# Help prove or disprove this math problem



## Dr.G. (Aug 4, 2001)

Someone at dinner this evening discussed a math problem that she contends could not be solved. I shall give you what she gave to us and ask if anyone can either support her contention that this problem has no solution, or come up with a solution that confirms the two answers that were suggested this evening. Here is all she gave us to work with :

There are two trains 120 miles apart. These two trains are on the same track and start at the exact same moment heading for each other at 60 miles an hour. There is a bee at the front of one of these trains who is able to fly at 120 miles per hour. As soon as the train starts to move, the bee flies in the direction of the other train. The bee will make as many trips as possible, back and forth between trains, in the time it takes for these two trains to travel down the track until they crash in one hour. How many trips will this bee make in this amount of time?

Her contention is that there is no answer and no way to answer this math problem. Any help to prove or disprove this contention would be greatly appreciated.


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## Beej (Sep 10, 2005)

What's the question?


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## AlephNull (Jan 28, 2005)

Hmm.. sounds like something that could maybe be solved through iterations.. We just finished relative motion in physics 12, perhaps I'll bring it up.


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## Dr.G. (Aug 4, 2001)

Thanks, Beej, for my oversight.

How many trips will this bee make in this amount of time?


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## Dr.G. (Aug 4, 2001)

Thanks, AN. Any help would be appreciated.


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## Beej (Sep 10, 2005)

Sounds like an infinite series solvable by google...

http://mathforum.org/library/drmath/view/56735.html


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## Dr.G. (Aug 4, 2001)

Beej, the difference between these two equations is that the speed of the trains are equal in what I was told, and different in that equation. However, the answer is similar to the one that I provided. Merci.


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## CN (Sep 3, 2004)

I agree that there is no solution. I was trying to solve it algebraically (using relative motion methods), but then I realized that even if we say the bee is a point particle, then the instant before collision, the bee is travelling between the two trains an infinite number of times (since it is touching both). Thus, it travels between the trains and infinite number of times when it touches both during this instant. 

Even if we consider the instant where it is touching both trains to not count as a trip (and thus that the infinite response is not valid), then not knowing the length of the bee, it would be impossible to determine how many trips it completes in the seconds before collision (before it is technically touching both trains and thus no longer completing any trips).


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## Beej (Sep 10, 2005)

Your welcome. Now about that bee...decelerate, stop, turnaround and accelerate in the opposite direction instantaneously. Science needs that bee!


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## CN (Sep 3, 2004)

Beej said:


> Your welcome. Now about that bee...decelerate, stop, turnaround and accelerate in the opposite direction instantaneously. Science needs that bee!


Yes, to even begin to consider this question, it must be assumed that these actions are all performed simultaneously (without being given any information on coefficients of friction, rate of acceleration etc.). If you take this question that far, then obviously there is no way to figure out the answer.

Instantaneous acceleration, ay yay yay XX) :lmao:


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## Beej (Sep 10, 2005)

CN said:


> Yes, to even begin to consider this question, it must be assumed that these actions are all performed simultaneously (without being given any information on coefficients of friction, rate of acceleration etc.). If you take this question that far, then obviously there is no way to figure out the answer.
> 
> Instantaneous acceleration, ay yay yay XX) :lmao:


With the instantaneous bee, the answer is, I believe it to be, infinity. And I'm a poet.


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## CN (Sep 3, 2004)

Beej said:


> With the instantaneous bee, the answer is, I believe it to be, infinity. And I'm a poet.


Agreed, but only if the bee can be assumed to be a point particle (no length given) and touching both trains constitutes a "trip". 

This problem would be easily solveable if the length of the bee was given, and if when the bee is touching both trains it does not count as a trip.


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## Beej (Sep 10, 2005)

We have to find this bee; it could unlock the secret to cheap and plentiful flying cars.


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## guytoronto (Jun 25, 2005)

KISS - Keep it simple stupid!

The answer is NOT infinite, because after 1 hour, the two trains would collide smushing the bee.

The bee has some length. No bee is a simple pinpoint, or single molecular entity.

The world's smallest bee is 2mm long. We will use this bee for our calculations.

The bee travels at 180mph in relation to the train (120mph speed of bee + 60mph speed of train).

At the point that the trains are less than 2mm apart, the bee would be squashed, and that would be your answer. These calculations are based on the theory that the bee turns instantly, never tiring in any of his journeys. (ok, we know this is impossible, but the question forces us to use these variables)

[Edit: See attached image for spreadsheet]

[Edit: Read my table wrong]

The distance between the trains at the start of each leg must be greater than the size of the bee. Since leg 18 is less than 2mm, the bee can never start the 18th leg of his journey. The answer is 17 legs back and forth.

If you assume the bee is the longest bee in the world, at 1.5" or 39mm, then he would have been sandwiched on trip 15.

15 to 17 - depending on the bee.


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## Beej (Sep 10, 2005)

guytoronto said:


> KISS - Keep it simple stupid!
> ....
> The bee has some length. No bee is a simple pinpoint, or single molecular entity.
> These calculations are based on the theory that the bee turns instantly, never tiring in any of his journeys.
> ....


Interesting bee. It can't be a simple pinpoint but can turn instantly and never tire.  

The simple result is from the theoretical bee because the complexities of reality (bee dimensions, turning, acceleration, energy etc...) make it more complex.


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## used to be jwoodget (Aug 22, 2002)

Stupid (but very fast) bee. 18 chances to get out of the way..... It's an impossible question though because the chances of two Via rail trains travelling for an hour at 60 mph without breaking down is close to zero.


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## Daktari (Feb 21, 2005)

guytoronto, I'm curious whats the sum of the distance travelled by the bee accordng to your calculations?


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## CanadaRAM (Jul 24, 2005)

It's just Zeno's Arrow with a lot of window dressing

http://en.wikipedia.org/wiki/Zeno's_arrow

It's a semantic argument that takes as a given that time and distance is infintely divisible and that the terms of the question encompass all of the governing laws -- which they don't so the question itself is meaningless.

It's like saying "you are either a liar or a thief, which one are you?" and allowing only the two answers. 

See also Logical Fallacies 
http://www.nizkor.org/features/fallacies/


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## MannyP Design (Jun 8, 2000)

I don't think it's a matter of semantics... window dressing, yes--but to give the person doing the problem some sort of connection to the problem by using simple objects to connect (visualize) the problem with.

It's a matter of figuring out how many times "X" can travel back and forth between "Point A" and "Point B" as they move towards each other. It's something I could illustrate and animate in a 3-D program, but that would be cheating. Each object has a direction and speed.

How many times can "X" travel back and forth between "A" and "B" taking their speeds and positions into consideration? I think GuyToronto has it right, but got a little too specific with the size of the Bee.


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## CanadaRAM (Jul 24, 2005)

I think it is a paradox question, because the answer ("how many") depends on whether you infinitely divide the time x between crash - x and crash - x/2
And it's like Zeno's questions which boil down to 
"if you shoot an arrow, and it travels half the distance to its destination in one second, it travels half the remaining distance in the next half second, so an arrow can never reach the target because it still has half of the remaining distance to travel in the next time interval/2"

In other words, the framing of the question restricts the answer to an unreal set of parameters.


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## MacDoc (Nov 3, 2001)

Here make your head hurt.

http://www.articlesfactory.com/arti...heoretical-resolution-of-zenos-paradoxes.html


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## Beej (Sep 10, 2005)

used to be jwoodget said:


> Stupid (but very fast) bee. 18 chances to get out of the way..... It's an impossible question though because the chances of two Via rail trains travelling for an hour at 60 mph without breaking down is close to zero.


:lmao: :lmao:


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## HowEver (Jan 11, 2005)

.


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## guytoronto (Jun 25, 2005)

Daktari said:


> guytoronto, I'm curious whats the sum of the distance travelled by the bee accordng to your calculations?


The bee travels 120 miles (less the size of the bee itself). Since the bee can fly at 120 miles per hour, and since it takes one hour for the trains to collide, the bee has exactly one hour to fly, meaning he can fly 120 miles.

Note: I read my table wrong last night. The answer is actually 15-17 trips back and forth. See my previous post for corrections, and a nicer table.


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## MACSPECTRUM (Oct 31, 2002)

guytoronto said:


> The bee travels 120 miles (less the size of the bee itself). Since the bee can fly at 120 miles per hour, and since it takes one hour for the trains to collide, the bee has exactly one hour to fly, meaning he can fly 120 miles.
> 
> Note: I read my table wrong last night. The answer is actually 15-17 trips back and forth. See my previous post for corrections, and a nicer table.


but the trains are also moving in opposite directions to the bee making the bee's relative speed 120 + 60 = 180 mph as mentioned earlier

Being VIA trains they should be travelling at 100 km / hr instead of 60 mph since Canada is a metric country and doesn't use outdated imperial(istic) measurements

of course the bee only has 1 hour to travel and after that he is splattered all over the windshield of one of the trains


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## guytoronto (Jun 25, 2005)

MACSPECTRUM said:


> but the trains are also moving in opposite directions to the bee making the bee's relative speed 120 + 60 = 180 mph as mentioned earlier


The bee's relative speed has nothing to do with the total distance he travels. Only his actual speed does.


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## Daktari (Feb 21, 2005)

guytoronto's solution looks right; the total distance travelled by the bee according to his calculations matches with the total distance the bee should travel. The two trains will meet in an hour, therefore the bee will have travelled a total of 120 miles. 

Furthermore, the problem is exactly the same as the as the one on Beej's link, ( the sum of the speed of the two trains is the same as the speed of the bee). The only thing that the different speeds of the two trains in Dr G. question will affect is where on the tracks the two trains meet.


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## RicktheChemist (Jul 18, 2001)

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## MACSPECTRUM (Oct 31, 2002)

RicktheChemist said:


> This is a very interesting question; could you consider when the bee is exactly at the distance between the two trains that it is travelling an infinite amount of times between the trains?
> 
> Trying to remember my limit equations right now from Calculus 1
> 
> ...


yeah, me too
but 25 years is a very, very long time....


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## RicktheChemist (Jul 18, 2001)

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## guytoronto (Jun 25, 2005)

RicktheChemist said:


> could you consider when the bee is exactly at the distance between the two trains that it is travelling an infinite amount of times between the trains?


No. Simple logic dictates that since the trains do collide, hence squishing the bee, that the bee is NOT travelling an infinite number of times.

The point at which the bee is exactly at the distance between the two trains is when this equation ends, as the very next moment, the bee is being compressed, and can no longer turn around.

Again, depending on the size of the bee, the answer is 15 to 17 trips. Sometimes the simplest answer is the right one. Some people are looking too deep into this problem. It doesn't require calculus or quantum theory. A simple Excel spreadsheet will do.


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## RicktheChemist (Jul 18, 2001)

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## Macfury (Feb 3, 2006)

So the simple answer is that, given the way your friend described the question, it cannot be answered with a high degree of certainty. Full disclosure of parameters is required.


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## CN (Sep 3, 2004)

Macfury said:


> So the simple answer is that, given the way your friend described the question, it cannot be answered with a high degree of certainty. Full disclosure of parameters is required.


Exactly. The whole point was to decide whether or not the problem could be solved in its first form. It could not (not without knowing the size of the bee). GuyToronto's answer most definitely demonstrates one correct method in which to solve this sort of problem, but when it comes down to it, he still had to "make up" an extra piece of data (the length of the bee) to solve it, so its not a true solution to the original question. The original question was ambiguous; it simply does not contain enough information to answer the question it poses (although he could have created an equation-although some of the same info is contained in his spreadsheet-that would demonstrate the number of trips in terms of a variable representative of the bee's length). 

In a purely academic question such as this, it isn't valid to research the average bee's size and subsitute that data into the question because the question itself is hypothetical and clearly is not meant to mirror true real world conditions (eg- no wind resistance etc. taken into account).

I would be interested in seeing this worked out purely in terms of variables rather than subsituting numbers in. I think some variables would cancel out, but the variable of the bee's length would not be cancelled out in such an equation so it remains a vital piece of information to solve the original problem.


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## guytoronto (Jun 25, 2005)

There is a definitive answer. Just because it is not one exact number doesn't mean it isn't definitive.

We know that the bee is somewhere between 2mm and 39mm. These are the lengths of bees.

By excluding this information in the question, we can include the variable in the final solution.

Somebody at the party should have grilled the person for all the details. As is, it's like asking "How much water is in a glass? Ha! It's unsolvable! There is no solution!". Of course not! You didn't give us enough information to figure it out.


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## Beej (Sep 10, 2005)

I think the purpose of the mathematical problem has been missed. 

Enough information was given to solve with a simple equation for mathematics students or enthusiasts to devise. More details or, the engineering problem, can always be added to force a more specific and arithemetic solution. Example: the trains may have 'cow-catchers' so they would 'crash' while still leaving some time for the bee to travel more, if the bee was flying at a level above the tips of the cow-catchers. An important detail for the arithmetic, but not for the abstract mathematical problem. 

A key piece in the math is that the bee (the magical turn-around instantly, very fast, untiring, non-dimensional bee...the impossible bee) is travelling at least as fast as the combined speeds of the trains. Once you have that, I think you can play with the distances and speeds all you want and come to the same answer. The 'magical bee' is a simplifying assumption to focus on the mathematics because it's a math problem, not biology, or engineering.


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## MLeh (Dec 23, 2005)

Of course, the real question is "Why would the bee want to do that anyway?"


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## Beej (Sep 10, 2005)

MLeh said:


> Of course, the real question is "Why would the bee want to do that anyway?"


It can fly 120 and stop, turnaround, and accelerate to 120 in 0 time. I'm inclined to think there's not much energy left for thought. It's a bee with a mission!


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## Macfury (Feb 3, 2006)

I can, with a great degree of certainty, say that Guytoronto loses his pants an average of 4.23 times per day--that is, provided he says it only about unique pants-losing incidents. If it refers only to a single, traumatic experience, we can at least deduce that Guytoronto feels compelled to share his experience with us an average of 4.23 times per day.


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## CN (Sep 3, 2004)

Macfury said:


> I can, with a great degree of certainty, say that Guytoronto loses his pants an average of 4.23 times per day--that is, provided he says it only about unique pants-losing incidents. If it refers only to a single, traumatic experience, we can at least deduce that Guytoronto feels compelled to share his experience with us an average of 4.23 times per day.


A-hah! Thats his post count! Now if we perform a regression of the data, let it be shown that...XX)

:lmao: This thread is turning into the Shang for geeks...


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