take a cyclic quadrilateral. there are two ways to divide it into two triangles, using a diagonal. prove that the sum of the two inradii of the triangles is the same, no matter what your choice of diagonal is. (you can easily generalize this to any cyclic polygon divided into triangles.)

it's such a simple fact, i can't believe i haven't seen it in any exposition of theorems. but i've only seen complex proofs of it. i hope it generalizes into some big theory, like inversion. that would be exciting.

i also wonder if the converse is true... if the sums of the two inradii are equal, is the quadrilateral cyclic necessarily?

landofnowhereMay 13 2004, 12:22:01 UTC 12 years ago

The key lemma is that the four incenters form a rectangle with sides parallel to the lines passing through the midpoints of the arcs of the circumcircle cut off by opposite sides (those points are very important in the proof, because they're how you get a hold on the incenter). I never would have thought of this lemma on my own, but I had to prove part of it for something else.

landofnowhereMay 13 2004, 12:23:27 UTC 12 years ago

landofnowhereMay 13 2004, 12:38:44 UTC 12 years ago