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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 circle meets the angle at two points: one on each leg. Using each of these points as a center, draw two circles of the same size. The intersection of the circles (two points) determines a line that is the angle bisector. The proof of the correctness of this construction is fairly intuitive, relying on the symmetry of the problem.
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).
A closed line segment includes both endpoints, while an open line segment excludes both endpoints; a half-open line segment includes exactly one of the endpoints. In geometry , a line segment is often denoted using an overline ( vinculum ) above the symbols for the two endpoints, such as in AB .
When two cells in the Voronoi diagram share a boundary, it is a line segment, ray, or line, consisting of all the points in the plane that are equidistant to their two nearest sites. The vertices of the diagram, where three or more of these boundaries meet, are the points that have three or more equally distant nearest sites.
Consider a triangle ABC.Let the angle bisector of angle ∠ A intersect side BC at a point D between B and C.The angle bisector theorem states that the ratio of the length of the line segment BD to the length of segment CD is equal to the ratio of the length of side AB to the length of side AC:
In general, the same inversion transforms the given line L and given circle C into two new circles, c 1 and c 2. Thus, the problem becomes that of finding a solution line tangent to the two inverted circles, which was solved above. There are four such lines, and re-inversion transforms them into the four solution circles of the Apollonius problem.
The twin circles (red) of an arbelos (grey) In geometry, the twin circles are two special circles associated with an arbelos.An arbelos is determined by three collinear points A, B, and C, and is the curvilinear triangular region between the three semicircles that have AB, BC, and AC as their diameters.