Would you expect an upper bound on the perimeter or diameter for sets containing just one lattice point?
No. It is possible to construct a long thin triangle withy horizontal base which contains just one lattice point, and with perimeter and diameter as large as we please.
2.
The extremal set for the maximal width is an isosceles triangle. Why isosceles?
For the extremal set, the maximal width is assumed in two places. If this is not so, we can pivot the base about the lattice point, and obtain a larger maximal width.
3.
Calculate the bound for the circumradius of sets containing one lattice point, assuming the extremal set is Q1.
The two arc angles are 5.47° and 20.23°. With some effort, R turns out to be approximately 1.593. Of course this ‘best’ result is only conjectured.
4.
Find the trivial upper bound for the inradius for sets containing just one lattice point.
Suppose K contains the origin O. Then K does not contain (1, 0), (0,1), (–1, 0), (0, –1) in its interior. The largest circle containing O but not these points has radius r = 1.
5.
Suggest a reason for the similarity of the A - d, and A - R inequalities.
The extremal set is the same, and in this case the diameter is twice the circumradius (of the bounding circle).
6.
Make a conjecture for the set of largest width containing just one point of the triangular lattice.
We expect the extremal set to be the equilateral triangle made up of 9 primitive triangles. If the edge length is 1, the width is 33/2. I don't think a proof of this has appeared: it should be easy!
3/2. I don't think a proof of this has appeared: it should be easy!