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I am an engineer not a mathematician. So, please accept my apology if this is a very simple problem for this site.

Given a coordinate $A(x_{1A},x_{2A},...,x_{nA})$ and a value d in n-dimension space. I would like to find farthest points to this centroid point $A$. So, a n-dimension hypercube should be built. I start from 2-dimension and then try to generalize the method to n-dimension

2-dimensional space:

Given a coordinate $A(x,y)$ and $d$. What is interested is to find $B(x_1,y_1)$ and $C(x_2,y_2)$ as farthest points from $A(x,y)$.

image is here

it is a easy task as they, $B$ and $C$, can be calculated as $$B(x_1,y_1)=A(x+d,y+d)$$ and $$C(x_2,y_2)=A(x-d,y-d)$$

It is worth to mention that we have to calculate coordinates of all vertices and then by using a distance measurement like Euclidean distance find which ones are farthest.

Now in 3-dimensional space:

Given the same point in 3-dimension space, i.e. $A(x,y,z)$ and we have,

image is here

So,

$$B(x_1,y_1,z_1)=A(x+d,y+d,z+d)$$ and $$C(x_2,y_2,z_2)=A(x-d,y-d,z-d)$$

and for 3-dimension space, is it correct to generalize it as following algorithm:

1- build a 3-face cube around point $A$ with $d$ distance from all faces.

2- find coordinates of all 8 vertices

3- find farthest two top points by using a distance measurement.

Generalized coordinates to the n-dimension are for antipodal points: $$B(x_{1B},x_{2B},...,x_{nB})=A(x_{1A}+d,x_{2A}+d,...,x_{nA}+d)$$ and $$C(x_{1C},x_{2C},...,x_{nC})=A(x_{1A}-d,x_{2A}-d,...,x_{nA}-d)$$

what I think is: 1- the method seems to be something like move from dimension $n$ to $n-1$. In other words, when I subtract $d$ from third dimension, i.e. $z$, I try to shift my point from midpoint of a cube to a 2-d rectangle. 2- all the vertices can be consider as a farthest point from $A$ as $A$ is the centroid and it has identical distance to all vertices. 3- I do not think that I can generalize the method by simply subtract or add $d$ from coordinates of $A$. 4- I found some pages in stackexchange like 1, 2, 3 and 4 but I cannot find the solution for what I am looking for.

How can I find those two farthest points.

Any help would be appreciated.

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David
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    If $A$ is a point in $n$-dimensional space, the vertices of the axis-oriented $n$-cube with side length $2d$ are the $2^{n}$ points obtained from $A$ by adding or subtracting $d$ from each coordinate. Two vertices $B$ and $C$ are "antipodal" if $B - A = A - C$ as vectors with $n$ components; the distance between any antipodal pair $B$ and $C$ is $2d\sqrt{n}$. Is that what you're asking, or are there other details? – Andrew D. Hwang Jul 10 '16 at 18:01
  • @AndrewD.Hwang, Thanks a lot. Is there any method to find coordinates of these points ,$B$ and $C$, in axis-oriented $n$-cube as well. – David Jul 10 '16 at 18:03
  • Every corner of your $n$-cube lies at the same distance, $d\sqrt{n}$, from the center, $A$. Every pair of antipodal corners is separated by twice this, and again, there is considerable freedom in choosing corners: You can add or subtract $d$ from each coordinate of $A$, with all $2^{n}$ possible choices of sign. :) – Andrew D. Hwang Jul 10 '16 at 18:08
  • @AndrewD.Hwang, you mean there is no farthest points from $A$ as all of the corners have the same distance from point $A$, i.e. $d\sqrt{n}$. and if I add $d$, I get $B$ and subtract $d$ to have $C$. and $B$ and $C$ are not unique. – David Jul 10 '16 at 18:16
  • @AndrewD.Hwang, you are telling me that my method is correct that $$B(x_{1B},x_{2B},...,x_{nB})=A(x_{1A}+d,x_{2A}+d,...,x_{nA}+d)$$ and $$C(x_{1C},x_{2C},...,x_{nC})=A(x_{1A}-d,x_{2A}-d,...,x_{nA}-d)$$ – David Jul 10 '16 at 18:18
  • Briefly, "yes": There is no unique farthest point from $A$ in the cube; mathematically speaking, each vertex is a farthest point. Your method does indeed generate a pair of "farthest points". – Andrew D. Hwang Jul 10 '16 at 18:29

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