4. In an equilateral triangle, a point $T$ is given whose distances from the sides of the triangle are 1, 2, and 3. What is the length of the side of this triangle?
Solution.

We see that triangle $ABC$ can be divided into three triangles $CAT, ABT$, and $BCT$. The area of triangle $ABC$ is equal to the sum of the areas of these three triangles:

$$
P(ABC) = P(CAT) + P(ABT) + P(BCT)
$$

![](https://cdn.mathpix.com/cropped/2024_05_30_460b0238d86ae9459a11g-4.jpg?height=534&width=571&top_left_y=650&top_left_x=761)

The area of triangle $CAT$ is equal to $\frac{1}{2}|CA| \cdot v_{b}$. Here, $v_{b}$, the length of the height of this triangle from vertex $T$, is the distance from point $T$ to side $\overline{CA}$.

(5 points)

Therefore, (see the figure) $P(CAT) = \frac{a \cdot 1}{2}$,

(2 points)

and similarly $P(ABT) = \frac{a \cdot 2}{2}, P(BCT) = \frac{a \cdot 3}{2}$,

(3 points)

so we get $\frac{a \cdot 1}{2} + \frac{a \cdot 2}{2} + \frac{a \cdot 3}{2} = \frac{a^2 \sqrt{3}}{4}$, and from this, finally, $3a = \frac{a^2 \sqrt{3}}{4}, a = 4 \sqrt{3}$.

(5 points)