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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 ...
The ten lines involved in Desargues's theorem (six sides of triangles, the three lines Aa, Bb and Cc, and the axis of perspectivity) and the ten points involved (the six vertices, the three points of intersection on the axis of perspectivity, and the center of perspectivity) are so arranged that each of the ten lines passes through three of the ...
Secant-, chord-theorem. For the intersecting secants theorem and chord theorem the power of a point plays the role of an invariant: . Intersecting secants theorem: For a point outside a circle and the intersection points , of a secant line with the following statement is true: | | | | = (), hence the product is independent of line .
Let A 1, A 2, B 1, B 2, C 1, C 2 be the six intersection points, with the same letter corresponding to the same line and the index 1 corresponding to the point closer to P. Let D be the point where the lines A 1 B 2 and A 2 B 1 intersect, Similarly E for the lines B 1 C 2 and B 2 C 1. Draw a line through D and E. This line meets the circle at ...
When the intersection is internal, the equality states that the product of the segment lengths into which E divides one diagonal equals that of the other diagonal. This is known as the intersecting chords theorem since the diagonals of the cyclic quadrilateral are chords of the circumcircle.
Although it may be embedded in two dimensions, the Desargues configuration has a very simple construction in three dimensions: for any configuration of five planes in general position in Euclidean space, the ten points where three planes meet and the ten lines formed by the intersection of two of the planes together form an instance of the configuration. [2]
This proves that all points in the intersection are the same distance from the point E in the plane P, in other words all points in the intersection lie on a circle C with center E. [5] This proves that the intersection of P and S is contained in C. Note that OE is the axis of the circle. Now consider a point D of the circle C. Since C lies in ...
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.