2N-Wing Butterfly Problem

This is an illustration of the 2N-Wing Butterfly problem that is discussed below.

2N-Wing Butterfly Theorem

Observe a 2N-wing butterfly cradle. Let M denote the nexus. Assume the spikes are crossed by the circles in points A_N, ..., A₁, (M), B₁, ..., B_N and C_N, ..., C₁, (M), D₁, ..., D_N, respectively. Let p be any permutation of the index set {1, ..., N}. Finally, for i = 1, ..., N, let X_i and Y_i denote the points where A_iD_p(i) and, respectively, C_iB_p(i) cross the classifier. Then

(1) 1/MX1 + ... + 1/MXN = 1/MY1 + ... + 1/MYN.

Note that for N = 1 the theorem reduces to (A). The proof depends on two lemmas.

Lemma 1

Let in RST, RU be a cevian through vertex R. Introduce angles a = SRU and b = URT. Then

(2) sin(a + b)/RU = sin(a)/RT + sin(b)/RS.

The proof is a two-liner that follows from the identity

Area(RST) = Area(RSU) + Area(RUT).

Lemma 2 is a curious result in its own right.

Lemma 2

Assume three rays C, B, and Y emanate from point M with Y between C and B. Assume also that, for N > 1, points C1, ..., CN have been marked on C, while points B1, ..., BN have been marked on B. Let p be any permutation of indices 1, ..., N. Denote the intersection of CiBp(i) with Y as Yi. Then the sum of the reciprocals 1/MY1 + ... + 1/MYN does not depend on p.

The proof builds on Lemma 1 which is applied to each of the triangles MC_iB_p(i):

(3)	sin(b + g)/MYi = sin(g)/Bp(i) + sin(b)/MCi,

g and b are the angles CMY and YMB, respectively. Add up all the identities (3). After rearranging the terms, we'll get

(4)	sin(b + g)(1/MY1 + ... + 1/MYN) = sin(g)(1/B1 + ... + 1/BN) + sin(b)(1/MC1 + ... + 1/MCN),

in which the right hand side does not depend on the permutation p. The same therefore is true for the left hand side.

Proof of Theorem

The proof easily follows from Lemma 2 by pidgin, or is it pigeon, induction (if a statement holds for N = 1, it holds for any N.) Apply the Butterfly theorem (A) to the i^th circle, i.e the triangles MA_iD_i and MC_iB_i, i = 1, ..., N. If the points of intersection of A_iD_i and C_iB_i with the classifier are denoted as, say, S_i and T_i, (A) assures us that MS_i = MT_i, and therefore 1/MS_i = 1/MT_i. Adding up we get

But due to Lemma 2,

and the result follows.

Remark

The theorem raises further questions. Each permutation can be represented as a product of irreducible cycles. It must be obvious that every such cycle is responsible for forming a unique butterfly. If the permutation is cyclic, the theorem yields a lone insect. However, in general, the N pairs of wings are split between a number of butterflies, one per an irreducible cycle that compose the permutation. This is an open question whether the butterflies are just stuck on top of each other and could be in principle separated, or whether their wings are so entangled that no separation is possible.

L. Bankoff was amazed with a fantastic variety of solutions to the Butterfly problem -- "some by Desargues' theorem on involution, some by the use of cross ratios, others stemming from Menelaus, analytic geometry, trigonometry, advanced Euclidean geometry, and various other modalities ..." May the braids theory hold the answer to the problem of separation of the entangled butterflies.

Butterfly Theorem and Variants

1. Butterfly theorem
2. 2N-Wing Butterfly Theorem
3. Better Butterfly Theorem
4. The Lepidoptera of the Circles
5. The Lepidoptera of the Quadrilateral
6. The Lepidoptera of the Quadrilateral II
7. Butterflies in Quadrilaterals and Elsewhere
8. Two Butterflies Theorem
9. Two Butterflies Theorem II
10. Two Butterflies Theorem III

Desargues' Theorem

• Desargues' Theorem
• 2N-Wing Butterfly Problem
• Cevian Triangle
• Do You Speak Mathematics?
• Desargues in the Bride's Chair (with Pythagoras)
• Menelaus From Ceva
• Monge from Desargues
• Monge via Desargues
• Nobbs' Points, Gergonne Line
• Soddy Circles and David Eppstein's Centers
• Pascal Lines: Steiner and Kirkman Theorems II
• Pole and Polar with Respect to a Triangle
• The Lepidoptera of the Circles

Copyright © 1996-2008 Alexander Bogomolny