Locate The Circumcircle

Let $\mathbf{a} = \overrightarrow{IA}$ , $\mathbf{b} = \overrightarrow{IB}$ , $\mathbf{c} = \overrightarrow{IC}$ , $\mathbf{q} = \overrightarrow{OI}$ . Then $|\mathbf{a}| = |\mathbf{b}| = |\mathbf{c}| = R$ , the outradius.

Then $\mathbf{a}+\mathbf{b}$ is a vector perpendicular to $\overrightarrow{AB}$ , and it follows that $\overrightarrow{IM} = \mathbf{a}+\mathbf{b}$ . Similarly $\overrightarrow{IN} = \mathbf{a}+\mathbf{c}$ . In addition $\tfrac13(\mathbf{a}+\mathbf{b} + \mathbf{c}) = \overrightarrow{IG}$ . Thus $\begin{aligned} \mathbf{a} + \mathbf{b} & = \; \binom{0}{1} - \mathbf{q} \\ \mathbf{a} + \mathbf{c} & = \; \binom{4}{1} - \mathbf{q} \\ \mathbf{a} + \mathbf{b} + \mathbf{c} & = \; \binom{8}{0} - 3\mathbf{q} \end{aligned}$ and hence $\mathbf{a} \; = \; \binom{-4}{2} + \mathbf{q} \hspace{1cm}\mathbf{b} \; = \; \binom{4}{-1} - 2\mathbf{q} \hspace{1cm}\mathbf{a} \; = \; \binom{8}{-1} - 2\mathbf{q}$ Since $(2,-1)$ lies on $BC$ , we deduce that $\binom{2}{-1} - \mathbf{q} \; = \; \lambda\mathbf{b} + (1-\lambda)\mathbf{c} \; = \; \binom{8-4\lambda}{-1} - 2\mathbf{q}$ for some $\lambda$ , and hence that $\mathbf{q} = \binom{\alpha}{0}$ for some $\alpha$ . Since $\mathbf{a},\mathbf{b},\mathbf{c}$ must all have the same length (the outradius), we deduce that $\alpha = 3$ .

Thus we deduce the coordinates $A\;(2,2)$ , $B\;(1,-1)$ , $C\;(5,-1)$ and $I\;(3,-0)$ , with outradius $\sqrt{5}$ . Thus the answer is $\boxed{305}$ .

Note: The above graph is now drawn to scale.

The equation of line $BC$ which contains point $(2;-1)$ and is parallel to line $MN$ is $y=-1$ .

Let point $H$ and $E$ be the midpoints of $MN$ and $BC$ respectively.

$H$ has the coordinate of $(2;1)$ .

We can easily prove that $AHEI$ is a parallelogram.

Let $F$ be the midpoint of $IE$ .

Because $AG=2GE$ , combined with $AHEI$ being a parallelogram, we can deduce that $HG=2GF$ .

Using $HG=2GF$ with coordinates of points $H$ and $G$ being known, we can find the coordinate of point $F$ .

We can find the the equation of line $EF$ which contains point $F$ and is perpendicular to line $BC$ .

Then find the coordinate of point $E$ which is the intersection of lines $EF$ and $BC$ .

Using the equation $IF=FE$ , we can find the coordinate of point $I$ as $(3;0)$ and $IA=HE=\sqrt{5}$ which is also the radius.

In conclusion, we can submit the answer as $\boxed{\overline{305}}$ .

Let A:(x1,y1), B:(x2,y2) & C:(x3,y3). Then x1+x2=a, y1+y2=b+1, x1+x3=a+4, y1+y3=b+1, x1+x2+x3=8, y1+y2+y3=0 These equations, yield A:[(2a-4),(2b+2)], B:[(4-a),(-b-1)] and C:[(8-a),(-b-1)]. Since I:(a,b) is the circum-centre with radius = √c, we can further write the following equations: ((2a-4)-a)^2+((2b+2)-b)^2=c -----(1) ((4-a)-a)^2+((-b-1)-b)^2=c --------(2) and ((8-a)-a)^2+((-b-1)-b)^2=c --------(3) **These yield, a=3, b=0 & c=5