9.3. (Sweden, 82). Find all values of $n \in \mathbf{N}$ for each of which there exists a number $m \in \mathbf{N}$, a triangle $A B C$ with sides $A B=33, A C=21, B C=n$ and points $D$, $E$ on sides $A B, A C$ respectively, satisfying the conditions $A D=D E=E C=m$.
9.3. Let the numbers $m, n \in \mathbf{N}$ satisfy the condition of the problem. Then $m = CE < AC = 21$ and from triangle $ADE$ (Fig. 3) we have $21 - m = AE < AD + DE = 2m$, hence

$$
7 < m < 21
$$

Furthermore, since $AD = DE$, for the angle $\alpha = \angle BAC$ we find

$$
\cos \alpha = AE / (2 \cdot AD) = (21 - m) / 2m
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_03a501298610b72904feg-173.jpg?height=413&width=522&top_left_y=120&top_left_x=12)

Fig. 3

Finally, by the cosine rule for triangle $ABC$ we obtain

$$
\begin{aligned}
& n^{2} = BC^{2} = AB^{2} + AC^{2} - \\
& - 2 AB \cdot AC \cdot \cos \alpha = \\
& = 33^{2} + 21^{2} - 2 \cdot 33 \cdot 21 \cdot \frac{21 - m}{2m} = \\
& = 2223 - \frac{27 \cdot 49 \cdot 11}{m}
\end{aligned}
$$

from which it follows that the number $m$ is a divisor of the number $27 \cdot 49 \cdot 11$. Of the two possible values $m = 9$ and $m = 11$, the first does not work (since $n^{2} \neq 606$), while the second gives the equation $n^{2} = 900$, i.e., $n = 30$. Verification shows that all conditions of the problem are satisfied for the found value of $n$.