The unit-radius circles $k_{1}$ and $k_{2}$ touch each other at point $P$. One of the common tangents of the two circles, which does not pass through $P$, is the line $e$. For $i>2$, let $k_{i}$ be the circle different from $k_{i-2}$ that touches $k_{1}$, $k_{i-1}$, and $e$. Determine the radius of $k_{1999}$.
If two circles and a line touch each other as shown in the first figure, the radii of the circles are $R$ and $r$, and the distance between the points of tangency on the line is $d$, then the centers of the circles and the points of tangency form a right trapezoid, for which - using the Pythagorean theorem - the relationship $d^{2}=4 R r$ holds.

Let the point of tangency of the circle $k_{i}$ and the line $e$ be $E_{i}$. From the definition of the circle $k_{i}$, it follows that for $i>2$, $E_{i}$ is an interior point of the segment $E_{1} E_{i-1}$. Therefore,

$$
E_{1} E_{i}=E_{1} E_{i+1}+E_{i+1} E_{i}
$$

Thus, the distance between the points of tangency, expressed using the radii of the circles, and using the result proven in the previous paragraph,

$$
\sqrt{4 r_{1} r_{i}}=\sqrt{4 r_{1} r_{i+1}}+\sqrt{4 r_{i+1} r_{i}}
$$

and therefore

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

We will prove by complete induction that this relationship implies $\sqrt{r_{i+1}}=\frac{1}{i}$. The statement is true for $i=1$, because $r_{2}=1$. Suppose the statement is true for $2 \leq i \leq k$. Then

$$
\sqrt{r_{k+1}}=\frac{\sqrt{r_{1} \cdot r_{k}}}{\sqrt{r_{1}}+\sqrt{r_{k}}}=\frac{1 \cdot \frac{1}{k-1}}{1+\frac{1}{k-1}}=\frac{1}{1+(k-1)}=\frac{1}{k}
$$

Therefore, $\sqrt{r_{i+1}}=\frac{1}{i}$ is true for all $i$, so $r_{1999}=\frac{1}{1998^{2}}$.

Gergely Kiss (Fazekas M. Főv. Gyak. Gimn., 12th grade)

![](https://cdn.mathpix.com/cropped/2024_05_02_25026db3e10a517736d3g-1.jpg?height=442&width=716&top_left_y=1134&top_left_x=726)

$d$

![](https://cdn.mathpix.com/cropped/2024_05_02_25026db3e10a517736d3g-1.jpg?height=460&width=727&top_left_y=1624&top_left_x=707)