# How many cubes can touch a point in space

Question: Let a point in 3D space. Let cubes with edges of arbitrary sizes. How many cubes with disjoint interiors can actually contain this point? The cubes of the solution can be of different sizes.

Ideas: There is a possible answer of 8 cubes of the same arbitrary size a. Make a cube where each edge is equal to $2∗a$ using 8 cubes. The center of this cube touches 8 cubes. Can you think of a better arrangement?

The problem seems somehow relevant to the kissing number, but it seems more general. Any feedback would be welcome

• One approach would be to center a sphere (of radius smaller than any of the cubes) about the given point $P$ and consider how the cubes disjoint interiors intersect the (surface of the) sphere. – hardmath Jan 31 '17 at 17:25
• Do the edges all have the same size? Otherwise you can have arbitrary many cubes. – Watson Jan 31 '17 at 17:44
• In $\Bbb R^d$, the kissing number of a convex compact set is $k(C) \leq 3^d - 1$. – Watson Jan 31 '17 at 17:45

8 is the max. Any cube touching the point (whether at a corner or an edge or interior) subtends a solid angle of at least $\pi/2$ from an arbitrarily small sphere around the point. Since $8 \cdot \pi/2 = 4\pi$ is the measure of the total solid angle available, $8$ is the max.
• What with hypercubes in n-dimensional space? Always $2^n$ ? – Widawensen Jan 31 '17 at 18:47
• Sure seems true: the "octant" defined by $x_1, \ldots, x_n \ge 0$ occupies $1/2^n$ of the volume of a sphere around the origin, hence so does any other octant. So at most $2^n$ of them can fit without mutual overlap of interiors. – John Hughes Jan 31 '17 at 19:43