Solving a ± b = 2c in elements of finite sets

We show that if A and B are finite sets of real numbers, then the number of triples (a, b, c) ∈ A x B x (A ∪ B) with a + b = 2c is at most (0.15 +o(1))(|A| + |B|)² as |A| + |B| → ∞. As a corollary, if is antisymmetric (that is, A ∩ (-A)) = θ), then there are at most (0.3 + o(1))|A|² triples (a, b, c) with a, b, c ∈ A and a - b =2c. In the general case where A is not necessarily antisymmetric, we show that the number of triples (a, b, c) with a, b, c ∈ A and a - b = 2c is at most (0.5+o(1))|A|². These estimates are sharp.