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Only way I can think of is to make a little pyramid from earth. Plant one seed on each corner. Three will be at ground level in a triangle and the other one will be at the top of the earth mound.

 

Is there another way to have them planted and equidistant?

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Only way I can think of is to make a little pyramid from earth. Plant one seed on each corner. Three will be at ground level in a triangle and the other one will be at the top of the earth mound.

 

Is there another way to have them planted and equidistant?

Yes. Plant as in corners of tetrahedron or triangular pyramid. You can also dig a cavity for the same.

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using your hands and other tools or containers, make a (globe) sphere of a quantity of soft, moist soil which, when compressed, coheres and holds its speherical shape.

 

 

 

lat-long-sphere.png

 

Then, place each of the four seeds equidistant (midway from center to surface) from the sphere's center and two equidistant mid-points in each (upper) and (lower) hemisphere.

 

In the model image above, a seed is planted along the radius --midway from the surface to the sphere's center point. Similarly, another seed in the same hemisphere is planted at 180° from the position of the first. Repeat this in the lower hemisphere, at right angles to the plane of the two seeds in the opposite hemisphere.

 

When finished, bury the sphere at any desired depth and water, fertilize, etc. :)

Edited by proximity1
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using your hands and other tools or containers, make a (globe) sphere of a quantity of soft, moist soil which, when compressed, coheres and holds its speherical shape.

 

 

 

lat-long-sphere.png

 

Then, place each of the four seeds equidistant (midway from center to surface) from the sphere's center and two equidistant mid-points in each (upper) and (lower) hemisphere.

 

In the model image above, a seed is planted along the radius --midway from the surface to the sphere's center point. Similarly, another seed in the same hemisphere is planted at 180° from the position of the first. Repeat this in the lower hemisphere, at right angles to the plane of the two seeds in the opposite hemisphere.

 

When finished, bury the sphere at any desired depth and water, fertilize, etc. :)

Wouldn't it be a square then?

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Wouldn't it be a square then?

No. Take the radius line segment described by the angle of phi from the center to the surface. Call the end-points of that radius "c" (at the sphere's center) and "N1" on the sphere's surface.

 

Now, from "c," mark the point on N1's hemisphere which is its chord (opposite) endpoint and call that "N2". So angle of phi + the angle of the N2, c equals ninety degrees.

 

Repeat these steps in the lower hemisphere, at a right-angle to the plane which passes through N1 & N2.

 

Now, taking each of the four points on the surface as a radius, relocate each midway to c.

 

the resulting four points are equidistant from both c and each other. They don't describe a square since they are not all on the same plane.

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No matter how you describe the process, the four seeds end up at the 4 corners of a regular tetrahedron. You can use the sphere to get there, but the answer has not changed.

If you think there's a "corner" (I.e. a ninety-degree angle) _anywhere_* in the result then you're not conceiving it correctly. In fact, there's no flat, rectilinear surface connecting any three points--only imaginary curved surfaces. The two points common to each hemisphere describe lines which nowhere intersect.

 

* The sphere's center is not one of the "seeds" --it's only a conceptual point of reference. Once buried, it has no part in the figure. It simply a common distance to each of the 4 points.

Edited by proximity1
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Wow - I didn't think it would be possible to misunderstand a correct answer so badly. Four equidistant points are at the vertices of a tetrahedron - exactly as DrKrettin described. Your purported answer is so needlessly complex that I doubt any have even bothered to check its validity; the first sentence itself is fairly meaningless and I stopped there.

 

 

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the rules don't require that the four points --or any three of them--share the same plane. I never disputed the validity of the tetrahedron reply at

 

http://www.scienceforums.net/topic/101796-a-simple-question/#entry962407

 

But that does _not_ correspond to my answer. If you took my four points, a plane described by the line of the two points in the "northern hemisphere" would bisect at a right angle the mid point of the plane of the line described by the two points in the "Southern hemisphere." That's not possible in a tetrahedron's four points--any three of which constitute a common plane surface.

 

 

Two put it differently, a regular tetrahedron can have either one or three points in either of two opposite hemispheres. My four are either of the following two configurations :

 

N hemisphere : 2 ; S hemisphere : 2; E hemisphere : 1 ; W hemisphere : 1

 

or N : 1; S :1; E : 2 & W : 2

 

I won't respond to any further smart-ass criticism but will answer good-faith questions or objections.

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This is arguing a silly point to death. No matter how you place your seeds on the surface of that sphere, the 4 seeds will form the vertices of a regular tetrahedron. What else are you arguing about? You are using a sledgehammer to crack a nut.

Maybe you think the whole puzzle is silly?

 

Are you seriously claiming what the plain meaning of these words say? : "No matter how you place your seeds on the surface of that sphere, the 4 seeds will form the vertices of a regular tetrahedron." ?

 

Suppose they're all planted at the North pole? Or two are planted at each pole? or all are planted along the "equatorial" line-- with one meter between each seed?

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Maybe you think the whole puzzle is silly?

 

Are you seriously claiming what the plain meaning of these words say? : "No matter how you place your seeds on the surface of that sphere, the 4 seeds will form the vertices of a regular tetrahedron." ?

 

Suppose they're all planted at the North pole? Or two are planted at each pole? or all are planted along the "equatorial" line-- with one meter between each seed?

 

*sigh* What I said was in the context of your unnecessarily complicated solution of placing the seeds on the sphere so that they were equidistant from each other. I shall now take the trouble to expand what I said so that even you might understand:

 

No matter how you place your seeds on the surface of that sphere such that they are all equidistant from each other, the 4 seeds will form the vertices of a regular tetrahedron.

 

Happy now?

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