<s>[INST] In a circle, parallel chords of length $2$, $3$, and $4$ I was helping my comrade answering some questions taken from review classes when I stumbled upon this question. It looks like this:

In a circle, parallel chords of length $2$, $3$, and $4$ determine central angles of $A$, $B$, and $A+B$ radians, respectively, where 
  $A+B < \pi$. Express the $\cos A$ into a fraction in lowest term and find the sum of the numerator and denominator of the fraction. 

My work
The problem sounds difficult for us, so we came here to seek help. How do you answer the above problem?
 [/INST] Note: this question is basically identical to the 1985 AIME Problem 9, so I'll summarise the solution below. 


Because all chords of a given length in a given circle subtend the same arc and therefore the same central angle, we can rearrange our chords into a triangle with the circle as its circumcircle. 

This triangle has a semiperimeter of $\frac{9}{2}$, and so by Heron's formula, this triangle has area $\sqrt{\frac{9}{2}\times\frac{5}{2}\times\frac{3}{2}\times\frac{1}{2}}=\frac{3}{4}\sqrt{15}$
Now, the area of a triangle with circumradius $R$ and sides $a,b,c$ is given by $\frac{abc}{4R}$, so $\frac{3}{4}\sqrt{15}=\frac{6}{R}$, which means $R=\frac{8}{\sqrt{15}}$. 
Now, consider the triangle formed by  two radii and the chord of length $2$ (see Fig. 2). Because this isosceles triangle has vertex angle $\alpha$, by the Law of Cosines, $$2^2=R^2+R^2-2R^2\cos\alpha\implies\cos\alpha=\frac{2R^2-4}{2R^2}=\frac{17}{32}$$ 
So $\alpha$ in lowest terms is $\frac{17}{32}$, and the sum of the numerator and denominator is equal to $17+32=49$

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