Homothetic circles and t10 quads
by math_explorer, Feb 3, 2012, 8:37 AM
Google has 43,100 hits for "homothety" (quoted) and 20,000 hits for "homothecy". AoPS has >1,000 and 208 with default search options. For some reason I can't get used to this.
If a homothety maps circle A to circle B then it maps the point on THAT *makes wild hand gesture* side of circle A to the point on THAT *repeats wild hand gesture* side of circle B. (Must learn to start noticing this.)
A lot of people call the center of a homothety which maps something to something a center of similitude, which is also confusing. Two circles in the plane have an internal and an external center of similitude. The centers lie on the internal/external resp. common tangents, except of course when they don't exist, and on the line connecting the centers. There are probably an awful lot of exceptions. I have no idea what I am writing.
The theorem which explains that sums of opposite sides of a "tangential/circumscribed/inscriptable" quadrilateral are equal is Pitot's. The converse is well-known (?) and nowhere as obvious (YMMV). This is also true for a concave quadrilateral, if the convex quad resulting from extending the sides is "t10/c13/i11".
A quad is "ex-tangential/exscriptable" if sums of some pairing of adjacent sides are equal; again, this is true for the concave and convex quads, with the equivalence called Urquhart's theorem.
If a homothety maps circle A to circle B then it maps the point on THAT *makes wild hand gesture* side of circle A to the point on THAT *repeats wild hand gesture* side of circle B. (Must learn to start noticing this.)
A lot of people call the center of a homothety which maps something to something a center of similitude, which is also confusing. Two circles in the plane have an internal and an external center of similitude. The centers lie on the internal/external resp. common tangents, except of course when they don't exist, and on the line connecting the centers. There are probably an awful lot of exceptions. I have no idea what I am writing.
The theorem which explains that sums of opposite sides of a "tangential/circumscribed/inscriptable" quadrilateral are equal is Pitot's. The converse is well-known (?) and nowhere as obvious (YMMV). This is also true for a concave quadrilateral, if the convex quad resulting from extending the sides is "t10/c13/i11".
A quad is "ex-tangential/exscriptable" if sums of some pairing of adjacent sides are equal; again, this is true for the concave and convex quads, with the equivalence called Urquhart's theorem.