smc

The semiperimeter $s$ of the triangle is $\frac{25+39+40}{2}=52$. Therefore, by Heron's Formula, we can find the area of the triangle as follows: $$\begin{array}{rl}A& =\sqrt{52(52-25)(52-39)(52-40)}\\ & =\sqrt{52(27)(13)(12)}\\ & =\sqrt{13\ast 4(9\ast 3)(13)(4\ast 3)}\\ & =13\ast 4\ast 3\ast 3\\ & =468\end{array}$$ Also, we know that the area of the triangle is $\frac{abc}{4R}$, where $a$, $b$, and $c$ are the sides of the triangle and $R$ is the triangle's circumradius. Thus, we can equate this expression for the area with $468$ to solve for $R$: $$\begin{array}{rl}\frac{(25)(39)(40)}{4R}& =468=39\ast 12\\ \frac{25\ast 40}{12}& =4R\\ R& =\frac{250}{12}=\frac{125}{6}\end{array}$$ Because the problem asks for the diameter of the circumcircle, our answer is $2R=\boxed{\frac{125}{3}}$.