Tetrahedron: Signed distance between circumcenter and face

Question

In a triangle, the signed distance between the edge $e_1$ and the circumcenter of the triangle can be written as $$ d_1 = \frac{\langle e_2, e_3\rangle}{\langle e_1\times e_2, e_1\times e_3\rangle}\cdot \frac{1}{2}\|e_2\times e_3\| \cdot \|e_1\| $$ and equivalently the area enclosed by edge $e_1$ and the circumcenter (relative to the total triangle area) $$ A_1 = \frac{1}{2}\frac{\langle e_1, e_1\rangle \cdot \langle e_2, e_3\rangle}{\langle e_1\times e_2, e_1\times e_3\rangle} = \frac{1}{2}\frac{\langle e_1, e_1\rangle \cdot \langle e_2, e_3\rangle}{\langle e_1, e_1\rangle \cdot \langle e_2, e_3\rangle - \langle e_1, e_2\rangle \cdot \langle e_1, e_3\rangle} $$ where $e_2$ and $e_3$ are the other edges.

The sign of $d_1$ and $A_1$ is negative if and only if the triangle circumcenter is on the "outside" of the triangle with respect to the edge.

I need compute the same value(s) for a tetrahedron, i.e., the signed distance of its circumcenter to a face circumcenter (which is where the connection line meets the face perpendicularly), or the signed relative volume of face $f_1$ and the tetrahedron circumcenter.

Right now, I'm doing a funny cross-product dance to find out what the sign $d$ should have, explicitly computing the tetrahedron circumcenter and the circumcenter of the face, then taking the distance $\|O-H\|$ and signing it accordingly.

Given the simple representation for triangles, I'm wondering if something like this exists for tetrahedra too. Any hints?

Answer 1

If we have a tetrahedron with the vertices $x_0$, $x_1$, $x_2$, $x_3$, let's consider the signed distance $d$ of the circumcenter to the face $x_0$, $x_1$, $x_2$. First, move the opposing vertex $x_3$ to the origin and call $A=x_0-x_3$, $B=x_1-x_3$, $C=x_1-x_3$.

The plane through $A$, $B$, $C$ is given by $$ \langle (B-A) \times (C-A), X\rangle = \langle A\times B, C\rangle, $$ or, more symmetrically $$ \langle A\times B + B\times C + C\times A, X\rangle = \langle A\times B, C\rangle \quad(=\langle B\times C, A\rangle = \langle C\times A, B\rangle). $$ Normalized, this will give a signed distance to the plane for any point $X$, $$ \tilde{d} = \frac{\langle A\times B + B\times C + C\times A, X\rangle - \langle A\times B, C\rangle}{\|A\times B + B\times C + C\times A\|}. $$ We want to assert that $d$ is positive if and only if $X$ is on the same side of the plane as the origin ($x_3$), in other words: $d$ should have the same sign as $- \langle A\times B, C\rangle$. We can thus multiply with $-\langle A\times B, C\rangle/|\langle A\times B, C\rangle|$. Considering $|\langle A\times B, C\rangle|=6|\text{tet}|$ and $\|A\times B + B\times C + C\times A\|=2|\text{face}|$, we get $$ d =\frac{1}{12\cdot|\text{tet}|\cdot|\text{face}|}(-\langle A\times B + B\times C + C\times A, X\rangle\langle A\times B, C\rangle + \langle A\times B, C\rangle^2). $$ Inserting the circumcenter $$ O=\frac{(B\times C)\langle A, A\rangle + (C\times A)\langle B, B\rangle + (A\times B)\langle C, C\rangle}{2\langle A\times B, C\rangle} $$ for $X$ simplifies things a little further to $$ d =\frac{1}{24\cdot|\text{tet}|\cdot|\text{face}|}(-\langle A\times B + B\times C + C\times A, (B\times C)\langle A, A\rangle + (C\times A)\langle B, B\rangle + (A\times B)\langle C, C\rangle\rangle + 2\langle A\times B, C\rangle^2). $$

Answer 2

If $H$ is the circumcenter of $BCD$, then: $$ O=H+{\vec{CD}\times\vec{CB}\over|\vec{CD}\times\vec{CB}|}h. $$ In the same way, if $H'$ is the circumcenter of $ABC$, then: $$ O=H'+{\vec{CB}\times\vec{CA}\over|\vec{CB}\times\vec{CA}|}h'. $$ By comparing these we get: $$ {\vec{CD}\times\vec{CB}\over|\vec{CD}\times\vec{CB}|}h= \vec{HH'}+{\vec{CB}\times\vec{CA}\over|\vec{CB}\times\vec{CA}|}h'. $$ To get rid of $h'$, dot multiply both sides by $\vec{CA}$, which is perpendicular to the last term: $$ {(\vec{CD}\times\vec{CB})\cdot\vec{CA}\over|\vec{CD}\times\vec{CB}|}h= \vec{HH'}\cdot\vec{CA}, $$ whence you can get $h$. Of course you must supply explicit formulae for $H$ and $H'$, but these can be derived from your formula for the triangle, or by repeating the above reasoning in the case of a triangle.

Answer 3

Change in direction. Let's not be to cute.

Translate the tetrahedron such that one vertex is on the origin.

the three remaining vertexes are at

$\mathbf v_1 = (x_1,y_1,z_1),\mathbf v_2 = (x_2,y_2,z_2), \mathbf v_3 = (x_3,y_3,z_3)$

Find $\mathbf c = (x,y,z)$ such that $d(\mathbf v_1,\mathbf c) = d(\mathbf v_2, \mathbf c) = d(\mathbf v_3, \mathbf c) = d(\mathbf 0,\mathbf c)$

$2x_1 x + 2y_1 y + 2 z_1 z - (x_1^2 +y_1^2+z_1^2) = 2x_2 x+ 2y_2 y + 2 z_2 z - (x_2^2 +y_2^2+z_2^2) = 2x_3 x + 2y_3 y + 2 z_3 z - (x_3^2 +y_3^2+z_3^2) = 0$

And that is a simple system of linear equations.

$2x_1 x + 2y_1 y + 2 z_1 z = x_1^2 +y_1^2+z_1^2\\ 2x_2 x+ 2y_2 y + 2 z_2 z = x_2^2 +y_2^2+z_2^2\\ 2x_3 x + 2y_3 y + 2 z_3 z = x_3^2 +y_3^2+z_3^2\\$