Let $T$ be a triangle with area 1 . We let $T_{1}$ be the medial triangle of $T$, i.e. the triangle whose vertices are the midpoints of sides of $T$. We then let $T_{2}$ be the medial triangle of $T_{1}, T_{3}$ the medial triangle of $T_{2}$, and so on. What is the sum of the areas of $T_{1}, T_{2}, T_{3}, T_{4}, \ldots$ ?
In general, the medial triangle has side length half the original triangle, hence $\frac{1}{4}$ the area. Thus, $T_{1}$ has area $(1 / 4)$; then $T_{2}$ has area $(1 / 4)^{2}$, and so on. Thus the answer is

$$
\frac{1}{4}+\left(\frac{1}{4}\right)^{2}+\left(\frac{1}{4}\right)^{3}+\cdots=\frac{1 / 4}{1-1 / 4}=\frac{1}{3}
$$