It is intuitive that at all times the ants will form the corners of a square, ever decreasing in size and rotating about the center.
Let C be the center of the square at point (0,0). Let the ants be at initial points (.5,.5), (.5,-.5), (-.5,-.5), and (-.5,.5). Let ant 1 bet at (.5,-.5) who is chasing ant 2 at (.5,.5). Consider the line from C to ant 1. Next consider the line from ant 1 to ant 2. The angle between these lines is 135 degrees.
Theorem: Let C be a fixed point, D a point on a curve, r the distance from C to D, x the polar angle formed by r, and y the angle formed by the angle between CD and the tangent line to the curve at point D. Then tan(y)=r/(dr/dx).
So the line of sight from the center to the initial point of ant 1 would be be at a 270 degree angle. The line of sight from ant 1 to ant 2 would be a vertical line of 90 degrees. The angle between these lines is 270 degrees. The tangent of 270 is -1. So we have r/(dr/dx) = -1. Thus r(x)=c*e-x.
At the moment the ants start walking x equals 0. At this moment the distance from the center to any of the ants is 2-1/2. So r(x)=2-1/2*e-x.
The arc length is given by the formula:
Integral from negative infinity to 0 of ((f(x))2+(f'(x))2)1/2 =
Integral from negative infinity to 0 of ((2-1/2*e-x)2 + (-2-1/2*e-x)2)1/2 =
Integral from negative infinity to 0 of (2-1*e-2x + 2-1*e-2x)1/2 =
Integral from negative infinity to 0 of e-x =
-e-x from negative infinity to 0 =
=0 - (-e0) =
0 -(-1) = 1.
The easy way to solve this problem is to first note that the direction of one ant will always be perpendicular to the path of the ant he is chasing. If ant A is chasing ant B then B has zero velocity in the direction of A's path. In other words B neither approaches nor recedes from A during the walk. So it is like A is walking directly towards an unmoving ant B, in which case the total distance is simply the intial separation of 1.
No we're ready to consider the pentagon case. Whether you solve the square case the easy or the hard way it should be clear that as ant A chases ant B what is important is the velocity of ant B relative to the path of ant A.
Before we go further we have to do some trigonometry. Consider an equilateral pentagon of side 1 with the first vertice at (0,0) and the second at (1,0). Where is the third vertice? Given that the interior angles of a pentagon are 108 degrees the next vertice is at (1+cos(72),sin(72)).
So as ant A inches 1 unit towards ant B, ant B will move cos(72) units directly away from ant A in the same direction and sin(72) in a perpenticular direction. We know form the four corners problem that the perpenticular direction can be ignorred. So for every micron ant A moves towards ant B, ant B will move away from ant A by cos(72) microns.
The total distance covered by A, or any ant, is the sum for i=0 to infinity of cos(72)i. We know from infinite sums that this equals 1/(1-cos(72)).
If you prefer to express the answer another way cos(72)=1/(1+51/2). So 1/(1-cos(72)) = 1/(1-1/(1+51/2)) = 1/((51/2)/(1+51/2)) = (1+51/2)/51/2 = 1 + 1/51/2 =~ 1.4472 .
For more information on this problem please see the November 1957 and July 1965 issues of Scientific American. The July 1965 issue notes that for the general case of an n-gon the total distance is as follows. Let A and B be the initial points of any two consecutive ants, where A is chasing B. Let C be the center of the n-gon. Consider the line segment AB extended to a point X, such that the angle XCA is a right angle. The total distance any ant will travel is the same as the distance from A to X.
Thus, if the interior angle of the n-gon is x then the total distance is 1/(1+cos(x)). So for an n-gon the total distance is 1/[1+cos(180-360/n)].
Michael Shackleford, A.S.A.