 A geometry problem from the Riddler blog. Here it goes:

A farmer owns a circular field with radius R. If he ties up his goat to the fence that runs along the edge of the field, how long does the goat’s tether need to be so that the goat can graze on exactly half of the field, by area?

Here is my solution:
[Show Solution]

## 4 thoughts on “Tether your goat!”

1. Adam says:

I reached the same solution, using basically the same approach. I worked out the equation for the area of the lens, where the field is centered at (R,0) and the goat is tied up at (0,0). The lens has two halves spreading out from the intersection of the two circles, the right side bounded by the goat arc and the left side bounded by the field arc. I couldn’t quite resolve the equation, so instead I plugged in values for a field radius R and a goat leash length of R+N. I iterated for a bit and found that N = R/6.3. This works out to the same value you found.

Here’s the equation for half the area of the field equal to the area of the lens:
(1/2)(pi * R^2) =
(1/2)(R^2)[2 * arcsin( sqrt((R+N)^2 – (((R+N)^2)/(2R))^2)/R )
– sin(2 * arcsin( sqrt((R+N)^2 – (((R+N)^2)/(2R))^2)/R ) ) ] +
(R+N)^2)[2 * arcsin( sqrt((R+N)^2 – (((R+N)^2)/(2R))^2)/(R+N) )
– sin(2 * arcsin( sqrt((R+N)^2 – (((R+N)^2)/(2R))^2)/(R+N) ) ) ]

2. James Jones says:

I also went for the area of the lens, but figured that WLOG, we could solve for R=1 and choose the coordinate system so that the stake is at (1,0). Then the x axis divides the lens into half, and the half above the x axis can in turn be split into the part bounded by the goat’s tether and that bounded by the field (aka unit circle). The switch comes at the place where those circles intersect, i.e. at the positive x for which sqrt(r^2 – (x – 1)^2) = sqrt(1 – x^2), which turns out to be (2 – r^2)/2.

I wimped out and calculated the integral numerically–shouldn’t have bothered to write an adaptive trapezoidal routine when there’s scipy.integrate.quad. (OTOH, there are online symbolic integrators; they’d be worth a try.) In any case, it does give a result consistent with Mr. Laurent’s.

3. Pierre Bierre says:

I solved the grazing radius problem using computational math. I have a 2D Algorithmic Geometry library in Java (high school students I teach develop this library and a more advanced 3D one).

One of the watershed 2D problems we solve algorithmically is the two intersection points of two overlapping circles. You can apply this algo to solve for the area of the football-shaped grazing region, starting with a reasonable guess for grazing radius r. As James Jones notes, the football cleaves down the middle using the line segment bridging the two circle intersection points. Each of these “lens” shapes is the minor segment of a circle+chord. The area of a minor segment can be solved by subtracting two areas…..the pizza-pie slice minus the central (isoscoles) triangle.

A converging iterative search for the value of grazing radius is not hard to set up, using Gauss’s method. Compute the football area A using the latest estimate for r. Compute the error in A
( A – 0.5 * pi). Flip the sign of this error to make useful as a correction to be added to r, and scale the correction by 1 / 2*pi*r which is the local derivative of pi*r_squared with respect to change in r.

r = 1.3 (starting guess)

loop:
compute A (football area)
A_error = A – 0.5 * pi
r = r – A_error / (2 * pi * r) // Gauss’ method
until (abs (A_error) < 0ne_billionth)

r = 1.158728473 * R

4. solved with pen and paper back in '88 says:

You forgot to take into account the length of the goat’s head which should be subtracted from the length of the rope, otherwise it eats to much.