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2024 AMC 12B Problems/Problem 19: Difference between revisions

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==Solution 1==
==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>
Easily get OF = <math>\frac{14\sqrt{3}}{3} </math>


2(area(OFC) + area (OCE)) = <cmath>  OF^2 * sin(\theta) + OF^2 * sin(120 - \theta)    </cmath>
2(area(<math>\triangle</math>OFC) + area (<math>\triangle</math>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{14^2 * 3}{9} (  sin(\theta)  +  sin(120 - \theta) )  </cmath>
<cmath> = \frac{196}{3}  (  sin(\theta)  +  sin(120 - \theta) )  </cmath>
<cmath> = \frac{196}{3}  (  sin(\theta)  +  sin(120 - \theta) )  </cmath>
Line 27: Line 27:
<cmath> \frac{169 }{49}  - \frac{26\sqrt{3} }{7} sin(  \theta)  + 4 sin(  \theta)^2 =0 </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>
<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 \leq 60^\circ </math>  
<math>\frac{4\sqrt{3} }{7} </math> is invalid given <math>\theta \leq 60^\circ </math> , <math>sin(\theta ) < sin( 60^\circ ) = \frac{\sqrt{3} }{2} = \frac{\sqrt{3} \cdot 3.5}{7} </math>
<cmath>  cos(  \theta)  = \frac{11 }{14}  </cmath>
<cmath>  cos(  \theta)  = \frac{11 }{14}  </cmath>
<cmath>  tan(  \theta)  = \frac{5\sqrt{3} }{11} \boxed{B } </cmath>  
<cmath>  tan(  \theta)  = \frac{5\sqrt{3} }{11} \boxed{B } </cmath>  

Revision as of 01:13, 16 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

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

2(area($\triangle$OFC) + area ($\triangle$OCE)) = \[OF^2 * sin(\theta) + OF^2 * sin(120 - \theta)\] \[= \frac{14^2 * 3}{9} ( sin(\theta) + sin(120 - \theta) )\] \[= \frac{196}{3} ( sin(\theta) + sin(120 - \theta) )\] \[= 2 * {\frac{1}{3} } * area (ABCDEF) = 2* \frac{91\sqrt{3}}{3}\]

\[sin(\theta) + sin(120 - \theta) = \frac{13\sqrt{3}}{14}\] \[sin(\theta) + \frac{ \sqrt{3}}{2}cos( \theta) +\frac{ \sqrt{1}}{2}sin( \theta) = \frac{13\sqrt{3}}{14}\] \[\sqrt{3} sin( \theta) + cos( \theta) = \frac{13 }{7}\] \[cos( \theta) = \frac{13 }{7} - \sqrt{3} sin( \theta)\] \[\frac{169 }{49} - \frac{26\sqrt{3} }{7} sin( \theta) + 4 sin( \theta)^2 =0\] \[sin( \theta) = \frac{5\sqrt{3} }{14} or \frac{4\sqrt{3} }{7}\] $\frac{4\sqrt{3} }{7}$ is invalid given $\theta \leq 60^\circ$ , $sin(\theta ) < sin( 60^\circ ) = \frac{\sqrt{3} }{2} = \frac{\sqrt{3} \cdot 3.5}{7}$ \[cos( \theta) = \frac{11 }{14}\] \[tan( \theta) = \frac{5\sqrt{3} }{11} \boxed{B }\]

~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 ~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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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