At the beginning of the exercise, three soldiers, named Ike, Jay and Kay, were at three different points equidistant from a target. Ike was 4 kilometers from Jay, and also 4 kilometers from Kay.

Then Jay started moving inward, directly toward the target. He stopped short of the target, at a point different from his original location, but again 4 kilometers from Ike.

At this point the distances between any two of Ike, Jay, Kay and the target were all whole numbers of kilometers.

In his new position, how far is Jay from Kay?

    Indeed, the distance between Ike and Jay depends on the original distance from the target or, depending on how you look at it, the distance traveled by Jay.

So let the distance traveled by Jay be d and the original distance from the target be r.

Construct a circle O of radius r with points K, I, and J equally spaced on its edge (I suggest letting all three lie in the same quadrant) Let KI = IJ = 4km. Construct j on JO such that jIJ is isoceles with jI =4 noting that Jj = d. Finally, construct jK. Call the point of intersection of IO and jK i.

We see from the properties of iscoceles triangles that IJO, JIO, KIO, IKO, and IjJ are congruent angles. Also, IjK and IKj are congruent. From standard geometric calculations of angles we see that ij is parallel to IJ. Thus Ki = 4 and Kj = 4 + 4(r-d)/r. We can put d in terms of r (or vice versa) by noting that r/4 = 4/d because of the similarity of IOJ and jIJ.

Thus the distance from Jay to Kay is 8 - 64/(r^2) km for r>4.

Posted by Eric on 2006-10-09 19:40:48