The cow in the field problem (intersecting circular areas) What length of rope should be used to tie a cow to an exterior fence post of a circular field so that the cow can only graze half of the grass within that field?
updated: To be clear: the cow should be tied to a post on the exterior of the field, not a post at the center of the field. 
 A: So, the area of the field is $\pi r^2$ and you want the cow to be able to graze an area equal to half of that.
All you need to do is set up the equation ($r_1$ is the radius of the field, $r_2$ is the length of the rope desired):
$$\frac{(\pi r_1^2)}{2} = \pi r_2^2$$
You can then simplify it down:
$$\frac{r_1^2 }{2} =r_2^2$$
and then taking roots:
$$r_2 =\frac{ r_1 }{\sqrt{2}}$$
So you need a rope that is equal to the radius divided by the square root of $2$, and the post can be no closer than this distance to the edge of the field.
A: Let the total area of the field = $A$.
We know $A = \pi R^2$ where $R$ = the radius of the field.
We want the cow to be able to graze half the area, so we solve for a length of rope $r$ such that $\pi r^2 = A / 2$.
This gives: $\pi r^2 = \pi R^2 / 2$, hence $r = R / \sqrt(2)$.
In words, the length of the cow's rope should be the radius of the field divided by sqrt(2).
A: 
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  $L=\text{ca. }1.159 R$ .
