## Re: Bug Problem

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• Here s a rigorous solution: Position the unit square in the plane so that it s center is at the origin and it s verticies lie on the axes. We can model each
Message 1 of 14 , May 1, 2002
Here's a rigorous solution:

Position the unit square in the plane so that it's center is at the
origin and it's verticies lie on the axes. We can model each bug's
path parametrically using polar coordinates. This provides us with

First, by a simple rotation of the plane we see that if p(r,\$) is a
solution [with r the radius and \$ the angle] for one bug, then p
(r,\$+pi/2), p(r,\$+pi) and p(r,\$+3pi/2) are solutions for the others.
Second, we have the restriction that the bug's speed is always one;
this corresponds to |p'(r,\$)| = (r')^2 + (r \$')^2 = 1. And third, the
condition that p'(r,\$) = q(p(r,\$+pi/2) - p(r,\$)), where q is some
function of t, leads to \$' = q and r' = - q r.

Consider the bug starting on the positive x-axis. Using the initial
conditions r(0) = 1/sqrt(2) and \$(0) = 0, we can solve the above
three equations to obtain r = (1-t)/sqrt(2) and \$ = -log(1-t).
Thus, the bug's coordinates are ( ((1-t)/sqrt(2)) cos(-log(1-t)), ((1-
t)/sqrt(2)) sin(-log(1-t)) ). The solutions for each
of the other three bugs can be found by adding multiples of pi/2 to \$.

This proves that it will take each bug 1 unit of time to reach the
origin and meet his or her mate. It also shows that each bug will
revolve about the origin infinitely many times (lim(t->0) -log(t) =
infinity).

The next question is how far will they travel. Of course, we already
have the condition that they move at a constant rate of 1 unit per
unit of time and that they reach the origin after 1 unit of time.
Thus they travel a distance of 1.
• I am reminded of some greek philosopher who reckoned that as some aincent Greek athlete ran against a tortoise and started from behind, he would never catch up
Message 2 of 14 , May 1, 2002
I am reminded of some greek philosopher who reckoned that as some
aincent Greek athlete ran against a tortoise and started from behind,
he would never catch up with it, because when the athlete reached the
spt where the tortoise was at the start, the tortoise had moved
further on, and when the athlete reached this new position, the
tortoise was yet further, and so on and so on.

Or else I'm not just using my imagination as well as I might. Is the
solution comparable to the dilemma of about a year.5 (!) ago, the
dilemma of a body with infinite surface area but finite volume?
• Something else. If bug 2 was stationary, then bug 1 travels in a straight line. But Bug 2 ain t, meaning there is an increased distance for Bug 1 to travel.
Message 3 of 14 , May 1, 2002
Something else. If bug 2 was stationary, then bug 1 travels in a
straight line. But Bug 2 ain't, meaning there is an increased
distance for Bug 1 to travel. Right? Or maybe not... Please
explain.
• The reason is perhaps that at no one point are the bugs on a collision course. Each bug is travelling towards a point which is instantly being vacated the the
Message 4 of 14 , May 1, 2002
The reason is perhaps that at no one point are the bugs on a
collision course. Each bug is travelling towards a point which is
instantly being vacated the the other bug.
• ... Actually that s not right. With Bug 2 moving, bug 1 could have a decreased, increased, or unchanged distance to travel. Examples: Bug 2 is moving towards
Message 5 of 14 , May 1, 2002
--- In mathforfun@y..., clooneman <no_reply@y...> wrote:
> Something else. If bug 2 was stationary, then bug 1 travels in a
> straight line. But Bug 2 ain't, meaning there is an increased
> distance for Bug 1 to travel. Right? Or maybe not... Please
> explain.

Actually that's not right. With Bug 2 moving, bug 1 could have a
decreased, increased, or unchanged distance to travel. Examples:

Bug 2 is moving towards bug 1 = decreased distance
Bug 2 is moving away from bug 1 = increased distance
Bug 2 is moving at a right angle = same distance.

Imagine if bug 2 stayed still. Then bug 1 would simply travel in a
straight line along the edge of the square a distance of one unit.
Now take that straight line and bend it around and around into a
spiral. Have bug 2 always move in such a way that bug 1's path veers
into and along the spiral. Also keep bug 2's distance such that it
sits in the middle of the spiral after one second (unit) of time.
Then bug one will exactly travel along this bent spiral path which we
already showed is 1 unit distance.
• Aaahhh right. So effectively, the effect of the constraints on the bugs travels is to take the straight lien route and spiral them into the center.
Message 6 of 14 , May 1, 2002
Aaahhh right. So effectively, the effect of the constraints on the
bugs' travels is to take the straight lien route and spiral them into
the center.
• What if you had only 3 bugs on an equilateral triangle? Or another type of triangle? What would happen then?
Message 7 of 14 , May 1, 2002
What if you had only 3 bugs on an equilateral triangle? Or another
type of triangle? What would happen then?
• ... Equilateral: The bug would spiral in faster. Relative to a bug, his mate would be moving to the left, but also towards him (60 degree angle). The left
Message 8 of 14 , May 1, 2002
--- In mathforfun@y..., clooneman <no_reply@y...> wrote:
> What if you had only 3 bugs on an equilateral triangle? Or another
> type of triangle? What would happen then?

Equilateral: The bug would spiral in faster. Relative to a bug, his
mate would be moving to the left, but also towards him (60 degree
angle). The left component would a speed of .5*root3, and
the "towards him" component would be .5 Combine that with the bugs'
own personal velocity of 1 towards the mate, and the rate of approach
becomes 1.5. They start out 1 unit apart, so the time until merging
= distance/speed = 1/1.5 = 2/3 seconds. (faster than the square.)

But I have to say Clooneman has brought up a good point with his
story about the tortoise and the runner. My argument about the bug
spiraling forever because he's always to the right of center is not
really valid. Just as in the tortoise runner case, the runner would
never seem to catch up with the tortoise, in this bug case, the
number of revolutions would seem to never end. (Just because he's
not aiming for center doesn't rigorously prove that the revolutions
will be infinite.) It turns out this is true (infinite #
revolutions) in the square case, as another poster
here has proved using polar coordinates, but I'm afraid I can't
verify whether it's true in the equilateral case. Maybe somebody
else would like a shot at the # revs in the equilateral case. I
think it's still infinite, but I'm not completely sure.
• Fix a bug. Let its coordinates be given by r(t) = (x(t), y(t)). If the bugs are on the corners of a regular polygon, then, by symmetry, its mate s position is
Message 9 of 14 , May 2, 2002
Fix a bug. Let its coordinates be given by r(t) = (x(t), y(t)). If
the bugs are on the corners of a regular polygon, then, by symmetry,
its mate's position is given by Ar(t), where A is the 2x2 matrix with
entries

a_11 = cos(T)
a_12 = -sin(T)
a_21 = sin(T)
a_22 = cos(T)

for some angle, T. Thus, its velocity, v(t) = r'(t), is proportional
to Ar(t) - r(t). Since speed is irrelevant in computing arclength or
number of revolutions (and, therefore, total time of travel as well),
we may assume

(*) r' = (A - I)r.

The eigenvalues of A - I are complex with negative real part, so the
solutions are inward spirals that circle the origin infinitely many
times. An exact formula for the solutions can be obtained rather
easily and integrated to find the arclength of travel.

An alternative solution is to work in the complex plane. Let the
bug's path be z(t), so that its mate is wz(t), where w = e^(iT). Then

z' = (w - 1)z
z = z(0)e^[(w - 1)t].

Arclength is then the integral from 0 to oo of

|z'|dt = |z(0)||w - 1|e^[(Re w - 1)t]dt,

which, since Re w < 1, is

|z(0)||w - 1|/(1 - Re w)

For z(0), we can take r, the radius of the polygon. For the case of
the unit square, w = i and r = 1/sqr(2), which gives 1. For the
equilateral triangle, w = e^(i*2pi/3) = -1/2 + isqr(3)/2 and r = 1/sqr
(3), which gives 2/3.

Writing everything in terms of n and simplifying, I get that the
total distance traveled if they start at the corners of a regular n-
gon with unit side length is 1/(1 - cosT), where T = 2pi/n.

By the way, starting at the corners of an arbitrary polygon (or even
a rectangle for that matter) looks interesting. Certainly, we lose
symmetry, so that the mate's position is not a simple rotation. That
doesn't seem too difficult, though, since we can make a larger
system of DE's to accomodate the paths of the other bugs. The more
difficult thing is speed. While it's not relevant that they move at
unit speed, it *is* relevant that they move at the same speed.
Assumption (*) will no longer preserve this in an asymmetric
situation.
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