Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2024 AMC 12B Problems/Problem 19

Revision as of 13:40, 18 November 2024 by Jlee2009 (talk | contribs) (→‎Solution 1)

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 )

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_12B_Problems/Problem_19&oldid=234918"