Maximizing the area!

Let $a = BC, b=CA, c=AB$ , as usual. It is a well-known property that $\angle AIO = \frac{\pi}{2}$ if and only if $2a=b+c$ . Therefore it follows that the area S of the triangle satisfying $\angle AIO = \frac{\pi}{2}$ is given by:

$16S^2 = (a+b+c)(a+b-c)(b+c-a)(c+a-b) = 3a^2(3a-2c)(2c-a)$

Moreover $a$ must satisfy $a> |b-c|$ , which is equivalent to requiring that $2a-2c<a, 2c-2a<a$ . It follows that:

$\dfrac{2}{3}c < a < 2c$

As $c=1$ , let $f(x) = x^2(3x-2)(2-x)$ . The derivative of $f$ is then:

$f'(x) = -12x \left(x - \dfrac{3+\sqrt{3}}{3} \right) \left(x - \dfrac{3-\sqrt{3}}{3} \right)$

Since $0<\dfrac{3-\sqrt{3}}{3}<\dfrac{2}{3}<\dfrac{3+\sqrt{3}}{3}<2$ , we see that $f$ reaches its maximum on $\left(\dfrac{2}{3},2 \right)$ when $x = \dfrac{3+\sqrt{3}}{3}$ .

As a result $S \leq S_m$ , where:

$S_m = \dfrac{\sqrt{3}}{4}\left(f\left(\dfrac{3+\sqrt{3}}{3}\right)\right)^{1/2} = \left( \dfrac{3+\sqrt{3}}{3} \right)\dfrac{\sqrt[4]{3}}{2\sqrt{2}} = \dfrac{\sqrt{9+6\sqrt{3}}}{6}$

To complete the solution, we will show that the value of $S_m$ is the area of the actual triangle $ABC$ constructed on the side $AB$ and satisfying $\angle AIO = \dfrac{\pi}2$ , with $AB = c= 1$ . Let $a = \dfrac{3+\sqrt{3}}{3}$ and $b=2a-1$ . Clearly $a<b-1$ and $a>|b-c|$ (because $a \in \left(\dfrac{2}{3},2 \right)$ ). Thus we can construct a triangle $\angle AIO = \dfrac{\pi}{2}$ ( because $2a = b+c$ ), and $S=S_m$ ( because $a = \dfrac{3+\sqrt{3}}{3}$ ). This triangle therefore achieves the maximal area $S_m$ while satisfying all the required constraints.

So, in $S_m = \boxed{\dfrac{\sqrt{9+6\sqrt{3}}}{6}}$ ,

$A=9, B=6, C=3, D=6, A+B+C+D=\boxed{24}$