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Suppose I have an inscribed simplex which has $(n+1)$ vertices, and the diameter of the hypersphere is $d$. I have a point $x$ inside this simplex, is it true that the distance between $x$ and $x$ is nearest to vertex is not greater than $\frac{d}{2}$ ?

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it should be not larger than d/2, thanks for your comments. –  cyliu Dec 21 '10 at 11:53

1 Answer 1

Center the hypersphere at the origin, and let $h$ be the (non-negative!) distance between the point $P$ and the center. We can position our hypersphere so that $P$ lies on the $x$ axis with $x$-coordinate $-h$. I'll write $r$ for $d/2$, the radius of the hypersphere.

Consider the case $n=3$.

Suppose that the distance from a vertex, $V(x,y,z)$, to $P$ is strictly greater than $r$. Then


But $V$ is on the (hyper)sphere, so

$$x^2 + y^2 + z^2 = r^2$$


$$\begin{align} 2hx+h^2 &> 0 \\ \Rightarrow h(x+\frac{h}{2}) &> 0 \end{align}$$

We see, then, that our non-negative $h$ must be strictly positive, and that $x > -\frac{h}{2} > -h$.

Now, if ALL vertices are further than $r$ away from $P$, then the above shows that $P$ is separated from those vertices --and the simplex they determine-- by the (hyper)plane $x=-\frac{h}{2}$. The point cannot lie inside the simplex.

Consequently, for a point within the simplex, the distance to at least one vertex (in particular, the closest) must be no greater than $r$. In the case where $P$ lies at the center of the hypersphere, the distance to any vertex is exactly $r$, so this upper bound on the minimum distance cannot be improved.

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great, thanks very much. –  cyliu Dec 22 '10 at 3:47

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