Let $ABCD$ be a rectangle with area $1$, and let $E$ lie on side $CD$. What is the area of the triangle formed by the centroids of triangles $ABE$, $BCE$, and $ADE$?
1. **Identify the centroids of the triangles:**
   Let \( G_1, G_2, G_3 \) be the centroids of triangles \( ADE, ABE \), and \( BCE \) respectively. The centroid of a triangle divides each median into a ratio of \( 2:1 \).

2. **Determine the coordinates of the centroids:**
   - The centroid \( G_1 \) of \( \triangle ADE \) is given by:
     \[
     G_1 = \left( \frac{A_x + D_x + E_x}{3}, \frac{A_y + D_y + E_y}{3} \right)
     \]
   - The centroid \( G_2 \) of \( \triangle ABE \) is given by:
     \[
     G_2 = \left( \frac{A_x + B_x + E_x}{3}, \frac{A_y + B_y + E_y}{3} \right)
     \]
   - The centroid \( G_3 \) of \( \triangle BCE \) is given by:
     \[
     G_3 = \left( \frac{B_x + C_x + E_x}{3}, \frac{B_y + C_y + E_y}{3} \right)
     \]

3. **Calculate the distances between the centroids:**
   - The distance \( G_1G_2 \) is:
     \[
     G_1G_2 = \sqrt{\left( \frac{A_x + D_x + E_x}{3} - \frac{A_x + B_x + E_x}{3} \right)^2 + \left( \frac{A_y + D_y + E_y}{3} - \frac{A_y + B_y + E_y}{3} \right)^2 }
     \]
     Simplifying, we get:
     \[
     G_1G_2 = \frac{1}{3} \sqrt{(D_x - B_x)^2 + (D_y - B_y)^2} = \frac{1}{3} BD
     \]
   - Similarly, the distance \( G_2G_3 \) is:
     \[
     G_2G_3 = \frac{1}{3} \sqrt{(C_x - A_x)^2 + (C_y - A_y)^2} = \frac{1}{3} AC
     \]

4. **Calculate the area of \( \triangle G_1G_2G_3 \):**
   - The angle \( \angle G_1G_2G_3 \) is the same as the angle between the diagonals \( AC \) and \( BD \) of the rectangle \( ABCD \).
   - The area of \( \triangle G_1G_2G_3 \) is given by:
     \[
     S_{\triangle G_1G_2G_3} = \frac{1}{2} G_1G_2 \cdot G_2G_3 \cdot \sin(\angle G_1G_2G_3)
     \]
     Substituting the values, we get:
     \[
     S_{\triangle G_1G_2G_3} = \frac{1}{2} \cdot \frac{1}{3} BD \cdot \frac{1}{3} AC \cdot \sin(\angle (AC, BD))
     \]
     Since \( S_{ABCD} = AC \cdot BD \cdot \sin(\angle (AC, BD)) \) and the area of the rectangle \( ABCD \) is 1, we have:
     \[
     S_{\triangle G_1G_2G_3} = \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot S_{ABCD} = \frac{1}{2} \cdot \frac{1}{9} \cdot 1 = \frac{1}{18}
     \]

The final answer is \(\boxed{\frac{1}{18}}\)