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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]
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]
Brahmagupta's theorem states that for a cyclic orthodiagonal quadrilateral, the perpendicular from any side through the point of intersection of the diagonals bisects the opposite side. [ 2 ] If an orthodiagonal quadrilateral is also cyclic, the distance from the circumcenter (the center of the circumscribed circle) to any side equals half the ...
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.
The value of the two products in the chord theorem depends only on the distance of the intersection point S from the circle's center and is called the absolute value of the power of S; more precisely, it can be stated that: | | | | = | | | | = where r is the radius of the circle, and d is the distance between the center of the circle and the ...
A chord (from the Latin chorda, meaning "bowstring") of a circle is a straight line segment whose endpoints both lie on a circular arc. If a chord were to be extended infinitely on both directions into a line, the object is a secant line. The perpendicular line passing through the chord's midpoint is called sagitta (Latin for "arrow").
Then the area of the arbelos is equal to the area of a circle with diameter HA. Proof : For the proof, reflect the arbelos over the line through the points B and C , and observe that twice the area of the arbelos is what remains when the areas of the two smaller circles (with diameters BA , AC ) are subtracted from the area of the large circle ...