Let $ABC$ be an equilateral triangle with side length $1$. This triangle is rotated by some angle about its center to form triangle $DEF.$ The intersection of $ABC$ and $DEF$ is an equilateral hexagon with an area that is $\frac{4} {5}$ the area of $ABC.$ The side length of this hexagon can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. What is $m+n$?

[i]Author: Ray Li[/i]
1. **Determine the area of the equilateral triangle \(ABC\):**
   The side length of \(ABC\) is given as 1. The area \(A\) of an equilateral triangle with side length \(s\) is given by:
   \[
   A = \frac{\sqrt{3}}{4} s^2
   \]
   Substituting \(s = 1\):
   \[
   A = \frac{\sqrt{3}}{4} \cdot 1^2 = \frac{\sqrt{3}}{4}
   \]

2. **Calculate the area of the hexagon:**
   The area of the hexagon is given as \(\frac{4}{5}\) of the area of \(ABC\):
   \[
   \text{Area of hexagon} = \frac{4}{5} \times \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{5}
   \]

3. **Express the side lengths of the smaller triangles:**
   Let the side lengths of the smaller triangles formed at the tips of the equilateral triangles be \(a\), \(b\), and \(1 - a - b\). The angle between sides \(a\) and \(b\) is \(60^\circ\).

4. **Apply the Law of Cosines:**
   Using the Law of Cosines in the triangle with sides \(a\), \(b\), and \(1 - a - b\):
   \[
   (1 - a - b)^2 = a^2 + b^2 - 2ab \cos(60^\circ)
   \]
   Since \(\cos(60^\circ) = \frac{1}{2}\):
   \[
   (1 - a - b)^2 = a^2 + b^2 - ab
   \]

5. **Calculate the area of the smaller triangles:**
   The area of each of the smaller triangles is given by:
   \[
   \text{Area} = \frac{1}{2} ab \sin(60^\circ)
   \]
   Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\):
   \[
   \text{Area} = \frac{1}{2} ab \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} ab
   \]
   Given that the total area of the hexagon is \(\frac{\sqrt{3}}{5}\), and there are 6 such smaller triangles:
   \[
   6 \cdot \frac{\sqrt{3}}{4} ab = \frac{\sqrt{3}}{5}
   \]
   Simplifying:
   \[
   \frac{3\sqrt{3}}{2} ab = \frac{\sqrt{3}}{5}
   \]
   \[
   3ab = \frac{2}{5}
   \]
   \[
   ab = \frac{2}{15}
   \]

6. **Solve for \(a + b\):**
   From the Law of Cosines equation:
   \[
   (1 - a - b)^2 = a^2 + b^2 - ab
   \]
   Substituting \(ab = \frac{2}{15}\):
   \[
   (1 - a - b)^2 = a^2 + b^2 - \frac{2}{15}
   \]
   The area of the hexagon is \(\frac{\sqrt{3}}{5}\), and the area of each small triangle is \(\frac{\sqrt{3}}{60}\):
   \[
   \frac{1}{2} ab \sin(60^\circ) = \frac{\sqrt{3}}{60}
   \]
   \[
   ab = \frac{1}{15}
   \]
   Thus, \(a + b = \frac{3}{5}\) and \(1 - a - b = \frac{2}{5}\).

7. **Determine the side length of the hexagon:**
   The side length of the hexagon is \(\frac{2}{5}\).

8. **Express the side length in the form \(\frac{m}{n}\):**
   The side length of the hexagon is \(\frac{2}{5}\), where \(m = 2\) and \(n = 5\).

9. **Calculate \(m + n\):**
   \[
   m + n = 2 + 5 = 7
   \]

The final answer is \(\boxed{7}\)