Given the length of the diagonals, what is the maximum perimeter of a quadrilateral? Suppose the length of the diagonal $d$. I'd like to maximize the perimeter of a quadrilateral containing two of these diagonals. I know that for a rectangle instead of a quadrilateral, the solution is the square, obtaining a perimeter of length $4d\frac{\sqrt{2}}{2} = 2\sqrt{2}d$. However when not constraint by right angles, the solution must more than the square's perimeter, since one could construct a triangle of side lengths $d$, $d$ and $\sqrt{2}d$ by coinciding two ends of the diagonals, and creating a right angle, giving a perimeter of $(2 + \sqrt{2})d$.
What is the largest perimeter possible?
EDIT: as pointed out by Henry, I'd like the shape to be convex, or else the perimeter is unbounded. Also, a flat quadrilateral, where the diagonals are basically aligned, has a perimeter of $4d$. The difficulty is to prove it optimal.
 A: The maximum perimeter is $4d$, or if you have diagonals of different lengths $d_1$ and $d_2$ then the maximum perimeter is $2d_1+2d_2$. This maximum is only reached in the degenerate case where they are end to end, i.e. when the sides of the quadrilateral are $d_1$, $0$, $d_2$, $d_1+d_2$ in order and the quadrilateral is actually a completely flat triangle. You can of course make a non-degenerate quadrilateral with a perimeter arbitrarily close to this maximum.
To prove this is the maximum, you can use the triangle inequality.
In any non-degenerate convex quadrilateral, the diagonals split the quadrilateral into four triangles. Use the triangle inequality on each of the four triangles; the outside edge is shorter than the two other legs together. Together those inner legs of the four triangles cover the two diagonals exactly twice. Therefore the perimeter is less than twice the length of the diagonals, i.e. less than $2d_1+2d_2$. If the intersection of the diagonals is internal to the quadrilateral (which is the case for any non-degenerate convex quadrilateral) then the inequality is strict.
