2024 AMC 12B Problems/Problem 19
Contents
[hide]Problem 19
Equilateral with side length is rotated about its center by angle , where , to form . See the figure. The area of hexagon is . What is ?
Solution #1
let O be circumcenter of the equilateral triangle
OF =
2(Area(OFC) + Area (OCE)) =
is invalid given <60
.
Solution #2
From 's side lengths of 14, we get OF = OC = OE = We let angle FOC = (\theta) and therefore angle EOC = 120 - (\theta)
The answer would be 3 * (Area + Area )
Which area = 0.5 * ^2 * sin(\theta)
And area = 0.5 * ^2 * sin(120 - \theta)
Therefore the answer would be 3 * 0.5 * ()^2 * (sin(\theta)+sin(120 - \theta)) = {91\sqrt{3}}
Which
So
Therefore
And
Which
(B)\frac{5\sqrt{3} }{11} $$ (Error compiling LaTeX. Unknown error_msg)
~mitsuihisashi14
See also
2024 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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 12 Problems and Solutions |
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