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Let $A$, $B$ and $C$ be three points in a disk, does $f\left(A,B,C\right)=\mbox{Area}\left(\mbox{triangle}\,ABC\right)/\mbox{Perimeter}\left(\mbox{triangle}\,ABC\right)$ have maximum on the boundary?

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By occurring on the boundary of $D^{3}$, do you mean that A,B and C are all points located on the boundary of the disc? – Epictetus Dec 5 '12 at 11:45

First note that if a triangle is subjected to a homothety by factor $r>1$ then the area multiplies by $r^2$ and the perimeter by $r$, so that area/perimeter gets multiplied by $r$. This means for the triangle $ABC$ with longest side say $AB$, that we may expand and move the triangle until vertices $A,B$ are on the boundary of $D$, while increasing the ratio area/perimeter.

If at this point the vertex $C$ happens to lie in the smaller part of $D$ cut by $AB$, reset $C$ to its reflection through $AB$, so that $C$ now lies in the larger part of $D$ cut by $AB$.

Now suppose the vertex $C$ is moved so that the perimeter remains constant. This means $C$ moves on an ellipse with foci at $A,B$; this ellipse will not entirely lie in $D$, however it is clear that $C$ may be moved until triangle $ABC$ becomes isosceles, and that during this mmovement the area of $ABC$ increases, since the altitude from $C$ increases. Thus the ratio area/perimeter increases at this step also.

Now move $C$ in the direction perpendicular to $AB$ and away from that line, until $C$ lies on the boundary of $D$. This will increase area more than perimeter: as a map it is an expansion in the direction perpendicular to $AB$ and thus multiplies area by some $k>1$, while since the sides $AC$ and $BC$ are on a slant to the perpendicular, they will each expand by a factor less than $k$. So again the ratio area/perimeter has increased.

We now have what is required, since we have the triangle $ABC$ with its vertices on the boundary of $D$, and during the process its ratio of area/perimeter has only increased.

With a little more work one can show that in fact the actual max ratio occurs when the triangle $ABC$ is equilteral, with vertices on the boundary of $D$.

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jpo: If the triangle starts out with its vertices in the interior of the disk, then it may be expanded and still be in the disk. If $AB$ is the longest side, this may be done at least until one of $A,B$ ends up on the boundary. Then if the other end $B$ is not on the boundary one can move the triangle a little (so after this move the ends $A,B$ have moved toward $B$ a bit) and expand some more. If you imagine the triangle simultaneously expanding and staying in the disk, it would bump into an edge and if further expanded the other parts would move. – coffeemath Dec 6 '12 at 0:42

$\triangle ABC=\dfrac{r(a+b+c)}{2}=rs$ $~$ $\Longrightarrow$ $~$ $\dfrac{\triangle ABC}{s}=r$
Euler theorem : $~$ $\underline{OI^{2}=R^{2}-2Rr}\geq 0$ $~$ $\Longrightarrow$ $~$ $r\leq\dfrac{R}{2}$ , which equality hold at $OI=0$ , $R=2r$
That is, it's equilateral.

Now, for the ratio $R$ ($\triangle ABC$ circumradius) to be maximum, the circumcircle ($\triangle ABC$) must be the boundary disk. $$$$ I add a little detailed.

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