[Monte314]

Consider the three-dimensional sphere of radius 1 centered at the origin. It consists of the set of ordered triples of real numbers (x,y,z) where:

x^2 + y^2 + z^2 = 1

We can embed this sphere in four-dimensional space by specifying a value for its fourth coordinate, w; to keep it centered at the origin, let’s perform the embedding by making w=0 for all points of the sphere. Then this embedded 3-sphere consists of the set of ordered quadruples of real numbers (x,y,z,w) where:

x^2 + y^2 + z^2 = 1, and w=0

Now, consider *any* point P that lies on the w-axis in 4-space (for example, P = (0,0,0,5) as in the diagram below).

I claim that any such P is closer to the center of the embedded sphere than it is to any point on the surface of the embedded sphere, even when P is not inside the sphere and its distance from the center of the sphere is greater than the sphere's radius!

If you are smarter than a dog, prove me wrong. Or, if you can, prove me right… either way, the Jedi Math Dog rules.

[Begoner]

[HIDE="show spoiler"]Yes: the distance between (0, 0, 0, w) and (x, y, z, 0) for some point on the w-axis and some point on the unit sphere is sqrt(x^2+y^2+z^2+w^2) = sqrt(1+w^2). On the other hand, the distance between (0, 0, 0, w) and the center of the sphere is sqrt(w^2), which is clearly less than sqrt(1+w^2).

Intuitively, this makes sense: the point (0, 0, 0, w) must be equidistant from all points on the sphere since the w-coordinate is unrelated to the x-, y-, or z- coordinates, just as the point (0, 0, z) is equidistant from all points on the circle x^2+y^2=1. In fact, the point (0, 0, z) is closer to the center of the circle than any point on the circle.[/HIDE]

[SirJac]

( Show Spoiler )

Considering that pythagoras theorem holds for N dimensions, and allowing the position of the point p to be denoted by the vector <0,0,0,w> considering that the center of the sphere is at the origin. Thus the distance from the center of the sphere to the point P will be w.

any point on or within the sphere will be given by the vector <x,y,z,0> given the contraint that x^2+y^2+z^2 = 1. From this the vector from the point p to any point on or within the sphere will be given by:

<x,y,z,0> - <0,0,0,w>= <x,y,z,-w>

the length of this vector by pythagoras is then:

|<x,y,z,-w>| = sqrt(x^2+y^2+z^2+(-w)^2) >= w

with equality if and only if x=y=z=0

Thus even without considering the restraints, any point on the w axis will be closest to the origin of the space defined by the x,y and z axis.

This problem is identical to a point on the z axis being closest to the center of a circle on the x,y plane centered at the origin but the number of dimensions increased.

[Malle]

( Show Spoiler )

Assuming we are working in the Euclidian space ℝ^4, the distance d between a point a = (x1,y1,z1,w1) and a point b = (x2,y2,z2,w2) is given as
d = |a-b| = √((x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2 + (w1-w2)^2)

Let's assume that point a is on the sphere and point b is on the w-axis, as the problem states. This gives us the following relations:

w1 = 0
x2 = y2 = z2 = 0
For a point on the surface: x1^2 + y1^2 + z1^2 = 1
For the centre of the sphere: x1 = y1 = z1 = 0

Substituting into the equation for the distance, we get:

d_surface = √(x1^2 + y1^2 + z1^2 + w2^2) = √(1 + w2^2) > √(w2^2) = d_centre

Thus we see that all points on the w-axis are closer to the centre of the sphere than to the surface of the sphere.

[Latro]

( Show Spoiler )

Just compute the distance. The result seems a little counterintuitive perhaps based on how we might try to geometrize higher dimensional space in our minds, but regardless:
Distance from point (0,0,0,e) to a point (a,b,c,f):
d = sqrt(a^2+b^2+c^2+(f-e)^2)
It is relatively simple to minimize this function with respect to a, b, and c, and if you do that you will arrive at a=b=c=0.

[BingeArtist]

Originally Posted by Begoner To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.

[HIDE="show spoiler"]Yes: the distance between (0, 0, 0, w) and (x, y, z, 0) for some point on the w-axis and some point on the unit sphere is sqrt(x^2+y^2+z^2+w^2) = sqrt(1+w^2). On the other hand, the distance between (0, 0, 0, w) and the center of the sphere is sqrt(w^2), which is clearly less than sqrt(1+w^2). Intuitively, this makes sense: the point (0, 0, 0, w) must be equidistant from all points on the sphere since the w-coordinate is unrelated to the x-, y-, or z- coordinates, just as the point (0, 0, z) is equidistant from all points on the circle x^2+y^2=1. In fact, the point (0, 0, z) is closer to the center of the circle than any point on the circle.[/HIDE]

Ding Ding Ding, looks like we have a winner. But why must we limit ourselves to points on the surface of the sphere?


As per the title, I believe we should truly "have a ball"...

[admittedheretic]

	P is not inside the sphere and its distance from the center of the sphere is greater than the sphere's radius!

How can P ever truly be inside the sphere? This doesn't seem like a coherent statement to me.

[rahdam]

( Show Spoiler )

no idea.

sqrt(5^2) = 5

if x,y,z are nonzero, sqrt(x^2+y^2+z^2+w^2) >= 5

Therefore, if x,y,z are nonzero, ie not on w, they are farther from the center than (0,0,0,w)

Any point on the surface of the sphere has w=0, so the MINIMUM distance from the point (0,0,0,w) to (x,y,z,0) is w, if x,y,z are 0 (degenerate sphere). This ALSO happens to be the distance between (0,0,0,w) and (0,0,0,0). Of course, if x,y,z are nonzero, then the distance from (0,0,0,w) to (x,y,z,0) becomes even greater than the distance from (0,0,0,w) to the origin.

sqrt(x^2+y^2+z^2+w^2) > sqrt(w^2) if x,y,z are nonzero.

[Monte314]

	Originally Posted by BingeArtist To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	Ding Ding Ding, looks like we have a winner. But why must we limit ourselves to points on the surface of the sphere? As per the title, I believe we should truly "have a ball"...

Mea culpa. But most people don't know the difference between a unit sphere and a unit ball... I guess I was pandering to the limitations of the ignorati...

*Shamed Dog begins running in circles*