<pre>
Let theta be the angle of the point's initial vector.  After traveling a
distance r, the point has moved r'''cos(theta) horizontally and r'''sin(theta)
vertically, and thus has struck r*(sin(theta)+cos(theta))+O(1) walls.  Hence
the average distance between walls will be 1/(sin(theta)+cos(theta)).  We now
average this over all angles theta:
 2/pi * intg from theta=0 to pi/2 (1/(sin(theta)+cos(theta))) dtheta
which (in a computation which is left as an exercise) reduces to
 2'''sqrt(2)'''ln(1+sqrt(2))/pi = 0.793515.
</pre>
