### 48th Putnam 1987

Problem A4

p(x, y, z) is a polynomial with real coefficients such that: (1) p(tx, ty, tz) = t2f(y - x, z - x) for all real x, y, z, t (and some function f); (2) p(1, 0, 0) = 4, p(0 ,1, 0) = 5, and p(0, 0, 1) = 6; and (3) p(α, β, γ) = 0 for some complex numbers α, β, γ such that |β - α| = 10. Find |γ - α|.

Solution

5 √30 /3.

Easy.

The key observation is that f must be a polynomial with each term of degree 2. This is well-known and could be quoted. It is also almost obvious. For put u = x, v = y - x, w = z - x. Then x = u, y = u + v, z = u + w, so we can write p(x, y, z) = q(u, v, w), where q is a polynomial. But we are given that f(v, w) = q(u, v, w) for all u, v, w, so u cannot appear in q (otherwise we could vary it to get a contradiction). Hence f is actually a polynomial. But now the relation with t implies that each term has degree 2.

So p(x, y, z) = a(y - x)2 + b(y - x)(z - x) + c(z - x)2 for some a, b, c. But we are given three values for p which allows us to solve for a, b, c: a = 5, b = -7, c = 6. We have a quadratic in (y - x)/(z - x) which has roots (7 ± √71 i)/10, so any roots α, β, γ of p must satisfy (β - α) / (γ - α) = (7 ± √71 i)/10, and hence |(β - α)| / |(γ - α)| = √120 /10. Hence if a root satisfies |β - α| = 10, then it also satisfies |γ - α| = 100/√120 = 5 √30 /3.