9.4. (Jury, SRV, 79). Find all triples of numbers $a, b, c \in \mathbf{N}$, which are the lengths of the sides of a triangle with the diameter of the circumscribed circle equal to 6.25.
9.4. Let $a, b, c \in \mathbf{N}$ be the lengths of the sides of a triangle with a diameter $2 R=6.25$ of the circumscribed circle, area $S$, and semiperimeter $p=(a+b+c) / 2$. Since $a, b, c \leqslant 2 R$, we have

$$
a, b, c \in\{1 ; 2 ; 3 ; 4 ; 5 ; 6\}
$$

Further, we have the equalities

$$
(a b c)^{2}=(4 S \cdot R)^{2}=p(p-a)(p-b)(p-c)(4 R)^{2}
$$

from which we obtain

$$
64 a^{2} b^{2} c^{2}=625(a+b+c)(c+b-a)(a+c-b)(a+b-c)
$$

Thus, the number $64 a^{2} b^{2} c^{2}$ is divisible by 625, which means that at least two of the three numbers $a, b, c$ are equal to 5. Let, for example, $a=b=5$, then

$$
64 c^{2}=(10+c) c^{2}(10-c)
$$

i.e., $64=100-c^{2}$, from which $c=6$. Therefore, the sought triplet can only consist of the numbers $5,5,6$, which, as verification shows, satisfy the condition of the problem.