Difference between revisions of "2024 AMC 12B Problems/Problem 19"
(→Solution #2) |
(→Solution #2) |
||
Line 48: | Line 48: | ||
Therefore the answer would be | Therefore the answer would be | ||
− | + | 3 * 0.5 * (<math>\frac{14\sqrt{3}}{3} </math>)^2 * (sin(\theta)+sin(120 - \theta)) = {91\sqrt{3}} | |
− | Which < | + | Which <cmath> sin(\theta)+sin(120 - \theta) = \frac{91\sqrt{3}}{98} </cmath> |
− | So < | + | So <cmath> \frac{1}{2}cos(\theta)+\frac{\sqrt{3}}{2}sin(\theta) = \frac{91}{98} </cmath> |
− | Therefore < | + | Therefore <cmath> sin(\theta + 30) = \frac{91}{98} </cmath> |
− | And < | + | And <cmath> cos (\theta + 30) = \frac{21\sqrt{3}}{98} </cmath> |
− | Which < | + | 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 < | + | tan(\theta) can be calculated using addition identity, which gives the answer of |
+ | |||
+ | <cmath>(B)\frac{5\sqrt{3}}{11}</cmath> | ||
(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) |
Revision as of 09:17, 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
tan(\theta) 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
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.