Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 41 (2000), No. 2, 401-409.

The Densest Packing of 12 Congruent Circles in a Circle

Ferenc Fodor

Department of Mathematics, Tennessee Technological University Box 5054, Cookeville, TN 38505, U.S.A. e-mail: ffodor@tntech.edu

Abstract. The densest packings of n congruent circles in a circle are known for n ≤ 11 and n = 19. In this paper we exhibit the densest packing of 12 congruent circles in a circle. In fact, we show that the optimal configuration is the same as the one Kravitz [10] conjectured. We use a technique developed from a method of Bateman and Erdős [1].

1. Preliminaries and results

We shall denote the points of the Euclidean plane 𝔼^{2} by capitals, sets of points by script capitals, and the distance of two points by d(P, Q). We use P Q for the line through P, Q, and \overline{P Q} for the segment with endpoints P, Q. ∠P OQ denotes the convex 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. A sector P OQ is a closed convex subset of 𝔼^{2} which is bounded by the rays OP and OQ. We also make use of the linear structure of 𝔼^{2} by identifying each point P with the vector \overrightarrow{OP}, where O is the origin. For a point P and a vector a⃗ by P + a⃗ we always mean the vector \overrightarrow{OP} + a⃗.

The problem of finding the densest packing of congruent circles in a circle arose in the 1960s. The question was phrased the following way. What is the smallest circle in which we can pack n congruent unit circles, or equivalently, what is the smallest circle in which we can place n points with mutual distances at least 1? In 1967 dense circle packings were given by Kravitz [10] for n = 2, …, 16. In 1969 Pirl [13] proved that in fact the arrangements given by Kravitz are optimal for n ≤ 9 and he also found the optimal configuration for n = 10. Pirl also conjectured dense configurations for 11 ≤ n ≤ 19. For n ≤ 6 proofs were given independently by Graham in response to a problem posed by Coxeter in the American Mathematical Monthly [3]. A proof for n = 6 and 7 was also given by Crilly and Suen [4]. Subsequent improvements were presented by Goldberg [7] for n = 14, 16 and 17. He also found a new packing with 20 circles. In 1975 Reis [14] used an ingenious mechanical contraption to generate remarkably good packings up to 25 circles. Recently, Graham et al. [8], [9] using modern computers established packings with more than 100 circles and improved the packing of 25 circles. In 1994 Melissen [11] proved Pirl’s conjecture for n = 11 and the author [6] proved it for n = 19. We shall find the optimal configuration for n = 12. Our goal is to prove the following theorem.

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix}2000 Heldermann Verlag

Theorem 1. The smallest circle C which can accommodate 12 points with mutual distances at least 1 has radius R = 1.5148 . . .. Moreover, the 12 points form the unique configuration shown in Figure 1.

Figure 1. The optimal configuration for n = 12

Note that two points are connected with a straight line segment on Figure 1 if and only if their distance is exactly 1. In Theorem 1, R = \frac{1}{\sqrt{3}x_{0}} , where x_{0} is the smallest positive root of 9x^{5} − 15x^{4} + 7x^{3} − 3x + 1 = 0.

The proof of Theorem 1 consists of three parts. We shall divide C(R) into a smaller circle C(S) and an annulus S < ρ ≤ R, choosing S such that there can be at most 5 points in C(S) and at most 9 points S < ρ ≤ R. This leaves us three different cases. Namely when there are 3, 4, 5 points in C(S). We deal with the latter two cases first showing that they cannot occur. Then we prove that the three points in C(S) form a regular triangle of unit side length centered at O.

In the course of our proof we shall utilize the following two statements. Lemma 1, slightly modified here, originates from a paper of Bateman and Erdős [1] where they used it to prove an improved version of a theorem of Besicovitch. Lemma 2 is used by the author [6] to show Pirl’s conjecture for n = 19.

Lemma 1. [1] Let r, s, (s ≥ \frac{1}{2}) be two positive 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 1. Then the minimum ϕ(r, s) of the angle ∠P OQ has the following values:

ϕ(r, s) = arccos \frac{s^{2} + r^{2} − 1}{2rs}, if 0 < s − 1 ≤ r ≤ s − 1/s;

ϕ(r, s) = 2 arcsin \frac{1}{2s}, if 0 < s − 1/s ≤ r ≤ s or s ≤ 1.

Lemma 2. [6] Let S be a set of n (n ≥ 2) points in the plane and C the smallest circle containing S. Let a⃗ be a vector. There exist two points P_{1}, P_{2} on the boundary of S, such that d(P_{1} + a⃗, O) + d(P_{2} + a⃗, O) ≥ 2r, where r is the radius and O is the center of C.

The problem of finding the densest packing of equal circles in a circle is also stated in the book of Croft, Falconer and Guy [5] on pp. 110–111. Packings of congruent circles in hyperbolic plane were treated by K. Bezdek [2]. Analogous results of packing n equal circles in an equilateral triangle and square can be traced down in the doctoral dissertation of Melissen [12].

2. Proofs

We prove Theorem 1 with a sequence of lemmas. Let S = R − 1/R = 0.854 . . ..

Lemma 3. There can be at most 9 points in the annulus S < ρ ≤ R.

Proof. This follows from the fact that ϕ(S, R) = ϕ(R, R) = 38^{∘}.5468 . . .. Thus, 10ϕ(R, R) > 360^{∘}. □

We know from [10] that the radius of the circumcircle of 6 points with mutual distances at least 1 is at least 1, which is larger than S. Therefore there cannot be fewer than 7 points in the annulus S < ρ ≤ R, or there would be at least 6 points in C(S). Thus, we have three cases to consider, namely, when there are 5, 4, 3 points in C(S). We shall deal with the three cases one-by-one.

Lemma 4. There cannot be exactly 7 points in the annulus S < ρ ≤ R.

Proof. If we have 7 points in the annulus S < ρ ≤ R there must 5 points in C(S). Consider the following assertion. There cannot be any of these 5 points in C(0.835). Because 2ϕ(0.835, S) = 145^{∘}.343 … and ϕ(S, S) = 71^{∘}.61 …, we have that 3ϕ(S, S) + 2ϕ(0.835, S) = 360^{∘}.175 … > 360^{∘}. Therefore we may assume that all 5 points are in the annulus 0.835 < ρ ≤ S.

Connecting the 5 points in C(S) to O we obtain 5 angular sectors. The minimal central angle of a sector is ϕ(S, S) = 71^{∘}.61 . . .. In one of these sectors, say P_{i}OP_{i+1}, there is a point P of the 7 points from the annulus S < ρ ≤ R. Then ∠P_{i}OP, ∠P_{i+1}OP ≥ ϕ(0.835, R) = 76^{∘}.1173 . . .. Therefore 4ϕ(S, S) + 2ϕ(0.835, R) = 362^{∘}.561 … > 360^{∘}. So the 5 points cannot all be in C(S). □

Lemma 5. There cannot be exactly 8 points in the annulus S < ρ ≤ R.

Proof. If we have 8 points in the annulus S < ρ ≤ R, then there must be 4 points in C(S). To accommodate 4 points in the plane with mutual distances at least 1 we need a circle of radius at least r_{4} = 1/\sqrt{2}. Suppose that these 4 points are in the minimum radius circle C′ with center O′. The radius of C′ is at least r_{4}. We know from Lemma 2 that there are two of the 4 points, P and Q on the boundary of C′, such that d(P, O)+d(Q, O) ≥ 2r_{4} = \sqrt{2}. The 8 points in the annulus S < ρ ≤ R connected to O divide C into 8 angular sectors. We can assume that P and Q are in the interior of two different sectors, say P is in P_{i}OP_{i+1} and Q in P_{j}OP_{j+1}, and the angle ∠P_{i}OP_{i+1} ≥ 2ϕ(d(P, O), R), and ∠P_{j}OP_{j+1} ≥ 2ϕ(d(Q, O), R).

Suppose that d(Q, O) ≥ d(P, O). We may assert then that one of the points in C(S) other than Q is at least 0.66 from O. To see this we show that the three points in C(S) other than Q do not fit in C(0.66). On the contrary let us suppose that they do. Connecting the three points to O we obtain three angular sectors with angles at least ϕ(0.66, 0.66) = 98^{∘}.5019 . . .. Q lies in one of these sectors, say in P_{1}OP_{2}. Then ∠P_{1}OP_{2} ≥ 2ϕ(0.66, S) = 163^{∘}.074 …, so 2ϕ(0.66, 0.66)+2ϕ(0.66, S) = 360^{∘}.079 … > 360^{∘}. Therefore the three points cannot all be in C(0.66).

Now the sum of the angles of the 8 sectors, determined by the points in the annulus S < ρ ≤ R, is at least 6ϕ(R, R) + 2ϕ(d(P, O), R) + 2ϕ(d(Q, O), R), where we assume that d(P, O) + d(Q, O) = \sqrt{2} and, by the above argument, that d(P, O) ≥ 0.66. It is easy to check that the minimum of the sum occurs when d(P, O) = 0.66 and it is 362^{∘}.16 … > 360^{∘}.

Therefore there cannot be exactly 8 points in the annulus S < ρ ≤ R. □

In the subsequent paragraphs we shall deal with the remaining case when we have only 3 points, P_{1}, P_{2}, P_{3}, in C(S). Let d_{i}, i = 1, 2, 3 denote the distance d(P_{i}, O), i = 1, 2, 3. In the following lemma we summarize several properties of these three points.

Lemma 6.

a) d_{i} ≥ \sqrt{3}/3 for some i and 0.43 ≤ d_{i} ≤ d = 0.6122 for all i = 1, 2, 3.

b) C(0.56) cannot contain two of P_{i}, i = 1, 2, 3.

c) A sector, determined by two points of the annulus S < ρ ≤ R, contains at most one of P_{i}, i = 1, 2, 3.

d) A sector, determined by P_{i} and P_{j}, contains exactly three points from the annulus S < ρ ≤ R.

e) If d_{i} ≤ 0.5635, then P_{i} does not contribute to the minimal angle of the sector, determined by two points of the annulus S < ρ ≤ R, in which it lies.

Proof. a) The radius of the circumcircle of three points with mutual distances 1 is \sqrt{3}/3, therefore at least one of d_{i} is not less than \sqrt{3}/3. Suppose that d_{i} ≥ 0.6122 for some i. Then adding up the angles of the nine sectors determined by the points in the annulus S < ρ ≤ R we obtain 8ϕ(R, R) + 2ϕ(d, R) = 308^{∘}.37 … + 51^{∘}.66 = 360^{∘}.83.

As d_{i} ≤ d for all i, no side of the triangle △P_{1}P_{2}P_{3} can be longer than 2d = 1.2244 …, so all side lengths are in between 1 and 2d. It is easy to see that such a triangle is always strictly acute and all three heights are at least sin arccos(0.6122) = 0.79 . . .. This shows that O must be inside △P_{1}P_{2}P_{3} otherwise it would be farther than d from one of the vertices. Now, suppose that d_{i} ≤ 0.43 for some i. Then one of the other two d_{j} is at least \sqrt{(0.79 − 0.43)^{2} + 1/4} = 0.616 … > d.

b) If d_{i}, d_{j} ≤ 0.56, then adding up the angles of the three sectors determined by P_{i}, I = 1, 2, 3 we obtain 2ϕ(0.56, d) + ϕ(0.56, 0.56) = 234^{∘}.06 + 126^{∘}.468 = 360^{∘}.52.

c) Let us assume, on the contrary, that in one sector there are two of P_{i}, i = 1, 2, 3. Then the central angle of this sector is at least ϕ(d, d) = 109^{∘}.516 . . .. Thus, if we sum up the angles of the 9 sectors determined by the points in the annulus S < ρ ≤ R we get 8ϕ(R, R) + ϕ(d, d) = 417^{∘}.8 … > 360^{∘}.

d) Now, we will show that each of the 3 sectors determined P_{i}, i = 1, 2, 3 contains exactly three of the points from the annulus S < ρ ≤ R. To see this suppose, on the contrary, that in the sector determined by P_{1} and P_{2} there are four points. We may suppose that d_{1} ≥ 0.56. If d_{2} ≥ 0.525, then the angle of the sector is at least 3ϕ(R, R) + ϕ(0.56, R) + ϕ(0.525, R) = 115^{∘}.64 … + 18^{∘}.57 … + 9^{∘}.16 … = 143^{∘}.37 . . .. Therefore the total angle of the three sectors is at least 2ϕ(d, d)+140^{∘}.87 … = 362^{∘}.4 . . .. If d_{2} < 0.525, then the total angle of the three sectors is 3ϕ(R, R) + ϕ(0.56, R) + ϕ(d, d) + ϕ(0.525, d) = 115^{∘}.64 … + 18^{∘}.57 … + 109^{∘}.516 … + 122^{∘}.9 … = 366^{∘}.5 . . .. Therefore there must be exactly three points in each of the three sectors.

e) Evaluating ϕ(0.5635, R) it turns out that it is less than \frac{1}{2}ϕ(R, R). □

Let us introduce the following function

F (d_{i}, d_{i+1}, s_{i+2}) = arccos \frac{d_{i}^{2} + d_{i+1}^{2} − s_{i+2}^{2}}{2d_{i}d_{i+1}} ,

which gives the central angle of the sector P_{i}OP_{i+1} under the assumption that the length of the side P_{i}P_{i+1} is s_{i+2}. If d_{i} = d_{i+1} = \sqrt{3}/3 and s_{i+2} = 1, then F (d_{i}, d_{i+1}, s_{i+2}) = 120^{∘}. Let us further note that the ϕ(d_{i}, R) are monotonically increasing functions of the d_{i}.

Suppose that two sides of △P_{1}P_{2}P_{3} are longer than 1. Then we move the vertex, say P_{1}, where these two sides meet toward O, while keeping all others points fixed, until one of the sides becomes 1.

Lemma 7. This operation is admissible, that is, the mutual distances remain at least 1 and the 12 points fit in C(R).

Proof. To see this draw circles of radius 1/2 around the 12 points. During the movement of P_{1} no circles can overlap. Clearly, as we move P_{1} toward O we only have to make sure that there is no circle which blocks the one around P_{1} before one of the sides shrinks to 1. Assume, on the contrary, that there is a circle around a point P which blocks the circle around P_{1}. Then d(P, O) ≤ \sqrt{d^{2} + 1} = 1.1725 . . .. Suppose that P is in the sector determined by P_{1} and P_{2}. One of d_{1} and d_{2} is at least 0.56. We distinguish three cases. If d(P, O) ≤ 1.1 and d_{1} ≥ 0.56, then total angle of the three sectors determined by the three points in C(S) is at least 2ϕ(d, d) + ϕ(0.56, 1.1) + 2ϕ(R, R) = 219^{∘}.03 … + 64^{∘}.849 … + 77^{∘}.08 … = 360^{∘}.96 . . .. If d_{1} < 0.56, then d_{2} ≥ 0.56, thus the total angle of the three sectors is 2ϕ(d, d)+ϕ(0.56, R)+2ϕ(R, R)+ϕ(0.43, 1.1) = 219^{∘}.03 …+18^{∘}.57 …+ 77^{∘}.08 … + 65^{∘}.52 … = 379^{∘}.08 … Finally, if d(P, O) > 1.1, then we will examine the 9 sectors determined by the points in the annulus S < ρ ≤ R. First we note the following.

If d_{i} ≤ 0.5635, then P_{i} does not contribute to the angle of the sector in which it lies, that is, 2ϕ(0.5635, R) < ϕ(R, R). By Lemma 2 there are two points, P_{i}, P_{i+1} for which d_{i} + d_{i+1} ≥ 2\sqrt{3}/3. The total angle of the two sectors they are in is 2ϕ(d_{1}, R) + 2ϕ(d_{2}, R). Supposing that d_{i} + d_{i+1} = 2\sqrt{3}/3 and 2\sqrt{3}/3 − 0.5635 ≤ d_{i}, d_{i+1} the minimum of the total angle occurs when one of d_{i} and d_{i+1} is 2\sqrt{3}/3−0.5635, namely 46^{∘}.61 . . .. Thus, the sum of the central angles of the 9 sectors is as follows, 6ϕ(R, R) + 2ϕ(1.1, R) + 46^{∘}.61 … = 231^{∘}.28 …+82^{∘}.54 …+46^{∘}.61 … = 360^{∘}.043 . . .. Therefore there cannot be a circle which blocks P_{1} as we move it toward O, so we can shrink one side to 1. From now on we always assume that at least two sides of △P_{1}P_{2}P_{3} are 1. □

Suppose we have a triangle △P_{1}P_{2}P_{3} with fixed sides s_{i}. We will call a position of this triangle admissible if F (d_{i}, d_{i+1}, s_{i+2}) ≥ 2ϕ(R, R) + ϕ(d_{i}, R) + ϕ(d_{i+1}, R) for all i. Notice that if d_{i} = \sqrt{3}/3 and all three sides are of unit length, then there is equality in each of the above inequalities. If any of these conditions fails, then we will call that position of the triangle not admissible.

First we will consider the case when s_{i} = 1 for all i.

Lemma 8. If the three points in C(S) form a regular triangle △P_{1}P_{2}P_{3} of side length 1, then the only admissible position is when O is at the center of this triangle.

Proof. Let d_{1} be the largest of the d_{i}, then d_{1} ≥ \sqrt{3}/3. Consider the position in which d_{2} = d_{3} and fix d_{1}. Note that in this position d_{2} is a function of d_{1}. Notice that the F (d_{1}, d_{i}, 1) are monotonically decreasing functions of the d_{i}. Thus, if we show that in this symmetric position the two sides P_{1}P_{2} and P_{1}P_{3} are not admissible, then in any other position, with the same fixed d_{1}, one of them will not be admissible.

It is enough to consider the following function of d_{1}, F (d_{1}, d_{2}, 1)−2ϕ(R, R)−ϕ(d_{1}, R)− ϕ(d_{2}, R) in the interval [\frac{\sqrt{3}}{3}, d]. Figure 2 shows its graph in [\frac{\sqrt{3}}{3}, d]. It can be seen that it is zero at \sqrt{3}/3 and negative elsewhere.

R := 1.5148

Plot[180/Pi(ArcCos[\frac{− \frac{3}{4}+( \frac{\sqrt{3}}{2}−x)^{2}+x^{2}}{2x\sqrt{\frac{1}{4}+( \frac{\sqrt{3}}{2}−x)^{2}}} ] −

(4ArcSin[\frac{1}{2R}] + ArcCos[\frac{− \frac{3}{4}+R^{2}+( \frac{\sqrt{3}}{2}−x)^{2}}{2R\sqrt{\frac{1}{4}+( \frac{\sqrt{3}}{2}−x)^{2}}} ] + ArcCos[\frac{−1+R^{2}+x^{2}}{2Rx} ])),

{x, Sqrt[3]/3, .6122}]

Figure 2.

Thus, we can conclude that a regular triangle of unit side length can only be placed in C(S) exactly one way, namely, when O is at the center of the triangle. □

Now, we return to the general case letting one side of △P_{1}P_{2}P_{3} be greater than 1.

Lemma 9. If one of the sides, say P_{1}P_{3}, is longer than 1, then the triangle △P_{1}P_{2}P_{3} cannot be admissible in any position.

Proof. Suppose that d_{1} ≥ d_{3}. Notice that d_{2} ≤ d_{1}, d_{3}. Otherwise d_{1} ≥ (1 + d_{2}^{2} −

2d_{2}\sqrt{1 − s_{2}^{2}/4} )^{1/2} which is longer than it would be if △P_{1}P_{2}P_{3} were a regular unit triangle. Therefore the side P_{1}P_{2} would not be admissible unless d_{2} = \sqrt{3}/3 by Lemma 8.

Suppose that d_{2} = d_{3}. Then d_{2}\sqrt{3}/3 or one of P_{1}P_{2} and P_{2}P_{3} would not be admissible. Moreover, as d_{1} ≤ d, it follows that d_{2} > 0.56. If we can show that in this symmetric position both sides P_{1}P_{2} and P_{1}P_{3} are not admissible, then by the same monotonicity argument as in Lemma 8, △P_{1}P_{2}P_{3} is not admissible in any other position.

For every d_{2} in [0.5635, \sqrt{3}/3] there is an f(d_{2}) defined by 6ϕ(R, R) + 4ϕ(d_{2}, R) + 2ϕ(f(d_{2}), R) = 360^{∘}, which means that \sqrt{3}/3 ≤ d_{1} ≤ f(d_{2}). For d_{2} in [0.56, 0.5635] define f(d_{2}) = d. Assume that △P′_{1}P_{2}P_{3} is a regular triangle of unit side length, then \sqrt{3}/3 ≤ d′_{1} < d_{1}. However, from the previous lemma we already know that the side P′_{1}P_{2} is not admissible unless d′_{1} = d_{2} = \frac{\sqrt{3}}{3}. Therefore P_{1}P_{2} is not admissible.

Now we have only to show that P_{1}P_{3} is not admissible as well. Notice that for a fixed d_{2} the ∠P_{1}OP_{2} + ∠P_{1}OP_{3} = const. Therefore \frac{∂}{∂d_{1}} F (d_{1}, d_{2}, s_{2}) = −\frac{∂}{∂d_{1}} F (d_{1}, d_{2}, 1). Using this fact simple calculus shows that

\frac{\frac{∂}{∂d_{1}} F (d_{1}, d_{2}, s_{2})}{\frac{∂}{∂d_{1}} ϕ(d_{1}, R)} = (\frac{(2d_{1}R)^{2} − (d_{1}^{2} + R^{2} − 1)^{2}}{(2d_{1}d_{2})^{2} − (d_{1}^{2} + d_{2}^{2} − 1)^{2}} )^{1/2} \frac{1 − d_{2}^{2} + d_{1}^{2}}{R^{2} − 1 − d_{1}^{2}} .

It can be easily seen that if d_{1} is in [\sqrt{3}/3, d] and d_{2} in [0.56, \sqrt{3}/3], then both the numerators and denominators are monotone functions of d_{1} as well as d_{2}. Thus, plugging in concrete values for d_{1} and d_{2} we can see that the expression is larger than one. So F (d_{1}, d_{2}, s_{2}) is increasing faster that ϕ(d_{1}, R). We already know that F (d′_{1}, d_{2}, 1) ≤ 2ϕ(R, R)+ϕ(d′_{1}, R)+ ϕ(d_{2}, R) with equality only if d_{2} = \sqrt{3}/3. Thus, if F (f(d_{2}), d_{2}, s_{2})−2ϕ(R, R)−ϕ(f(d_{2}), R)− ϕ(d_{2}, R) < 0, then the same relation holds everywhere in [d′_{1}, f(d_{2})]. Figure 3 shows the graph of the above function for d_{2} ∈ [0.5635, \sqrt{3}/3] drawn by Mathematica. It can be seen that it is zero at d_{2} = \sqrt{3}/3 and negative elsewhere.

R := 1.5148

F[x_{−}] := −RCos[2ArcCos[\frac{−1+R^{2}+x^{2}}{2Rx} ] + 6ArcSin[\frac{1}{2R}]] −

√(1 − R^{2} + R^{2}Cos[2ArcCos[\frac{−1+R^{2}+x^{2}}{2Rx} ] + 6ArcSin[\frac{1}{2R}]]^{2})

Plot[180/Pi(ArcCos[(xˆ 2 + F[x]ˆ 2 − (2 − 2Cos[ArcCos[1/(2x)] + ArcCos[(xˆ 2 − F[x]ˆ 2 + 1)/(2x)]]))/(2xF[x])] − (4ArcSin[1/(2R)] + ArcCos[(xˆ 2 + Rˆ 2 − 1)/(2xR)] + ArcCos[(F[x]ˆ 2 + Rˆ 2 − 1)/(2F[x] R)])), {x, .5635, Sqrt[3]/3}]

Figure 3.

If d_{2} ∈ [0.56, 0.5635] it is not hard to check that F (d, d_{2}, s_{2}) takes on its maximum if d_{2} = 0.5635 and ϕ(d_{2}, R) attains its minimum if d_{2} = 0.56. Thus, by plugging in the actual numbers it turns out that the same relation holds as in [0.5635, \sqrt{3}/3].

Therefore we can conclude that the triangle △P_{1}P_{2}P_{3} is not admissible in any position if any of the sides is longer than 1. Thus, we have finished the proof of Theorem 1. □

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Received October 20, 1998