the side length of the hexagon be <math>s</math>.  
 +
 
 +
The area of each isosceles triangle is <math>\frac{1}{2}s^2\sin{30}=\frac{1}{4}s^2</math>.  
 +
 
 +
By the [[Law of Cosines]], the square of the side length of equilateral triangle BDF is <math>2s^2-2x^2\cos{30}=(2-\sqrt{3})s^2</math>. Hence, the area of the triangle is <math>\frac{\sqrt{3}}{4}(2-\sqrt{3})s^2=\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2</math>.  
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So, the total area of the hexagon is the area of the equilateral triangle plus thrice the area of each isosceles triangle: <math>\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2+3(\frac{1}{4}s^2)=\frac{\sqrt{3}}{2}s^2=6\sqrt{3}</math>. The perimeter is <math>6s=\boxed{12\sqrt{3}}</math>.

Revision as of 18:06, 23 November 2021

Solution (Law of Cosines and Equilateral Triangle Area)

Isosceles triangles $ABE$, $CBD$, and $EDF$ are identical by SAS similarity. $BF=BD=DF$ by CPCTC, and triangle $BDF$ is equilateral.

Let the side length of the hexagon be $s$.

The area of each isosceles triangle is $\frac{1}{2}s^2\sin{30}=\frac{1}{4}s^2$.

By the Law of Cosines, the square of the side length of equilateral triangle BDF is $2s^2-2x^2\cos{30}=(2-\sqrt{3})s^2$. Hence, the area of the triangle is $\frac{\sqrt{3}}{4}(2-\sqrt{3})s^2=\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2$.

So, the total area of the hexagon is the area of the equilateral triangle plus thrice the area of each isosceles triangle: $\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2+3(\frac{1}{4}s^2)=\frac{\sqrt{3}}{2}s^2=6\sqrt{3}$. The perimeter is $6s=\boxed{12\sqrt{3}}$.

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