Solving $a\pm b=2c$ in the 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 $(0.15+o(1))(\mathrm{\mid }A\mathrm{\mid }+\mathrm{\mid }B\mathrm{\mid }{)}^2$ as $\mathrm{\mid }A\mathrm{\mid }+\mathrm{\mid }B\mathrm{\mid }\to \mathrm{\infty }$. As a corollary, if $A$ is antisymmetric (that is, $A\cap (-A)=\text{est}$), then there are at most $(0.3+o(1))\mathrm{\mid }A{\mathrm{\mid }}^2$ 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 $(0.5+o(1))\mathrm{\mid }A{\mathrm{\mid }}^2$. These estimates are sharp.