Minimising $GI$

The distance $\overline{GI}$ is given by the formula $9\overline{GI}^2 \; = \; 5r^2 - 16Rr + s^2$ where $r,R,s$ are the inradius, outradius and semiperimeter respectively. If we set $u = sin\tfrac12A$ and $x = \cos\tfrac12(B-C)$ , then we have $0 < u < x < 1$ and $\begin{aligned} r & = \; 4R\sin\tfrac12A\sin\tfrac12B\sin\tfrac12C \; = \; 2Ru\big[\cos\tfrac12(B-C) - \cos\tfrac12(B+C)\big] \; =\; 2Ru(x-u) \\ s & = \; R(\sin A + \sin B + \sin C) \; = \; R\big(\sin A + 2\sin\tfrac12(B+C)\cos\tfrac12(B-C)\big) \; =\; R\big(\sin A + 2\cos\tfrac12A x\big) \; = \; 2R\sqrt{1-u^2}(x+u) \end{aligned}$ and hence, since $R=1$ in this case, $9\overline{GI}^2 \; =\; F_u(x) \; =\; 20u^2(x-u)^2 - 32u(x-u) + 4(1-u^2)(x+u)^2$ We wish to minimize $F_u(x)$ over the range $u < x < 1$ . Since $F_u'(x) \; =\; 8\big[(1 + 4u^2)x - 3u(1 + 2u^2)\big]$ we see that $F_u$ is minimized globally at $x = \hat{u}$ , where $\hat{u} \; = \; \frac{3u(1 + 2u^2)}{1 + 4u^2}$ It is easy to see that $\hat{u} > u$ for all $0 < u < 1$ . There is a unique real root $\beta \approx 0.420381$ of the cubic equation $6X^3 - 4X^2 + 3X - 1 = 0$ , and we deduce that $\hat{u} < 1$ if $0 < u < \beta$ , while $\hat{u} > 1$ if $\beta < u < 1$ . Thus $\min_{u < x < 1} F_u(x) \; = \; \left\{ \begin{array}{lll} F_u(\hat{u}) \; = \; {\displaystyle \frac{16u^4(1 - 5u^2)}{1 + 4u^2}} & \hspace{1cm} & 0 < u < \beta \\[2ex] F_u(1) \; = \; 4(1-u)^2(2u-1)^2 & & \beta < u < 1 \end{array}\right.$ Thus the minimum value of $\overline{GI}$ is $\overline{GI}_{\mathrm{min}} \; = \; \left\{ \begin{array}{lll} {\displaystyle\tfrac43 \sin^2\tfrac12A \sqrt{\frac{1 - 5\sin^2\frac12A}{1 + 4\sin^2\frac12A}}} &\hspace{1cm} & 0 < A < 2\sin^{-1}\beta \\[2ex] \tfrac23(1 - \sin\tfrac12A)\big|2\sin\frac12A - 1\big| & & 2\sin^{-1}\beta < A < \pi \end{array} \right.$ Thus we deduce that $k=3$ , $\ell = 1$ , $m = 5$ , $n=4$ and $\alpha = 2\sin^{-1}\beta \approx 0.86773$ . This makes the answer $3+1+5+4=\boxed{13}$ .