Where's The Centre?

Here is a solution using complex numbers.Let point A be the origin in the argand plane.We denote cos(x)+isin (x)=cis (x).In triangle ABC ,rotation about A gives c=bcis (pi/3) where b and c are complex numbers corresponding to points B and C in the complex plane.Note that we take the angle of rotation to be positive in case of counter clockwise rotation.Suppose circumcenter of triangle BIG is denoted by z.We have angle BOG=2 times angle BIG=60 degrees.In triangle BOG,rotation about O gives (g-z)=(b-z)cis (pi/3).Rearranging and using c=bcis (pi/3),we get (g-c)=z (1-cis (pi/3)).Taking modulus on both sides,we get |g-c|=|z||(1-cis (pi/3)|.But |1-cis (pi/3)|=1.Hence,we get |z|=|g-c|=length of line segment CG.But |z|=length of line segment AO as A is the origin.Therefore,we conclude that lengths of line segments AO and CG are congruent.Now ,triangle BGC is a 30-60-90 triangle,hence we get BG=half of BE=4.5.Hence CG=BC-BG=6-4.5=1.5.Therefore AO=1.5.Hence we have the answer as 8 times 1.5=12.

Draw a line passes point $I$ and perpendicular to $IG$ , it cuts the line that passes point $B$ perpendicular to $BG$ at $D$ . Then $BGID$ is concylic, and the center $O$ of the circumcircle is the midpoint of $GD$ . Looks like $O$ lies on $AB$ , we're going to prove it.

$\angle OBD = \angle GDB = \angle GIB = 30^\circ$ ; therefore, $\angle GBA = 60^\circ$ , which means $AOB$ is a straight line (prove complete!)

$\angle GDB = \angle GEB = 30^\circ$ ; also, $EG$ is parallel to $BD$ , so we have $EGDB$ is a parallelogram. Then, $DG = BE = 9$ or $OB = 4.5$ . 'Cause $AOB$ is a straight line, $AO = AB - OB = 6 - 4.5 = 1.5$ , which gives the answer $\boxed{12}$ .

Bonus: In the diagram, looks like circle $(O)$ and circle $(A,AB)$ are tangent at $B$ . Can you prove it?

$\angle BIG=30^{\circ} \Rightarrow \angle BOG=60^{\circ}$ (angle at centre is twice angle at radius)

$OG=OB \Rightarrow \angle OGB=\angle OBG=60^{\circ}$ so $\triangle OBG$ is equilateral.

$\angle GBO=60^{\circ} \Rightarrow O$ is on line $AB$ because $\angle CBA=60^{\circ}$

$\angle EBG=60^{\circ} \Rightarrow GB=9 \cos{(60^{\circ})}=\frac{9}{2}$ (external bisector+sine rule)

$BO=GB=\frac{9}{2}$ as $\triangle BOG$ is equilateral.

$AO=AB-BO=6-\frac{9}{2}=\frac{3}{2}$ because $AOB$ is a straight line.

$8 \times AO=8 \times \frac{3}{2} =12$

$Fig. ~30-60-90~shows~the ~relations ~in ~a ~30-60-90 ~\Delta, ~used ~below.\\ \angle ~GBE=\frac 1 2 *(180 - 60)=60. ~\Delta ~ is ~ 30-60-90 . ~\therefore ~ GB=9/2 ~and ~CG=3/2,\\ and ~\color{#3D99F6}{ \angle ~ CFG =30}. \\ But~ GCF~is ~ also ~30-60-90 .~~\therefore~CF=3,~ \implies~F ~ is ~ the ~midpoint~ of ~ AC.\\ \therefore ~BF ~ is~ \bot ~bisector ~ as ~ well ~ as ~ angle ~ bisector ~ of ~ AC. \\ \implies ~ \angle ~ IFG=\angle ~ IFC + \angle ~ CFG=120 ~ and ~ \angle ~ IBG=30.\\ \implies ~\Delta ~BIG ~ is ~ isosceles. ~ Its~circumradius ~ R=\dfrac{product~ of~ sides}{4*area.}.\\ R=\dfrac{(\frac 9 2)^2*(\frac 9 2 *\sqrt3) }{4*\frac 1 2 *(\frac 9 2 *\sqrt3)*(\frac 1 2 *\frac 9 2)}= 9/2=GO.\\ GO=GB, ~ \angle ~GBA=60, ~\implies~\Delta ~ GOB ~ is ~ equilateral ~\therefore ~OB=9/2.\\ \implies ~AO=3/2. ~~\therefore~8*AO=\Large ~~~~~~\color{#D61F06}{12}$