Being as people were bored and someone asked for a puzzle....

[MyUserName]

Depends. Which part of the goat is attached to the inflexible chain, how big is the goat, how long is its neck etc. If it's a very small field that I supposed it's feasible.

You can assume that the goat is a particle.

[pjclarke]

So, circle 1 in the field.

Given the mythical goat is tether at the inside edge then the length of the tether defines another circle.

However it is bounded by the first circle, so defines an arc.

So the question is

"What is radius of a circle which is centered on the edge of another circle where the overlapping arc is equal to half the entire area of the overlapped circle"

At least I think that is the question.

So it is, and there is a formula for overlapping circles, that I CBA to look up but which involves a square root and at least one cosine. The intersecting arcs define an area consisting of two segments the triangular height of which can be derived from the radiuses.

At this point I hand over Bob to code up the solution.

[MyUserName]

So it is, and there is a formula for overlapping circles, that I CBA to look up but which involves a square root and at least one cosine. The intersecting arcs define an area consisting of two segments the triangular height of which can be derived from the radiuses.

At this point I hand over Bob to code up the solution.

He got back to me and said that the answer was "Helium" but when I tried running the program myself it threw an exception.

[ASB]

Yes, that is the question. Or is it?

Tether needs to be > radius of the field otherwise clearly it would reach less than half the field area. It needs to be < √2 or clearly the goat would reach more than half the field. So somewhere between 1.0 and 1.4ish.

However the maths seems so impossibly difficult that it must be a trick question. Is the tether fixed to a single point? I'm not sure that was stated. Anyway, off to google now...

Yes. I was looking at the ratio the wrong way round.

Root 2 looks like it might have been the right answer (it isn't).

The 4 relevant areas are 2 segments with a chord of length r2 (on circle1). An isoceles triangle with sides of r2 and an angle unknown. The final piece being a chord with length the base of the isoceles triangle (on circle2).

It will involve using a cosine and an arc sine.

Maybe I should have listened a bit more in my geometry and trig all those years ago when I was about 13.

[xoggoth]

Oh, inside edge! Didn't read it properly, thought it looked too easy.

[Doggy Styles]

Does this involve that funny old calculus stuff?

[Bacchus]

The 4 relevant areas are 2 segments with a chord of length r2 (on circle1). An isoceles triangle with sides of r2 and an angle unknown. The final piece being a chord with length the base of the isoceles triangle (on circle2).

I think there are just two areas? If you draw a straight line between where the goat's arc of nibbling intersects with the edge of the garden (a chord) you will have two segments.

I can't remember the formula for the area of a segment, but you can be pretty sure it involves pi and the radius, the radius of the garden is fixed, and the radius of the goat's arc of nibbling is what we are trying to find.

If only I still smoked I could draw it on the back of a fag packet (c:

[BrilloPad]

Goat curry anyone?

[Bacchus]

You can assume that the goat is a particle.

Goat curry anyone?

There won't be enough to go round if it's a particle