IMO shortlist 1983/19 solution

24th IMO 1983 shortlist

Problem 19

k is a positive real. Solve the equations:
x₁ | x₁ | = x₂ | x₂| + (x₁ - k) | x₁ - k |
x₂ | x₂ | = x₃ | x₃| + (x₂ - k) | x₂ - k |
...
x_n | x_n | = x₁ | x₁| + (x_n - k) | x_n - k |.

Solution

Solution by Lauri Ahlroth

Adding we get &sum (x_i-k)|x_i-k| = 0 (*), so for some m we must have xm ≥ k > 0 (otherwise all terms would be negative). Now the equation for m gives x_m+1|x_m+1| = 2kx_m - k² = k(2x_m-k) ≥ k². Hence x_m+1 ≥ k also. Hence x_i ≥ k for all i. So all terms in (*) are non-negative and hence all must be zero. So the unique solution is x_i = k for all i.

24th IMO shortlist 1983