8. [5] The incircle $\omega$ of equilateral triangle $A B C$ has radius 1. Three smaller circles are inscribed tangent to $\omega$ and the sides of $A B C$, as shown. Three smaller circles are then inscribed tangent to the previous circles and to each of two sides of $A B C$. This process is repeated an infinite number of times. What is the total length of the circumferences of all the circles?
Answer:
$5 \pi$
Solution: One can find using the Pythagorean Theorem that, in each iteration, the new circles have radius $1 / 3$ of that of the previously drawn circles. Thus the total circumference is $2 \pi+3 \cdot 2 \pi\left(\frac{1}{1-1 / 3}-1\right)=$ $5 \pi$.