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Consider the circles centered on the midpoints of the sides of a reference triangle and passing though the orthocenter H. These circles cut the sides in six points lying on a ...

The radical circle of the Stammler circles has center at the nine-point center N, which is Kimberling center X_5. The radius is given by R_S = sqrt(R^2+ON^2) (1) = ...

Taking the locus of midpoints from a fixed point to a circle of radius r results in a circle of radius r/2. This follows trivially from r(theta) = [-x; 0]+1/2([rcostheta; ...

The radical circle of the mixtilinear incircles has a center with trilinear center function alpha_(999)=cosA-2, (1) which is Kimberling center X_(999). It has radius (2) ...

The orthoptic circle of the Steiner inellipse is the circle with center at alpha_2=1/a, (1) corresponding to the triangle centroid G and radius ...

The pedal curve of circle involute f = cost+tsint (1) g = sint-tcost (2) with the center as the pedal point is the Archimedes' spiral x = tsint (3) y = -tcost. (4)

Let ABCD be a quadrilateral with perpendicular polygon diagonals. The midpoints of the sides (a, b, c, and d) determine a parallelogram (the Varignon parallelogram) with ...

The radical circle of the McCay circles has center (1) which is not a Kimberling center, and radius (2) where (3) Its circle function is (4) where (5) which is also does not ...

The radical circle of the Neuberg circles has circle function l=(a^2b^4-b^4c^2+a^2c^4-b^2c^4)/(bc(a^2b^2+a^2c^2+b^2c^2)), (1) which does not correspond to any Kimberling ...

If P is any point on a line TT^' whose orthopole is S, then the circle power of S with respect to the pedal circle of P is a constant (Gallatly 1913, p. 51).