inscribed simplex. 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 $\frac{d}{2}$ ?
 A: 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
$$\text{dist}(V,P)^2=(x+h)^2+y^2+z^2>r^2$$ 
But $V$ is on the (hyper)sphere, so 
$$x^2 + y^2 + z^2 = r^2$$
Therefore,
$$\begin{align}
2hx+h^2 &> 0 \\
\Rightarrow h(x+\frac{h}{2}) &> 0
\end{align}$$
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.
