Comments on: Yoak: Mountain Climbing

By: Shawn

Shawn — Sun, 11 Mar 2012 02:25:33 +0000

The answer is yes.

By: jyoak

jyoak — Sat, 28 Mar 2009 03:35:41 +0000

czarandy, that’s wonderful. Thanks. I originally envisioned the answer with a graph as mathphan pointed out, but this is a wonderful rigorous explanation.

By: czarandy

czarandy — Fri, 27 Mar 2009 03:44:52 +0000

More math-y solution:
Since he takes the same path both ways, you can assume he is traveling from 0 to 1 and that the total time is 1 unit. Say f(t) gives you his position when going up and g(t) his position when going down. Both of those must be continuous. So h(t) = f(t) – g(t) is also continuous. Since h(0) = -1 and h(1) = 1, by the IVT h(x) = 0 for some x, so at that point he is in the same position at the same time.

By: mathphan

mathphan — Thu, 26 Mar 2009 23:54:01 +0000

You can model this as a graph. Let’s imagine you had a graph of height vs. time. [spoiler]So the vertical axis would be the distance up the mountain and the horizontal axis would go from 6am to 6pm. The trip up would look like a meandering line, perhaps flat at some points, perhaps going down at some points, but it will definitely be continuous from 6am to 6pm. It will connect the bottom point (6am, bottom) to the top point (6pm, top). Think about the return trip. It will have a similar meandering path from (6am, top) to (6pm, bottom). There’s no way to draw these two graphs without them intersecting in at least one point. The answer is, yes, there must be a time when you are at the exact same point up/down the mountain at the exact same time of day.[/spoiler]

By: philhart

philhart — Wed, 25 Mar 2009 05:30:01 +0000

I have to be very guarded at the moment, otherwise I will simply end up spoiling. Comments such as “even backtrack to see something interesting” serve to distract the solver’s attention from using a limiting case to answer the question that was posed. Another distraction is casting the puzzle in the context of the human experience: a Saturn V rocket meanders around the launch pad in the first few moments after lift-off, but the payload can reach escape velocity.

By: jyoak

jyoak — Tue, 24 Mar 2009 17:31:22 +0000

yanmi, that’s an excellent way to reveal the answer. It’s also the one I was going to post myself eventually. :-)

By: jyoak

jyoak — Tue, 24 Mar 2009 17:29:57 +0000

philhart, It’s funny that you mention that. When this problem was originally presented to me “sunrise” and “sunset” on two days were used instead of time. Because those happen at slightly different times, I answered in the negative thinking that the person had intended to trick me with this detail. It appears that I overthought the problem. :-)

That said, I’m not sure what you mean about distracting detail or what special significance making the trip in one second has. Can you explain a bit more?

By: philhart

philhart — Tue, 24 Mar 2009 03:54:06 +0000

Interesting use of distracting detail. A consideration of limit cases is often useful in puzzles like this [spoiler] , think about staying within 1 metre of each starting point for 11:59:59, and covering the remaining distance in 1 second [/spoiler].

By: yanmi

yanmi — Mon, 23 Mar 2009 08:02:14 +0000

I think there is one point .If there were two men who all started at 6:00AM,but one from the foot of the mountain and the other from the top .They would meet each other at some point .