A circle of radius $1$ is internally tangent to two circles of radius $2$ at points $A$ and $B$, where $AB$ is a diameter of the smaller circle. What is the area of the region, shaded in the picture, that is outside the smaller circle and inside each of the two larger circles?
(source: https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_25 )
This is my work can someone please tell me what I did wrong Call the center of the circle with radius 1 o. Call the intersection point between the 2 circles c and d. Now let oc equal h. Since ac equals 2 we use pythag and get h^2+1=4 so h=sqrt(3) Now we know that the desired area is the difference of AOC(the ellipse sector) and the quarter circle. Now we do arithmetic and get sqrt(3)pi-pi
Label the intersections $C$ and $D$. Obviously, $\triangle CAB$ and $\triangle DAB$ are equilateral. Meaning sector $BCD$ and $ACD$ has an angle of $120^\circ$. The area of the shaded region plus the area of the circle in the middle is twice the area of the sector less the area of the triangle: $$\begin{align} A&=2(A_{\text{sector } ACD}-A_{\triangle ACD})\\ &=2\left(\frac{4\pi}3-\sqrt3\right) \end{align}$$ Finally, subtract the area of the circle in the middle: $$A_{\text{shaded}}=\frac{8\pi}3-2\sqrt3-\pi=\frac{5\pi}3-2\sqrt3$$