that hill,tl;dr

Kevin Duffy

mackg wrote: »

When they travel at varying speeds of course there is no way to tell where they would meet but that isn't the question. The brothers method does however answer the question in the OP

If you can't tell where they would meet, then you can't tell if there is a place where the one person making the journey would be at the same place at the same the following day.

Journey up -point A at 1000, B at 1200 C at 1400, D at 1600

Journey down - Point D at 1003, point C at 1227, point B at 1440, point A at 1723

Two brothers passing at each other at some unknown point is not the same as someone being in the same place at the same time on a different journey when reversing the course.

pickarooney

Kevin Duffy wrote: »

If you can't tell where they would meet, then you can't tell if there is a place where the one person making the journey would be at the same place at the same the following day.

Journey up -point A at 1000, B at 1200 C at 1400, D at 1600

Journey down - Point D at 1003, point C at 1227, point B at 1440, point A at 1723

Two brothers passing at each other at some unknown point is not the same as someone being in the same place at the same time on a different journey when reversing the course.

The meeting point in this case would be roughly three-quarters way between B and C at about 1327.

Emiko

pickarooney wrote: »

But the half-way point is only a single example of an infinite number of possible meeting points and times. In this case, the actual meeting point would be something like a few metres higher up and at 15.00.30

If you don't accept the two brothers example as an easy way to illustrate the answer, make it harder - Draw (or visualise) a red line going up the hill, marking each second spent climbing, then do the same with a green line coming down. They'll meet somewhere. This is just an example of one pair of speeds travelling. The meeting point can be shunted up or down the line according to how fast each one travels.

The mistake to avoid in this is to consider the speeds and the time since starting as important for each leg of the journey. They're irrelevant, just included to confuse you.

Of course the speed matters.

Imagine the hill is 300 mile up.

On the first day he runs at 30 mph, for 6 hours, and walks at 20 mph for the last 6 hours.

On the return journey he runs at 30 mph for 4 hours, then walks at 20 mph for 6 hours, then runs the last 60 miles at 30 mph.

Is he at the same point, at the same time, at any point/time?

The answer to the question is maybe.

Edit;I've only checked at the hourly point, so maybe there is a correlation at smaller time increments

pickarooney

Emiko wrote: »

Of course the speed matters.

Imagine the hill is 300 mile up.

On the first day he runs at 30 mph, for 6 hours, and walks at 20 mph for the last 6 hours.

On the return journey he runs at 30 mph for 4 hours, then walks at 20 mph for 6 hours, then runs the last 60 miles at 30 mph.

Is he at the same point, at the same time, at any point/time?

The answer to the question is maybe.

Edit;I've only checked at the hourly point, so maybe there is a correlation at smaller time increments

There's a good, solid example. IF you actually map that out on paper you'll find the exact point where the guy is at the same time on successive days.

04KY

Emiko wrote: »

Of course the speed matters.

Imagine the hill is 300 mile up.

On the first day he runs at 30 mph, for 6 hours, and walks at 20 mph for the last 6 hours.

On the return journey he runs at 30 mph for 4 hours, then walks at 20 mph for 6 hours, then runs the last 60 miles at 30 mph.

Is he at the same point, at the same time, at any point/time?

The answer to the question is maybe.

Edit;I've only checked at the hourly point, so maybe there is a correlation at smaller time increments

I think the graph in the link explains it. The distance to be covered each day is the same as is the time of leaving and the time of arriving. The speed can vary, as shown by lines on the graph.

The intersection of the lines in the graph shows the point where the guy will be at the same place at the same time on both days.

http://postimage.org/image/1r1fm0res/

Oh an apologies for the crappy Y axis!

Emiko

04KY wrote: »

I think the graph in the link explains it. The distance to be covered each day is the same as is the time of leaving and the time of arriving. The speed can vary, as shown by lines on the graph.

The intersection of the lines in the graph shows the point where the guy will be at the same place at the same time on both days.

http://postimage.org/image/1r1fm0res/

Oh an apologies for the crappy Y axis!

I thought the premise was he left at 8am on one day, and 8 am on the next day?

Edit. makes no odds as to your graph, i suppose.

I'm still not convinced though.

It BeeMee

pickarooney wrote: »

If you don't accept the two brothers example as an easy way to illustrate the answer, make it harder - Draw (or visualise) a red line going up the hill, marking each second spent climbing, then do the same with a green line coming down. They'll meet somewhere.

But the red and green lines don't meet?

04KY

Emiko wrote: »

I thought the premise was he left at 8am on one day, and 8 am on the next day?

He does, sorry, not the best graph in the world!

Take it as the blue line represents the guy walking up the hill. So he starts at 08:00 and finishes at 20:00, the blue line is going left to right.

The red line represents the guy coming back. He starts at 08:00 at the top of the hill (which I've taken as distance of 15) and the graph goes from right to left.

I've taken the bottom of the hill to be 0 on the X axis and the top of the hill as 15.

Emiko

04KY wrote: »

He does, sorry, not the best graph in the world!

Take it as the blue line represents the guy walking up the hill. So he starts at 08:00 and finishes at 20:00, the blue line is going left to right.

The red line represents the guy coming back. He starts at 08:00 at the top of the hill (which I've taken as distance of 15) and the graph goes from right to left.

I've taken the bottom of the hill to be 0 on the X axis and the top of the hill as 15.

So at some point their paths meet. That's a given.

But is it at the same hour of the day?

This would make a great After Hours field trip.

MungBean

Twin example seems to hold true. I was adamant that it was wrong and that the speed mattered but now I see the light. You have to cross paths at some point and given that you leave at the same time whatever point you cross on HAS to be the exact time you were there the day before, if not you wouldn’t be crossing paths at all.

04KY

Emiko wrote: »

So at some point their paths meet. That's a given.

But is it at the same hour of the day?

This would make a great After Hours field trip.

Yea it will be at the same time. If you drew a horizontal line from the intersection point to the Y axis it will tell you the time that he would be at the same point.

pickarooney

Emiko wrote: »

So at some point their paths meet. That's a given.

But is it at the same hour of the day?

This would make a great After Hours field trip.

That's really what meeting is - if you show up an hour after someone is gone, you don't meet them. You're not aiming for a fixed point at a fixed time - as both are flexible they have to meet somewhere. In your example it's at a little before 13.30 about 162 miles up from the bottom.

looksee

Emiko wrote: »

I thought the premise was he left at 8am on one day, and 8 am on the next day?

Edit. makes no odds as to your graph, i suppose.

I'm still not convinced though.

NO! it does not say that he left at 8AM it says 8o'clock.

enda1

Jesus people are reading WAY too much into this. A riddle like this has to have a definite answer so the crossing paths nonsense doesn't hold. Its clearly just a word trick made to dupe you into assuming a.m. or p.m.

Looksee is right and answered this ages ago. Read the OP and then Looksee's first answer and you'll see the light.

(Its a crap riddle by the way).

28064212

enda1 wrote: »

Jesus people are reading WAY too much into this. A riddle like this has to have a definite answer so the crossing paths nonsense doesn't hold.

Actually, it does. The question is "Is there a place where you were at the the same place at the same time on both days?". The crossing paths explanation most definitely answers this with a yes.

If the question asked where you were at the same place at the same time, you'd be right, since the only way to know that would be if it was at the top

pickarooney

enda1 wrote: »

Jesus people are reading WAY too much into this. A riddle like this has to have a definite answer so the crossing paths nonsense doesn't hold. Its clearly just a word trick made to dupe you into assuming a.m. or p.m.

Looksee is right and answered this ages ago. Read the OP and then Looksee's first answer and you'll see the light.

(Its a crap riddle by the way).

Nonsense - that's just a weak attempt to deflect from the (rather interesting) question.

Kevin Duffy

04KY wrote: »

I think the graph in the link explains it. The distance to be covered each day is the same as is the time of leaving and the time of arriving. The speed can vary, as shown by lines on the graph.

The intersection of the lines in the graph shows the point where the guy will be at the same place at the same time on both days.

http://postimage.org/image/1r1fm0res/

Oh an apologies for the crappy Y axis!

No it doesn't. The graph shows that two people on the same path at the same time would meet at some point. That neither needed a graph nor answers the question. The person in the question is not on the same path as another person, there is only one person. It matters not that two people would meet, it matters that you know if there's a particular point you be at at the same time on both days. To answer the question requires knowledge of the schedule/pace up and down, which is not given.

InTheTrees

This is like the aeroplane on the conveyor belt isnt it?

I. Refuse. To. Get. Drawn. In.

:mad:

28064212

Kevin Duffy wrote: »

No it doesn't. The graph shows that two people on the same path at the same time would meet at some point. That neither needed a graph nor answers the question. The person in the question is not on the same path as another person, there is only one person. It matters not that two people would meet, it matters that you know if there's a particular point you be at at the same time on both days. To answer the question requires knowledge of the schedule/pace up and down, which is not given.

I thought the same thing at first, but it really is true. Take this scenario:

• On the 1st day, starting at 8am, you walk up the hill, taking 12 hours at varying speeds. Your good friend Joe tracks the exact speeds and times you were walking
• On the 2nd day, starting at 8am, you walk down the hill, taking 12 hours at (different) varying speeds
• Also on the second day, Joe starts at the bottom of the hill and exactly replicates what you did on the first day

At some point during the day, you must meet Joe going up the hill. That's when you will be at the same place at the same time as yesterday

pickarooney

Kevin Duffy wrote: »

No it doesn't. The graph shows that two people on the same path at the same time would meet at some point. That neither needed a graph nor answers the question. The person in the question is not on the same path as another person, there is only one person. It matters not that two people would meet, it matters that you know if there's a particular point you be at at the same time on both days. To answer the question requires knowledge of the schedule/pace up and down, which is not given.

If you're intent on not trying to understand it this way, at least try to prove it wrong. Come up with just one single schedule for days one and two where the guy is not in the same place at the same time and we'll all admit you're right and we're wrong.

04KY

Kevin Duffy wrote: »

No it doesn't. The graph shows that two people on the same path at the same time would meet at some point. That neither needed a graph nor answers the question. The person in the question is not on the same path as another person, there is only one person. It matters not that two people would meet, it matters that you know if there's a particular point you be at at the same time on both days. To answer the question requires knowledge of the schedule/pace up and down, which is not given.

Ok the graph isn't anyway self explanitory! So I'll try write what I mean it to show!

I have assumed that the distance from the bottom of the hill to the top of the hill is 15 in my graph. distance 0 is the bottom, distance 15 is the top.

The blue line on the graph represents the speed at which the person walks up the hill, as speed is distance over time. The person starts their walk at 08:00 at,on my graph, distance 0, or the foot of the hill (graph position 0,08:00). He arrives at the top of the hill at 20:00.

The next morning at 08:00 the person leaves from the top of the hill, which is the right end of the red line, (graph position (15, 08:00)), walking towards the bottom of the hill, distance 0. The person reaches the bottom of the hill at 20:00, the left tip of the red line, or (0,20:00). The red line represents the speed at which the person goes down the hill. The fact that the line is curved shows that the person decended at a varying speed.

Both the start and end points of both lines are fixed. But the lines can curve at any other points to represent various speeds at which the person is going. The two lines will always intersect no matter what speed the person goes. The value on the Y axis at the intersection point is the time that the person will be at the same point on both days.

Kevin Duffy

pickarooney wrote: »

If you're intent on not trying to understand it this way, at least try to prove it wrong. Come up with just one single schedule for days one and two where the guy is not in the same place at the same time and we'll all admit you're right and we're wrong.

I'd already given my answer to the problem long ago, but thanks anyway.

28064212

Kevin Duffy wrote: »

I'd already given my answer to the problem long ago, but thanks anyway.

Your schedule? That was answered by pickarooney in the very next post.

I'm reposting this. Point out the bit you aren't following:

28064212 wrote: »

Take this scenario:

• On the 1st day, starting at 8am, you walk up the hill, taking 12 hours at varying speeds. Your good friend Joe tracks the exact speeds and times you were walking
• On the 2nd day, starting at 8am, you walk down the hill, taking 12 hours at (different) varying speeds
• Also on the second day, Joe starts at the bottom of the hill and exactly replicates what you did on the first day

At some point during the day, you must meet Joe going up the hill. That's when you will be at the same place at the same time as yesterday

Orando Broom

im invisible wrote: »

but, but BUT...

or imagine the hill as a 1 lap race in mariocart time trial whatever, do one lap, save your 'ghost' start another lap, and see your ghost start off up the hill, but do a 180 at the start, (you have put yourself at the 'top' of the hill) and go the wrong way around the track, , at some point you will meet your ghost at whatever time, in the same place of the track, the same amount of time since setting out

If he met his ghost on the hill at what point does it prove that ouija boards work?

J K

Is the ouija board on the plane or on the conveyor belt

sheesh

housetypeb wrote: »

One day,starting at Eight o clock,you go up a hill ,sometimes running and sometimes walking and it takes 12 hours, you spend the night on the hill top.
Next day,starting at Eight o clock, you start back down, sometimes running ,some times walking and it takes 12 hour again
Is there a place where you were at the the same place at the same time on both days?

I know a place where you could possible do such a thing

sliabh mish mountains in kerry there is a minor road up the mountain take a right and hike for 6- 8 hours to bartregaum
you could also do a ring of mountains that would include carrantouhil that takes about 12 hours to do

sorry thought it was a real question

Ficheall

Brouwer's fixed point theorem

im invisible

Ficheall wrote: »

Brouwer's fixed point theorem

balls to that

pickarooney

Ficheall wrote: »

Brouwer's fixed point theorem

Or the closely related Hairy balls theorem.

Edit: I didn't see the post above for some reason

General General

Thread is truly wonderful.