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Two Classic Mountain-Climbing Problems

Problem 1: One man started climbing a mountain at 5:00 AM. He spent the night up on the peak. Then, on Day 2, he set off down the mountain at 5:00 AM. Prove that at some point of the day, he was at the exact same altitude as he was at that same time the previous day.

Problem 2: Two men are planning to climb a mountain. One is climbing from the North side, the other from the South. Prove that the two men can climb the mountain in such a way that at all times, their altitudes are the same.

These problems are harder than some of you may claim. You can't simply say he can do it; you have to rigorously prove he can do it. Also, you have to prove that the conditions must be satisfied no matter how he takes his hike, not just that there exists a way for him to satisfy the conditions.

Note by Daniel Liu
2 years, 6 months ago

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Problem 1 Solution Assuming that the man travels at a constant rate, when he is exactly halfway down the mountain, it will be at the same time as yesterday when he was halfway up the mountain. Thus the condition is satisfied.

Problem 2 Solution This one is much easier. If they go at the same speed relative to the inclination and terrain, they will stay level with each other.

Why are these classic? They're not difficult! :P Finn Hulse · 2 years, 6 months ago

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@Finn Hulse Your proofs are not rigorous enough... You can't simply assume he is traveling at a constant rate. He could be traveling fast at some periods, slow at some other periods, maybe even going back up at other periods. For the second problem, it's not as easy as you state it to be; although it isn't hard to see that it is possible, you have to prove it.

Also, you must remember that the terrain of the mountain is completely random. It might be a straight incline, or it might have multiple ups and downs. Daniel Liu · 2 years, 6 months ago

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@Daniel Liu Oh... in that case, I'm clueless. :D Finn Hulse · 2 years, 6 months ago

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Problem 1: As we know that v=s/t (assume that mid point is the point where altitude is same) 1:when he was moving upwards he took more times to reach mid point due to gravity factor and velocity is lesser from the relation v=s/t. 2:when he was moving downward he take minimum time and velocity is greater at this moment. Result:the product of velocity and time is same hence distance or Altitude is same at mid point. Now the question is how the time is same? Ans:Assume a total distance of a mountain is S.When he move upward distance traveled S/2 where altitude is same.and When he move downward distance covered by 3S/2.hence time is also same at this situation. Muhammad Arslan · 2 years, 6 months ago

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For problem 1,the solution is given in "The Art and Craft of Problem Solving". Problem 2 is a bit harder but I figured it out : Since the two climbers start from the same altitude,then the solution is:When one of the climbers descends(perhaps because of a downward slope on the way)the other also descends.Then,they will be at the same altitude at all times. Q.E.D Tan Li Xuan · 2 years, 5 months ago

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At 5:00AM the first day he was at the top of the mountain.The second day at 5:00AM he is at the top.So this satisfies the question and the conditions. Adarsh Kumar · 2 years, 6 months ago

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@Adarsh Kumar Nope, he was at the bottom of the mountain on the first day at 5:00 am. Sreejato Bhattacharya · 2 years, 6 months ago

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@Sreejato Bhattacharya So sorry I realised that after I posted the comment. Adarsh Kumar · 2 years, 6 months ago

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The first is in the classic " The art and crat of problem solving" . 1) To prove think of their positions or altitudes as being function of time let h(t) it then is obviously is continuous .Similarly for descent lets say function is g(t) . plot these both on the same graph.now consider new fn. Y(x) = h(t) - g(t) so h( 5 am) is negative and uf Karttikeya Mangalam · 2 years, 6 months ago

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Already knew problem 1.

How do you solve 2? Mursalin Habib · 2 years, 6 months ago

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Problem 1:

Consider the time v/s altitude plot in the \(x-y\) plane, where time is denoted by the \(x\) axis and the altitude is denoted by the \(y\) axis. For the first day, at time 5:00 AM, the altitude is zero, and for the second day, at time 5:00 AM, the altitude is the height of the mountain. After some time, the altitude for the first day equals the height of the mountain and the altitude for the second day equals zero. The functions which input the time and return the altitude for both days are continuous, so their graphs must intersect at some point. This is where he was at the exact same altitude on both days.

I don't get the challenge in problem 2. Why can't the men just climb at the same rate? Sreejato Bhattacharya · 2 years, 6 months ago

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@Sreejato Bhattacharya What if one side of the mountain slopes back down? Daniel Liu · 2 years, 6 months ago

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Problem 1 is straightforward to "prove". Combine both Day 1 and Day 2, so that the same climber leaves from both the top and the bottom of the mountain at 5:00 AM. At some point in time, they're going to meet. Not exactly a rigorous mathematical proof, but good enough.

Problem 2 is much harder to prove "easily". Let us first assume that both climbers are each following a route which altitude is a function of the distance from the bottom. Then for each time either route hits a peak or a valley en route to the summit, we enumerate possible and distinct states as follows: For climber N at turnaround point A, all the other places the other climber S can be on his route that share the same altitude are numbered, e.g. A-1, A-2, A-3, and likewise for climber S at turnaround point B, we have B-1, B-2,... etc. The initial state is 0-0, and the final state at the summit is S-S. Without loss of generality, we'll assume that none of the peaks and valleys in both routes are at exactly the same altitude. Then whenever the 2 climbers N & S are at any particular state, there are exactly only 2 other states they can go to from there, being that at any particular state, only one of them is at a peak or a valley, and therefore there are 2 choices to go from there. The only exceptions are at the initial state 0-0 and final state S-S, which is linked to only 1 other state. The only way all the states can be linked is either directly from 0-0 to S-S, or some cycle of states somewhere, because otherwise there'd exist a state with more than 2 links to it, which isn't possible. It is not possible for the chain starting from 0-0 not ultimately reach S-S, because either the chain would have to 1) terminate somewhere other than S-S, or 2) terminate in a state already passed in the chain, both contradicting the fact all states other than 0-0 and S-S are linked to exactly only 2 other states. Thus, regardless if there exists cycles of states, there is at least a linked chain from 0-0 to S-S.

The key to this proof is carefully choosing which states to bother to enumerate. There are other states that can be enumerated, but they tend to over-complicate the proof. Michael Mendrin · 2 years, 6 months ago

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