Task 6. (30 points) Three compressor stations are located not on the same straight line, but are connected by straight roads. The distance from the first station to the third through the second is four times longer than the direct route between them; the distance from the first station to the second through the third is longer by a km than the direct route; the distance from the second station to the third through the first is 85 km. Determine all values of $a$ for which the specified arrangement of compressor stations would be possible. Calculate the distances between the compressor stations when $a=5$.
Solution.

Let $x$ be the distance between the first and second compressor stations, $y$ the distance between the second and third, and $z$ the distance between the first and third (Fig. 6).

![](https://cdn.mathpix.com/cropped/2024_05_06_6d5fca905c76f828a3d8g-04.jpg?height=337&width=596&top_left_y=1005&top_left_x=730)

Fig. 6

According to the conditions of the problem, we form a system of equations:

$$
\left\{\begin{array}{l}
x+y=4 z \\
z+y=x+a \\
x+z=85
\end{array}\right.
$$

From the third equation of the system, express $z: z=85-x$, we get the system $\left\{\begin{array}{l}x+y=4(85-x) \\ 85-x+y=x+a ;\end{array}\right.$ $\left\{\begin{array}{l}5 x+y=340 \\ 2 x-y=85-a ;\end{array}\right.$ $\left\{\begin{array}{l}5 x+y=340, \\ 7 x=425-a .\end{array}\right.$

Then, $x=\frac{425-a}{7}, y=\frac{255+5 a}{7} z=\frac{170+a}{7}$.

Determine all values of $a$ for which the found values of $x$, $y$, and $z$ would be possible. Using the triangle inequality (any side of a triangle is less than the sum of the other two sides), we form a system of inequalities to estimate the value of the parameter $a$:

$$
\left\{\begin{array}{l}
x<y+z \\
y<x+z \\
z<x+y
\end{array}\right.
$$

$$
\left\{\begin{array}{l}
\frac{425-a}{7}<\frac{255+5 a}{7}+\frac{170+a}{7} \\
\frac{255+5 a}{7}<\frac{425-a}{7}+\frac{170+a}{7} \\
\frac{170+a}{7}<\frac{425-a}{7}+\frac{255+5 a}{7}
\end{array}\right.
$$

Solving this system, we get that $0<a<68$.

Calculate the distances between the compressor stations when $a=5$: $x=\frac{425-5}{7}=60$ km, $y=\frac{255+25}{7}=40$ km, $z=\frac{170+5}{7}=25$ km.

Answer. $0<a<68 ; 60$ km, 40 km, 25 km.