This is not intended to be a complete list -- people are apparently rather industrious by nature, at least when it comes to Borromean rings, and they seem to have been getting rendered a plenty.

This is just a list of the highlights from my browsings for Borromean rings in early 1998, in no particular order.

Ned Seeman builds Borromean rings out of DNA in his lab.

John Robinson makes beautiful real-life Borromean rings, thanks to his admirable sponsors:

- a square, a bent triangle, and a bent circle (no, those aren't his sponsors!)
- three large wood squares
- three rhombi
- three triangles -- also shown on that page are borromean triangles in a worrisome 9th century scandinavian rune... a completely unrelated web page also mentioned this, saying that:

- There is another interesting historical context in which the rings
arise. The diagram was found in picture-stones on Gotland, an island in
the Baltic sea off the southeast coast of Sweden. These are dated around
the ninth century and are thought to tell tales from the Norse myths. To
the Norse people of Scandinavia, a drawing of the Borromean rings using
triangles is known as "Odin's triangle" or the "Walknot" (or "valknut"
-- the knot of the slain). The symbol was also carved on the bedposts used
in their burials at sea.

Paul Bourke has several brightly colored diagrams and pictures

A cat looking at borromean rings

For a grand finale, here's
a page with 3 rings, 4 rings, 6 rings, mpeg animation, stereoscopic views,
raytraced pictures, ...

**Proof #1:**

(probably originally by Mike Friedman)

Suppose you have a link in R^{3} made of circles such that
any two circles have linking number 0. Then the link is trivial.
Why? Consider the 4-D unit ball B^{4}. First put R^{3}
stereographically onto the ball's surface S^{3} (this preserves
circles). Each circle now bounds a (flat) 2-disk in B^{4}.
When two circles have linking number 0, the corresponding disks do not
intersect (Can you picture this well enough to believe this claim?
I have trouble visualizing this). Now consider the family of concentric
3-spheres centered at the origin. The intersections of the spheres
with the family of disjoint disks gives a homotopy of the link, except
that at certain strategic moments, certain of the circles vanish to a point
and disappear. Anyway this homotopy shows that the link is the trivial
link.

**Proof #2:**

(this is a 3D-ization by me of the above proof)

Given any arrangement of circles in 3-D, you can steadily shrink the
areas of the circles without any of the circles ever touching each other,
until each circle is reduced to a point (therefore nothing "went through"
it), whereupon you can remove that circle from the others (therefore it
wasn't linked to anything in any way).

(This can easily be made fully rigorous, if we specify that the circles should shrink so that each remains in a constant plane with a constant center, and the amount of area lost by each circle should equal the time T that has passed so far. If the disks bounded by two rings don't intersect, then the rings clearly won't touch as they shrink, so the only interesting case is when the disks do intersect. If the disks are in the same plane, then one is inside the other, and its radius is shrinking faster, so they won't meet. If the disks are not in the same plane, then, looking at the line of intersection, one can easily check that a shrinking circle's two points on this line are shrinking just as if they were themselves the diameter of a shrinking circle, so this case can be transformed in this way to a corollary of the case where the circles are in the same plane.)

**Proof #3:**

(by Conway and myself)

The disc bounded by any ring (A) must intersect another (B) (since
otherwise it could be shrunk to a point without hitting the others), and
therefore must intersect it twice. I'll say A swallows B. Then
plainly B can't also swallow A, so B swallows C and C swallows A.
The planes of the A and B rings intersect in a line L -- if this is parallel
to the plane of the C ring, then looking along it we see a picture:

`
\ /`
`
\`
`
/ \`
` B------/------`
`
/ \`
`
/ C`
`
A`

in which we can see that one "end" of any ring can be "shrunk" until it is no longer "swallowed" (eg, the lower end of the C ring), so that the object plainly isn't Borromean.

The same argument works if L hits the plane of C in a point outside
the disc of C - we just project from this point (which is just the intersection
of the planes of all three rings. The only case left is therefore
that in which the three discs have a common point. Call it P.
The power of the point P (the product of the two distances to a given circle
on any line through P and the circle) will always be larger with respect
to the swallower than with respect to the swallowee, since P is inside
both discs. Chasing this inequality around the three circles gives
a contradiction, so it is impossible for three (or more) discs to share
a common point if each one swallows the next. So such a case in fact
cannot occur, so we are done.

**Conjecture:**

**I conjecture that** the converse is also true:

That is, given any three simple unknotted closed curves in 3-D, they
can always be arranged in a Borromean arrangement *unless* they are
*all* *circles*.

For example, try making a Borromean arrangement with:

- a medium-size circle, a very small circle, and a very large square
- three congruent circles but one of them is stretched very slightly into an ellipse
- three "tennis-ball-seam" shapes