Let $G$ be the centroid of triangle $ABC$ with $AB=13,BC=14,CA=15$. Calculate the sum of the distances from $G$ to the three sides of the triangle.

Note: The [i]centroid[/i] of a triangle is the point that lies on each of the three line segments between a vertex and the midpoint of its opposite side.

[i]2019 CCA Math Bonanza Individual Round #11[/i]
1. **Identify the centroid and its properties**:
   The centroid \( G \) of a triangle divides each median into a ratio of \( 2:1 \). This means that the distance from the centroid to a side is one-third the length of the altitude from the opposite vertex to that side.

2. **Calculate the area of triangle \( ABC \)**:
   We use Heron's formula to find the area of the triangle. First, compute the semi-perimeter \( s \):
   \[
   s = \frac{AB + BC + CA}{2} = \frac{13 + 14 + 15}{2} = 21
   \]
   Then, the area \( \Delta \) of triangle \( ABC \) is:
   \[
   \Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{7056} = 84
   \]

3. **Calculate the altitude from vertex \( A \) to side \( BC \)**:
   The altitude \( AD \) from vertex \( A \) to side \( BC \) can be found using the area:
   \[
   AD = \frac{2 \Delta}{BC} = \frac{2 \cdot 84}{14} = 12
   \]

4. **Calculate the distance from the centroid \( G \) to side \( BC \)**:
   Since \( G \) divides the altitude \( AD \) in the ratio \( 2:1 \):
   \[
   GG_A = \frac{1}{3} AD = \frac{1}{3} \cdot 12 = 4
   \]

5. **Repeat for the other sides**:
   Similarly, we calculate the distances from \( G \) to sides \( CA \) and \( AB \):
   - For side \( CA \):
     \[
     BE = \frac{2 \Delta}{CA} = \frac{2 \cdot 84}{15} = 11.2
     \]
     \[
     GG_B = \frac{1}{3} BE = \frac{1}{3} \cdot 11.2 = 3.7333
     \]
   - For side \( AB \):
     \[
     CF = \frac{2 \Delta}{AB} = \frac{2 \cdot 84}{13} = 12.9231
     \]
     \[
     GG_C = \frac{1}{3} CF = \frac{1}{3} \cdot 12.9231 = 4.3077
     \]

6. **Sum the distances**:
   The sum of the distances from \( G \) to the three sides is:
   \[
   GG_A + GG_B + GG_C = 4 + 3.7333 + 4.3077 = 12.041
   \]

7. **Recalculate using the given formula**:
   Alternatively, using the given formula:
   \[
   \text{Sum of distances} = \frac{2}{3} \Delta \left( \frac{1}{BC} + \frac{1}{CA} + \frac{1}{AB} \right)
   \]
   \[
   = \frac{2}{3} \cdot 84 \left( \frac{1}{14} + \frac{1}{15} + \frac{1}{13} \right)
   \]
   \[
   = \frac{2}{3} \cdot 84 \left( \frac{195 + 182 + 210}{2730} \right)
   \]
   \[
   = \frac{2}{3} \cdot 84 \cdot \frac{587}{2730}
   \]
   \[
   = \frac{2 \cdot 84 \cdot 587}{3 \cdot 2730}
   \]
   \[
   = \frac{2 \cdot 84 \cdot 587}{8190}
   \]
   \[
   = \frac{98424}{8190}
   \]
   \[
   = \frac{2348}{195}
   \]

The final answer is \(\boxed{\frac{2348}{195}}\).