Consider a n-simplex. For each edge (i,j), consider a n-ball, such that vertices i and j are antipodal on this ball. Is the simplex covered by the union of these balls? Thank you.



If $P$ is our point, then it will be contained in the ball corresponding to edge $(i,j)$ if and only if the angle $\angle iPj$ is greater than or equal to $\frac{\pi}{2}$. If there is no such edge $(i,j)$, then every vertex $j$ is on a fixed side of the hyperplane through $P$ orthogonal to the line connecting vertex $1$ with $P$, so $P$ is outside the simplex.

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    $\begingroup$ So it seems the point is in at least n/2 of the balls. Can it be shown that it is in n-1 balls at least? $\endgroup$ – The Masked Avenger Aug 22 '13 at 9:08
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    $\begingroup$ Can you please expand on the logic behind the last "so..."? Every vertex $j \neq 1$ is to one side of that hyperplane. How does this imply $p$ is outside? $\endgroup$ – Joseph O'Rourke Aug 22 '13 at 12:37
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    $\begingroup$ Draw a picture with P in the plane so dictated and vertex 1 on one side. If 1Pj is less than 90 degrees, j is on the same side as 1. If this is true for all j, P lies in a half space on the wrong side of a tangent plane to the (convex) simplex . $\endgroup$ – The Masked Avenger Aug 22 '13 at 16:36
  • $\begingroup$ Zeb's proof shows that for every vertex j, there is i with ij the diameter of a ball containing P (all j exists i). Is the (possibly) stronger statement (Exist j forall i) true? (Or even an intermediate, saying there are at least n-1?) $\endgroup$ – The Masked Avenger Aug 22 '13 at 16:41

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