Puzzle about a monk walking to his neighbourmonastery

Question

A monk wants to make a journey to the neighbourmonastery where he wants to meditate. So he begins his trip one day at 8 am at his homemonastery per pedes to the neighbourmonastery.

He walks with irregular speed, makes here and there a pause of different length and finally reaches the monastery at 8 pm.

There he lingers 3 days and at the morning of the 4th day he sets out at 8 am. He walks the same way, again with irregular speed and arbitrary pauses, and reaches his homemonastery at 8 pm.

Now the question:

Is it necessary, that the monk was at a certain daytime at his outward-/return journey at the same place?

I would say yes, but I can't prove it.

I guess this problem is already posted with another story but how would I find it then?

If there where two monks going on the same day they would certainly met each other at the same place sometime, but I can't argue that way or?

Answer 1

If we say that the home monastery is at $0,$ and the visiting monastery is at a distance $d.$

Lets make hour $0 = 8:00$ AM the time the monk leaves and hour $12 =8:00$ PM the time he arrives at the other monastery.

We can say $f(t)$ with $f(0)= 0$ and $f(12) = d$ represents the monks position as a distance from home as a function of time, and $f$ is a continuous function.

On the return trip $g(t)$ represents the monks distance from home with $g(0) = d$, and $g(12) = 0.$ $g$ is also a continuous function.

Then $g(t) - f(t)$ is continuous, $g(0)- f(0) = d, g(12) - f(12) = -d$

By the intermediate value theorem there exists a $t\in (0,12)$ such that $g(t)- f(t) = 0$

Answer 2

Yes, indeed, the proof is, as you said: imagine if it was the same day and it was two monks.

A (mostly) rigorous mathematical proof would represent the movement of a monk by a function (time in hr $\mapsto$ distance from homemonastery in miles) and so you would have one function $f, f(8)=0, f(20)=d$ and another $g, g(8)=d, g(20)=0$, where $d$ is the distance between the monasteries. Also, we must assume both $f$ and $g$ are continuous on $[8,20]$ (as all functions modelling nature, pretty much, are). Thus, $h=g-f$ is also continuous on $[8,20]$ and $h(8)=-d<0$ and $h(20)=d>0$ so there must be time $t$, $8<t<20$ where $h(t)=f(t)-g(t)=0$, as a consequence of Intemediate Value Theorem (https://en.wikipedia.org/wiki/Intermediate_value_theorem). At (time) $t$, therefore, we have $f(t)=g(t)$, so the two monks are at the same place.