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, 2021

Combinatorics Difficulty 5.6 AIME, harder Prove it United States

Problem:
A chord is drawn on a circle by choosing two points uniformly at random along its circumference. This is done two more times to obtain three total random chords. The circle is cut along these three lines, splitting it into pieces. The probability that one of the pieces is a triangle is $\frac{m}{n}$ , where $m, n$ are positive integers and $\operatorname{gcd}(m, n)=1$ . Find $100 m+n$ .

Solution

Solution:
Instead of choosing three random chords, we instead first choose 6 random points on the circle and then choosing a random pairing of the points into 3 pairs with which to form chords. If the chords form a triangle, take a chord $C$ . Any other chord $C^{\prime}$ must have its endpoints on different sides of $C$ , since $C$ and $C^{\prime}$ intersect. Therefore, the endpoints of $C$ must be points that are opposite each other in the circle:

Conversely, if each point is connected to its opposite, the chords form a triangle unless these chords happen to be concurrent, which happens with probability 0. Therefore, out of the pairings, there is, almost always, exactly only one pairing that works. Since there are $\frac{1}{3!}\binom{6}{2}\binom{4}{2}\binom{2}{2}=15$ ways to pair 6 points into three indistinguishable pairs, the probability is $1 / 15$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.