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

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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>.
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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=MH+HB=\frac{7\sqrt{3}}{3}+2\sqrt{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
 
-Magnetoninja

Revision as of 21:24, 16 April 2025

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)

[asy] // Modified Asymptote diagram with correct right angle mark and alpha label. import olympiad; defaultpen(fontsize(13)); size(200); // Circumradius R = 14*sqrt(3)/3 real R = 14*sqrt(3)/3; // Define original points with scaling pair O = (0,0); pair A = R * dir(225); pair B = R * dir(-15); pair C = R * dir(105); pair D = rotate(38.21, O) * A; pair E = rotate(38.21, O) * B; pair F = rotate(38.21, O) * C; // Point H: Foot of altitude from B to DE pair DE_vec = E - D; pair perp_DE = (DE_vec.y, -DE_vec.x); // Rotate -90 degrees perp_DE = perp_DE / length(perp_DE); // Unit vector pair H = B - (2*sqrt(3)) * perp_DE; // Corrected direction // Point M: Extend BH to M, BM = 13*sqrt(3)/3 pair BH_vec = H - B; pair BM_unit = BH_vec / length(BH_vec); // Unit vector along BH pair M = B + (13*sqrt(3)/3) * BM_unit; // BM = 13*sqrt(3)/3 // Point P: Midpoint of BE pair P = (B + E) / 2; // Draw original triangles and polygon 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); // Draw new dashed segments draw(B--H, black+0.7+dashed); draw(H--M, black+0.7+dashed); draw(O--M, black+0.7+dashed); draw(O--P, black+0.7+dashed); draw(P--E, black+0.7+dashed); draw(O--B, black+0.7+dashed); // Draw right angle marker at OMB draw(rightanglemark(B, M, O, 15)); // Right angle at M, size 15 // Draw dots and labels dot(O); dot("$O$", O, dir(180)); 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)); dot("$H$", H, dir(270)); dot("$M$", M, dir(90)); dot("$P$", P, dir(0)); // Label angle POB as alpha, to the right of O draw(anglemark(B, O, P, 50)); // Angle mark at O label("$\alpha$", O + (1.5, -.23), dir(0)); // Position to the right of O [/asy]

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 $MB=MH+HB=\frac{7\sqrt{3}}{3}+2\sqrt{3}=\frac{13\sqrt{3}}{3}$. $OB=\frac{14\sqrt{3}}{3}$ 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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