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A quadrilateral such as BCEF is called an adventitious quadrangle when the angles between its diagonals and sides are all rational angles, angles that give rational numbers when measured in degrees or other units for which the whole circle is a rational number. Numerous adventitious quadrangles beyond the one appearing in Langley's puzzle have ...
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.
More generally, if the quadrilateral is a rectangle with sides a and b and diagonal d then Ptolemy's theorem reduces to the Pythagorean theorem. In this case the center of the circle coincides with the point of intersection of the diagonals. The product of the diagonals is then d 2, the right hand side of Ptolemy's relation is the sum a 2 + b 2.
One more interesting line (in some sense dual to the Newton's one) is the line connecting the point of intersection of diagonals with the vertex centroid. The line is remarkable by the fact that it contains the (area) centroid. The vertex centroid divides the segment connecting the intersection of diagonals and the (area) centroid in the ratio 3:1.
If the incircle is tangent to the sides AB, BC, CD, DA at T 1, T 2, T 3, T 4 respectively, and if N 1, N 2, N 3, N 4 are the isotomic conjugates of these points with respect to the corresponding sides (that is, AT 1 = BN 1 and so on), then the Nagel point of the tangential quadrilateral is defined as the intersection of the lines N 1 N 3 and N ...
The interior angle concept can be extended in a consistent way to crossed polygons such as star polygons by using the concept of directed angles.In general, the interior angle sum in degrees of any closed polygon, including crossed (self-intersecting) ones, is then given by 180(n–2k)°, where n is the number of vertices, and the strictly positive integer k is the number of total (360 ...
For a cyclic quadrilateral that is also orthodiagonal (has perpendicular diagonals), suppose the intersection of the diagonals divides one diagonal into segments of lengths p 1 and p 2 and divides the other diagonal into segments of lengths q 1 and q 2. Then [28] (the first equality is Proposition 11 in Archimedes' Book of Lemmas)
where P is the point of intersection of the diagonals. From this equation it follows almost immediately that the diagonals of a convex quadrilateral are perpendicular if and only if the projections of the diagonal intersection onto the sides of the quadrilateral are the vertices of a cyclic quadrilateral. [6]