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

Latest revision as of 15:56, 23 December 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 $\angle EOC = 120 - \theta$

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

@@ Line 61: / Line 61: @@
 ~mitsuihisashi14
 ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] (fixed Latex error )
+==Solution 3 (No Trig Manipulations)==
+Let the circumcenter of the circle inscribing this polygon be <math>O</math>. The area of the equilateral triangle is <math>\frac{\sqrt{3}}{4}*196=49\sqrt{3}</math>. The area of one of the three smaller triangles, say <math>\triangle{DBE}</math> is <math>14\sqrt{3}</math>. Let <math>BH</math> be the altitude of <math>\triangle{DBE}</math>, so if we extend <math>BH</math> to point <math>M</math> where <math>MO\perp{BM}</math>, we get right triangle <math>\triangle{OMB}</math>. Note that the height <math>BH=2\sqrt{3}</math>, computed given the area and side length <math>14</math>, so <math>MB=BH+HB=2\sqrt{3}+\frac{7\sqrt{3}}{3}=\frac{13\sqrt{3}}{3}</math>. <math>OB=\frac{14\sqrt{3}}{3}</math> so Pythag gives <math>OM=\sqrt{OB^2-MB^2}=3</math>. This means that <math>HE=7-OM=4</math>, so Pythag gives <math>BE=2\sqrt{7}</math>. Let <math>\frac{\theta}{2}=\alpha</math> and the midpoint of <math>BE</math> be <math>P</math> so that <math>BP=PE=\sqrt{7}</math>, so that Pythag on <math>\triangle{OPE}</math> gives <math>OP=\sqrt{OE^2-PE^2}=\sqrt{\frac{175}{3}}</math>. Then <math>\tan{\alpha}=\frac{\sqrt{\frac{175}{3}}}{\sqrt{7}}=\frac{\sqrt{3}}{5}</math>. Then <math>\tan{2\alpha}=\tan{\theta}=\frac{\frac{2\sqrt{3}}{5}}{1-\frac{3}{25}}=\boxed{\frac{5\sqrt{3}}{11}}</math>.
+-Magnetoninja
 ==Video Solution by SpreadTheMathLove==

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