G18 (6-3, Yugoslavia) In a triangle $\triangle A B C$ with side lengths $a, b, c$, construct the incircle and draw three tangents to this circle, each parallel to one of the sides of the given triangle. These tangents intersect the given triangle, forming three new triangles. Construct the incircles of these three new triangles. Find the sum of the areas of the four circles.

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This translation preserves the original format and structure of the text, including the problem statement and the specific details about the geometric construction.
Consider Figure 4-18, let the radii of the four inscribed circles be $r, r_{a}, r_{b}, r_{c}$. Considering $\triangle A P Q$, since $\triangle A P Q \backsim \triangle A B C$, we have
$$
\frac{h_{a}-2 r}{h_{a}}=\frac{r_{a}}{r},
$$

where $h_{a}$ is the altitude of $\triangle A B C$ on side $B C$.
Therefore, the sum of the areas is
$$
\begin{aligned}
& \pi r^{2}+\pi r^{2} \sum\left(\frac{h_{a}-2 r}{h_{a}}\right)^{2} \\
= & \pi r^{2}\left(1+\sum\left(\frac{\frac{2 S_{\triangle}}{a}-\frac{2 S_{\triangle}}{p}}{\frac{2 S_{\triangle}}{a}}\right)^{2}\right) \\
= & \pi r^{2}\left(1+\sum\left(\frac{p-a}{p}\right)^{2}\right) \\
= & \frac{\pi r^{2}}{p^{2}}\left(a^{2}+b^{2}+c^{2}\right) \\
= & \frac{\pi}{p^{3}}(p-a)(p-b)(p-c)\left(a^{2}+b^{2}+c^{2}\right) .
\end{aligned}
$$

where $S_{\triangle}$ is the area of $\triangle A B C$, and $p=\frac{1}{2}(a+b+c)$.