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 = (<math>\theta</math>) |
− | And therefore angle EOC = 120 - (\theta) | + | And therefore angle EOC = 120 - (<math>\theta</math>) |
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>) | ||
− | Which area <math>\triangle FOC</math> = 0.5 * <math>\frac{14\sqrt{3}}{3} | + | Which area <math>\triangle FOC</math> = 0.5 * <math> (\frac{14\sqrt{3}}{3} )^2 * sin(\theta)</math> |
− | And area <math>\triangle COE</math> = 0.5 * <math>\frac{14\sqrt{3}}{3} | + | And area <math>\triangle COE</math> = 0.5 * <math> (\frac{14\sqrt{3}}{3} )^2 * sin(120 - \theta)</math> |
Therefore the answer would be | Therefore the answer would be | ||
− | 3 * 0.5 * (<math>\frac{14\sqrt{3}}{3} | + | 3 * 0.5 * (<math>\frac{14\sqrt{3}}{3} )^2 * (sin(\theta)+sin(120 - \theta)) = {91\sqrt{3}}</math> |
Which <cmath> sin(\theta)+sin(120 - \theta) = \frac{91\sqrt{3}}{98} </cmath> | Which <cmath> sin(\theta)+sin(120 - \theta) = \frac{91\sqrt{3}}{98} </cmath> | ||
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Which <cmath> tan(\theta + 30) = \frac{91}{21\sqrt{3}} = \frac{91\sqrt{3}}{63} </cmath> | Which <cmath> tan(\theta + 30) = \frac{91}{21\sqrt{3}} = \frac{91\sqrt{3}}{63} </cmath> | ||
− | tan(\theta) can be calculated using addition identity, which gives the answer of | + | <math>tan(\theta)</math> can be calculated using addition identity, which gives the answer of |
<cmath>(B)\frac{5\sqrt{3}}{11}</cmath> | <cmath>(B)\frac{5\sqrt{3}}{11}</cmath> | ||
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(I would really appreciate if someone can help me fix my code and format) | (I would really appreciate if someone can help me fix my code and format) | ||
− | ~mitsuihisashi14 | + | ~mitsuihisashi14 |
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] (fixed Latex error ) | ||
==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 09:34, 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 = () And therefore angle EOC = 120 - ()
The answer would be 3 * (Area + Area )
Which area = 0.5 *
And area = 0.5 *
Therefore the answer would be 3 * 0.5 * (
Which
So
Therefore
And
Which
can be calculated using addition identity, which gives the answer of
(I would really appreciate if someone can help me fix my code and format)
~mitsuihisashi14 ~luckuso (fixed Latex error )
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.