The cow in the field problem (intersecting circular areas)

The field is the smaller/left circle, centered at A. The cow is tied to the post at E. The larger/right circle is the grazing radius. Let the radius of the field be R and the length of the rope be L.

The grazable area is the union of a segment of the circular field and a segment of the circle defined by the rope length. (A segment of a circle is a sector of a circle less the triangle defined by the center of the circle and the endpoints of the arc.) The area of a segment of a circle of radius $R$ with central angle $t$ is $\frac{1}{2}R^2(t-\sin(t))$, where $t$ is measured in radians.

In order to express the grazable area in terms of $R$ and one angle, we consider the angles ∠CED and ∠CAD (which define the segments of the circles; call these α and β for convenience) and the triangle CEF. Let $\theta$ be ∠EFC. $2\theta$ is an inscribed angle for the central angle $\beta$ over the same arc, making $\beta = 4\theta$. The sum of angles in triangle CEF is $\theta + \pi/2 +\alpha/2=\pi$ or $\alpha =\pi-2\theta$.

The grazable area is $\frac{1}{2}L^2(\alpha-\sin\alpha)+\frac{1}{2}R^2(\beta-\sin\beta)=R^2(\frac{1}{2}(L/R)^2((\pi-2\theta)-\sin(\pi-2\theta))+\frac{1}{2}(4\theta-\sin(4\theta)))$, where $a = CE = L/R=2\sin(\theta)$. We want that to be equal to half the area of the field, $\frac{1}{2}\pi R^2$.

That is, the equality of areas is $$R^2(2(\sin(\theta))^2((\pi-2\theta)-\sin(\pi-2\theta))+\frac{1}{2}(4\theta-\sin(4\theta)))=R^2\frac{\pi}{2}$$

Simplifying:

$$R^2(\pi+(2\theta-\pi)\cos(2\theta)-\sin(2\theta)=\frac{\pi}{2})$$

(The grazable area seems to be $\pi+\alpha\cos\alpha-\sin\alpha$; can this be seen easily?)

The desired equality of areas is obtained for $\theta = \text{ca. } 0.618$ or $L=\text{ca. }1.159 R$ .