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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 three chords of three <math>d</math>-degree arcs and one chord of a <math>3d</math>-degree arc. The diagonals of this trapezoid turn out to be 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 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.
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=== Solution 1 ===
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<center>[[File:2001AIME13.png]]</center>
  
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Using Ptolemy's theorem,
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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.
  
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<cmath>AB(CD) + AC(BD) = AD(BC)</cmath>
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Using [[Ptolemy's theorem]],
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<cmath>22(22) + 22x = (x + 20)^2</cmath>
 
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<cmath>484 + 22x = x^2 + 40x + 400</cmath>
 
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<cmath>0 = x^2 + 18x - 84</cmath>
 
  
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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>AB \cdot CD + AC \cdot BD = AD \cdot BC</cmath>
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<cmath>x = \[\frac{-18 + \sqrt{18^2 + 4(84)}}{2}}</cmath>
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<cmath>\iff 22x + 22 \cdot 22  = (x + 20)^2</cmath>
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<cmath>x = \[\frac{-18 + \sqrt{660}}{2}}</cmath>
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<cmath>\iff 22x + 484 = x^2 + 40x + 400</cmath>
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<cmath>\iff x^2 + 18x - 84 = 0.</cmath>
  
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<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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We can then apply the quadratic formula to find the positive root of this equation (since polygons obviously cannot have sides of negative length):
 +
<cmath>x = \frac{-18 + \sqrt{18^2 + 4 \cdot 84}}{2} = \frac{-18 + \sqrt{660}}{2}.</cmath>
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 +
This simplifies to <math>x = \frac{-18 + 2\sqrt{165}}{2} = -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, \quad\text{and}</cmath> <cmath>2R(\sin 2z-\sin 3z)=20.</cmath> Dividing the latter equation by the former gives <cmath>\frac{2\sin z\cos z-\left(3\cos^2z\sin z-\sin^3 z\right)}{\sin z}=2\cos z-\left(3\cos^2z-\sin^2z\right)=1+2\cos z-4\cos^2z=\frac{10}{11}</cmath> <cmath>\iff 4\cos^2z-2\cos z-\frac{1}{11}=0. \qquad (*)</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>(*),</math> this is equivalent to <math>44\cos z-20</math>. Using the quadratic formula, we deduce that this expression equals <math>-9+\sqrt{165}</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 a <math>3d</math>-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), \quad\text{and}</cmath>
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<cmath>a=2R\sin(3z).</cmath>
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Dividing the second equation by the first, and using the double angle formula, we obtain <math>\cos(z)=\frac{20+a}{44}</math>.
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Now, using the triple angle formula, we can rewrite the third equation as follows: <cmath>a=2R \sin(3z)=\frac{22}{\sin(z)}\left(3\sin(z)-4\sin^3(z)\right) = 22\left(3-4\sin^2(z)\right) = 22\left(4\cos^2(z)-1\right) = \frac{(20+a)^2}{22}-22,</cmath>
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and solving this quadratic equation gives the answer as <math>\boxed{174}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 23:01, 1 June 2025

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,

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

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

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

Solution 2

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

Solution 3

Let $z=\frac{d}{2}$, $R$ be the circumradius, and $a$ be the length of a $3d$-degree chord. Using the extended sine law, we obtain: \[22=2R\sin(z),\] \[20+a=2R\sin(2z), \quad\text{and}\] \[a=2R\sin(3z).\] Dividing the second equation by the first, and using the double angle formula, we obtain $\cos(z)=\frac{20+a}{44}$. Now, using the triple angle formula, we can rewrite the third equation as follows: \[a=2R \sin(3z)=\frac{22}{\sin(z)}\left(3\sin(z)-4\sin^3(z)\right) = 22\left(3-4\sin^2(z)\right) = 22\left(4\cos^2(z)-1\right) = \frac{(20+a)^2}{22}-22,\] and solving this quadratic equation gives the answer as $\boxed{174}$.

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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