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2001 AIME I Problems/Problem 13: Difference between revisions

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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.  
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 \cdot CD + AC \cdot BD = AD \cdot BC</cmath>
<cmath>22x + 22(22) = (x + 20)^2</cmath>
<cmath>\iff 22x + 22 \cdot 22  = (x + 20)^2</cmath>
<cmath>22x + 484 = x^2 + 40x + 400</cmath>
<cmath>\iff 22x + 484 = x^2 + 40x + 400</cmath>
<cmath>0 = x^2 + 18x - 84</cmath>
<cmath>\iff x^2 + 18x - 84 = 0.</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 of this equation (since polygons obviously cannot have sides of negative length):
<cmath>x = \frac{-18 + \sqrt{18^2 + 4(84)}}{2}</cmath>
<cmath>x = \frac{-18 + \sqrt{18^2 + 4 \cdot 84}}{2} = \frac{-18 + \sqrt{660}}{2}.</cmath>
<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>.
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>.


=== Solution 2 ===
=== Solution 2 ===


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>
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>.


===Solution 3===
===Solution 3===


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:
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:
<cmath>22=2R\sin(z)</cmath>
<cmath>22=2R\sin(z),</cmath>
<cmath>20+a=2R\sin(2z)</cmath>
<cmath>20+a=2R\sin(2z), \quad\text{and}</cmath>
<cmath>a=2R\sin(3z)</cmath>
<cmath>a=2R\sin(3z).</cmath>
Dividing the second from the first we get <math>\cos(z)=\frac{20+a}{44}</math>
Dividing the second equation by the first, and using the double angle formula, we obtain <math>\cos(z)=\frac{20+a}{44}</math>.
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>
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>
Solving the quadratic equation gives the answer to be <math>\boxed{174}</math>.
and solving this quadratic equation gives the answer as <math>\boxed{174}</math>.


== See also ==
== See also ==

Latest revision as of 22:01, 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

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
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