One morning, exactly at sunrise, a Buddhist monk began to climb a tall mountain. The narrow path, no more than a foot or two wide, spiraled around the mountain to a glittering temple at the summit.

The monk ascended the path at varying rates of speed, stopping many times along the way to rest and to eat the dried fruit he carried with him. He reached the temple shortly before sunset. After several days of fasting and meditation, he began his journey back along the same path, starting at sunrise and again walking at variable speeds with many pauses along the way. His average speed descending was, of course, greater than his average climbing speed.

Prove that there is a spot along the path that the monk will occupy on both trips at precisely the same time of day.

Originally posted by HandyAndy One morning, exactly at sunrise, a Buddhist monk began to climb a tall mountain. The narrow path, no more than a foot or two wide, spiraled around the mountain to a glittering temple at the summit.

The monk ascended the path at varying rates of speed, stopping many times along the way to rest and to eat the dried fruit he carried with him. He reached th ...[text shortened]... a spot along the path that the monk will occupy on both trips at precisely the same time of day.

Originally posted by HandyAndy Proof? Bear in mind that his average speed descending was greater than his average climbing speed.

I think the easiest way to see the answer is to draw a graph of the monk's path starting at (sunrise, 0) and proceeding to (sunset, H). The points define a box, which gets cut into two pieces by the monk's path. It's impossible to draw the return trip starting at (sunrise, H) and ending at (sunset, 0) without crossing the first line unless you travel down the sides of the box, but this means the monk fell off the mountain instantaneously.

Not quite a proof, but instructive. I'll work on something more rigorous later.

Originally posted by PBE6 I think the easiest way to see the answer is to draw a graph of the monk's path starting at (sunrise, 0) and proceeding to (sunset, H). The points define a box, which gets cut into two pieces by the monk's path. It's impossible to draw the return trip starting at (sunrise, H) and ending at (sunset, 0) without crossing the first line unless you travel down the ...[text shortened]... neously.

Not quite a proof, but instructive. I'll work on something more rigorous later.

You're close. Think about the rope burning at both ends.

Imagine we have two monks and when one is travelling down the other is travelling up (both starting at sunrise).

At some point the two monks must pass each other. When they do they have both occupied the same point at the same time. And this is obviously true no matter what speed (constant or varying) the two monks travel at.

Originally posted by HandyAndy [snip]
Prove that there is a spot along the path that the monk will occupy on both trips at precisely the same time of day.

Imagine there's some twilight-zone stuff going on and that having spent several days with his dried apricots and deep thoughts the Monk is still at the top of the mountain but has been transported back in time a couple of days. At sunrise the original Monk begins climbing the mountain at the same time as twilight-zone Monk begins going back down.
At some point Monk and T-ZMonk will meet.

QED.

[Edit - in the time it took me to start typing this, go get ice cream from the fridge, finish it (the ice cream and the typing) XanthosNZ typed a similar answer. Darn!]

Originally posted by XanthosNZ Imagine we have two monks and when one is travelling down the other is travelling up (both starting at sunrise).

At some point the two monks must pass each other. When they do they have both occupied the same point at the same time. And this is obviously true no matter what speed (constant or varying) the two monks travel at.

But the problem explicitely states that it is the very same monk that starts the trip, gets to the temple, rests there and begins his journey back, doesn't it?

let F(x) = distance from base camp with respect to time on day one (ranging from 0 at t=0 to H at t=sunset. clearly this is a continuous function, provided the monk hasn't learned to teleport.

let G(x) = distance from base camp with respect to time on day two (ranging from H at t=0 to 0 at t=sunset). also continuous.

now consider the function F(x) - G(x). this is a composition of continuous functions, along the same domain, and thus is also continuous. Now, it ranges from -H to +H and must achieve every value in between at some point along the way (by the definition of continuity). But the moment when [F(x) - G(x)] achieves a height of 0 is exactly the moment when the monk was equidistant to the base camp on subsequent days.

note that the monk must not teleport, and the time interval must be the same so as to not allow for the possibility that he could roll down the mountain really quick after sunset. 🙂

p.s. i like the idea of two monks doing it on the same day - i've always thought of it as running the tape of the monk on the first day, superimposed over the tape of his trip on the second day, like the "ghost car" in the original super mario kart? anybody know that reference?

I dont like the variable speeds part of the question.
Is this a budhist reference to 'emptiness'...not - 'you can't step in the same river [mountin] twice' but that you can't step in the same river once?
1 and 0 are the same....the points of the journey are the same no matter which direction [life] you travel.