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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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2 Answers 2

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

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