Goat problem - Alchetron, The Free Social Encyclopedia

The goat problem is a problem in recreational mathematics known since the 18th century. It was first published in 1748 in England, in the yearly publication The Ladies Diary: or, the Woman’s Almanack.

Problem statement

How big must r be chosen in the diagram, in order for the red area to equal one half of the area of the circle? Illustration: A goat/bull/horse is tethered at point Q . How long needs the line to be, to allow the animal to graze on exactly one half of the circle area?

Solution by calculating the lens area

The area reachable by the animal is in the form of a single-line symmetric lens, limited by the two circular arcs.

The area A of a lens with two cirles of radii R , r and distance between centers d is:

A = r 2 cos − 1 ⁡ ( d 2 + r 2 − R 2 2 d r ) + R 2 cos − 1 ⁡ ( d 2 − r 2 + R 2 2 d R ) − 1 2 ( d + r − R ) ( d − r + R ) ( − d + r + R ) ( d + r + R ) ,

which simplifies in case of R = d = 1 and one half of the circle area to

1 2 π = r 2 cos − 1 ⁡ ( 1 2 r ) + cos − 1 ⁡ ( 1 − 1 2 r 2 ) − 1 2 r 4 − r 2 .

The equation can only be solved iteratively and results in r = 1.1587 … (sequence A133731 in the OEIS).

Solution using integration

By integrating over the right half of the lens area with

1 4 π = ∫ 0 r 2 − r 4 4 ( r 2 − t 2 + 1 − t 2 − 1 ) d t

the transcendent equation

r = π π r − 4 − r 2 + ( 4 r − 2 r ) sin − 1 ⁡ ( 1 2 r )

follows, with the same solution.

The goat in space

In the three-dimensional case, point Q lies on the surface of a unit sphere, and the problem is to find radius r 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.

The volume V of a lens with two spheres of radii R , r and distance between the centers d is:

V = π ( R + r − d ) 2 ( d 2 + 2 d r − 3 r 2 + 2 d R + 6 r R − 3 R 2 ) 12 d

which simplifies in case of R = d = 1 and one half of the sphere volume to

1 2 ⋅ 4 3 π = − 1 4 π r 4 + 2 3 π r 3

leading to a solution of r = 1,228 5 …

It can be demonstrated that with increasing dimensionality, r approaches the value 2 .

The goat and the silo

In the two-dimensional case, the question about the reachable area outside the red circle may be asked. This concerns a situation where the animal is tethered to a silo.

The area consists of a half-circle (light blue) with radius r and of two areas, which are bordered by the red circle and the circle involute (dark blue). Using Leibniz’s sector formula, the size of one of the dark blue areas can be calculated. The entire reachable area (light and dark blue) then equals

A = 1 2 π r 2 + 1 3 r 3

assuming that r ≤ π (otherwise, the two dark blue areas would intersect behind the silo).

References

Goat problem Wikipedia

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