What is the area of the marked region?



We have, by further zooming, the following qualitative picture:

The marked region

Coordinate diagram with the three circles

Choosing coordinates as in the figure, the three circles are described by the equations \[ \mathcal{C}_1 : x^2 + (y-1)^2 = 1, \qquad \mathcal{C}_2 : (x-1)^2 + (y-1)^2 = 1, \qquad \mathcal{C}_3 : \left(x-\tfrac{1}{2}\right)^2 + \left(y-\tfrac{1}{16}\right)^2 = \tfrac{1}{256}. \] The point \(A\) is the intersection point of \(\mathcal{C}_1\) and \(\mathcal{C}_3\) that lies within the square, so we have \[ A = \left(\tfrac{8}{17},\, \tfrac{2}{17}\right). \] By symmetry at the line \(x=\frac{1}{2}\) we get \[ B = \left(\tfrac{9}{17},\, \tfrac{2}{17}\right). \] The point \(C\) is the intersection point of \(\mathcal{C}_1\) and \(\mathcal{C}_2\) that lies within the square so it is \[ C = \left(\tfrac{1}{2},\, \tfrac{2-\sqrt{3}}{2}\right). \] From these coordinates we may compute the lengths \[ AB = \tfrac{1}{17}, \] \[ AC = BC = \dfrac{\sqrt{442 - 255\sqrt{3}}}{17}. \]
The area of the isosceles triangle \(ABC\) is \[ \frac{1}{2} \cdot \frac{1}{17} \cdot \sqrt{\left(\frac{\sqrt{442-255\sqrt{3}}}{17}\right)^{\!2} - \left(\frac{1}{34}\right)^{\!2}} = \frac{30 - 17\sqrt{3}}{1156}. \]
To compute the area of the three circular segments within the triangle \(ABC\) we apply the following formula.
Here \(\ell\) is the length of the chord and \(r\) is the radius of the circle: \[r^2 \arcsin\!\left(\dfrac{\ell}{2r}\right)- \dfrac{\ell}{2}\sqrt{r^2 - \left(\dfrac{\ell}{2}\right)^{\!2}}\].
The circular segments at the sides \(AC\) and \(BC\) have radius \(1\) hence area \[\arcsin\!\left(\dfrac{\sqrt{442-255\sqrt{3}}}{34}\right) - \dfrac{15 - 8\sqrt{3}}{68}\,.\] The circular segment at the side \(AB\) has radius \(\tfrac{1}{16}\), see , hence area \[\dfrac{1}{256}\arcsin\!\left(\dfrac{8}{17}\right) - \dfrac{15}{9248}\,.\]
The area of the marked region can be obtained by subtracting from the triangle area the areas of the three circular sectors: \[ \frac{30-17\sqrt{3}}{1156} - 2\left(\arcsin\!\left(\frac{\sqrt{442-255\sqrt{3}}}{34}\right) - \frac{15-8\sqrt{3}}{68}\right) - \frac{1}{256}\arcsin\!\left(\frac{8}{17}\right) + \frac{15}{9248}\,. \] This expression can be simplified to \[ \frac{15}{32} - \frac{\sqrt{3}}{4} - \frac{\pi}{6} + \frac{255}{256}\arcsin\!\left(\frac{8}{17}\right)\,. \]