On equilateral $\triangle{ABC}$ point D lies on BC a distance 1 from B, point E lies on AC a distance 1 from C, and point F lies on AB a distance 1 from A. Segments AD, BE, and CF intersect in pairs at points G, H, and J which are the vertices of another equilateral triangle. The area of $\triangle{ABC}$ is twice the area of $\triangle{GHJ}$. The side length of $\triangle{ABC}$ can be written $\frac{r+\sqrt{s}}{t}$, where r, s, and t are relatively prime positive integers. Find $r + s + t$.

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1. **Using Routh's Theorem**: Routh's Theorem is a useful tool for finding the ratio of areas of triangles formed by cevians intersecting at a point. For an equilateral triangle, if the cevians divide the sides in the ratio \( x : 1 \), \( y : 1 \), and \( z : 1 \), the area of the inner triangle formed by the cevians is given by:
   \[
   \frac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(xz+x+1)}
   \]
   Given that \( x = y = z \), we simplify the expression:
   \[
   \frac{(x^3-1)^2}{(x^2+x+1)^3}
   \]

2. **Equating the area ratio**: We know that the area of \( \triangle{ABC} \) is twice the area of \( \triangle{GHJ} \). Therefore:
   \[
   \frac{(x^3-1)^2}{(x^2+x+1)^3} = \frac{1}{2}
   \]

3. **Solving the equation**: We need to solve for \( x \):
   \[
   (x^3-1)^2 = \frac{1}{2} (x^2+x+1)^3
   \]
   Taking the square root of both sides:
   \[
   x^3 - 1 = \frac{\sqrt{2}}{2} (x^2 + x + 1)^{3/2}
   \]
   This equation is quite complex, so we simplify it by assuming a simpler form:
   \[
   \frac{(x-1)^2}{x^2+x+1} = \frac{1}{2}
   \]
   Multiplying both sides by \( 2(x^2+x+1) \):
   \[
   2(x-1)^2 = x^2 + x + 1
   \]
   Expanding and simplifying:
   \[
   2(x^2 - 2x + 1) = x^2 + x + 1
   \]
   \[
   2x^2 - 4x + 2 = x^2 + x + 1
   \]
   \[
   x^2 - 5x + 1 = 0
   \]

4. **Solving the quadratic equation**: Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   x = \frac{5 \pm \sqrt{25 - 4}}{2}
   \]
   \[
   x = \frac{5 \pm \sqrt{21}}{2}
   \]
   Since \( x \) must be positive:
   \[
   x = \frac{5 + \sqrt{21}}{2}
   \]

5. **Finding the side length of \( \triangle{ABC} \)**: The side length of \( \triangle{ABC} \) is \( x + 1 \):
   \[
   x + 1 = \frac{5 + \sqrt{21}}{2} + 1 = \frac{5 + \sqrt{21} + 2}{2} = \frac{7 + \sqrt{21}}{2}
   \]

6. **Summing the integers**: The side length is given by \( \frac{r + \sqrt{s}}{t} \), where \( r = 7 \), \( s = 21 \), and \( t = 2 \). Therefore:
   \[
   r + s + t = 7 + 21 + 2 = 30
   \]

The final answer is \( \boxed{30} \).