Puzzle #1

[KotysDad]

Imagine a hiker is on a 2-day hike: day 1 up the mountain and day 2 down the same mountain. On day 1 he begins up the mountain at exactly 6am on a certain path. He arrives at his destination point at the top of the mountain at exactly 6pm. His pace is totally random just as long as he starts at exactly 6am and finishes at exactly 6pm.

Now, on day 2 he goes down the exact same path as he came up. Begins his trek down again at exactly 6am and reaches the bottom at exactly 6pm. Pace again totally random. Prove or disprove the following statement:

*There is a point somewhere on the path that he will reach at EXACTLY the same time each day.

 

[Palmetto Pimp]

Well false would be the easy answer so the statement must be true

 

[Blitz]

the finishing point?

 

[Hypatia]

I agree with njsf...

I agree with njsf...

Assuming the path traveled was continuous and unbroken, there would be at least one time he would be in the exact same spot, exact same time, on the same mountain path. I think it would be harder to prove or disprove exactly where or when, but I think there would at least be a "crossing point" so-to-speak. No?? :shrug:

 

[KotysDad]

Re: I agree with njsf...

Re: I agree with njsf...

There is a point (and it isnt the starting or finishing point). You cant state exactly where it is, but there is a relatively short proof that the point actually exists. By relatively short, my answer will take about 3-4 sentences.

 

[yak merchant]

Okay for the sake of proofs let say on day 1 going up the mountain he leaves at 6am. Goes the speed of light for 1 second and stops(or if stopping isn't allowed moves at a pace of 1 cm per day) 10 feet from the top arriving at 6:00.01 AM Waiting until 5:59.50 and walking the final 10 feet in 10 seconds.

Now on day 2 he starts off at 6Am walks 5 feet stops until 6:00.05 and then continues down the path crossing the 10 foot mark at 6:00.10 and to the bottom at whatever pace.

ON the way up all of the points on the path besides the last 10 feet were crossed before 6:00.01 am, and the last 10 feet were crossed during 5:59:50PM to to 6:00 PM. Now on the way down
the first 10 feet were crossed before 5:59:50 PM and the rest of the way was crossed after 6:01 AM.

Go ahead make me look like a dumbass

 

[KotysDad]

Yak,

Honestly, you gave almost the same exact answer I gave when I first heard this question. They still cross......here is the proof.


Imagine at a distance that someone was videotaping this trek up the mountain on the first day. The video recorder sets its clock at the bottom of the screen at 6am and starts taping at the same time the man starts up the mountain. When the man reaches his destination, the clock obviously reads 6pm.

On day 2, the video recorder does the same thing. Tapes the trek down the mountain setting the time at 6am when the man starts down the mountain. Now, imagine if you could superimpose these two days one over the other. What you would see is a tape of the same man going up and down the mountain with the clock starting at 6am and stopping at 6pm. Assuming he travels the exact same path going up as coming down then he must "meet himself" on the path. The "when" and "where" is determined by the pace he walks, run, or whatever he does. But he must cross his own path at some time between 6am and 6pm when you view the superimposed videos.



ok, so maybe its a few more than 3-4 sentences. lol

 

[dr. freeze]

this might be the stupidest puzzle i have ever heard......it makes no sense when you first read it, and the "solution" to the puzzle is also stupid.....in fact, the solution's answer doesn't even follow the rules asked in the problem......the guy will not be in the same place at the exact time on both days unless he travels at the same velocity to the midpoint OR his velocities vary in such a way that say he goes 20% of the distance the first day to the exact same spot it takes him 80% of the distance in an equal time....the concept of the video "superimposed" thing is bogus and does not apply here.....

 

[yak merchant]

Makes sense now. If it were two guys that left at 6am on the same day and had to be at the other end at 6Pm. Seeing how they are on the same trail, they will cross at some point, and it would have to be the same time for both guys regardless of how fast they are going.

 

[fatdaddycool]

not correct in the wording
Distance traveled is related speed and velocity adn as the velocity will automatically be changed by the gravity constant 9.80665 m/s2 (mass divided by speed squared) when the person is descending the hill the time that he passes a certain point on the hill will not be reflected in the time he passes it the next morning coming down as v2 will be changed

 

[PerpetualCzech]

dr. freeze,

What are you talking about? The wording of the puzzle is fine and so is the solution. See KotysDad's answer for totally valid explanation.

Fatdaddycool,

v2 being different doesn't matter. It's the *average* velocity at the time he "meets himself" that matters. All that extra stuff you you wrote is nonsense (velocity will be changed by the gravity constant??? Please ...)


An alternative solution would be to plot the 2 treks on a graph with distance on one axis and time on the other. You can easily see that it is impossible to plot the 2 hikes without the lines crossing at some point. The point that they cross at is the same time of the day where he reaches the same point.

 

[dr. freeze]

what i was trying to say is that there has to be a point on the path where the velocity vector's reach the same on both days....i don't really consider this to be a puzzle.....

 

[TheShrimp]

Some else posted this, but didn't state it so succintly:

Imagine one person leaves the top of the mountain at 6am and arrives at the bottom at 6pm. Imagine another person leaves the bottom at 6am on the same day and arrives at the top at 6pm. Those two people have to be at the same location at the same time at some point during the day. Now imagine those two people are, in fact, our original man on consecutive days.

4 sentences. I hope that's clear.

 

[KotysDad]

Dr. Freeze,

I am sorry you find this puzzle stupid. Given that you are a student, I would think you would have the ability to think "outside of the box". I am not a doctor, but I am pretty sure that medicine isnt always "by the book".

Czech,

Yes, your alternate solution is also valid - the more mathematical solution. Thanks.

 

[PerpetualCzech]

dr. freeze,

Now it's your turn to post silliness. Velocity vectors don't "reach" anywhere on the path. The solution to the puzzle has nothing to do with velocity vectors - these are tools used to describe an object's speed and direction, which you don't need to solve it.

For some reason you've taken the time to be totally arrogant about the puzzle, yet you aren't showing that you understand it very well.

 

[fatdaddycool]

PerpetualCzech,
I think you are ablsolutely incorrect you are relating this to a simple Distance traveled over time which is not what he is asking.

His pace is totally random

the riddle itself takes the question into speed and velocity for instance:
on a graph you have a line made up of a series of points, I know you know this I am just stating my complete point from a physics standpoint, on line AB is an infinite number of points. right?
So taking into account that the speed is variable, along with direction of line AB there is no possible way that you can prove that starting from opposite directions and altering speed and velocity and direction, and force due to gravity and wind , that the you will land on the exact same point on the line every time with the only constant being distance over time. I am sorry I do not agree with you. But that doesn't mean anything I sucked in physicsI & II......lol

 

[PerpetualCzech]

I agree my solution wasn't much of a "proof"; it was more of a graphical way of showing the statement had to be true. Here it is in more detail:

Use your vertical axis for distance and horizontal axis for time. We will use ground level as distance point 0 (zero) and the top of the mountain as 100. Time can be measured in hours, from 0 (6am) to 12 (6pm). Now the graph on day 1 must start at the point (0,0) and must end at the point (100,12). The graph for day 2 must start at (100,0) and end at (0,12). Both lines must be continuous (this is important). Because it is impossible to draw the graphs so that they do not cross at some point, this shows that at some point he must be at the same place at the same time of day.

The random pace given in the question is just represented by the fact you can draw an infinite number of shapes of the line on the graph, that's all. My answer was inductive, not deductive, but relating distance over time is indeed exactly what you need to do to solve the problem.

(There probably is a way to give a deductive answer by setting 2 equations with 2 variables and showing there must exist a solution. Those 2 variables would again be time and distance)

 

[fatdaddycool]

PC,
you are correct only if speed remains a constant therby giving you the exact straight line you are talking about but, if speed is not constant then the line is no longer straight and although I agree the lies MUST cross at some point it will not be at the exact same point on the graph at the exact same time on the graph the line will not be linear. You are not taking into account the other variables as this now becomes a where is hat at what time of day if speed does not remain constant. If distance is constant say 1000 meters, time is constant, twelve hours, you can deduct the average speed, but when taking into the equation ACTUAL speed as related into time of day and distance traveled in that time, the equation changes...but what ever

 

[PerpetualCzech]

I never talked about the line needing to be straight, in fact I said it can be an infinite number of shapes, as long as it's continuous. Speed does NOT have to be constant, the function does NOT have to be linear.

As long as you agree the lines must cross at some point then you've found the solution. Get a piece of paper out and try it.

 

[fatdaddycool]

PC.....LMAO....I know but the riddle said the EXACT SAME POINT everytime and its not..lol..anyway you are probably sick of this anyway...thanks for the help though, I think I may be trying to interpet a little too literally.
FDC