## May 29, 2006

### Priest on a hill

This is one of the oldest puzzles I know, one I've always liked. Give it a shot. (If you know the answer, please hold off telling us so as not to spoil it for those who want to think it over).

***

A priest lives at the bottom of a middling hill. Once a year, he makes a pilgrimage by foot to a temple at the top of the hill. He makes it a point to set out at daybreak, walks up leisurely, stops for snacks and lunch and an occasional rest ... and reaches the temple just as the sun is setting.

He spends a few days at the temple, in prayer and meditation. Then he makes his return trip in much the same way. He sets out at daybreak, walks down leisurely, stops for snacks and lunch and so forth ... and reaches his home a little while before sunset, because walking downhill is faster than walking uphill.

Question: is there a point on the hill at which the priest can be found at the same time on both days? Yes, no or maybe? Whatever your answer, prove it.

(Note: If your answer is yes, I'm not interested in where the point is, or at what time he is there. Just prove that there is such a point).

Anonymous said...

If the priest climbs down the same path he climbs up, and if the path is a continuous function of time, then the answer is yes.

Dilip D'Souza said...

Vish, you gotta prove it. (Yes, same path both ways, should have made that clear).

Varun Singh said...

Lets plot a graph (Position=x vs Time=y). While climbing, priest starts with - t=0 & d=0
when he reaches the top, t=12 hrs (assume) & d=D.
When he starts from the top, t=t1 & d=D
When he reaches bottom, t=t2 & d=0.
t1,t2 < 12 hrs

The Y positions grow from 0 to D while going up and come down from D to 0 while coming down. The graph for the upward journey would go from (0,0) to (t=12, D), it need not be linear. t=12 & D are the maximum values. The graph for the downward journey would always intersect the first graph at one point (and *only* one point if we assume it to be linear)

FifthBeatle said...

Answer yes. Simple application of Brouwer's Fixed Point Theorem.

Anonymous said...

The problem of the priest going up
one day and down another day is the same as if two people started at the same time, one from the top and one from the bottom. The question now is whether these two people will meet somewhere....

Anonymous said...

Imagine a rope between the house and the temple. Now imagine the priest at the bottom of the hill and a "clone" at the top. If they walk towards each other at daybreak holding onto the rope, then there will be a point where the priest and his clone meet. Note that this does not require the priest and the clone to start simultaneously: the priest or the clone could continue to be at his starting point for some time after the other has started to move. Nonetheless, since both hold onto the rope, there is a point where they meet.

As Arnold noted, there is a fixed point theorem hidden somewhere. However, Brouwer's theorem concerns a mapping from a domain D to itself. I am not sure what the relevant function is here: care to fill in the details, Arnold? Clearly, however, continuity (the "rope") is necessary for the result.

Suresh.

Anonymous said...

doing the undesirable of following up my own post, i should note that no fixed point theorem is required - all it needs is something far more elementary, known I think as the intermediate value theorem in real analysis. Brouwer's theorem is more deep and is not needed here.

S.

Dilip D'Souza said...

Thanks for your interesting responses, guys.

Anonymous 846 and Suresh (nothing undesirable!), you have the answers that I had in mind. It's best to think of it as two simultaneous journeys -- then the two men have to meet somewhere. The rope helps make it even clearer.

Arnold, like Suresh, I'm not sure why you would need Brouwer's theorem (a thought-provoking theorem by itself, of course). Please do explain.

Varun, thanks too. Uma, that's right! Damn. We'll have to do it another time.

Anonymous said...

they will meet at a time that is the harmonic mean of the times to go up and come down.