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Difference between revisions of "2001 AIME I Problems/Problem 13"

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== Solution ==
 
== Solution ==
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                                                      [[File:2001AIME13.jpg]]
 
  
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Note that a cyclic quadrilateral in the form of an isosceles trapezoid can be formed from the three chords of the three <math>d</math>-degree arcs and the chord of the <math>3d</math>-degree arc. The diagonals of this trapezoid turn out to be the two chords of the <math>2d</math>-degree arcs. Let <math>AB</math>, <math>AC</math>, and <math>AD</math> be the chords of the <math>d</math>-degree arcs, and let <math>CD</math> be the chord of the <math>3d</math>-degree arc. Also let <math>x</math> be equal to the chord of the <math>3d</math>-degree arc. Hence, the length of the chords of the <math>2d</math>-degree arcs can be represented as <math>x + 20</math>, as given in the problem.  
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=== Solution 1 ===
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<center>[[File:2001AIME13.png]]</center>
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Note that a cyclic quadrilateral in the form of an isosceles trapezoid can be formed from three chords of three <math>d</math>-degree arcs and one chord of one <math>3d</math>-degree arc. The diagonals of this trapezoid turn out to be two chords of two <math>2d</math>-degree arcs. Let <math>AB</math>, <math>AC</math>, and <math>BD</math> be the chords of the <math>d</math>-degree arcs, and let <math>CD</math> be the chord of the <math>3d</math>-degree arc. Also let <math>x</math> be equal to the chord length of the <math>3d</math>-degree arc. Hence, the length of the chords, <math>AD</math> and <math>BC</math>, of the <math>2d</math>-degree arcs can be represented as <math>x + 20</math>, as given in the problem.  
  
 
Using Ptolemy's theorem,
 
Using Ptolemy's theorem,
  
 
<cmath>AB(CD) + AC(BD) = AD(BC)</cmath>
 
<cmath>AB(CD) + AC(BD) = AD(BC)</cmath>
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<cmath>22(22) + 22x = (x + 20)^2</cmath>
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<cmath>22x + 22(22) = (x + 20)^2</cmath>
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<cmath>484 + 22x = x^2 + 40x + 400</cmath>
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<cmath>22x + 484 = x^2 + 40x + 400</cmath>
 
<cmath>0 = x^2 + 18x - 84</cmath>
 
<cmath>0 = x^2 + 18x - 84</cmath>
  
 
We can then apply the quadratic formula to find the positive root to this equation since polygons obviously cannot have sides of negative length.
 
We can then apply the quadratic formula to find the positive root to this equation since polygons obviously cannot have sides of negative length.
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<cmath>x = \[\frac{-18 + \sqrt{18^2 + 4(84)}}{2}}</cmath>
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<cmath>x = \frac{-18 + \sqrt{18^2 + 4(84)}}{2}</cmath>
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<cmath>x = \[\frac{-18 + \sqrt{660}}{2}}</cmath>
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<cmath>x = \frac{-18 + \sqrt{660}}{2}</cmath>
  
 
<math>x</math> simplifies to <math>\frac{-18 + 2\sqrt{165}}{2},</math> which equals <math>-9 + \sqrt{165}.</math> Thus, the answer is <math>9 + 165 = \boxed{174}</math>.
 
<math>x</math> simplifies to <math>\frac{-18 + 2\sqrt{165}}{2},</math> which equals <math>-9 + \sqrt{165}.</math> Thus, the answer is <math>9 + 165 = \boxed{174}</math>.
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=== Solution 2 ===
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Let <math>z=\frac{d}{2},</math> and <math>R</math> be the circumradius. From the given information, <cmath>2R\sin z=22</cmath> <cmath>2R(\sin 2z-\sin 3z)=20</cmath> Dividing the latter by the former, <cmath>\frac{2\sin z\cos z-(3\cos^2z\sin z-\sin^3 z)}{\sin z}=2\cos z-(3\cos^2z-\sin^2z)=1+2\cos z-4\cos^2z=\frac{10}{11}</cmath> <cmath>4\cos^2z-2\cos z-\frac{1}{11}=0 (1)</cmath> We want to find <cmath>\frac{22\sin (3z)}{\sin z}=22(3-4\sin^2z)=22(4\cos^2z-1).</cmath> From <math>(1),</math> this is equivalent to <math>44\cos z-20.</math> Using the quadratic formula, we find that the desired length is equal to <math>\sqrt{165}-9,</math> so our answer is <math>\boxed{174}</math>
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===Solution 3===
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Let <math>z=\frac{d}{2}</math>, <math>R</math> be the circumradius, and <math>a</math> be the length of 3d degree chord. Using the extended sine law, we obtain:
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<cmath>22=2R\sin(z)</cmath>
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<cmath>20+a=2R\sin(2z)</cmath>
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<cmath>a=2R\sin(3z)</cmath>
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Dividing the second from the first we get <math>\cos(z)=\frac{20+a}{44}</math>
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By the triple angle formula we can manipulate the third equation as follows: <math>a=2R\times \sin(3z)=\frac{22}{\sin(z)} \times (3\sin(z)-4\sin^3(z)) = 22(3-4\sin^2(z))=22(4\cos^2(z)-1)=\frac{(20+a)^2}{22}-22</math>
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Solving the quadratic equation gives the answer to be <math>\boxed{174}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 18:25, 27 November 2022

Problem

In a certain circle, the chord of a $d$-degree arc is $22$ centimeters long, and the chord of a $2d$-degree arc is $20$ centimeters longer than the chord of a $3d$-degree arc, where $d < 120.$ The length of the chord of a $3d$-degree arc is $- m + \sqrt {n}$ centimeters, where $m$ and $n$ are positive integers. Find $m + n.$

Solution

Solution 1

2001AIME13.png

Note that a cyclic quadrilateral in the form of an isosceles trapezoid can be formed from three chords of three $d$-degree arcs and one chord of one $3d$-degree arc. The diagonals of this trapezoid turn out to be two chords of two $2d$-degree arcs. Let $AB$, $AC$, and $BD$ be the chords of the $d$-degree arcs, and let $CD$ be the chord of the $3d$-degree arc. Also let $x$ be equal to the chord length of the $3d$-degree arc. Hence, the length of the chords, $AD$ and $BC$, of the $2d$-degree arcs can be represented as $x + 20$, as given in the problem.

Using Ptolemy's theorem,

\[AB(CD) + AC(BD) = AD(BC)\] \[22x + 22(22) = (x + 20)^2\] \[22x + 484 = x^2 + 40x + 400\] \[0 = x^2 + 18x - 84\]

We can then apply the quadratic formula to find the positive root to this equation since polygons obviously cannot have sides of negative length. \[x = \frac{-18 + \sqrt{18^2 + 4(84)}}{2}\] \[x = \frac{-18 + \sqrt{660}}{2}\]

$x$ simplifies to $\frac{-18 + 2\sqrt{165}}{2},$ which equals $-9 + \sqrt{165}.$ Thus, the answer is $9 + 165 = \boxed{174}$.

Solution 2

Let $z=\frac{d}{2},$ and $R$ be the circumradius. From the given information, \[2R\sin z=22\] \[2R(\sin 2z-\sin 3z)=20\] Dividing the latter by the former, \[\frac{2\sin z\cos z-(3\cos^2z\sin z-\sin^3 z)}{\sin z}=2\cos z-(3\cos^2z-\sin^2z)=1+2\cos z-4\cos^2z=\frac{10}{11}\] \[4\cos^2z-2\cos z-\frac{1}{11}=0 (1)\] We want to find \[\frac{22\sin (3z)}{\sin z}=22(3-4\sin^2z)=22(4\cos^2z-1).\] From $(1),$ this is equivalent to $44\cos z-20.$ Using the quadratic formula, we find that the desired length is equal to $\sqrt{165}-9,$ so our answer is $\boxed{174}$

Solution 3

Let $z=\frac{d}{2}$, $R$ be the circumradius, and $a$ be the length of 3d degree chord. Using the extended sine law, we obtain: \[22=2R\sin(z)\] \[20+a=2R\sin(2z)\] \[a=2R\sin(3z)\] Dividing the second from the first we get $\cos(z)=\frac{20+a}{44}$ By the triple angle formula we can manipulate the third equation as follows: $a=2R\times \sin(3z)=\frac{22}{\sin(z)} \times (3\sin(z)-4\sin^3(z)) = 22(3-4\sin^2(z))=22(4\cos^2(z)-1)=\frac{(20+a)^2}{22}-22$ Solving the quadratic equation gives the answer to be $\boxed{174}$.

See also

2001 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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