Three into two puzzle

[ceptimus]

Three brothers decide to exhibit their two vehicles at a vintage car and motorcycle show twelve miles from their house. They jointly own a single-seat car and a very early single-seat motorcycle. They are discussing how to get themselves and their vehicles to the show.

"Our car goes at 60mph and our bike at 15mph", says Archie. "Whoever drives the car will only take twelve minutes to get there, the bike will take forty-eight minutes. One of us will have to walk - at four miles an hour that will take three hours."

"Why not share the driving?", says Brian. "If one of us drives the car for four miles, then parks it, and another of us does the same with the bike, then we can swap over, and park them again after another four miles. If we work it out right we can each walk, ride and drive for one third of the distance - that will be fair and we'll all arrive together and faster than one of us having to walk the whole way."

"I see what you mean", says Charlie, the youngest and smartest brother. "But there will be some waiting around involved because the bike won't be at the final swapping point early enough. I have a better idea..."

What is Charlie's idea and how quickly can all three brothers be together at the show, exhibiting their vehicles, if they all set off together and share the driving, riding and walking?

[davidpbrown]

If you hadn't suggested walking as a limit, then I'd have them tow the motorcycle at speed and one runs at 6mph - half marathon speed or slightly faster, if they're all fit and shared the run. Could have been there in an hour and a half but they're unfit and must walk so..

[davidpbrown]

Tow the bike.. 36 mins and three trips of 12 mins.. if the bike can freewheel. Tell me I'm wrong, I want it to be hard!

[ceptimus]

No towing allowed! And no, they can't sit on the back of either vehicle! Both vehicles strictly one person at a time!

:crackwhip:

[David B]

Brian is wrong, AFAICS - his method doesn't speed it up at all.

The thing to do is to minimise the walking time.

I reckon Archie has to take the car to the point where he has to walk the rest.

Brian has to take the bike, and leave it at a point where he can walk to the car and drive, to get there at the same time as Archie, and at the same time as Charlie walks to the bike and cycles the rest, if there us such a point.

I'd start trying to write equations if my dinner wasn't boiling away.

David

[davidpbrown]

The thing to do is to minimise the walking time.

I can't see how that works any faster. :drinking:

If at any one point one of them is walking, the full distance gets walked. At 4mph that's 3hrs and the use of the car and the bike is incidental.

Now if they get the bus instead of walking.. :)

[David B]

The thing to do is to minimise the walking time.

I can't see how that works any faster. :drinking:

If at any one point one of them is walking, the full distance gets walked. At 4mph that's 3hrs and the use of the car and the bike is incidental.

Now if they get the bus instead of walking.. :)

Then they leave the car and the bike behind

But it doesn't follow that if at any one point one of them is walking then the whole distance gets walked. Depends on how long it takes to get there

Now back to poker

David

[ceptimus]

Imagine for a moment that it were possible for each of the brothers to walk one third of the total distance (four miles), ride the motorcycle for four miles and drive the car the remaining four miles. Then each would take an hour for the walk, sixteen minutes for the ride and four minutes for the drive. A total of eighty minutes.

Of course, that may not be possible, but it's the idea that Brian vaguely had in mind.

[davidpbrown]

how quickly can all three brothers be together at the show, exhibiting their vehicles

I'm thinking the whole event, from first setting off to the last arriving takes 3hrs. That is, the full effect on the first of them to set off, as they're not all there until 3hrs. The overlap is 80mins but that's looking incidental.

Eitherway, I can't see how that can be reduced.


Disclaimer: I'm tired..

[David B]

I'm registered for another poker game, and am way down a bottle.

It can be done in less than 3 hours though.

Example - motorist drives all the way. Bikist cycles half way then walks, inititial walker picks up bike half way.

That would not be the optimum, though.

David

[davidpbrown]

That's still framing the overlap. The 3hrs total event, is noting that the initial walker would have started out earlier than the biker (bikist!?) to be there when the bike is dropped off. 6miles initial walker -> 1.5hrs and the second walker similarly doesn't then arrive until at least 1.5hrs later -> 3hrs.

[David B]

That's still framing the overlap. The 3hrs total event, is noting that the initial walker would have started out earlier than the biker (bikist!?) to be there when the bike is dropped off. 6miles initial walker -> 1.5hrs and the second walker similarly doesn't then arrive until at least 1.5hrs later -> 3hrs.

No. The question says they start at the same time.

In the case I state the time taken is 1 hour 54 minutes, if I'm still sober enough to do mental arith.

The motorist just goes there.

The initial walker walks for an hour and half, then cycles for 24 mins.

The initial cyclist cycles for 24 mins, then walks for an hour and a half.

Not optimum, though.

David

[David B]

I initially wondered if you could just add all the travelling times and divide by 3 to get to an answer of 80 mins.

But I thought that the initial conditions would have to be pretty carefully chosen to make that the case, and it would need a lot of stops, I think. And the somewhat unrealistic assumption that no time is lost in braking or accelerating.

Maybe whatever the initial speeds you just have to add up the times and divide by 3, but I'd have to work it through to know that.

Too late at night, too drunk.

David

[Pandora]

I can set it up... but my brain's not working right now for the necessary math.

Archie drives to some point X, then walks the remainder of the distance.
Brian bikes to point Y, then walks to point X and picks up the car and drives the remainder of the distance.
Charlie walks to point Y, picks up the bike and bikes the remaining distance.

We know the velocities (3 values), and we know the total distance to be covered. We've only got two unknowns: X and Y.

I think there might be some calculus involved, but my brains not fully functional right now, so I can't figure out how to set up the equations. But there's your starting point, anyway.

[David B]

I can set it up... but my brain's not working right now for the necessary math.

Archie drives to some point X, then walks the remainder of the distance.
Brian bikes to point Y, then walks to point X and picks up the car and drives the remainder of the distance.
Charlie walks to point Y, picks up the bike and bikes the remaining distance.

We know the velocities (3 values), and we know the total distance to be covered. We've only got two unknowns: X and Y.

I think there might be some calculus involved, but my brains not fully functional right now, so I can't figure out how to set up the equations. But there's your starting point, anyway.

Yeah, that's as far as I got. Including wondering if the sums would lead to the answer being 80 mins, but not confident enough to so assert.

David

[Pandora]

Okay. I come out with 80 minutes.

Archie drives 50/7 miles at 60 mph, then walks 34/7 miles at 4 mph.
Brian bikes 32/11 miles at 15 mph, walks till he reaches the 50/7 miles marker, then picks up the car and drives 34/7 miles at 4 mph.
Charlie walks 32/11 miles at 4 mph, picks up the bike and bikes the remainder at 15 mph.

They all leave at the same time, and they arrive at the same time, 80 minutes later.

[David B]

Okay. I come out with 80 minutes.

Archie drives 50/7 miles at 60 mph, then walks 34/7 miles at 4 mph.
Brian bikes 32/11 miles at 15 mph, walks till he reaches the 50/7 miles marker, then picks up the car and drives 34/7 miles at 4 mph.
Charlie walks 32/11 miles at 4 mph, picks up the bike and bikes the remainder at 15 mph.

They all leave at the same time, and they arrive at the same time, 80 minutes later.


OK I believe you, apart from what I take to be a glitch between ears at the end of the Brian bit.

Now can you prove that for all speeds of the 3 modes of transport that there will be a solution that will be the sum of the times needed to do the distance by all three modes of transport divided by 3?

David

[Rie]

Er, I'll give this one to my grandson and Sapphyre who is good at this sort of thinking. I used to be but left it all for music and Art.

[davidpbrown]

The question says they start at the same time.

Understanding the question is key.. I never expected them to leave a vintage vehicle unattended. Simpler times. :D

[ceptimus]

There is at least one 80-minute solution where the brothers leave home at the same time and arrive at the show at the same time. Each brother uses all three modes of transport and travels one-third of the distance using each mode.

[davidpbrown]

Or infinite variations of that.. six iterations x infinite fragmented lengths. Since it's not a real world.

[ceptimus]

Warning! Spoiler! Click the link to see one solution.

http://ceptimus.co.uk/jref/images/three_brothers.gif

[davidpbrown]

:dunno:

Am I missing something or does the trivial solution not just work?..

A takes Bike 16, Walks 76, Car 80
B takes Car 4, Bike 24, Walks 80
C takes Walk 60, Car 64, Bike 80

The bike and car are in the right place at the right time and again 80 mins each

[Pandora]

Okay. I come out with 80 minutes.

Archie drives 50/7 miles at 60 mph, then walks 34/7 miles at 4 mph.
Brian bikes 32/11 miles at 15 mph, walks till he reaches the 50/7 miles marker, then picks up the car and drives 34/7 miles at 4 mph.
Charlie walks 32/11 miles at 4 mph, picks up the bike and bikes the remainder at 15 mph.

They all leave at the same time, and they arrive at the same time, 80 minutes later.


OK I believe you, apart from what I take to be a glitch between ears at the end of the Brian bit.

Now can you prove that for all speeds of the 3 modes of transport that there will be a solution that will be the sum of the times needed to do the distance by all three modes of transport divided by 3?

David

he he he... probably not. I strongly suspect it would require me to do linear algebra,,, and I've managed to forget everything about that topic except that it fits into pretty squares :p

[davidpbrown]

If the trivial solution above is valid then I think it obviously follows that 'for all speeds of the 3 modes of transport that there will be a solution that will be the sum of the times needed to do the distance by all three modes of transport divided by 3?' so long as the order of usage iterates the same for each.

[Pandora]

Hmm. I think I made the algebra more difficult by having Archie and Charlie only use one of the two of the vehicles. I think it's actually easier math if they all three use each vehicle and walk some... or at least prettier ending numbers :p

[Pandora]

Interestingly... you can also get a nice easy solution is Archie bikes the whole way, and Brian and Charlie swap the car halfway. Still works out to 80 minutes total.

[ceptimus]

Interestingly... you can also get a nice easy solution is Archie bikes the whole way, and Brian and Charlie swap the car halfway. Still works out to 80 minutes total.

No. Walking 6 miles at 4mph takes 90 minutes.

[Rie]

Got to run this by my granson or either of my daughters. I've resigned myself that my grey cells are damaged in this area. And, not to offend ... I gloried in solving algebraic stuff once... my mind is now exploring its music section.

I'm going to employ Sapphy as my technological advisor and my grandson for maths.

[rcscwc]

I am an engineer. Why trouble show case such lousy contraption?

[ceptimus]

:dunno:

Am I missing something or does the trivial solution not just work?..

A takes Bike 16, Walks 76, Car 80
B takes Car 4, Bike 24, Walks 80
C takes Walk 60, Car 64, Bike 80

The bike and car are in the right place at the right time and again 80 mins each

When B gets out of the car, the bike won't have arrived. So he'll have to wait for 12 minutes to continue his journey. Consequently, even if there are no other delays, he will take 92 minutes total.