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Two diagonals of a regular nonagon (a $9$-sided polygon) are chosen. What is the probability that their intersection lies inside the nonagon?

This is similar to a previous problem that I posted about a dodecahedron. However, this provides more difficult, as a nonagon is (in my opinion) more difficult to work with than a dodecahedron. A dodecahedron, in every sense of the object, is even (even edges, even vertices...). This cannot be said for the $2D$ Nonagon.

Here is what I'm thinking. Draw a nonagon. Find all of the intersection points of the nonagon. A quick glance at online diagram of a nonagon (with diagonals drawn) shows that this method would fall flat quickly. Finding all of the intersection points can be found with $\binom{9}{4}=126$. This means that $126$ is the numerator. If only I could find the denominator....

Help is greatly appreciated!

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3 Answers

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There are $\frac{9*(9-3)}{2}=27$ diagonals in a nonagon. There are 27C2 ways to choose two diagonals, which is $351$. This is our denominator. The only way the diagonals can intersect inside the nonagon is if they share an endpoint. For each diagonal, there are $5$ other diagonals that share one endpoint, and 5 that share the other for a total of $10$ ways for a certain diagonal to share an endpoint with another. $27$ diagonals means $\frac{10*27}{2}=135$ ways to have adjacent diagonals. The probability for the diagonals to intersect is $\frac{135}{351}=\frac{5}{13}$. Taking the complement of this gives us an answer of $\frac{8}{13}$.

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There are $27$ diagonals in a convex nonagon $P$, and you can choose $2$ of them in ${27\choose 2}=351$ ways. In order to intersect in the interior of $P$ the two diagonals have to have different endpoints, a total of four. Conversely any four vertices of $P$ determine exactly one pair of diagonals intersecting in the interior of $P$, hence there are ${9\choose4}=126$ such pairs. The probability in question therefore is ${126\over351}={14\over39}$.

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Call the points $1$ through $9$ clockwise. The first point can be $1$ by symmetry and the second can be $3,4,5$. If it is $3$ the diagonals cross if $2$ is a point of the other diagonal, which is $6$ of the $26$ choices. If the second point is $4$ the diagonals cross if one point is $2,3$ and the other is $5-9$, which is $10$ of the $26$. If the second point is $5$ the diagonals cross if one is $2-4$ and the other is $6-9$, which is $12$ cases of $26$. The overall probability is $\frac 13\cdot \frac {6+10+12}{26}=\frac {28}{78}=\frac {14}{39}$

This agrees with your $\frac {126}{351}$, which is correct.

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