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My problem is the following: I want to construct $n$ rays all starting at a point $v$ that is not in the $n$-dimensional ball around $0$ such that the following is true:

  • The $n$-dimensional ball is a subset of the convex cone spanned by those $n$ rays together.
  • Each ray intersects the ball in only one point, so they are basically tangents to the boundary.

Is this possible for $n>2$? I know certainly that it is possible for $n=2$ and maybe even $n=3$. However I am not quite sure for even higher $n$.

Edit: Here is a picture of the situation I mean when $n=2$. Also I forgot to mention the cones top is $v$.

enter image description here

When $n>3$, replace the triangle with a Simplex. Is it still possible to get n rays touching the ball and forming a Simplex?

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    $\begingroup$ for $n\ge 3$ only rays that are convexly dependent to the others may touch the ball (these rays can always be removed or added without changing the convex cone). Consider the section by the hyperplane for the center of the ball, orthogonal to the line through v. You see an $n-1$ ball inside a convex polytope: vertices don't touch the ball $\endgroup$
    – Pietro Majer
    Commented Aug 8 at 6:32
  • $\begingroup$ Of course, there's a sort of dual way to generalize the $n=2$ case: we take hyperplanes passing through $v$ and tangent to the ball, and ask if as few as $n$ such hyperplanes can contain the ball in the intersection of their halfspaces. In other words, the "front-on" view of the two approaches looks like en.wikipedia.org/wiki/Method_of_exhaustion $\endgroup$
    – usul
    Commented Aug 10 at 5:38

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It is not possible for $n>2$.

Pick any point $v$ and consider the cone spanned by all tangent rays. Clearly that cone has to be a subset of the solution, as otherwise you would miss part of the ball. At the same time it is also the largest cone you can get by such vectors, as it already includes all of them.

So it is the solution. But for $n>2$, it is circular and not spanned by a finite number of rays.

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