Problem 798

Solution:
Let $x$, $y$, $z$ be the integers. We have
$$\begin{aligned} & \frac{x+y}{2} + z = 42 \\ & \frac{y+z}{2} + x = 13 \\ & \frac{x+z}{2} + y = 37 \end{aligned}$$
Adding these three equations yields $2(x+y+z) = 92$, so $x+y+z = 46$.

Now, from $\frac{x+y}{2} + z = 42$:
$$\frac{x+y}{2} + z = 42 \\ \Rightarrow x+y = 2(42 - z) \\$$
But $x+y+z = 46$, so $x+y = 46 - z$.

Set equal:
$$46 - z = 2(42 - z) \\ 46 - z = 84 - 2z \\ 2z - z = 84 - 46 \\ z = 38$$

Now $x+y = 46 - 38 = 8$.

From $\frac{x+z}{2} + y = 37$:
$$\frac{x+z}{2} + y = 37 \\ x+z = 2(37 - y) \\$$
But $x+z = (x+y+z) - y = 46 - y$.

Set equal:
$$46 - y = 2(37 - y) \\ 46 - y = 74 - 2y \\ 2y - y = 74 - 46 \\ y = 28$$

Now $x = 46 - 28 - 38 = -20$.

Therefore, the three integers are $-20$, $28$, $38$.