Analyzing the problem we get:
$$S = a(2-b), \quad S = b(2-c), \quad S = c(2-a),$$
where $S$ -- positive integer which is equal to the distance.

We may assume that $a \ge b$. If $a > b$, then $2-b < 2-c$ or $b > c$. Analogously, $2-c < 2-a$ or $c > a$. Contradiction. So $a = b = c$. Then we have that $S = a(2-a)$. Let us show that $S = a(2-a) \le 1$. Indeed, $2a - a^2 \le 1 \Leftrightarrow (a-1)^2 \ge 0$. Hence, since $S$ is a positive integer which is less or equal than $1$, $S=1$.