Let $P$ be a convex polytope in $\mathbb{R}^3$ whose every vertex lies in the $\mathbb{Z}^3$ lattice.


**Question:** If $P$ contains exactly one lattice point in its interior, what is the maximum possible volume of $P$? 


Notice that a convex lattice polytope with no interior lattice point can have arbitrarily large volume, but if the number of interior lattice points in it is a given finite number, then its volume cannot be arbitrarily large. In two dimensions the answer is known: A convex lattice polygon with exactly one interior point has volume at most $9/2$, which follows from a more general inequality by P.R. Scott (see *On Convex Lattice Polygons*, Bull. Austral. Math. Soc., vol 15; 1976, 395-399).


The best example I know in that respect is the tetrahedron with vertices $(0,0,0),\ (4,0,0),\ (0,4,0)$ and $(0,0,4)$, with volume ${32}\over3$. Is this perhaps the maximum volume? Of course, the same question can be asked in higher dimensions as well, with an analogous example to consider as a possible candidate for an answer.


In view of several examples presented in answers and comments below, it seems that the optimal polytope should be a simplex, in every dimension. Has this been conjectured already?



Volume of convex lattice polytopes with one interior lattice point
This is addressed in the 2013 paper (appeared in Advances in 2015) by [Averkov, Krumpelmann, Nill](https://arxiv.org/abs/1309.7967). The give a sharp bound for the volume of a lattice simplex with one interior lattice point (Theorem 2.2 in the paper), and an improved bound for a general lattice polyhedron with the same property (theorem 2.7) (the two bounds are not the same, indicating that Wlodek's conjecture is either still open or false). The results are stated in terms of the [Sylvester sequence:](https://www.wikiwand.com/en/Sylvester%27s_sequence)


$$s\_1 = 2; s\_i = 1 + \prod\_{j=1}^{i-1} s\_j.$$


With that, the volume of tbe biggest simplex in $d$ dimension with one lattice point is bounded (with equality achieved) by


$$ \frac2{d!}(s\_d-1)^2,$$


while for arbitrary polytopes, the bound is $$(s\_{d+1}-1)^d,$$ so a lot worse.


