Difference between revisions of "1991 AIME Problems/Problem 14"

Latest revision as of 03:46, 1 December 2024

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

@@ Line 20: / Line 20: @@
 [[Ptolemy's Theorem]] on <math>ABCD</math> gives <math>81y+31\cdot 81=xz</math>, and Ptolemy on <math>ACDF</math> gives <math>x\cdot z+81^2=y^2</math>.
 Subtracting these equations give <math>y^2-81y-112\cdot 81=0</math>, and from this <math>y=144</math>. Ptolemy on <math>ADEF</math> gives <math>81y+81^2=z^2</math>, and from this <math>z=135</math>. Finally, plugging back into the first equation gives <math>x=105</math>, so <math>x+y+z=105+144+135=\boxed{384}</math>.
+== Solution 2 ==
+Let <math>\theta</math> be the inscribed angle in each of the 5 sides of length 81, so <math>d \sin \theta = 81</math>. Since the inscribed angles sum to <math>\pi</math>, <math>d \sin 5\theta = d \sin (\pi - 5\theta) = 31</math>.
+Now consider the Chebyshev polynomials that put <math>\dfrac{\sin n\theta}{\sin \theta}</math> in terms of <math>\cos \theta</math>:
+<math>\dfrac{\sin 2\theta}{\sin \theta} = 2 \cos \theta, \dfrac{\sin 3\theta}{\sin \theta} = 4 \cos^2 \theta - 1</math>
+<math>\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</math>
+The sum of the diagonals is <math>d\sin 2\theta + d\sin 3\theta + d\sin 4\theta</math>, which becomes <math>(d \sin \theta)(8\cos^3 \theta + 4\cos^2 \theta - 2\cos \theta - 1)</math>, and we're given <math>16 \cos^4 \theta - 12 \cos^2 \theta + 1 = \dfrac{31}{81}</math>
+Solve for <math>\cos \theta</math>: <math>16 \cos^4 \theta - 12 \cos^2 \theta + \dfrac{9}{4} = \dfrac{9}{4} - \dfrac{50}{81}</math>
+<math>\left(4 \cos^2 \theta - \dfrac{3}{2}\right)^2 = \dfrac{729}{324} - \dfrac{200}{324} = \left(\dfrac{23}{18}\right)^2</math>
+<math>8 \cos^2 \theta - 3 = \dfrac{23}{9}</math> or <math>-\dfrac{23}{9}</math>, so <math>8 \cos^2 \theta = \dfrac{50}{9}</math> or <math>\dfrac{4}{9}</math>
+<math>\cos^2 \theta = \dfrac{25}{36}</math> or <math>\dfrac{1}{18}</math>, which means <math>\cos \theta</math> must be <math>\dfrac{5}{6}</math> if <math>5 \theta < \pi</math>.
+Now <math>(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)</math>
+<math>= 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}</math>
 == Video Solution by OmegaLearn ==

1991 AIME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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