### 2N-Wing Butterfly Problem

This is an illustration of the 2N-Wing Butterfly problem that is discussed below. BTW, the only browser that runs Java is Forefox. Please declare the site cut-the-knot.org as trusted in Java setup

 What if applet does not run?

### 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 AN, ..., A1, (M), B1, ..., BN and CN, ..., C1, (M), D1, ..., DN, respectively. Let p be any permutation of the index set {1, ..., N}. Finally, for i = 1, ..., N, let Xi and Yi denote the points where AiDp(i) and, respectively, CiBp(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 ΔRST, 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( ΔRST) = Area( ΔRSU) + Area( ΔRUT).

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 MCiBp(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 ith circle, i.e the triangles MAiDi and MCiBi, i = 1, ..., N. If the points of intersection of AiDi and CiBi with the classifier are denoted as, say, Si and Ti, (A) assures us that MSi = MTi, and therefore 1/MSi = 1/MTi. Adding up we get

 1/MS1 + ... + 1/MSN = 1/MT1 + ... + 1/MTN.

But due to Lemma 2,

 1/MS1 + ... + 1/MSN = 1/MX1 + ... + 1/MXN, while 1/MT1 + ... + 1/MTN = 1/MY1 + ... + 1/MYN,
 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 ### Desargues' Theorem 