8th APMO 1996

Given four concyclic points. For each subset of three points take the incenter. Show that the four incenters from a rectangle.

Solution

Take the points as A, B, C, D in that order. Let I be the incenter of ABC. The ray CI bisects the angle ACB, so it passes through M, the midpoint of the arc AB. Now ∠MBI = ∠MBA + ∠IBA = ∠MCA + ∠IBA = (∠ACB + ∠ABC)/2 = 90^o - (∠CAB) /2 = 90^o - ∠CMB/2 = 90^o - ∠IMB/2. So the bisector of ∠IMB is perpendicular to IB. Hence MB = MI. Let J be the incenter of ABD. Then similarly MA = MJ. But MA = MB, so the four points A, B, I, J are concyclic (they lie on the circle center M). Hence ∠BIJ = 180^o - ∠BAJ = 180^o - ∠BAD/2.

Similarly, if K is the incenter of ADC, then ∠BJK = 180^o - ∠BDC/2. Hence ∠IJK = 360^o - ∠BIJ - ∠BJK = (180^o - ∠BIJ) + (180^o - ∠BJK) = (∠BAD + ∠BDC)/2 = 90^o. Similarly, the other angles of the incenter quadrilateral are 90^o, so it is a rectangle.

8th APMO 1996

© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002