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 = | + | From <math>\triangle ABC</math>'s side lengths of 14, we get |
− | We let angle FOC = | + | <cmath>OF = OC = OE = \frac{14\sqrt{3}}{3}.</cmath> |
− | And | + | We let <math>\angle FOC = \theta</math> |
+ | And <math>\angle EOC = 120 - \theta</math> | ||
− | The answer would be | + | The answer would be <math>3([\triangle FOC] + [\triangle COE])</math> |
− | Which area <math>\triangle FOC</math> = | + | Which area <math>\triangle FOC</math> = <math>\frac{1}{2}\left(\frac{14\sqrt{3}}{3}\right)^2 \sin(\theta)</math> |
− | And area <math>\triangle COE</math> = | + | And area <math>\triangle COE</math> = <math>\frac{1}{2}\left(\frac{14\sqrt{3}}{3}\right)^2 \sin(120 - \theta)</math> |
− | + | So we have that | |
− | 3 | + | <cmath>3\cdot \frac{1}{2}\cdot \left(\frac{14\sqrt{3}}{3}\right)^2 (\sin(\theta)+\sin(120 - \theta)) = 91\sqrt{3}</cmath> |
− | Which <cmath> sin(\theta)+sin(120 - \theta) = \frac{91\sqrt{3}}{98} </cmath> | + | Which means |
+ | <cmath>\sin(\theta)+\sin(120 - \theta) = \frac{91\sqrt{3}}{98}</cmath> | ||
+ | <cmath>\frac{1}{2}\cos(\theta)+\frac{\sqrt{3}}{2}\sin(\theta) = \frac{91}{98}</cmath> | ||
+ | <cmath>\sin(\theta + 30) = \frac{91}{98}</cmath> | ||
+ | <cmath>\cos (\theta + 30) = \frac{21\sqrt{3}}{98}</cmath> | ||
+ | <cmath>\tan(\theta + 30) = \frac{91}{21\sqrt{3}} = \frac{91\sqrt{3}}{63}</cmath> | ||
− | + | Now, <math>\tan(\theta)</math> can be calculated using the addition identity, which gives the answer of | |
− | + | <cmath>\boxed{\text{(B) }\frac{5\sqrt{3}}{11}}.</cmath> | |
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~mitsuihisashi14 | ~mitsuihisashi14 |
Latest revision as of 14:42, 15 December 2024
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
Easily get
is invalid given ,
Solution #2
From 's side lengths of 14, we get We let And
The answer would be
Which area =
And area =
So we have that
Which means
Now, can be calculated using the addition identity, which gives the answer of
~mitsuihisashi14 ~luckuso (fixed Latex error )
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=akLlCXKtXnk
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.