IMO 1982 Problem B2 The diagonals AC and CE of the regular hexagon ABCDEF are divided by inner points M and N respectively, so that: AM/AC = CN/CE = r. Determine r if B, M and N are collinear.
For an inelegant solution one can use coordinates. The advantage of this type of approach is that it is quick and guaranteed to work! Take A as (0,√3), B as (1,√3), C as (3/2,√3/2, D as (1,0). Take the point X, coordinates (x,0), on ED. We find where the line BX cuts AC and CE. The general point on BX is (k + (1- k)x,k√3). If this is also the point M with AM/AC = r then we have: k + (1-k)x = 3r/2, k√3 = (1-r)√3 + r√3/2. Hence k = 1 - r/2, r = 2/(4-x). Similarly, if it is the point N with CN/CE = r, then k + (1-k)x = 3(1- r)/2, k√3 = (1-r)√3/2. Hence k = (1-r)/2 and r = (2-x)/(2+x). Hence for the ratios to be equal we require 2/(4-x) = (2-x)/(2+x), so x 2 - 8x + 4 = 0. We also have x < 1, so x = 4 - √12. This gives r = 1/√3. A more elegant solution uses the ratio theorem for the triangle EBC. We have CM/MX XB/BE EN/NC = -1. Hence (1-r)/(r - 1/2) (-1/4) (1-r)/r = -1. So r = 1/√3. s are also available in Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48- X. 23rd IMO 1982 © John Scholes jscholes@kalva.demon.co.uk 14 Oct 1998