Exercise 55Calculus › Single-variable calculus › Continuity › Intermediate Value TheoremHigh schoolAdded 2 Nov 2026

The monk on the mountain

#intermediate value theorem#classic puzzle

Statement

A monk climbs a mountain path from the foot to the summit, leaving at 7:00 and arriving at 19:00. He spends the night there. The next day he comes down the same path, leaving at 7:00 and arriving at the foot at 19:00. He may walk fast or slowly, and even stop or turn back at times, on either day.

Prove that there is a point on the path where he is at exactly the same time of day on both days.

Hint

Superimpose both days. Introduce the functions $u(t)$ and $d(t)$ giving his distance from the foot at time $t$ on the way up and on the way down.

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Solution

Model. Measure the time $t$ in hours after 7:00, so $t\in[0,12]$ , and the position as the distance $s$ along the path from the foot. Let $u(t)$ be the position on the first day and $d(t)$ the position on the second day. Walking is continuous, so $u$ and $d$ are continuous.

Values at the ends. Let $L$ be the length of the path.
- $u(0)=0$ and $u(12)=L$ ,
- $d(0)=L$ and $d(12)=0$ .

Auxiliary function. Let $h(t)=u(t)-d(t)$ , continuous on $[0,12]$ . Then $h(0)=0-L=-L<0$ and $h(12)=L-0=L>0$ .

Conclusion. By the intermediate value theorem there is $t_0\in(0,12)$ with $h(t_0)=0$ , that is $u(t_0)=d(t_0)$ : at the same time of day $t_0$ the monk is at the same point on both days.

Picture. Draw the two graphs $u$ and $d$ in the same $(t,s)$ plane: one goes from the bottom-left to the top-right, the other from the top-left to the bottom-right. They must cross.

Remark. Equivalently, two walkers starting at opposite ends of a path on the same day and walking towards each other's end must meet.

The monk on the mountain

Statement

Solution

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