Place the vertices of the equilateral triangle at the points $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$, and denote the radius of the semicircles by $r$. The equilateral triangle formed by the diameters of the semicircles has side length $2\sqrt2$, so its area is $2\sqrt3$. To find the distance from a vertex where the diameter corresponding to that vertex intersects the semicircle corresponding to another vertex, take the diameter
$$
\pmatrix{1\\0\\0}+\lambda\pmatrix{0\\1\\-1}
$$
and find the point on it at distance $r$ from the vertex $\pmatrix{0\\0\\1}$:
$$
1+\lambda^2+(1+\lambda)^2=r^2\;,
\\
2\lambda^2+2\lambda+2=r^2\;,
\\
\lambda=-\frac12+\sqrt{\frac{r^2}2-\frac34}\;.
$$
Thus the side length of the three smaller equilateral triangles that are "missing" is
$$
\sqrt2\left(1-\left(-\frac12+\sqrt{\frac{r^2}2-\frac34}\right)\right)=\sqrt2\left(\frac32-\sqrt{\frac{r^2}2-\frac34}\right)\;,
$$
so their total area is
$$
3\cdot\frac{\sqrt3}4\cdot2\left(\frac94+\frac{r^2}2-\frac34-3\sqrt{\frac{r^2}2-\frac34}\right)=\frac34\sqrt3\left(3+r^2-3\sqrt{2r^2-3}\right)\;.
$$
Subtracting this from $2\sqrt3$ leaves
$$
\frac{\sqrt3}4\left(9\sqrt{2r^2-3}-3r^2-1\right)\;.
$$
Now we need to add back the three circular segments that extend into the three equilateral triangles we subtracted. Their radius is $r$, and their angle $\alpha$ is twice the angle between $\pmatrix{1\\\lambda\\-\lambda}-\pmatrix{0\\0\\1}$ and $\pmatrix{1\\1\\-1}-\pmatrix{0\\0\\1}$ and thus
$$
\alpha=2\arccos\frac{3(\lambda+1)}{\sqrt{6\cdot(2\lambda^2+2\lambda+2)}}=2\arccos\left(\sqrt{\frac38}\frac{1+\sqrt{2r^2-3}}r\right)\;.
$$
With the formula for the area of a circular segment, the total area is then
$$
\frac{\sqrt3}4\left(9\sqrt{2r^2-3}-3r^2-1\right)+\frac32r^2\left(\alpha-\sin\alpha\right)\;.
$$
Here's a plot of this total area for $r$ in the relevant range $[\sqrt2,\sqrt6]$, with the area ranging from $\pi-\sqrt3$ to $2\sqrt3$.
In this calculation the side length of the equilateral triangle was fixed at $\sqrt2$; for a different side length $a$, scale the radius by $\sqrt2/a$ and then scale the resulting area by $a^2/2$.
Assuming that tha motion starts in the fixed point A(a,0), we can write the coordinate of the center B of the moving circle as
\begin{equation}
((a-b) \cos(t), (a-b) \sin(t))
\end{equation}
The coordinate of the moving point P relative to B are:
\begin{equation}
(b \cos(\phi), -b \sin(\phi))
\end{equation}
where the minus sign is due to the fact that $\phi$ is measured clockwise. Thus the coordinate of P relative to O are:
\begin{equation}
((a-b) \cos(t)+b \cos(\phi), (a-b) \sin(t)-b \sin(\phi))
\end{equation}
In order to avoid misunderstanding I added a picture.
The key point is that the angle $t$ and $\phi$ are related to each other. As a matter of fact, the arcs of the fixed and moving circles that came in contact (arcs AA' and A'P) must be of equal length. The arc length is
\begin{equation}
AA'= a t
\end{equation}
while
\begin{equation}
A'P= b(t+\phi)
\end{equation}
We can then conclude that
\begin{equation}
\phi=(\frac{a}{b}-1)t
\end{equation}
The coordinates of the point P respect to the origin O are:
\begin{equation}
x=(a-b) \cos(t)+b \cos((\frac{a}{b}-1)t)
\end{equation}
\begin{equation}
y=(a-b) \sin(t)-b \sin((\frac{a}{b}-1)t)
\end{equation}
Note that if the rolling circle move at constant angular velocity, t will be proportional to the elapsed time. For the case of an epicycloid you can derive the equation in a similar way. You should get something like:
\begin{equation}
x=(a+b) \cos(t)-b \cos((\frac{a}{b}+1)t)
\end{equation}
\begin{equation}
y=(a+b) \sin(t)-b \sin((\frac{a}{b}+1)t)
\end{equation}
the minus sign in the second term of the equation of the x coordinate is due to the fact that the rotation of the rolling circle and the motion of its center are in the same direction.
The relationship between the angle $t$ and $\phi$ can be written as
\begin{equation}
\phi=(\frac{2 \pi a}{2 \pi b}-1)t
\end{equation}
Than the ratio $a/b$ tell us how many times the circumference of the stationary circle exceed that of the rolling one. For sake of simplicity we call this ratio $n$. The stationary circle has a radius $a= n b$. When $t$ ranges from 0 to $2 \pi$ the angle $\phi$ ranges from 0 to $2 \pi(n-1)$, i.e. $n-1$ times less and not $n$ times less. In this sense the smaller circle has lost a rotation.
Best Answer
I suggest that you first regard the case with "total slipping".
Let a coin slide around another coin while having always the same point attached to the central coin. Note that the outer coin makes exactly one rotation in this scenario.
Now, roll out the inner circle and let the outer coin slide along. Clearly, it does not rotate at all.
So, there is one rotation of the outer coin that comes just from going around the inner coin. The rest of the rotation can then be found by looking at the rotation along the line and combining the two.