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A hiker starts to climb up from base B to the summit S on sat 6am one day,spends the night at S and starts to climb down at 6am the next day.Prove that there is a point on the path B-S (there is only one path connecting B-S) which hiker crosses at the same time both days. Give the complete & reasonable proof.

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Should we assume that the hiker makes it back to B before the end of the second day? –  Peter Taylor Jan 6 '13 at 19:34
No need. Just needs to take the same route. There is a continuity requirement too. Hint: consider two hikers on the same day ... –  Mark Bennet Jan 6 '13 at 19:38

1 Answer 1

Let $x$ describe his hike up from $B$ to $S$, and $y$ his hike down. Assume continuity of the paths. Normalize the time to $0$ for the respective departure, and 1 for arrival up, $A$ for arrival hiking down, and assume without loss of generality that $A \leq 1$, and that after arriving on his hike down he stays at $B$.  

Them we have: $x(0) = B = y(A) = y(1)$, and $x(1) = S = y(0)$. Let $f(t) := x(t) - y(t)$, a continuous function satisfying $f(0) = B - S$, $f(1) = S - B$. As $S \neq B$, one of $f(0), f(1)$ will be positive, one negative, so by the intermediate value theorem, there is a $t_0$ such that  

$0 = f(t_0) = x(t_0) - y(t_0)$,  

so at $t_0$ he is at the same point $x(t_0) = y(t_0)$.

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I believe the tricky part of this question is establishing a homeomorphism between $[0,1]$ and the subset of $\mathbb{R}^3$ that describes the path from $B \in \mathbb{R}^3$ to $S \in \mathbb{R}^3$. –  copper.hat Jan 6 '13 at 21:05
@copper.hat: Isn't the path in $\mathbb{R}$, stretching it out along its trajectory, and so The imbedding into $\mathbb{R}^{3}$ doesn't matter? This sounds like an intro analysis/calculus question, not like topology. –  gnometorule Jan 6 '13 at 21:15
@copper.hat: if we don't assume there is only 1 path, there won't necessarily a point where the two trips will meet; so its an essentially one-dimensional problem. –  gnometorule Jan 6 '13 at 21:19
I may be assuming the problem is more technical than intended. (There is only one path by assumption.) –  copper.hat Jan 6 '13 at 21:26

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