Zoran Skoda tetrahedron1

We may consider orthogonal projection to lower dimensional subspaces in a Euclidean space, not only to hyperplanes.

For a fixed vector $\vec{a}$ in a Euclidean 3-space, consider its orthogonal component, a plane $M = M_{\vec{a}}$. Any vector $\vec{x}$ then decomposes as $\vec{x} = \vec{x}_{\|} + \vec{x}_M$ where $\vec{x}_{\|}$ is a vector parallel to $\vec{a}$ and $\vec{x}_M\in M$. Clearly, $\vec{x}_{\|} + \vec{y}_{\|} = (\vec{x} + \vec{y})_{\|}$ and $\vec{a}_{\|} = \vec{a}$.

Problem. Let $ABCD$ be an arbitrary (nondegenerate) tetrahedron in a Euclidean space. Then for some vertex of the tetrahedron, the three edges from this vertex may form a triangle. (Bulgarian national competition, 1966)

Solution. Take one vertex, say $A$ and compare sides $A B, A C, A D$. These sides can be assembled into a triangle if neither of them is bigger than or equals the sum of the other two. If this is the case vertex $A$ is the solution. If not, then there is a vertex other than $A$, say $D$, such that $A D\geq A B + A C$. We want to show that $D$ is the solution vertex, that is we may form a triangle with sides $A D, B D, C D$.

There are two possible cases.

Case I. $A D$ is the maximum length among the lengths $A D, B D, C D$. Let us first show that $A D \lt B D + C D$ (the other two inequalities are automatic in this case). $B,C$ and $D$ are not colinear, hence by the strict triangle inequality, it is sufficient to show $A D \leq |\vec{B D} + \vec{C D}|$. That means $A D \leq | \vec{B A} + \vec{A D} + \vec{C A} + \vec{A D}|$. It is enough to observe that

$$A D \leq |(\vec{B A} + \vec{C A} + \vec{A D} + \vec{A D})_{\|}| = A D + |\vec{B A}_{\|} + \vec{C A}_{\|} + \vec{A D}_{\|}|,$$

where $\|$ denotes the orthogonal projection to the direction of $\vec{A D} = \vec{A D}_{\|}$. For the second equality, notice that the orientation of $\vec{B A}_{\|} + \vec{C A}_{\|} + \vec{A D}_{\|}$ (if nonzero) is the same as $\vec{A D}_{\|}$, because by assumption, $|\vec{B A}_{\|} + \vec{C A}_{\|}| \leq |\vec{A D}_{\|}| = A D$.

Case II. $A D$ is not the maximum length among $A D, B D, C D$, say $B D$ is the maximum. It is then sufficient to show that $B D \lt A D + C D$.

Regarding that $\vec{A D}$ and $\vec{C D}$ are not colinear, there is a strict inequality $|\vec{A D} + \vec{C D}|\lt A D + C D$. It is therefore enough to show that

$$B D \leq |\vec{A D} + \vec{C D}| = |\vec{A B}_{\|} + \vec{B D}_{\|} + \vec{C B}_{\|} + \vec{B D}_{\|}|,$$

where this time we consider the projection $(-)_{\|}$ to the line of $\vec{B D} = \vec{B D}_{\|}$. Now $|(\vec{A B} + \vec{C B})_{\|}|\leq |\vec{A B}+ \vec{C B}| \leq B D = |(\vec{B D})_{\|}|$, hence $\vec{A B}_{\|} + \vec{B D}_{\|} + \vec{C B}_{\|}$ is in the direction of $\vec{B D}$ or equals $\vec{0}$, hence

$$|\vec{A B}_{\|} + \vec{B D}_{\|} + \vec{C B}_{\|} + \vec{B D}_{\|}| = |\vec{A B}_{\|} + \vec{B D}_{\|} + \vec{C B}_{\|}| + |\vec{B D}_{\|}|\geq B D.$$

Problem. Does a generalization in $n$-dimensions hold ? In other words, consider an arbitrary nondegenerate $n$-simplex $A_0 A_1\cdots A_n$ in a Euclidean $(n+1)$-space, where $n\geq 3$. Is there always a vertex $A_i$ such that the $(n-2)$-simplices whose one vertex is $A_i$ can be each one isometrically embedded into a Euclidean space so that their union is a boundary of an $n$-simplex ? For example for $n = 4$, for each vertex we have $4$ triangles incident with this vertex; from these 4 triangles we want to make the boundary of a tetrahedron.

To see where the difficulty is consider the following picture related to the problem for $n=4$. Imagine 4 triangles, one as drawn below and for each of the other two take one edge from the main triangle and for the other two take the radii of the circles drawn around each of the vertices of the edge. Clearly, we can not form a tetrahedron as the balls around the vertices of the main triangle of the radii sharing with the circles do not have a common triple intersection in 3-space.