11th Chinese 1996 Problem B3 ABC has AB = 2, BC = 1 and angle C a right-angle. Take D on the segment BC, E on the segment CA, and F on the segment AB. What is the smallest possible value of max(DE, EF, FD)? Solution
: √(3/7). We suspect that an equilateral triangle gives the minimum value. Put BD = x, BF = y, CE = z. Then using the cosine formula, DE 2 = x 2 + z 2 - 2x + 1, DF 2 = x 2 + y 2 - xy, EF 2 = y 2 + z 2 - √3 yz - y + 1. A little experimentation shows that x = 4/7, y = 5/7, z = (2√3)/7 gives DE 2 = DF 2 = EF 2 = 3/7. However, that does not prove that there is not another smaller equilateral triangle or that an equilateral triangle is best. So put x = 4/7 + X, y = 5/7 + Y, z = (2√3)/7 + Z. Then DE 2 = X 2 + Z 2 - 6/7 Z + (4√3)/7 Z + 3/7, DF 2 = X 2 + Y 2 + 3/7 X + 6/7 Y + 3/7, EF 2 = Y 2 + Z 2 - 3/7 Y - √3 YZ - (√3)/7 Z + 3/7. Hence (DE 2 - 3/7) + 2(DF 2 - 3/7) + 4(EF 2 - 3/7) = 3X 2 + 6Y 2 + 5Z 2 - 4√3 YZ = 3X 2 + 3Y 2 + Z 2 + (√3Y - 2Z) 2 >= 0 with equality iff X = Y = Z. Hence is any of X, Y, Z are non-zero, then at least one of DE 2 , DF 2 , EF 2 exceeds 3/7, which proves the result. 11th Chinese 1996 © John Scholes jscholes@kalva.demon.co.uk 22 June 2002