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A regular polygon with $100$ sides is inscribed in a circle. What is the probability that three randomly chosen vertices of this polygon form a right angled triangle?

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    $\begingroup$ What did you try so far? $\endgroup$
    – Quimey
    May 16 '14 at 18:05
  • $\begingroup$ I think you can "fix" a vertex and workout some condition for the other two. $\endgroup$
    – Quimey
    May 16 '14 at 18:06
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There are $100\choose 3$ possible choices. There are $50\cdot 98$ way to pick a diagonal that is a diameter of the circumscribed circle and a point off the diagonal.

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An inscribed triangle is a right triangle if its largest side is a diameter of the circle. If the vertices of the regular $100$-gon are numbered $0$ through $99$ then a diameter connects vertices whose numbers differ by $50$. Without loss of generality we may choose to label the first vertex chosen as $0$. Given two numbers $x$ and $y$ selected uniformly, randomly, without replacement from $1$-$99$ we seek the probability that either number is $50$, or that $\lvert x-y\rvert = 50$. Note that these are disjoint events they cannot both be true.

I will end at this point in case this is an unlabeled homework assignment. The rest is rather straightforward.

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