Let $O$ be the center of an equilateral triangle $ABC$ of area $1/\pi$.  As shown in the diagram below, a circle centered at $O$ meets the triangle at points $D$, $E$, $F$, $G$, $H$, and $I$, which trisect each of the triangle's sides.  Compute the total area of all six shaded regions. 
[asy]
unitsize(90);
pair A = dir(0);
pair B = dir(120);
pair C = dir(240);
draw(A -- B -- C -- cycle);
pair D = (2*A + B)/3;
pair E = (A + 2*B)/3;
pair F = (2*B + C)/3;
pair G = (B + 2*C)/3;
pair H = (2*C + A)/3;
pair I = (C + 2*A)/3;
draw(E -- F);
draw(G -- H);
draw(I -- D);
draw(D -- G);
draw(E -- H);
draw(F -- I);
pair O = (0, 0);
real r = 1/sqrt(3);
draw(circle(O, r));
fill(O -- D -- E -- cycle, gray);
fill(O -- F -- G -- cycle, gray);
fill(O -- H -- I -- cycle, gray);
fill(arc(O, r, -30, 30) -- cycle, gray);
fill(arc(0, r, 90, 150) -- cycle, gray);
fill(arc(0, r, 210, 270) -- cycle, gray);
label("$A$", A, A);
label("$B$", B, B);
label("$C$", C, C);
label("$D$", D, unit(D));
label("$E$", E, unit(E));
label("$F$", F, unit(F));
label("$G$", G, unit(G));
label("$H$", H, unit(H));
label("$I$", I, unit(I));
label("$O$", O, C);
[/asy]



1. **Determine the side length of the equilateral triangle:**
   Given the area of the equilateral triangle \(ABC\) is \(\frac{1}{\pi}\), we use the formula for the area of an equilateral triangle:
   \[
   \text{Area} = \frac{s^2 \sqrt{3}}{4}
   \]
   where \(s\) is the side length. Setting this equal to \(\frac{1}{\pi}\), we have:
   \[
   \frac{s^2 \sqrt{3}}{4} = \frac{1}{\pi}
   \]
   Solving for \(s^2\):
   \[
   s^2 = \frac{4}{\pi \sqrt{3}}
   \]

2. **Determine the radius of the circle:**
   The circle is centered at \(O\) and trisects each side of the triangle. The radius \(r\) of the circle is \(\frac{1}{3}\) of the side length \(s\):
   \[
   r = \frac{s}{3}
   \]
   Substituting \(s^2 = \frac{4}{\pi \sqrt{3}}\):
   \[
   r^2 = \left(\frac{s}{3}\right)^2 = \frac{s^2}{9} = \frac{\frac{4}{\pi \sqrt{3}}}{9} = \frac{4}{9 \pi \sqrt{3}}
   \]

3. **Calculate the area of the circle:**
   The area \(A\) of the circle is given by:
   \[
   A = \pi r^2 = \pi \left(\frac{4}{9 \pi \sqrt{3}}\right) = \frac{4}{9 \sqrt{3}}
   \]

4. **Determine the total area of the shaded regions:**
   The shaded regions are formed by rearranging the lunes to fill in the sectors of the circle. The total area of the shaded regions is equal to half the area of the circle:
   \[
   \text{Shaded Area} = \frac{1}{2} \times \frac{4}{9 \sqrt{3}} = \frac{2}{9 \sqrt{3}}
   \]

5. **Simplify the final expression:**
   Simplifying \(\frac{2}{9 \sqrt{3}}\):
   \[
   \frac{2}{9 \sqrt{3}} = \frac{2 \sqrt{3}}{27}
   \]

The final answer is \(\boxed{\frac{2 \sqrt{3}}{27}}\).