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
Area of a circle the excess amount. Inscribe a square in the circle, so that its four corners lie on the circle. Between the square and the circle are four segments. If the total area of those gaps, "G", is greater than "E", split each arc in half. This makes the inscribed square into an inscribed octagon, and produces eight segments with a smaller total gap, "G". Continue splitting until the total gap area, "G", is less than "E". Now the area of the inscribed polygon, "P" = "C" − "G", must be greater than that of the triangle. But this forces a contradiction,