Let $ABCD$ be a square with side length $1.$ A laser is located at vertex $A,$ which fires a laser beam at point $X$ on side $\overline{BC},$ such that $BX = \frac{3}{4}.$ The beam reflects off the sides of the square, until it ends up at another vertex; at this point, the beam will stop. Find the length of the total path of the laser beam.


Let $ A(0, 1) $, $ B(0, 0) $, $ C(1, 0) $, and $ D(1, 1) $ be the coordinates of the vertices of square $ ABCD $.

The point $ X $ on side $ BC $ is given by $ BX = \frac{3}{4} $, thus the coordinates of point $ X $ are $ (0.75, 0) $.

The laser beam starts at $ A(0, 1) $ and hits $ X(0.75, 0) $. We can calculate the slope of the line from $ A $ to $ X $:

\[
\text{slope} = \frac{0 - 1}{0.75 - 0} = -\frac{4}{3}
\]

The equation of the line from $ A $ to $ X $ is:

\[
y - 1 = -\frac{4}{3}(x - 0)
\]
\[
y = -\frac{4}{3}x + 1
\]

To find where this line intersects the sides of the square, we will evaluate it at the sides:

1. **Intersection with $ BC $ (where $ x = 0 $)**:
   \[
   y = -\frac{4}{3}(0) + 1 = 1
   \]
   This point $ (0, 1) $ is actually $ A $, which we already know.

2. **Intersection with $ CD $ (where $ y = 0 $)**:
   \[
   0 = -\frac{4}{3}x + 1
   \]
   Solving for $ x $:
   \[
   \frac{4}{3}x = 1 \implies x = \frac{3}{4}
   \]
   The intersection point is $ \left(\frac{3}{4}, 0\right) $.

3. **Now, we reflect the square across the sides.**
   To visualize the path of the laser, we will consider "unfolding" the square. Reflect the square across the line $ BC $ to create a new square where the line continues in a straight path. 

   The new vertex is at $ C'(1, 0) $, and since the laser hits $ X(0.75, 0) $ and travels down to $ D' $ (reflection across $ x = 1 $):
   - Reflect across line $ y = 0 $ to find $ D' = (1, -1) $.

The laser now travels from $ A(0, 1) $ to $ X(0.75, 0) $, then it continues from $ X $ to $ D' $.

Now, we find the total distance:

1. Distance from $ A(0, 1) $ to $ X(0.75, 0) $:
   \[
   d_1 = \sqrt{(0.75 - 0)^2 + (0 - 1)^2} = \sqrt{(0.75)^2 + (-1)^2} = \sqrt{0.5625 + 1} = \sqrt{1.5625} = \frac{5}{4}
   \]

2. Distance from $ X(0.75, 0) $ to $ D'(1, -1) $:
   \[
   d_2 = \sqrt{(1 - 0.75)^2 + (-1 - 0)^2} = \sqrt{(0.25)^2 + (-1)^2} = \sqrt{0.0625 + 1} = \sqrt{1.0625} = \frac{15}{16}
   \]

Now, we calculate the total path length:
\[
d_{\text{total}} = d_1 + d_2 = \frac{5}{4} + \frac{15}{16}
\]

To add these fractions, convert $ \frac{5}{4} $ into sixteenths:
\[
\frac{5}{4} = \frac{20}{16}
\]
Thus,
\[
d_{\text{total}} = \frac{20}{16} + \frac{15}{16} = \frac{35}{16}
\]

Finally, the length of the total path of the laser beam is 

\[
\boxed{\frac{35}{16}}
\]

Let $ABCD$ be a square with side length $1.$ A laser is located at vertex $A,$ which fires a laser beam at point $X$ on side $\overline{BC},$ such that $BX = \frac{3}{4}.$ The beam reflects off the sides of the square, until it ends up at another vertex; at this point, the beam will stop. Find the length of the total path of the laser beam.

Explore the problem of a laser beam reflecting within a square. This detailed solution calculates the total path length of the laser from vertex A to point X on side BC, demonstrating key geometric principles and the distance formula. Ideal for high school geometry students.

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