Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

closed as off-topic by 900 sit-ups a day, Claude Leibovici, hardmath, gekkostate, azimut Jul 4 at 22:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 900 sit-ups a day, Claude Leibovici, hardmath, gekkostate, azimut
If this question can be reworded to fit the rules in the help center, please edit the question.

    
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
2  
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 $x(t_0) = y(t_0)$.

share|improve this answer
    
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

Not the answer you're looking for? Browse other questions tagged or ask your own question.