On some interlocking problems

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On some interlocking problems

Citation for published version (APA):

Post, K. A. (1982). On some interlocking problems. (Eindhoven University of Technology : Dept of Mathematics : memorandum; Vol. 8206). Technische Hogeschool Eindhoven.

Document status and date: Published: 01/01/1982

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..

,

EINDHOVEN UNIVERSITY OF TECHNOLOGY

Department of Mathematics and Computing Science

Memorandum 1982-06 April 1982

On some interlocking problems by

K.A. Post

University of Technology

Department of Mathematics and Computing Science PO Box 513, Eindhoven

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O.

INTRODUCTION

ON SOME INTERLOCKING PROBLEMS

by

K.A. Post

Department of Mathematics and Computing Science University of Technology, Eindhoven

The Netherlands

In a recent paper ([4], problem 48) W. MOSER resutrected a problem that was posed by L. MOSER around 1963 and that was formulated as follows:

"Can every set of 6 points in general position in space (not co-planar, no 4 concyclic, not all cospheric etc.) be partitioned into two sets of 3 points each so that the circles determined by the triples, are interlocked? Same question with "circles" replaced by "triangles"."

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2

-I. FIVE POINTS IN SPACE

We shall prove the following lemma.

LEMMA 1.

Let

A. (1

~ i $

5)

be

5

points in

3-space,

no

4

being coplanar. 1

Then

either there is a unique point Ai' which is interior point of the convex hull of the remaining 4 points,

or the convex hull of a unique pair {A.,A.} and the conveX hull of the 1 J

remaining 3 points have a (unique) point in common.

Proof. Let A. be given by the vector a. (t $ i $ 5). The non-coplanarity

-

1 -1

of the 4-tup1es implies that all 4 x 4 subdeterminants of the matrix

a_{1 t} a₂₁ a₃₁ a₄₁ 3 51 a l2 a22 a32 a42 a52 a 13 a23 a33 a43 a53 I

are non-zero, so that the system of linear equatitms

5

I

i=1 ' i - 0 } 5

has a one-dimensional set of solutions

A

E ~, that can be written as

?l ..

t X (t € It),

where y is a fixed element of (E\{O})5.

Without 108s of generality there are two cases.

(a) ~ component of

I'

Yi say, is negative, the other four components are -I

positive. Choosing t ..

y.

we find that

a.

is a convex combination of

(5)

....

3

-the remaining vectors.

(b) Two components of y, y. and y. say, are negative, the other three

com-- com-- com-- - 1 J

-I -I

ponents are positive. Now we choose t = (Yi + Yj) = -(Yk + Y!), + _Ym) and conclude that a (unique) convex combination of {a.,a.} and a

-1 -J (unique) convex combination of {~'~£'~m} coincide.

Corollary 1. Given 5 points A. (1 ~ i ~ 5) in 3-space, no 4 being coplanar,

1

there exists a line A.A. that contains an interior point of triangle

1 J

~A!),Am (i,j,k,!1.,m distinct).

2. THE CIRCLE PROBLEM

The following theorem will be shown.

THEOREM 2. Every set {B₁, ••• ,B

6} of 6 points in 3-space, no 4 collinear, no 4 concyclic, not all coplanar, not all cospheric, can be partitioned into two sets of 3 points each so that the circles/lines, determined by these triples, are interlocked.

Proof. One of the points B. can be chosen, that is not on the sphere/plane,

- 1

determined by any of the 4-tuples from the remaining 5 points. Call this point B₆• Since our assertion is inversion-invariant when we consider the one-point compactification of the 3-space (i.e. the 4-sphere) we apply inversion with respect to this point B6 and obtain as image points 5 points A. (I S i ~ 5) no 4 being coplanar. Hence, by corollary I there is a line

1

AiAj intersecting triangle ~A!),Am and therefore circular disk ~A!),Am

o

({i,j,k,!1.,m} - {1,2,3,4,S}). In other words, by inversion circle/line B.B.B6 and circle/line BkB~B are interlocked.

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4

-3. THE TRIANGLE PROBLEM

Here we have the following theorem

THEOREM 3. Every set of 6 points A. (1 $ i $ 6) in 3-space, no 4 being

1

coplanar, can be partitioned into two sets of 3 points each, so that the

two triangles, determined by these triples, are interlocked.

Proof. LetA. be given by the vector a. (1 $ i $ 6). As in section 1 all

- 1 -1

4 x 4 subdeterminants of the matrix

are nonzero, so that the system of linear equations

I

A.=o}

i= I 1

has a two-dimensional space of solutions

~

in R6. We are interested in

the six special solutions in which one of the components A. is zero.

1

Remark. Two components A. and A. cannot be zero simultaneously, unless

- 1 - - J

-all components vanish!

So let us have the two basic special solutions

" (1) (0 1 )

"

=

"p,q, r.B

\ (2)

= (

I 0 )

1\ - , ,t,u,v,w

All other solutions we are looking for are linear combinations of these solutions and have a zero in some position.

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5

-l\ way III ml:lly:dng IIII'IH' sollli ion:: i!~ to introdlll'l' 6 :'-dil11(,IlHioll.:tl v('c-tors related to the basic special solutions as:

~I

- [ _: ].

~2

[ :J.

~6

v

=

-5 [ 0 ].

~3

- [ : ].

~4

= [ : ].

[

:

].

First of all these vectors are pairwise linearly independent (cf. the remark).

Secondly we find A(i)

=

_{(det(~i'~k)k=l}6 _{for the basic special solutions}

(i=I,2).

Finally, as a matter of verification we observe that all special solutions are given by

(i 1 .... ,6).

Summing up, in view of lemma I case 2 we now have to study the sign

distri-bution of det(v.,v

k) for six pairwise linearly independent vectors in the

1

-plane.

Recall that det(Yj'Y~) is positive if and only if ~£ is in the left half

plane of v .. Now for each of the twelve directions : v. we indicate by two

-J -J

indices how many vectors ~k are in its left and its right half plane.

Since opposite directions v. and -v. get complementary indices there is a

-J -J

pair of adjacent directions that get indices (2,3) and (3,2), respectively. Let this situation occur for the adjacent directions d. and d .• It is again

-1 -J

a matter of verification that the distribution of vectors v., v., Vk' v_n'

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6 -two vectors: v t v -m -n , / ' d.

3/

1.

/2 2

__ ...

3

two vectors:

Yk'

y~

V t V must be as sketched in the figure, from which the sign distribution

-m -n

follows:

i j k m n

i

o

+ +

J +

o

+ +

Hence, by lemma I case 2 it follows that segment A A intersects triangle _mn

Aj~At and that segment ~At intersects the triangle AiAmAn' In other words:

o

4. GENERALISATtONS. FINAL REMARKS.

The following generalisation of the triangle theorem to odd-dimensional spaces is evident when one uses the same method:

THEOREM 4. Every set of (2n + 2) points in (2n - l)-space, no 2n being

situated in a hyperplane, can be partitioned into two sets of (n + 1) points

each, so that their convex hulls C

r

and CrI have the following property:

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7

-For even-dimensional space we can also use the method of section 3 to prove

THEOREM 5. Every set of (2n + 3) points in (2n)-space, no (2n + I) in a

hyperplane, has two disjoint (n + I)-subsets so that their convex hulls

have exactly one point in common.

For n ~ 1 this theorem states that from 5 points in a plane, no 3 on a

line, always 4 are the vertices of a convex quadrilateral. This is a special case of a theorem by ERDOS and SZEKERES. (cf. [1], [2] p. 55,

[3], [4] problem 29).

References.

[I] ERDOS, P. and G. SZEKERES .. A combinatorial problem in geometry. Compo-sitio Math. 2 (1935) 463-470.

[2] HALL, M. Combinatorial Theory, Blaisdell, Waltham, Massachusetts, (1967). [3] LEWIN, M. A new proof of a theorem of Erdos and Szekeres, The Math.

Gazette, 60 (412) (1976) 136-138.

[4] MOSER, W. Research problems in discrete Geometry (1981) Mc.Gill University, Montreal, Canada.