Zeno's Mice

Suppose there is a mouse in each of the six corners of a regular
hexagonal room, and at a given instant each mouse begins to run
with a constant speed directly toward the mouse on his right.  What
paths will the mice follow, and what distance will they have run
when they meet?  Also, what will be each mouse's total angular
travel around the center of the room?

It's clear that the situation is perfectly symmetrical with respect
to the 6 mice, and it remains so throughout the process.  Thus, at
each instant we have 6 mice located at the verticies of a hexagon, 
and they are always following the same rules about how to move, so 
we always have just a rescaled and rotated version of the original 
condition.

From this it's clear that the mice must spiral in along similar 
logarithmic spirals, i.e., spirals with polar coordinates [r,theta] 
satisfying the equation 

                  ln(r) = k*theta

because this is the general form of a spiral that is geometrically
similar to itself on all scales.  Of course, this implies that the 
mice just keep sprialing in a smaller and smaller circle, going 
infinitely many times around the center of the room, as illustrated
in the figure below.



Notice that, even though a logarithmic spiral circles the center 
infinitely many times, the total arc length of the spiral is 
finite.  In fact, the log spiral was the first "transcendental" 
curve to be "rectified".  This was done by Torricelli in 1645 
(before calculus was invented) using Archimedes "method of 
exhaustion" for the proof.

The answer is surprisingly simple, and can be seen intuitively by
placing a segment of the spiral from the center C to a point P
on a flat surface so the curve is tangent to the surface at P, and
then "unrolling" the curve.  Since the figure is geometrically
similar at each stage, it's clear that the center point C linearly
approaches the flat surface as the curve unwinds.  Therefore, if
we let phi denote the angle that the line PC makes with the flat
surface, the unwound spiral will have a length of R/cos(phi),
where R is the length of the segment PC.  (In other words, R is 
the distance from the center of the room to a corner.)

Since the spiral is tangent to the wall at the corner, we know
that phi is simply the angle that a line from the center of the
room to a corner makes with the adjacent wall.  For a hexagonal
room this angle is 60 degrees (pi/3), which implies that the length
of the logarithmic spiral is R/cos(pi/3) = 2R.  On the other hand, 
if we had chosen a square room with four mice, the angle phi would 
be 45 degrees (pi/4), so the mice would have to travel a distance
of R/cos(pi/4) = sqrt(2)R.

In general the angle between an edge of an n-sided polygon and
a line from the center to a vertex of that edge is (pi - 2pi/n)/2,
so the general length of a logarithmic spiral inscribed within
an n-sided polygon of "radius" R is

       L  =  R / cos(pi/2 - pi/n)   =   R / sin(pi/n)

On the other hand, as mentioned above, the total angular travel of
a logarithmic spiral from any finite point to the center is infinite.
Therefore, since the mice are moving at a constant speed and they
reach the center in a finite amount of time, it's clear that their
angular velocity becomes infinite as they approach the center.  Now,
the centripital acceleration at a distance r from the center is 
v^2 / r where v is the tangential components of the mouse's velocity,
and this will be a constant fraction of his total speed, which is
constant.  Thus his acceleration goes to infinity as r goes to zero.

(It would be interesting to figure out what central force law applied 
to point particles would yield this kind of logarithmic spiraling 
orbit.  Of course, it wouldn't be an inverse-square law, because that 
yields only elliptical bound orbits.)