A line through the incenter of a triangle meets the circumcircle and incircle in the points A, B, C, D (in that order). Show that AB·CD ≥ BC2/4. When do you have equality?
Solution
Answer: equality when the line passes through the circumcenter.
Let M be the midpoint of AD, r the inradius, O the circumcenter and R the circumradius. We have AB.CD - r2 = (MA + MI - r)(MA - MI - r) - r2 = (MA - r)2 - MI2 - r2 = MA2 - 2r MA - MI2. Now MI2 = OI2 - OM2. By Euler's formula (a well-known result, see IMO 62/6 for a proof) we have OI2 = R2 - 2rR. Also OM2 = R2 - MA2. Hence AB.CD - r2 = (R2 - OM2) - 2r MA - (R2 - 2rR - OM2) = 2r(R - MA) ≥ 0, with equality iff R = MA and hence the line passes through O.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002