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Let $P$ be a parallelepiped with all of its vertices lattice points. Define $A,B,C$ and $D$ as follows:

$A$ = the number of lattice points strictly inside $P$.

$B$ = the number lattice points which are on the faces of $P$ but not on the edges.

$C$ = the number of lattice points which are on the edges of $P$ but not on the vertices.

$D$ = the number of lattice points which are on the vertices of $P$ (i.e., the number of vertices)

Something interesting (at least for me) that I have read somewhere recently is that the volume of $P$ is equal to $A+\frac 12B + \frac 14C + \frac 18D$.

I can not prove it. So if you know a proof, please let me know.

Silvi
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1 Answers1

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The statement is true for parallelepiped. A simple way to see this is pack $n \times n \times n$ copies of original parallelepiped into a big one.

Let $V$ be the volume of the original parallelepiped. Let $A', B', C', D'$ and $V'$ be the corresponding number of lattice points and volume for the big parallelepiped.

Two things that are obvious are:

$$B' + C' + D' = O(n^2)\quad\quad\text{ and }\quad\quad V' = n^3 V = A' + O(n^2)$$

If you look at what contributes to $A'$, it is clear:

  • $n^3 A$ lattice points comes from lattice points in interior of the smaller parallelepipeds.
  • $\frac{n^3}{2} B + O(n^2)$ lattice points come from lattice points on faces of smaller parallelepipeds. The factor $\frac12$ comes from the fact every faces in the "interior" of the big parallelepiped is shared by two smaller parallelepipeds.
  • $\frac{n^3}{4} C + O(n^2)$ lattice points come from lattice points on edges of smaller parallelepipeds because every "interior" edge is shared by 4 small parallelepipeds.
  • $\frac{n^3}{8} D + O(n^2)$ lattice points come from lattice points on corners of smaller parallelepipeds.

As a result, we get

$$n^3 V = n^3 A + \frac{n^3}{2} B + \frac{n^3}{4} C + \frac{n^3}{8} D + O(n^2)$$

Divide both sides of $n^3$ and send $n \to \infty$, we obtain the formula:

$$V = A + \frac{B}{2} + \frac{C}{4} + \frac{D}{8}.$$

achille hui
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