In the plane, there are two unit-radius, tangent circles: $k$ and $k_{1}$. One of their common external tangents is the line $e$. Then we draw the circles $k_{2}, k_{3}, \ldots, k_{n}$ in sequence such that each of them is tangent to $k$, $e$, and the circle with the number one less. What is the radius of the $k_{n}$ circle?
Let $k$ and $k_{1}$ have centers $O, O_{1}$, and their points of tangency with $e$ and with each other be $E, E_{1}$, and $T_{1}$, respectively. It is clear that $k_{2}$ is located in the region bounded by the segment $E E_{1}$ and the quarter-circle arcs $E T_{1}$ and $E_{1} T_{1}$, and it is externally tangent to $k$ and $k_{1}$. The center $O_{2}$ of $k_{2}$ lies on the perpendicular bisector of the segment $E_{1} E$. Denoting the radius of $k_{2}$ by $r_{2}$, the points of tangency of $k_{2}$ with $k$ and $e$ by $T_{2}$ and $E_{2}$, respectively, and the projection of $O_{2}$ onto $O E$ by $O_{2}^{\prime}$, from the right triangle $O_{2} O_{2}^{\prime}$ we have

$$
O O_{2}^{2}=O_{2} O_{2}^{\prime 2}+O O_{2}^{\prime 2}=E_{2} E^{2}+\left(O E-O_{2}^{\prime} E\right)^{2}
$$

which simplifies to

$$
\left(1+r_{2}\right)^{2}=1+\left(1-r_{2}\right)^{2}
$$

and thus $r_{2}=1 / 4$.

It is clear that for $k_{3}$, we need to consider the circle located in the region bounded by the segment $E E_{2}$ and the arcs $E T_{2}$ and $E_{2} T_{2}$, and that the circles $k_{4}, k_{5}, \ldots, k_{n}$ are all different from the previous ones. Let the two consecutive terms of the sequence of circles be $k_{i}$ and $k_{i+1}$, with radii $r_{i}, r_{i+1}$, points of tangency with $k$ and $e$ being $T_{i}, T_{i+1}, E_{i}, E_{i+1}$, centers $O_{i}, O_{i+1}$, and their projections onto $O E$ being $O_{i}^{\prime}$ and $O_{i+1}^{\prime}$, respectively, and the projection of $O_{i+1}$ onto $O_{i} E_{i}$ being $O_{i+1}^{\prime \prime}$ (the case $i=2$ is shown in the diagram). From the right triangles $O O_{i} O_{i}^{\prime}$, $O O_{i+1} O_{i+1}^{\prime}$, and $O_{i} O_{i+1} O_{i+1}^{\prime \prime}$, we have:

$$
E E_{i}=O_{i}^{\prime} O_{i}=\sqrt{O O_{i}^{2}-\left(O E-O_{i}^{\prime} E\right)^{2}}=\sqrt{\left(1+r_{i}\right)^{2}-\left(1-r_{i}\right)^{2}}=2 \sqrt{r_{i}}
$$

Similarly,

$$
E E_{i+1}=2 \sqrt{r_{i+1}}
$$

Thus,

$$
E_{i+1} E_{i}=O_{i+1} O_{i+1}^{\prime \prime}=\sqrt{O_{i} O_{i+1}^{2}-O_{i} O_{i+1}^{\prime \prime 2}}=\sqrt{\left(r_{i}+r_{i+1}\right)^{2}-\left(r_{i}-r_{i+1}\right)^{2}}=2 \sqrt{r_{i} r_{i+1}}
$$

From $E E_{i}=E E_{i+1}+E_{i+1} E_{i}$, we get

$$
2 \sqrt{r_{i}}=2 \sqrt{r_{i+1}}+2 \sqrt{r_{i} r_{i+1}}
$$

Dividing both sides by $2 \sqrt{r_{i} r_{i+1}}$, we obtain

$$
\frac{1}{\sqrt{r_{i+1}}}=\frac{1}{\sqrt{r_{i}}}+1
$$

From $r_{2}=1 / 4$, we get $r_{3}=1 / 9$, and from this, $r_{4}=1 / 16$. The conjecture

$$
r_{i}=\frac{1}{i^{2}}, \quad \frac{1}{\sqrt{r_{i}}}=i
$$

can be easily proven by induction; if (4) holds, then from (3) we have

$$
\frac{1}{\sqrt{r_{i+1}}}=i+1, \quad \text{so} \quad r_{i+1}=\frac{1}{(i+1)^{2}}
$$

This completes the solution to the problem.