Maximum Circumcircle

Call the triangle's area $T$ and its semiperimeter $s$ . Let its sidelengths be $a,b,c$ (as usual opposite vertices $A,B,C$ respectively).

We know $T$ and $s$ ; let's make $a$ a parameter. By definition, $b+c=2s-a$

Now, from Heron's formula, $\begin{aligned} s(s-a)(s-b)(s-c) &=T^2 \\ s(s-a)\left( s^2-(b+c)s+bc\right) &= T^2 \\ s(s-a)\left( s^2-(2s-a)s+bc\right) &= T^2 \\ s(s-a)( bc - s(s-a)) &= T^2 \\ bc &= s(s-a)+\frac{T^2}{s(s-a)} \end{aligned}$

Knowing $b+c$ and $bc$ means we can find $b$ and $c$ as the roots of a quadratic (see below).

The circumradius is given by $R=\frac{abc}{4T}=\frac{1}{4T} \left(as(s-a)+\frac{aT^2}{s(s-a)} \right)$

In this case, $s=8$ and $T=10$ ; so $R=\frac{1}{40} \left(8a(8-a)+\frac{25a}{2(8-a)} \right)$

Setting the derivative with respect to $a$ equal to zero (and tidying up) we get $4 a^3 - 80 a^2 + 512 a - 1049 =0$

This has three positive roots; the maximum value of $R$ is found at the root $a\approx 6.3792$ which gives $R=3.551815\ldots$ , and the answer $\boxed{3551815}$ .

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Additional consideration: OK, this is a "maximum"; but when $a=7.5$ , we get $R=5.4375$ ; so what's going on?

Recall we have a quadratic to find the sides $b$ and $c$ . Explicitly, $b$ and $c$ are the roots of $x^2-(2s-a)x+s(s-a)+\frac{T^2}{s(s-a)}=0$

We need both roots to be real; so the discriminant of this quadratic needs to be non-negative, ie $(2s-a)^2-4\left(s(s-a)+\frac{T^2}{s(s-a)}\right)\ge 0$

This simplifies to $a^2\ge \frac{4T^2}{s(s-a)}$

Now, if $a\ge s$ , we get $b+c=2s-a \le a$ , which breaks the triangle inequality (we can't form a triangle with these sidelengths). So $s-a$ is positive and $a^2 s(s-a) \ge 4T^2$

Solving this inequality in this case we find (approximately) $3.24<a<6.97$

If we let $a$ be the larger root here we find the same value of $R$ as above . So it is indeed the maximum (but I'm a bit surprised by this particular result!)

If the sides of the triangle are $a = v+w$ , $b = u+w$ and $c = u+v$ for $u,v,w > 0$ , then we have that $u+v+w = 8$ . Since $10 = \tfrac12bc\sin A$ we deduce that $\sin A = \tfrac{20}{(u+v)(u+w)}$ , and since $2R = \tfrac{a}{\sin A}$ we deduce that $R \; = \; \tfrac{1}{40}(u+v)(u+w)(v+w)$ Also note that $10 = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{uvw(u+v+w)} = \sqrt{8uvw}$ , so we have that $uvw = \tfrac{25}{2}$ . But then $R \; = \; \tfrac{1}{40}\big[8^3 -8^2(u+v+w) + 8(uv + uw + vw) - uvw\big] \; = \; \tfrac15(uv +uw + vw) - \tfrac{5}{16}$ and hence we want to minimize $R = \tfrac15(uv + uw + vw) - \tfrac{5}{16}$ subject to the constraints $u+v+w=8$ and $uvw = \tfrac{25}{2}$ . Using the method of Lagrange multipliers we are looking to solve the equations $\begin{aligned} \tfrac15(v+w) - \lambda - \mu vw & = \; 0 \\ \tfrac15(u+w) - \lambda - \mu uw & = \; 0 \\ \tfrac15(u+v) - \lambda - \mu uv & = \; 0 \end{aligned}$ and these equations imply that $(u-v)(\tfrac15 - \mu w) = (u-w)(\tfrac15 - \mu v) = (v-w)(\tfrac15 - \mu u) = 0$ . If $u,v,w$ were distinct, we would deduce that $u=v=w=\tfrac{1}{5\mu}$ , which is absurd. Since $(\tfrac83)^3 \neq \tfrac{25}{2}$ , we cannot have $u=v=w$ . Thus, without loss of generality, we have $v=u$ , $w = 8 - 2u$ , $\lambda =\tfrac15u$ , $\mu = \tfrac15u^{-1}$ , where $u^2(8-2u) = \tfrac{25}{2}$ .

There are two positive real roots to this cubic equation, $1.62077230$ and $3.48555973$ . The first corresponds to the minimum possible value of $R_{\mathrm{min}} = 3.29782965$ , while the second corresponds to the maximum possible value of $R_{\mathrm{max}} = 3.55181516$ , and hence $\lfloor 10^6 R_{\mathrm{max}}\rfloor = \boxed{3551815}$ .