How to construct six points $ABCDEF$ on a plane so that the distance between any two of them is an integer, and no three are collinear?

Question

How to construct six points $ABCDEF$ on a plane so that the distance between any two of them is an integer, and no three are collinear?

I tried with some right angled triangles with pythagorean triples and you get 3 points. and i am stuck with 3 points

Answer 1

See the attached image. The idea is to find Pythagorean triples $(a,b,c)$, $(a,d,e)$, and $(2a,b+d,f)$ such that $b<d$. Then, let $$A=(-b-d,-2a),$$ $$B=(b+d,-2a),$$ $$C=(b+d,2a),$$ $$D=(-b-d,2a),$$ $$E=(b-d,0),$$ and $$F=(-b+d,0).$$ Therefore, $$AB=CD=2(b+d),$$ $$BC=DA=4a,$$ $$AC=BD=2f,$$ $$EA=ED=FB=FC=2c,$$ $$EB=EC=FA=FB=2e,$$ and $$EF=2(d-b).$$ I found $(a,b,c)=(40,9,41)$, $(a,d,e)=(40,30,50)$, and $(2a,b+d,f)=(80,39,89)$.

If you want a convex hexagon, you can find Pythagorean triples $(a,b,c)$, $(a,d,e)$, and $(2a,d-b,f)$ such that $b<d$. Then, let $$A=(b-d,-2a),$$ $$B=(-b+d,-2a),$$ $$C=(-b+d,2a),$$ $$D=(b-d,2a),$$ $$E=(-b-d,0),$$ and $$F=(b+d,0).$$ Therefore, $$AB=CD=2(d-b),$$ $$BC=DA=4a,$$ $$AC=BD=2f,$$ $$EA=ED=FB=FC=2c,$$ $$EB=EC=FA=FB=2e,$$ and $$EF=2(b+d).$$ I found $(a,b,c)=(12,9,15)$, $(a,d,e)=(12,16,20)$, and $(2a,d-b,f)=(24,7,25)$. See the image below.

The last picture is very nice. This hexagon is also inscribed in a circle with integer radius. However, $AC$, $BD$, and $EF$ are all diameters of this circle. This makes me think: is it possible to find a cyclic hexagon $ABCDEF$ such that all sides and diagonals have integer lengths, the circumradius is also an integer, and no diagonal or side of the hexagon is a diameter of the circumcircle?

Answer 2

The OP does not impose any upper bound on how many of the points may be on the same circle, or any requirement that their Cartesian coordinates must be integer (or even rational). So let me offer an alternative approach: all the points are on a circle of radius $R=n/\sqrt3$ where $n$ is a Loeschian number with $3\nmid n$. For example, we may take $n=7, R=7/\sqrt3$.

The distances between the points are $AB=CD=EF=3$; $BC=DE=FA=5$; $AC=CE=EA=BD=DF=FB=7$; $AD=BE=CF=8$.

We may construct configurations of even more concyclic points at integer distances by taking $n$ as the product of $k$ distinct primes $p_i=1\mod 6$. The greater $k$, the more points.