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1991 AIME Problems/Problem 14

Revision as of 03:46, 1 December 2024 by Eightyearletterman (talk | contribs) (solution 2)
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Problem

A hexagon is inscribed in a circle. Five of the sides have length $81$ and the sixth, denoted by $\overline{AB}$, has length $31$. Find the sum of the lengths of the three diagonals that can be drawn from $A_{}^{}$.

Solution

[asy]defaultpen(fontsize(9)); pair A=expi(-pi/2-acos(475/486)), B=expi(-pi/2+acos(475/486)), C=expi(-pi/2+acos(475/486)+acos(7/18)), D=expi(-pi/2+acos(475/486)+2*acos(7/18)), E=expi(-pi/2+acos(475/486)+3*acos(7/18)), F=expi(-pi/2-acos(475/486)-acos(7/18)); draw(unitcircle);draw(A--B--C--D--E--F--A);draw(A--C..A--D..A--E); dot(A^^B^^C^^D^^E^^F); label("\(A\)",A,(-1,-1));label("\(B\)",B,(1,-1));label("\(C\)",C,(1,0)); label("\(D\)",D,(1,1));label("\(E\)",E,(-1,1));label("\(F\)",F,(-1,0)); label("31",A/2+B/2,(0.7,1));label("81",B/2+C/2,(0.45,-0.2)); label("81",C/2+D/2,(-1,-1));label("81",D/2+E/2,(0,-1)); label("81",E/2+F/2,(1,-1));label("81",F/2+A/2,(1,1)); label("\(x\)",A/2+C/2,(-1,1));label("\(y\)",A/2+D/2,(1,-1.5)); label("\(z\)",A/2+E/2,(1,0)); [/asy]

Let $x=AC=BF$, $y=AD=BE$, and $z=AE=BD$.

Ptolemy's Theorem on $ABCD$ gives $81y+31\cdot 81=xz$, and Ptolemy on $ACDF$ gives $x\cdot z+81^2=y^2$. Subtracting these equations give $y^2-81y-112\cdot 81=0$, and from this $y=144$. Ptolemy on $ADEF$ gives $81y+81^2=z^2$, and from this $z=135$. Finally, plugging back into the first equation gives $x=105$, so $x+y+z=105+144+135=\boxed{384}$.

Solution 2

Let $\theta$ be the inscribed angle in each of the 5 sides of length 81, so $d \sin \theta = 81$. Since the inscribed angles sum to $\pi$, $d \sin 5\theta = d \sin (\pi - 5\theta) = 31$.

Now consider the Chebyshev polynomials that put $\dfrac{\sin n\theta}{\sin \theta}$ in terms of $\cos \theta$:

$\dfrac{\sin 2\theta}{\sin \theta} = 2 \cos \theta, \dfrac{\sin 3\theta}{\sin \theta} = 4 \cos^2 \theta - 1$

$\dfrac{\sin 4\theta}{\sin \theta} = 8 \cos^3 \theta - 4 \cos \theta, \dfrac{\sin 5\theta}{\sin \theta} = 16 \cos^4 \theta - 12 \cos^2 \theta + 1$

The sum of the diagonals is $d\sin 2\theta + d\sin 3\theta + d\sin 4\theta$, which becomes $(d \sin \theta)(8\cos^3 \theta + 4\cos^2 \theta - 2\cos \theta - 1)$, and we're given $16 \cos^4 \theta - 12 \cos^2 \theta + 1 = \dfrac{31}{81}$

Solve for $\cos \theta$: $16 \cos^4 \theta - 12 \cos^2 \theta + \dfrac{9}{4} = \dfrac{9}{4} - \dfrac{50}{81}$

$\left(4 \cos^2 \theta - \dfrac{3}{2}\right)^2 = \dfrac{729}{324} - \dfrac{200}{324} = \left(\dfrac{23}{18}\right)^2$

$8 \cos^2 \theta - 3 = \dfrac{23}{9}$ or $-\dfrac{23}{9}$, so $8 \cos^2 \theta = \dfrac{50}{9}$ or $\dfrac{4}{9}$

$\cos^2 \theta = \dfrac{25}{36}$ or $\dfrac{1}{18}$, which means $\cos \theta$ must be $\dfrac{5}{6}$ if $5 \theta < \pi$.

Now $(d \sin \theta)\left(8\cos^3 \theta + 4\cos^2 \theta - 2\cos \theta - 1\right) = 81\left(8 \cdot \dfrac{125}{216} + 4 \cdot \dfrac{25}{36} - 2 \cdot \dfrac{5}{6} - 1\right)$

$= 3 \left(8 \cdot \dfrac{125}{8} + 4 \cdot \dfrac{75}{4} - 2 \cdot \dfrac{45}{2} - 27\right) = 3(125 + 75 - 45 - 27) = \boxed{384}$

Video Solution by OmegaLearn

https://youtu.be/DVuf-uXjfzY?t=522

~ pi_is_3.14

See also

1991 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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