31st Eötvös 1927

r is the inradius of the triangle ABC and r' is the exradius for the circle touching AB. Show that 4r r' ≤ c², where c is the length of the side AB.

Solution

Let the incenter be I and the relevant excenter be E. Then ∠IAE = ∠IBE = 90^o, so AIBE is cyclic. Let the center of the circle through A,I,B,E be C. Then r = IP, r' = EQ. But IP·EQ ≤ XZ·ZY, where XY is the diameter perpendicular to AB. We have XZ·ZY = AZ·ZB = c²/4.

31st Eötvös 1927

© John Scholes
jscholes@kalva.demon.co.uk
1 Nov 2003
Last corrected/updated 1 Nov 03