It is a simple geometry question... if you hold the perimeter of a convex polynomial constant, what are the possible limits on the sum of the diagonals, in particular, what is the maximum. I realized I was pretty unsure about the solution, even for a quadrilateral, the simplest case. I thought about a rectangle, and realized that the diagonals get longer as the sides become less equal. If we assume the perimeter is p, then any two sides of the rectangle would be x, and p/x-2. The two equal diagonals would each be the square root of (x^2 + (p/2 -x)^2)) which is largest when the sum of the squares is largest. But the sum is 2x^2 - px + p^2/4 . This is a positive quadratic, and so it is greatest at the ends of its domain, and smallest at the vertex (p/4)... a square has the smallest sum of the diagonals of any rectangle. This is easy to confirm if you take a 4x4 square which has two diagonals of 4 sqrt(2) for a total of 8 sqrt(2) (more or less 11.314), but if we put make it more oblong, say a 7x1 rectangle, the diagonals are each sqrt(50) or 5 sqrt(2) for a total of 10 sqrt(2) (a larger 14.142)... if we extend this to the limiting behavior, we see that each of the diagonals would be close to p/2 so the total sum of the diagonals would approach the perimeter as a limit.. Trying different shapes led me to think (but not prove) that p is probably the limit for a quadrilateral.

But what happens if we let the number of sides go beyond four.... I had no idea how to approach the problems except to experiment. I began by drawing five points on a circle and taking the ratio of the sum of the diagonals to the perimeter. I quickly realized that if I moved two of the points close together on one place and three others near the opposite side of the circle the ratio approached two.. the diagonals were nearly twice the perimeter. This made sense, three of the edges would be nearly zero, and two would be almost the length of the diagonal of the circle, so the total perimeter would approach 2d. And what of the diagonals? Well, there would be four diagonals that went from the two points on one side to the opposite three points that would each be approximately the length of the diagonal, for a total of 4d.

Would this be better with four on one side and one on the other? The perimeter would be the same, but there would only be two long diagonals. It seems that splitting the points up evenly increased the total sum of the diagonals. It seems the diagonals would sum to 2p in the limit. I couldn't imagine exceeding this (correct me if I have failed to visualize something here) for a pentagon...

So, what about a hexagon......? six edges, and 9 diagonals.. . Using our previous insights, we could try putting 3 points close together on opposite sides of the circle, This would mean that would mean that the total perimeter was again approaching 2d, but there would be 7 diagonals which were also approaching the limit of a diameter. The other two diagonals would approach zero, and the total sum of the diagonals would be 7d, for a ratio of 3.5......

Hmmmmm could I see a pattern here? for n=4, the ratio of diagonals to perimeter was 1, for n=5 sides, the ratio was 2, and at n=6 the sum would be 3.5. Well, for a four sided figure, there were two diagonals (and essentially all the perimeter was in two sides, 2/2 =1 ) . For a pentagon, there were 5 diagonals, but one of them went to essentially zero... the ratio was 4/2 or 2. With a hexagon there were 9 -2 long diagonals, so a ratio of 7/2. Could I extend this?

If we went to seven sides, there would be 7 choose 2 - 7 diagonals, or 14 of them. We would put four on one side and three on the other. By my count there woud be 10 long diagonals and four that diminished to zero (one on the end with three points connecting the outside two, and three on the end with four points) so the ratio would grow to 5 times the perimeter

lets tabulate what we know (or think we know)

n-sides... total diag.. short diag...long diag.. ratio --- change

4----------2----=======--0-------------2------------2/2=1---------0

5------------5------------1--------------4-----------4/2=2--------1

6------------9-------------2-------------7-------------7/2--------1.5

7-----------14-------------4------------10------------10/2=5------1.5

8-----------20------------6-------------14-----------14/2=7 -----2

9-----------27------------9 ------------18-----------18/2=9-------2

10----------35-----------12-------------23----------23/2=11.5----2.5

Ok, the recursive pattern seems to be r(n+1) = r(n) + floor(n/2)/2... I think I did that right....

and it is late, so I will let you write the explicit function...I'm just thankful that the holiday is at hand. I'm off to London with my sweetheart for the weekend to see a west-end show and have a good Japanese meal... and hold hands and walk through the neighborhood markets... hope you have a great holiday too.