#### Answer

There is a time on both days when the monk was at exactly the same point on the path.

#### Work Step by Step

Let $t$ be the time in hours where $0 \leq t \leq 12$
Let position 0 be the bottom of the path and let position Y be the top of the path.
Let $g(t)$ be the monk's position on the path on the way up.
Then $g(0) = 0$ and $g(12) = Y$
Let $h(t)$ be the monk's position on the path on the way down.
Then $h(0) = Y$ and $h(12) = 0$
Note that both $g$ and $h$ are continuous functions.
Let $f = g-h$
Then $f(0) = -Y$ and $f(12) = Y$
Since $f$ is the difference of two continuous functions, $f$ is a continuous function.
By the intermediate value theorem, there is a time $c$ in $(0,12)$ such that $f(c) = 0$. Then $g(c) = h(c)$.
This means that at time $c$ on both days, the monk was at exactly the same point on the path.