For n = 4 the following trick works, but does not really have the same flavor as the n = 3 case:
If the soldiers land in a convex position (they are the corners of a convex quadrilateral) then meet at the intersection of the straight lines connecting the opposite soldiers. If they land in a triangle with one soldier in the interior, meet at the soldier in the middle.
Here the soldiers are forced to walk in straight lines.
It would not surprise me if there are tricks like these for some more n, but I don't see a general strategy. I think there is something you can do for odd n that is quite different from the other solutions so far, but I have not checked it fully yet and worked out what it requires from the radar.
For n = 3 you can exploit that the soldiers land on a circle for a neat solutions (meet at the soldier opposite to the longest circle segment walking along the arcs).
I see I should have read the instructions more carefully. I don't quite understand all of your complaints though, if they can notice if they are on the same spot at the same time (they can't I missed that) the strategy would work: the two soldiers would meet as they will not leave the line connecting them (walking towards some random point on it sure but still in the right direction)?
Everyone uses the radar once they wake up. The two closest soldiers go towards each other on a straight line. Once they meet up and possibly wait for the other one to wake up they are free to visit all other soldiers in any order, when they are all together they can detonate. The important property is that the initial shortest distance is unique almost surely.
Thinking about how to prove the multilinearity of the volume of a parallelepiped definition I like this sketched approach:
The two dimensional case is a “cute” problem involving rearranging triangles and ordinary areas (or you solve this case in any other way you want). The general case then follows from linearity of integrals (you get the higher dimensional cases by integrating the two dimensional case appropriately).
Nice theorem and write up. Already the one dimensional case is something interesting called “The second symmetric derivative”. And I think it might be used to prove the general case directly: If you add up that result for n orthogonal directions then the left hand side is the Laplacian. The right hand side is a sum of a n limits that at first glance seems to depend heavily on the picked direction, but they can’t as the left hand side is independent of the picked directions. We are free to pick arbitrary many different directions and take the average. In the limit it becomes the average over a sphere but the order of limits is wrong, some standard theorem might apply to resolve that?
That’s pretty cool, I wonder why I did not remember this concept.