Cover of n-simplex with balls.

Question

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.

Answer 1

It seems the following.

Let $\Delta=\operatorname{conv}\{x_0,x_1,\dots,x_n\}\subset\mathbb R^n$ be a $n$-dimensonal simplex with the vertices $x_0,x_1,\dots, x_n$. Let $x\in\Delta$ be an arbitrary point. Then there exist a non-negative real numbers $\lambda_0,\lambda_1,\dots,\lambda_n$ such that $\sum_{i=0}^n \lambda_i=1$ and $x=\sum_{i=0}^n \lambda_i x_i$. We are going to show more: for every vertex $x_i$ there is a $n$-ball $B_{ij}$ with antipodal vertices $x_i$ and $x_j$ covering the point $x$. Indeed, suppose the converse. The point $x$ is not covered by a $n$-ball $B_{ij}$ with antipodal vertices $x_i$ and $x_j$ iff

$$\left|x-\frac{x_i+x_j}2\right|>\left|\frac {x_i-x_j}2\right|.$$

After the routine equivalent transformations this condition magically turns into

$$(x-x_i,x -x_j)>0,$$

which means it is equivalent to the angle between the vectors $x_i-x$ and $x_j-x$ is less than $\pi/2$. :-)

Suppose that the point $x$ is not covered by every $n$-ball $B_{ij}$. Then

$$0<\sum_{j=0}^n \lambda_j (x-x_i,x - x_j)=\sum_{j=0}^n (x- x_i, \lambda_j x - \lambda_j x_j)= (x- x_i, x - x)=0,$$

a contradiction.

Moreover, similar proof should show that for every vertex $x_i$ of a convex $n-$polyhedron $P$ the family of $n$-balls $\{B_{ij}$ with antipodal vertices $x_i$ and $x_j$: $x_j$ is a vertex of the polyhedron $P$ $\}$ covers $P$.