THE DENSE PACKING OF 13 CONGRUENT CIRCLES IN A CIRCLE

Transcription

1 DEPARTMENT OF MATHEMATICS TECHNICAL REPORT THE DENSE PACKING OF 3 CONGRUENT CIRCLES IN A CIRCLE Ferenc Fodor March 00 No. 00- TENNESSEE TECHNOLOGICAL UNIVERSITY Cookeville, TN 38505

2 THE DENSEST PACKING OF 3 CONGRUENT CIRCLES IN A CIRCLE FERENC FODOR Abstract. The densest packings of n congruent circles in a circle are known for n» and n = 9. In this paper we exhibit the densest packings of 3 congruent circles in a circle. We show that the optimal configurations are identical to Kravitz's [] conjecture. We useatechnique developed from a method of Bateman and Erd}os [] whichproved fruitful in investigating the cases n = and 9 by the author [6, 7].. Preliminaries and Results We shall denote the points of the Euclidean plane E by capitals, sets of points by script capitals, and the distance of two points by d(p; Q). We usepq for the line through P, Q, andpq for the segment with endpoints P, Q. POQ denotes the angle determined by the three points P,O,Q in this order. C(r) means the closed disc of radius r with center O. By an annulus r<ρ» s we mean all points P such that r<d(p; O)» s. We utilize the linear structure of E byidentifying each point P with the vector OP ~,whereo is the origin. For a point P and a vector ~a by P + ~a we always mean the vector OP ~ + ~a. The problem of finding the densest packing of congruent circles in a circle arose in the 960s. The question was to find the smallest circle in which we can pack n congruent unit circles, or equivalently, the smallest circle in which we can place n points with mutual distances at least. Dense circle packings were first given by Kravitz [] for n =;:::;6. Pirl [4] proved that the these arrangements are optimal for n» 9 and he also found the optimal configuration for n = 0. Pirl also conjectured dense configurations for» n» 9. For n» 6 proofs were given independently by Graham [3]. A proof for n = 6 and 7 was also given by Crilly and Suen [4]. Subsequent improvements were presented by Goldberg [8] for n = 4; 6 and 7. He also found a new packing with 0 circles. In 975 Reis [5] used a mechanical argument togenerate remarkably good packings up to 5 circles. Recently, Graham et al. [9, 0] using computers established packings with more than 00 circles and improved the packing of 5 circles. In 994 Melissen [] proved Pirl's conjecture for n = and the author [6, 7]proved it for n =9and n =. The problem of finding the densest packing of equal circles in a circle is also mentioned as an unsolved probleminthebookofcroft,falconer and Guy [5]. Packings of congruent circles in hyperbolic plane were treated by K. Bezdek []. Analogous results of packing n equal circles in an equilateral triangle and square can be traced down in the doctoral dissertation of Melissen [3]. 99 Mathematics Subject Classification. Primary5C5. Key words and phrases. circle packings, density, optimal packing.

3 FERENC FODOR In this article we shall find the optimal configurations for n =3. We are going to prove the following theorem. Theorem. The smallest circle C in which we can pack 3 points with mutual distances at least has radius R =(sin36 ffi ) = +p 5. The 3 points form the following two configuration as shown on Figure. Figure. The optimal configurations for n = 3. We shall prove Theorem in the following way. We are going to show that it is possible to divide C(R) into a smaller circle C(S) and an annulus S <ρ» R, choosing S such that there can be at most 4 points in C(S) and at most 0 points S<ρ» R. Then we prove that the 4 points in C(S) form the configurations shown on Figure. In the course of our proof we shall use the following two statements. Lemma, slightly modified here, originates from a paper of Bateman and Erd}os []. Lemma was used by the author in [6, 7]. Lemma ([]). Let r;s;(s ) be twopositive real numbers and suppose that we have two points P and Q which lie in the annulus r» ρ» s and which have mutual distance at least. Then the minimum ffi(r;s) of the angle POQ has the following values: ffi(r;s) = arccos s +r rs, if 0 <s» r» s =s; ffi(r;s) = arcsin s, if 0 <s =s» r» s or s» : Lemma ([6]). Let S be a set of n (n ), points in the plane and C the smallest circle containing S. Let ~a be avector. There exist two points P ;P on the boundary of S, such that d(p + ~a; O)+d(P + ~a; O) r, where r is the radius and O is the center of C.. Proofs Let S = R =R =. It was proved by Bateman and Erd}os [] that 7 points with mutual distances at least can be packed into a unit circle in a unique way; one point is at the center of the circle, and the other 6 form a regular hexagon of unit side length with vertices on C(). There cannot be any further points situated in the annulus <ρ» R. The argument for 6 points is very similar. The radius of the circumcircle of 6 points with mutual distances at least is, see []. Furthermore,

4 THE DENSEST PACKING OF 3 CONGRUENT CIRCLES IN A CIRCLE 3 the 6 points either form a regular hexagon of unit side length with vertices on C(), orapentagon with all 5 vertices on C() and a sixth point ato. In both cases it is clear that there cannot be 7 points in the annulus <ρ» R. Note that if there are exactly 9 points in the annulus <ρ» R, then the 3 points must form the first configuration shown on Figure. Lemma 3. There cannot be exactly 5 points in C(). Proof. Suppose, on the contrary, that there are 5 points in C(). Bateman and Erd}os [] proved that the radius of the circumcircle of 5 points with mutual distances at least is d 5 = ( cos 54 ffi ) = 0:85 :::. The minimal radius is realized by a regular pentagon of unit side length. According to Lemma there must be two points P and Q of the 5 such that dp = d(p; O) dq = d(q; O) anddp +dq d 5. Furthermore, the 4 points other than P cannot all be in C(0:8). To see this suppose that they all are in C(0:8). By adding up the five central angles we obtain 3ffi(0:8; 0:8)+ffi(; 0:8) = 360 ffi : :::. Therefore wemay suppose that dp» d 5 0:8 = 0:8894 :::. Simple calculus shows that ffi(dp ;R)+ffi(dQ;R) takes on its minimum, under these circumstances, when dp = d 5 0:8 and dq =0:8, and the minimum is 6:97 ffi. Note further that not all 3points other than P and Q can be in C(0:73). To see this add up the 5 central angles, ffi(; ) + ffi(0:73; 0:73) + ffi(; 0:73) = 370 ffi ::: if P and Q are adjacent. If they are not adjacent, then the sum is ffi(0:73; 0:73) + 4ffi(; 0:73) = 360 ffi :8 :::. Hence, there is a point S such thatds = d(s; O) 0:73. Let us also observe that ds» 0:76 else the sum of the 8 central angles determined by the points in the annulus is 80 ffi +6 ffi :+ffi(0:76;r) = 360 ffi :035 :::. We can also conclude that it is not possible that the remaining points are both in C(0:7). This follows from the fact that the sum of the 5 central angles would be larger than 360 ffi. We must check two different scenarios. One is when P and Q are adjacent. Then the sum is either at least ffi(; ) + ffi(; 0:76) + ffi(0:7; 0:7) + ffi(0:76; 0:7) + ffi(0:7; ) = 60 ffi +67 ffi :6+89 ffi :5+85 ffi :5+69 ffi : = 37 ffi :::, or ffi(; )+ffi(0:76; 0:7) + ffi(0:7; ) = 60 ffi + 85 ffi :5+ 69 ffi :=369 ffi :::. If P and Q are not adjacent, then the sum of the 5 central angles is either ffi(; 0:76) + ffi(0:7; 0:7) + ffi(0:7; ) = 67 ffi :6+89 ffi :5+ 69 ffi : = 363 ffi : :::, or ffi(; 0:76) + ffi(0:7; 0:76) + 3ffi(0:7; ) = 67 ffi :6+85 ffi : ffi : = 360 ffi :7 :::. Now, the sum of the 8 central angles is 4 36 ffi +ffi(dp ;R)+ffi(dQ;R)+ ffi(0:73;r)+ffi(0:7;r)=70+45 ffi +48 ffi :8 = 363 ffi :::. Thus we have finished the proof of the lemma. Proposition. Let f(r) =ffi(r;r) +ffi(r;s) and» s» fixed. If R» r» minf0:77;sg, then f(r) is an increasing function of r. Proof. To see this evaluate the derivative off(r). Our goal is to show that R r r r () f 0 (r) = p s + p (rr) (r + R) r (rs) (s + r ) > 0 () After rearrangement of the terms we obtain s (rs) (s + r ) (rr) (r + R) p > r s + R r

5 4 FERENC FODOR Notice that r s + R r takes on its maximum if r = s =0:77, and the maximum is less than. On the left hand side we can see that we decrease (rs) (s + r ) if we replace s by r because it is an increasing function of s. Now, we are going to show that (r ) (r ) (rr) (r + R) > : After simplification weobtain (r ) (r ) ((rr) (r + R) )=r 4 Rr + R: This polynomial is zero if r = p R =0:78 :::, therefore for r [R ; 0:77] it is positive. Thus, the inequality holds. Let the 4 points in C() be labeled as P ;:::;P 4 in clockwise direction. Let di = d(pi;o) and assume that d is the largest of di, i =;:::;4. Proposition. In each sector determined by two consecutive points in C(), there must be atleast points from the annulus <ρ» R. Proof. First, note that di» 0:77, i =; 3; 4 beacuse 4ffi(0:77;R)+7 36 ffi > 36 ffi. Suppose, on the contrary, that in PiOPi+ there is only one point. The sum of the 9central angles is not less than 5 ffi + ffi(di;r)+ffi(di;di+)+ffi(di+;r). If i =,thenweknow that d» d. Therefore by Proposition the total angle is not less than 5 ffi + ffi(d ;R)+ffi(d ;R ) + ffi(r ;R). This function takes on its minimum at d =, where it is exactly 360 ffi. If i =4,thend 4» d and so we may repeat the previous argument such that we obtain the same formula for the total angle. If i = or 3, then assume that di di+. Then, by Proposition the total angle is not less than 5 ffi + ffi(di;r)+ffi(di;r ) + ffi(r ;R). This function takes on its minimum at di = R and di =. It means that di = R whichgives 360 ffi for the total angle, plus ffi(d ;R) 36 ffi which makes the total larger than 360 ffi. Now we will examine the positions of the points P ;:::;P 4. We are going to prove that d =. Let the 9 points in the annulus < ρ» R be labeled by P 5 ;:::;P 3 in clockwise direction such that P 5 is adjacent to the segment P O. Let ci denote the unit circle centered at Pi. Furthermore, let Qi = ci ci and Q = c P O, Q 0 = c 9 P O be points in C(). Let R be the region of C(d ) which is not covered by the circular discs Ci. This is where P ;P 3 ;P 4 can be situated. The vertical line P O cuts R into two subregions, R = R [R. We are going to examine the diameter of R and R. We willshowthatifd <, then R i, i =; cannot accomodate two points. We are going to demonstrate this by proving that the diameter of R i, i =; is less than. now, we try to maximize the diameter of R. We assume that P 5 ;:::;P 3 are on the circle C(R). Lemma 4. For a fixed value of d [ p ; ], the arc of c between Q 0 and Q 6 is the longest if P OP 6 =36 ffi + ψ, where ψ = ffi(d ;R). Proof. Notice that of two unit circles centered on C(R), the one whose center makes the smaller central angle with P O provides the longer arc on c if their intersection is in c. Let P OP 6 = ff. For a fixed d, ff [36 ffi + ψ; 08 ffi ψ]. If d is fixed it is enough to show that the intersection of the two unit circles in the extreme positions is in c. Note that this intersection point T (d ) is always such that P OT(d ) = 7 ffi. Let S = c OT(d ) and let s = d(s; O). It is

6 THE DENSEST PACKING OF 3 CONGRUENT CIRCLES IN A CIRCLE 5 P P6 O Q 6 Q7 Q 0 P 7 Q 9 P 8 P 9 Figure enough to show that d (S; P 6 ) = R + s Rs cos(36 ffi ψ)». The equation R + s Rs cos(36 ffi ψ) = can, after some transformations, be written as an algebraic equation for d as follows. (3 p 5+7)x +( 3 3 p 5)x 0 +(6 p 5+8)x 8 +( p 5+)x 6 +(45+03 p 5)x 4 + ( p 5)x +47+ p 5=0 The above polyomial has only one root in [ p ; ] and it is equal to. This proves our claim. We assume that P 9 OP =6 ffi ψ, where ψ = ffi(d ;R). This ensures that the part of P O which bounds R is maximal in length. We will also assume that the angles are P 6 OP 7 = P 8 OP 9 =7 ffi ψ. This guarantees that the arcs of c 7 and c 9 are the longest possible. However, notice that in such apositiond(p 7 ;P 8 ) becomes less than. Lemma 5. The diameter of R does not exceed if d [0:745; ] Proof. Notice that P OQ 7 =90 ffi and P OQ 9 =6 ffi,andd(q 7 ;O)=d(Q 9 ;O) is a decreasing function of d. Also, d(q 0 ;O) is a monotonically decreasing functions of d. R is bounded by arcs of c ;c 6 ;:::;c 9, and the line P O. Its vertices are Q 0 ;Q ;Q 6 ;:::Q 9 as shown on Figure. Under these circumstances, the diameter of R can only be realized by a pair of the vertices Q 0 ;Q 6 ;Q 7 ;Q 9. Note that d(q 7 ;Q 9 ) and d(q 0 ;Q 7 ) are both decreasing functions of d, and by direct substitution we can see that they do not exceed if d [0:745; ]. Using the coordinates of Q 0 and Q 6 we may writed (Q 0 ;Q 6 )=asanalgebraic equation for d. Furthermore, after a sufficient number of transformations,

7 6 FERENC FODOR x Figure 3 it may be written as a polynomial equation p(q 0 ;Q 6 ) = 0 for d. The Cartesian coordinates of Q 0 and Q 6 are the following. Q 6 =(R(sin(36 ffi + ψ) sin ψ);d+ R(cos(36 ffi + ψ) cos ψ)); Q 0 =(0; R cos(36 ffi ψ)+ p R cos (36 ffi ψ) R) The polynomial equation is as follows. p(q 0 ;Q 6 )=( 0 4 p 5)d 4 +( 5 7 p 5)d 6 +4 p 5d 8 +(55+3 p 5)d 0 +(5+ 5 p 5)d +(0+4 p 5)d 4 =0 This equation has two roots in the [ p ; ] interval, 0:744 ::: and. By direct substitution we cancheck that p(q 0 ;Q 6 ) < in(0:7448; ). In a similar manner we may write d (Q 6 ;Q 9 ) = as a polynomial equation for d and check for roots in the designated interval. Note that d(o; Q 9 )=Rcos(36 ψ) p ffi ( R sin (36 ffi ψ)). The graph of the function d(q 6 ;Q 9 )isshown on Figure. This function has no zeros in the interval [0:745; ]. Lemma 6. If P 9 ;P 0 ;P P 3 OP 4, then is not possible that d [ p =; 0:745]. Proof. For every value of d there is a dm such that none of the three points P ;P 3 ;P 4 can be closer to O than dm. We may obtain dm from the the follwing equation. ffi(d ;d )+ffi(d ;dm) = 80 ffi Easy calculation shows that dm = d d. Clearly, dm is a monotonically p decreasing function of d. Furthermore, let dm = d. We claim that it

8 THE DENSEST PACKING OF 3 CONGRUENT CIRCLES IN A CIRCLE 7 is not possible that both d 3 and d 4 are less than or equal to dm. This may be shown simply by examining the function that describes the sum of the four central angles determined by the points P ;:::;P 4. It is f(d )=ffi(d ;d )+ffi(dm ;d )+ ffi(dm ;dm ). Simple calculus shows that (3) df(d ) dd = d d p + 4 =d ( d q4 )3= d It is a simple exercise to show that (3) is larger than or equal to 0 in the interval [ p ; ]. In particular, f( p ) = 360ffi, which proves our claim. Also note that ffi(dm ;d )=90 ffi. Now, we are going to add up the four central angles, PiOPi+. The sum of the angles is as follows ffi(d ;d )+ffi(d ;d 3 )+ffi(d 3 ;R)+7 ffi + ffi(d 4 ;R)+ffi(d 4 ;d ). Note that ffi(d ;d ) is not less than ffi(d ;d ) and that by Lemma, ffi(d ;d 3 )+ffi(d 3 ;R) takes on its minimum when d = d and d 3 = dm. The same is true for ffi(d 4 ;R)+ ffi(d 4 ;d ) so we may assume that d 4 = dm. Whence, the sum of the angles is ffi(d ;d )+ffi(dm;d )+ffi(dm;r)+7 ffi + ffi(dm ;R)+ffi(dM ;d ). After simplification we obtain F (d )=ffi(dm ;R)+ffi(dm;R)+34 ffi : F (d ) is a decreasing function of d and it is larger than 360 ffi on the interval [ p =; 0:735]. In the interval [0:735; 0745] we will write the total angle differently. First, notice that if d R+p 6 R =0:7376 :::, then dm» R, so we may omit ffi(dm;r). Case. d 3 0:6 The sum of the angles is not less than ffi(d ;d )+ffi(d ; 0:6) + ffi(0:6;r)+ffi(dm ;R)+ffi(dM ;d )+7 ffi, which is equal to 6 ffi + ffi(0:6;r)+ ffi(0:6;d )+ffi(d ;d ). This is a decreasig function of d and its value at 0:745 is 360: ffi. Case. d 3 < 0:6 Notice that one of d and d 4 has to be larger than or equal to 0:73 or ffi(0:6; 0:73)+ffi(0:73; 0:745) > 360 ffi. Moreover, none of d and d 4 can be less than 0:7 or else ffi(0:745; 0:745) + ffi(0:745; 0:6) + ffi(0:6; 0:7) + ffi(0:7; 0:745) > 360 ffi. In this case the total angle is not less than ffi(0:7;r)+ffi(0:73;r)+ ffi(d ;R) + 6 ffi 36 ffi. Lemma 7. If P 0 ;P ;P P 4 OP,thend cannot be in[ p =; 0:745]. Proof. The total angle is not less than ffi(d ;d )+ffi(dm;d )+ffi(dm;r)+7 ffi + ffi(d ;R) which is larger than or equal to 7 ffi +ffi(d ;d )+ffi(d ;R)+ffi(R ;d ) in the [0:737; 0:745] interval. This function is decreasing and its value exceeds 360 ffi :atd =0:745. Note that ffi(d ;d )+ffi(d ;R) is a decreasing function in the designated interval. In [ p =; 0:737], the total angle is larger than or equal to 5 ffi + ffi(d ;d )+ ffi(dm;r)+ffi(d ;R) which is also a decreasing function of d and its value at d = 0:737 is 366 ffi Toseethatthisnotethatffi(dm;R)+ffi(d ;R) is a decreasing function of d. Now, the only possibility is that d =. We saw in Lemma 5, that in this case the diameter of R and R is equal to and this diameter is realized by Q 0 and

9 8 FERENC FODOR Q 6. Therefore the four points in C() must be in the configuration shown in the second part of Figure. References [] P. Bateman, P. Erd}os, Geometrical extrema suggested by a lemma of Besicovitch, Amer. Math. Monthly, 58, (95), [] K. Bezdek, Ausfüllungen eines Kreises durch kongruente Kreise in der hyperbolischen Ebene, Studia Sci. Math. Hungar, 7, (98), [3] H.S.M. Coxeter, M.G. Greening and R.L. Graham, Sets of points with given maximum separation (Problem E9), Amer. Math. Monthly, 75, (968), [4] T. Crilly, S. Suen, An improbable game of darts, Math. Gazette, 7, (987), [5] H.T. Croft, K.J. Falconer, R.K. Guy, Unsolved Problems in geometry, SpringerVerlag New York, Berlin, Heidelberg, 99. [6] F. Fodor, The densest packing of 9 congruent circles in a circle, Geom. Dedicata 74, (999), [7] F. Fodor, The densest packing of congruent circles in a circle, Beitrage Alg. Geom. 4, (000), [8] M. Goldberg, Packing of 4; 6; 7 and 0 circles in a circle, Math Mag., 44, (97), [9] R.L. Graham, B.D. Lubachevsky, Dense packings of 3k(k +)+ equal disks in a circle for k =; ; 3; 4 and 5, Proc. First Int. Conf., "Computing and Combinatorics" COCOON'95, Springer Lecture Notes in Computer Science, 959 (996), [0] R.L. Graham, B.D. Lubachevsky, Curved hexagonal packings of equal disks in a circle, Discrete Comp. Geom., 8, (997), [] S. Kravitz, Packing cylinders into cylindrical containers