Cubes inscribed into a sphere Suppose n ≥ 3 unit cubes are inscribed into a sphere, such that every
three of them have a common vertex. Prove that all n cubes have a common vertex.
Could anyone please help me the approach?
Thanks in advance.
P.S. I edited out what I have tried since it was pointed out my approach was wrong.
 A: Suppose four unit cubes A, B, C and D are inscribed in a common sphere. This means that not only are they the same size, they also have the same centre.
Suppose:
B, C, D share a common vertex P.
A, C, D share a common vertex Q.
If P and Q coincide, then obviously all four cubes share a common vertex. The same is true if P and Q are diametrically opposite each other because every vertex is diametrically opposite another vertex of the same cube. Therefore we can assume that P and Q are distinct non-opposite vertices.
Then the two cubes C and D share two distinct non-opposite vertices (that is to say an edge or a face diagonal), as well as their centre. The only way this is possible is for them to fully coincide. Since the triplet A, B, and C share a common vertex, this vertex must also be shared by D.
For larger n you can use induction using essentially the same proof. The base case, n=4, is above. For induction, suppose it is true for n-1. Given cubes $C_1,...C_n$ satisfying the requirements. First exclude $C_1$ to get a vertex P that is shared by the other cubes. Then exclude $C_2$ to get a vertex Q that is shared by the other cubes. If P and Q coincide or are opposite, then we are done. If not, then $C_3$ and $C_4$ share both vertices P and Q, so must coincide. Then exclude $C_4$ to find a vertex that must be shared by all the cubes.
A: The following are all my thoughts on the problem:
It looks like as you say, the result possibly follows directly from Helly's theorem (although perhaps it is necessary that $n > 3$?).
As a sphere is convex, and there $n <\infty$ cubes inscribed inside, then we have a finite collection of convex subsets of $\mathbb{R}^{3}$. The
intersection number or Helley number must be $4$, but since we only have that the Helley number is $3$ I am not sure about this problem.
As I said in a comment, one might translate this to a graph theory problem using a tree where for $n = 3$, all three nodes (cubes) connect to the same point (vertex) like so: $0---Vertex(0)---0$. (It will help to draw pictures here).
Then when you increase this to $n=4$, you may notice that if this new node does not meet at the same vertex as the other $3$, you get a contradiction of the intersection number $k=3$.
Perhaps it is possible (for you or in general) to finish this off by induction. I have hopefully correctly translated this problem to something I think is more  manageable, and attempting to keep the spirit of the question.
To explain better what I am thinking of, we will consider the $n=4$ case of the graph theory problem. As above, for $n=3$, we have $0---Vertex(0)---0$. Add another node to increase the case to $n=4$ and attach this new node to either of the  two nodes that are not the $3$-vertex. Now think of what this means for vertices of cubes instead of nodes. In our graph theory example, the fourth node will be too far away to share a vertex of its cube with one of them, which implies that we must then have a $4$-vertex. If this process is repeatable, we will achieve an $n$-vertex for any $n$.
It would help others as well as myself if you gave more context to the problem, and what you have learned about geometry up to this point.
