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

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==Solution 1==
 
==Solution 1==
−
let O be circumcenter of the equilateral triangle  
+
Let O be circumcenter of the equilateral triangle  
  
−
OF = <math>\frac{14\sqrt{3}}{3} </math>
+
Easily get <math>OF = \frac{14\sqrt{3}}{3}</math>
  
−
2(area(OFC) + area (OCE)) = <cmath> OF^2 * sin(\theta) + OF^2 * sin(120 - \theta)   </cmath>
+
<math>2 \cdot \triangle(OFC) + \triangle(OCE) =  OF^2 \cdot \sin(\theta) + OF^2 \cdot \sin(120 - \theta)</math>
−
<cmath> = \frac{14^2 * 3}{9} (  sin(\theta)  +  sin(120 - \theta) )  </cmath>
+
<cmath> = \frac{14^2 \cdot 3}{9} (  \sin(\theta)  +  \sin(120 - \theta) )  </cmath>
−
<cmath> = \frac{196}{3}  (  sin(\theta)  +  sin(120 - \theta) )  </cmath>
+
<cmath> = \frac{196}{3}  (  \sin(\theta)  +  \sin(120 - \theta) )  </cmath>
−
<cmath> = 2 * {\frac{1}{3}  } * area  (ABCDEF) = 2* \frac{91\sqrt{3}}{3} </cmath>
+
<cmath> = 2 \cdot {\frac{1}{3}  } \cdot(ABCDEF) = 2\cdot \frac{91\sqrt{3}}{3} </cmath>
−
 
 
−
<cmath> sin(\theta)  +  sin(120 - \theta) = \frac{13\sqrt{3}}{14}  </cmath>
 
−
<cmath> sin(\theta)  +  \frac{ \sqrt{3}}{2}cos(  \theta) +\frac{ \sqrt{1}}{2}sin(  \theta) = \frac{13\sqrt{3}}{14}  </cmath>
 
−
<cmath> \sqrt{3} sin(  \theta) + cos( \theta) = \frac{13 }{14}  </cmath>
 
−
<cmath> cos(  \theta) = \frac{13 }{14}  - \sqrt{3} sin(  \theta) </cmath>
 
−
<cmath> \frac{169 }{49}  - \frac{26\sqrt{3} }{7} sin(  \theta)  + 4 sin(  \theta)^2 =0 </cmath>
 
−
<cmath> sin(  \theta)  = \frac{5\sqrt{3} }{14}  or \frac{4\sqrt{3} }{7}  </cmath>
 
−
<math>\frac{4\sqrt{3} }{7} </math> is invalid given <math>\theta \leq 60^\circ </math>
 
−
<cmath>  cos(  \theta)  = \frac{11 }{14}  </cmath>
 
−
<cmath>  tan(  \theta)  = \frac{5\sqrt{3} }{11} </cmath>
 
−
 
 
−
<math>\boxed{B -34} </math>. 
 
  
 +
<cmath> \sin(\theta)  +  \sin(120 - \theta) = \frac{13\sqrt{3}}{14}  </cmath>
 +
<cmath> \sin(\theta)  +  \frac{ \sqrt{3}}{2}\cos(  \theta) +\frac{ \sqrt{1}}{2}\sin(  \theta) = \frac{13\sqrt{3}}{14}  </cmath>
 +
<cmath> \sqrt{3} \sin(  \theta) + \cos(  \theta) = \frac{13 }{7}  </cmath>
 +
<cmath> \cos(  \theta)  = \frac{13 }{7}  - \sqrt{3} \sin(  \theta) </cmath>
 +
<cmath> \frac{169 }{49}  - \frac{26\sqrt{3} }{7} \sin(  \theta)  + 4 \sin(  \theta)^2 = 1 </cmath>
 +
<cmath> \sin(  \theta)  = \frac{5\sqrt{3} }{14}  or \frac{4\sqrt{3} }{7}  </cmath>
 +
<math>\frac{4\sqrt{3} }{7} </math> is invalid given <math>\theta \leq 60^\circ </math> , <math>\sin(\theta ) < \sin( 60^\circ ) = \frac{\sqrt{3} }{2} = \frac{\sqrt{3} \cdot 3.5}{7} </math>
 +
<cmath>  \cos(  \theta)  = \frac{11 }{14}  </cmath>
 +
<cmath>  \tan(  \theta)  = \frac{5\sqrt{3} }{11} \boxed{B } </cmath>
  
 
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
 
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
  
 
==Solution #2==
 
==Solution #2==
−
From <math>\triangle ABC</math>'s side lengths of 14, we get OF = OC = OE =<cmath>\frac{14\sqrt{3}}{3} </cmath>
+
From <math>\triangle ABC</math>'s side lengths of 14, we get
−
We let angle FOC = (<math>\theta</math>)
+
<cmath>OF = OC = OE = \frac{14\sqrt{3}}{3}.</cmath>
−
And therefore angle EOC = 120 - (<math>\theta</math>)
+
We let <math>\angle FOC = \theta</math>
 +
And <math>\angle EOC = 120 - \theta</math>
  
−
The answer would be 3 * (Area <math>\triangle FOC</math> + Area <math>\triangle COE</math>)
+
The answer would be <math>3([\triangle FOC] + [\triangle COE])</math>
  
−
Which area <math>\triangle FOC</math> = 0.5 * <math> (\frac{14\sqrt{3}}{3} )^2 * sin(\theta)</math>
+
Which area <math>\triangle FOC</math> = <math>\frac{1}{2}\left(\frac{14\sqrt{3}}{3}\right)^2 \sin(\theta)</math>
  
−
And area <math>\triangle COE</math> = 0.5 * <math> (\frac{14\sqrt{3}}{3} )^2 *  sin(120 - \theta)</math>
+
And area <math>\triangle COE</math> = <math>\frac{1}{2}\left(\frac{14\sqrt{3}}{3}\right)^2 \sin(120 - \theta)</math>
  
−
Therefore the answer would be
+
So we have that
−
3 * 0.5 * (<math>\frac{14\sqrt{3}}{3} )^2 * (sin(\theta)+sin(120 - \theta)) = {91\sqrt{3}}</math>
+
<cmath>3\cdot \frac{1}{2}\cdot \left(\frac{14\sqrt{3}}{3}\right)^2 (\sin(\theta)+\sin(120 - \theta)) = 91\sqrt{3}</cmath>
  
−
Which <cmath> sin(\theta)+sin(120 - \theta) = \frac{91\sqrt{3}}{98} </cmath>
+
Which means
 +
<cmath>\sin(\theta)+\sin(120 - \theta) = \frac{91\sqrt{3}}{98}</cmath>
 +
<cmath>\frac{1}{2}\cos(\theta)+\frac{\sqrt{3}}{2}\sin(\theta) = \frac{91}{98}</cmath>
 +
<cmath>\sin(\theta + 30) = \frac{91}{98}</cmath>
 +
<cmath>\cos (\theta + 30) = \frac{21\sqrt{3}}{98}</cmath>
 +
<cmath>\tan(\theta + 30) = \frac{91}{21\sqrt{3}} = \frac{91\sqrt{3}}{63}</cmath>
  
−
So <cmath> \frac{1}{2}cos(\theta)+\frac{\sqrt{3}}{2}sin(\theta) = \frac{91}{98} </cmath>
+
Now, <math>\tan(\theta)</math> can be calculated using the addition identity, which gives the answer of
  
−
Therefore <cmath> sin(\theta + 30) = \frac{91}{98} </cmath>
+
<cmath>\boxed{\text{(B) }\frac{5\sqrt{3}}{11}}.</cmath>
  
−
And <cmath> cos (\theta + 30) = \frac{21\sqrt{3}}{98} </cmath>
+
~mitsuihisashi14
−
 
+
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] (fixed Latex error )
−
Which <cmath> tan(\theta + 30) = \frac{91}{21\sqrt{3}} = \frac{91\sqrt{3}}{63} </cmath>
 
−
 
 
−
<math>tan(\theta)</math> can be calculated using addition identity, which gives the answer of
 
  
−
<cmath>(B)\frac{5\sqrt{3}}{11}</cmath>
+
==Solution 3 (No Trig Manipulations)==
  
−
(I would really appreciate if someone can help me fix my code and format)
+
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>.
  
−
~mitsuihisashi14
+
-Magnetoninja
−
~[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==
 
==See also==
 
{{AMC12 box|year=2024|ab=B|num-b=18|num-a=20}}
 
{{AMC12 box|year=2024|ab=B|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

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

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

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