Definitions: The diameter of a convex region is the greatest distance between any pair of points in the region. The least width of a $2$D convex region can be defined as the least distance between any pair of parallel lines that touch the region. Let us refer to a polygon as 'fat' if the ratio between its diameter and least width (call this the 'thinness ratio') is low.

General Question: Given a general $n$, which $n$-gon is the fattest? For which values of $n$ (if at all) is the fattest n-gon a regular $n$-gon?

2 Special cases: -For $n =3$, the equilateral triangle is the fattest. The thinness ratio is $2/\sqrt(3) \approx 1.15$.

-For $n = 4$, the square is not the fattest - its thinness ratio is $\sqrt{2}$. As pointed out by Yakov Baruch below, if we form a quad by adding a vertex arbitrarily close to one of the vertices of an equilateral triangle, its thinness ratio is only about $1.15$.

Some Further Queries: Thanks very much to Prof. O'Rourke for the pointer to Bezdek and Fodor's work (answer below) that answers just the above question. One could ask a further question: Given values for area and perimeter, which convex shape (not necessarily polygonal) is the most/least fat?

For fixed $A$ and $P$ within a suitable range (from perimeter equal to that of circle of area $A$ to perimeter of a Reuleux triangle with area $A$), it appears that fatness is maximized by curves of constant width (https://nandacumar.blogspot.com/2012/11/maximizing-and-minimizing-diameter-ii.html). For $P$ values greater than this range, I don't know.

For specified $A$ and $P$, diameter is maximized by a 'convex lens' shape as given in above linked page. But this shape does not seem to minimize fatness.

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    $\begingroup$ For $n=4$ an equilateral triangle seems to be fatter than the example given. Then also any non-degenerate quadrilateral close to it. $\endgroup$ Mar 8, 2021 at 9:13
  • $\begingroup$ Thanks for pointing this out. Shall correct it! $\endgroup$ Mar 8, 2021 at 9:17
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    $\begingroup$ Notice that in general the regular $(2n-1)$-agon beats the regular $(2n)$-agon. I suspect the odd regular polygons are optimal. $\endgroup$ Mar 8, 2021 at 10:50

1 Answer 1


The paper

Audet, Charles, Pierre Hansen, and Frédéric Messine. "Extremal problems for convex polygons-an update." J Global Optimization 55 (2009): 1-16. Springer link.

discusses this problem, which I believe is equivalent to your question:

Find for all $n$ which convex polygon with unit diameter has the largest width.

They cite Bezdek and Fodor for these results:

(a) The maximum width satisfies $W_n \le \cos \frac{\pi}{2n}$ for $n \ge 3$ and equality holds if $n$ has an odd divisor greater than $1$.

(b) If $n$ has an odd prime divisor then a polygon $V_n$ is extremal if and only if it is equilateral and is inscribed in a Reuleaux polygon of constant width $1$, so that the vertices of the Reuleaux polygon are also vertices of $V_n$.

(c) Results are also obtained for quadrilaterals: $W_4 = \sqrt{3}/2 \; (=W_3)$. Moreover, all extreme quadrilaterals have the property that three of their vertices form a regular triangle and the fourth vertex is contained in the Reuleaux triangle determined by the three vertices.

A. Bezdek and F. Fodor, On convex polygons of maximal width, Archiv der Mathematik, Vol. 74, No. 1, pp. 75–80, 2000.


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