In a [circle](https://artofproblemsolving.com/wiki/index.php/Circle), [parallel](https://artofproblemsolving.com/wiki/index.php/Parallel) [chords](https://artofproblemsolving.com/wiki/index.php/Chord) of lengths 2, 3, and 4 determine [central angles](https://artofproblemsolving.com/wiki/index.php/Central_angle) of $\alpha$, $\beta$, and $\alpha + \beta$ [radians](https://artofproblemsolving.com/wiki/index.php/Radian), respectively, where $\alpha + \beta < \pi$. If $\cos \alpha$, which is a [positive](https://artofproblemsolving.com/wiki/index.php/Positive) [rational number](https://artofproblemsolving.com/wiki/index.php/Rational_number), is expressed as a [fraction](https://artofproblemsolving.com/wiki/index.php/Fraction) in lowest terms, what is the sum of its numerator and denominator?
All chords of a given length in a given circle subtend the same [arc](https://artofproblemsolving.com/wiki/index.php/Arc) and therefore the same central angle.  Thus, by the given, we can re-arrange our chords into a [triangle](https://artofproblemsolving.com/wiki/index.php/Triangle) with the circle as its [circumcircle](https://artofproblemsolving.com/wiki/index.php/Circumcircle).  


This triangle has [semiperimeter](https://artofproblemsolving.com/wiki/index.php/Semiperimeter) $\frac{2 + 3 + 4}{2}$ so by [Heron's formula](https://artofproblemsolving.com/wiki/index.php/Heron%27s_formula) it has [area](https://artofproblemsolving.com/wiki/index.php/Area) $K = \sqrt{\frac92 \cdot \frac52 \cdot \frac32 \cdot \frac12} = \frac{3}{4}\sqrt{15}$.  The area of a given triangle with sides of length $a, b, c$ and circumradius of length $R$ is also given by the formula $K = \frac{abc}{4R}$, so $\frac6R = \frac{3}{4}\sqrt{15}$ and $R = \frac8{\sqrt{15}}$.
Now, consider the triangle formed by two radii and the chord of length 2.  This [isosceles triangle](https://artofproblemsolving.com/wiki/index.php/Isosceles_triangle) has vertex angle $\alpha$, so by the [Law of Cosines](https://artofproblemsolving.com/wiki/index.php/Law_of_Cosines),
\[2^2 = R^2 + R^2 - 2R^2\cos \alpha \Longrightarrow \cos \alpha = \frac{2R^2 - 4}{2R^2} = \frac{17}{32}\]
and the answer is $17 + 32 = \boxed{049}$.