A train line is divided into 10 segments by stations $A, B, C, D, E, F, G, H, I, J$ and $K$. The distance from $A$ to $K$ is equal to $56 \mathrm{~km}$. The route of two consecutive segments is always less than or equal to $12 \mathrm{~km}$ and the route of three consecutive segments is always greater than or equal to $17 \mathrm{~km}$. Determine the distances:

a) from $J$ to $K$;

b) from $D$ to $H$;

c) from $B$ to $G$.

#
Solution

a) We can write $A K=56$ and $A K=A D+D G+G J+J K$. Since $A D, D G$ and $G J \geq 17$, then $J K \leq 5$. Therefore, for $H K \geq 17$, we must have $H J \geq 12$. But, we know that $H J \leq 12$, so $H J=12$. From $H K \geq 17$ and $H J=12$, we conclude $J K \geq 5$ and the only possibility is $J K=5$.

b) By symmetry, we will have $A B=5$ and $B D=12$. Now, we proceed with

$$
\begin{aligned}
D H & =A K-A B-B D-H J-J K \\
& =56-5-12-5-12=22
\end{aligned}
$$

c) Finally, $G J \geq 17$, but $H J=12$. Thus $G H \geq 5$. From $D G \geq 17$ and $D H=D G+G H=$ 22, we get $D G=17$ and $G H=5$. Therefore

$$
\begin{aligned}
B G & =B D+D G \\
& =12+17 \\
& =29
\end{aligned}
$$