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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 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.
Construct a pentagon in a circle by one of the methods shown in constructing a pentagon. Extend a line from each vertex of the pentagon through the center of the circle to the opposite side of that same circle. Where each line cuts the circle is a vertex of the decagon.
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.
Following Archimedes' argument in The Measurement of a Circle (c. 260 BCE), compare the area enclosed by a circle to a right triangle whose base has the length of the circle's circumference and whose height equals the circle's radius. If the area of the circle is not equal to that of the triangle, then it must be either greater or less.
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.
The area of an isosceles (or any) trapezoid is equal to the average of the lengths of the base and top (the parallel sides) times the height. In the adjacent diagram, if we write AD = a , and BC = b , and the height h is the length of a line segment between AD and BC that is perpendicular to them, then the area K is