# Unit diameter pentagons with maximum area

In the euclidean plane, if one considers the set of quadrilaterals having unit diameter (maximum distance between two points in the convex envelope), it is quite easy to give a description of the elements that have maximum area: since the area of a convex quadrilateral is half the product of the length of the diagonals and the sine of the angle between them, a quadrilateral with unit diameter has maximum area, equal to $\frac{1}{2}$, iff its diagonals have unit length, are perpendicular and all the sides have length $\leq 1$. The question now is: how can we generalize this to pentagons? Id est:

Describe all the pentagons having unit diameter and maximum area.

The next step is to describe the unit-diameter $n$-gons having maximum area $A(n)$, then the behaviour of the function $A(n)$ - for example it is natural to expect that $\lim_{n\to +\infty}A(n)=\frac{\pi}{4}$ (the area of the circle having radius $\frac{1}{2}$), but, quite surprisingly, $A(n)$ does not seem to be monotonic, since the area of the regular pentagon with unit-diameter $\left(\frac{5}{8\sin(2\pi/5)}\right)$ is greater than the area of the regular hexagon with unit diameter $\left(\frac{3\sqrt{3}}{8}\right)$.

• Just curious; By any chance, were you reading this?. I was reading this and am surprised to find a related question here. – user17762 Dec 8 '13 at 2:17
• This paper may be relevant: MR2128992 (2006b:51016), Maley, F. Miller; Robbins, David P.; Roskies, Julie; On the areas of cyclic and semicyclic polygons, Adv. in Appl. Math. 34 (2005), no. 4, 669–689. – Gerry Myerson Dec 8 '13 at 2:23
• I didn't say that, and I don't know that it's true. It seems plausible to me that you could "expand" a polygon to be cyclic without increasing its diameter, but plausible is as far as I'll go. In any event, even if the solutions aren't cyclic, you never know what you might find of use in a paper on a related topic, and its references. – Gerry Myerson Dec 8 '13 at 2:46
• $A(n)$ should be monotonic, because, for example, a pentagon, where we have added a vertex on an edge, is an hexagon – francis-jamet Dec 16 '13 at 18:01
• @francis-jamet: Yes! And then you can make it a little bigger by moving the new vertex outside the pentagon $-$ but not as far as the circumcircle. So perhaps optimal solutions will not in general be cyclic polygons. – TonyK Dec 17 '13 at 9:14

Michael J. Mossinghoff, "A \$1 Problem." Amer. Math. Monthly 113 (2006), 385–402; jstor. I quote: The isodiametric problems for polygons were first studied by Karl Reinhardt, Bieberbach's first student, in 1922 [22]. He solved the area problem for odd values of$n$, showing that the regular$n$-gon is best possible. Then, in an appendix that seems to have been missed in some of the later literature, he proved that the regular$n$-gon is never optimal when$n$is an even number and$n\ge 6$. Reinhardt's proof for the case of$n$odd, and his example for$n$even, is given in the article; it also states and cites the solutions for the cases$n=6$(Bieri 1961, Graham 1975) and$n=8$(Audet et al. 2002), and some information about the asymptotics as$n\to\infty$(another Mossinghoff paper). Just to add something concrete, here's a straightforward proof that$\lim_{n\to\infty} A(n) = \frac\pi4$. Let$K$be any set with diameter at most 1; then$K-K\subseteq C$, where$C$is the unit circle, so by the Brunn–Minkowski inequality,$\text{Area}(K) \le \frac14 \text{Area}(K-K) \le \frac14 \text{Area}(C) = \frac\pi4$. On the other hand, a regular$n$-gon with circumradius$\frac12$has diameter at most 1 and contains a circle of radius$\frac12 \cos\frac\pi n$, and so$A(n)\ge\frac\pi4 \cos^2\frac\pi n$. Circumscribe the polygon by a circle. If n is even, The longest diagonals pass through the center of the circle.Therefore$r=\frac{1}{2}$. $$\text{Area of the polygon =} \frac{1}{2}nr^2 \sin\frac{2\pi}{n}=\frac{1}{8}n \sin\frac{2\pi}{n}$$ If n is odd, The longest diagonals will not pass through the center of the circle,angle subtended by the diagonal at the center =$\frac{n-1}{2}\frac{2\pi}{n}=\left(1-\frac{1}{n}\right)\pi$Therefore,$2r\sin\frac{(n-1)\pi}{2n}=1\rightarrow r = \large\frac{1}{2\sin\frac{(n-1)\pi}{2n}}$$$\text{Area of the polygon =} \frac{1}{2}nr^2 \sin\frac{2\pi}{n}=\frac{1}{8}n \sin\frac{2\pi}{n}\csc^2\frac{(n-1)\pi}{2n}$$ The red line shows the area in cases when n is even and the green line shows the odd cases. As expected there is a very small difference between the 2 lines and they converge to$\pi/4$.It can also be observed that it is not absolutely monotonous.There are certain exceptions. • I still think the they should be monotonic in$n$, considering @francis-jamet comment above. Have you check the values at each$n\$ to say that "there are certain exceptions"? – MoonKnight Dec 19 '13 at 18:56