Great Circles

We consider $m$ and $M$ as functions of $k$ , the number of great circles drawn on the sphere. Let $M_{k}$ ( $m_{k}$ ) denote the maximum (minimum) number of regions in which the sphere can be dissected by k great circles. Then $M_{1}$ = $m_{1}$ =2.

Assume that there k great circles drawn on the sphere, and the sphere is dissected into $a_{k}$ regions. When the $(k+1)^{st}$ great circle is drawn, it will cut through a certain number of regions. One more region will be produced by each region it cuts through. The $(k+1)^{st}$ great circle intersects the other k great circles, and every pair of adjacent points forms an arc that cuts through an existing region.

Note that two distinct great circles must intersect at two points. Hence any number of great circles have a minimum of two intersection points, and $m_{k+1} >= m_{k}+2$ . Thus, $m_{5} >= m_{4}+2 >=m_{3}+2+2 . . . >=m_{1}+8=10$ . It is easy to see that we can obtain 10 regions by drawing five great circles through two diametrically opposite points, so m = 10. On the other hand, the $(k+1)^{st}$ great circle will have at most 2k intersection points with the first k great circles, leading to 2k new regions. By rotating the circle slightly along any axis, we can indeed obtain 2k new intersections. Hence $M_{k+1} = M_{k}+2k$ , and we obtain M = $M_{5}$ = $M_{1}+2(1+2+3+4) = 22$ .

Hence $M+m=32$ , and the required answer is 5.

Nice problem ... 10 min. And 22 max. Same way but I did in a non-mathematical way! For min have a 2-D projection. For max rotation along axis n d no. F intersection!