IMO Hk TST, 1997

The product is $9360$.
Let $a$, $b$, $c$, $r$, $s$ be the lengths of $BC$, $CA$, $AB$, the inradius and the semiperimeter of $\triangle ABC$ respectively. Let $N$ be the midpoint of $AC$, and let $P$ be the intersection point of $EF$ and $MN$. It is well-known that $\angle BPC = 90^\circ$. Therefore, we have $MP = \frac{a}{2}$. (Alternatively, one can prove this by noting $MN = \frac{c}{2}$ and $PN = NF = \frac{c-a}{2}$.)

Note that $NM \parallel AB$ by the midpoint theorem. Since $ADMI$ is a parallelogram, we have
$$AI = DM = PM \sin \angle DPM = \frac{a}{2} \sin \angle FEA = \frac{a}{2} \cos \frac{A}{2}.$$
On the other hand, we have
$$AI = \frac{r}{\sin \frac{A}{2}} = \frac{bc \sin A}{2s \sin \frac{A}{2}} = \frac{bc}{s} \cos \frac{A}{2}.$$
Equating these, we obtain $2bc = as$. As it is given that $2s = 65$, this yields $4bc = 65a$. Then the given condition becomes $\frac{4bc}{65} + b + c = 65$. This can be factorized as
$$(4b + 65)(4c + 65) = 5 \cdot 65^2 = 5^3 \cdot 13^2.$$
As $4b + 65$ and $4c + 65$ are larger than $65$, we must have $4b + 65 = 125$ and $4c + 65 = 169$ up to permutation. This gives $(a, b, c) = (24, 15, 26)$ or $(24, 26, 15)$. Therefore, $abc = 24 \times 15 \times 26 = 9360$.