1

1000 alien spaceships meet in a 4-dimensional battlefield. At an agreed time (ignoring relativistic effects on their clocks) every spaceship fires its laser to the spaceship which is closest (assume that all distances between the spaceships are different). What is the maximum number of hits a single spaceship can suffer?

This problem is equivalent to this question: "How many points can you place on the surface of a 4D-sphere, such that all distances between these points (which must all be different) are longer than the radius of the sphere?"

I assume I solved this problem for a 3D-sphere: If you add a small $\epsilon$ to the positions of the 12 vertices of an icosahedron such that all distances are different, then all these 12 "spaceships" will fire to the centre of the icosahedron. In 4D, the solution might be derived from the 24 vertices of an icositetrachor but at this point I am stuck. Any suggestions?

Rüdi Jehn
  • 624
  • Comparing the total volume of the $(r/2)$-disks around each point with the total volume of the sphere, you get an upper bound of $\lfloor 12 \pi \rfloor = 37$, I don't know how sharp (the same idea gives an upper bound of $16$ in 3D and $6$ in 2D). Also a matter of convention, I would call the sphere in the question either a sphere in 4D or a 3D sphere. – ronno May 11 '23 at 09:22
  • According to https://mathworld.wolfram.com/KissingNumber.html the answer is $24$ but I haven't looked through the references to find the construction – ronno May 11 '23 at 09:27

0 Answers0