A hiker starts to climb up from base B to the summit S on sat 6am one day,spends the night at S and starts to climb down at 6am the next day.Prove that there is a point on the path B-S (there is only one path connecting B-S) which hiker crosses at the same time both days. Give the complete & reasonable proof.
Let $x$ describe his hike up from $B$ to $S$, and $y$ his hike down. Assume continuity of the paths. Normalize the time to $0$ for the respective departure, and 1 for arrival up, $A$ for arrival hiking down, and assume without loss of generality that $A \leq 1$, and that after arriving on his hike down he stays at $B$.
Them we have: $x(0) = B = y(A) = y(1)$, and $x(1) = S = y(0)$. Let $f(t) := x(t) - y(t)$, a continuous function satisfying $f(0) = B - S$, $f(1) = S - B$. As $S \neq B$, one of $f(0), f(1)$ will be positive, one negative, so by the intermediate value theorem, there is a $t_0$ such that
$0 = f(t_0) = x(t_0) - y(t_0)$,
so at $t_0$ he is at the same point $x(t_0) = y(t_0)$.