Distance between two centers

The legs of a $30-60-90$ right $\triangle$ with hypotenuse of $2$ are $\sqrt{3}$ and $1$ .

The radius of the incircle can be computed by the formula $r=\dfrac{2A}{P}$ where: $A$ = area of the $\triangle$ and $P$ = perimeter of the $\triangle$ , we find $r=\dfrac{\sqrt{3}-1}{2}$ .

In this given triangle, the location of the circumcenter is at the midpoint of the hypotenuse. From the figure, $x=\dfrac{1}{2}-r=\dfrac{2-\sqrt{3}}{2}$ . It follows that $y=\dfrac{\sqrt{3}}{2}-r=\dfrac{1}{2}$ . Applying the pythagorean theorem, we obtain

$h^2=x^2+y^2=\left(\dfrac{2-\sqrt{3}}{2}\right)+\left(\dfrac{1}{2}\right)^2=2-\sqrt{3}$

$\large\therefore$ $a=2$ and $b=3$

Finally, $1007a+b=1007(2)+3=2017$

Let $\triangle ABC$ be the right-triangle, with $C (0,0)$ , $A(1,0)$ and $B(\sqrt 3, 0)$ .

Let the incenter and inradius of $\triangle ABC$ be $I$ and $r$ respectively. $I$ is the meeting point of the internal angle bisectors. Therefore, we have:

$\begin{aligned} \frac r{\tan 30^\circ} + \frac r{\tan 45^\circ} & = AC \\ \frac r{\frac 1{\sqrt 3}} + \frac r1 & = 1 \\ \implies r & = \frac 1{\sqrt 3 + 1} = \frac {\sqrt 3-1}2 \end{aligned}$

Therefore, $I \left(\frac {\sqrt 3-1}2, \frac {\sqrt 3-1}2\right)$ .

Let the circumcenter be $O$ . It is the meeting point of the perpendicular line bisectors of $\triangle ABC$ . Therefore, $O \left(\frac {\sqrt 3}2, \frac 12\right)$ .

The square of the distance between $I$ and $C$ is:

$\begin{aligned} L^2 & = \left(\frac {\sqrt 3}2 - \frac {\sqrt 3-1}2\right)^2 + \left(\frac 12 - \frac {\sqrt 3-1}2\right)^2 \\ & = \left(\frac 12 \right)^2 + \left(1 - \frac {\sqrt 3}2\right)^2 \\ & = \frac 14 + 1 - \sqrt 3 + \frac 34 \\ & = 2 - \sqrt 3 \end{aligned}$

$\implies 1007a + b = \boxed{2017}$