4. The radii of the excircles of a certain triangle are 2, 3, and 6 cm. Find the radius of the circle inscribed in this triangle.
3.II.4. First, let's derive the formula for the radius of the excircle. Let $P$ be the center of the circle with radius $r_{3}$, which touches the side $AB$ and the extensions of the sides $CA$ and $CB$ of triangle $ABC$. From the equality $S_{PAB} + S_{ABC} = S_{PAC} + S_{PBC}$, it follows that

$$
S_{ABC} = S = \frac{a r_{3}}{2} + \frac{b r_{3}}{2} - \frac{c r_{3}}{2}
$$

from which $r_{3} = \frac{2 S}{a + b - c}$. The radii of the other two excircles are found by analogous formulas: $r_{1} = \frac{2 S}{b + c - a}$ and $r_{2} = \frac{2 S}{a + c - b}$. Since the radius of the circle inscribed in the given triangle is found by the formula $r = \frac{2 S}{a + b + c}$, the following relationship holds:

$$
\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} = \frac{b + c - a}{2 S} + \frac{a + c - b}{2 S} + \frac{a + b - c}{2 S} = \frac{a + b + c}{2 S} = \frac{1}{r}
$$

According to the problem, $r_{1} = 2$, $r_{2} = 3$, and $r_{3} = 6$, from which $r = 1$. A triangle with the given radii of the excircles exists; it is a right triangle with legs 3 and 4, known as the "Egyptian" triangle.