Triangle Centers

Using the circumcentre $0$ as the origin, if $A,B,C$ have position vectors $\mathbf{a},\mathbf{b},\mathbf{c}$ respectively, then the position vectors of the centroid $G$ and the incentre $I$ are $\overrightarrow{OG} = \tfrac13(\mathbf{a}+\mathbf{b}+\mathbf{c}) \hspace{2cm} \overrightarrow{OI} \; = \; \tfrac{1}{a+b+c}(a\mathbf{a} + b\mathbf{b} + c\mathbf{c})$ and hence $\overrightarrow{OI} \times \overrightarrow{OG} \; = \; \frac{1}{3(a+b+c)}\Big[(a-b)\mathbf{a} \times \mathbf{b} + (b-c)\mathbf{b} \times \mathbf{c} + (c-a)\mathbf{c} \times \mathbf{a}\Big]$ Thus the area of $OIG$ is (here $R$ is the radius of the circumcircle): $\begin{aligned} |OIG| & = \; \left|\frac{R^2}{6(a+b+c)}\big[(a-b)\sin2C + (b-c)\sin2A + (c-a)\sin2B\big]\right| \\ & = \; \frac{R}{6(a+b+c)}\Big|(a-b)c\cos C + (b-c)a\cos A + (c-a)b \cos B\Big| \\ & = \; \frac{abc}{24(a+b+c)\Delta}\left|(a-b)c\frac{a^2+b^2-c^2}{2ab} + (b-c)a\frac{b^2 + c^2 - a^2}{2bc} + (c-a)b\frac{a^2 + c^2 - b^2}{2ac}\right| \\ & = \; \frac{1}{48(a+b+c)\Delta}\Big| (a-b)c^2(a^2+b^2-c^2) + (b-c)a^2(b^2+c^2-a^2) + (c-a)b^2(a^2+c^2-b^2)\Big| \\ & = \; \tfrac{1}{12}\big|(b-a)(c-a)(c-b)\big| \sqrt{\frac{a+b+c}{(-a+b+c)(a-b+c)(a+b-c)}} \end{aligned}$ where $\Delta = \tfrac14\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$ is the area of the triangle $ABC$ . Thus $|OIG|$ has been expressed in terms of the sides of the triangle $ABC$ . Maximising this expression over the (finite) set of possible edge lengths $\big\{(a,b,c) \in \mathbb{N}^3 \, \big| \, 1 \le a \le b \le c \le 10\,,\,c < a+b\big\}$ , we see that the largest area of $OIG$ is $\frac{7\sqrt{7}}{6}$ , when $a=3$ , $b=8$ , $c=10$ .

This makes the answer $7-6 = \boxed{1}$ .