For the question see Puzzle No 8 - The Gobbling Goat in issue 8.

Let the circular field have radius $r$. \par Let the length of the rope, which is anchored at point $A$ on the circumference of the field, be $R$. \par Now, with the rope at full stretch, the goat will be able to move in an arc from point $B$ on the circumference to point $C$.

Let $O$ be the centre of the field. \par Clearly, the angle $OAB$ is equal to the angle $OAC$. Let the magnitude of $OAB$ be $x$ radians. \par Thus, the area accessible to the goat will be a circle sector with radius $R$ and angle $2x$ (yellow), plus two circle segments (pink) from a circle of radius $r$, cut off by the chords $AB$ and $AC$ respectively.

Now, the area of the circle sector is: $$\frac{1}{2}(R^2.2x)=R^{2}x$$ The area of each circle segment is: $$(1/2)r^2(\pi -2x)-(1/2)r^2\sin (\pi -2x)$$ (because each is a sector of a circle minus a triangle) and so the total area accessible by the goat is: $$R^{2}x+r^2[\pi -2x-\sin (2x)]$$ (the yellow sector plus the two pink segments).

Currently, we have two different variables in our area equations: $r$ and $R$. Let's try to eliminate one. \par Obviously, the length of the line segment $AO$ is $r$, the radius of the field. Similarly, the radius of the line segment $OC$ must be $r$. \par Therefore, by similar triangles, if we drop a perpendicular from $O$ to the line segment $AC$, the perpendicular will bisect $AC$. Therefore the length of $AP$ (and $PC$, of course) is $R/2$. \par We now have a right-angled triangle and enough information to calculate the relationship between $r$ and $R$: $$\begin{array}{rcl}\cos x& =& R/(2r)\\ R& =& 2r\cos x\end{array}$$ \par So the total area accessible to the goat is: $$(4r^2{\cos }^{2}x)x+r^2[\pi -2x-\sin (2x)].$$ \par We wish for this area to be half the area of the total field; therefore we have: $$\begin{array}{rcl}4r^{2}x{\cos }^{2}x+r^2[\pi -2x-\sin (2x)]& =& \pi r^2/2\\ 4x{\cos }^{2}x+\pi -2x-\sin (2x)& =& \pi /2\\ 4x{\cos }^{2}x+\pi /2-2x-\sin (2x)& =& 0\end{array}$$

We can't easily solve the equation but we can use a graphical calculator or numerical method such as Newton-Raphson to find an approximate solution.

Using the Newton-Raphson method as described in the Coda, we find that $x$ is approximately $0.953$, and therefore $\cos x=0.579$. \par Now, since $R=2r\cos x$ and $r$, the radius of the field, is 100m, we have $R=200\cos x$ and thus the required length of rope is approximately 116m.

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Coda: Solving the equation using Newton-Raphson

The basic idea

In the goat puzzle, we were left with the following equation to solve: $$4x{\cos }^{2}x+\pi /2-2x-\sin (2x)=0$$ The Newton-Raphson method is an approximate method for finding roots of equations that are differentiable. \par Let $f(x)$ be a differentiable function. Since $f(x)$ is differentiable, every point on the graph of $f(x)$ must have a gradient and a unique tangent line. \par Now, the tangent at $x_0$ is an approximation to the graph of $f(x)$ near the point $(x_0,f(x_0))$. \par Therefore the zero of the tangent line (the point where the tangent line crosses the $x$-axis) is an approximation (perhaps a very bad one, however!) of the zero of $f(x)$ (the point where $f(x)$ crosses the $x$-axis, i.e. the root of $f(x)$). It's like we're pretending that $f(x)$ is really a straight line, like the tangent line, and therefore crosses the $x$-axis at the same place the tangent does.

In the Newton-Raphson method, we start with a "best guess" $x_0$ as to the zero of $f(x)$. We then calculate the first approximation, $x_1$, as the zero of the tangent line to $f(x)$ at $x_0$. \par We then calculate the second approximation, $x_2$, as the zero of the tangent line crossing the $x$-axis at $x_1$, and so forth. \par The diagram above shows the initial guess $x_0$, the first approximations $x_1$ and the relevant tangents. The second approximation $x_2$ is the coordinate where the second tangent crosses the $x$-axis. As you can see, the approximations are getting closer to the actual zero point of $f(x)$. If we continue iterating like this, we will get better and better estimates for the zero point of $f(x)$.

How do we do it?

We wish to solve $4x{\cos }^{2}x+\pi /2-2x-\sin (2x)=0$. Obviously, plotting $f(x)=4x{\cos }^{2}x+\pi /2-2x-\sin (2x)$ and drawing tangents is not going to be very much fun! However, we can perform Newton-Raphson numerically. \par Our initial point is $x_0$. The gradient of $f(x)$ at $x_0$ is given by $f^{\prime }(x_0)$, and the tangent line to $f(x)$ at $x_0$ is therefore given by: $$y-f(x_0)=f^{\prime }(x_0)(x-x_0)$$ To find $x_1$, we must find the point where this tangent crosses the $x$-axis, i.e. to let: $$0-f(x_0)=f^{\prime }(x_0)(x_1-x_0)$$ and therefore $$x_1-x_0=\frac{-f(x_0)}{f^{\prime }(x_0)}$$ so that $$x_1=x_0-\frac{f(x_0)}{f^{\prime }(x_0)}$$ Similarly, in the general case we obtain: $$x_{n+1}=x_n-\frac{f(x_n)}{f^{\prime }(x_n)}$$ \par Now, our function is $f(x)=4x{\cos }^{2}x+\pi /2-2x-\sin (2x)$. Via standard differential calculus, the gradient $f^{\prime }(x)$ of this function is $$4{\cos }^{2}x-8x\cos x\sin x-2-2\sin (2x)\cos (2x).$$ Therefore, to find the approximate root of $f(x)$ we can use the following: $$x_{n+1}=x_n-\frac{4x_n{\cos }^2{x}_n+\pi /2-2x_n-\sin (2x_n)}{4{\cos }^2{x}_n-8x_n\cos x_n\sin x_n-2-2\sin (2x_n)\cos (2x_n)}$$ \par So, we know how to calculate $x_{n+1}$ from $x_n$. But how do we find our starting value, $x_0$? Well, in this particular case we know that the magnitude of $x$ must be between 0 and $\pi /2$ radians (go back to the second diagram and think about it if you're not sure why!). So a good initial guess might be (for example) $\pi /4$. \par As it turns out, all sorts of values will do: here's a table of the iterative steps of Newton-Raphson on our function $f(x)$ for a range of initial values of $x_0$. As you can see, they all converge quite rapidly to the same twelve-significant-digit approximation.

	$x_0=\pi /4$	$x_0=\pi /6$	$x_0=\pi /3$	$x_0=\pi /2$
$x_0$	0.785398163397	0.523598775598	1.047197551200	1.570796326790
$x_1$	0.967088277214	1.254847487960	0.956164730983	0.785398163397
$x_2$	0.953058379193	0.966611488070	0.952884951928	0.967088277214
$x_3$	0.952849994306	0.953049230238	0.952848237620	0.953058379193
$x_4$	0.952847886046	0.952849901115	0.952847868401	0.952849994306
$x_5$	0.952847864870	0.952847885110	0.952847864693	0.952847886046
$x_6$	0.952847864657	0.952847864860	0.952847864655	0.952847864870
$x_7$	0.952847864655	0.952847864657	0.952847864655	0.952847864657
$x_8$	0.952847864655	0.952847864655	0.952847864655	0.952847864655
$x_9$	0.952847864655	0.952847864655	0.952847864655	0.952847864655