Minimising $HI$

The distance $\overline{HI}$ between the orthocentre $H$ and the incentre $I$ is given by the formula $\overline{HI}^2 \; = \; 2r^2 - 4R^2\cos A \cos B \cos C$ where $r,R$ are the inradius and the outradius, respectively. Putting $u = \sin\tfrac12A$ , $x= \cos\tfrac12(B-C)$ , we see that $0 < u < x < 1$ , and $\begin{aligned} \overline{HI}^2 & = \; 4R^2\big[(1-\cos A)(1-\cos B)(1-\cos C) - \cos A \cos B \cos C\big] \\ & = \; 4R^2\big[(1 - \cos A) - (1 - \cos A)(\cos B + \cos C) + (1 - 2\cos A)\cos B \cos C\big] \\ & = \; 4R^2\big[2u^2 - 4u^2 \cos\tfrac12(B+C)\cos\tfrac12(B-C) + \tfrac12(4u^2-1)\big(\cos(B+C) + \cos(B-C)\big)\big] \\ &= \; 4R^2\big[2u^2 - 4u^3x + (4u^2-1)(x^2 + u^2 - 1)\big] \end{aligned}$ In this case, $R=1$ , and hence $\overline{HI}^2 \; = \; F_u(x) \; = \; 4\big[(4u^2-1)(x^2 + u^2 - 1) - 4u^3x + 2u^2\big]$ We thus want to minimize $F_u(x)$ over $u < x < 1$ for each $0 < u < 1$ . Note that $F_u'(x) \; =\; 8\big[(4u^2-1)x - 2u^3\big]$ Note that there is a unique positive root $0 < \beta < 1$ , with $\beta \approx 0.596968283237$ , of the cubic equation $2X^3 - 4X^2 + 1 = 0$ . Now

• If $0 < u < \tfrac12$ then $F_u'(x) < 0$ for all $u < x < 1$ , and so the infimum over $u < x < 1$ of $F_u(x)$ is $F_u(1)$

• If $\tfrac12 < u < \beta$ , then $F_u'(x) = 0$ where $x = \hat{u} = \tfrac{2u^3}{4u^2-1}$ , but the fact that $u < \beta$ means that $\hat{u} > 1$ , and hence the infimum of $F_u(x)$ for $u < x < 1$ is $F_u(1)$ again.

• If $\beta < u < \tfrac{1}{\sqrt{2}}$ , then $F_u$ has a turning point at $\hat{u}$ defined as above, and $u < \hat{u} < 1$ , so the infimum of $F_u$ over $u < x < 1$ is $F_u(\hat{u})$ .

• If $\tfrac{1}{\sqrt{2}} < u < 1$ then $\hat{u} < u$ , and so the infimum of $F_u(x)$ over $u < x < 1$ is $F_u(u)$ .

Thus we deduce that $\overline{HI}^2_{\min} \; = \; \left\{ \begin{array}{lll} F_u(1) \; = \; 4u^2(1 - 2u)^2 & \hspace{1cm} & 0 < u < \beta \\[2ex] F_u(\hat{u}) = {\displaystyle \frac{4(1-2u^2)^2(3u^2-1)}{4u^2-1}} & & \beta < u < \tfrac{1}{\sqrt{2}} \\[2ex] F_u(u) \;=\;4(2u^2-1)^2 & & \tfrac{1}{\sqrt{2}} < u < 1 \end{array}\right.$ and hence $\overline{HI}_{\min} \; = \; \left\{ \begin{array}{lll} 2\sin\tfrac12A\big|1 - 2\sin\tfrac12A\big| & \hspace{1cm} & 0 < A < 2\sin^{-1}\beta \\[2ex] 2\cos A{\displaystyle \sqrt{\frac{3\sin^2\frac12A - 1}{4\sin^2\frac12A - 1}}} & & 2\sin^{-1}\beta < A < \tfrac12\pi \\[2ex] 2|\cos A| & & \tfrac12\pi < A < \pi \end{array} \right.$ Thus we calculate (numerically) $\alpha = \int_0^{2\sin^{-1}\beta} 2\sin\tfrac12A\big|1 - 2\sin\tfrac12A\big|\,dA + \int_{2\sin^{-1}\beta}^{\frac12\pi} 2\cos A{\displaystyle \sqrt{\frac{3\sin^2\frac12A - 1}{4\sin^2\frac12A - 1}}}\,dA \; =\; 0.2468923537$ which makes $\lfloor 10^6\alpha \rfloor = \boxed{246892}$ .