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/sci/ - Science & Math

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3239929 No.3239929 [Reply] [Original]

explain to me why you can't put 5 equidistant points in 3 dimensional space.

>> No.3239935

you can. you can put infinitely many points in 3 dimensional space = sphere.

>> No.3239933

Because it's not possible?

>> No.3239940

>>3239935
>equidistant

>> No.3239938

try and do it in your head, won't ever work

>> No.3239941

>>3239935
you're fucking retarded. He means 5 points equidistant from each other, not one individual point

>> No.3239943

oh! hahah. my bad, didn't read properly!

>> No.3239944

>>3239935
you need to rework that
all points are equidistant from the center with a circle..

there is no central point here

>> No.3239949

>>3239935
The points on a sphere aren't equidistant. The distance from a point on the equator to the north pole is less than the distance from the north pole to the south pole.

The reason its not possible should be obvious if you just try to imagine the geometry. I'm sure there is a mathematical proof, but I crammed for those classes and forgot everything after the exam.

>> No.3239950
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3239950

What if I put all the points on the same point? They would all be the same distance from each other; 0.

>> No.3239952

>>3239950
i just thought of that as well...creepy.

>> No.3239955

>>3239950
haha, dam you!
no, you can't do that, I'll just rephrase the question :}

>> No.3239957

>>3239950 all the points on the same point?
Then you have a single point, not five points.

>> No.3239961

not possible. equidistance requires triangles.

>> No.3239966

>>3239957

You have 5 variables, v, w, x, y, and z

All of them occupying a spot in 3 dimensional space. Travelling towards each other until they meet. Then they stop.

5 points, all equidistant. Limitations of 3 dimensional space have now been smashed.

>> No.3239970
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3239970

>>3239961

>I have never in my life heard of, witnessed or experienced a square.

Shit son, what are you doing?

>> No.3239972

>>3239966
Prettysure zero distance spans a null space

>> No.3239973

>>3239966

when they meet, they will not be on top of each other, since they will then be one point, only able to be described by one co-ordinate, thus one point.

If they are not occupying the same position, then they are not equidistant

>> No.3239974

>>3239966
that sounds like you're missing some information about the variables

>> No.3239977

>>3239973

All 5 points will have the same co-ordinate but you can't suddenly say, v, w, y and z no longer exist because x = 0.

>> No.3239982

>>3239961
60 daaa ggrrrreeeeeeeeeeeeeees

>> No.3239984

>>3239970

a square, you say?
let me get this right, you say the points in a squre are equidistant?

you sure about that?

>> No.3239992

>>3239984

What are you trying to imply? the diagonal line through the middle is longer than an edge? Yes all points are equidistant.

>> No.3240005

.(a) .(b)

.(c) .(d)

ab =ab =! ad

thusthe distance from a to each of the othe points is not equal, thus not equidistant.

dj. 1. equidistant - the same distance apart at every point

>> No.3240006

>>3239984

Why do you think OPs question is 5 points, not 4?

>> No.3240009

>>3239977
I can call myself sally, john, nick and sarah but that doesn't make me 5 people occupying the same place

>> No.3240010

ab = ac =! ad
sorry typo

>> No.3240017

>>3240006
http://en.wikipedia.org/wiki/Tetrahedron
goddammit /sci/

>> No.3240022

>>3240017
>http://en.wikipedia.org/wiki/Tetrahedron

Sorry I am from /g/ trying to be more /sci/ but quite clearly failing.

>> No.3240028

>>3239992
yes, diagonal > edge

>> No.3240029

>>3240009

Now you're getting into the philosophy behind multiple personality disorders. I do not approve of philosophy. It cannot be proven.

>> No.3240031
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3240031

lol, someones been trolled.

>> No.3240033

>>3240006
With four points you could make a tetrahedron, and the points would be equidistant. NOT a squre though.

>> No.3240036

>>3240009

No you're right, it doesn't make you 5 people.

HOWEVER

it does make you 4 people.

>> No.3240038
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3240038

it'd be possible in 4th dimensional space...

>> No.3240040

>>3240036

and thus a square, because that is OBVIOSLY the ONLY shape you can make with 4 points. ever.

>> No.3240047

>>3240040

What are you talking about? You're trying to mix the two conversations in this thread?

>> No.3240050 [DELETED] 

>>3240040
3 sided pyramid, not square.

>> No.3240053

>>3240040
4 sided pyramid including bottom, not square.

>> No.3240061

>>3240038
no 4th dimension 4 u!

>> No.3240062

>>3240038
yes, and fairies exist in magic land.

>> No.3240070

>>3240062
not in practise, tsch, just in theory.
you can have n+1 equidistant points in n dimensions.

for example, on 2d, like on a sheet of paper, an equilateral triangle works. but no 4 pointed shape works on 2d paper.

>> No.3240071

>>3240029
so basically you wouldn't agree with multiple point disorder either... I wouldn't blame you.

>> No.3240082

>>3240036
no I was also including myself which you could call meticulous which you didn't count.

>> No.3240092

OP didn't specify which metric … So I hereby define <span class="math">d(x,y)=0[/spoiler] if <span class="math">x=y[/spoiler] and <span class="math">d(x,y)=1[/spoiler] if <span class="math">x \neq y[/spoiler].

Enjoy an infinite amount of equidistant points.

>> No.3240111

make a line. place 3rd point in center. drag 3rd point into 2nd dimension. you now have a triangle.

place 4th point in center of triangle. drag 4th point up into 3rd dimension. you now have a tetrahedron.

place 5th point dead center. drag 5th point up into 4th dimension.

>> No.3240163

you'd need to embed (n+1)-simplex in <span class="math">\mathbb{R}^n[/spoiler].

>> No.3240167

you can, the distance has to be 0 though.

>> No.3240200

Because a triangle defines a plane.
In a given plane there can't be more than three points (proof possible through co-ordinate geometry).

Now draw this equilateral triangle and for its every vortex plot another equilateral triangle. Again through co-ordinate geometry you can prove that there exists a unique point in space where the three vertices will have a point of for forming their equilateral triangles.

So essentially its axiomatic property of 3D space.

The required figure is a regular tetrahedron.

>> No.3240202

Try pulling your foreskin back.