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The number of points (n), chords (c) and regions (r G) for first 6 terms of Moser's circle problem. In geometry, the problem of dividing a circle into areas by means of an inscribed polygon with n sides in such a way as to maximise the number of areas created by the edges and diagonals, sometimes called Moser's circle problem (named after Leo Moser), has a solution by an inductive method.
The difference between the perimeter of the inscribed megagon and the circumference of this circle comes to less than 1/16 millimeters. [3] Because 1,000,000 = 2 6 × 5 6, the number of sides is not a product of distinct Fermat primes and a power of two. Thus the regular megagon is not a constructible polygon.
The area of a regular myriagon with sides of length a is given by A = 2500 a 2 cot π 10000 {\displaystyle A=2500a^{2}\cot {\frac {\pi }{10000}}} The result differs from the area of its circumscribed circle by up to 40 parts per billion .
He gives d (diagonal) with reflection lines through vertices, p with reflection lines through edges (perpendicular), and for the odd-sided pentadecagon i with mirror lines through both vertices and edges, and g for cyclic symmetry. a1 labels no symmetry. These lower symmetries allows degrees of freedoms in defining irregular pentadecagons.
More generally, if the quadrilateral is a rectangle with sides a and b and diagonal d then Ptolemy's theorem reduces to the Pythagorean theorem. In this case the center of the circle coincides with the point of intersection of the diagonals. The product of the diagonals is then d 2, the right hand side of Ptolemy's relation is the sum a 2 + b 2.
The area of a regular chiliagon with sides of length a is given by A = 250 a 2 cot π 1000 ≃ 79577.2 a 2 {\displaystyle A=250a^{2}\cot {\frac {\pi }{1000}}\simeq 79577.2\,a^{2}} This result differs from the area of its circumscribed circle by less than 4 parts per million .
The four sides can be split into two pairs of adjacent equal-length sides. [7] One diagonal crosses the midpoint of the other diagonal at a right angle, forming its perpendicular bisector. [9] (In the concave case, the line through one of the diagonals bisects the other.) One diagonal is a line of symmetry.
The area of a bicentric quadrilateral can be expressed in terms of two opposite sides and the angle θ between the diagonals according to [9] = = . In terms of two adjacent angles and the radius r of the incircle, the area is given by [9]