M is any point on the side AB of the triangle ABC. r, r1, r2 are the radii of the circles inscribed in ABC, AMC, BMC. q is the radius of the circle on the opposite side of AB to C, touching the three sides of AB and the extensions of CA and CB. Similarly, q1 and q2. Prove that r1r2q = rq1q2.
Solution
We need an expression for r/q. There are two expressions, one in terms of angles and the other in terms of sides. The latter is a poor choice, because it is both harder to derive and less useful. So we derive the angle expression.
Let I be the center of the in-circle for ABC and X the center of the external circle for ABC. I is the intersection of the two angle bisectors from A and B, so c = r (cot A/2 + cot B/2). The X lies on the bisector of the external angle, so angle XAB is 90o - A/2. Similarly, angle XBA is 90o - B/2, so c = q (tan A/2 + tan B/2). Hence r/q = (tan A/2 + tan B/2)/(cot A/2 + cot B/2) = tan A/2 tan B/2.
Applying this to the other two triangles, we get r1/q1 = tan A/2 tan CMA/2, r2/q2 = tan B/2 tan CMB/2. But CMB/2 = 90o - CMA/2, so tan CMB/2 = 1/tan CMA/2. Hence result.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
6 Oct 1998
Last corrected/updated 14 Mar 03