Poly-gone; So try-angles!

Any right triangle inscribed in a circle must have as its hypotenuse a diameter of the circle. There are $50$ pairs of diametrically opposed vertices in the given polygon, and for each of these pairs a right triangle can be formed when they are joined to any of the remaining $98$ vertices. The desired probability is then

$\dfrac{50*98}{\dbinom{100}{3}} = \dfrac{50*98}{\dfrac{100*99*98}{6}} = \dfrac{1}{33}.$

Thus $a + b = 1 + 33 = \boxed{34}.$

Since $100$ is even there is in fact a non zero probability to have a right triangle. It happens if and only if two vertices of the triangle are diametrically opposed (this is a very basic geometry fact on the circle).

We can start our thinking with the first vetrice $A$ arbitrarily fixed, since it is not a loss of generality. We can choose the other vertices $B$ e $C$ among the $99$ vertices that remains.

Choose $B$ first.

If $B$ is the vertice diametrically opposed to $A$ then for every choice of $C$ , $ABC$ is right on $C$ . The chance to have a right triangle is in this case the chance to choose $B$ as the only one vertice diametrically opposed to $A$ among the $99$ possibilities. So the probability is $\frac{1}{99}$ .

On the contrary $B$ is not diametrically opposed to $A$ with a probability of $\frac{98}{99}$ , but in this latter case we can have still a right triangle if $C$ is opposed (diametrically) to $A$ or to $B$ . So we have $2$ possibilities among $98$ for a choice of $C$ that gives a right triangle. So the probability in this case (that is obviously DISJOINT for the previous case) is $\frac{98}{99} \frac{2}{98} = \frac{2}{99}$ .

Summing up the two probability we get the probability to choose $A,B,C$ in a as vertices of a right triangle is

$\frac{1}{99} + \frac{2}{99} = \frac{3}{99} = \frac{1}{33}$

and the answer is $34$ .