Minimum sum of areas

Let the center and radius of the left circle be $O$ and $r$ and those of the right circle be $P$ and $R$ . $OD$ and $PE$ be perpendicular to $BC$ , and $OF$ perpendicular to $PE$ . Then the length of $BC$ is given by:

$\begin{aligned} BD+DE+EC & = BC \\ OD \cdot \cot \frac B2 + OF + PE \cdot \cot \frac C2 & = BC \\ OD \cdot \cot \frac B2 + \sqrt{OP^2-PF^2}+ PE \cdot \cot \frac C2 & = BC \\ r \cot \frac B2 + \sqrt{(R+r)^2 - (R-r)^2} + R\cot \frac C2 & = 15 \\ R \cot \frac C2 + 2\sqrt{rR} + r\cot \frac B2 - 15 & = 0 \end{aligned}$

The above equation is a quadratic equation for $\sqrt R$ , and we can express $R$ in terms of $r$ as follows.

$\begin{aligned} \sqrt R & = \frac {\sqrt{4r - 4 \cot \frac C2 \left(r\cot \frac B2 - 15\right)}-2\sqrt r}{2\cot \frac C2} \end{aligned}$

Given the side lengths of $\triangle ABC$ , we can find the measures of angles $A$ , $B$ , and $C$ . Therefore we can compute $R$ for a particular $r$ and hence $S(r) =\pi(r^2+R^2)$ , and find $S_{\min} = \min (S(r))$ . I used an Excel spreadsheet to compute $R$ and $S(r)$ and found $S_{\min} \approx 24.46086344$ , when $r \approx 1.650890917$ . Therefore $10^5S_{\min} = \boxed{2446086}$ .