Will Euler help?

Let the circumcentre of the triangle be the origin of coordinates. Then $A,B,C$ have position vectors $\mathbf{a},\mathbf{b},\mathbf{c}$ , all of modulus $R$ , the outradius, and the centroid and incentre have position vectors $\mathbf{g} \; = \; \tfrac13(\mathbf{a} + \mathbf{b} + \mathbf{c}) \hspace{2cm} \mathbf{i} \; = \; \tfrac{1}{a+b+c}(a\mathbf{a} + b\mathbf{b} + c\mathbf{c})$ so that $3(a+b+c)(\mathbf{g} - \mathbf{i}) \; = \; (b+c-2a)\mathbf{a} + (a+c-2b)\mathbf{b} + (a + b - 2c)\mathbf{c}$ Note that $\mathbf{a} \cdot \mathbf{a} = \mathbf{b} \cdot \mathbf{b} = \mathbf{c} \cdot \mathbf{c} = R^2$ , while $\mathbf{a} \cdot \mathbf{b} \; = \; R^2 \cos2C \; = \; R^2(1 - 2\sin^2C) \; = \; R^2 - \tfrac12c^2$ and, similarly $\mathbf{a} \cdot \mathbf{c} = R^2 - \tfrac12b^2$ and $\mathbf{b} \cdot \mathbf{c} = R^2 - \tfrac12a^2$ . Thus $\begin{aligned} 9(a+b+c)^2\Vert \mathbf{g} - \mathbf{i}\Vert^2 & = \; \begin{array}{l} (b+c-2a)^2R^2 + (a+c-2b)^2R^2 + (a+b-2c)^2R^2 \\ {} + (b+c-2a)(a+c-2b)(2R^2 - c^2) + (b+c-2a)(a+b-2c)(2R^2 - b^2) + (a+b - 2c)(a+c-2b)(2R^2 - a^2) \end{array} \\ & = \; \begin{array}{l} R^2\big(a + b - 2c + a + c - 2b + b + c - 2a\big)^2 \\ {} - a^2(a+b-2c)(a+c-2b) - b^2(a+b-2c)(b+c-2a) - c^2(a+c-2b)(b+c-2a) \end{array} \\ & = \; - a^2(a+b-2c)(a+c-2b) - b^2(a+b-2c)(b+c-2a) - c^2(a+c-2b)(b+c-2a) \\ & = \; -(a + b + c)\Big(\big(a^3 + b^3 + c^3\big) - 2\big(a^2b + a^2c + ab^2 + b^2c + ac^2 + bc^2\big) + 9abc\Big) \\ & = \; \tfrac13(a+b+c)\big[2(a+b+c)^3 - 5(a^3 + b^3 + c^3) - 39abc\big] \end{aligned}$ using the identity $(a+b+c)^3 \; = \; (a^3+b^3+c^3) + 3(a^2b+a^2c+ab^2+b^2c+ac^2+bc^2) + 6abc$ Thus $\Vert \mathbf{g} - \mathbf{i} \Vert \; = \; \sqrt{\frac{2(a+b+c)^3 - 5(a^3 + b^3 + c^3) - 39abc}{27(a+b+c)}}$ making the answer $2 + 5 + 39 + 27 = \boxed{73}$ .