Approximation of convex discs by polygons
Discrete Comput Geom
Approximation of Convex Discs by Polygons*
A. Florian 0 1
0 8A(x, Y) = a , X A Y
1 lnstitut fiir Mathematik, Universit/it Salzburg , Petersbrunnstrasse 19, A5020 Salzburg , Austria
We consider the class of all convex discs with areas and perimeters bounded by given constants. Which disc of this class has the least possible area deviation from a kgon? This and related questions are the subject of the present paper.
1. Introduction
By a convex disc we mean a convex compact subset of the Euclidean plane with
interior points. In this paper we shall deal with the approximation o f convex
discs by convex polygons. There are several methods of measuring the deviation
between two convex discs. The following are two of the most usual methods. We
write a ( M ) and p ( M ) for the area, i.e., the Lebesgue measure, and the perimeter
of the set M. If X and Y are convex discs, the area deviation between X and Y
is defined by
and the perimeter deviation by
Equation (
1
) may also be written in the form
&~(X, Y ) = a ( X u Y )  a ( X c~ Y),
Be(X, Y ) = p ( X u
Y )  p ( X n
Y).
* Dedicated to Professor E. Hlawka on the occasion of his birthday.
(
1
)
(
2
)
X A Y = ( X \ Y) w ( Y ~ X )
is the symmetric difference of the two sets. Note that 8 A makes the class of all
convex discs into a metric space, but 8 P does not. Various concepts of deviation
o f two convex bodies are discussed in [17]. Gruber [14] gives an uptodate
review o f the results concerning the approximation of a convex body by polytopes.
The approximation of convex discs by convex polygons is of interest by itself.
Moreover, it is important in its application to problems of packing and covering
(see [3, 5, 6, 811, 13]).
Throughout this paper let a and p be positive numbers satisfying the
isoperimetric inequality
Let qg(a, p) be the class ofaU convex discs with area not less than a and perimeter
not greater than p. A convex polygon with at most k sides is simply called a
kgon. Let ~k denote the class of all kgons. Two measures for the closeness of
the approximation of kgons to discs from ~ ( a , p) are given by
Ae(a,p, k) = inf 8P(C, P),
p2 "~ 77"
"~ 4k tan ~,
where
and
where the infimum is taken over all C E ~(a, p) and all P E ~k. Both functions
are interesting only in the case when
which means that p is less than the perimeter of a regular kgon of area a.
Otherwise we have Cg(a,p) r~ ~k # 0 , SO that AA(a, p, k) = Ae(a, p, k) = O.
By combining some results from [2], [5] we shall obtain AP(a, p, k) in Section
2. Using ideas of Besicovitch [1], Eggleston [2], and Fejes T6th and Florian [5]
we will find the supremum of a(C n P) taken over all C E ~(a, p) and all kgons
P of a given area; this is the subject of Section 3 and the main part of this paper.
In Section 4 we will apply the results of Section 3 to determine those members
of qg(a, p) and ~k for which 8A(C, P) is minimal.
A. FIorian
(
3
)
(
4
)
(
5
)
According to a remarkable result by Eggleston [2], the perimeter deviation of a
given convex disc from an arbitrary convex kgon is minimal for a kgon inscribed
in the disc. Thus we have
A e ( a , p , k ) = i n f ~ e ( C , P )
( C ~ ( a , p ) , P ~ k ,
P C C ) .
(
6
)
Before we state the result, we need to describe a certain geometrical
configuration. Let P* be a regular kgon. We join each two consecutive vertices of P* by
congruent circular arcs o f radius not less than the circumradius of P*. These
arcs form the boundary of a convex disc C* which we call a regular arcsided
kgon with kernel P* (Fig. 1). If (
5
) is satisfied it can be shown (see [5]) that
(i) there is exactly one regular arcsided kgon C* with area a and perimeter
p, and
(ii) the infimum on the righthand side of (
6
) is attained only for C* and its
kernel P*.
Hence
Ae(a, p, k) = 8 P ( c *, P*).
A simple expression for ~P(C*, P*) can be obtained in the following way
(see [5]). Let 2 a be the central angle o f the circular arcs bounding C*, where
0 < a  7r/k. We define a function O(q) by the parametric equations
2. T h e P e r i m e t e r D e v i a t i o n
a 2
~ ( q ) = sin 2 a '
q
a  s i n o t c o s a
sin 2 a
[ 0 < a < ~ r \
for 0 < q  t ] , where # corresponds to a = ~'/k, and put O ( 0 ) = 1. Elementary
calculation yields the equation
p2
  = 4k sin
a
~r
qb(q)
k cos(~'/k)+q sin(~r/k)'
(
7
)
(
8
)
(
9
)
which has a single root q e (0, 4]. Since p(P*) = (p sin a)/a, we finally conclude
that
8P(C*, P*) = p(1  (~(q))~/2).
(
10
)
(
12
)
(
13
)
(
14
)
3. A Maximum Problem
Let ~k(ao) denote the class of all kgons o f given area ao. In this section we
shall deal with the following
Problem. Find a member of ~ ( a , p) and a member of ~ak(a0) such that their
intersection has the greatest possible area.
Accordingly we introduce the function
M(a, p; k, ao) = m a x a( C c~P),
(
11
)
where the maximum is to be taken over all discs C from ~ ( a , p) and all kgons
P from ~k(aO).The existence o f the maximum follows from the Blaschke selection
theorem.
In the particular case when p2/a = 4~, the class CO(a,p) consists only of the
circle of area a. Fejes T6th [7] showed that the intersection o f a given circle C
and a kgon P of given area has maximal area if P is regular and concentric
with C; for alternative proofs of this " m o m e n t u m lemma" see [4], [12], [15].
In certain cases the solution to our problem can be deduced from two previous
results which we now recollect.
Let the function fi(a, p, k) be defined by
I
p2 1
A ( a , p , k ) [4p£2~t~(q)
where t = t(k) = (k/1r) tan(~r/k), and ~ ( q ) is given by (
8
). Let C be a disc from
qg(a, p) and P a kgon with P c C. It was proved in [5] that
a(P)<f~(a,p, k),
with equality if and only if either C is a regular arcsided kgon of area a and
perimeter p, and P is the kernel o f C (p2/a <4~rt), or C = P is a regular kgon
o f perimeter p (p2/a >=4~rt). By (
13
) we have
M(a,p;k, ao)=ao
if ao<fl(a,p,k),
and
M(a,p;k, ao)<ao
if fl(a,p,k)<ao.
Hence in the following we can assume that ao>f~(a, p, k).
Let x be a convex disc with in radius r. X_p (0 < p < r) denotes the inner
parallel domain and Xp (0 < p) the outer parallel domain of X at distance p. If
/5 is a regular kgon, we call a set of the form (/5_p)p a smooth regular kgon with
case /5 (Fig. 2). It will be convenient to consider both /5 and its incircle as
degenerate smooth regular kgons with case /5. The corresponding values of p
are 0 and r.
We define the function F(p, k, ao) by
ao
p2
47r
F ( p , k, ao) =
p a#ff~o~tp2/4  aolr
( t  1 ) ~ "
p2
if ao < 41rt'
2 <p2 t
if 4 ~ t  a o  4~'
if P2t<
4~r
ao.
Let C be a convex disc of perimeter not greater than p and P a kgon from
~k(ao) with C c P. Fejes T6th ([6] or [8, p. 175]) proved that
Let p2/4~rt <ao<p2t/47r. Then equality holds in (
16
) if and only if C is a (possibly
degenerate) smooth regular kgon with perimeter p, and P is the case of C.
Let C be a disc from ~ ( a , p) and P a kgon with C c P. Since F is a strictly
increasing function of ao for 0 < ao<p2t/47r, it follows from (
15
) and (
16
) that
a( C) <F(p, k, ao).
a(P)>f~(a,p, k),
where
f2(a, p, k) =
l r ( p  x / ( p 2 ' 4 a T r ) ( 1 ' i : i ) ) 2 if P<47rt,a
2
if P__ 4~rt.
a
(
15
)
(
16
)
(
17
)
(18)
and
If p2/a <4~rt and a ( P ) = f 2 ( a , p, k), then C is a (possibly degenerate) smooth
regular kgon of area a and perimeter p, and P is the case of C. Thus, by (18)
we have
fE(a, p, k) >a
M ( a , p ; k , ao)>a
if ao>f2(a,p,k).
Returning to our problem, we distinguish the cases p2/a <~4rrt and p2/a >47rt,
and begin with the simpler
Case (a). p2/a >4,n't
From (
12
) and (18) we see that
Let ao>fl(a, p, k), and let C and P be members of ~(a, p) and ~k(ao) such that
f2(a, p, k) <fl(a, p, k).
a( C c~P) = M ( a , p; k, ao).
a( C c~P) >>a.
(19)
(20)
(21)
(22)
Inequalities (
13
) and (20) imply that P C C and
We shall see in the proof o f Theorem 1 (Lemma 4) that, whenever (21) together
with the assumptions P C C and C ¢ P are satisfied, then a ( C ) = a , whence
a ( C c~P ) < a. By (22) this last inequality is impossible. Thus we conclude that
C c p, and from (
16
) and (21) it follows that
a( C c~P) = F(p, k, ao).
(23)
Equations (
14
) and (23) can be summarized in
Remark 1. If p 2 / a > 4~rt we have
M ( a , p; k, ao) = F(p, k, ao).
From the suppositions (21) and p2/4~rt<<ao<<p2t/4~r it follows that C is a
(possibly degenerate) smooth regular kgon of perimeter p, and P is the case of C.
Case (b). p2/ a < 41rt
Because of (
12
) and (19) we obtain the inequality
f~(a, p, k) <f2(a, p, k),
(24)
which is contrary to case (a).
Let ao>fE(a, p, k), and let C e CO(a,p) and P ~ ~k(ao) satisfy (21). By repeating
the argument used in case (a) we again come to the conclusion that C is contained
in P and that (23) holds in case (b) as well.
Remark 2. I f p 2 / a < 4 ~ ' t we have
ao
M ( a , p ; k , ao)=
F ( p , k , ao)
if ao <fl( a, p, k),
if f 2 ( a , p , k ) < a o.
From (21) and ao =fl(a, p, k) it follows that C is a regular arcsided kgon of
area a and perimeter/7, and P is the kernel o f C. From (21) and f2(a, p, k) < ao
<p2t/47r it follows that C is a (possibility degenerate) smooth regular kgon o f
perimeter p, and P is the case of C.
We now proceed to find the m a x i m u m o f a ( C n P ) with C ~ ( a , p )
PE ~k(aO) in the more difficult case when
and
f~(a,p, k)<ao<f2(a,p, k).
To describe the extremal configuration we consider the outer parallel domain C
of a regular arcsided kgon at some distance p. We first assume that C is not a
circle. Then C is b o u n d e d by k equal circular arcs of radius F and k equal circular
arcs of radius r, where ~ < r. The lines joning the endpoints of every arc of radius
r enclose a regular kgon P which we call the central kgon of C (Fig. 3). By a
central kgon of a circle C we mean any regular kgon concentric with C.
Theorem 1. Let
(i)
fl(a, p, k) < ao <f2(a, p, k)
and let C and P be such members o f ~ ( a , p) and ~k(ao) that
a ( C n P) = M ( a , p; k, ao).
Then C is an outer parallel domain o f a regular arcsided kgon, and P is
the central kgon o f C. Furthermore, C has area a and perimeter p.
We shall, in fact, prove Theorem 1 by making the weaker assumptions (ii) and
(iii)
C ¢ P,
P C C
instead of (i) and (ii). Theorem 1 together with Remarks 1 and 2 solve the problem
set at the beginning o f this section. Observe that a regular arcsided kgon and
its kernel as well as a smooth regular kgon and its case may be regarded as a
degenerate parallel domain of a regular arcsided kgon and its central kgon.
Proof of Theorem 1. Let C and P satisfy suppositions (ii) and (iii). We will
develop the properties o f C and P in the following 12 lemmas. The last lemma
shows that C and P correspond with the statement o f our theorem.
First, we remark that by (ii) and (iii) there is a vertex of P outside C, and
there is a side of P intersecting the interior o f C
Lemma 1. P has exactly k vertices.
Proof. Suppose that P has fewer than k vertices. Let A~ be a vertex o f P outside
C, and let A,A~+t be a side o f P containing interior points of C. We cut off from
P a sufficiently small triangle with vertex At and displace A~A~+I toward the
exterior o f P such that the new kgon P ' obtained from P by this process has
area ao. Then we have
a ( C n P ' ) > a ( C n P)
in contradiction to assumption (ii).
We denote the vertices o f P in the anticlockwise sense by A~, A2. . . . , Ak and
set Ak+ 1= A1, A o = Ak.
Lemma 2.
N o vertex o f P lies in the interior o f C
Proof. Suppose that A~ is outside C and A2 is an interior point of C. Let the
side A~A2 rotate in the clockwise sense about a point between AI and A: such
that the new kgon P' = A~A~A3 • •. Ak has area ao. I f the angle of rotation is
small, A~ is exterior to C and A[ is an interior point o f C. This again leads to
inequality (25).
(25)
[]
Approximationof ConvoxDiscs by Polygons 249 Suppose that A~ lies on the boundary and A2 in the interior of C. Let the side
A1A2 rotate about A~ in the clockwise sense through a small angle into the new
position A~A~ with A~ in the interior o f C. If A~ is one o f the vertices outside C
we displace A~A~+~toward the interior of P such that the kgon P' finally obtained
has area a0. Since P ' satisfies (25), the supposition was wrong, and Lemma 2 is
proved. []
We shall use the symbol AB to denote both the segment A B and its length.
Lemma 3. (a) A side A1A 2 of P that contains interior points of C meets the
boundary of C at points Ull, U12, where Ul2 is between Ull and A2, such
that
(b) Any side A~Ai+~ of P intersects C at the points o f a segment U~ Ui2 such that
Proof. (a) Suppose that, contrary to (26),
This implies that A~ is outside C and
(26)
(27)
(28)
(29)
(31)
A~ UH = UlzA2.
UilUi2~ UIIUI2
A f A H AIA2
A1 Ull > Ut2A2.
MU1~ < MU~2,
From (29), (30) and (31) we see that P' satisfies (25) if 9 is sufficiently small.
Because (25) is impossible, part (a) of Lemma 3 is proved.
(b) It suffices to show (27) for 2_<i_< k  1 . We displace the side AIA2 of P
outward parallel to itself through a small distance 7/t > 0. Let P ' = A i A ~ . . . A'k
be the new kgon, where A~ = Aj for j = 3. . . . . k; F r o m P' we obtain the further
kgon P" by displacing the side A~A~+~inward parallel to itself through a small
distance ~ 2 < 0 , such that a(P")=ao. Write P " = A ~ . . . A ' ~ with A'f=A~ for
j # i, i + 1. We shall use the notations
C c~ P = S,
where M is the midpoint o f A~A2. Let AlA2 rotate in the clockwise sense through
a small angle ~ into the new position A~A;, such that P ' = AiA~A3 • • • A k has
area ao. The segments A~A2 and AiA~ intersect at, say, M'. Denoting the triangle
with vertices X, Y, Z by X Y Z , we have
a ( C c ~ P ' )  a ( C c ~ P ) = a ( C n M ' A 2 A ~ )  a ( C n M ' A I A i ) .
(30)
If M ' is outside or on the boundary o f (7, then a(Cc~M'A2A~)>O and
a ( C n M ' A ~ A i ) = 0 , so that P ' satisfies (25). Thus we can assume that M ' is an
inner point o f C. Since M ' approaches M as ~ tends to zero, M belongs to C
and we obtain
lim a( C c~M'AIA~) MU21
~,o a( C c~M'A2A~) = MU22"
and note that S c S' and S " c S'. From the definitions o f P ' and S' we obtain the
relations
lim
rtl,O
a ( P ' )  a ( P )
"t]l
= A1A2,
lim
nl.O
a ( S ' )  a ( S )
T/I
 U n U12.
(33)
Because A~A~+~ > A~A~+1, we have
a ( P )  a ( P ' ) = a ( P " )  a ( P ' )  ~l 't~a''~ai'+ 1 + A i"A~"+l)772
< ½AiAi+l rl2 < O.
Together with limn,_,o a ( P ' ) = a ( P ) this implies that
(34)
D e c o m p o s i n g P ' \ P " into two disjoint sets according to
a ( S')  a( S") = a( S ' \ S") = a( C ra ( P'k P")).
P'\e" = [(e'\P")
e l u [ ( e ' \ P") ( e ' \ e ) ] ,
we distinguish two cases. In both we assume, as we clearly may, that ~t is
sufficiently small.
( a ) If i = 2 then ( P ' \ P " ) n ( P ' \ P ) is a parallelogram with sides A2A'2 and
A[A~. Hence
(41)
(42)
(/3) If i > 2 then
a[(P'\P") ~ ( P ' \ P ) ] = O(n,n2).
(P'\P") n ( P ' \ P ) = 0 .
Let I be the line joining A~ and A~+~, and let L(~72)be the"parallel strip bounded
by ! and the line through A~' and A,"÷~.Let s(x) denote the length of the intersection
of C n P with the line parallel to l at distance x > 0 from /. Then we have
a[ C c~P n (P'\P") ] = a[ C n P c~ L(n2)] =  n 2 s ( x )
(43)
for some x between 0 and ~72. Because lim s(x) = U~IUi2 as x tends to zero, the
required equation (38) follows from (35) and (39) to (43).
Supposing the contrary to (27), we obtain by (33), (37) and (38) that
a( S')  a( S")
lim
~,~o a ( S ' )  a ( S )
Ui1U~2/A, Ai+I
UIIUt2/AtA2 < 1 ,
which implies by (32) that
a( C c~ P")> a( C n P).
(44)
The contradiction o f (44) to assumption (ii) of Theorem 1 completes the p r o o f
of part (b). []
Corollary I. Any side of P intersects C at the points of a segment of positive length.
A side of P which intersects the interior of C has its endpoints outside C.
Following the notation used in Lemma 3 we shall constantly denote the points
at which the side A~Ai+t meets the boundary of C by Ui~ and U~2. The points
Ul~, U t 2 , . . . , Uk~, Uk2 are round the boundary of C in the anticlockwise sense.
In the proofs o f the above lemmas C is a given convex disc and P runs through
the class ~k(ao). In the proofs of the following seven lemmas P is a given kgon
while C varies on Ca(a, p).
Lemma 4. p ( C ) = p and a ( C ) = a.
Proof If p ( C ) < p we choose p > 0 such that p(Cp)<p. Thus Cp belongs to
qg(a, p). Since the boundary of C intersects the interior of P we have
in contradiction to the maximum property o f C.
Suppose that a ( C ) > a. Since there are interior points of C outside P, we can
find a proper convex subset of C, say C', such that C' is a member of C~(a,p) and
a ( C p n P ) > a(C n P )
a(C' n P) = a(C n P).
Because p ( C ' ) < p this is impossible.
Lemma 5. Suppose that the side AIA2 contains interior points of C. Let sl be the
arc on the boundary of C between U~I and U12, which is outside P. s~ is a circular
arc.
Proof. Let c be the circular arc on the same side of the line AIA2 as s~ that has
endpoints U1~ and U~2 and the same length as s~. Let us assume that s~ ~ c.
Denoting the convex hull of the set M by conv M and the (not necessarily
convex) disc ( C \ c o n v s l ) u cony c by D we have
p(D) = p(C),
a(O n P) = a(C n P).
(45)
We now refer to the wellknown fact that the area of a convex disc which is
bounded by a given straight segment and an arc of given length attains its
maximum if and only if the disc is a circular segment. Thus
a(conv sl) < a(conv c),
a(D) > a(C).
a(C' n P ) > a ( C n P ) .
[]
(46)
(47)
whence
and
From (45) and (47) it follows that C' = c o n v D belongs to ~(a, p) and
By (47), however, we have a ( C ' ) > a, which contradicts Lemma 4. Thus the
assumption sl ~ c was wrong and Lemma 5 is proved. []
Lemma 6. C is strictly convex.
Proof. Suppose that the straight segment V~V2 is part of the boundary of C. Let
S be a circular segment o f C\P, the chord o f which has length less than V~V2.
We cut S off from C and join it to V~V2, obtaining a nonconvex disc D with
a(D) = a(C), p(D) = p ( C ) and a a(D n P) > a(C n P). Then C'= conv D has
the properties
p(C')<p,
a ( C ' ) > a
a ( C ' n P) >a(C n P)
in contradiction to Lemma 4. Thus Lemma 6 is proved.
Approximationof ConvexDiscsby Polygons From Lemma 3, Corollary 1 and Lemma 6 we infer
Corollary 2. (i) Every vertex o f P is exterior to C.
(ii) Every side o f P intersects the interior o f C.
(iii)
(iv)
A,U/I = U~2A~+I
for i= 1 , . . . , k.
U,tU,~
. A. IA. 2 . .
Uk,Uk~
AkAI
Lemma 8.
radius, say r.
Lemma 7. Let s~ be the circular arc on the boundary o f C between Ui~ and U~2,
which is outside P. s~ is less than a semicircle, for i = 1 , . . . , k.
Proof. Suppose that s~ is greater than or equal to a semicircle. Let tl and t2 be
the tangent lines to sl at Utt and U~2,respectively. By (48) and (49) the segments
Ut2U2~ and UtIU22 are parallel. Since C is contained in the set b o u n d e d by st,
tt and t2, s~ is at most a semicircle. In this case U2t lies on t2, and U22 on tl.
Thus A~ and A3 are separated from C by tt, showing that the sides A 3 A 4 , . . . , AkAt
do not intersect C. But by Corollary 2(ii) this is impossible. []
The circular arcs sl . . . . , Sk considered in Lemma 7 have the same
Proof Suppose that st and s2 have different radii. Let ct and c2 be two chords
of the arcs st and s2, other than Utt U12 and U21U22, and having equal lengths.
Let s~ and s~ be the respective subarcs of st and s2. From C we obtain a new
disc, say D, by exchanging the positions of the circular segments conv s~ and
conv s~. This means that we cut them off from C and join them to c2 and Cl,
respectively. Obviously,
The contradiction to Lemma 4 proves the lemma.
Lemma 9. Let s~ be the arc on the boundary o f C between U/t.2 and Ull, which
is contained in P, for i = 1. . . . , k.
(i) ~i is a circular arc;
(ii) the arcs ~ , . . . , sk have the same radius, say F;
(iii) i f r is the radius o f the arcs considered in Lemma 8, then
F< r.
(48)
(49)
(51)
A. Flofian
Thus C ' = c o n v D s a t i s ~ s (51) which is impossible. Hence each o f the subarcs
is circular, and so is V~V2.
(ii) Suppose that ~ and s2 have different radii. We obtain a nonconvex disc
D satisfying (50) by exchanging the positions of two small circular segments,
the arcs o f which are part of g~ and s~, respectively. Since the p r o o f is quite
similar to that o f Lemma 5 we omit the details.
(iii) Suppose that P> r. Similarly, as in the p r o o f o f Lemma 5, we exchange
the positions of two small circular segments, the arcs of which are part of ~t and
s~, respectively. We obtain a nonconvex disc D with p ( D ) = p ( C ) , a ( D ) = a ( C ) ,
and a ( D n P) > a(C r~P). Hence C ' = cony D satisfies (51), and the p r o o f o f
Lemma 9 is complete. []
We next prove that C has a smooth boundary.
L e m m a 10.
Through every boundary point of C there passes exactly one support line.
Proof Let t and t' be the tangent lines to the arcs sl and s2 at U~2. To prove
Lemma 10 we have to show that t = ~'. Suppose that t # t~. Let X Y be a chord o f
C parallel to i', where X lies on s2 and Y on st (Fig. 4). We denote the convex
?
r

S4
A2o
Al
hull of X U t 2 Y by S1, the angle between X T and the arc XU~2 by AX, and
the angle between YX and the arc Y'012 by ~.Y. Let ' ~ ¢ be a subarc of 82
contained in the interior of P and such that VW = XY. Write $2 for the convex
hull of VW, and ~_V for the angle of $2 at the vertex V. By exchanging the
positions of $1 and $2 we obtain the sets Tt and T2, which are congruent to S~
and $2, respectively.
If the segment X Y has a sufficiently small distance from i', the following
conditions are satisfied:
(i) ~ X < z ~ I~
(ii) S i n S 2 = T~n T2=~;
(iii) T~c p.
Since t ~ t', we have N V < L X , and by (i) ~ . V < ~ Y . Hence
T2c Sl.
(53)
By cutting S~ and $2 off from C and replacing them by T2 and T~ we obtain a
nonconvex disc D with
By using (ii), (iii), and (53) we find
a( D n P) = a( C c~ P)  a( Si c~ P)  a( S2 n P) + a( T i n P) + a ( T2 n P)
= a ( C n P) + a ( S l \ T2)  a ( ( S l \ T:) n P)
>  a ( C n P ) ,
which shows that C' = c o n v D satisfies (51). Thus the supposition t ~ ~'was wrong,
and Lemma 10 is proved. []
In the case when : = r all the arcs st, st (i = 1 , . . . , k) have the same radius.
Lemma 10 shows that C is a circle. By (48) P is inscribed in a circle concentric
with C, and from (49) it follows that P is regular, as stated in Theorem 1. By
(52) we can from now on suppose that
: < r.
(54)
We shall denote the center of the circle to which the arc s~ belongs by Mr, for
i = l . . . . , k .
Lemma 11. Po = M ~ . . . Mk is a convex kgon inscribed in a circle that has its
center 0 in the interior o f Po. Let Co be the convex disc obtained f r o m Po by joining
each two consecutive vertices by circular arcs o f radius r  ~ . Then C is the outer
parallel domain o f Co at distance ~. P and Po are homothetic with respect to O.
Proof. Let O 1 be the center of the circle to which s~ = Ull U12 belongs. By (48),
O1 lies on the perpendicular bisector b o f the segment AIA2. In view of Lemma
7, O1 and the kgon P are on the same side of the line AIA2. From Lemma 10
and (54) it follows that M1 is between O1 and UH, and M2 between O1 and U12.
Because MI Ull = M2U12 = ~, we see that M1M2 is parallel to A1A2 and
Writing Ull U12/A1A2 = q, we have by (49)
and by (55)
MtM:

U11U12
 1
r
.
Uil U~2
A~Ai+I q
for i = 1 , . . . , k
MIME= 1
A1A2
Since MI and M2 are symmetric with respect to b, the lines ArM1 and A2M2
intersect at a point, say O, on b. Hence OAI = OA2. Because M1M2 < AIA2, 0
and P are on the same side of the line AIA2, and the line MIM2 separates O
and AIA2. Equation (57) implies that
In the same way it can be shown that the lines A2M2 and A3M3 intersect at a
point, say O', on the same side of the line AeA 3 as P such that O'A3 = O'A2 and
OM~ = OM2 = ( 1  r ~ ) q • OA2.
O'M3= O ' M 2 = ( 1  r ~ ) q • O'A2.
By (58) Po is obtained from P by homothety o f center O and ratio (1  ~/r)q. In
view o f the construction, O is an inner point of Po. Since O M I < 01M1, the
circular arcs of radius r  ~ joining each two consecutive vertices of Po form the
boundary o f a convex disc Co. C is the outer parallel domain o f Co at distance
as required. []
From (58) and (59) we infer that O ' = O, OMI = OM2 = OM3, and OAI = OA2 =
OA3. By applying this argument to A3A4,... we conclude that
and
OMI . . . . .
OMk
The following lemma completes the p r o o f of Theorem 1. We shall use the
same notation as in Lemma 11.
Lemma 12. P is regular.
Proof. It suffices to show that
AIA2 = A2A3.
The segments OA~ and OA3 meet the boundary of C at two points, say A~
and A~, belonging to ~ and ~3, respectively. Observe that by (60) OA~ = OA'3,
and by (61) OA! = OA3. Thus AIA3 and A~A~ have the same perpendicular
bisector t passing through (9. Write X ' for the convex set obtained from the
convex set X by Steiner symmetrization about the line t. If T denotes the triangle
A~A3A2, then P' = ( P \ T ) u T' is a kgon which is, by (61), convex. Since a ( T ' ) =
a ( T ) , P' is a member o f ~k(aO).
The chord A~A'~ dissects C into two convex subsets. Let C~ be that subset
which contains the arc s2. Since a(C~) = a(CO and p( C~) <p( CO, the (possibly
nonconvex) set D = ( C \ C I ) u Ctl has the properties
We proceed to show that
(62)
(63)
(64)
(65)
(66)
Let I be any line perpendicular to t. We have to consider three possible eases:
(i) l meets the interiors of C1 and T. Denoting the length of the segment s
by Isl we have
[ I n C n P [ = [ / n C1n T [  min{]ln Cd, ]In TI}
= l l n C~n T'] = l l n D n
P'I(ii) l meets neither the interior o f C1 nor that o f T. Then
l l n C n P l = l l n D n e'l.
(iii) l meets either the interior o f C1 or that o f T. Let t~ and t3 be the tangents
to sA~ and s3 at A1, and A~ respectively, h, t3, A~A3 and A1' A3' enclose a
(possibly degenerate) trapezium S that is symmetric with respect to t
A. Fiorian
Is A2
t!
a 3 ~
(67)
(68)
(69)
(see Figs. 5 and 6). Because OA~As is contained in P, so is S. Since tl
and t3 are support lines of C, S contains the intersection of C with the
parallel strip bounded by the lines AIA3 and A~A~. Thus, if I meets the
interior of C~ we have
It n C n PI = tl n Cl = I1n C d = II :', C~I = II n D n P'I,
and if I meets the interior of T
Now (64) follows from (65) to (68).
tl m C n P l = l l n C l = l l m C
n T ' I = l l n D n P ' I .
By (63) and (64), C' = c o n v D is a member of C~(a,p) satisfying
a ( C ' n P') > a ( C c~ P).
In view o f supposition (ii) of Theorem 1 equality holds in (69). Using Lemma 4
we obtain from (63) that
p = p ( C ' ) <p ( O ) <p ( C ) = p,
ts
As
A2
tI
Since C1 is not contained in a line perpendicular to t we conclude from (70)
(see [16, p. 208]) that C1 is symmetric with respect to a line t' parallel to t. Since
A~A'3 is symmetric with respect to t we see that t ' = t. By (49) the lines U~2U2~
and UH U22 are perpendicular to t. Hence U~2 and U2~ as well as Uu and U22
are pairs of symmetric points, so that
Equation (62) follows from (71) and (49). This completes the p r o o f o f Lemma
12 and that o f Theorem 1. []
and p ( D ) = p( C) implies that
(70)
(71)
(72)
(73)
(74)
(75)
(76)
q and ~ ( q ) are given by (
8
), and u = c o t ( I t / k ) . The discs C form an array joining
the smooth regular kgon, corresponding to a = 0 , with the regular arcsided
kgon, corresponding to a = a*, where q = q(a*) is determined by (
8
) and (
9
).
By applying Steiner's formulas to (Cl)p we obtain from (72) that
p(a)=~(p~/(p24aTr)qb
¥ ~S ~    ~  ~ ' k ] .
Differentiation yields
p ' ( a ) =
x/p2 4a~r(v  u)
2k[d~_Tr(u+q)/k]3/2
sin a  a cos a
sin2 a
,
where v = cot a. Since a < ~r/k, we can remark for later use that
By using Steiner's formula and (73) we find the inradius o f the kernel o f C1
p ' ( a ) < 0 .
r , ( ) =
u
and observe that
For the inradius o f P
we obtain by (74) and (77)
r ~ ( a ) =
ux/p24a~r(a  at~k) sin a  a cos a
2k[dpTr(u+q)/k] 3/2 sin 3 a
> 0 .
re(a) = r~+ p cos a
2k sin ~
k
sin 3 a(r'v+p sin a )
= _~/p2_ 4alr(sin a  a cos a ) cos a cos
 a
x [ t a n ( k  a )  ( k  a ) + ( ~  a ) t a n ( ~  a ) t a n a
].
Hence
Thus a(P) is a strictly decreasing function of a, and a is uniquely determined
by a(P) = ao. This proves the above statement.
4. The Area Deviation
We now turn to the problem of finding such members o f ~(a, p) and ~k for
which 6A(C, P) is minimal. In view of a remark m a d e in Section 1, we have to
consider only such values o f a, p, and k that
2
P<4ka tan k"
For a disc C f r o m ~g(a,p) and a kgon P from ~k(ao) we have b y (
1
)
6 A ( c , P) = a( C) + a o  2a( C c~P).
Because a ( C ) > a, it follows from (80) and (
11
) that
I f ao<f~(a, p, k), R e m a r k 2 implies that
/~A(C, P ) >  a + a o  2 M ( a , p ; k, ao).
8 a ( C , P ) >  a  f l ( a , p , k),
(77)
(78)
(79)
(80)
(81)
with equality if and only if C is a regular arcsided kgon of area a and perimeter
p, and P is the kernel of C. Let P' be the kgon obtained from P by displacing
a side of P outward parallel to itself through a sufficiently small distance. By
using (33) it follows easily that
~A(c~,P ' ) ( ~A(c~ P),
which shows that 8A(c, P) is not minimal. Thus we can assume in the following
that ao> fl(a,p, k).
Since a(C)>a(C n P), we conclude from (80) and (
11
) that
If ao>f2(a,p, k) we have by (82) and Remark 2
8A(c, P) >ao M (a, p;k, ao).
8A( c, P) >ao F(p, k, ao),
(82)
(83)
(84)
where F is given by (
15
). As can be shown by differentiation, the function o f ao
on the righthand side o f (83) is strictly increasing for ao>f2(a, p, k). This function
thus attains its minimum for ao =f2(a, p, k). Therefore, we need to consider only
such values of ao for which
fl(a,p, k)<ao <f2(a,p, k).
We shall again make use of (81) and observe that, by Theorem 1 and Remark 2,
equality occurs in (81) if and only if C is an outer parallel domain of a regular
arcsided kgon of area a and perimeter p, and P is the central kgon of C. If
ao =f2(a, p, k) C is degenerate, which means that C is a smooth regular kgon
with case P.
Let us first assume that 8A(c, P) is minimal for some a0 from the interior of
the interval (84). Resuming the notation used in Section 3, we can state that (see
also [2, p. 363])
U11U12 1
A1A2 2
if A(a,p, k ) < ao<f2(a,p, k).
(85)
Otherwise we could reduce 8A(C, P ) by displacing AIA2 parallel to itself through
a small distance. This follows from (33). Second, if we assume that 8A(C, P ) is
minimal for ao =f2(a, p, k), the same argument as above shows that
ull
U12>_1
AIA2 2
if ao=f2(a,p, k).
(86)
Using the notation introduced at the end of Section 3 we have
A1A2= 2rp tan
~rr
Ull UI2=2(rt tan
+ p sin t~).
UI1UI2
A~A2
rl tan( ~r/k)+ p sin a
r~+ p cos a
~r
cot ~ = g ( a ) ,
and
Hence by (78)
(87)
(88)
where p ( a ) and rl(a) are given by (73) and (76). From
a
+ r l p c o s a
+ rip sin
 a
+p2 cos ~ ,
and (75) and (77) we see that
g'(a) > o.
Thus we have to consider two cases.
(i) If g(0)<½, (86) is impossible, and the minimum of 5A(C, P) is attained
in the case indicated by (85). C is a parallel domain of a (proper) regular
arcsided kgon, and P is the central kgon of C.
(ii) If g(0)>½, (85) is impossible and the minimum of 8A(C, P) is attained
in the case indicated by (86). C is a smooth regular kgon, and P is the
case of C.
g(0) can easily be evaluated by (87), (73), and (76). Writing (k/~r) tan(Tr / k ) = t
and referring to (
3
) we can summarize the result of this section in
T h e o r e m 2. Suppose that p2/4acr < t. There is exactly one disc C from C¢(a,p)
and one kgon P such that
8A(C, P)=AA(a,p, k).
C and P are characterized by the following properties:
(i) a ( C ) = a , p ( C ) = p .
(ii) if p2/ 4aTr < ( l + t)2/ ( l + 3t), C is a parallel domain of a regular arcsided
kgon, and P is the central kgon o f C. Any side o f P, say A1A2, meets the
boundary of C at points U1~, U~2 such that
A1 Ull = U12A2= ¼AIAz.
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