InversionProduct

Inversions product

Here we study the product of two inversions F(A, a) and G(B, b). In parenthesis are the center and the label of the inversion circles. The properties are:
[1] The product H = G*F of the invertions is a map leaving invariant each member (c) of the circle bundle (I) of circles simultaneously orthogonal to circles (a) and (b).
[2] The product H of the inversions restricted on each member (c) of bundle (I) defines a homography H' of (c), whose homography axis is line e = AB.
[3] For points X₂ of (c) Z₂ = H(X₂) = H'(X₂) define lines X₂Z₂ which envelope a conic (f) belonging to the family of conics generated by the circle (c) and the (double) line (e). [4] The figure below corresponds to the case of non-intersecting inversion circles a and b. In this case, conic (c) is tangent to the sides of triangle CDE. Later is formed by the tangents CY₁, CY₂ to (c) at the limit points Y₁, Y₂ of bundle (I) and line AZ₁, where Z₁ is the intersection point of (c) with (b).

All the properties follow easily from the general considerations on the product of two involutions, discussed in InvolutionsProduct.html .
Here the involutions are induced on (c) by the corresponding inversions, see InversionAsInvolution.html for a discussion of this case.
Notice that the procedure can be inverted. Given a triangle like CDE and an inscribed conic like (f), such that |CY₁| = |CY₂|, one can reconstruct the above figure as follows:
[1] Define circle (c) tangent at CD, CE at Y₁, Y₂ respectively.
[2] Define circle (a) centered at the intersection point A of line Y₁Y₂ and DE and orthogonal to (c).
[3] Define circle (b) by the requirement to be orthogonal to (c), with center on Y₁Y₂ and pass through Z₂, inverse of X₂ with respect to circle (a).
[4] This "reversing" of the figure makes it interesting to investigate the location of points Q relative to a triangle CDE, such that the corresponding cevians intersect on the sides two equal segments, like CY₁, CY₂ above.

Inversions product

See Also