What is the area of the marked region?



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

Coordinate diagram with the three circles

Choosing coordinates such that the origin is the top left square vertex, the three circles giving the circular arcs 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 then \[ \frac{1}{2} \cdot \frac{1}{17} \cdot \sqrt{\left(\frac{\sqrt{442-255\sqrt{3}}}{17}\right)^{\!2} - \left(\frac{1}{34}\right)^{\!2}} = \frac{30 - 17\sqrt{3}}{1156}. \]
To compute the area of the three circular segments within the triangle \(ABC\) we apply a known formula involving the arcsin function.
The area of a circular segment with radius \(r\) and chord length \(\ell\) is \[r^2 \arcsin\!\left(\dfrac{\ell}{2r}\right)- \dfrac{\ell}{2}\sqrt{r^2 - \left(\dfrac{\ell}{2}\right)^{\!2}}\].
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 \[ \frac{15}{32} - \frac{\sqrt{3}}{4} - \frac{\pi}{6} + \frac{255}{256}\arcsin\!\left(\frac{8}{17}\right)\,. \]