* A monk starts to climb a mountain at 8:00 am and reaches the summit at noon. He spends the rest of the day and that night on the summit. The next morning he leaves the summit at 8:00 am and descends by the same route he used the day before, reaching the bottom at noon. Prove that there is time between 8:00 am and noon at which the monk was at exactly the same spot on the mountain on both days. Note that the monk can walk at different speeds, rest, or even go backward whenever he wants. *

A nice problem. Source unknown.

I agree, it’s an excellent problem with a brilliant solution. It was published in New Scientist in the 60s so is at least 40 years old.

What a practical application of the Intermediate Value Theorem! Assuming that you need to know that this might occur— which, well, you probably wouldn’t.

I am just wondering, cann’t the proof for this problem be stated in english language like this:

Assuming the monk starts climbing up at 8:00 am and another monk (clone) starts descending down at 8:00 am the same-day, they are supposed to meet somewhere if it is the same route. So there should be a place on the mountain where both the monks meet. In other words, is this not a solution? Just curious.

Thanks

~Venkat

Dear Venkat,

The suggested solution is exactly what you described. But it seems that not everyone can think of things in this way.

And of course, we being maths geeks, must always using scary things like intermediate value theorem.

http://unimodular.net/blog/?p=86

Nice to know the french-connection:

http://unimodular.net/blog/?p=86.

I was able to grasp the formalization as stated in Intermediate Value Theorem. I think whoever wrote the Wikipedia article on “Continuous function” did a nice job in explaining the essence to a layman.

An excerpt from the article:

“If a child undergoes continuous growth from 1m to 1.5m between the ages of 2 years and 6 years, then, at some time between 2 years and 6 years of age, the child’s height must have been 1.25m.”