(X) ON THE PROJECTION OF A PLANE SET OF FINITE LINEAR MEASURE *

ON THE PROJECTION OF A PLANE SET OF FINITE LINEAR MEASURE *

BY

H. G. E G G L E S T O N in Cambridge

Introduction

F o r a n y set X w e d e n o t e b y A ( X ) t h e l i n e a r m e a s u r e ( l ) of X a n d b y A ( X 0 ) t h e l i n e a r m e a s u r e of t h e p r o j e c t i o n of X o n t o a h n e p e r p e n d i c u l a r t o t h e d i r e c t i o n 0. W e w r i t e /~ (X) for t h e g r e a t e s t lower b o u n d of A

(Xo)

t a k e n o v e r all d i r e c t i o n s 0. W e shall consider t h r e e classes of p l a n a r sets, n a m e l y m e a s u r a b l e sets, c o n n e c t e d sets, a n d arcs.

F o r e a c h class we shall f i n d t h e u p p e r b o u n d of t h e r a t i o lu ( X ) / A (X).

F o r t h e class of m e a s u r a b l e sets t h e r e s u l t is c o n n e c t e d w i t h t h e p r o p e r t i e s of r e g u l a r a n d i r r e g u l a r sets a n d is a c o n s e q u e n c e of t h e p r o p e r t i e s of t h e s e sets e s t a b l i s h e d b y B e s i c o v i t c h . F o r t h e class of c o n n e c t e d sets a n d for t h e class of arcs ;u (X) is t h e m i n i m a l w i d t h of t h e c o n v e x cover of X or its c o n v e x hull as i t is s o m e t i m e s called. T h e p r o b l e m of t h e r e l a t i o n s h i p b e t w e e n t h i s f u n c t i o n a n d A

(X)

is one b e t w e e n a set a n d its c o n v e x cover. T h e r e ' a r e of course a large n u m b e r of such p r o p e r t i e s a n d a f u r t h e r r e s u l t of t h i s t y p e is g i v e n in S e c t i o n 5.

A n i n t e r e s t i n g f e a t u r e of t h i s p r o b l e m is t h e d i f f i c u l t y of d e t e r m i n i n g c o m p l e t e l y t h e class of e x t r e m a l figures. F o r t h e class of m e a s u r a b l e sets t h e u p p e r b o u n d of # ( X ) / A (X) is n e v e r a t t a i n e d , b u t we give e x a m p l e s t o s h o w t h a t t h e u p p e r b o u n d w h i c h we e s t a b l i s h is in f a c t t h e l e a s t u p p e r b o u n d . On t h e o t h e r h a n d b o t h t h e u p p e r b o u n d s for t h e class of c o n n e c t e d sets a n d for t h e class of a r c s a r e a t t a i n e d ; in t h e first class b y a set c o m p o s e d of t h r e e e q u a l s e g m e n t s e q u a l l y i n c l i n e d t o one a n o t h e r a n d in t h e s e c o n d c a s e b y a n a r c

* Editor's n o t e . - - T h i s paper was received on January 4, 1957. Without our knowledge it has appeared during 1957 as part of the book Problems i n Euclidean Space, London 1957, by the same author.

(1) Hausdorff one.dimensional measure. See [1], where it is referred to as Carath6odory measure.

5 4 H. G. EGGLESTON

composed of four linear segments a n d two circular arcs (which will be specified more e x a c t l y later). To simplify t h e proofs we shall consider in b o t h cases the subclasses of connected sets or of arcs whose convex covers are polygons with at m o s t n vertices. Since, in fact, one c x t r e m a l figure for the class of arcs is n o t of this n a t u r e we h a v e no hope, b y this means, of specifying all the extremal figures. B u t even for the case of connected sets when the only k n o w n extremal figure is of this kind I h a v e n o t been able to specify com- pletely all the extremal figures. Some f u r t h e r remarks a b o u t this point will be given later (see Section 6).

The actual results p r o v e d in the following p a r a g r a p h s are (i) for a n y measurable set E with A (E) > 0,

(ii) for a n y connected set E

/~ (E) < 2 A(E),

Y~

/~ (E) ~< 89 (E), (iii) for a n y simple arc E

F~(E) ~<A(E)/(sec ~ + 2 t a n cr § - 4 f l - 2 a ) where ~ a n d fl are defined b y

89 § sin a = 4 cos ~ a / ( 1 § 4 cos 2 a)

a n d t a n fl = 89 see a.

T h e results p r o v e d in Section 5 are s t a t e d in t h a t p a r a g r a p h . I a m indebted to t h e referee for suggesting simplifications of some of t h e properties established in Section 3.

w 1. E any measurable plane set of finite positive linear measure

W e can write E = E 1 U E 2 where E I is a regular a n d E 2 an irregular set (see [1], p. 304). F u r t h e r E 1 = El 0 E~' where A (El') = 0 a n d El is a measurable subset of the union of an enumerable infinity of rectifiable arcs (see [1], pp. 324 a n d 304). A n o t h e r p r o p e r t y t h a t we require is t h a t the projection of an irregular set is of zero measure in almost all directions (see [2], p. 357). Since we do n o t require m a n y other properties of regular a n d irregular sets I shall n o t give their definitions n o r the derivation of the prop- erties s t a t e d above. T h e y can be f o u n d in t h e papers [1] a n d [2].

W r i t e P ( X , O) for the set which is t h e projection of X in t h e direction 0. T h e following l e m m a s are needed.

O N T H E P R O J E C T I O N O F A P L A N E S E T O F F I N I T E L I N E A R M E A S U R E 55 LEMMA 1. [P (E~, 0)] depends continuously on O.

L e t A~ be a sequence of arcs each of finite linear measure such t h a t 5 A i D E~ and

i = l

let ~ be a sequence of positive numbers decreasing to zero. F o r each integer i there exists a closed subset Fi of E~ and a positive integer N~ such t h a t

A(F~) > A(E~) - el

Ni

A(F~ n U As) > A (F~) - e~.

./=I

The set A s fl F~ is a closed subset of the arc As and its complement in As is an at most enumerable infinity of open subintervals of As say Bs,1, Bs,2, . . . These subintervals of As are open relative to As and there m a y of course be only a finite n u m b e r of them. We can choose an integer M~. s such t h a t

A(

U

Bs~)< - - 1 k~Mi] Ni e i~

The complement of

U

Bs~ in A s consists of a finite n u m b e r of arcs or points, say AS.l, ...,

l ~ k < M i j

As h, where h depends on both i and ?'. We omit the points in this set and rename the set

L i

of all these arcs for all ] from 1 to N~ as Ci, C~ . . . Cr.~. Write C for U Ct, then if H denotes

1=1

the set of points omitted in renaming the As.k as Cl, we have

F r o m (2) it follows t h a t

C U H D F ~ N U As, (1)

1~1_<N t

C - F ~ c

U U

Bs~. (2)

I<:I<_N t k>.Mtj

A (C - F,) < e,. (3)

Hence A ( C - E~) + A ( E ~ - C) < 3~,.

Thus I A[P(E~, 0)] - A [ P ( C , 0)][ < 3 e,. (4)

B u t A [P(C, 0)] depends continuously on 0, and (4) shows t h a t A [ P ( E ~ , 0)] is the uniform limit of a sequence of continuous functions. Thus A [ P ( E ~ , 0)] is continuous and the l e m m a is proved.

2~

LEMMA 2. f A [P (E~ , O)]dO-~.4A(E1). -< '

: 0

As in l e m m a I there is a sequence of sets {C~) each of which is a union of a finite n u m b e r of rectifiable arcs and such t h a t A[P(Ct, 0)] tends to A [ P ( E ~ , 0)] uniformly in 0 and

5 6 H . G . EGGLESTON

A (Ct) tends to A (E~). T h u s we need only prove L e m m a 2 when E~ is a union of a finite n u m b e r of rectifiable arcs. Clearly this case will follow if we can establish t h e inequality for one are. B u t we can a p p r o x i m a t e to an arc .4 b y a polygonal line R such that, given e > 0 every point of .4 is within a distance 89 e of some point of R a n d A (R) ~<A (A). T h e n A [ P ( R , 0)] >~A[P(A, 0)] - e a n d it is sufficient t o prove t h e inequality for a polygonal line. Finally this case will follow if the inequality is true for a single segment. B u t the t r u t h of the inequality i n this last case is easily verified. T h e l e m m a is proved.

I n the n e x t L e m m a we need to consider t h e relationship between the set E~ a n d the union of an enumerable infinity of rectifiable arcs of which E~ is a measurable subset.

There are of course m a n y such sets of arcs. We select one A a n d call the arcs of which it is t h e union A1, A s . . . . L e t p be a p o i n t of E~ lying on arc Al of A. The densities of A t a n d of E~ n At at p are defined to be

lim A (A~ N C (p, r)) lira A (E~ N At n C (p, r))

T-,0 2 r ,-,0 2 r

respectively, when these limits exist where C (p, r) is t h e closed set of points whose distance f r o m p is less t h a n or equal to r. I t is k n o w n t h a t at almost all(1) points p of A~ the first density exists a n d is equal to u n i t y a n d at almost all points p of E~ n A t the second density exists a n d is equal to u n i t y (see [1], p. 303-304). F u r t h e r , since A t is a rectifiable arc it is k n o w n t h a t at almost all points of it there is a t a n g e n t to it. T h u s finally at a l m o s t all points p of E~, the densities of At a n d E~ N At are u n i t y a n d t h e t a n g e n t t o A~ exists.

There is of course a certain a m b i g u i t y in this since p m a y belong to more t h a n one arc A 4.

B u t in this case we simply select one A~ corresponding to each p a n d consider this arc At associated with p t h r o u g h o u t w h a t follows. T h e t a n g e n t to p will be d e n o t e d b y t (p) a n d a n y point p of E~ with the a b o v e properties will be called an R-point.

LEMMA 3. Either

(a) almost all points o/ E~ lie on one straight line or

(b) there are two R-points o/ E~, say Pl, P~, such that p~ does not lie on t (Pl) and Pl does not lie on t (P2).

I f (a) is false we c a n select an R - p o i n t of E~, ql a n d a second R - p o i n t q2 t h a t does n o t lie on t (ql). I f ql does n o t lie on t (q~) t h e n ql, q~ h a v e the properties required. I f ql lies on t (q2) we select, if possible, a t h i r d R - p o i n t qa n o t on t (ql) nor t (q2). N o w t (qa) c a n n o t contain b o t h ql a n d q2 since if it did q3 would lie on qlq~, i.e. t(q~). T h u s one of the pairs qlq3 or

(x) "almost all" means "all but a set of zero linear measure".

O N T H E P R O J E C T I O N O F A P L A N E S E T O F F I N I T E L I N E A R M E A S U R E 5 7

q2qa

has the required properties. I f we cannot select a point such as qa then almost all of E~ lies on t (ql) U t (q~), and there are points of E1 other t h a n ql or q2 on each of these lines.

L e t q4 be an R-point of E on

t(ql)

distinct from ql. Since

f(q4)

m u s t coincide with

t(ql)

we can take the pair q2, q4 as the pair Pl, P2-

The l e m m a is proved.

We are now in a position to prove the main result. I f A(E~) = (~ > 0, then for almost all 0

A [P(E~,

0)] = 0. (5)

By Lemma 2 we can choose an angle 0 such t h a t

A [ P ( E l , 0)] < 2_ ( h (El) + 8),

(6)

and since b y L e m m a 1 A

[P(E~,

0)] is a continuous function of 0 we m a y suppose t h a t both (5) and (6) hold for the same value of 0. Since A (El') = 0 we have A

[P(E~',

0)] = 0 for all 0. Thus finally

[P (E, 0)] < _2 A (E).

A

I f A (E2) = 0, and (a) of L e m m a 3 holds for E~, then almost all points of E lie on one straight line and projecting parallel to this line we see t h a t / u ( E ) --0. This implies the required result.

I f A (E2) = 0 and (a) of L e m m a 3 is false for E~ let Pl and p~ be two R-points of E1 for which (b) holds. We now require the p r o p e r t y t h a t if p, an R-point of E~ projects onto the point q of the set

P(EI, O)

a n d the direction of projection is not parallel to t(p), then the set

P(E~, O)

has unit density at q. We suppose A, is the arc associated with

p, C(p, 8)

is the closed disc centre p and radius ($ (as above), and write

l(q, 8)

for the linear closed interval perpendicular to the direction of projection with q as mid-point and of length 2 8.

Given a positive n u m b e r e we can find a positive n u m b e r 80 such t h a t

A(E~ flA, n c(p, 5)) >(I -D28 (7)

A(A~ N

C(p,

~))< (I

+e)25

for all 8 < 80. Now write A* for A, n

C(p, 8),

t h e n

A [P (E; N A*, 0) N I (q, 8)]/> A [P (A*, 0) N I (q, 8)] - A [ P (A~ - E~, 0) N I (q, 6)], (8) and

A [ P ( A * - E ; , O)N I(q,

8)]~< A [ ( A ~ - E;) N

C(p,

8 ) ] < 4 e ~ (9)

5 - 665064 )4cta m a t h e m a t i c a . 99. I m p r i m ~ le 19 a v r i l 1958

5 8 H . G ~. E G G L E S T O N

if '~ < 50. But if (~ is sufficiently small, say ~ < 51, then

P(A*, O)~I (q, ~).

Thus from (8) and (9)

A[P(E~ N

A?, O) N I (q,

5 ) ] ~ > 2 6 ( 1 - 2 e ) for 6 < min (~o, 61). Since obviously

A [P(E~

N A*, O) N I(q,

~)] ~<25, it follows that q is a point of unit density of

P(E~, 0).

Now suppose that the direction of the line joining the two R-points

PlP~

of E~ is 00.

We divide E~ into two sets, E~ formed from those points of E~ whose distance from Pl is less than one half the distance of Pl from P2 and E~* defined by E~* = E~ - E~. Then since

P(E*, 0o)

and P(E~*, 00) have a common density point,

A [ P (E~, 0o) ] < A [P (E~, 00) ] + A [P (E~*, 0o)].

By continuity established in lemma 1 and by lemma 2 applied to E~ and E~* it follows t h a t

2n

f A [P(E~,

0)] d 0 < 4 A (E~) + 4 A (E~*) = 4 A (E~).

0

Since we have A ( E ) = A (E;) we conclude t h a t for some 0 A [P (E, 0)] < 2 h (E).

Yg

Thus in all cases we have

/~ (E) < ~ A (E). 2

Example.

We next construct an example to show t h a t this result is the best possible.

Let e be a given positive number and n a large positive integer, the actual lower bound of which will be specified later. Let M1, M~ . . . M4n be 4n points such t h a t all the lines M~ Mj have different directions. Let Lt be a segment of length 5//4n in a direction making an angle 2

zi//4n

with a fixed direction, and with mid-point at M~. Choose (~ so small t h a t if we project the segments in any direction at mose ~w, , f ~h~ ~gements overlap.

4n

Denote L; Lt by E. Then

~ 1

A [P(E, 0)]1> ~ . \ 4 n 4 n '

ON THE PROJECTION OF A PLANE SET OF F I N I T E L I N E A R MEASURE 5 9

a n d since t h e expression on t h e right h a n d side of (10) is periodic in 0 with period z / 2 n we m a y assume' t h a t 0 ~< 0 ~< ~ / 2 n . S u b s t i t u t e those v a l u e s of 0 which lie in this r a n g e a n d reduce t h e right h a n d side t e r m s to their least values, i.e. for 0 < j < n a n d 2 n ~< j < 3n,

= 4n, p u t 0 = ~ / 2 n ; for n ~ ~ < 2 n a n d 3 n ~ ~ < 4 n p u t 0 = 0. T h e n

2~

A [ P ( E , O ) ] ~ - n ~ ~ c O S ~ n - ~ n = 5 l e o s 2 z t x l d x + o ( 1 ) , (11)

0

as n - + c~. T h u s choosing first n sufficiently large, a n d t h e n points M 1 ... M4n, a n d 5 we h a v e for all 0

A [ P ( E , 0 ) ] > 6 ( 2 - e ) .

T h u s ~t (E) 1> 2

A (E) zt

and this shows t h a t t h e result o b t a i n e d is the best possible.

(12)

w 2. Some preliminary results

I n t h e following two t h e o r e m s t h e containing space is R z.

THEOREM 3.1. Let T be a closed connected set o / / i n i t e linear measure and let H ( T) be its convex cover. Then either there is a tree T 1 contained in T such that the convex cover o / T 1 coincides with that o / T , or there is a simple closed convex curve K contained in T such that its convex cover coincides with that o / T .

W e use H ( X ) to d e n o t e t h e c o n v e x cover of t h e set X .

T h e r e is a subset K of T which is irreducible with respect to t h e t h r e e properties, (i) K c~T,

(ii) H ( K ) = H ( T ) ,

(iii) K is closed a n d connected.

There certainly exist sets with these three properties since T is one such set. I f possible f o r m a sequence of sets K , such t h a t each K~ has properties (i), (ii), (iii) a n d K j is a p r o p e r subset of K s if j > i. I f it is only possible to define a finite sequence of such sets t h e n t h e last m e m b e r of t h e sequence i s irreducible. I f t h e sequence has infinitely m a n y m e m b e r s it can contain a t m o s t a n e n u m e r a b l e infinity of members(1) (since t h e sequence of sets

(1) T h e s e q u e n c e K t m a y of c o u r s e be t r a n s f i n i t e b u t s i n c e t h e c a r d i n a l is l e s s t h a n ~] w e c a n a l w a y s f i n d a n e n u m e r a b l e s e q u e n c e of o r d i n a l s a s s t a t e d .

60 H.G. EGGLESTON

complementary to K~ in T form a strictly increasing sequence of sets open in T). B u t then N Kt = K* has properties (i) and (iii). We shall show t h a t it also has property (ii).

i

I f p E H ( T ) then p EH(K~) and therefore, since K i is connected by Bunt's refinement of Carath~odory's theorem (see [5]) there exist two points k~, kt' of K~ such t h a t p is a point of the segment k~k[. We can select a subsequence of ordinals n~ such t h a t for every ?" of the

I t

sequence there exists an i with n~ > j and such t h a t kn~-> k, and knt-> k . Then since k i EK~

if i > ?" and Kj is closed k U k' EKj, all j. Thus/r U k' EK*. Also p is a point of the segment kk'. Hence pEH(K*) and this means t h a t H(T)cH(K*): since the reverse inclusion is trivial (ii) is proved. Clearly K* is irreducible and the statement is proved.

If K* is a tree we have the desired result. If K* is not a tree, then there are two points Pl, P2 of K such t h a t two arcs exist ~1, ~ both contained in K* and having in common only their end points Pl and p~. (K* is of finite linear measure and therefore both locally connected and arc-wise connected.) If these two arcs lie in F r (H(T)) = Fr (H(K*)) then they comprise the whole of t h a t frontier and form a closed convex curve with the properties stated in the theorem. Otherwise there is a point say p on them which is an interior point of H (T). Let the distance of p from Fr H (T) be ~.

Now every component of K* - (~1 U ~ ) meets ~1 U ~ in a single point, for if this were not the case we could join two distinct points of ~1 t) ~ say p and q by an arc t h a t lies in K* - (~1 U ~ ) . This are cannot lie in ~1 U ~ since K* is locally connected, and thus this arc contains a subarc meeting ~1 U ~ only at its end points Pl and ql. But this means t h a t in K* there are three distinct arcs joining Pl to ql and intersecting only in their end points.

Then one of these arcs lies in the bounded domain of which the other two form the frontier.

Denote this open domain by D. K* - D has the same convex cover as K*, is closed con- nected and is a proper subset of K*. This is impossible by the irreducibility property of K*.

Let ~ be a subset of ~1 U ~ contained in C(p, 89 Since K* is irreducible every compo- nent of K* - (~1 U ~2) meets Fr H(K*). If it also meets ~ such a component must have linear measure of at least 89 ~. Since K* is of finite linear measure there are at most a finite number of such components and hence a subarc ~1 of ~ which is disjoint from K* - (~1 U ~ ) . B u t then K* - ~ 1 is a closed connected set with the same convex cover as K* (since ~1 is interior to this convex cover) and is a proper subset of K*. This is impossible since K* is irreducible.

Thus arcs such as ~1, ~2 do not exist and Theorem 3.1 is proved.

DEFINITION: A polygonal tree is a tree formed from a finite number of linear segments.

We always consider such a tree to have a simplicial decomposition into linear segments. So

O N T H E P R O J E C T I O N O F A P L A N E S E T O F F I N I T E L I N E A R M E A S U R E 61 t h a t if t w o s e g m e n t s m e e t t h e y d o so o n l y in a c o m m o n e n d p o i n t , a n d e v e r y e n d p o i n t of a s e g m e n t is e i t h e r a n e n d p o i n t of t h e t r e e or a n e n d p o i n t of a t l e a s t one o t h e r s e g m e n t of t h e tree. A p o i n t w h i c h belongs t o m o r e t h a n one s e g m e n t of t h e t r e e is c a l l e d a s i n g u l a r p o i n t of t h e tree.

THEOREM 3.2. Let / ( X ) be an increasing continuous/unction o / t h e convex set X , i.e.

X I ~ X 2 i m p l i e s / ( X 1 ) >~ /(X~). Let ff be the class o/connected closed sets o / f i n i t e positive linear measure. Let ~ (n) be the subclass o / • o/those polygonal trees whose convex covers are polygons with at most n sides. Then

sup [ ( H (T)) = s u p s u p / ( H ( P ) ) . r~z A ( T ) ~ P ~ ( . ) A ( P )

B y t h e p r e v i o u s r e s u l t t h e r e is a t r e e K c o n t a i n e d in T s u c h t h a t t h e c o n v e x c o v e r of K coincides w i t h t h a t o f T, or a s i m p l e closed c u r v e K c o n t a i n e d in T for w h i c h t h e c o n v e x c o v e r s of T a n d K coincide.

L e t kl, k2 . . . kn, b e a sequence of p o i n t s d e n s e in K a n d consider t h e class of p o l y g o n a l t r e e s w h i c h c o n t a i n k 1 .. . . . k.. A m o n g s t t h e s e we select one w i t h l e a s t l e n g t h a n d d e n o t e i t b y K , . T h e n , w h e t h e r / t : is a t r e e o r a s i m p l e closed c u r v e ,

A (K.) ~<A(K)

H ( K ) ~ ( J H ( K ~ ) ~ ( H ( K ) ) ~ .

n

T h u s g i v e n e > 0 t h e r e e x i s t s a n i n t e g e r n s u c h t h a t / ( H ( K n ) ) / ( H ( K ) )

A (K.) A (K)

B u t K n E ~)(m) for s o m e m, t h u s

[ ( H ( T ) ) s u p s u p /(H(P)__) > / s u p - - 9

. P~(.) A ( P ) r~ A ( T )

T h e i n e q u a l i t y in t h e r e v e r s e d i r e c t i o n is t r i v i a l . T h u s t h e t h e o r e m is p r o v e d . T h e r e is a s i m i l a r r e s u l t for t h e class of arcs.

THEORE~ 3.3. L e t / ( X ) be an increasing continuous/unction o/ the convex set X . Let CI be the class o/arcs o/finite positive linear measure and .,4(n) be the subclass o/those members

o / a / o r m e d ]rom at most n segments. Then

/ ( H ( A ) ) / ( H ( A * ) ) s u p - - = s u p s u p

A~a A ( A ) n A ~A(-) A ( A * ) T h e p r o o f is o m i t t e d .

62' H.G. EGGLESTON,

w 3. E a closed c o n n e c t e d plane set o f finite positive l i n e a r m e a s u r e , DenOte b y s (n) the class' of closed connected plane sets ~ which are of finite positive linear measure and sUch thag their convex Covers are polygOnS with at most n vertices:

Since the subclass of I~ (n) contained in a bounded part of the plane forms a Compact, space under the closed-set metric (see [1], p. 316, a n d [3]) it follows t h a t there is a member

T of s (n) such t h a t

( T ) (E)

= sup # (13)

"E,~(n) A (E)

We shall show that / x ( T ) = 89 This will imply that for any connected set E /x(E) ~< 89 (E), for the general case when the convex cover of E is not a polygon can be dealt with b y Theorem 2, Section 2.

Our argument will be such t h a t we can specify the e x t r e m a l figures T exactly, in so far as T is a m e m b e r of some s (n) b u t not when T is n o t a member of some IZ (n), When E is composed of three equal segments equally inclined to one a n o t h e r , / x ( E ) = 89 (E).

Thus we h a t e /x(T) ~> 89 (T), (*)

and our aim in the rest of this paragraph is to show tha t # (T) ~< 89 A (T). One method is to assume the contrary,(1) namely t h a t ~u(T) > 89 (T) and show t h a t this leads to a con- tradiction. I have not followed t h a t method here because it is not then possible to par- ticularize the extremal figures. The method is to use (13) and (*) to establish b y variational arguments a number of properties of T which will specify it more a n d more exactly Until"

finally we can assert t h a t # ( T ) < 89 (T).

Denote the polygon which is the convex cover of T by P . / x (T) is the minimal width of P. A support line of P which is at a distance/x (T) fro m the parallel support line will be referred to as a minimal support line. A vertex of P which lies on a minimal support line of P will be referred to as a minimal v e r t e x . T h e r e are two properties of minimal support lines of which we shall make frequent use.

(A) A pair o f minimal ~ support' lines i s such' t h a t a t least one of the lines: meets P i n a segment. Otherwise we could give each of the lines ,an equal r o t a t i o n a b o u t t h e vertices of P through which they passed and reduce the distance apart of the two lines. This would contradict the fact t h a t ' t h e y are a pair Of minimal Support lines of P.

(B) If the lines l 1 and 11 are a pair Of minimal sUpport lines mid ~ e e t P ' i n X 1 aiid ~72 ~ respectively, then the projection of X 1 onto 12 b y means' of lines perpendicular ~0 both l~

and 12 is a set YI which intersects X2.

(1) I feel no aversion to this t y p e of a r g u m e n t b u t I find it r e p u g n a n t to have to illustrate a hypothetical argument b y drawing a diagram which cannot exist! (See Fig. 3 later'.)

O N T H E P R O J E C T I O N O F A P L A N E S E T O F F I N I T E L I N E A R M E A S U R E 6 3

F o r if t h i s were n o t t h e case t h e r e w o u l d be a line m p e r p e n d i c u l a r t o 11 a n d t~ sepa- r a t i n g X 1 f r o m X 2. S u p p o s e m m e e t s 11 in L 1 a n d l~ in L 2. I f we give t o l I a r o t a t i o n a b o u t L 1 a n d t o 12 a n e q u a l r o t a t i o n a b o u t L~ we s h o u l d r e d u c e t h e d i s t a n c e b e t w e e n t h e t w o p a r a l l e l lines. B u t since X 1 a n d X 2 lie on o p p o s i t e sides of m we can choose t h i s r o t a t i o n t o b e in such a sense a n d of such a m a g n i t u d e t h a t t h e r o t a t e d s t r i p still c o n t a i n s P . B u t t h i s c o n t r a d i c t s t h e f a c t t h a t t h e d i s t a n c e a p a r t of ll a n d 12 is t h e m i n i m a l w i d t h of P . W e shall l a t e r r e q u i r e t h e following l e m m a : i t is i n s e r t e d here ;[or c o n v e n i e n c e of reference.

LEMMA. Let A B C be a triangle every angle o/which is less than ~ re. Let K be the unique point such that / A K B = / _ B K C = ~ . C K A = ~ z~. On A B erect the triangle A D B which is equilateral and such that D lies on the side o / A B opposite to C, then

(i) o / a l l connected sets containing A, B and C the t r e e / o r m e d / r o m the three segments A K , B K , CK, has the least length,

(ii) the sum o/the lengths A K + B K + C K is equal to the length CD.

L e t ~ be a c o n n e c t e d set j o i n i n g A , B a n d C. I f ~q has i n f i n i t e l i n e a r m e a s u r e we n e e d n o t consider i t f u r t h e r . I f ~O h a s f i n i t e l i n e a r m e a s u r e t h e n i t c o n t a i n s a n a r e ~21 j o i n i n g A t o B a n d a n arc ~ j o i n i n g A t o C. L e t K 1 be t h e l a s t p o i n t of 71 N ~ on ~ in t h e o r d e r A t o C. T h e n arc A K 1 of ~1 has l e n g t h g r e a t e r t h a n or e q u a l t o s e g m e n t A K 1 : a r c K 1 B of )p, h a s l e n g t h ' g r e a t e r t h a n or ~ eq u a l to t h a t of s e g m e n t K I B : arc KIG of~] 2 has l e n g t h g r e a t e r t h a n 0r e q u a l t 0 ~ t h a t of K~C. T h u s

A (~q~ ~ > A K I rk K 1 B + K1C.

W e n e x t consider a v a r i a b l e p o i n t X a n d t h e f u n c t i o n X A + X B + X C = F ( X ) . ( 1 ) T h e r e is a p o s i t i o n of X for w h i c h F ( X ) a t t a i n s its l e a s t v a l u e . L e t t h i s p o s i t i o n be X 0.

I t is e a s y t o see t h a t X 0 does n o t coincide w i t h a n y of A o r B or C since e a c h a n g l e of t r i a n g l e A B C is less t h a n ] re. I f we mo,ve X f r o m X o in t h e d i r e c t i o n p e r p e n d i c u l a r t(x A X o t h e n A X = A X o + O(XXo) ~ a n d t h e r e f o r e B X + C X = B X o + C X o + O(XXo) 2, i.e.

X X o is p e r p e n d i c u l a r t o t h e i n t e r n a l b i s e c t o r of / BXoC. Thus / A X o C ~ Z_ AX~o B a n d s i m i l a r l y b o t h t h e s e angles a r e e q u a l t o / BXoC., i.e: X o c o i n c i d e s w i t h t h e p o i n t K . T h u s

K~A q: K 1 B + K1C >~ K A + K B + K C a n d (i) is p r o v e d .

T o , p r o v e (ii) we, h a v e ~ A K B ~ / _ A D B = re so t h a t A D B K is a c y c l i c q u a d r i l a t e r a l (see F i g . 1~, A l s o ~ A K D - fl_ABD~--fxare so t h e p o i n t s C,: K , D a r e col!inear, a n d !we

(1) Here X A denotes the length of the segment joining X to A,

6 4 H. G. EGGLESTON

g

B

D Fig. I.

need only show t h a t K D = A K + K B . T a k e E on K D so t h a t L K E B = L K B E . Then, since L E K B = ~ ~, K E B is an equilateral triangle and K E = K B. Since L K B E

= 89 z~, L K B A = L E B D . Hence in triangles A K B , D E B , L K A B = L E D B since A D B K is a cyclic quadrilateral, A B = D B since A B D is an equilateral triangle, L K B A = L E B D proved above.

Thus triangle A K B is congruent to triangle D E B a n d E D --- A K .

Hence A K + K B = E D + K E = K D a n d the proof of (ii) is complete.

Properties of

the extremal figure T 1. Every vertex o / P belongs to T.

2. O] all connected sets containing the vertices o / P , T has the least length.

3. T is a polyqonal tree/ormed/rom a / i n i t e number o/linear segments.

4. Every end-point o / T and every singular point o / T is a vertex o / P .

I f a n end-point t of T was not a vertex of P we could remove from T a small segment with one end point a t t a n d obtain T I E s ). Since A ( T ~ ) < A ( T ) , and (if the segment r e m o v e d is sufficiently small) T1 contains the vertices of P, we have a contradiction with 2.

Thus every end point of T is a vertex of P.

ON T H E P R O J E C T I O N OF A P L A N E S E T O F F I N I T E L I N E A R M E A S U R E 6 5

Similarly if q is the end point of the t w o segments of T, p q, q r a n d q is n o t a vertex of P , t h e n we can select a point ql on p q near to q a n d replace p q, q r b y p q~, qlr to o b t a i n T 1.

Again T 1 E C(n), A (T1) < A (T), a n d we h a v e a contradiction with 2.

5. The angle between two adjacent segments o/ T is not less than ~ ~.

F o r if tit 2 a n d t~t 3 are two a d j a c e n t segments of T a n d / t l t 2 t 3 < ~ ~ we can replace these segments b y a c o n n e c t e d set containing tl, t2, t 3 a n d of less length. B y 2 this is im- possible.

6. Every node o / T is o/ order 3 and is an interior point o / P . The three segments o/ T which abut at a node o/ T are inclined to one another at an angle o / ~ ~r.

This follows i m m e d i a t e l y f r o m 5.

7. T has either 3 or 4 end-points.

Suppose t h a t T has r end-points a n d t h a t ~ is a positive n u m b e r less t h a n the least length of a segment of T. L e t T 1 be the subtree of T o b t a i n e d from T b y r e m o v i n g r segments each of length ~ a n d such t h a t each of these segments has one end point at a n end point of T a n d each end point of T is a n end point of one of these r segments. T h e n

A (T1) : A (T) - r~,

a n d since every point of T is d i s t a n t at m o s t (~ from some point of T1, /~ (T~) ~>/~(T) - 2 5

( T 1 is n o t void because every node of T is a point of T1, a n d if T has no nodes it is an arc a n d m u s t contain at least t w o segments for otherwise / ~ ( T ) = 0 ) . I f (~ is small t h e c o n v e x cover of T 1 has t h e same n u m b e r of vertices as P . N o w if r 1> 5,

#(T~) ~> # ( T ) - 2 ~ /~(T) (14)

A ( T I ~ A ( T ) - 5 5 > A ( T ~ '

since we k n o w t h a t A ( T ) ~ 2 # ( T ) . B u t (14) is in c o n t r a d i c t i o n with (13). T h u s r = 2, 3 or 4.

I f r = 2, projection in the direction of the line joining t h e end points of T shows t h a t

# ( T ) < 8 9

in contradiction with (*). T h u s r ~= 2, a n d p r o p e r t y 7 is proved.

8. I / T h a s / o u r end points then P is a quadrilateral with these ]our points as vertices.

I f P has more t h a n four vertices t h e n one of them, s a y p, is n o t a n e n d - p o i n t of T.

L e t t h e t w o segments p ql, Pq~ of T meet a t p a n d let p ' be a point on p q~, d i s t a n t ~ from

6 6 H . G. EGGLESTON

p. I n T replace p ql, P q2 by, P'ql, P'q2 a n d remove segments of length: ~ from each end-point of T as in 7. We obtain a connected set T 1 with

A ( T 1 ) < A ( T ) - 4 ( $ /~(T~)>~ # ( T ) - 2 5 , a n d since

/~(T1) :>/~(T) , A (T 1) A (T)

we again have a contradiction with 13. Thus P has at most four vertices. B u t b y 4, P has at least four vertices and these vertices are end-points of T. P r o p e r t y 8 is established.

9. T has exactly three end-points.

Otherwise b y 7, 8 and 2, P is a quadrilateral and T is the connected set of least length joining the vertices of P. I n this case T i s a polygonal tree with two third-order nodes and is formed from five segments. L e t the vertices of P be a, b, c, d (in order round F r P) a n d the nodes of T be klk~ with the notation chosen so t h a t the segments of T are ak.1, b k 2,

c ks, d kl and k2.

The line through a perpendicular to a/c 1 is a support line of P. F o r otherwise / b a k 1 > 89 re (since / d a k 1 <~ ~ re). Suppose / b a k 1 > 1 ~r. L e t a 1 be the foot of the per- pendicular from /c 1 to the, line a b . I n T replace segment a k 1 b y the s e g m e n t a i]c 1 to obtain the tree T 1. The convex cover of T 1 contains P and A (T1) < A (T). Thus we have

/u (T1) > /~ (T)

in contradiction with (13). Thus the line through a perpendicular to a k~ is a support line of P.

Since clc 2 is parallel to ak~ we have a pair of parallel support lines, one each through a and c. Thus, projecting the polygonal line a k , k2c perpendicular to a k , we have

a k ~ §

89 + k2C ~ ( T ) . (15)

Similarly, by,projecting blcjcid perpendicular: t o bk2,

die I -q- 89 ~- k~b >~ /a(T). (16)

Adding, we obtain A (T) >i 2# (T). (I7)

N o w ' s t r i c t inequality in ~(17) is impossible (by (*)). Thus equality m u s t hold in (17) and t h e r e f o r e in each of (15) and (16)~ Hence t h e lines t h r o u g h a and: c perpendicular to a k l a n d those: through d and b perpendicular to dk i a r e all minimal support lines.

O N T H E P R O J E C T I O N O F A P L A N E S E T O F F I N I T E L I N E A R M E A S U R E 67

b

F i g . 2.

B y p r o p e r t y (A) of minimal support, hnes applied to the p a i r of minimal s u p p o r t lines perpendicular to a k 1, one of t h e segments ab, ad, bc, ed is perpendicular t o a k 1. Clearly this is not true o f ad and bc since , L d a k 1 and / b c k ~ are both less t h a n ~ re. Thus e i t h e r ab or cd is perpendicular to a k 1. B u t in an exactly similar w a y we see from the pair of minimal support lines perpendicular to bk~ t h a t either ab or cd is perpendicular to bk 2.

Since a k 1 and b k S are not parallel we conclude t h a t either a b is perpendicula r t o a kl and cd is perpendicular to bk2 or ab is perpendicular to bk~ and cd is perpendicular to c k 2.

The arguments in the two cases a r e the same and we shall consider the first case only.

R e m o v e from a k 1 a segment of length 6 with end point a t a and similarly f r o m dkl a segment of length 5 with end point at d. Denote the resulting tree b y T~. Then

A ( T 1 ) = A(T) " 2 ~ ,

~ ( T ~ ) > ~ ( T ) - ~.

B u t this is impossible since it implies a contradiction with (13).

Thus T has not got four end-points and b y 7 m u s t have exactly three end-points.

REMARK. T has one node and it is of order three. We shall 9 i t b y k and t h e three arcs of T which terminate at k b y a, fl and ?. 9 the vertices of P on a, fl, ? b y al, a 2 .. . . . a~; bl, b 2 .. . . . b i and Cl, c2, ..., c s where a is a 1 .. . . . ah, k and this order is the order i n which t h e s e points l i e on a. S i m i l a r l y for fl and 0-

68 H. G. EGGLESTON 10. Every vertex o/ P is a minimal vertex.

Suppose t h a t the vertex p of P is not minimal. If p is an end-point of T we can remove a small segment one of whose end-points is p from T to obtain a subtree T l, for which

/l(T1)

=/~(T), A(T~) < A ( T ) ,

which leads to a contradiction with (13). Similarly if p is a point common to two segments Pq, Pq2 of T we could move it into a new position p ' on the internal bisector of the angle of these two segments in such a w a y t h a t A (T) is reduced but/~ (T) remains unaltered.

This again leads to a contradiction with (13).

I) E F I N I T I O ~ . Two vertices of P joined b y a single segment lying in the frontier of P (belonging to T or not) are said to be P-adjacent. Two singular points of T joined b y a single segment of T are called T-adjacent.

11. To each pair o/ end-points o / T say al, b I there corresponds a pair o/ parallel minimal support lines

11

and l~ such that l i contains a 1 and l~ contains b r

Suppose t h a t this is not the case. R e m o v e length (~ from the segment of T terminating a t a~ and another equal length from the segment of T terminating a t b~ to obtain the tree T r Let the new end-points be a~ in place of a i a n d b~ in place of

bl,

and let the convex cover of T i be

P r

We shall assume t h a t (~ is a small number. Then b y (13),

a n d b y construction,

/~ (T~) /~ (T) (18)

h (T l) ~< A - ( ~ '

A (T~) = A (T) - 2 6. (19)

Since # ( T ) > ~ 8 9 (18) and (19) imply ( T ) , ,

# (T1) ~< A - - ~ (l~ (T) - 26) ~</~ (T) - (~.

Further, if/~ (T) > 89 A (T), then

~ ( T 1 ) < ~ ( T ) - 6.

Now b y the method of construction of T 1 from T,

fl(T1)

~>#(T) - ($.

For of the two lines which form a pair of minimal support lines of P, one is a support line

ON T H E P R O J E C T I O N O F A P L A N E S E T OF F I N I T E L I N E A R M E A S U R E 69 of P1 and the other is distant at most 6 from a parallel support line of P~. By the inequality for # (T1) proved above it follows t h a t

/~(T1) : / ~ ( T ) - 6.

Thus ,u (T) - 89 A (T) and therefore

/~ (T1) /~ (T) - 6 = /~ (T) A (T1) = A (T) - 2 6 A (T) /~(T1)

Thus T 1 is also an extremal connected set for which A ~ ) I ) assumes its least upper bound.

The results proved about T apply equally well to T l.

:By 10 every vertex of P1 is a minimal vertex and since/~(TI) =/~ (T) - 6 a n y pair of minimal support lines of P1 are obtained from a pair of minimal support lines of P by keeping one line of the pair fixed and moving the other line a distance 6 into a parallel position. There are at most two support lines of P for which the parallel corresponding support line of P1 is distant 6, and these are the two lines perpendicular respectively to a 1 a'l and to

bibS;

further this is so only if these lines contain no points of P apart from a 1 and bl respectively. Now P1 must have at least two pairs of parallel minimal support lines. For otherwise, a small affine contraction orthogonal to the single pair of parallel minimal sup- port ~ines would reduce A (T1) without altering/~ (T1).

Thus the lines through a I and b I perpendicular respectively to

ala~

and

blb~

are minimal support lines of P and contain no points of P apart from a 1 and b 1. Let the line through a 1 perpendicular to ala ~ be m 1 and the parallel support line of P be m~. Let the line through b 1 perpendicular to b 1 b'l be m S and the parallel support line be m~. Since every vertex of P is a vertex of P1 apart from a 1 and b 1 (assuming that 6 is sufficiently small) it follows from 10 t h a t every vertex of P apart from a 1 and b 1 lies on m~ or m~.

Denote the rhombus bounded by

mlm~m2m~

b y R, let its vertices be

A B C D

in order where a I lies on

A B

and b 1 on

BC.

^Let

ala~

produced meet

blb~

produced in s. Let

als

produced meet

DC

in a~ and

bls

produced meet

A D

in b*. Property (B) of minimal support lines implies t h a t a~ is a point of the segment

DC

and b~' a point of the segment

AD.

Then

ala~

and

blb~

lie inside R. Thus

als

and

blS

contain no vertices of T. (Every vertex of T is a vertex of P, see 4.) If a singular point of T lay on

als

apart from a 1 and s, it would have to be a node k. Of the segments of T terminating at k, one has points"

interior to the quadrilateral a I

Bbls.

This segment cannot meet a~ B or

Bb i

since no points of P lie on these segments apart from al and b 1. Nor can it terminate in the interior of this quadrilateral for such a termination would be a singular point of T, therefore a vertex

7 0 H . G. E G G L E S T O N

ai \ \ b,

/ \

/

/ k

/ \

/ \

t

?/$1

a~

Fig. 3.

of P . B u t there are no such vertices of P . T h u s this segment m u s t m e e t segment b 1 s in s a y k l , B u t t h e n segment

bls

contains a singular point of T a n d this singular point m u s t be a node of T. Since T has only one node it follows t h a t it lies at s. I n the n o t a t i o n which we h a v e a d o p t e d s is k. L e t t h e t h i r d segment of T at k = s m e e t t h e frontier of R in d.

B y a similar a r g u m e n t to t h a t used a b o v e

kd

is a s e g m e n t of

T.

N o w t h e three segments

kal, kblkd

divide R into three domains one of which d e n o t e d b y D 1 contains a~ a n d another, d e n o t e d b y D 2 contains b~. B y p r o p e r t y (B) a~ a n d b~ are points of P a n d neither a I n o r b I are vertices of R (since t h e y do n o t lie One each on a pair of parallel minimal s u p p o r t lines of P). Thus b o t h D 1 a n d D~ contain points of T on F r R other t h a n d. Since T is a tree with one node of order three a n d since a 1 a n d b 1 a r e e n d - p o i n t s of T we have a contradiction, If for e x a m p l e t h e third end point of T l a y in D 1 so would the whole of the are of T joining this point to d a n d would t h u s h a v e no points in D 2.

T h u s we are led to a contradiction. T h e original a s s u m p t i o n is false a n d 11 is proved.

ON T H E P R O J E C T I O N O F A P L A N E , S E T O F F I N I T E L I N E A R M E A S U R E 71 R E M A R K. O f a n y two parallel support lines of P at least one passes 9 through a n end point of T. For if two parallel support lines exist neither of which contains an end-point of T we can take two points of the frontier of P one on each of these lines. B u t then these two points divide ,the frontier of P into two arcs one of which must contain two end-points of T. Through these two end-points there is then no pair of parallel support lines. This contradiction with 11 establishes the above statement. Then the three end-points of T , al, b I and cl, divide the frontier of P into three non-overlapping ares which are denoted by A (al, bl), A (b l, cl), and A (el, al) where arc A (al, b 0 does not contain e t etc. Then a n y support line to P at a point of A

(al, bl)

is parallel to a support line of P at c 1 and there are similar relations for A (bl, el) and A (el, al)-

12. The angle between two end-segments o/ T that lie i n t h e / r o n t i e r o / P m u s t be greater than 89 ~.

For if it were less than or equal to 89 Jr then the removal of segments of length b from the two end-segments concerned to produce a new tree T 1 would imply ju (T1) ~> ~u (T) - ( $ and the only directions in which T 1 can have minimal support lines are those orthogonal to the end-segments and (when the angle is equal to 89 zr) t h a t parallel to the bisector of the angle between the end-segments. This last case is impossible by (A) and the argument of 11 can then be used t o establish property 12.

13. Each o/ the three arcs ~, fl, ? has length less t h a n / ~ ( T ) .

If, for example, A (~) ~>#(T) then from 11 there are a pair of parallel support lines to P through the end points of fl U?. Thus

A (flU ?) > ~ ( T ) and

in contradiction with (*).

A (T) > 2~(T)

14, I / a s exists then al, a 2 lies i n t h e / r o n t i e r o / P , i.e. i / a 1 is not T-adjacent to k then al a ~ lies i n t h e / r o n t i e r o/ P .

The points al, a s belong to the frontier of P and thus if a l a 2 does not lie in the frontier of P it divides P into two non-empty domains. Thus there is a vertex of P on each side of the line containing alas. B y 1 there are points of T on each side of the line containing a l a ~.

These points are joined by an arc of T inside P. Since these points tie on opposite sides of a l a s this are meets a l a 2. B u t this is not so since there is no node of T on alas, a contradiction

which establishes 14.

72 H . G . E G G L E S T O N

15. 1/ three vertices o/ o~, say as, as+l, as+s, are such that as, as+ 1 and as+l, as+ ~ are P - adjacent (as well as T adjacent) then asss+ 1 and as+l as+ 2 are tangent to a circle whose radius is # ( T ) and whose centre is b 1 or c I.

The three points ax, bl, c 1 divide the frontier of P into three non-overlapping arcs which we shall denote, as before, by A (al, bi) , A (bl, Ca) and A (c1, al). The three vertices asas+las+2 cannot belong to A (blCl), since if they did, the support line parallel to asas+ 1 would pass through a~ (by the remark after 11) and this implies A (~) ~>/~ (T) in contra- diction with 13. Thus asas+l, as+ 2 belong entirely either to A (al, bl) or to A (Cl, al). Suppose t h a t they belong to A (albs). The argument in the alternative case is similar. The support line of P parallel to the line asas+l passes through c 1. Thus asas+ 1 is either tangent to the circle whose centre is c 1 and radius /~(T), C(Cl,#(T)), or the line containing asas+ 1 lies outside this circle. I n the second case select a point a~+l on as+las+ ~ near to as+ 1 such t h a t asa~+l lies outside the circle c ( c l , / ~ ( T ) ) . I n T replace segments asas+l, as+las+2 by a s a ~ + i ,

i t

as+las+2. If as+l is n o t coincident with as+ 1 the effect is to reduce A (T) without altering /~(T). This is impossible by (13).

Property 15 is proved.

16. I / t h e vertex a 2 exists and i / p is the other vertex o / P which is P-adjacent to al, then either the line through a 1 perpendicular to a l a 2 is a m i n i m a l support line o / P or the line containing a l p is a m i n i m a l support line o / P .

We assume, without a n y real loss of generality t h a t the points a2asp are in the clock- wise sense round the frontier of P. Let the class of minimal support lines through a 1 be denoted by 7. A n y member l of ff together with the line a l a ~ divides the plane into four sectors of which one contains k. The angle of this sector is denoted by r (1).

The set of values •(l) is closed. I f the line containing a l p is not a minimal support line of P and if there is an l of ff with ~b (l) < ~ g, this line 1 meets P in the single point a 1.

For it cannot coincide with a l a 2 since r < 89 g, nor with a l p since by assumption this is not a minimal support line. B y (B) the line through a s perpendicular to l must meet P in a segment of length /~ (T). B u t in fact this line meets P in the single point a 1. Thus if a l p is not a minimal support line then for every l of 7, r 1 8 9

If ~b {l) > 89 g for all l of ~T then there exists a small positive number e such that r (1) > 89 g + e for all 1 of 7. Thus we can rotate a l a 2 about a~ in the anti-clockwise sense so t h a t a I becomes a~. Replace a l a 2 b y a~a 2 to obtain the tree T ' . Now if asp is not a minimal support line and if the rotation is sufficiently small I~(T') = # (T). B u t A ( T ' ) = A (T) so t h a t T ' is an extremal figure. B y 10 a'l is an extremal vertex of T': but by the construction a~ is not an extremal vertex of T ' .

O N T H E P R O J E C T I O N O F A P L A N E S E T O F F I N I T E L I N E A R M E A S U R E 73 This c o n t r a d i c t i o n shows t h a t e i t h e r t h e line c o n t a i n i n g a l p is a m i n i m a l s u p p o r t line or t h e line 1 w i t h r = 8 9 is a m i n i m a l s u p p o r t line of P . T h u s p r o p e r t y 16 is es- t a b l i s h e d .

17. A n y two vertices o / P w h i c h are T - a d j a c e n t are also P - a d j a c e n t , i.e. the p o i n t s a l a 2 . . . a h are i n order r o u n d t h e / r o n t i e r o/ P a n d so are b 1 . . . b~ a n d c a, c 2 . . . cj.

B y 14 a 1 a n d a 2 a r e P - a d j a c e n t . W e shall show first t h a t a 2 a n d a a a r e also P - a d j a c e n t . ( 1 ) T h e v e r t e x a a is P - a d j a c e n t e i t h e r t o a 2 or t o a 1. F o r o t h e r w i s e t h e s e g m e n t a~aa s d i v i d e s P i n t o t w o d o m a i n s e a c h of w h i c h c o n t a i n s v e r t i c e s of P , s a y p, q such t h a t n e i t h e r p n o r q is a n y one of al, a 2 or a s. T h e r e is a n arc in T j o i n i n g p t o q. This arc m u s t c u t aea a w h i c h t h e r e f o r e c o n t a i n s a n o d e of T. B u t t h i s is n o t so.

W e shall a s s u m e t h a t a a is P - a d j a c e n t to a 1 a n d show t h a t t h i s l e a d s to a c o n t r a d i c t i o n . W e a s s u m e for d e f i n i t e n e s s t h a t t h e o r d e r a 2 a l a 3 r o u n d t h e f r o n t i e r of P is clockwise.

Consider t h e m i n i m a l s u p p o r t lines t h a t p a s s t h r o u g h a~. W e shall show t h a t a l a a is n o t a m i n i m a l s u p p o r t line. L e t q be t h e v e r t e x of P t h a t is P - a d j a c e n t a n d n o t T - a d j a c e n t t o a 2. L e t a~ b e a p o i n t on t h e line q a 2 such t h a t a s lies b e t w e e n a~ a n d q, a n d l e t T ' be t h e t r e e o b t a i n e d f r o m T b y r e p l a c i n g s e g m e n t s a l a 2 a n d a3a 2 b y ala~ a n d aaa~. Since t h e c o n v e x cover of T ' i n c l u d e s P i t follows f r o m (13) t h a t A

(T')

> / A

(T).

T h i s in t u r n is t r u e

f '

for a n y choice o a2 as d e s c r i b e d a b o v e o n l y if

/ qa2a3 < A a 2 a 2 a 1.

B u t b y 5 / a l a s a 3 > ~ 7 ~ a n d t h e r e f o r e

~_ q a2 a a <~ ~ 7e.

NOW if a 1 a 3 is a m i n i m a l s u p p o r t line of P t h e r e is a p o i n t of P on t h e p a r a l l e l s u p p o r t line inside t h e s t r i p w h i c h is b o u n d e d b y t h e lines t h r o u g h a 1 a n d a a p e r p e n d i c u l a r t o a 1 a a.

B y 5 a g a i n ~ a s a l a a ~< 89 r thus, if we p r o d u c e asq t o m e e t t h e line t h r o u g h a a p e r p e n d i c u l a r t o alaa, i t will do so in a p o i n t r on t h e s a m e side of ala3 as a 2 (see F i g . 4). T h u s i t follows t h a t t h e lines t h r o u g h p o i n t s of s e g m e n t a l a a p e r p e n d i c u l a r t o a l a a i n t e r s e c t t h e q u a d r i - l a t e r a l a l a s r a a in s e g m e n t s of w h i c h t h e l a r g e s t h a s l e n g t h g r e a t e r t h a n or e q u a l t o / z (T).

T h e l a r g e s t s e g m e n t (or one of t h e m ) is e i t h e r t h e p e r p e n d i c u l a r f r o m a2 t o axa 3 or i t is t h e s e g m e n t aar. I n t h e first case A (a) is g r e a t e r t h a n t h e l e n g t h of t h e s e g m e n t a l a 2 a n d is t h e r e f o r e g r e a t e r t h a n / z (T). T h i s is i m p o s s i b l e b y 13. I n t h e s e c o n d case we c o n s i d e r t r i a n g l e a 2 r a a. W e h a v e / _ a 2 a a r < 89 a n d t h u s ~ . a 3 a 2 r + ~ a a r a s > 89 B u t we h a v e (1) I t is assumed that such vertices as az, a 3 etc. exist. Otherwise there is nothing to prove.

6 - 6 6 5 0 6 4 A e t a m a t h e m a t i e a . 99. I m p r i m ~ le 19 a v r i l 1 9 5 8

74 I t . G. EGGLESTO:N"

Fig. 4.

already seen t h a t fl~qaea a = / r a t a a < ~ ,~. Thus L a 3 r a 2 > ~ z a n d hence / aara 2

> / ' r a 2 a ~. This implies t h a t a a a 2 > r a a > ~ # ( T ) . F i n a l l y we again obtain A ( ~ ) > a a a ~

> # ( T ) . B y 13 this is impossible. Thus a l a 3 is n o t a minimal s u p p o r t line of P.

B u t b y 16 this implies t h a t the line perpendicular to a l a 2 t h r o u g h a I is a minimal

1 above)

s u p p o r t line of P. This line meets P only in the point a 1 (since J a2ala 3 ~ 3 ~, see and thus the line t h r o u g h a l a 2 meets P in a segment of length/~ (T), i.e. a l a 2 is of length at least a (T). B y 14 this is n o t so since it implies t h a t A (~) > u (T). Thus finally aa is n o t P - a d j a c e n t to a 1 and a 3 m u s t therefore be P - a d j a c e n t to a2.

N e x t we suppose t h a t there is a first integer m such t h a t a,~ a n d am+ 1 are n o t P - a d j a c e n t . T h e n m ~> 3 and, b y an a r g u m e n t similar to t h a t used for a a above, it can be seen t h a t a~

a n d a~+ 1 are P - a d j a c e n t . The points a 2 .. . . . a~ all belong to A (a D bl) or to A (al, Cl), Suppose t h a t t h e y b e l o n g to A (al, bl) t h e n by 15 each segment a l a ~, a2a 3, ~...,'a,~_la,~ is p a r t of a t a n g e n t to the circle centre c I a n d r a d i u s / ~ ( T ) a n d b y (B) the segment am_lain a c t u a l l y touches this circle. T h u s ~ a m - 1 amcl < 89 ft. Since b y 5 / a m _ l a ~ a ~ + l >1 ~ ~ a n d since am+ 1 a n d c 1 lie on the same side of am_lain (they are points of P a n d amam_l is p a r t of a s u p p o r t line of P), it. follows t h a t a~_ 1 a n d a~+ 1 lie on opposite sides of t h e line amc 1. H e n c e a 1 a n d am+r lie on opposite sides of the line a ~ c~. B u t a~ a n d cx are b o t h vertices of P . T h u s a 1 a n d am+ 1 are n o t P - a d j a c e n t .

This contradiction establishes ~the required result.

18. 1] the vertex a 2 exists and i] the vertices a 2 a l b l p are i n Order round the /rontier o/

P (i.e. a2 a~, a~ b~, b~ p are P-ad~acent), then the line a l a 2 is not parallel to the line b l p . ~ :

R e m o v e a small segment of length (~ f r o m a l a 2 at a I a n d from t h e end b 1 o f t h e s e g m e n t of T t h a t t e r m i n a t e s at b 1. D e n o t e the new tree b y T ' with end points,a~ in place of a 1 a n d b~ in place of b~. N o w if a~a2 is parallel to b~p t h e n a n y pair of parallel s u p p o r t lines of the convex cover of T ' are such t h a t a t m o s t one goes t h r o u g h a~ or b~l {except when

O N T H E P R O J E C T I O N O F A P L A N E S E T O F . F I N I T E L I N E A R M E A S U R E 75 b 1 and p are T-adjacent, in which case the pair of parallel lines alag. and b l p are support lines of the convex cover of T ' and go through a~ and b~ respectively). B u t in a n y case

A (T') = A (T) - 2 ~ , /~ (T') >/# (T) - 6.

As in 11 it follows t h a t T ' is an extremal figure, t h a t / ~ ( T ' ) =/,~(T) - 6 and t h a t the line through a~ perpendicular to a l a 2 is a minimal support line of P meeting P in the single point a 1. B y P r o p e r t y (B) of minima! support lines the line ala2 meets P in a segment of length # (T). Thus the length of ala2 is/~(T) and

A (a) > ~ ( T ) in contradiction with 13.

Thus the assumption t h a t a l a 2 is parallel to blp is false and 18 is Proved.

We n e x t consider the various cases t h a t might arise according to the different orders of a I . . . . , ah; bl . . . b~ and cl, . . . , cj on the frontier of P, and according as.~, fl, y are formed from one segment o~ more t h a n one segment.

Case I. Each are ~, fi, ~ is made up o / m o r e than one segment and the orders a l , : , , , ah;

b x . . . . b~; c 1 .. . . . cj on/rontier P are all the same.

There is no real loss of generality in supposing t h a t the vertices of P in clockwise order are a 1 .. . . . ah, bl . . . bi, cl . . . cj. The other cases are obtained either b y a change of notation or b y an argument similar to the following.

Produce blah to d (see Fig. 5). Then

/ k a h b l <~ ~ d a h a h - 1

for if this was not the case we could replace ah b y a~ on bla~ such t h a t aa lies between a~

and b r and s u c h t h a t t h e new tree o b t a i n e d from T h y replacing kah a n d ah.~ah b y ka'a.

a n d ah_la'a has less length t h a n T (see the a r g u m e n t i n 17). T h u s we h a v e A kahbl <~ ~ :t

a n d f u r t h e r / _ k ah b ~ ~ :~. S i n c e l_ b, k a a = ~ 7~ i t follows t h a t k b~ < kaa.

Similarly k a h < kcj, kcr < kb,. Thus w e a r e led to the contradiction

and this case c a n n o t occur:.

k a h < k a h

76 ^{H . G .} E G G L E S T O N

\

ah-1

/ / / /

\

\ ah i

/

/

z~ ^/

\ \

\

\

\

\ ,,

\ \

j \ \ \

~ \ \ \\\,,

Fig. 5.

Case I I . T w o arcs ~, fl, 7, say o~ and fl, are e a c h / o r m e d / r o m more than one segment and the orders al, ..., aa; b~ . . . . , b i are such that the sector o/ angle ~ ~ bounded by the hall lines containing lcah, kb ~ respectively and terminating at k, is void o / t h e points al, ..., ah_l bl, 9 hi_ I.

B y a n a r g u m e n t of t h e s a m e t y p e as t h a t u s e d in Case I we h a v e L k a h b i =

= ~ k b ~ a h = ~ a n d fl_kahah i = ~ k b i b i - 1 = ~ . T a k e k' on cjk d i s t a n t ~ f r o m /c a n d a~, on ahah_1, b[ o n b~bi-1 so t h a t k ' a ~ is p a r a l l e l t o kah a n d /c'b[ t o /cbi. I n T

i t i t r

r e p l a c e cjlc, kah, kbi, ahah-1 bib~-i b y c j k , Ic ah, Ic'b[, ahah i, b~bi-1 t o o b t a i n T ' . T h e n

A ( T ' ) = A ( T ) - c$, # ( T ' ) > ~ # ( T ) - 8 9

a n d b y a n a r g u m e n t s i m i l a r t o t h a t in 11 T ' is a n e x t r e m a l figure. B u t t h i s is n o t so since, for e x a m p l e , a I is n o t a m i n i m a l v e r t e x of T ' . T h i s case c a n n o t occur.

O N T H E P R O J E C T I O N O F A P L A N E S E T OF F I N I T E L I N E A R M E A S U R E

.d

77

/

/ \

/ \

f / x

z ~ \ . e

c:~ b~

c1 bl

Fig. 6.

Case I I I . One arc, say ~, is a single segment and the other two arcs each contain more than one segment. The orders o / b 1 . . . b~ and c I . . . cj are such that all these points belong to the sector whose angle is ~ ~ and which is bounded by halprays containing k b i and k c j respec- tively and terminating at k.

I n t h i s case t h e v e r t i c e s b 1 a n d c 1 a r e P - a d j a c e n t .

W e r e m a r k first t h a t t h e line bib 2 m e e t s t h e line clc~ a t a p o i n t d w h i c h lies on t h e s a m e side of blc 1 as k a n d t h a t , of t h e f o u r sectors i n t o w h i c h t h e lines bib 2 a n d ctc ~ d i v i d e t h e p l a n e , t h e s e c t o r c o n t a i n i n g k has a n a n g l e g r e a t e r t h a n or e q u a l t o ~ g (from 12).

B y t h e a r g u m e n t in 15 t h e p e r p e n d i c u l a r d i s t a n c e f r o m c I t o bib 2 is e q u a l t o # ( T ) a n d so is t h a t f r o m b 1 t o clc ~. T h u s in t r i a n g l e d b l c 1, / b t d c I > ~ ~ a n d A d b l c I = / d c l b ~ < ~ z~.

B u t t h i s i m p l i e s t h a t t h e d i s t a n c e f r o m d t o blc I is less t h a n # (T). Since t h i s is i m p o s s i b l e t h i s case c a n n o t occur.

Case I V. One arc, s a y ~, is a single segment and the other two arcs are not single segments.

The vertices b I . . . b~, c 1 .. . . . cj are i n order on t h e / r o n t i e r o / P .

E i t h e r t h e t h r e e p a i r s al, cj; Cl, b~; bi, a I a r e P - a d j a c e n t or t h e t h r e e p a i r s al, cl; cj, bl;

b~, a I a r e P - a d j a c e n t . W e s u p p o s e t h a t t h e first a l t e r n a t i v e holds: t h e a r g u m e n t w h e n t h e s e c o n d a l t e r n a t i v e h o l d s is t h e s a m e w i t h b's a n d c's i n t e r c h a n g e d .