So Many Great Circles

The first circle creates two areas.

After that, every great circle added intersects every great circle already there in two points. The number of intersections is maximized if none of these points coincide.

In that case adding circle number $n$ results in $2(n-1)$ new intersections. These intersections are connected by the same number of segments and each of these segments creates a new area. So adding a circle number $n$ adds $2(n-1)$ new areas.

The answer then is the $2$ areas generated by the first circle plus twice the sum of all numbers from 1 to $29$ .

Using the formula for the sum of first $n$ integers as $\frac{n(n+1)}2$ , we get $2+2\cdot\frac{29\cdot30}2=2+29\cdot30=\boxed{872}$ .

Relevant wiki: Linear Recurrence Relations - Problem Solving

Suppose that $n$ great circles can be used to form a maximum of $X_n$ regions on sphere. We will be able to create $X_n$ regions with no more than two great circles meeting at any point on the sphere.If we consider adding an extra great circle, it can be drawn so that no more than two great circles meet at any point on the sphere. The new great circle will meet each of the previous $n$ great circles twice.

Each time the new great circle intersects an old great circle, it will create a new region on the sphere. Thus we deduce that $X_{n+1} = X_n + 2n$ . Solving this recurrence relation, together with the initial condition $X_1 = 2$ , we obtain $X_n = n^2 - n + 2$ . Thus $X_{30} = \boxed{872}$ .

Relevant wiki: Regular Polyhedra

Since $2$ great circles can meet at $2$ different points, $n$ great circles can create at most $2 \cdot \binom{n}{2} = n(n - 1)$ intersection points.

Since each intersection point has $4$ arcs emanating from them (as long as the number of regions is being maximized), and each arc between intersection points shares $2$ points, there are $\frac{4}{2} \cdot n(n - 1) = 2n(n - 1)$ arcs between intersection points.

The intersection points, the arcs between the intersection points, and regions can be projected to the vertices, edges, and faces of a polyhedron, and so by Euler's formula $F + V = E + 2$ . Since $V = n(n - 1)$ and $E = 2n(n - 1)$ , $F + n(n - 1) = 2n(n - 1) + 2$ or $F = n^2 - n + 2$ .

Therefore, $30$ great circles will cut the sphere into $30^2 - 30 + 2 = \boxed{872}$ regions.

Consider any arrangement of $n$ circles on the sphere.
Take a point that doesn't lie on this sphere.
Take the stereographic projection of the sphere onto the plane.
We thus get the equivalent question of:

Given $n$ circles on the plane, what is the maximum number of regions that there can be? (Include the outside region)

The answer to this (much more familiar) problem is $n^2 - n + 2$ .

The proof is very similar to those above, where you count the number of intersection points. (Which shouldn't be surprising, because these are equivalent problems).

2 great circles can divide the sphere into at most 4 sections. In the case of 3 great circles, the 3rd is able to pass through at most 4 of the regions, and divides each one into 2, which adds 4 new regions for a total of 4+4=8 regions. Add a 4th great circle, and it's able to pass through at most 6 regions, dividing each into 2 and adding 6 new regions for a total of 8+6=14 regions. A 5th great circle can pass through at most 8 for total of 14+8=22 regions, and so on, adding the next even number to the total until 30 great circles, where you reach a total of 872 regions. I'm honestly not sure if this is a valid technique, but it yielded the correct answer.