The sum of unit vectors from a point in a polyhedron to its $n$ vertices has magnitude $< n-2$

Question

There is a polyhedron with $n$ vertices and a point $O$ inside it. Let $e_i$ be a unit vector directed from the point $O$ to the $i$-th vertex of the polyhedron. Prove that $$ |e_1+\ldots+e_n|<n-2. $$

This is an exercise from a book in Russian How to solve non-standard problems (page 11). Also, I have encountered this problem on several forums, but nowhere have I seen a solution to this problem.

It seems to me that a similar but somewhat simpler problem for the plane can be formulated as follows:

Let us give an $n$-gon and the point $O$ inside of it. Let $e_i$ be a unit vector directed from $O$ to the $i$-th vertex of the $n$-gon. Prove that $$ |e_1+\ldots+e_n|<n-2. $$

My hope was that these problems could be solved with the help of the statements from the "About four points lying on a sphere" post.

In the plane, for example, the reasoning could be as follows. Let $e_i=\vec{OA_i}$, $i=1,\ldots,n$. There are three points $A,B,C\in\{A_1,\ldots,A_n\}$ such that $O$ lies inside triangle $\triangle ABC$. We can assume without loss of generality that $A=A_1$, $B=B_1$, $C=C_1$. Then we get $$ |e_1+\ldots+e_n|\leq|e_1+e_2+e_3|+(|e_4|+\ldots+|e_n|)<1+(n-3)=n-2. $$ Understandably similar reasoning in 3D.

All is well and wonderful.

But why do the above three vertices exist? And in the case of 3D, four vertices?

How to justify this? That's actually my question.

However, perhaps there is another solution to these problems using other considerations?

I will gladly accept such answers too.

Adding. (2024/01/11)
It seems to me that whether a polyhedron ($n$-gon) is convex or not does not matter. I have posted a picture that clarifies this point. We know that the vertices of the $n$-gon are located on the rays with origin at $O$. There are many such $n$-gons, but the sum of the vectors depends only on the direction of the rays.

Answer 1

Note that if $O$ is in the interior of a polyhedron, then $O$ is also in the interior of the convex hull of the polyhedron.

Thus we don't need to assume that the polyhedron is convex.

In fact, we don't even need a polyhedron.

We'll use the following specifications . . .

With $n\ge 3$, let $v_1,...,v_n\in\mathbb{R}^m$, not necessarily distinct, but not all collinear.

It follows that $m\ge 2$.

Let $H$ be the convex hull of $\{v_1,...,v_n\}$ and let $O$ be an interior point of $H$ with $O$ distinct from $v_1,...,v_n$.

Note that $H$ need not be $m$-dimensional, so an open subset of $H$ need not be open in $\mathbb{R}^m$.

For each $i\in\{1,...,n\}$ let $e_i$ be the unit vector directed from $O$ to $v_i$.

Claim:$\;|e_1+\cdots+e_n| < n-2$.

Proof:

Assume $v_1,...,v_k$ are the distinct extreme points of $H$.

Since $v_1,...,v_n$ are not all collinear, it follows that $k\ge 3$, so $3\le k\le n$.

Choose points $A,B,C$ in the interior of $H$ such that $A,B,C$ are the vertices of a non-degenerate triangle whose centroid is $O$.

Since $A,B,C$ are in the interior of $H$, we can write \begin{align*} A&=\sum_{i=1}^k a_iv_i\\[4pt] B&=\sum_{i=1}^k b_iv_i\\[4pt] C&=\sum_{i=1}^k c_iv_i\\[4pt] \end{align*} expressing each of $A,B,C$ as a convex combination of $v_1,...,v_k$.

Since $O$ is the centroid of triangle $ABC$, we get $$ \sum_{i=1}^k r_iv_i=0 $$ where $r_i=a_i+b_i+c_i$, hence $$ \sum_{i=1}^k s_ie_i=0 $$ where $s_i=r_i|v_i|$.

Then letting $$ S=\sum_{i=1}^k s_i $$ we have the convex combination $$ \sum_{i=1}^k t_ie_i=0 $$ where $t_i=s_i/S$.

Without loss of generality, assume $t_1\ge\cdots\ge t_k$.

Suppose $t_3=0$.

Then for all $i\in\{3,..,k\}$, we have $t_i=0$.

But from the implications $$ t_i=0\implies s_i=0\implies r_i=0\implies a_i,b_i,c_i=0 $$ it would follow that $A,B,C$ are linear combinations of $v_1,v_2$.

But then $A,B,C$ would be collinear, contradiction, since triangle $ABC$ is non-degenerate.

Hence $t_3 > 0$, so $t_1,t_2,t_3 > 0$.

Since $v_1,v_2,v_3$ are distinct extreme points of $H$, they are not collinear, hence the vectors $$ -t_1e_1,\;t_2e_2,\;t_3e_3 $$ are not all parallel.

Then we get \begin{align*} 2t_1 &= |-2t_1e_1| \\[4pt] &= \left| -2t_1e_1 + \sum_{i=1}^k t_ie_i \right| \\[4pt] &= \left| -t_1e_1 + \sum_{i=2}^k t_ie_i \right| \\[4pt] & < \sum_{i=1}^k |t_ie_i| \qquad \bigl(\text{strict inequality}\bigr) \\[4pt] &= \sum_{i=1}^k t_i \\[4pt] &= 1 \\[4pt] \end{align*} hence $2t_i < 1$ for all $i\in\{1,...,k\}$.

Then we get \begin{align*} \left| \sum_{i=1}^k e_i \right| &= \left| \sum_{i=1}^k e_i - 2\sum_{i=1}^k t_ie_i \right| \\[4pt] &= \left| \sum_{i=1}^k (1-2t_i)e_i \right| \\[4pt] & < \sum_{i=1}^k |(1-2t_i)e_i| \qquad \bigl(\text{strict inequality}\bigr) \\[4pt] &= \sum_{i=1}^k (1-2t_i) \\[4pt] &= k-2 \\[4pt] \end{align*} so then \begin{align*} \left| \sum_{i=1}^n e_i \right| &= \left| \sum_{i=1}^k e_i + \sum_{i=k+1}^n e_i \right| \\[4pt] &\le \left| \sum_{i=1}^k e_i \right| + \left| \sum_{i=k+1}^n e_i \right| \\[4pt] & < (k-2)+(n-k) \\[4pt] &= n-2 \\[4pt] \end{align*} as was to be shown.

Answer 2

I am going to try to outline a solution that uses an idea for the solution of the tetrahedron problem.

There is a polyhedron $\mathcal{P}$ with $n$ vertices and a point $O$ inside it. Let $e_i=\vec{OA_i}$ be a unit vector directed from the point $O$ to the $i$-th vertex $A_i$ of the polyhedron. We need to prove that $$ |e_1+\ldots+e_n|<n-2. $$

By contradiction. Let $\vec{ON}=e_1+\ldots+e_n$ and $|\vec{ON}|\geq n-2$.

Let $P$ be a plane passing through $O$ and perpendicular to $\vec{ON}$. If the vertices of $\mathcal{P}$ all lie above the plane $P$, then the point $O$ is not an interior point of $\mathcal{P}$. Therefore, there is at least one point $A_i$ that lies below $P$. The lowest point among those below $P$ is denoted by $A$ (see the figure).

Let $P_A$ be a plane passing through the point $A$ parallel to $P$ and $P'_A$ be a plane centrally symmetric to $P_A$ with respect to the center $O$. Let us call the line of intersection of the sphere and the plane $P_A$ (resp. $P'_A$) South (resp. North) circle.

Consider an axis $L$ passing through $O$ and parallel to $\vec{ON}$. We can look at $L$ as a number line, $1$ corresponds to the point of intersection of the ray $ON$ and the sphere. Let $\pi$ be the orthogonal projection of $\mathbb{R}^3$ onto the $L$. If $\pi(A_i)=a_i,\pi(N)=t$, then $|a_i|\leq1$, $t=|\vec{ON}|$ and $t=a_1+\ldots+a_n$. Let $\pi(A)=a$. If $A=A_i$, then $a=a_i$ and we have $a<0$.

If there is a point $B\in\{A_1,\ldots,A_n\}$ south of North circle besides the point $A$ and $\pi(B)=b$, then $|b|<|a|$. Now, if $C\in\{A_1,\ldots,A_n\}\setminus\{A,B\}$ and $\pi(C)=c$, then $a+b+c<1$ and $$ t=\sum_{A_i\not\in\{A,B,C\}}a_i+(a+b+c)\leq (a+b+c)+(n-3)<n-2. $$ The latter contradicts our assumption.

Consequently, all points $A_i$ other than $A$ lie on or north of North circle. In this case, all vertices of the polyhedron $\mathcal{P}$ lie on or above the plane $P_{OA}$ passing through $O$ and $A$ and perpendicular to the plane $P$. So $O$ does not lie inside $P$. The contradiction proves our statement.

Adding.
If a vertex $V$ of $\mathcal{P}$ lies below the plane $P_{OA}$, then the point of intersection of the ray $OV$ and the sphere is one of our points $A_k$ which lies south of North circle.