1.23 In triangle $ABC$, $AB=33$, $AC=21$, and $BC=m$, where $m$ is a positive integer. If there exists a point $D$ on $AB$ and a point $E$ on $AC$ such that $AD=DE=EC=n$, where $n$ is a positive integer, what value must $m$ take?
(Swedish Mathematical Competition, 1982)
［Solution］In $\triangle A D E$,

then
$$
\begin{array}{c}
A D+D E>A E, \\
n+n>21-n .
\end{array}
$$

Thus $\quad 7<n<21$.
Since $\triangle D E A$ is an isosceles triangle, we have
$$
\cos A=\frac{\frac{A E}{2}}{n}=\frac{21-n}{2 n} \text {. }
$$

In $\triangle A B C$, using the cosine rule
$$
m^{2}=21^{2}+33^{2}-2 \cdot 21 \cdot 33 \cdot \frac{21-n}{2 n} \text {. }
$$

This gives
$$
m^{2}=2223-\frac{3^{3} \cdot 7^{2} \cdot 11}{n} .
$$

Thus $n$ is a divisor of $3^{3} \cdot 7^{2} \cdot 11$, and satisfies $7<n<21$, so $n=9$ or $n=11$.
When $n=9$, we get
$$
m^{2}=606,
$$

but 606 is not a perfect square.
When $n=11$, we get
$$
m^{2}=900, \quad m=30 .
$$