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This is a self-answered question, after some playing around. I would be happy to see alternative solutions.

Let $x_1,x_2,x_3,x_4 \in \mathbb{R}^3$ be the vertices of a regular tetrahedron, which lie on the unit sphere, i.e. $|x_i-x_j|$ is constant for $i \neq j$, and $|x_i|=1$.

Claim: $x_4$ is perpendicular to the opposite face, i.e. $x_4 \perp x_1-x_2, x_4 \perp x_1-x_3$.

Question: How to prove this? I would like to find a slick elegant proof.

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Asaf Shachar
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1 Answers1

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It is "geometrically obvious" that on a sphere $\mathbb{S}^2 \subseteq \mathbb{R}^3$, the Euclidean (or spherical) distance between two points is determined by the angle between them, so it follows that $x_4$ has equal angles with $x_1,x_2$. Since $|x_1|=|x_2|$, this implies that $x_4$ is perpendicular to the line connecting $x_1$ and $x_2$, i.e. to $x_1-x_2$.

In shorter (modern) phrasing:

Since $|x_i-x_j|^2=2-2\langle x_i,x_j \rangle$ does not depend on $i,j$, $\langle x_1,x_4 \rangle=\langle x_2,x_4 \rangle$, thus $\langle x_1-x_2,x_4 \rangle=0$ as required.

Asaf Shachar
  • 25,967
  • I believe, shorter phrasing is elegant enough. Other proof may be more geometric: projection of both $x_4$ and $O$ on face $(x_1-x_2-x_3)$ are lying at the center of $(x_1-x_2-x_3)$ then $x_4O$ is perpendicular to $(x_1-x_2-x_3)$. – Ivan Kaznacheyeu Apr 28 '22 at 14:59