You should be able to solve this

You should be able to solve this.

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true

In R7.5
Arrange spheres in pairs on opposite sides of central spheres on orthogonal axes such that they touch the center sphere

0) following the above, all peripheral spheres touch the center sphere
1) 7.5 *2 = 13 spheres, requirement satisfied
2) distance between closest spheres' centers (disregarding center sphere) is sqrt(2*2^2) > 2, ergo no spheres overlap

Ergo op sucks cocks

Qed

>7.5 *2 = 13

Arithmetic isn't math and as such beneath me.

It still works so fuck off.

The trick is that the 13th sphere is the central sphere. The remaining 12 are peripheral.

>7.5 *2 = 13 spheres
Brainlet detected

Because if more than 12 were possible, FCC/ABC atomic arrangement would not have the densest packing ratio.

t. chem/phys brainlet

Proof:
1) 12 spheres can all touch the central sphere without overlap. (Obvious axiom)
2) 13 spheres cannot (Trivial principle)
QED

That's really easy if you know about how chemistry structures works
Maximum number of spheres around one sphere is 12 (6 around the xy plane, 3 on +z axis, and 3 on -z axis)
About the math demostration, heh, i got no idea

Working on a legit proof, nobody prove before me pls

Do they have to be evenly distributed and of the same size?

1) Let each ball be represented by a vertex on a polyhedron.
2) The edge length of the shape must be AT LEAST 2 because the closest two unit spheres can be is 2*r.
3) let us inscribe the polyhedron in a sphere, such that the radius of the sphere, R, is the length of the central point to any vertex.
4) The radius of this sphere must be exactly 2, as all spheres must touch the central sphere, and the shortest distance between spheres is 2r = 2.
6) Suppose this polyhedron is an icosahedron (12 vertices)
5) The formula for the resulting radius of an Icosahedron is [math]\dfrac{\phi \sqrt{5}}{2}a[/math] where [math]\phi[/math] is the golden ratio and [math]a[/math] is the length of a side. Subbing in the radius, this expression evaluates to roughly 2.1, meaning that the representation is valid for 12 tangential balls.
13) Now let us move on to 13 and see if it still holds...

>prove that you cannot place 13 peas on planet earth so that they do not overlap

I'm tired and want to go to bed. My next steps would involve one of the following:
1) Use Euler's formula (V-E+F=2) and prove that such a shape is impossible with a radius of 2 and a side length > 2
2) use the .1 "wiggle room" from the previous example to push the balls away from one spot, allowing me to show that under the best conditions, a 13th ball added would intersect its neighboring balls if it touched the central sphere.
etc.

If anybody would like to finish my proof, feel free

Learn to read nigger

it said ontop of a sphere you dunce is the earth a sphere?

what would one have to learn to be able to prove this? Is this like arcane russian geometry or american math?

based on quarters on a table in 2d. I think it is possible.

trigonometry, dumbass

im not writing out equations.

>draw like 7 congruent circles like a flower
>???
>profit

The inner circle will always be larger, the closest you can get the ratio of the size of the inner circle and any of the others to 1/1 is 1.866/1.

>arithmetic isn't math

user, you're only digging your brainlet hole deeper. I'm going to prescribe Serre's "A Course in Arithmetic" to combat this. And because your condition is clearly so severe, I'm also going to recommend Apostols "Analytic Number Theory"

Proof;
>this proof is left to the reader as an excersize

unit peas are the size of the earth though

Unit sphere touching a central unit sphere i.e. they all have to be the same size including the center one

yes

It said a unit sphere, ie. a sphere with radius 1.

why can't makise be real Veeky Forums bros
i love her so goddamn much

en.wikipedia.org/wiki/Sphere_packing

the wiki article you want is kissing number not sphere packing you cocky twit