2024 AMC 12B Problems/Problem 19

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)) = $OF^2 * sin(\theta) + OF^2 * sin(120 - \theta)$

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

$\boxed{B -34}$ .

~luckuso

2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2024 AMC 12B Problems/Problem 19

Problem 19

Solution #1

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