The sides of a triangle, as well as the radii of the inscribed circle $r$, and the exscribed circles $r_{1}, r_{2}, r_{3}$, are integers, and the radii are even. Determine the sides if

$$
r r_{1} r_{2}+r r_{2} r_{3}+r r_{3} r_{1}+r_{1} r_{2} r_{3}=r r_{1} r_{2} r_{3}
$$
None of the radii is zero, so the given relationship can be written as:

$$
\frac{1}{r}+\frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{r_{3}}=1
$$

Furthermore, the following relationships are known for these radii in terms of the sides, semiperimeter, and area of the triangle:

$$
r=\frac{t}{s}, \quad r_{1}=\frac{t}{s-a}, \quad r_{2}=\frac{t}{s-b}, \quad r_{3}=\frac{t}{s-c}
$$

Substituting these into (1) and rearranging,

$$
4 s-(a+b+c)=2 s=t
$$

and combining this with the equality $r s=t$ from (2), we get $r=2$. Furthermore, from (2) and using Heron's formula,

$$
r r_{1} r_{2} r_{3}=\frac{t^{4}}{s(s-a)(s-b)(s-c)}=t^{2}
$$

which is a perfect square because, according to (3), $t$ is an integer along with the sides.

From $r=2$ and (1),

$$
\frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{r_{3}}=\frac{1}{2}
$$

Assuming $r_{1} \leq r_{2} \leq r_{3}$, substituting the non-larger $r_{1}$ for $r_{2}$ and $r_{3}$ in the left-hand side of (5) increases or keeps it unchanged:

$$
\frac{3}{r_{1}} \geq \frac{1}{2}, \quad \text { from which } \quad r_{1} \leq 6
$$

On the other hand, considering their geometric meaning and the evenness assumption, $r_{1}>r$, so $r_{1} \geq 4$, thus either $r_{1}=6$ or $r_{1}=4$.

The first case is not possible because (5) leads to $r_{2}=r_{3}=6$, and with these, the left-hand side of (4) is $2 \cdot 6^{3}=432$, which is not a perfect square. In the second case,

$$
\frac{1}{r_{3}}+\frac{1}{r_{4}}=\frac{1}{4}
$$

From this, $r_{3}>4$, otherwise $r_{4}$ cannot be positive, and similarly $r_{3} \leq 8$, so either $r_{3}=8$ or $r_{3}=6$. In the first case, $r_{4}=8$, and the left-hand side of (4) is not a perfect square, so there is no solution in this case.

In the case $r_{3}=6$, from (6) $r_{4}=12$, so from (4) $t=\sqrt{2 \cdot 4 \cdot 6 \cdot 12}=24$ (since $t=-24$ does not fit), and from (2):

$$
s=12, \quad s-a=6, \quad s-b=4, \quad s-c=2
$$

Thus, we have a system of four equations for the three sides. This system is solvable and not contradictory because the sum of the last three equations leads to the first equation:

$$
3 s-(a+b+c)=s=6+4+2=12
$$

Furthermore, it provides integer values for the sides, because subtracting the last three equations from the first:

$$
a=s-(s-a)=6, \quad \text { and similarly } \quad b=8, \quad c=10 .
$$

The solution is a right-angled triangle, specifically twice the well-known "Egyptian triangle." It is clear from the above that there are no other solutions.

Ferenc Rába (Budapest, István I. Gymnasium III. grade)

Note. The value $r=2$ can be determined without considering the sides, the area, and (2), using only the geometric meaning of the radii and the fact that $r1, rr$ are positive, and taking $t$ to be positive. This ensures that $s-a, s-b$, and $s-c$ are all positive, meaning no side reaches half the perimeter, and thus is smaller than the sum of the other two sides.