13.412 Three state farms are not located on the same straight line. The distance from the first to the third via the second is four times longer than the direct route between them; the distance from the first to the second via the third is $a$ km longer than the direct route; the distance from the second to the third via the first is 85 km. In what range are the values of $a$ for which the specified arrangement of the state farms not on one straight line would be possible? Calculate the distances between the state farms.
Solution. Let $x$ be the distance between the first and second collective farms, $y$ be the distance between the second and third, and $z$ be the distance between the first and third. Then, according to the condition,

$$
\left\{\begin{array} { l } 
{ x + z = 85, } \\
{ x + y = 4 z, } \\
{ z + y = x + a }
\end{array} \Rightarrow \left\{\begin{array} { l } 
{ z - y = 85 - 4 z, } \\
{ z + 2 y = x + a }
\end{array} \Rightarrow \left\{\begin{array}{l}
z=\frac{170+a}{7} \\
y=\frac{255+5 a}{7} \\
x=\frac{425-a}{7}
\end{array}\right.\right.\right.
$$

Obviously, $y>z$ and $x<z+y$. Therefore, the collective farms will not be located on the same straight line if and only if $y<x+z$ (the triangle inequality), from which $a<68$.

Answer: $\frac{425-a}{7}, \frac{255+5 a}{7}, \frac{170+a}{7}$ for $0<a<68$.