Read new chestnut: a fly and two bicycles, if you (like the anonymous softlinker whose handiwork you may see below) haven't a clue what this node is all about...

gorgonzola (writing below, but actually before; don't ask, it's a long story) gives the right answer, but works fairly hard for it. So let's start with the answer: the fly could be anywhere, or, in other words, the problem is ill posed and has no answer. The true math geeks among you will see that the proposal to take δ->0 doesn't work (the starting point doesn't converge as δ->0, unfortunately).

The time-reversed case is just what's described in old chestnut: two bicycles and a fly, for varying initial positions of the fly.

So how can we see this immediately? Well, as gorgonzola does, we should reverse time. But we can cut through the icky mathematics and do it in one fell swoop. Reversing time, we know the bicycles start 10km from the centre point. So let's pick any starting (ending) position between them for the fly, and let time flow backwards. After 1 hour, the bicycles crash together (start out), and the fly has to be between them. So the fly too is at the centre position. But this is true no matter which starting position we chose for the fly!

We see that in the reversed case, information (the fly's starting/ending position) gets destroyed. Since information cannot get created, there is no solution, as gorgonzola correctly states.


Also, note that gorgonzola is truly in illustrious company here!
In the end, we have to throw up our hands and say there isn't an "answer" answer.   Some will be tempted to say "but wait, this could really happen! There has to be a final outcome!"

Sorry, that's wrong.   We have idealized our experiment to a great extent:

  • The bicycles and the fly all start at the same point.
  • The bicycles travel at a constant speed and never waver.
  • The fly travels at a constant speed, and never wavers. Not only that, there is no turnaround time between trips.
If we're going to accept all of these idealized conditions, we have to accept an idealized set of equations to solve it, with a singularity just where we were expecting an answer. So:

Mu.

And don't ask me if the fly has a buddha-nature or I will knock you into the river.

Of course, anything worth doing is worth expending far too much effort on, and making yourself a fool over.   Here's the rest of my original writeup, just so you know what ariels is on about.

First of all, let's dispense with all of these silly "kilometer" thingies.   I will measure everything in a distance unit of my own invention, the "mu", equivalent to 10 km.    This simplifies our thought experiment a bit:
  • The track extends from -1 mu to +1 mu.
  • The bicycles travel at 1 mu/hr.
  • The fly travels at 2 mu/hr.
At the beginning of the experiment, the track looks like this:

  +--------oo*oo--------+
 -1          0          1

So let's try ariels' suggestion and pick a small time t that the fly waits before starting its flying. In our system, time is equivalent to position, so we can also say that the fly starts at a positive position d=2t at time 0.  Now let's assume that the fly passes the bicycle traveling to the right and flies all the way to the bicycle traveling to the left (well, we can say the fly alighted on the right-traveling bicycle and began its next trip, so we aren't really assuming anything).

              *
  <--------oo+oo------->
           -d d

The fly starts out d from the left-traveling bicycle, and catches it at -d after an elapsed time of d, traveling 2d.

           *
  <-------oo-+-oo------->
           -d d

For the second trip, the fly is 2d from the right-traveling bicycle, and catches it at +3d after an other 2d, traveling 4d.
For the third trip, the fly is 6d from the left-traveling bicycle, and catches it at -9d after an elapsed time of 6d, traveling 12d.
For the fourth trip, the fly is 18d from the right-traveling bicycle, and catches it at +27d after an elapsed time of 18d, traveling 36d.

For a particular d, we add up the trips 2d + 4d + 12d + ... + 4*3t-2 and find where the next trip would exceed 2 mu, the remainder giving us our final position.

We have a problem!   If you try different values for d, you will see that for a d=3-2n the fly will end up at one bicycle, for d=3-2n+1 the fly will end up at the other bicycle! Not only that for d=3-2n/2, the fly will end up in the middle!   As d gets closer and closer to 0, the ending point travels continuously back and forth across the entire track from -1 to +1 and back again.   A particularly strange attractor which fills our entire problem space!



Ok, so let's work the problem backwards:  We need to find the final position, where if you backtrack to the beginning, the fly will have traveled 2 mu.  We pick a final position and measure the distance d to the bicycle the fly left on the last trip. We can see right off that no matter where the fly ends up, the last trip will be 2/3*d mu, and will have taken 1/3*d hr.

So if the fly ends up at one of the bicycles, all the way at -1,

oo*----------+----------oo
 -1          0          1

the previous trip will have begun 1/3 of the way into the trip, at +1/3.  The fly travels 4/3 during this trip, and is 2/3 from the other bicycle:

  +-----oo---+---*oo----+
 -1          0          1

The trip before that will have begun 1/9 of the way into the trip, at -1/9.  The fly travels 4/9 during this trip, and is 2/9 from the other bicycle.

  +-------oo*+-oo-------+
 -1          0          1

So each trip will begin at -3-t (t being the number of trips until the end), will take 2/3t hr, and causes the fly to travel 4/3t mu.

Thus:

4/3 + 4/9 + 4/27 + 4/81 + .... = 2.

So the fly could end up at -1.  But there's a problem!  Assume the fly ends up in the middle:

oo+----------*----------+oo
 -1          0          1

The previous trip will have begun 2/3 of the way into the trip, at +2/3.  The fly travels 2/3 during this trip, and is 4/3 from the other bicycle:

  +--oo------+------*oo-+
 -1          0          1

The trip before that will have begun 2/9 of the way into the trip, at -2/9.  The fly travels 8/9 during this trip, and is 4/9 from the other bicycle.

  +------oo*-+--oo------+
 -1          0          1

So, the fly travels 2/3 + 8/9 + 8/27 + 8/81 + 8/243 + ... = 2.  Oh crap.  In fact, no matter where the fly ends up, it has traveled exactly 2 mu since the beginning.  Well, we probably could have predicted that from our previous results.

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