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Difference between revisions of "2024 AMC 12B Problems/Problem 19"

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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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So <cmath> \frac{1}{2}cos(\theta)+\frac{\sqrt{3}}{2} = \frac{91}{98} </cmath>
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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} </cmath>
 
Therefore <cmath> sin(\theta + 30) = \frac{91}{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>
  
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<cmath> tan(\theta) can be calculated using addition identity, which gives the answer of </cmath> (B)\frac{5\sqrt{3} }{11} <math></math>
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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 09:13, 14 November 2024

Problem 19

Equilateral $\triangle ABC$ with side length $14$ is rotated about its center by angle $\theta$, where $0 < \theta < 60^{\circ}$, to form $\triangle DEF$. See the figure. The area of hexagon $ADBECF$ is $91\sqrt{3}$. What is $\tan\theta$? [asy] // Credit to shihan for this diagram. defaultpen(fontsize(13)); size(200); pair O=(0,0),A=dir(225),B=dir(-15),C=dir(105),D=rotate(38.21,O)*A,E=rotate(38.21,O)*B,F=rotate(38.21,O)*C; draw(A--B--C--A,gray+0.4);draw(D--E--F--D,gray+0.4); draw(A--D--B--E--C--F--A,black+0.9); dot(O); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); [/asy]

$\textbf{(A)}~\frac{3}{4}\qquad\textbf{(B)}~\frac{5\sqrt{3}}{11}\qquad\textbf{(C)}~\frac{4}{5}\qquad\textbf{(D)}~\frac{11}{13}\qquad\textbf{(E)}~\frac{7\sqrt{3}}{13}$

Solution #1

let O be circumcenter of the equilateral triangle

OF = $\frac{14\sqrt{3}}{3}$

2(Area(OFC) + Area (OCE)) = \[OF^2 * sin(\theta) + OF^2 * sin(120 - \theta)\] 

\[= \frac{14^2 * 3}{9} (   sin(\theta)  +  sin(120 - \theta) )\]
\[= \frac{196}{3}  (   sin(\theta)  +  sin(120 - \theta) )\]
\[= 2 * {\frac{1}{3}  } * Area  (ABCDEF) = 2* \frac{91\sqrt{3}}{3}\]

 \[sin(\theta)  +  sin(120 - \theta) = \frac{13\sqrt{3}}{14}\]
 \[sin(\theta)  +   \frac{ \sqrt{3}}{2}cos(  \theta) +\frac{ \sqrt{1}}{2}sin(  \theta) = \frac{13\sqrt{3}}{14}\]
\[\sqrt{3} sin(  \theta) + cos(  \theta) = \frac{13 }{14}\]
\[cos(  \theta)  = \frac{13 }{14}  - \sqrt{3} sin(  \theta)\]
\[\frac{169 }{49}  - \frac{26\sqrt{3} }{7} sin(  \theta)  + 4 sin(  \theta)^2 =0\]
\[sin(  \theta)  = \frac{5\sqrt{3} }{14}  or \frac{4\sqrt{3} }{7}\]

$\frac{4\sqrt{3} }{7}$ is invalid given $\theta$<60 \[cos( \theta) = \frac{11 }{14}\] \[tan( \theta) = \frac{5\sqrt{3} }{11}\]

$\boxed{B -34}$.


~luckuso

Solution #2

From $\triangle ABC$'s side lengths of 14, we get OF = OC = OE =\[\frac{14\sqrt{3}}{3}\] We let angle FOC = (\theta) And therefore angle EOC = 120 - (\theta)

The answer would be 3 * (Area $\triangle FOC$ + Area $\triangle COE$)

Which area $\triangle FOC$ = 0.5 * $\frac{14\sqrt{3}}{3}$^2 * sin(\theta)

And area $\triangle COE$ = 0.5 * $\frac{14\sqrt{3}}{3}$^2 * sin(120 - \theta)

Therefore the answer would be 3 * 0.5 * ($\frac{14\sqrt{3}}{3}$)^2 * (sin(\theta)+sin(120 - \theta)) = {91\sqrt{3}}

Which \[sin(\theta)+sin(120 - \theta) = \frac{91\sqrt{3}}{98}\]

So \[\frac{1}{2}cos(\theta)+\frac{\sqrt{3}}{2}sin(\theta) = \frac{91}{98}\]

Therefore \[sin(\theta + 30) = \frac{91}{98}\]

And \[cos (\theta + 30) = \frac{21\sqrt{3}}{98}\]

Which \[tan(\theta + 30) = \frac{91}{21\sqrt{3}} = \frac{91\sqrt{3}}{63}\]

tan(\theta) can be calculated using addition identity, which gives the answer of \[(B)\frac{5\sqrt{3} }{11}\]

(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

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