Difference between revisions of "2024 AMC 12B Problems/Problem 19"
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<math>\textbf{(A)}~\frac{3}{4}\qquad\textbf{(B)}~\frac{5\sqrt{3}}{11}\qquad\textbf{(C)}~\frac{4}{5}\qquad\textbf{(D)}~\frac{11}{13}\qquad\textbf{(E)}~\frac{7\sqrt{3}}{13}</math> | <math>\textbf{(A)}~\frac{3}{4}\qquad\textbf{(B)}~\frac{5\sqrt{3}}{11}\qquad\textbf{(C)}~\frac{4}{5}\qquad\textbf{(D)}~\frac{11}{13}\qquad\textbf{(E)}~\frac{7\sqrt{3}}{13}</math> | ||
− | ==Solution | + | ==Solution 1== |
let O be circumcenter of the equilateral triangle | let O be circumcenter of the equilateral triangle | ||
OF = <math>\frac{14\sqrt{3}}{3} </math> | OF = <math>\frac{14\sqrt{3}}{3} </math> | ||
− | 2( | + | 2(area(OFC) + area (OCE)) = <cmath> OF^2 * sin(\theta) + OF^2 * sin(120 - \theta) </cmath> |
− | + | <cmath> = \frac{14^2 * 3}{9} ( sin(\theta) + sin(120 - \theta) ) </cmath> | |
− | + | <cmath> = \frac{196}{3} ( sin(\theta) + sin(120 - \theta) ) </cmath> | |
− | + | <cmath> = 2 * {\frac{1}{3} } * area (ABCDEF) = 2* \frac{91\sqrt{3}}{3} </cmath> | |
− | + | <cmath> sin(\theta) + sin(120 - \theta) = \frac{13\sqrt{3}}{14} </cmath> | |
− | + | <cmath> sin(\theta) + \frac{ \sqrt{3}}{2}cos( \theta) +\frac{ \sqrt{1}}{2}sin( \theta) = \frac{13\sqrt{3}}{14} </cmath> | |
− | + | <cmath> \sqrt{3} sin( \theta) + cos( \theta) = \frac{13 }{14} </cmath> | |
− | + | <cmath> cos( \theta) = \frac{13 }{14} - \sqrt{3} sin( \theta) </cmath> | |
− | + | <cmath> \frac{169 }{49} - \frac{26\sqrt{3} }{7} sin( \theta) + 4 sin( \theta)^2 =0 </cmath> | |
− | + | <cmath> sin( \theta) = \frac{5\sqrt{3} }{14} or \frac{4\sqrt{3} }{7} </cmath> | |
− | <math>\frac{4\sqrt{3} }{7} </math> is invalid given <math>\theta</math> | + | <math>\frac{4\sqrt{3} }{7} </math> is invalid given <math>\theta \leq 60^\circ </math> |
<cmath> cos( \theta) = \frac{11 }{14} </cmath> | <cmath> cos( \theta) = \frac{11 }{14} </cmath> | ||
− | <cmath> tan( \theta) = \frac{5\sqrt{3} }{11} | + | <cmath> tan( \theta) = \frac{5\sqrt{3} }{11} \boxed{B } </cmath> |
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~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] |
Latest revision as of 21:38, 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
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.