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

Revision as of 14:54, 19 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 $OF = \frac{14\sqrt{3}}{3}$

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

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

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=akLlCXKtXnk

@@ Line 66: / Line 66: @@
 ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] (fixed Latex error )
+==Video Solution by SpreadTheMathLove==
+https://www.youtube.com/watch?v=akLlCXKtXnk
 ==See also==
 {{AMC12 box|year=2024|ab=B|num-b=18|num-a=20}}
 {{MAA Notice}}

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