Question 1210201
Let the square \( ABCD \) be oriented in the coordinate plane with points defined as follows:
- \( A = (0, 1) \)
- \( B = (0, 0) \)
- \( C = (1, 0) \)
- \( D = (1, 1) \)

The laser beam is fired from point \( A \) at an angle towards point \( X \), where \( X \) is the point on side \( BC \) such that \( BX = \frac{2}{3} \). The coordinates of point \( X \) are given by:
\[
X = (0, \frac{2}{3})
\]

To find the direction of the laser beam from \( A \) to \( X \), we calculate the vector from \( A \) to \( X \):
\[
\overrightarrow{AX} = X - A = (0, \frac{2}{3}) - (0, 1) = (0, -\frac{1}{3})
\]
The slope of this line is
\[
\text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\frac{2}{3} - 1}{0 - 0} \quad \text{(vertical line)}.
\]

For the moment, let us compute the direct distance from \( A \) to \( X \):
\[
d_{AX} = \sqrt{(0 - 0)^2 + \left(\frac{2}{3} - 1\right)^2} = \sqrt{0 + \left(-\frac{1}{3}\right)^2} = \frac{1}{3}.
\]

The laser beam will reflect off of side \( BC \) and into side \( CD \). The point X is at a vertical location of \( \frac{2}{3} \) from the bottom of the square.

Now let's consider what happens when it reflects off side \( BC \). The line from \( A \) to \( X \) intersects \( BC \) at \( X \), and then heads towards side \( CD \). 

The angle of incidence equals the angle of reflection. After reflecting off side \( BC \), the beam will change its direction but maintain its angle of slope towards the other vertical wall, \( CD \).

The horizontal direction can be preserved by extending or 'unfolding' the square. The strategy involves reflecting the square:
1. From \( A \) to \( X \), the laser hits \( BC \).
2. It then reflects and heads towards \( D \), calculating where it hits before potentially hitting at \( D \).

With this reflection, the new direction of travel can be calculated using the symmetry of the reflections:
- Upon reflecting, after reaching \( BC \) it would travel back upwards towards \( CD \) (at point \( D \)), moving rightward.

The distance until it would again meet the line \( AD \):
1. From \( X \) to \( D \) directly makes up the next segment. The coordinates for \( D \) are \( (1, 1) \) when revisiting from \( (0, \frac{2}{3}) \):

This results in:
\[
d_{XD} = \sqrt{(1 - 0)^2 + (1 - \frac{2}{3})^2} = \sqrt{1^2 + \left(\frac{1}{3}\right)^2} = \sqrt{1 + \frac{1}{9}} = \sqrt{\frac{10}{9}} = \frac{\sqrt{10}}{3}.
\]

Total distance:
\[
d_{AX} + d_{XD} = \frac{1}{3} + \frac{\sqrt{10}}{3} = \frac{1 + \sqrt{10}}{3}.
\]

Thus, the total path of the laser beam from A to D via reflections is
1 + \sqrt{10}.