Dividing a circle into areas lines (there are two points on one side and one on the other). So the recurrence can be expressed as which can be easily reduced to using the sums of the first formula_3 natural numbers and the first formula_3 squares, this combines to Finally which yields The lemma asserts that the number of regions is maximal if all "inner" intersections of chords are simple (exactly two chords pass through each point of intersection in the interior). This will be the case if the points on the circle are chosen "in general position". Under this assumption of "generic intersection", the number
Dividing a circle into areas cross. In the second case (b), the new line crosses each of the old lines in a different point. It will be useful to know the following fact. Lemma. We can choose the new point "A" so that case "b" occurs for each of the new lines. Proof. Notice that, for the case "a", three points must be on one line: the new point "A", the old point "O" to which we draw the line, and the point "I" where two of the old lines intersect. Notice that there are "n" old points "O", and hence finitely many points "I"