Difference between revisions of "2024 AMC 12B Problems/Problem 19"
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==Solution #2== | ==Solution #2== | ||
From <math>\triangle ABC</math>'s side lengths of 14, we get OF = OC = OE =<cmath>\frac{14\sqrt{3}}{3} </cmath> | From <math>\triangle ABC</math>'s side lengths of 14, we get OF = OC = OE =<cmath>\frac{14\sqrt{3}}{3} </cmath> | ||
− | We let angle FOC = (\theta) | + | We let angle FOC = (\theta) |
+ | And therefore angle EOC = 120 - (\theta) | ||
The answer would be 3 * (Area <math>\triangle FOC</math> + Area <math>\triangle COE</math>) | The answer would be 3 * (Area <math>\triangle FOC</math> + Area <math>\triangle COE</math>) | ||
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<cmath> tan(\theta) can be calculated using addition identity, which gives the answer of </cmath> (B)\frac{5\sqrt{3} }{11} <math></math> | <cmath> tan(\theta) can be calculated using addition identity, which gives the answer of </cmath> (B)\frac{5\sqrt{3} }{11} <math></math> | ||
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+ | (I would really appreciate if someone can help me fix my code and format) | ||
~mitsuihisashi14 | ~mitsuihisashi14 | ||
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==See also== | ==See also== | ||
{{AMC12 box|year=2024|ab=B|num-b=18|num-a=20}} | {{AMC12 box|year=2024|ab=B|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 09:12, 14 November 2024
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)
(I would really appreciate if someone can help me fix my code and format)
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.