3.2. The radius of the circumscribed circle of a triangle is 2, and the lengths of all altitudes are integers. Find the sides of the triangle.
3.2. Answer: $2 \sqrt{3}, 2 \sqrt{3}, 2 \sqrt{3}$.

Firstly, note that the height of the triangle must be less than the diameter of the circumscribed circle, so all heights have a length of 1, 2, or 3.

From the triangle inequality $a < \frac{1}{2} + \frac{1}{3}$, which contradicts the inequality we have derived. Therefore, among the heights, there will be pairs of equal heights, meaning the triangle is isosceles. Consider the height $h$ at the vertex of the isosceles triangle. By the Pythagorean theorem, the base will be $2 \sqrt{R^{2}-(R-h)^{2}} = 2 \sqrt{4 h - h^{2}}$. Depending on $h$, there are three possible cases:

$h=1$. It is not difficult to verify that this is a triangle with angles $120^{\circ}, 30^{\circ}, 30^{\circ}$ and a side length of 2. The height dropped to the side is $\sqrt{3}$, which is not an integer.

$h=2$. This is a triangle with angles $90^{\circ}, 45^{\circ}, 45^{\circ}$ and a side length of $2 \sqrt{2}$, the height to the side is also $2 \sqrt{2}$, which is not an integer.

$h=3$. This is an equilateral triangle with a side length of $2 \sqrt{3}$, all its heights are 3, which fits the condition of the problem.