Each circle in the shape has radius $1$. The circles are placed around a center, and distance between the center of a circle and the placement center is $3$ ($2+1$ radius).
From a certain angle, tangents from each circle to another is drawn.
What is the gray area?
EDIT. I guess I should mention what my approach was, it was to try and find the length of the segments near the gray area, but to do that I used lot's of Law of Cosines and it became a bit mess by the time I got to the segment itself!
Coordinate System approach seems way cleaner.
This is quite convoluted. But here is one way to solve it:
I assume you can complete the rest if you know $\angle BXD$.
Look at $\triangle BXC$. It is right angled at $B. BX = 1$. By assuming a coordinate system where $O$ is the origin and the line connecting $O$ to the center of the horizontal circle as $x$-axis, you can calculate coordinates of $C = \left(2.5, \frac{\sqrt{3}}{2}\right)$ and $X = \left(-\frac{3}{2}, \frac{3\sqrt{3}}{2}\right).$ Thus you can find $XC$. From this you can find BC using Pythagoras theorem and from that also find $\angle BXC$ using inverse sine.
Now look at $\triangle OXC$. You know XC (from $1$), $OX = 3$ and $OC = \sqrt{6.25 +\frac{3}{4}}$ from the coordinate system in step $1$. Thus you can find $\angle CXO$.
$\angle BXD = \angle BXC + \angle CXO - 60$
Edit. Completing the calculations.
$XC = \sqrt{19}$
$\angle BXC = \cos^{-1}\left(\frac{1}{|XC|}\right) = 76.73^{\circ}$
$OC = \sqrt{7}$
$\angle CXO = \cos^{-1} \left(\frac{|CX|^2 + |XO|^2 - |CO|^2}{2|CX||XO|}\right) = \cos^{-1}\left(\frac{19+9-7}{2\cdot 3 \cdot \sqrt{19}}\right) = \cos^{-1}(0.8029) = 36.59^{\circ}$
$\angle BXD = 53.32^{\circ}$
$\angle BXA = 120 - \angle BXD = 66.68^{\circ}$
$\text{grey area }= 6 \cdot ((\text{area of }\triangle \text{BXA }) - (\text{area of sector } BXA))$
$\text{area of }\triangle \text{BXA } = 0.5 \cdot |BX|^2 \cdot \tan{\angle BXA} = 2.32$
$\text{area of sector }BXA = 1.16$
Finally: $\text{grey area }= 6 \cdot (2.32-1.16) = 6.96$