In a triangle with sides $|AB| = 33$, $|AC| = 21$ and $|BC| = n$ we place point $D$ on $AB$ and $E$ on $AC$ so that $|AD|=|DE|=|EC|=m \in \mathbb{N}$.

Question

Find all natural numbers n for which: in a triangle with sides $|AB| = 33$, $|AC| = 21$ and $|BC| = n$ we can place point $D$ on $AB$ and point $E$ on $AC$ such that $|AD| = |DE| = |EC| = m$, where $m$ is also a natural number.

Attempt: I saw that $m=21$ satisfied the condition, so $n$ can be any natural number between $13$ and $53$ by Triangle Inequality theorem. Is this correct?

Answer 1

Just in case, I found another possible solution: $n=30$ and $m=11$.

Let $\angle BAC=\alpha$. Then from the cosine rule applied to triangles $ABC$ and $ADE$ we have (for $m\ne21$): $$ \cos\alpha={21-m\over 2m}={33^2+21^2-n^2\over2\cdot 33\cdot 21}, $$ that is: $$ m={14553\over2223-n^2}. $$ It's easy to check that the only integer solution for $13\le n\le53$ is the one given above.