3 Jan
2011
3 Jan
'11
1:23 p.m.
P.S. This problem certainly didn't stump many people! In greater generality let a "unit slab" denote all points at a distance <= 1 from a hyperplane through the origin in R^N. Then if {S_k : k in Z+} is any countable family of unit slabs in R^N, their intersection has surface area = N * volume. (E.g., if in R^3 the countable family consists of only 3 distinct, mutually perpendicular unit slabs, one gets a cube of volume 8 and surface area 24.) --Dan Those who sleep faster get more rest.