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.
On
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.
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.