Assume we have this is an equilateral triangle, that the smaller circle isn' tangent to the sides of the triangle, that the larger circle goes through the three vertices of the triangle, and that the sides of the triangle have length three.
Prove the two shaded regions have the same area as the smaller circle. ![Figure][1]
On
The small circle has radius equal to half the radius of the big one
If this holds, $Area_{big circle} = Area_{small circle} + P = \frac{1}{4}Area_{big} + P$.
And $Area_{small}=\frac{1}{4}Area_{big}$, so $P=\frac{3}{4}Area_{big}$ Now the shadowed area is P/3. So you have $\frac{P}{3} = \frac{1}{3} \frac{3}{4} Area_{big} = Area_{small}$
edit: I was hoping to leave that part up to you, well nevermind. It is a well known fact that given the medians of an equilateral triangle they intersect at one point such that one part is two times the other. So draw the medians. Obviously they intersect at the center of the circle, so calling the center $O$, a vertex $A$ and the middle point of $\overline{BC}$ $M$, we have that $radius_{big}=\overline{AO}=2 radius_{small}$
Let the larger shaded area be $S_1$ and the smaller be $S_2$. Denote by $A$ the area of the smaller circle and by $T$ the area of the triangle.
First, the area of the larger circle is $4A$, because, an equilateral triangle inscribed in it is two times bigger than the similar triangle inscribed in the small circle (place it so that it touches both the circle and the larger triangle).
Now, consider the equalities
\begin{align} 4A &= 3S_1+T\\ T &= 3S_2 + A \end{align}
Together these imply $3A = 3(S_1+S_2)$, that is $$S_1+S_2 = A$$ which is exactly what we wanted to prove.
I hope this helps $\ddot\smile$