2021 AMC 10A Problems/Problem 21

Problem 21

Let $ABCDEF$ be an equiangular hexagon. The lines $AB, CD,$ and $EF$ determine a triangle with area $192\sqrt{3}$ , and the lines $BC, DE,$ and $FA$ determine a triangle with area $324\sqrt{3}$ . The perimeter of hexagon $ABCDEF$ can be expressed as $m +n\sqrt{p}$ , where $m, n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m + n + p$ ?

$\textbf{(A)} ~47\qquad\textbf{(B)} ~52\qquad\textbf{(C)} ~55\qquad\textbf{(D)} ~58\qquad\textbf{(E)} ~63$

Solution (Misplaced problem?)

Note that the extensions of the given lines will determine an equilateral triangle because the hexagon is equiangular. The area of the first triangle is $192\sqrt{3}$ , so the side length is $\sqrt{192\cdot 4}=16\sqrt{3}$ . The area of the second triangle is $324\sqrt{3}$ , so the side length is $\sqrt{4\cdot 324}=36$ . We can set the first value equal to $AB+CD+EF$ and the second equal to $BC+DE+FA$ by substituting some lengths in with different sides of the same equilateral triangle. The perimeter of the hexagon is just the sum of these two, which is $16\sqrt{3}+36$ and $16+3+36=\boxed{55~\textbf{(C)}}$

Video Solution by OmegaLearn (Angle Chasing and Equilateral Triangles)

https://youtu.be/ptBwDcmDaLA

~ pi_is_3.14

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2021 AMC 10A Problems/Problem 21

Problem 21

Solution (Misplaced problem?)

Video Solution by OmegaLearn (Angle Chasing and Equilateral Triangles)