9th APMO 1997

ABC is a triangle. The bisector of A meets the segment BC at X and the circumcircle at Y. Let r_A = AX/AY. Define r_B and r_C similarly. Prove that r_A/sin²A + r_B/sin²B + r_C/sin²C >= 3 with equality iff the triangle is equilateral.

Solution

AX/AB = sin B/sin AXB = sin B/sin(180 - B - A/2) =sin B/sin(B + A/2). Similarly, AB/AY = sin AYB/sin ABY = sin C/sin(B + CBY) = sin C/sin(B + A/2). So AX/AY = sin B sin C/sin²(B + A/2). Hence r_A/sin²A = s_A/sin²(B + A/2), where s_A = sin B sin C/sin²A. Similarly for r_B and r_C. Now s_As_Bs_C = 1, so the arithmetic/geometric mean result gives s_A + s_B + s_C ≥ 3. But 1/sin k ≥ 1 for any k, so r_A/sin²A + r_B/sin²B + r_C/sin²C ≥ 3.

A necessary condition for equality is that sin²(B + A/2) = sin²(B + A/2) = sin²(B + A/2) = 1 and hence A = B = C. But it is easily checked that this is also sufficient.

9th APMO 1997

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