Goat grazing problem

If the goat were instead tethered to a post on the edge of a circular path of pavement that did not obstruct the goat (rather than a fence or a silo), the interior and exterior problem would be complements of a simple circular area.

The original problem was the exterior grazing problem and appeared in the 1748 edition of the English annual journal The Ladies' Diary: or, the Woman's Almanack, designated as Question CCCIII attributed to Upnorensis (an unknown historical figure), stated thus: Observing a horse tied to feed in a gentlemen's park, with one end of a rope to his fore foot, and the other end to one of the circular iron rails, enclosing a pond, the circumference of which rails being 160 yards, equal to the length of the rope, what quantity of ground at most, could the horse feed?

The related problem involving area in the interior of a circle without reference to barnyard animals first appeared in 1894 in the first edition of the renown journal American Mathematical Monthly.

The solutions in both cases are non-trivial but yield to straightforward application of trigonometry, analytical geometry or integral calculus.

Both problems are intrinsically transcendental – they do not have closed-form analytical solutions in the Euclidean plane.

Three-dimensional analogues and planar boundary/area problems on other shapes, including the obvious rectangular barn and/or field, have been proposed and solved.

The complication here is that the grazing area overlaps around the silo (i.e., in general, the tether is longer than one half the circumference of the silo): the goat can only eat the grass once, he can't eat it twice.

The answer to the problem as proposed was given in the 1749 issue of the magazine by a Mr. Heath, and stated as 76,257.86 sq.yds.

If it matters, there is a constructive way to obtain a quick and very accurate estimate of

Find the area between a circle and its involute over an angle of 2π to −2π excluding any overlap.

In Cartesian coordinates, the equation of the involute is transcendental; doing a line integral there is hardly feasible.

Because the "sweep" of the area under the involute is bounded by a tangent line (see diagram and derivation below) which is not the boundary (

(see diagram, above) as the origin with the circle representing the circumference of the pond below the x-axis, and

on the y-axis below the circle representing the point of intersection of the tether when wound clockwise and counterclockwise, let

be the point of intersection of the circumference of the pond on the y-axis (opposite to

That is a transcendental equation that can only be solved by trial-and-error, polynomial expansion, or an iterative procedure like Newton–Raphson.

This sum is The bounds of the integral represent the area under the involute in the fourth quadrant between

Unfortunately, there is no way to simplify the latter term representing the lower bound of the eval expression because

The area reachable by the animal is in the form of an asymmetric lens, delimited by the two circular arcs.

and one half of the circle area to The equation can only be solved iteratively and results in

and integrating over the right half of the lens area with the transcendental equation follows, with the same solution.

[5] Assuming the leash is tied to the bottom of the pen, define

as the angle the taut leash makes with upwards when the goat is at the circumference.

then taking the cosine of both sides generates extra solutions even if including the obvious constraint

Using trigonometric identities, we see that this is the same transcendental equation that lens area and integration provide.

By using complex analysis methods in 2020, Ingo Ullisch obtained a closed-form solution as the cosine of a ratio of two contour integrals:[6] where C is the circle

The three-dimensional analogue to the two-dimensional goat problem is a bird tethered to the inside of a sphere, with the tether long enough to constrain the bird's flight to half the volume of the sphere.

lies on the surface of a unit sphere, and the problem is to find radius

The volume of the unit sphere reachable by the animal has the form of a three-dimensional lens with differently shaped sides and defined by the two spherical caps.

It can be demonstrated that, with increasing dimensionality, the reachable area approaches one half the sphere at the critical length