$N$ circles in a plane all intersect each other such that every circle intersects every other circle at exactly 2 points. Find in terms of $N$ the minimum and maximum number of disjoint closed regions that can be formed. Hint: You may wish to check your answers (visually) for greater $N,$ e.g. $N=4.$
  1. Given that $a_1=1$ and $a_n=2a_{n-1}$ find $a_{20}.$

Try to think about adding circles one at a time.

Consider the number of different points at which circles intersect.

Try to achieve the maximum and the minimum number of intersection points.

How many existing regions can a newly added circle pass through?

Note that a newly added circle does not (necessarily) pass through every previous region. In fact, after the $3^{rd}$ circle that’s impossible.

How does the number of new intersection points relate to the number of new regions?

Denote with $N_k$ the number of disjoint regions after $k$ circles have been added. The base case is $N_1=1.$

The minimum occurs when the 2 points are the same for all pairs of circles, and each new circle is bigger than the last one added. Adding a new circle slices one internal region in two and adds a new region outside of the previous union. Thus we have $N_{k+1} = N_k+2.$ Hence $N_{k}=2k-1.$

The maximum occurs when each new added circle intersects each existing circle at two different points for every circle (maximum number of intersection points). Note that a newly added circle does not necessarily pass through every previous region; in fact, after the 3rd circle that’s impossible. In this case, adding a new circle will add as many new regions as twice the number of existing circles. In other words, $N_{k+1}=N_k+2k.$ We have $N_{k+1}=1+2\sum_{i=1}^{k}i=1+k(k+1)$ or $N_k=1+k(k-1).$