Generalising Thales theorem for points on a sphere to form a 3-orthoscheme (tetrahedron.)

Question

I am trying to find the condition that four points $p_1,p_2,p_3,p_4$ on the unit sphere $\mathbb{S}^1$ need to statisy in order to form a 3-orthoscheme (Tetrahedron with all faces as right angled triangles), preferably phrased in the language of vectors and rotations.

A 3-orthoscheme is basically determined by choosing 3 mutually orthogonal edges, $e_1 = \overline{p_1p_2}, e_2=\overline{p_2p_3}, e_3=\overline{p_3p_4}$. The center of the circumscribing sphere is thus the midpoint of the longest edge of the tetrahedron. Assuming that the center of the sphere is the origin, I can take unit vectors $\boldsymbol{v_i}$, corresponding to the points $p_i$. Actually all the four points lie on a hemisphere and points $p_1$ and $p_4$ can be taken to be antipodal.

The orthogonality condition means that $(\boldsymbol{v}_2-\boldsymbol{v}_1)\cdot(\boldsymbol{v}_3-\boldsymbol{v}_2) = 0$ and $(\boldsymbol{v}_3-\boldsymbol{v}_2)\cdot(\boldsymbol{v}_4-\boldsymbol{v}_3) = 0$. Decoding these gives a condition on the cosine of the angles between the edges.

If I know the lengths of the three edges, say $\ell_{i}$ which are mutually orthogonal, then it is easy to find the lengths of the remainining edges using the right angle condition. The longest edge is known to be of length 2. So this means that $$ \Sigma_{i=1}^3\ell_i^2 = \Sigma_{i=1}^{3} ||\boldsymbol{v}_{i+1}-\boldsymbol{v}_i||^2 = 4.$$ In fact if the $\ell_i$ are all same, then this precisely means $\ell_i = \dfrac{2}{\sqrt{3}}$, which corresponds to the orthoscheme that determines a cube.

Are the above set of conditions enough to say that four points form a 3-orthoscheme? Alternatively can something similar be deduced as in the case of this question for a regular tetrahedra? I am attempting to see this as a 3d generalisation of Thales theorem. Thales theorem tells us how three points on a circle need to be distributed for them to form a 2-orthoscheme, which is a right angled triangle.

Answer 1

This is a partial answer to my own question although it is not very satisfactory to me. The conditions boil down to the expressions: \begin{align} \boldsymbol{v}_1\cdot\boldsymbol{v}_2 + \boldsymbol{v}_2\cdot\boldsymbol{v}_3 +\boldsymbol{v}_3\cdot\boldsymbol{v}_4 = 1 \\ \boldsymbol{v}_1\cdot\boldsymbol{v}_2 + \boldsymbol{v}_2\cdot\boldsymbol{v}_3 = 1 + \boldsymbol{v}_1\cdot\boldsymbol{v}_3 \\ \boldsymbol{v}_2\cdot\boldsymbol{v}_3 + \boldsymbol{v}_3\cdot\boldsymbol{v}_4 = 1 + \boldsymbol{v}_2\cdot\boldsymbol{v}_4 \end{align}

But I am wondering if this determines an orthoscheme upto even similarity. Maybe a clever usage of spherical law of cosines will help, I am not sure.

Answer 2

Take antipodes $P$ and $Q$, and any distinct $R$ on the sphere. The plane through $Q$ orthogonal to $PR$ cuts the sphere in a circle having $QR$ as a diameter. (Why?) For any distinct $S$ on this circle, $PQRS$ is a $3$-orthoscheme.

After all, $R$ and $S$ being on the sphere with diameter $PQ$ guarantees (by Thales Theorem in great circles $\bigcirc PQR$ and $\bigcirc PQS$) that $\angle PRQ=\angle PSQ=90^\circ$. Taking $S$ on the plane orthogonal to $PR$ necessarily implies $\angle PRS=90^\circ$, and since $S$ is in particular on a circle with diameter $QR$ implies (by Thales again) that $\angle QSR=90^\circ$. Each face is a right triangle.

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Vectorially, take antipodes $P$ and $Q$, and consider additional points $R$ and $S$ (such that all four points are distinct). Defining vectors $u:=R-P$ and $v:=S-Q$, the equations in OP's answer, with $v_1=P$, $v_2=R$, $v_3=S$, $v_4=Q=-P$ all reduce to $u\cdot v=0$. Thus, all we require is that vectors $u$ and $v$ are orthogonal.

Looking at this slightly differently, consider antipodes $P$ and $Q$, and orthogonal vectors $u$ and $v$. Let $R$ is the point where the line through $P$ in direction $u$ meets the sphere; likewise define $S$ from $Q$ and $v$. If $P$, $Q$, $R$, $S$ are distinct, then they determine a $3$-orthoscheme.

(Non-distinctness arises when $u$ or $v$ are orthogonal to $PQ$, which would make them tangent to the sphere, so that $P=R$ or $Q=S$; or, when $u$, $v$, and $PQ$ are linearly dependent, which would make $P$, $Q$, $R$, $S$ coplanar, which in turn requires $R=S$.)