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 $r\geq4/\sqrt{\pi}$ .
 A: 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...
A: $\bullet$ 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.
