Problem 1393

1. Let the equilateral triangle be denoted as $\triangle ABC$ with circumradius $R = 1$. The side length of the equilateral triangle can be calculated using the formula for the circumradius of an equilateral triangle:
$$R = \frac{a}{\sqrt{3}}$$
where $a$ is the side length. Given $R = 1$, we have:
$$a = \sqrt{3}$$

2. Let $P$ be a point inside the triangle. Denote the perpendicular distances from $P$ to the sides $BC$, $CA$, and $AB$ as $a_P$, $b_P$, and $c_P$ respectively. The sum of these distances is a constant for any point inside an equilateral triangle and is equal to the altitude of the triangle:
$$a_P + b_P + c_P = \text{altitude of } \triangle ABC$$
The altitude $h$ of an equilateral triangle with side length $a$ is given by:
$$h = \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{2} \cdot \sqrt{3} = \frac{3}{2}$$
Therefore:
$$a_P + b_P + c_P = \frac{3}{2}$$

3. We need to determine the point $P$ such that the total lengths of the sides (extended), which lie inside the circle with center $P$ and radius $1$, is maximized. We consider two cases:

Case 1: $a_P, b_P, c_P \leq 1$
- In this case, the total length of the sides lying inside the circle is:
$$2\sqrt{1 - a_P^2} + 2\sqrt{1 - b_P^2} + 2\sqrt{1 - c_P^2}$$
- The function $x \mapsto 2\sqrt{1 - x^2}$ is concave on $[0, 1]$. By Jensen's inequality, for concave functions, we have:
$$2\sqrt{1 - a_P^2} + 2\sqrt{1 - b_P^2} + 2\sqrt{1 - c_P^2} \leq 3 \cdot 2\sqrt{1 - \left(\frac{a_P + b_P + c_P}{3}\right)^2}$$
- Substituting $a_P + b_P + c_P = \frac{3}{2}$:
$$2\sqrt{1 - a_P^2} + 2\sqrt{1 - b_P^2} + 2\sqrt{1 - c_P^2} \leq 3 \cdot 2\sqrt{1 - \left(\frac{\frac{3}{2}}{3}\right)^2} = 3 \cdot 2\sqrt{1 - \left(\frac{1}{2}\right)^2} = 3 \cdot 2\sqrt{\frac{3}{4}} = 3 \cdot \sqrt{3} = 3\sqrt{3}$$
- Equality is obtained when $P$ is the circumcenter of the triangle.

Case 2: One of $a_P, b_P, c_P$ is greater than $1$
- Without loss of generality, assume $c_P > 1$. In this case, the total length of the sides lying inside the circle is:
$$2\sqrt{1 - a_P^2} + 2\sqrt{1 - b_P^2}$$
- Since $a_P + b_P + c_P = \frac{3}{2}$ and $c_P > 1$, it follows that $a_P + b_P < \frac{1}{2}$. Therefore:
$$2\sqrt{1 - a_P^2} + 2\sqrt{1 - b_P^2} \leq 2 + 2 = 4$$
- This is less than the maximum obtained in Case 1.

4. Therefore, the required maximum total length of the sides lying inside the circle is $3\sqrt{3}$.

The final answer is $\boxed{ 3\sqrt{3} }$.