What shape is the locus of a 3D corner with a circular ring that touches the sides of the corner?

The locus of the apex is an ellipsoid.

Let the ring be represented as a circle in the $xy$ plane, with center $(0,0,0)$ and radius $1$, and let $$ A=(1,0,0),\quad B=(\cos\beta,\sin\beta,0),\quad C=(\cos\gamma,\sin\gamma,0), $$ be the points where the edges touch the ring. If $P$ is the apex, then by Pythagoras' theorem we have $$ PA^2={1\over2}(AB^2+AC^2-BC^2)=1-\cos\beta-\cos\gamma+\cos(\beta-\gamma),\\ PB^2={1\over2}(BA^2+BC^2-AC^2)=1-\cos\beta+\cos\gamma-\cos(\beta-\gamma),\\ PC^2={1\over2}(CA^2+CB^2-AB^2)=1+\cos\beta-\cos\gamma-\cos(\beta-\gamma).\\ $$ Point $P$ is then the intersection of three spheres, centered at $A$, $B$, $C$ and with radii $PA$, $PB$, $PC$ given above. A little algebra gives then: $$ P=\left(1+\cos\beta+\cos\gamma, \sin\beta+\sin\gamma, \sqrt{-1-\cos\beta-\cos\gamma-\cos(\beta-\gamma)}\right) $$ and you can check that the coordinates of $P$ satisfy the equation of the ellipsoid $$x^2+y^2+2z^2=1.$$

WOLOG, we will assume the radius of ring is $1$.

Choose a coordinate system co-moving with the ring so that the ring always coincides with the unit circle centered at origin on $xy$-plane.

Let $v_1$, $v_2$, $v_3$ be the contact points of the table edges with the ring. Let $u = (x_u,y_u,z_u)$ be the apex. Decompose $u$ into $c + h\hat{z}$ where $c = (x_u,y_u,0)$ lies on the $xy$-plane and $h = z_u$.

Since the $3$ vectors $v_1 - u$, $v_2 - u$, $v_3 - u$ are orthogonal to each other, we have

$$ (v_1 - c)\cdot(v_2 - c) + h^2 = (v_2 - c)\cdot(v_3 - c) + h^2 = (v_3 - c)\cdot(v_1 - c) + h^2 = 0$$ Given $v_1, v_2, v_3$, in order for this to be possible, we need to find a $c$ such that $$(v_1 - c)\cdot(v_2 - c) = (v_2 - c)\cdot(v_3 - c) = (v_3 - c)\cdot(v_1-c) = -h^2 < 0$$ Subtracting the equations, we need

$$(v_1 - c)\cdot(v_2 - v_3) = (v_2 - c)\cdot(v_3 - v_1) = (v_3-c)\cdot(v_1-v_2) = 0$$ As a consequence of the fact $|v_1|^2 = |v_2|^2 = |v_3|^2 = 1$, above equation has a trivial solution $c = v_1 + v_2 + v_3$ (this solution is unique if $v_1, v_2, v_3$ is on general position).

For this choice of $c$, we find

$$-h^2 = (v_1 - c)\cdot(v_2 - c) = v_1\cdot v_2 - c\cdot(v_1+v_2) + |c|^2 = v_1\cdot v_2 - c\cdot( c- v_3) + |c|^2\\ = v_1\cdot v_2 + (v_1 + v_2 + v_3)\cdot v_3 = 1 + v_1\cdot v_2 + v_2\cdot v_3 + v_3 \cdot v_2$$

Together with the identity: $$|c|^2 = |v_1 + v_2 + v_3|^2 = 3 + 2(v_1\cdot v_2 + v_2\cdot v_3 + v_3 \cdot v_1)$$

We obtain

$$|c|^2 + 2h^2 = 1\quad\iff\quad x_u^2 + y_u^2 + 2z_u^2 = 1$$

This is the equation of an oblate ellipsoid with unit semi-major axis along the equator and semi-minor axis $\frac{1}{\sqrt{2}}$ along the poles.

Update

Above formulas offer us a geometric way to locate possible contact points given an apex.

• Take any point $u$ on the ellipsoid above the plane as apex.
• Project $u$ to $c$ on the $xy$-plane.
• Take $v_1$ to be any point on unit circle.
• Find the mid point $m$ of $c$ and $-v_1$, the antipodal point of $v_1$.
• Draw a line $\ell$ through $m$ perpendicular to line joining $c$ and $v_1$.
• Let $v_2$ and $v_3$ be the intersection of $\ell$ with unit circle.

The three points $v_1$, $v_2$, $v_3$ will be a possible choice of contact points.