Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $r\geq4/\sqrt{\pi}$ .

share|improve this question
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 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 at 0:47
a region $C$ that minimizes $r$ would probably have this property: mathoverflow.net/questions/78165/… –  Yoav Kallus Mar 1 at 2:02
Can you tell us which shape achieves the extremal $2/\sqrt{\pi}$ segment? –  Joseph O'Rourke Mar 1 at 15:15
@JosephO'Rourke: the circle of radius $1/\sqrt{\pi}$ does. –  Benoît Kloeckner Mar 1 at 16:29

2 Answers 2

$\bullet$ 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.

share|improve this answer
... and the same is true for the $n$-gon, for every $n\ge3$, in place of the triangle. –  Wlodek Kuperberg Mar 2 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...

share|improve this answer
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 at 2:42
So $r(0.821)=2.974$, as you calculated. And $r(a)$ is minimal at $a=1$. –  Yoav Kallus Mar 2 at 2:43
Oh, sorry; that was me who miscalculated... –  Ilya Bogdanov Mar 2 at 9:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.