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!

  • $\begingroup$ I also know that there are $27$ diagonals in a regular nonagon. I think that the denominator would be $\binom{27}{2}=351$. This means that the answer would be $\frac{14}{39}$. Can someone verify this or correct me? $\endgroup$ – A Piercing Arrow Feb 27 '18 at 5:21

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 ${126\over351}={14\over39}$.

  • $\begingroup$ Note that this method generalizes easily to an arbitrary $n$-gon: it has ${n \choose 2} - n = \frac{1}{2}n(n-3)$ interior diagonals, and there are ${n \choose 4}$ pairs of diagonals that intersect. This means that the OP's concerns about the number of vertices being even or odd is moot. $\endgroup$ – Michael Seifert Feb 27 '18 at 15:37

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 $\frac{8}{13}$.

  • $\begingroup$ This is not correct. If you number the vertices $1$ to $9$, the diagonals $1-3$ and $4-8$ do not intersect inside the nonagon. $\endgroup$ – Ross Millikan Feb 27 '18 at 5:31
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    $\begingroup$ This is still not correct. Diagonals $1-5$ and $2-6$ intersect within the nonagon and do not share an endpoint. If they share an endpoint they cannot intersect. $\endgroup$ – Ross Millikan Feb 27 '18 at 5:42
  • $\begingroup$ Ross. Can you provide a solution that patches up on Jeffrey's? $\endgroup$ – A Piercing Arrow Feb 27 '18 at 5:47
  • $\begingroup$ You made me realize that I was assuming all the endpoints were distinct, which OP did not specify. $\endgroup$ – Ross Millikan Feb 27 '18 at 5:48

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.

  • $\begingroup$ Unfortunately, this is incorrect. $\endgroup$ – A Piercing Arrow Feb 27 '18 at 6:04
  • $\begingroup$ @APiercingArrow: It would be more helpful if you would point out what you think is the problem. I could then consider your argument. That is what I did in my comments to Jeffrey H. and my first paragraph. $\endgroup$ – Ross Millikan Feb 27 '18 at 6:09
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    $\begingroup$ Your edited answer is a bit confusing; the first and last sentence currently contradict each other. $\endgroup$ – Michael Seifert Feb 27 '18 at 15:34

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