Question on a function of triangles

Question

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?

Answer 1

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 $D$.

Answer 2

$\triangle ABC=\dfrac{r(a+b+c)}{2}=rs$ $~$ $\Longrightarrow$ $~$ $\dfrac{\triangle ABC}{s}=r$
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.