Cyclic Evasion in the Three Bug Problem

In this note, we present a simple proof that three bugs involved in cyclic evasion converge to an equilateral triangle configuration. The approach relies on an energy-type estimate that makes use of a new inequality for the triangle. The problem of the cyclic pursuit or n−bug problem is a classical one; see e.g., an article by Klamkin and Newman, “Cyclic pursuit or the three bugs problem” [5]. Because of applications in robotics and availability of more powerful computers, there has recently been some revival of interest in the n-bug problem and various generalizations [3, 4, 6, 7, 9]. The related problem of cyclic evasion, where each bug runs with the unit velocity directly away from one other bug, corresponds to reversing the time. This problem has also received some attention, especially in the computer science literature; see e.g., [2]. It was observed numerically and verified with heuristic arguments, that asymptotically, the n bug configuration converges either to a regular (convex or star) polygon or to a line configuration. In particular, it is widely believed that in the case of three bugs, the limiting configuration is that of an equilateral triangle; however, no proof is available in the literature (to our knowledge). The goal of this expository note is to give a complete and short proof in this simplest case of three bugs1. An apparently new inequality for a triangle is proved and used to establish the convergence. To fix the notation, let each bug be represented by ri(t) = (xi(t), yi(t)) ∈ R2, i = A, B,C , where t is the time parameter. The bugs’ velocities are then given by dri dt = ri − ri+1 |ri − ri+1| , where i + 1 is understood as cyclic shift A → B → C → A. Using elementary geometry (see the Figure 1), one obtains the following equations; see e.g., [1, 5] for the angles and side lengths: ȧ = 1 + cosβ, ḃ = 1 + cos γ, ċ = 1 + cosα, (1) α̇ = sin γ b − sinα c , β̇ = sinα c − sinβ a , γ̇ = sinβ a − sin γ a . (2) If we denote the perimeter by P = a + b + c, then V (a, b, c) = Ṗ = 3 + cosα + cosβ + cos γ 0, http://dx.doi.org/10.4169/amer.math.monthly.122.04.377 MSC: Primary 51N35, Secondary 51N05 1The case n = 2 is completely trivial as two points always belong to a line. April 2015] NOTES 377 Figure 1. The three bug configuration and V̇ = sin 2 α c + sin 2 β a + sin 2 γ b − sinα sinβ c − sinβ sin γ a − sin γ sinα b . Using the law of sines, we replace angles by lengths, and clearing denominators we obtain: V̇ = S 2 4a3b3c3 ( ab + bc + ca − ab − bc − c2a2) , where S is the area of the triangle. It turns out that the fourth order homogeneous polynomial in the brackets is non-negative, provided the three variables are the side lengths in a triangle. Thus, we will have V̇ 0. Proposition. Let a > 0, b > 0 and c > 0 be sides of a triangle, then W (a, b, c) = ab + bc + ca − (ab + bc + ca) 0 (3) and W (a, b, c) = 0 if and only if a = b = c. Proof. Since a, b and c are sides of a triangle, we have a + b c, b + c a and c + a b. Introducing change of variables a = x + y, b = y + z, c = z + x , using that x, y, z 0, and substituting in the equation, we obtain W = x y + yz + zx − xyz(x + y + z). Next, we use the weighted inequality for arithmetic and geometric mean [8]: w1 F1 + w2 F2 + w3 F3 w1 + w2 + w3 w √ F1 1 F w2 2 F w3 3 , (4) 378 c © THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 122 where w = w1 + w2 + w3 and equality occurs only if F1 = F2 = F3, for positive weights. We first prove that w1x y + w2 y3z + w3zx w1 + w2 + w3 x 2 yz using the appropriate weights: ⎧⎨ ⎩ 3w1 + w3 = 2w1 + 2w2 + 2w3 w1 + 3w2 = w1 + w2 + w3 w2 + 3w3 = w1 + w2 + w3. Hence, for w1 = 4, w2 = 1, w3 = 2, we get 4 7 x y + 1 7 yz + 2 7 zx x yz. Permuting the variables twice, we obtain two more inequalities. Adding all three inequalities, we obtain the desired one. The equality occurs only if x3 y = y3z = z3x , which is equivalent to x = y = z, provided xyz = 0. Note that if at least one variable vanishes, say x = 0, and equality occurs, then y3z = 0 and either a = x + y = 0 or c = z + x = 0, contradicting our assumption. Clearly, V̇ = 0 for the line configuration because S = 0, and V̇ = 0 for the equilateral triangle because a = b = c and W vanishes. Now, we prove convergence to the equilateral triangle. Theorem 1. Suppose the initial configuration of the bugs A(0), B(0),C(0) is not a line configuration, then as t → ∞, A(t), B(t),C(t) approach equilateral configuration. Proof. If the initial configuration is an equilateral triangle a = b = c, α = β = γ , then we are done. Thus, we assume the triangle is nondegenerate and not equilateral. The function V is defined in the set D = {α > 0, β > 0, γ > 0, α + β + γ = π}, which is itself a triangle. In the interior of D, the function V has a single extremum α = β = γ = π/3 as can be verified by the method of Lagrange multipliers, which leads to the relation sinα = sinβ = sin γ ⇒ α = β = γ = π/3. We used the constraint α + β + γ = π and that the angles do not vanish. It is easy to check that V (π/3, π/3, π/3) = 9/2 is the only maximum of V in D. We claim that an orbit starting in the interior of D, will stay away from the boundary of D. Substituting γ = π − α − β, we have V (α, β) = 3 + cosα + cosβ − cos(α + β). April 2015] NOTES 379 Because of the threefold symmetry of the domain D, we only need to consider one segment of the boundary of D, e.g., γ = 0. At γ = 0, we have ∂V ∂α (γ = 0) = − sinα, ∂V ∂β (γ = 0) = − sinβ, (5) which implies that V increases in any direction inside D from any point on the boundary segment γ = 0 provided α, β = 0, π . Near the corners of D, such degeneracy actually takes place. Consider then a neighborhood of the corner α = 0, β = 0. Using Taylor expansions, we have V (α, β) = 4 + αβ + o(α + β), which together with (5) proves that V increases in any direction inside D from any point on the boundary. Since V̇ > 0 in the interior of D away from α = β = γ = π/3, no orbit can approach the boundary. Thus, V (t) increases and V (t) → V0 ≤ Vmax. Then we must also have V̇ (t) → 0, and there exists a sequence of times tn → ∞ such that (α(tn), β(tn), γ (tn)) → (π/3, π/3, π/3). Then V (t) → Vmax and the orbit (α(t), β(t), γ (t)) must converge to (π/3, π/3, π/3). ACKNOWLEDGMENT. This work was partially supported by a grant from the Simons Foundation (#278840 to Vadim Zharnitsky) and AFOSR MURI grant FA9550-10-1-0567. The authors would like to thank Yuliy Baryshnikov for several helpful discussions.