# 2001 AIME I Problems/Problem 13

## 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, $$AB(CD) + AC(BD) = AD(BC)$$ $$22x + 22(22) = (x + 20)^2$$ $$22x + 484 = x^2 + 40x + 400$$ $$0 = x^2 + 18x - 84$$

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 $\boxed{174.}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 