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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.
If M, N are the midpoints of the diagonals, and E, F are the intersection points of the extensions of opposite sides, then the area can also be expressed as = ¯ ¯ ¯ where Q is the foot of the perpendicular to the line EF through the center of the incircle. [9]
Four line segments, each perpendicular to one side of a cyclic quadrilateral and passing through the opposite side's midpoint, are concurrent. [ 23 ] : p.131, [ 24 ] These line segments are called the maltitudes , [ 25 ] which is an abbreviation for midpoint altitude.
The area bounded by one spiral rotation and a line is 1/3 that of the circle having a radius equal to the line segment length; Use of the method of exhaustion also led to the successful evaluation of an infinite geometric series (for the first time);
The area of a regular polygon is half its perimeter multiplied by the distance from its center to its sides, and because the sequence tends to a circle, the corresponding formula–that the area is half the circumference times the radius–namely, A = 1 / 2 × 2πr × r, holds for a circle.
The radius of a circle is perpendicular to the tangent line through its endpoint on the circle's circumference. Conversely, the perpendicular to a radius through the same endpoint is a tangent line. The resulting geometrical figure of circle and tangent line has a reflection symmetry about the axis of the radius.
Equal chords are subtended by equal angles from the center of the circle. A chord that passes through the center of a circle is called a diameter and is the longest chord of that specific circle. If the line extensions (secant lines) of chords AB and CD intersect at a point P, then their lengths satisfy AP·PB = CP·PD (power of a point theorem).
Owing to the Pythagorean theorem, the diagonal dividing one half of a square equals the radius of a circle whose outermost point is the corner of a golden rectangle added to the square. [1] Thus, a golden rectangle can be constructed with only a straightedge and compass in four steps: Draw a square