2021 Fall AMC 12A Problems/Problem 14

Revision as of 18:04, 23 November 2021 by Ehuang0531 (talk | contribs) (Solution (Law of Cosines and Equilateral Triangle Area))

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 equilateral triangle is $\frac{1}{2}s^2\sin{30}=\frac{3}{4}s^2$. 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 $\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2+3(\frac{3}{4}s^2)=\frac{\sqrt{3}}{2}s^2=6\sqrt{3}$ and $6s=\boxed{12\sqrt{3}}$.