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

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 $\angle EOC = 120 - \theta$

The answer would be $3([\triangle FOC] + [\triangle COE])$

Which area $\triangle FOC$ = $\frac{1}{2}\left(\frac{14\sqrt{3}}{3}\right)^2 \sin(\theta)$

And area $\triangle COE$ = $\frac{1}{2}\left(\frac{14\sqrt{3}}{3}\right)^2 \sin(120 - \theta)$

So we have that \[3\cdot \frac{1}{2}\cdot \left(\frac{14\sqrt{3}}{3}\right)^2 (\sin(\theta)+\sin(120 - \theta)) = 91\sqrt{3}\]

Which means \[\sin(\theta)+\sin(120 - \theta) = \frac{91\sqrt{3}}{98}\] \[\frac{1}{2}\cos(\theta)+\frac{\sqrt{3}}{2}\sin(\theta) = \frac{91}{98}\] \[\sin(\theta + 30) = \frac{91}{98}\] \[\cos (\theta + 30) = \frac{21\sqrt{3}}{98}\] \[\tan(\theta + 30) = \frac{91}{21\sqrt{3}} = \frac{91\sqrt{3}}{63}\]

Now, $\tan(\theta)$ can be calculated using the addition identity, which gives the answer of

\[\boxed{\text{(B) }\frac{5\sqrt{3}}{11}}.\]

~mitsuihisashi14 ~luckuso (fixed Latex error )

Solution 3 (No Trig Manipulations)

Let the circumcenter of the circle inscribing this polygon be $O$. The area of the equilateral triangle is $\frac{\sqrt{3}}{4}*196=49\sqrt{3}$. The area of one of the three smaller triangles, say $\triangle{DBE}$ is $14\sqrt{3}$. Let $BH$ be the altitude of $\triangle{DBE}$, so if we extend $BH$ to point $M$ where $MO\perp{BM}$, we get right triangle $\triangle{OMB}$. Note that the height $BH=2\sqrt{3}$, computed given the area and side length $14$, so $MB=BH+HB=2\sqrt{3}+\frac{7\sqrt{3}}{3}=\frac{13\sqrt{3}}{3}$. $OB=\frac{14\sqrt{3}}{3}$ so Pythag gives $OM=\sqrt{OB^2-MB^2}=3$. This means that $HE=7-OM=4$, so Pythag gives $BE=2\sqrt{7}$. Let $\frac{\theta}{2}=\alpha$ and the midpoint of $BE$ be $P$ so that $BP=PE=\sqrt{7}$, so that Pythag on $\triangle{OPE}$ gives $OP=\sqrt{OE^2-PE^2}=\sqrt{\frac{175}{3}}$. Then $\tan{\alpha}=\frac{\sqrt{\frac{175}{3}}}{\sqrt{7}}=\frac{\sqrt{3}}{5}$. Then $\tan{2\alpha}=\tan{\theta}=\frac{\frac{2\sqrt{3}}{5}}{1-\frac{3}{25}}=\boxed{\frac{5\sqrt{3}}{11}}$.

-Magnetoninja

Video Solution by SpreadTheMathLove

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

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=237658"