Approximation of convex discs by polygons

Discrete & Computational Geometry, Sep 1986

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 ak-gon? This and related questions are the subject of the present paper.

A PDF file should load here. If you do not see its contents the file may be temporarily unavailable at the journal website or you do not have a PDF plug-in installed and enabled in your browser.

Alternatively, you can download the file locally and open with any standalone PDF reader:

https://link.springer.com/content/pdf/10.1007%2FBF02187698.pdf

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, A-5020 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 k-gon? 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 up-to-date 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, 8-11, 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 k-gon. Let ~k denote the class of all k-gons. Two measures for the closeness of the approximation of k-gons 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 k-gon 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 k-gons 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 k-gon is minimal for a k-gon 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 k-gon. We join each two consecutive vertices of P* by congruent circular arcs o f radius not less than the circum-radius of P*. These arcs form the boundary of a convex disc C* which we call a regular arc-sided k-gon with kernel P* (Fig. 1). If ( 5 ) is satisfied it can be shown (see [5]) that (i) there is exactly one regular arc-sided k-gon C* with area a and perimeter p, and (ii) the infimum on the right-hand 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 k-gons 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 k-gons 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 k-gon 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 k-gon 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 arc-sided k-gon of area a and perimeter p, and P is the kernel o f C (p2/a <4~rt), or C = P is a regular k-gon 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 k-gon, we call a set of the form (/5_p)p a smooth regular k-gon with case /5 (Fig. 2). It will be convenient to consider both /5 and its in-circle as degenerate smooth regular k-gons 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~t-p2/4 - 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 k-gon 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 k-gon with perimeter p, and P is the case of C. Let C be a disc from ~ ( a , p) and P a k-gon 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 k-gon 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 k-gon 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 arc-sided k-gon 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 k-gon 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 arc-sided k-gon 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 k-gon P which we call the central k-gon of C (Fig. 3). By a central k-gon of a circle C we mean any regular k-gon 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 arc-sided k-gon, and P is the central k-gon 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 arc-sided k-gon and its kernel as well as a smooth regular k-gon and its case may be regarded as a degenerate parallel domain of a regular arc-sided k-gon and its central k-gon. 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 k-gon 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 k-gon 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 k-gon 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 k-gon, where A~ = Aj for j = 3. . . . . k; F r o m P' we obtain the further k-gon 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 k-gon 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 well-known 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 A-X, 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 k-gon 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 k-gon 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 k-gon 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 k-gon, corresponding to a = 0 , with the regular arc-sided k-gon, 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-~/(p2-4aTr)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 in-radius o f the kernel o f C1 p ' ( a ) < 0 . r , ( ) = u and observe that For the in-radius o f P we obtain by (74) and (77) r ~ ( a ) = ux/p2-4a~r(a - at~k) sin a - a cos a 2k[dp-Tr(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 k-gon 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 arc-sided k-gon of area a and perimeter p, and P is the kernel of C. Let P' be the k-gon 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 right-hand 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 arc-sided k-gon of area a and perimeter p, and P is the central k-gon of C. If ao =f2(a, p, k) C is degenerate, which means that C is a smooth regular k-gon 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 arc-sided k-gon, and P is the central k-gon 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 k-gon, 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 k-gon 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 arc-sided k-gon, and P is the central k-gon 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. 1. A. S. Besicovitch , Variants of a classical isoperimetric problem , Quart. J. Math. Oxford ser. 2 3 ( 1952 ), 42 - 49 . 2. H. G. Eggleston , Approximation to plane convex curves, I. Dowker-type theorems , Proc. London Math. Soc. (3) 7 ( 1957 ), 351 - 377 . 3. G. Fejes T6th , Covering the plane by convex discs , Acre Math. Acad. Sei. Hungar . 23 ( 1972 ), 263 - 270 . 4. G. Fejes Ttth , Sum of moments of convex polygons , Acta Math. Aead. Sci. Hungar . 24 ( 1973 ), 417 - 421 . 5. G. Fejes T t t h and A. Florian, Covering of the plane by discs , Geom. Dedicata 16 ( 1984 ), 315 - 333 . 6. L. Fejes T6th , Filling of a domain by isoperimetric discs , Pubt. Math. Debrecen , 5 ( 1957 ), 119 - 127 . 7. L. Fejes T6th , On the isoperimetric property of the regular hyperbolic tetrahedra , Magyar Tud. Akad. Mat. Kut. Int. Kfzl. 8A ( 1963 ), 53 - 57 . 8. L. Fejes T6th , Regular Figures , Pergamon Press, Oxford, 1964 . 9. L. Fejes T6th , Lagerungen in der Ebene, auf tier Kugel und im Raum , 2nd ed., Springer-Verlag, Berlin-Heidelberg-New York, 1972 . 10. L. Fejes T6th and A. Florian, Packing and covering with convex discs , Mathematika 29 ( 1982 ), 181 - 193 . 11. L. Fejes T6th and A. Heppes, Filling of a domain by equiareal discs , PubL Math. Debrecen 7 ( 1960 ), 198 - 203 . 12. A. Florian , Integrale auf konvexen Mosaiken, Period. Math. Hungar. 6 ( 1975 ), 23 - 38 . 13. A. Florian , Packing and coveting with convex discs , Studia Sci. Math . Hungar., to appear. 14. P. M. Gruber , Approximation of convex bodies, Convexity and lts Applications , Birkhiiuser-Verlag, Basel-Boston-Stuttgart, 1983 . 15. G. Haj6s , 0 b e t den Durchschnitt eines Kreises und eines Polygons , Ann. Univ. Sci. Budapest. Eiitvfs Sect. Math. 11 ( 1968 ), 137 - 144 . 16. K. Leichtweiss , Konvexe Mengen, Springer-Verlag, Berlin-Heidelberg-New York, 1980 . 17. G. C. Shephard and R.J. webster Metrics fr sets f convex bdies Mathematika 2 ( 9 65 ) 73 - 88 


This is a preview of a remote PDF: https://link.springer.com/content/pdf/10.1007%2FBF02187698.pdf

A. Florian. Approximation of convex discs by polygons, Discrete & Computational Geometry, 1986, 241-263, DOI: 10.1007/BF02187698