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# Monk and the mountain

## Question

A monk started out at 7:00 am at the bottom of the mountain and climbed to the top arriving at 7:00 pm  He prayed all thru the night and at 7:00 am started down the mountain arriving at the bottom at 7:00 pm.  Is there a point on the mountain trail where the monk was at the same time going up and coming down  Proof please

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Spoiler

Assuming that the path the monk takes is exactly the same going up and down (no forks in the path, no loops unless taken in the reverse order and direction), and there is no teleportation involved, then there is a point on the path where the time would be the same for the monk going up or down.

Let P1(t) be the progress along the path from 0=bottom to 1=top of the mountain for the monk on day 1 at time t.  Similarly P2(t) for day 2.  Then F(t) = P1(t) - P2(t) is the difference in the monk's location at time t between the two days.  Both P1(t) and P2(t) are continuous (since the monk doesn't teleport ), so their difference F(t) is continuous.  F(7am)=0-1=-1, and F(7pm)=1-0=1.  Using the intermediate value theorem we know there exists a time t such that F(t)=0.  It would be at this time that the position the monk is on the two days would be the same.

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17 hours ago, EventHorizon said:
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I agree with @EventHorizon.

Let P1(t) be the progress along the path from 0=bottom to 1=top of the mountain for the monk on day 1 at time t.  Similarly P2(t) for day 2.  Then F(t) = P1(t) - P2(t) is the difference in the monk's location at time t between the two days.  Both P1(t) and P2(t) are continuous (since the monk doesn't teleport ), so their difference F(t) is continuous.  F(7am)=0-1=-1, and F(7pm)=1-0=1.  Using the intermediate value theorem we know there exists a time t such that F(t)=0.  It would be at this time that the position the monk is on the two days would be the same.

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Spoiler

Monk M1 starts  at 7:00 am at the bottom going up.

Monk M2 starts at 7:00 am at the top going down.

If there is just one path, they must meet.

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