All the black squares of a chessboard are enclosed with circumscribed circles.
Assume that each cell is of side 2 and area 4. Each circle has a radius of
sqrt(2) and an area of 2*pi. The area in each white cell that is left
white (not enclosed by a circle) is 8 - 2 * pi. In the analogous situation
for a cubical chessboard, the black cubical cells of edge 2 are surrounded
by spheres. The volume of each black cell is 8 and the volume of each sphere,
which has a radius of sqrt(3), is 4*pi*sqrt(3), but the volume of the
unenclosed portion of each white cube is not so easy to calculate because
the six surrounding spheres intersect one another. Consider now the
four-dimensional lattice of hypercubes of edge 2 with cells alternately
colored as before so that each cell is surrounded by eight hypercubes of
opposite color. Around each black hypercell is circumscribed a hypersphere.
What is the hypervolume of the unenclosed portion within each white cell?
The surprising answer can be determined quickly without knowing the formula
for the volume of a hypersphere.