Solving $a\pm b=2c$ in elements of finite sets
We show that if $A$ and $B$ are finite sets of real numbers, then the number of triples $(a,b,c)\in A\times B\times (A\cup B)$ with $a+b=2c$ is at most $(0.15+o(1))(|A|+|B|)^2$ as $|A|+|B|\to \infty $. As a corollary, if $A$ is antisymmetric (that is, $A\cap (-A)=\emptyset $), then there are at most $(0.3+o(1))|A|^2$ triples $(a,b,c)$ with $a,b,c\in A$ 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 $a,b,c\in A$ and $a-b=2c$ is at most $(0.5+o(1))|A|^2$. These estimates are sharp.