# Triangle with largest perimeter in a convex region

What is the largest value of $r$ such that the following statement is always true?

"Let $C$ be a convex region with area $1$. There must exist a triangle contained in $C$ whose perimeter is at least $r$."

I don't need the actual largest value of $r$, but a lower bound would be nice. Using the fact that any convex region with unit area must contain a line segment of length $2/\sqrt{\pi}$, it is clear, for example, that $r\geq4/\sqrt{\pi}$ .

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A bit of a tangent, but: The maximum perimeter triangle inscribed in a convex $n$-gon can be found in $O(n \log n)$ time: Boyce, James E., David P. Dobkin, Robert L. Drysdale III, and Leo J. Guibas. "Finding extremal polygons." SIAM Journal on Computing, 14, no. 1 (1985): 134-147. – Joseph O'Rourke Mar 1 '14 at 0:27
A quick observation: If triangle $PQR$ contained in $C$ is of maximum perimeter, then $C$ is contained in the intersection of three ellipses, each having foci at two of the points $P$, $Q$, $R$ and passing through the third. – Wlodek Kuperberg Mar 1 '14 at 0:47
a region $C$ that minimizes $r$ would probably have this property: mathoverflow.net/questions/78165/… – Yoav Kallus Mar 1 '14 at 2:02
Can you tell us which shape achieves the extremal $2/\sqrt{\pi}$ segment? – Joseph O'Rourke Mar 1 '14 at 15:15
@JosephO'Rourke: the circle of radius $1/\sqrt{\pi}$ does. – Benoît Kloeckner Mar 1 '14 at 16:29

## 2 Answers

$\bullet$ Every convex region of area $1$ contains a triangle of area at least as large as the area of the equilateral triangle inscribed in the circle of area $1$. Moreover, if the region does not contain a triangle of larger area, then it is an ellipse. [E. Sas, Über eine Extremaleigenschaft der Ellipsen, Compositio Math. 6 (1939), 468–470]. (Actually, Sas proves a more general statement, for $n$-gons with $n\ge3$.)

$\bullet$ Among all triangles of a given area the equilateral one, and no other, is of minimum perimeter.

$\bullet$ If the maximum area triangle inscribed in an ellipse is equilateral, then the ellipse is a circle.

Putting it all together:

Theorem. If $C$ is a convex region with area $1$, then there exists a triangle contained in $C$ whose perimeter is at least as large as the maximum perimeter of a triangle inscribed in the circle of area $1$, and if $C$ does not contain a triangle of perimeter larger than that, then $C$ must be a circle.

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... and the same is true for the $n$-gon, for every $n\ge3$, in place of the triangle. – Wlodek Kuperberg Mar 2 '14 at 3:15

Here is Ilya's $0.821$-ellipse (if I interpreted his intention correctly), discretized to an $180$-point polygon at $2^\circ$-degree angular increments:

His point, if I may editorialize, is that the naive lower bound of $4/\sqrt{\pi} \approx 2.26$ is not so far from the bound for the optimal ellipse. My computations for this discrete version yield $\approx 2.97$, in fact, slightly larger than $r=3 \sqrt{3/\pi} \approx 2.93$ for the circle. Perhaps I miscalculated...

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I get the following expression for the perimeter $r$ as a function of the aspect ratio $a$: $\frac{2}{(1-a^2)\sqrt{a\pi}} (a^2\sqrt{2s-1-a^2}+(1-a^2)\sqrt{2s+2-a^2})$, where $s=\sqrt{1-a^2+a^4}$. – Yoav Kallus Mar 2 '14 at 2:42
So $r(0.821)=2.974$, as you calculated. And $r(a)$ is minimal at $a=1$. – Yoav Kallus Mar 2 '14 at 2:43
Oh, sorry; that was me who miscalculated... – Ilya Bogdanov Mar 2 '14 at 9:57