IMO 1973 SL 3

Origin: CZS

Problem

Prove that the sum of an odd number of unit vectors passing through the same point O and lying in the same half-plane whose border passes through O has length greater than or equal to 1.

Solution

We use induction on odd numbers n. For n = 1 there is nothing to prove. Suppose that the result holds for n−2 vectors, and let us be given vectors v1, v2, . . . , vn arranged clockwise. Set v′ = v2+v3+\cdot \cdot \cdot+vn−1, u = v1+vn, and v = v1 + v2 + \cdot \cdot \cdot + vn = v′ + u. By the inductive hypothesis we have |v′| \geq1. Now if the angles between v′ and the vectors v1, vn are \alpha and \beta respectively, then the angle between u and v′ is |\alpha −\beta|/2 \leq90◦. Hence |v′ + u| \geq|v′| \geq1. Second solution. Again by induction, it can be easily shown that all possible values of the sum v = v1 + v2 + \cdot \cdot \cdot + vn, for n vectors v1, . . . , vn in the upper half-plane (with y \geq0), are those for which |v| \leqn and |v −ke| \geq1 for every integer k for which n −k is odd, where e is the unit vector on the x axis.