There are two $d$-dimensional unit spheres centred at $A$ and $B$ in $\mathbb{R}^d$. They are touching at location $I$. How can I find the centre of a third $d$-dimensional unit sphere $C$ that touches the first two spheres? I know that:

  • |A-B|=|A-C|=|B-C|=2,
  • |A-I|=|B-I|=1.

I don't know how to proceed from here. I know there can be multiple locations for $C$. In fact the set of possible locations forms a $(d-1)$-sphere. But how to find its radius and centre?

  • $\begingroup$ Hint: Reduce it to two circles touching in $\mathbb{R}^2$. $\endgroup$ Apr 19, 2021 at 12:04
  • $\begingroup$ In 2 dimensions there are two locations for C and I know how to find them. But I don't know what to do in higher dimensions. $\endgroup$ Apr 19, 2021 at 12:05
  • $\begingroup$ Is $d$ the dimension of ambient space $\mathbb R^d$ ? $\endgroup$
    – Jean Marie
    Apr 19, 2021 at 12:08
  • 1
    $\begingroup$ @JeanMarie yes it is. $\endgroup$ Apr 19, 2021 at 12:09
  • $\begingroup$ Why the minus? :( $\endgroup$ Apr 19, 2021 at 12:25

1 Answer 1


From the fact that

$$ \lvert A - B\rvert = \lvert A - C\rvert = \lvert B - C\rvert = 2 $$

you know that $A, B, C$ are vertices of an equilateral triangle of side $2$. It doesn't matter how many dimensions you are working in, three non-collinear points still make the vertices of a triangle and if all three sides are equal then the triangle is equilateral.

Given that $\triangle ABC$ is an equilateral triangle of side $2$, what is the relationship of $C$ to the midpoint of side $AB$?

The point $C$ must be on the perpendicular bisector of $AB$ at a distance $\sqrt3$ from $AB.$

Noting that $I$ is the midpoint of $AB,$ we have $CI = \sqrt3$ and $C$ lies on the $(d-1)$-plane through $I$ orthogonal to the line $AB,$ since that $(d-1)$-plane contains all lines through $I$ perpendicular to $AB$ and all lines through $I$ perpendicular to $AB$ are in that plane.

In summary, the sphere you are looking for has center $I$ and radius $\sqrt3$ and lies in the $(d-1)$-plane through $I$ orthogonal to $AB.$

  • $\begingroup$ I believe the midpoint of $AB$ is $I$ and $|C-I|=\sqrt 3$. $\endgroup$ Apr 19, 2021 at 13:44
  • $\begingroup$ That's most of the information you need to construct the $(d-1)$-sphere. $\endgroup$
    – David K
    Apr 20, 2021 at 0:23
  • $\begingroup$ I think I found the answer, but not sure if it is correct. I need to construct a $(d-1)$-plane between the spheres going through point $I$. The location of $C$ can be all points on that plane that have a distance of $\sqrt 3$ from $I$. In other words the intersection of the plane and a sphere. By the way, this is not a homework assignment as it may seem. In fact I am working on the kissing number problem. $\endgroup$ Apr 20, 2021 at 6:42
  • 1
    $\begingroup$ OK, I see this is not a question that needs to be drawn out in hints, so I've finished describing the locus of $C.$ Anyway it just confirms what you already figured out. $\endgroup$
    – David K
    Apr 20, 2021 at 12:19
  • $\begingroup$ Thank you David. I really appreciate your help. $\endgroup$ Apr 20, 2021 at 12:30

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