# 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.

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I'm guessing that most fence posts are at the edge of a field, which makes this a far more interesting problem. – FordBuchanan Jul 20 '10 at 19:48
@FordBuchanan: Correct. I've updated the question to (hopefully) make that clear. Sorry for the confusion! – e.James Jul 20 '10 at 19:51

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 $L=\text{ca. }1.159 R$ .

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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):

(pi*r[1]^2) / 2 = pi*r[2]^2

You can then simplify it down:

r[1]^2 / 2 = r[2]^2

and then taking roots:

r[2] = 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.

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The post must be on the boundary of the field. – augurar Sep 2 '14 at 2:36

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).

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Like myself, you assumed the post should be in the center of the field. I deleted my answer because of this mistake. The problem has been reformulated. – BBischof Jul 20 '10 at 20:03