2021 AMC 10A Problems/Problem 21

Problem

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 $m + n + p$ ?

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

Diagram

500px

~MRENTHUSIASM (by Geometry Expressions)

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

Video Solution by TheBeautyofMath

https://youtu.be/8qcbZ8c7fHg

~IceMatrix

Video Solution by MRENTHUSIASM (English & Chinese)

https://www.youtube.com/watch?v=0n8EAu2VAiM

~MRENTHUSIASM

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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