Yoak: Mountain Climbing
Here’s a quick puzzle that I don’t think has appeared here previously.
A man leaves at exactly 6:00 AM to climb a mountain. He may not climb at a constant rate. In fact, he may stop to rest, or even backtrack a bit to see something interesting. He arrives at the summit at exactly 6:00 PM and camps for the night.
The next day, he starts down at exactly 6:00 AM, again meandering unevenly, but following exactly the same path and reaching the bottom at exactly 6:00 PM.
The question is, was there some point on the path he was sure to have visited at exactly the same time on the two trips?
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yanmi said,
March 23, 2009 at 2:02 am
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 .
philhart said,
March 23, 2009 at 9:54 pm
Interesting use of distracting detail. A consideration of limit cases is often useful in puzzles like this Show Spoiler ▼
.
jyoak said,
March 24, 2009 at 11:29 am
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?
jyoak said,
March 24, 2009 at 11:31 am
yanmi, that’s an excellent way to reveal the answer. It’s also the one I was going to post myself eventually. :-)
philhart said,
March 24, 2009 at 11:30 pm
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.
mathphan said,
March 26, 2009 at 5:54 pm
You can model this as a graph. Let’s imagine you had a graph of height vs. time. Show Spoiler ▼
czarandy said,
March 26, 2009 at 9:44 pm
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.
jyoak said,
March 27, 2009 at 9:35 pm
czarandy, that’s wonderful. Thanks. I originally envisioned the answer with a graph as mathphan pointed out, but this is a wonderful rigorous explanation.