Dividing a circle into areas "F" can also be derived, which will solve the problem. Its vertices include the "n" points on the circle, referred to as the exterior vertices, as well as the interior vertices, the intersections of distinct chords in the interior of the circle. The "generic intersection" assumption made above guarantees that each interior vertex is the intersection of no more than two chords. Thus the main task in determining "V" is finding the number of interior vertices. As a consequence of the lemma, any two intersecting chords will uniquely determine an interior vertex. These chords are in turn uniquely determined by
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,