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!
Best Answer
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 $\frac{8}{13}$.