This is problem 9.90 from A path to combinatorics for undergraduates.
Let $n$ be a positive integer. A convex polygon $A_1A_2...A_n$ is inscribed in the circle $\omega$. All of the diagonals of the polygon are drawn. Determine the maximum number of regions that are enclosed by the sides and diagonals of the polygon and the arcs of $\omega$.
Here is my attempt at the problem.
Define $a_n$ to be the maximum number of regions that are enclosed by the sides and diagonals of a convex polygon $A_1A_2...A_n$ that is enclosed in a circle $\omega$ and the arcs of $\omega$.
By drawing a picture, we clearly have $a_3=4$.
Let $n\ge 3$. Suppose we have a convex polygon $A_1A_2...A_n$, with all of its diagonal drawn, that is enclosed in a circle $\omega$. Add the point $A_{n+1}$ in the arc $\alpha$ with end points $A_1$ and $A_n$ that contains none of the points $A_2,A_3,...,A_{n-1}$. Join $A_1A_{n+1}$ and $A_nA_{n+1}$. This divides the region enclosed by $A_1A_n$ and $\alpha$ into three regions, the region that is the triangle $A_1A_nA_{n+1}$, the arc with end points $A_1,A_{n+1}$ that contains no other vertices of the polygon and the arc with end points $A_n,A_{n+1}$ that contains no other vertices of the polygon. So we now have two more regions than before. We draw the rest of the diagonals of the polygon $A_1A_2...A_{n+1}$ by joining $A_jA_{n+1}$ for any $2\le j\le n-1$. Note that $A_jA_{n+1}$ divides the circle into two arcs, one containing the points $A_1,A_2,...,A_{j-1}$ and the other one containing the points $A_{j+1},...,A_n$. The diagonals that $A_jA_{n+1}$ intersect are exactly the diagonals formed by joining a point from $A_1,A_2,...,A_{j-1}$ and another point from $A_{j+1},...,A_n$, for a total of $(j-1)(n-j)$ diagonals. Each time $A_jA_{n+1}$ intersects with a diagonal, two new region is formed, so there is one more region than before. Note that we also have to account for one more region that is formed when we finally finish joining $A_jA_{n+1}$. We have $$\begin{align}a_{n+1}&=a_n+2+\sum_{j=2}^{n-1}((j-1)(n-j)+1)\\&=a_n+2-\sum_{j=2}^{n-1}j^2+(n+1)\sum_{j=2}^{n-1}j+\sum_{j=2}^{n-1}(-n+1)\\&=a_n+2-(\frac{(n-1)(n)(2(n-1)+1)}{6}-1)+(n+1)(\frac{(n-1)(n)}{2}-1)+(n-2)(-n+1)\\&=a_n+\frac{1}{6}n^3-\frac{1}{2}n^2+\frac{4}{3}n\\&=a_3+\frac{1}{6}(\sum_{j=1}^{n}j^3-9)-\frac{1}{2}(\sum_{j=1}^{n}j^2-5)+\frac{4}{3}(\sum_{j=1}^{n}j-3)\\&=4+\frac{1}{6}(\frac{n^2(n+1)^2}{4}-9)-\frac{1}{2}(\frac{n(n+1)(2n+1)}{6}-5)+\frac{4}{3}(\frac{n(n+1)}{2}-3)\\&=\frac{1}{24}n^4-\frac{1}{12}n^3+\frac{11}{24}n^2+\frac{7}{12}n+1\end{align}$$
Now I don't know under what geometric condition is the maximum achieved or if the above is value is achievable and how $a_1$ and $a_2$ should be defined.
