I discovered part of the following curiosity of inversive geometry a week ago, when I was setting homework problems for my computational geometry class, and the rest of it today when the students discussed their solutions.

Claim: let $$abc$$ be a right angle, and invert $$b$$ and $$c$$ through a circle centered on $$a$$ to get points $$b'$$ and $$c'$$. Then triangle $$abc$$ is similar to triangle $$ac'b'$$. Note the permuted vertices: the right angle in the new triangle is at $$c'$$!

Proof sketch: line $$ab$$ inverts to itself and contains $$b'$$; line $$ac$$ inverts to itself and contains $$c'$$. Therefore, angle $$bac$$ equals angle $$c'ab'$$. Line $$bc$$, which is at right angles to $$ab$$, inverts to a circle $$ab'c'$$, also at right angles to $$ab$$. Thus $$ab'$$ is the diameter of the circle and any point $$c'$$ on the circle forms a right angle to this diameter. With two angles equal, the triangles are similar.

...which I thought was pretty impressive until I realized it had nothing to do with diameters and right angles. More generally, $$abc$$ will always be similar to $$ac'b'$$, no matter whether any angle is right, because the two triangles have one equal angle and one equal ratio of side lengths: $|ab|/|ac| = \frac{R^2/|ac|}{R^2/|ab|} = \frac{|ac'|}{|ab'|}.$

Oh well, despite being such a trivial piece of geometry I had fun making diagrams for it.

Comments:

eccemathematica:
2006-04-22T21:40:09Z

Trivial? Pshaw. I think it's adorable. And there's nothing nicer than a little discovery every so often. Yay!