Source: 2017 Canadian Open Math Challenge, Problem A2
-----
An equilateral triangle has sides of length $4$cm. At each vertex, a circle with radius $2$cm is drawn, as shown in the figure below. The total area of the shaded regions of the three circles is $a\cdot \pi \text{cm}^2$. Determine $a$.

[center][asy]
size(2.5cm);
draw(circle((0,2sqrt(3)/3),1));
draw(circle((1,-sqrt(3)/3),1));
draw(circle((-1,-sqrt(3)/3),1));
draw((0,2sqrt(3)/3) -- arc((0,2sqrt(3)/3), 1, 240, 300) -- cycle);
fill(((0,2sqrt(3)/3) -- arc((0,2sqrt(3)/3), 1, 240, 300) -- cycle),mediumgray);
draw((1,-sqrt(3)/3) -- arc((1,-sqrt(3)/3), 1, 180, 120) -- cycle);
fill(((1,-sqrt(3)/3) -- arc((1,-sqrt(3)/3), 1, 180, 120) -- cycle),mediumgray);
draw((-1,-sqrt(3)/3) -- arc((-1,-sqrt(3)/3), 1, 0, 60) -- cycle);
fill(((-1,-sqrt(3)/3) -- arc((-1,-sqrt(3)/3), 1, 0, 60) -- cycle),mediumgray);
[/asy][/center]
1. **Determine the area of one circle:**
   Each circle has a radius of \(2\) cm. The area \(A\) of a circle is given by the formula:
   \[
   A = \pi r^2
   \]
   Substituting \(r = 2\) cm:
   \[
   A = \pi (2)^2 = 4\pi \text{ cm}^2
   \]

2. **Calculate the area of a 60-degree sector of one circle:**
   An equilateral triangle has internal angles of \(60^\circ\). Therefore, each shaded region is a \(60^\circ\) sector of a circle. The area of a sector of a circle is given by:
   \[
   \text{Area of sector} = \frac{\theta}{360^\circ} \times \text{Area of circle}
   \]
   For \(\theta = 60^\circ\):
   \[
   \text{Area of sector} = \frac{60^\circ}{360^\circ} \times 4\pi = \frac{1}{6} \times 4\pi = \frac{4\pi}{6} = \frac{2\pi}{3} \text{ cm}^2
   \]

3. **Calculate the total area of the shaded regions:**
   There are three such sectors, one at each vertex of the equilateral triangle. Therefore, the total area of the shaded regions is:
   \[
   \text{Total shaded area} = 3 \times \frac{2\pi}{3} = 2\pi \text{ cm}^2
   \]

4. **Determine the value of \(a\):**
   The problem states that the total area of the shaded regions is \(a \cdot \pi \text{ cm}^2\). From the previous step, we have:
   \[
   2\pi = a \cdot \pi
   \]
   Dividing both sides by \(\pi\):
   \[
   a = 2
   \]

The final answer is \(\boxed{2}\).