Kanel-Belov A.Y.

On a sphere, five points are marked, no three of which lie on a great circle (a great circle is a circle formed by the intersection of the sphere and a plane passing through its center). Two great circles that do not pass through the marked points are called equivalent if one can be continuously moved to the other on the sphere without passing through the marked points.

a) How many great circles can be drawn that do not pass through the marked points and are not equivalent to each other?

b) The same problem for $n$ marked points.
a) Let's move on to dual objects: each circle corresponds to a pair of opposite points on the sphere such that the diameter connecting them is perpendicular to this circle; conversely, each point corresponds to a great circle. Then the problem reduces to a dual one: points are considered equivalent if one can be translated into another without touching five given great circles (no three of which intersect at a single point). Clearly, a point can be moved within the regions into which the great circles divide the sphere. Thus, the number of equivalence classes is half the number of parts into which the great circles divide the sphere (opposite parts correspond to one class, since the original great circle in the dual problem corresponds to two diametrically opposite points).

According to problem 60324, $n$ of our great circles divide the sphere into $n^{2}-n+2$ parts. In particular, five circles will divide the sphere into 22 parts. And the answer, as shown above, is half of that.

b) See a).

## Answer

a) 11, b) $1 / 2\left(n^{2}-n+2\right)$ circles.

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