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 wrong. Suppose that the area enclosed by the circle is less than the area "T" of the triangle. Let "D" denote the deficit amount. Circumscribe a square, so that the midpoint of each edge lies on the circle. If the total area gap between the square and the circle, "G", is greater than "D", slice off the corners with circle tangents to make a circumscribed octagon, and continue slicing until the gap area is less than "D". The area of the polygon, "P", must be less than "T". This, too, forces a contradiction. For, a perpendicular to the midpoint of