Thanks to Abhinav Jain for this wonderful puzzle!

*One morning at **6 am**, a monk began climbing a tall mountain, which happened to only have one path to the top. He ascended the path at his leisure, taking some stops along the way. He reached the top at **8 pm**.*

*The next morning at **6 am**, the monk descended the mountain along the same path. He took several breaks along the way, and reached the bottom at **8 pm**.*

*The amazing result: there is some spot on the path that the monk occupied at precisely the same time of day for both trips. Why is this?*

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy mountain climbing!

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A good way of thinking of this is by superimposing the two journeys into one journey. Suppose on the day of the second journey at around 4am is struck by lightning. Luckily he doesn’t die, but unbeknownst to his knowledge he has actually traveled back exactly 1 day in time! The monk doesn’t realize this and he hikes down the hill as if nothing had happened at all. This means that he’d start at 6am and reach down by 8pm. Now at 6am there are two monks – one monk travelling uphill, and one monk from the future travelling downhill. They have to reach their destinations at the same time. Since there is only one path and they must reach their destinations at the same times, the two monks (the present day monk and the monk from the future) must cross each other at some point of time.This proves the problem (since the time of crossing is the point of time they have the same position.

Drawing the graphs on the same chart where y axis represents the height of the monk and x axis represents the time during the day.

First graph (f1) is a continuous graph from (6am, 0) to (8pm, h) and the second graph (f2) is a continuous graph from (6am, h) to (8pm, 0) where h>0

Let f be f1- f2

f(6am) = -h

f(8pm) = h

Both f1 and f2 are continuous, so f is also continuous.

By intermediate value theorem, f(t) = 0 for some t between 6 am and 8 pm.

For both the days, lets draw a graph between his height and time. On day 1, it will start from 0 at 6am and end at h at 8pm. It will be a continuous graph. On day 2, the graph will start from h and go till 0. We can see that there will be a point where both the lines intersect. This is the point where the monk was on both the days at same time.