Difference between revisions of "2024 AMC 12B Problems/Problem 19"
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And area <math>\triangle COE</math> = 0.5 * <math>\frac{14\sqrt{3}}{3} </math>^2 * sin(120 - \theta) | And area <math>\triangle COE</math> = 0.5 * <math>\frac{14\sqrt{3}}{3} </math>^2 * sin(120 - \theta) | ||
− | Therefore the answer would be 3 * 0.5 * (<math>\frac{14\sqrt{3}}{3} </math>)^2 * (sin(\theta)+sin(120 - \theta)) = {91\sqrt{3}} | + | Therefore the answer would be |
+ | <math></math> 3 * 0.5 * (<math>\frac{14\sqrt{3}}{3} </math>)^2 * (sin(\theta)+sin(120 - \theta)) = {91\sqrt{3}} <cmath> | ||
− | Which <cmath> sin(\theta)+sin(120 - \theta) = \frac{91\sqrt{3}}{98} < | + | Which </cmath> sin(\theta)+sin(120 - \theta) = \frac{91\sqrt{3}}{98} <cmath> |
− | So <cmath> \frac{1}{2}cos(\theta)+\frac{\sqrt{3}}{2}sin(\theta) = \frac{91}{98} < | + | So </cmath> \frac{1}{2}cos(\theta)+\frac{\sqrt{3}}{2}sin(\theta) = \frac{91}{98} <cmath> |
− | Therefore <cmath> sin(\theta + 30) = \frac{91}{98} < | + | Therefore </cmath> sin(\theta + 30) = \frac{91}{98} <cmath> |
− | And <cmath> cos (\theta + 30) = \frac{21\sqrt{3}}{98} < | + | And </cmath> cos (\theta + 30) = \frac{21\sqrt{3}}{98} <cmath> |
− | Which <cmath> tan(\theta + 30) = \frac{91}{21\sqrt{3}} = \frac{91\sqrt{3}}{63} < | + | 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 <cmath> (B)\frac{5\sqrt{3} }{11} </ | + | tan(\theta) can be calculated using addition identity, which gives the answer of </cmath> (B)\frac{5\sqrt{3} }{11} <math></math> |
(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 09:15, 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 $$ (Error compiling LaTeX. Unknown error_msg) 3 * 0.5 * ()^2 * (sin(\theta)+sin(120 - \theta)) = {91\sqrt{3}} sin(\theta)+sin(120 - \theta) = \frac{91\sqrt{3}}{98} \frac{1}{2}cos(\theta)+\frac{\sqrt{3}}{2}sin(\theta) = \frac{91}{98} sin(\theta + 30) = \frac{91}{98} cos (\theta + 30) = \frac{21\sqrt{3}}{98} tan(\theta + 30) = \frac{91}{21\sqrt{3}} = \frac{91\sqrt{3}}{63} (B)\frac{5\sqrt{3} }{11} $$ (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.