# Maximal sum of diagonals in convex quadrilateral

Among all convex quadrilaterals with given sides(and their order), which one does have maximal sum of diagonals? Does it have any other interesting properties?

I thought about something like this: let $A = (0; 0), C = (x; 0)$ (i. e. fix the length of one diagonal), then we can find points $B$ and $D$, so the desired sum is a function of $x$ with a lot of roots, and we can just find its maximum. But:

1. Derivative is very cumbersome.
2. We need to ensure not only that $x \le \min(a + b, c + d)$ but also that resulting quadrilateral is convex, I'm not sure what limits it puts on x.

Any thoughts? Maybe it is just cyclic quadrilateral, but it doesn't look so.

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This is not an answer (sorry!), but may point toward an answer. A quadrilateral with all four side lengths given forms what is known as a four-bar linkage. The motions of these linkages have been heavily studied. The paper "Simple Proofs of a Geometric Property of Four-Bar Linkages" by Godfried Toussaint (Amer Math Monthly 2003) says in its abstract

For convex and crossing linkages, one diagonal increases if and only if the other decreases.

I cannot access the paper now, but it may point to an answer.

The cyclic quadrilateral maximizes area, but I agree, not the diagonal sum.

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If the side lengths are given to you, you can plot a circle of the respective side lengths as radius from the two points and see where the intersect.

So you would plot circles of length $BC$ and $CD$ from $C$, and circles of length $AB$ and $DA$ from $A$, and see where these intersect with the circles from $C$ respectively, to get $B$ and $D$.

To get a convex quadrilateral, you would have to keep the points $B$ and $D$ on opposite sides of the diagonal $AC$.

This should give you the length of the second diagonal in terms of $x$, and then you can use calculus.

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I'm really sorry but I made an error in describing a convex quadrilateral. I mistook it for concave... Silly me. So the change in my answer is that $B$ and $D$ should lie on opposite sides of the diagonal $AC$. –  udiboy1209 Jul 5 '13 at 15:12