Geometry With Calculus 2? Maybe

$d = 2 \sqrt{3} \implies r = \sqrt{3}$

Now, let's say, you have a isosceles $\triangle$ with sides $a \ , \ a \ , \ b$ .

Denote $P = a + a + b = 2a + b$

Then, $Area_{ \triangle } = \sqrt{3} \cdot \cfrac{P}{2} = \sqrt{\cfrac{P}{2} \cdot \left( \cfrac{P}{2} - a \right ) \cdot \left( \cfrac{P}{2} - a \right ) \cdot \left( \cfrac{P}{2} - b \right )}$

$\implies \cfrac{3P^2}{4} = \cfrac{P}{2} \cdot \left( \cfrac{P}{2} - a \right ) \cdot \left( \cfrac{P}{2} - a \right ) \cdot \left( \cfrac{P}{2} - b \right )$

By A.M. - G.M. ,

$\implies \cfrac{\cfrac{P}{2} + \left( \cfrac{P}{2} - a \right ) + \left( \cfrac{P}{2} - a \right ) + \left( \cfrac{P}{2} - b \right )}{4} = \cfrac{2P - P}{4} = \cfrac{P}{4} \geq \sqrt[4]{\cfrac{P}{2} \cdot \left( \cfrac{P}{2} - a \right ) \cdot \left( \cfrac{P}{2} - a \right ) \cdot \left( \cfrac{P}{2} - b \right )} = \sqrt[4]{\cfrac{3P^2}{4}}$

raising to fourth-power both sides, $\implies \cfrac{P^4}{256} \geq \cfrac{3P^2}{4}$

multiply both sides by $256 \cdot \cfrac{1}{P^2} \implies P^2 \geq 192$

$\therefore P \geq 8 \sqrt{3} \implies \boxed{A = 8 , B = 3 \implies A + B = 11 }$