TITLE: Packing obtuse vectors in $\mathbb{R}^d$
QUESTION [11 upvotes]: I came across this attractive theorem:

Theorem. In $\mathbb{R}^d$, there can be at most $d+1$ vectors that
form an obtuse angle with one another.

This was proved1 as a corollary of a lemma about irreducible matrices.
I am wondering if anyone knows of an alternative, more geometric proof that somehow
more directly captures the sense that one cannot "pack" more than
$d+1$ obtuse vectors in $\mathbb{R}^d$.

         


1Lipeng Ning, Tryphon T. Georgiou, Allen Tannenbaum, Stephen P. Boyd.
"Linear models based on noisy data and the Frisch scheme." 
SIAM Review. 57(2) 2015.
arXiv preprint.

REPLY [3 votes]: The proof I know of this result is essentially the same as Vladimir Dotsenko's, but in terms of Radon's lemma.:
Assume such a set exists with at least $d+2$ points.  By Radon's lemma, given any $d+2$ points in $\mathbb{R}^d$ there is a partition of them into two sets $A$, $B$ such that $conv(A) \cap conv(B) \neq \emptyset$.  Let $p$ be a point of this intersection.  If we write $p$ as a convex combination of $A$ and one of $B$, and using the fact that the dot product of any two points in the original set is negative, we immediately obtain $p \cdot p < 0$, a contradiction.