The goat grazing problem is either of two related problems in recreational mathematics involving a tethered goat grazing a circular area: the interior grazing problem and the exterior grazing problem. The former involves grazing the interior of a circular area, and the latter, grazing an exterior of a circular area. For the exterior problem, the constraint that the rope can not enter the circular area dictates that the grazing area forms an involute. 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 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 160yards, 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. Attributed to Charles E. Myers, it was stated as: A circle containing one acre is cut by another whose center is on the circumference of the given circle, and the area common to both is one-half acre. Find the radius of the cutting circle. 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. The numerical answers must be obtained by an iterative approximation procedure. The goat problems do not yield any new mathematical insights; rather they are primarily exercises in how to artfully deconstruct problems in order to facilitate solution.

Three-dimensional analogues and planar boundary/area problems on other shapes, including the obvious rectangular barn and/or field, have been proposed and solved. A generalized solution for any smooth convex curve like an ellipse, and even unclosed curves, has been formulated.

Exterior grazing problem

The question about the grazable area outside a circle is considered. This concerns a situation where the animal is tethered to a silo. 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. which was arrived at partly by "trial and a table of logarithms". The answer is not so accurate as the number of digits of precision would suggest. No analytical solution was provided.

A useful approximation

Let tether length R=160yds. and silo radius r=R/(2). Let c be the length of an arc segment of the involute between the y-axis and a vertical line tangent to the silo at θ=3/2; it is the arc subtended byΦ (while the arc is minutely longer thanr and θ here is , r. The area of the involute excludes half the area of the silo (1018.61) in the fourth quadrant, so its approximate area is 18146, and the grazable area including the half circle of radius R. That is 249 sq.yds. greater than the correct area of 76256, an error of just 0.33%. This method of approximating may not be quite so good for angles <3/2 of the involute.

Solution by integrating with polar coordinates

Find the area between a circle and its involute over an angle of 2 to −2 excluding any overlap. In Cartesian coordinates, the equation of the involute is transcendental; doing a line integral there is hardly feasible. A more felicitous approach is to use polar coordinates(z,θ) between overlapping areas, the decomposition of the problem results in four computable areas: a half circle whose radius is the tether length (A1); the area "swept" by the tether over an angle of 2 (A2); the portion of area A2 from θ= (A3); and the wedge area qFtq (A4). So, the desired area A is A1 + (A2−A3+A4)·2. The area(s) required to be computed are between two quadratic curves, and will necessarily be an integral or difference of integrals.

(see diagram, above) as the origin with the circle representing the circumference of the pond below the x on the y.

.

First, the area A1

Next compute the area between the circumference of the pond and involute. Compute the area in the tapering "tail" of the involute, i.e. the overlapped area (note, on account of the tangent tF, that this area includes the wedge section, area A4. This sum is

.

times the central angle don't significantly affect the final result.

A4 by reduction.

(80,425sq.yds.), keep 6 digits of precision (5 digits plus a guard digit):

1. ; the length of the tether (given)

2. ; the radius of the pond

3. and the Y axis

4. and the Y axis

is the area bounded by the involute and pond in the 4th quadrant.

; the grazing area

sq.yds.rounded up to the nearest square yard.) of the silo, or no overlap to account for. The goat can eat all but a bit more than 5% of the area of the great circle defined by its tether length, and almost half the area it cannot eat is within the perimeter of the pond/silo.

Solution by ratio of arc length

.

Interior grazing problem

need to be to allow the animal to graze on exactly one half of the circle's area (white area in diagram, in plane geometry, called a lens)?

Solution by calculating the lens area

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

is

and one half of the circle area to

The equation can only be solved iteratively .

Solution using integration

and integrating over the right half of the lens area with

the transcendental equation

follows, with the same solution.

, the transcendental equation derived from lens area can be obtained.

Solution by sector area plus segment area

The area can be written as the sum of sector area plus segment area.

as the angle from downwards to the same location, but from the center of the pen instead of from the center of the larger circle. The sum of angles of a triangle for the resulting isosceles triangle.

. Using the circular sector and circular segment area formulae gives

1. ,

.

Combining into a single equation gives

1. .

.

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

Closed-form solution

By using complex analysis methods in 2020, Ingo Ullisch obtained a closed-form solution as the cosine of a ratio of two contour integrals:

where C.

A solution can also be written using Bell polynomials, which follows from the Lagrange inversion theorem:

where

.

3-dimensional extension

lies on the surface of a unit sphere of the second sphere so that the volume of the intersection body equals exactly half the volume of the unit sphere.

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.

is

and one half of the sphere volume to

, the area covered approaches the sphere's entire area.

See also

• Mrs. Miniver's problem, another problem of equalizing areas of circular lunes and lenses

References

External links

• The Goat Problem - Numberphile video about the goat problem.