3. Circles $k, l$ are both located inside rectangle $A B C D$, whose area is $72 \mathrm{~cm}^{2}$. Circle $k$ touches sides $C D, D A$ and $A B$, while circle $l$ touches sides $A B$ and $B C$. Determine the radii of circles $k$ and $l$, given that the radius of circle $k$ is expressed in centimeters as a whole number.
3. Let $r, s$ be the radii of the circles $k, l$ (in centimeters) and $K, L$ their points of tangency with side $AB$ (Fig. 2). Then $|AK| = r, |LB| = s$, and as we can easily calculate from the Pythagorean theorem (see also the 3rd problem of the school round in category $\mathrm{C}$)

$$
|KL| = \sqrt{(r+s)^2 - (r-s)^2} = 2 \sqrt{rs}.
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_05fdfd338c92e7f3acbdg-3.jpg?height=462&width=646&top_left_y=911&top_left_x=705)

Fig. 2

For the lengths of the sides of the rectangle $ABCD$, we have $|AD| = 2r, |AB| = r + 2\sqrt{rs} + s = (\sqrt{r} + \sqrt{s})^2$. According to the problem, it is given that

$$
2r(\sqrt{r} + \sqrt{s})^2 = 72
$$

or after dividing by two and taking the square root,

$$
r + \sqrt{rs} = 6.
$$

From this, it follows that $r < 6$, and for the size of the radius $s$, we get the expression

$$
\begin{aligned}
rs & = (6 - r)^2 \\
s & = \frac{(6 - r)^2}{r}
\end{aligned}
$$

From the conditions of the problem, it further follows that $s$ cannot be greater than $r$, otherwise the circle $l$ would not fit in the given rectangle, and since the circle $k$ must also fit in the given rectangle, it must be $|AB| \geq |AD| = 2r$. From the inequality $s \leq r$ according to (1), we get the condition $36 - 12r + r^2 \leq r^2$, i.e., $r \geq 3$. From the inequality $|AB| \geq 2r$, it follows that $72 = |AB| \cdot |AD| \geq 4r^2$, or $r^2 \leq 18$, which for integer $r$ means that $r \leq 4$. For the radius $r$, we thus get only two possibilities, $r \in \{3, 4\}$, and the corresponding values of the radius $s$ can be calculated from relation (1).

The problem has exactly two solutions: $r = s = 3 \, \text{cm}$ and $r = 4 \, \text{cm}, s = 1 \, \text{cm}$.

For a complete solution, award 6 points regardless of whether the solver determined the only two possibilities by calculation or eliminated the ones that do not meet the given conditions from five possibilities. For each incorrect solution (e.g., if the student states $r = 5, s = \frac{1}{5}$ as a possible result), deduct one point.