Suppose three points are chosen at random in a circle. A triangle is made with these three points as vertices.
What's the probability that the triangle contains the origin of the circle?
(Although I have an answer, I thought I would throw it out there and see what others make of it!)
Answer
First, observe
That the triangle does not contain the center if and only if all three points are contained in a half-circle. So, only the polar angles of the points matter, not their distance to the center.
Next, let's
Label the points $A$, $B$, $C$. If we starting at point $A$ and go clockwise for half a circle, there's a $1/4$ chance that both point $B$ and point $C$ are in that half-circle, putting all the points on a half-circle. Likewise for starting at point $B$ and starting at point $C$.
These events are
Disjoint (well, overlap with probability 0), and comprise all the way the three points can be on a half-circle, since there's always exactly one for the other two are within half a circle clockwise. So, the probability of them being on a half-circle is $$3/4 = 1/4 + 1/4 + 1/4 $$and so the probability the triangle contains the center is the complement $$1 - 3/4 = 1/4$$ By the same argument, if we generalize to choosing $n$ points and connecting them in a polygon in cyclic order of polar angle, the probability that it contains the center is $$ 1 - \frac{n}{2^{n-1}}$$.
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