| They are crawling-type flies only, so they move only on the
surface of the cube, not on its interior.
Perhaps I should have made them spiders, or roaches (they don't
have wings do they?) or tiny rhinoceroses with suckers on the ends of
their feet.
Of course, if they *did* fly around the interior, then the
problem would be simpler, since the three initial positions determine a
plane that no fly can ever leave.
Andrew
|
| Ok, I'll have a crack at it.
Each of our hymenopterous friends heads off in the direction of a
geodesic in the surface of the cube passing through its own location
and that of its target. Initially, this geodesic is simply the diagonal
of the face common to the two flies. Each one heads off down a
diagonal. After that first instant, each has moved onto one of the
three faces all of whose coordinates are positive. The geodesic now
runs across two faces of the cube.
For a given fly, we can find the geodesic simply by unfolding the cube
to that the two faces involved lie adjacent to each other in a plane.
The geodesic is a straight line between them. In fact, we can unfold
another face as well, so that we have a picture like this:
__________________________
| ,.|'Z |
| ,.' | |
| ,.' | |
| ,.' | |
|Y | |
__________________________
| \ |
| \ |
| \ |
| \ |
| X |
______________
Of course, this does not show us the way Z moves after X. However, this
is easily remedied: We imagine a fourth, virtual fly, Vera, and set Z
chasing V while V chases X. Z, and indeed V, then move in an identical
way to X and Y.
__________________________
| ,.|'Z |
| ,.' | \ |
| ,.' | \ |
| ,.' | \ |
|Y | \ |
__________________________
| \ | V |
| \ | ,.' |
| \ | ,.' |
| \ |,.' |
| X'| |
__________________________
We therefore have four flies at the corners of a square on the plane, each
moving towards the current instantaneous location of the next corner. By
symmetry, the flies will always remain at the corners of a square centered
on the origin. Further, each moves such that its current direction is at
pi/4 to its radius vector from the origin. [The resultant motion is, of
course, a logarithmic spiral r = exp(theta). This quickly gives all the
answers. We are not allowed to know this, however.]
(a)
Observing that each fly has a radial component of velocity of v*sin(pi/4)
towards the origin, we deduce that the flies meet at the origin, ie the
(0,0,0) corner of the cube.
(b)
Initial radial distance from the origin is _/2 (diagonal of a face). Speed
towards origin is 1/(_/2). Time taken is 2 units.
(c)
We see that the actual path of the flies is independent of the speed at
which they travel. Also, the picture at any given time can be obtained from
the starting position by a rotation through a given angle, and a uniform
scale reduction. Indeed, since the path is determined purely geometrically,
we can deduce that each successive rotation through an angle alpha, say,
will result in a scale reduction by a factor of m, say.
Certainly the flies move though some angle greater than zero because their
tangential velocity is v*cos(pi/4) > 0. After moving through the initial
alpha, we have the same picture reduced by a scale factor m. So, they
certainly move through angle alpha again, to give a picture reduced by
factor m�. Repeating this, we see that they continue to spin round the
origin an infinite number of times before finally meeting.
[Did you notice some slight waving of hands in that last paragraph? Well, I
couldn't immediately see how to tighten it up without using nasty words
like "converge" or "unbounded". Does anyone have a more elegant solution?]
Andy.
|
| > We therefore have four flies at the corners of a square on the plane, each
> moving towards the current instantaneous location of the next corner. By
> symmetry, the flies will always remain at the corners of a square centered
> on the origin. Further, each moves such that its current direction is at
> pi/4 to its radius vector from the origin. [The resultant motion is, of
> course, a logarithmic spiral r = exp(theta). This quickly gives all the
> answers. We are not allowed to know this, however.]
> ... we see that they continue to spin round the
> origin an infinite number of times before finally meeting.
> [Did you notice some slight waving of hands in that last paragraph? Well, I
> couldn't immediately see how to tighten it up without using nasty words
> like "converge" or "unbounded". Does anyone have a more elegant solution?]
Actually if you work the problem backwards starting with theta = 0 you will
see that you get out to radius r for a fairly small theta, so (in the
original problem) you don't spin around very far before the whole thing
converges to the center. Both the total distance and the angle converge
very rapidly.
|
| Use the fact that the angle between the trajectory and the radius vector is
always pi/4. Let the fly move a small amount, so that the radius vector sweeps
out a small angle da. The path of the fly forms the hypotenuse of an
isosceles right triangle. Its two sides are dr and r * da. Since they are
equal
dr = r * da
This says that as r becomes smaller, a larger change in angle is necessary to
effect the same small change in r. As r decreases without limit, the change
in a increases without limit. Thus the flies must cross the edges an infinite
number of times.
I'm not sure that this is really 'without calculus', but at least it looks
like it would be acceptable to one who did not know calculus.
The equation above works for positive and negative change, so I think it is a
counter example to .6. If you run the film backwards from zero, I think that
some small dr will correspond to a very large change in angle.
|
| Using Andy's fine diagram:
__________________________
| ,.|'Z |
| ,.' | \ |
| ,.' | \ |
| ,.' | \ |
|Y | \ |
__________________________
| \ | V |
| \ | ,.' |
| \ | ,.' |
| \ |,.' |
| X'| |
__________________________
The first thing to say is that the radial component of the walk is
-1/sqrt(2). So after time sqrt(2), all the flies will be together in the
middle, this answers (a) & (b) without tears.
Let's put ourselves in X's shoes. He starts off with a positive
Westward component in his Yvetteward walk. He will continue to have a
+ve W. cmpt, for a finite time, for *at* *least* time � (and in fact more than
that.) The point is that he has moved some finite distance westwards. At
some first point, the Westward component of his velocity must become zero.
We know this since we know he has to get back east to the origin.
At the moment at which his Westward cmpt is zero, he must be heading
due North towards Yvette. But Yvette has travelled as far East as Xavier
has North, by symmetry. So at that moment each fly finds itself on a main
diagonal of the large square, each fly at the same distance from the origin.
Now, we know that the flies cannot have reached the origin yet,
because this was the very first time that Xavier's Westward cmpt is zero.
That's all we need. Say that each fly is situated at distance � from the
origin. By redrawing, etc, when each fly has moved another pi/4 in angle,
the distance from the origin will have shrunk to ߲ (�<1, since we know that
the flies do reach the origin.)
So after any finite angle traversed by the fly, say n*pi/4, he will
still be �^n from the origin. He must therefore traverse an infinite angle
(albeit in a finite time), in order to reach the origin.
Regrds,
Andrew.
|