Tangential Equidiagonal (Irregular) Quadrilateral with integer coordinates

Question

Playing with mathematics this week-end, I ended up with this question I couldn't solve, so I'm asking here. I try to find a Quadrilateral ABCD with integers (cartesian) coordinates (points are on a regular square grid) such that :

1. ABCD is convex
2. ABCD is Tangential (it has an inner circle, hence $|AB|+|CD|=|AD|+|BC|$)
3. ABCD is Equidiagonal ($|AC|=|BD|$)
4. It is otherwise irregular : it's not a kite, a trapezoid or a square. Hence all sides must be of different lengths.

I did not find a solution to this problem nor a way to prove there are no solution. Without the grid constraint, there are plenty of such quadrilaterals, but with the grid...

Note that the lengths of sides or diagonals are not required to be integers.

EDIT :

After a good night sleep, I realize that if a solution exists, a solution with integer length sides exists. So it's a property on pythagorean triplets that we are looking for. I added the tag.

Answer 1

Certain it is that any set of permissible side lengths will give a diagonal length that meets the criteria. You just have to solve for the diagonal length. However, the development below does NOT address setting up integer coordinates, because it is based on elementary geometry instead of coordinate geometry.

I demonstrate with quadrilateral $ABCD$ with $AB=2, BC=3, CD=5, DA=4$. We install $l$ as the common lengths of the diagonals, for which we will solve.

When we draw the diagonals, we have four overlapping triangles. Two of these share a diagonal and cover the whole area of the quadrilateral. The other two share the other diagonal and these also cover the same area. Obviously the two sums of triangle areas must agree, whereupon Heron's Formula applied to all the triangles will produce an equation for $l$.

With the side lengths I have chosen, after clearing some fractions I get

$|\triangle ABD|+|\triangle CDA|=|\triangle BCA|+|\triangle DAC|$

$\sqrt{(5-l)(l+5)(l+1)(l-1)}+\sqrt{(9-l)(l+9)(l+1)(l-1)}=\sqrt{(8-l)(l+8)(l+2)(l-2)}+\sqrt{(6-l)(l+6)(l+2)(l-2)}$

The Triangle Inequality indicates that the required root should be between $2$ (the largest difference between any two adjacent sides) and $5$ (the smallest sum between any two adjacent sides) for the diagonal length. Numerical calculation renders a root of approximately $4.66$ for my quadrilateral.