Dealing with centres

By cosine rule: $\cos{(\frac{\pi}{6})}=\frac{2^2+(\sqrt{3})^2-a^2}{2 \times 2 \times \sqrt{3}} \Rightarrow a=1$

Area of triangle is $\frac{1}{2} b c \sin{(A)}=\frac{\sqrt{(3)}}{2}$

The area of a triangle is also $\frac{abc}{4R}=(\frac{a+b+c}{2})r=\frac{\sqrt{(3)}}{2} \Rightarrow R=1,r=\frac{\sqrt{3}}{3+\sqrt{3}}$

By Euler's Formula: $OI^2=R(R-2R)=1 \times (1-\frac{2\sqrt{3}}{3+\sqrt{3}})=\frac{3-\sqrt{3}}{3+\sqrt{3}}$

$OI^2=\frac{12-6 \sqrt{3}}{6}=2-\sqrt{3} \Rightarrow OI=\sqrt{(2-\sqrt{3})}$

So $a=2,b=3 \Rightarrow a+b=5$

Note : $1^2+(\sqrt{3})^2=2^2$ so the triangle is right-angled. This means the circumcentre lies at the midpoint of one of the sides. I wonder if this could have a more elegant solution.

This is a 30-60-90 triangle, with legs a=1, c= $\sqrt3$ , and the hypotenuse b=1. So circumradius, R = 1. Area = $\dfrac{\sqrt3} 2$ . Semiperimeter, s = $\dfrac{1+2+\sqrt3} 2.~~\therefore ~ inradius, r= \dfrac{\frac{\sqrt3} 2}{\dfrac{1+2+\sqrt3} 2}=\dfrac{\sqrt3 - 1} 2.$
By Euler's Formula: the distance between the centers, $d=\sqrt{R(R - 2r)}=\sqrt{1(1 - 2r)}= \sqrt{1 - (\sqrt3 - 1)}\\ \sqrt{2 - \sqrt3}.~~So~~a+b=5$