### 60th Putnam 1999

**Problem A1**

Find polynomials a(x), b(x), c(x) such that |a(x)| - |b(x)| + c(x) = -1 for x < -1, 3x + 2 for -1 ≤ x ≤ 0, -2x + 2 for x > 0.

**Solution**

Note that we do not have to find *all* solutions, just a solution.

We look first for linear polynomials. The points of discontinuity in the derivative are evidently where a(x) and b(x) change sign. At -1 the function jumps by 3x + 3, so the function changing sign must be (3x + 3)/2. Moreover, it must be a(x) that changes sign, since the jump is +(3x + 3) not -(3x + 3). So a(x) = (3x + 3)/2. Similarly, at 0, the function jumps by -5x, so b(x) = 5x/2. Hence c(x) = -1 + |b(x)| - |a(x)| = -1 - 5x/2 + (3x + 3)/2 = -x + 1/2 (for x < -1 and hence for all x).

Finally, notice that the sign of a(x) and b(x) is not determined uniquely.

60th Putnam 1999

© John Scholes

jscholes@kalva.demon.co.uk

14 Dec 1999