In a certain [circle](https://artofproblemsolving.com/wiki/index.php/Circle), the [chord](https://artofproblemsolving.com/wiki/index.php/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 1
[2001AIME13.png](https://artofproblemsolving.com/wiki/index.php/File: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 $\boxed{174}$.