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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 area within a circle is equal to the radius multiplied by half the circumference, or A = r x C /2 = r x r x π.. Liu Hui argued: "Multiply one side of a hexagon by the radius (of its circumcircle), then multiply this by three, to yield the area of a dodecagon; if we cut a hexagon into a dodecagon, multiply its side by its radius, then again multiply by six, we get the area of a 24-gon; the ...
The area formula is intuitive: start with a circle of radius (so its area is ) and stretch it by a factor / to make an ellipse. This scales the area by the same factor: π b 2 ( a / b ) = π a b . {\displaystyle \pi b^{2}(a/b)=\pi ab.} [ 18 ] However, using the same approach for the circumference would be fallacious – compare the integrals ...
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 quotients formed by the area of these polygons divided by the square of the circle radius can be made arbitrarily close to π as the number of polygon sides becomes large, proving that the area inside the circle of radius r is πr 2, π being defined as the ratio of the circumference to the diameter (C/d).
The four line segments between the center of the incircle and the points where it is tangent to the quadrilateral partition the quadrilateral into four right kites. If a line cuts a tangential quadrilateral into two polygons with equal areas and equal perimeters, then that line passes through the incenter. [4]
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 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.