Is the minimal distance from a point inside a paralellopiped to its 8 vertices always less than the maximal length of all edges?

Question

Is the minimal distance from a point inside a paralellopiped to its 8 vertices is always less than the maximal length of all edges?

Note that for an extreme case, the center of a unit cube has distance $\frac{\sqrt 3}{2}<1$ to its vertices.

I can prove that if a point $X$ from the interior of a paralellopiped lies in a "corner" tetrahedron say, in the following picture, $ABCF$, then $BX$ is less than $\max(BA,BC,BF)$, which is the same as the maximum of lengths all edges. But not every point lies in one of these eight "corner" tetrahedra, for example the center.

Answer 1

Attempt:

Here is the 2d case of a parallelogram. We draw the Voronyi cells for the set of vertices $V$. The magenta points are the the points inside the parallelogram where the function $$X \mapsto d(X, V)$$ achieves its maximum. They are the centers of the circumscribed circles for the the two acute triangles. This maximum value is $R$, the radius of the circles. It is easy to see that $2 R < l + l'$, the sum of two sides.

In the 3d case, we could imagine that the points of maximum are again the centers of the acute tetrahedrons, two of them.

$\bf{Added:}$ In 3d looking at the Voronyi cells could be difficult. However, there is another method that would work say for (semi-)regular polytopes. Use the fact that

$$\sum XA_i^2 - n X C^2$$ is constant, where $C$ is the center of mass of the set of points $X_i$. This allows us to show that the center of the regular polytope is where the function $d$ achieves its maximum ( intuitive).

$\bf{Added:}$ Consider a finite set of points $V$ in space, and $D$ a closed bounded domain. The function

$$X\mapsto d(X,V)$$ achieves it maximum on $D$. Let's show that if $X_0$ is a point of maximum inside $D$, then the distance $d(X_0, V)$ is achieved for at least $4$ points in $V$. Indeed, otherwise write the distances from $X_0$ to points in $V$ in increasing order

$$d_1 = d_2 = d_3 < d_4 \cdots$$

Now, we can move $X_0$ ever so slightly (along an axis) to $X_0'$ so that the distances become

$$d_1' = d_2' = d_3'< d_4'\cdots$$

and $d_1' > d_1$, contradiction with maximality.

Conclusion: $X_0$ is either on the boundary of $D$, or is inside $D$, and is the center of a circumscribed sphere of $4$ points in $V$.

Answer 2

Problem : Assume that $X$ is an interior point in the parallelepiped $P=A_1A_2A_3A_4$-$A_5A_6A_7A_8 $ in $\mathbb{R}^3$

Prove that the minimum distance from $X$ to vertices is strictly smaller than maximum length of edges in $P$.

Proof : (a) Consider a triangle $\Delta XAB, \ A=A_1,\ B=A_2$ with side lengths $$x=|XA|,\ y=|XB|,\ z=|AB|$$ and $\theta = \angle \ AXB$. Assume that $$ x \geq y \geq z $$

Then prove that $$ \theta \leq \frac{\pi}{3}$$

Proof : By 2nd Cosine Law, we have $$ x^2+y^2-2xy\cos\ \theta = z^2 $$

If $\theta > \frac{\pi}{3}$, then $\cos\ \theta <\frac{1}{2}$ so that $$\frac{ x^2+y^2-z^2}{2xy} =\cos\ \theta < \frac{1}{2} $$

which is a contradiction.

(b) If there is $\theta$ with $\theta > \frac{\pi}{3}$, then $z>y$. Hence we complete the proof.

We will assume that $$ \angle A_iXA_j \leq \frac{\pi}{3}$$ for all edges $[A_iA_j]$ in $P$.

We define $$ a_i = \frac{A_i-X}{|A_i-X|} \in \mathbb{S}^2$$

Hence if $|\ast - \ast |$ is a distance function on $\mathbb{S}^2$, then $|a_i-a_j| \leq \frac{\pi}{3}$ where $[A_iA_j]$ is an edge.

If $A_1,\ A_7$ are in diagonal position, then $$|a_1-a_7| \leq |a_1-a_2| + |a_2-a_6|+|a_6-a_7| \leq \pi $$

If $M$ is a mid point between $a_1$ and $a_7$ (We have at least one $M$), then we consider an open hemisphere $H$ s.t. $\partial H$ is a great circle $M^\perp$ and $H$ contains $a_1,\ a_7$.

In $\mathbb{S}^2$, we consider a broken geodesic for example $\alpha = [a_1a_4]\cup [a_4a_8] \cup [a_8a_7]$ whose length is $\leq \pi$. If $\alpha'$ is $\pi$-rotation-image of $\alpha$ wrt $M$, then $\alpha\cup\alpha'$ is a loop of length $\leq 2\pi$.

If this loop escapes the closed hemisphere $\overline{H}$, then it intersects $\partial H$ at some point $h$. Clearly, if $h'$ is $\pi$-rotation-image of $h$ wrt $M$, then the loop contains both points $h,\ h'$, which implies that this loop is in the great arc between $h,\ h'$.

Hence points $a_4,\ a_8$ are also in the hemisphere $\overline{H}$. In the long run all points $a_i$ are in the hemisphere $\overline{H}$. Note that this is possible only when $X$ is not an interior point. Hence we complete the proof.