The great circle

Let $m_n$ , $M_n$ be the minimum and maximum, respectively, number of regions that a sphere can be divided into by $n$ great circles.

Suppose that a number of great circles have been drawn (which divides the sphere into a number of regions), and that we now draw an additional circle. The number of intersections of this new circle with the previous circles is equal to the number of additional regions created (by splitting previous regions into two) by the new circle. Since every great circle meets any other great circle at two points, the number of intersections between the new great circle and the previous ones can be as small as $2$ and could be as great as $2(n-1)$ .

The smallest number of regions will be created if there are exactly $2$ intersections of each new great circle with the previous circles. This is possible if all the circles pass through the North and South poles (so are circles of longitude). Thus $m_n = 2 + 2\times(n-1) = 2n$ .

The largest number of regions will be be created if the $j$ th great circle has $2(j-1)$ intersections with the previous $j-1$ circles, in which case $M_n = 2 + \sum_{j=2}^n 2(j-1) = n^2 - n + 2$ . This is always possible. Suppose that we have drawn $j$ great circles to form $M_j = j^2 - j + 2$ regions. These $j$ circles will intersect with each other a total of $j^2-j$ times (a finite number). Since there are uncountably many great circles, it is bound to be possible to draw a great circle which misses all these intersections, and hence meets the $j$ circles at $2j$ new points. This will mean that we have drawn $j+1$ great circles which subdivide the sphere into $(j^2 - j + 2) + 2j = j^2 + j + 2 = M_{j+1}$ regions. The result follows by induction, since the ground step $M_1 = 2$ is trivial.

Thus $M_n + m_n = n^2 + n + 2$ , which equals $\boxed{32}$ when $n=5$ .

Relevant wiki: Euler Characteristic

You can use Euler's formula for this: $f=e-v+2$ , where $v$ is the number of vertices, $e$ is the number of edges and $f$ the number of faces/areas (which is what we need).

We can solve the problem for $n$ circles.

If we make sure the circles all have different intersections, the resulting graph on the sphere has $n^2-n$ vertices (the sum $2+4+\ldots+2(n-1)$ : the $n$ -th circle has $2$ intersections with each of the existing $n-1$ circles) and $n\cdot2(n-1)$ edges (there are $2(n-1)$ vertices on each circle).

For $n=5$ this results in $f=5\cdot8-(25-5)+2=22$ .

Comments to the other solution already show that the situation where all circles have different intersections is obtainable (in fact almost all configurations of circles have all different intersections).

This situation is also easily seen to cause the largest number of areas: for equal intersections the number of vertices decreases, but the number of edges decreases twice as fast, so the number of faces/areas decreases.

In the extreme case all circles have the same intersection, resulting in $2$ vertices and $2n$ edges, so for $n=5$ we get $f=2\cdot5-2+2=10$ .

The final answer for $n=5$ is $22+10=32$ .

Here's another way to visualize the cuts of the sphere. Imagine the first great circle cut is the equator, and you're looking down the axis. Then 4 more great circles cut the surface of the sphere as the sequence shows. The last one shows 11 pieces, with 11 more on the other side, making it 22 in all. Add to that 10 pieces which is the minimum that can be cut with 5 great circle planes.

The diagram is not an accurate representation of how great circles would cut the sphere, which would form half ellipses instead of straight lines, as seen from the top. Then the other side will look the same as the first.

This problem is related to the Lazy Caterer's Sequence

Hey Man , I do not know much about Mathematics, but it seems to me that A space can be considered as a sphere(indeed a Hypersphere) and vice-versa. That's the juice of polar coordinates. Now the Question is, in how many parts atmost plane can divide a space. To solve this we use the following general argument. Consider n planes in space intersecting in such a way that the space is divided into maximum number of parts. Indeed 3 planes intersect at a vertex and there are ${n\choose 3}$ such vertices. Each vertex is considered in exactly one part of space. Thus there are ${n\choose 3}$ parts .But that's not all the parts. The planes which are not bounded below ( i.e don't form a vertex) cut a reference plane P in n+1 lines. These lines divide the plane P in ${P_n}$ parts. In a Similar line of argument we can show that $P_n = {n\choose 2}$ + ${n\choose 1}$ + ${n\choose 0}$ . Thus Maximum number of space parts is $M= {n\choose 3}$ + ${n\choose 2}$ + ${n\choose 1}$ + ${n\choose 0}$ Minimum number of space parts is $n+1$ . Put n=5, And $M +m= 32$

The minimum number of intersections is 2. This is all five circles intersecting at opposite points on the sphere (like lines of longitude on a globe).

To create the maximum number of intersections, every circle would need to go through every other circle. With only 2 circles this is 2 times. With 3 circles this is 4 times. 2 Times from the previous 2 circles and 2 more times from the new circle intersecting the old circles. With 4 circles this is 10 times. 4 Times from the previous 3 circles and 6 more times from the new circle intersecting the old circles. Continuing this pattern gives us 2+4+6+8+10=30 times as M.

m+M=32