# Recurrence formula for division of a sphere by n great circles

Find the recurrence relation satisfied by $$R_n$$, where $$R_n$$ is the number of regions into which the surface of a sphere is divided by n great circles (which are the intersections of the sphere and planes passing through the center of the sphere), if no three of the great circles go through the same point

From: Rosen Discrete Mathematics textbook

My attempt: I was unable to visualize these greater circles for values of $$n \geq 4$$. So, I launched GeoGebra and made the following plot:

My initial planes were $$x=y$$, $$x=-y$$, and the $$z=0$$ plane. Now, I added a fourth yellow plane $$y=z$$. However, I still find it hard to understand how to calculate what number of new sections are formed. I am able to calculate the number of new sections in this specific case, but, I am unable to arrive at a general formula for the new number of sections, through recurrence.

I visited the following website - Slader - which happens to have a solution, where they mention the addition of $$2(n-1)$$ regions to the existing regions, after the addition of the $$n$$-th plane. However, I am unable to understand why exactly that is true.

• It's $2(n-1),$ not $3(n-1)$. Also, "splitting up into $2(n-1)$ regions" sounds like you end up with just $2(n-1)$ regions after the "splitting," whereas actually they are adding $2(n-1)$ new regions to the ones that already existed. – David K Oct 12 '18 at 10:53
• @DavidK Sorry you're right, I've made the necessary correction. – Gaurang Tandon Oct 12 '18 at 10:56

Why, that's simple: the $$n$$'th great circle intersects each and every one of the previous circles, right? OK, then how many intersection points are on it? Obviously, $$2(n-1)$$, because each pair of circles intersects twice.
Now, the $$2(n-1)$$ intersection points break our circle into as many pieces. Look at one such piece. What good does it do? See, it breaks some region of the sphere in two. In other words, it increments the number of regions by 1.