$ABC$ is an equilateral triangle and $l$ is a line such that the distances from $A, B,$ and $C$ to $l$ are $39, 35,$ and $13$, respectively. Find the largest possible value of $AB$. 

[i]Team #6[/i]
To solve this problem, we need to use the properties of an equilateral triangle and the distances from its vertices to a given line. Let's denote the side length of the equilateral triangle \(ABC\) as \(s\).

1. **Set up the problem with distances:**
   Given that the distances from vertices \(A\), \(B\), and \(C\) to the line \(l\) are 39, 35, and 13 respectively, we can use the fact that the sum of the perpendicular distances from the vertices of an equilateral triangle to any line is constant and equal to the altitude of the triangle.

2. **Calculate the altitude of the equilateral triangle:**
   The altitude \(h\) of an equilateral triangle with side length \(s\) is given by:
   \[
   h = \frac{\sqrt{3}}{2} s
   \]

3. **Sum of distances:**
   The sum of the distances from the vertices to the line \(l\) is:
   \[
   39 + 35 + 13 = 87
   \]
   This sum must be equal to the altitude of the triangle:
   \[
   \frac{\sqrt{3}}{2} s = 87
   \]

4. **Solve for \(s\):**
   \[
   s = \frac{2 \cdot 87}{\sqrt{3}} = \frac{174}{\sqrt{3}} = 58\sqrt{3}
   \]

5. **Verify the largest possible value:**
   Since the distances given are positive and the sum matches the altitude, the side length \(s = 58\sqrt{3}\) is valid and represents the largest possible value for \(AB\).

The final answer is \(\boxed{58\sqrt{3}}\).