borromean circles are impossible

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Borromean Circles Are ImpossibleAuthor(s): Bernt Lindstrom and Hans-Olov ZetterstromSource: The American Mathematical Monthly, Vol. 98, No. 4 (Apr., 1991), pp. 340-341Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/2323803 .

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340

BERNT

LINDSTROM

AND HANS-OLOV

ZETTERSTROM

[April

form

verthe

complex

numbers s all that s

needed, the use of

Proposition may

simplify alculations.

As we

have seen here, that is the

case

with

the

proofsof

[1,2]. The existenceproofof [3] can also be shortened y use ofProposition .

REFERENCES

1. A. F.

Filippov,

A

shortproof f the theorem

n the reduction f a

matrix o

Jordan orm,Moscow

Uniu.

Math. Bull.,26 (1971) 70-71.

2. James

M. Ortega,

MatrixTheory, lenum

Press,New

York,1987.

3.

H. Valiaho,

An

elementary pproach to the Jordanform

f a matrix, his

MONTHLY,

93

(1986)

711-714.

BorromeanCircles

Are

mpossible

BERNT LINDSTR6M

Departmentf Mathematics, oyal nstitutef Technology,-10044

Stockholm, weden

HANS-OLOV

ZETTERSTR6M

NationalDefenceResearch stablishment,-171290

Sundbyberg,

weden

The arrangement f three topological ircles n R3 depicted n FIG. 1 is called

the Borromean inks in knot

theory cf.

1].

Such an arrangementf 'circles' is

possible

with

curves,

which

are

homeomorphic

o

geometriccircles.

To our

knowledge

t

has not

been

observed

before

that

the arrangements impossible

n

R3 ifthe circles re

geometric ven when the radii are arbitrary.

(,

2

FIG.

1.

Our

proof

is indirect.

We assume

that the

arrangement

s

possible using

ordinary

ircles.Lift

C2 (cf.

FIG.

1)

towards he

spectator

uch

that

t will meet C1

in two points n its

new position CI). Observe

that C, will

not meet

C3

since

we

lift

C2 away

from

C3.

There are now two cases: either

C,

and C,

belong

to

a

plane v

or

they

do not

belong

to the

same

plane.

In

the first ase

C3

will

be

partly

bove the

plane, partly


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3/3

1991]

NOTES

341

below.

If we

follow the circle C3 in clockwise directionwe are first bove the

plane, then below, then above, then below, then back again and above the plane.

This means thatthe circle C3 meets the plane at least 4 times,which s impossible

since the circledoes not belong to the plane.

In the othercase, when C1 and C, do not lie in the same plane, we will prove

that theybelong to a sphere. The circle

C3

will meet this sphere at least 4 times,

again a contradiction.We will now provethat C1 and C, belong to a spherewhen

they re

not

coplanar. Let

P

and Q be the points n which C1 and

C,

meet see

FIG.

2).

Let R

be

the

midpoint

f the line

segment PQ.

Let

S,

and

S2 be the

centres

f

the circles

C1 and

C1, respectively.

he

plane

7'

containingR,

S1 and

S2

is orthogonal o the ine PQ. Let

li

be the normal o the plane of

Ci

through

i

(i = 1,2). The line li is orthogonal o the ine PQ and contains hepoint

Si

inthe

plane ST'. t follows hat

i

lies

in

7T'

i

=

1,

2). Since

11

nd

12are not parallel

ines

theywill

meet

n

a

point

T. The distancebetween

T and

anypoint

n

C1 u C,

is a

constantr. Therefore

C1

and C, belong to

the

spherewith entre

T and radius r.

It

is

not

hard to

see that he

circle

3 goes

in

and out of this

phere

4

times,

which

is impossiblebecause a circle and a sphere have at mosttwo points n common

when the circle

does not

belong

to the

sphere. Therefore,

orromean circles re

impossible

p

,<

S

FIG. 2.

It is

interesting

o make a model

of the Borromean

rings

n

some elastic

material

ike

thin

ron wire.

Try

to make the circles'

as

perfect

s

possible

The

arrangement

f

rings

will

then attain

nfinitely any

table

positions

n

space.

REFERENCE

1. D. Rolfsen,

Knots and Links,Publish

or Perish,Berkeley, 976.

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