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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)==
 
==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.  
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Isosceles triangles <math>ABE</math>, <math>CBD</math>, and <math>EDF</math> are identical by [[SAS_Similarity|SAS similarity]]. <math>BF=BD=DF</math> by [[CPCTC]], and triangle <math>BDF</math> is equilateral.  
  
 
Let the side length of the hexagon be <math>s</math>.  
 
Let the side length of the hexagon be <math>s</math>.  

Revision as of 19:11, 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 fourth formula here.

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 equilateral 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 or $\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2+3(\frac{1}{4}s^2)=\frac{\sqrt{3}}{2}s^2=6\sqrt{3}$. The perimeter is $6s=\boxed{12\sqrt{3}}$.

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