What shape is the locus of a 3D corner with a circular ring that touches the sides of the corner? Hard for me to draw this, but I happen to have my wedding ring, and the corner of a table. I can put the ring over the corner, the ring touching the three edges of the table, with the apex of the corner sticking through.
If I slide the ring around while three points on it touch the edges, the apex moves around and there's a locus of points that the apex can be on.
Relative to the circle, is it part of a sphere?
Also, is there another name for this problem?
 A: 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 
$$x^2+y^2+2z^2=1.$$
A: 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
$$-h^2 = 
(v_1 - c)\cdot(v_2 - c) = v_1\cdot v_2 - c\cdot(v_1+v_2) + |c|^2
= v_1\cdot v_2 - c\cdot( c- v_3)  + |c|^2\\
= v_1\cdot v_2 + (v_1 + v_2 + v_3)\cdot v_3 
= 1 + v_1\cdot v_2 + v_2\cdot v_3 + v_3 \cdot v_1$$
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.
A: Others have posted the exact answer -- but, without using equations, there's an easier way visualize that it is, at least, not a sphere.
Imagine dragging the bottom of the ring downward against the vertical edge of the table, so the opposite side of the ring is dragged toward the corner where it's almost slipping off.  The plane of the ring is almost vertical, and so, relative to the ring, the corner of the table is moving almost perpendicular to the ring.  So the locus of the corner, where it meets with the ring, must be perpendicular to the plane of the ring.  So IF it's a section of a sphere, it has to be a hemisphere.
But it can't be a hemisphere, because you can also picture a cube inscribed inside a sphere, with one of the corners touching the top of the sphere, and the three nearest corners lying in a circle. This is the position that the ring would be in when the corner is in the center.  But the section of the sphere cordoned off by the ring is not a hemisphere; the corners don't come halfway down the sphere (their altitude is 1/3 of the way down from the top of the sphere to the bottom).
