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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 upper horizontal line segment in the above figure appears to be longer than the lower line segment despite the fact that both are the same length.

The horizontal line segment in the above figure appears to be shorter than the vertical line segment, despite the fact that it has the same length.

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 ...