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 Post subject: Cars around a track.
PostPosted: Tue, 1 Nov 2011 09:54:38 UTC 
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Hi, here's a math problem that some of you may find interesting :).

Problem:
There are n identical cars on a circular track. Together they have just enough gas for one car to complete a lap. Show that there is a car that can complete a lap by collecting gas from the other cars on its way round.

Solution:
Spoiler:
An additional car with a sufficiently large tank starts somewhere on the circle, and is initially half full. At each car, it buys up all its gas. At some point A, the level of gas in his tank must be the lowest. The car at this point can complete a lap.


My explanation:
Spoiler:
It can be assumed without loss of generality that the cars travel clockwise around the track. A occurs at a car, for if it did not, the car could use up more gas by travelling further along the track, contradicting the assumption of the lowest level of fuel. Assuming the additional car starts at the point B, the arc AB is traversable by car A. Lets call the initial level of fuel in the additional car when it's at point A as F. If arc AB is not traversable by car A, it would mean that at some point along AB, call this point C, the car A would reach 0 fuel but still has some distance to cover before reaching the next car. Similarly, at point C, the additional car would have a fuel level of F, but still have distance to cover before reaching the next car. This again leads to a contradiction of the assumption of the lowest fuel point being F. Say the initial amount of fuel in the additional car was E, and we call the deficit of fuel from E to F as D, where D = E - F. We know that the additional car returns to point B with the same amount of fuel as it started, E. As such, along the arc AB, it must collect excess fuel of magnitude equivalent to D. Applying this to the car A, after traversing arc AB, it has excess fuel of magnitude equal to D. If it is unable to complete arc BA, it must mean that at some point along BA, there is a point where the fuel needed to travel from B to that point is of larger magnitude than D. Applying this back to the additional car which starts at point B, this would contradict the initial assumption of the lowest point being A.


I would like it very much if someone could give a better explanation, thanks!



Arthur , Engel. Problem Solving Strategies. Germany: Springer, 1998. eBook. <http://www.scribd.com/doc/26521262/Arthur-Engel-Problem-Solving-Strategies>.


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 Post subject: Re: Cars around a track.
PostPosted: Tue, 1 Nov 2011 11:58:56 UTC 
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Hmmm...this "seems" ok:
say there's 4 cars; place 'em at noon, 3, 6 and 9 (clock face);
each car can go 1/4 of track; car at noon leaves; others don't move!
Moving car takes all gas from each car...

Similarly, any number can be handled.
Too easy...what did I miss?

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 Post subject: Re: Cars around a track.
PostPosted: Tue, 1 Nov 2011 12:30:22 UTC 
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Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
lentan wrote:
Hi, here's a math problem that some of you may find interesting :).

Problem:
There are n identical cars on a circular track. Together they have just enough gas for one car to complete a lap. Show that there is a car that can complete a lap by collecting gas from the other cars on its way round.

Solution:
Spoiler:
An additional car with a sufficiently large tank starts somewhere on the circle, and is initially half full. At each car, it buys up all its gas. At some point A, the level of gas in his tank must be the lowest. The car at this point can complete a lap.


My explanation:
Spoiler:
It can be assumed without loss of generality that the cars travel clockwise around the track. A occurs at a car, for if it did not, the car could use up more gas by travelling further along the track, contradicting the assumption of the lowest level of fuel. Assuming the additional car starts at the point B, the arc AB is traversable by car A. Lets call the initial level of fuel in the additional car when it's at point A as F. If arc AB is not traversable by car A, it would mean that at some point along AB, call this point C, the car A would reach 0 fuel but still has some distance to cover before reaching the next car. Similarly, at point C, the additional car would have a fuel level of F, but still have distance to cover before reaching the next car. This again leads to a contradiction of the assumption of the lowest fuel point being F. Say the initial amount of fuel in the additional car was E, and we call the deficit of fuel from E to F as D, where D = E - F. We know that the additional car returns to point B with the same amount of fuel as it started, E. As such, along the arc AB, it must collect excess fuel of magnitude equivalent to D. Applying this to the car A, after traversing arc AB, it has excess fuel of magnitude equal to D. If it is unable to complete arc BA, it must mean that at some point along BA, there is a point where the fuel needed to travel from B to that point is of larger magnitude than D. Applying this back to the additional car which starts at point B, this would contradict the initial assumption of the lowest point being A.


I would like it very much if someone could give a better explanation, thanks!



Arthur , Engel. Problem Solving Strategies. Germany: Springer, 1998. eBook. <http://www.scribd.com/doc/26521262/Arthur-Engel-Problem-Solving-Strategies>.


The "solution" fails when you lump all n cars together (say at 12 o'clock position), the extra car starts at 1 o'clock position, say, and forced to go clockwise --- this car will run out of fuel at 7 o'clock position, before it can have any chance of refuelling.

Denis wrote:
Hmmm...this "seems" ok:
say there's 4 cars; place 'em at noon, 3, 6 and 9 (clock face);
each car can go 1/4 of track; car at noon leaves; others don't move!
Moving car takes all gas from each car...

Similarly, any number can be handled.
Too easy...what did I miss?


No, although the cars are identical, how much fuel they each hold (can be different) is not the same, and they are fixed (you don't have any freedom in putting them where you like, except there is a rotational symmetry).

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: Cars around a track.
PostPosted: Tue, 1 Nov 2011 16:37:18 UTC 
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Joined: Fri, 28 Oct 2011 11:24:34 UTC
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Oh the cars do not need to be evenly spaced. The additional car is simply used to solve the question, it has no relation to the other n cars. The solution is likely to be correct, as this was found in a mathematics Olympiad training book. However my explanation of the solution seems very lengthy, I feel that I may have misunderstood the solution in some way. Perhaps it is not necessary to divide the track into arcs AB and BA?

The solution's claim that the car at point A is able to complete a lap does not seem obvious to me.. Can anyone explain why it would be obvious?


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