# Cut The Knot!

An interactive column using Java applets
by Alex Bogomolny

# A Case of Similarity

December 2001

In his marvelous In Pólya's Footsteps (MAA, 1999), Ross Honsberger describes and proves a theorem due to the nineteenth-century British mathematicians William Kingdom Clifford and Arthur Cayley. The theorem, although called the Fundamental Theorem of 3-Bar Motion, is apparently little known. Indeed, Honsberger credits his friend Mike McKierman for bringing this gem to his attention. Moreover, the theorem relates to a more fortunate statement popularized by M. Gardner as the Asymmetric Propeller. In progressively more general forms, the latter was offered at the 1968 Putnam Competition, discussed by Leon Bankoff, Paul Erdös and Murray Klamkin in 1973 and by M. Gardner in 1999. Gardner dates the theorem to the early 1930s, but indicates that its origins are unknown. Nowhere the Clifford-Cayley theorem is mentioned. Which is amazing because the two theorems are equivalent — a fact which is impossible to miss when the diagrams for the two theorems are placed side by side.

The Fundamental Theorem of 3-Bar Motion (I can only guess at why it is called that. Because of having 3 bars per side?) describes a linkage of 15 rods and 10 swivel-joints. The simplest configuration (it can of course be distorted by moving the rods) is that of a triangle ABC with a point 7 through which three lines are drawn parallel to the sides of the triangle. The points of intersection are denoted 1, 2, 3, 4, 5, and 6. The lines form three triangles 127, 347, 567, all similar to ABC, and three parallelograms. While the whole configuration may change, triangles 127, 347, 567 remain the same in shape and size. In the deformation, the parallelograms always remain parallelograms, because their opposite sides remain equal, but the angles change.

The Fundamental Theorem of 3-Bar Motion asserts that the triangle ABC, while it may change in size, always keeps its original shape. (In particular, this means that if two of the vertices ABC are fixed, the third one can't move either.) In other words, the triangle ABC always remain similar to the small triangles 127, 347, 567.

The Asymmetric Propeller theorem appeared over time in several variants, successively more general. The first dealt with three congruent equilateral triangles OAB, OCD, OEF. It claimed that the midpoints of segments BC, DE, FA form another equilateral triangle.

In 1973, Bankoff, Erdös and Klamkin noticed that the triangles need not be congruent. As long as the three triangles are equilateral, the "midpoint" triangle is also equilateral.

Bankoff later proved two further generalizations (see [Gardner]). First he showed that the triangles need not be equilateral. Suffice it to request that they be similar — of the same shape. This is the variant equivalent to the Clifford-Cayley theorem. (In the latest generalization he removed the condition that the triangles share a point. Does that correspond to a more fundamental analogue of the Clifford-Cayley theorem? I do not know.)

The equivalence of the two statements is immediate. Indeed, let X, Y, Z denote the midpoints of segments 61, 23, 45, respectively. In the parallelogram A176, X is the midpoint of the diagonal 61 and hence that of the diagonal 7A. In other words, X is midway between A and 7. Similarly, YB = 7Y and ZC = 7C. It follows, that ABC is similar to XYZ. But XYZ is exactly the triangle that appears in the Asymmetric Propeller theorem.

The variants 1 [AMM] and 2 [Bankoff, Erdös, M. Klamkin] of the Asymmetric Propeller are easily proven with complex variables. Synthetic proofs were given by Bankoff, Erdös and M. Klamkin for the variant 2 and by Bankoff for the variant 4, and hence 3.

Honsberger's proof of the Clifford-Cayley theorem makes an elegant use of spiral similarities (see also [Yaglom]).

Let S be the spiral similarity that maps, say, 56 onto 57. S is an affine mapping of the plane that rotates the plane as a whole such that all lines form the same angle with their respective images, namely, 657, which is equal to the original angle A. All segment lengths are modified by S by the same factor. Thus S induces an operator S on vectors in the plane.

We have the following vector identities: AB = A5 + 56 + 6B and AC = A4 + 43 + 3C. Apply S to AB:

 S(AB) = S(A5 + 56 + 6B) = S(A5) + S(56) + S(6B) = S(47) + 57 + S(71) = 43 + A4 + 72 = A4 + 43 + 3C = AC,

which exactly means that the angle BAC at A is equal to 657 and, also, 56/57 = AB/AC, or that the triangles 567 and ABC are similar.

At this point the two problems seem to part ways. They are equivalent, but not quite. Bankoff's latest generalization of the Asymmetric Propeller problem

does not seem to have an analogue with the Fundamental Theorem of 3-Bars Motion. However, we may adapt the technique used in the proof of the latter to establish Bankoff's generalization. The simplicity (if not the elegance) of Honsberger's proof is carried over.

As a matter of fact, the proof appears notationally simpler if vectors are identified with complex numbers. Spiral similarities are obtained as products by a complex number. The proof differs from the known proofs of the variants 1 and 2, in that the latter were specifically designed to demonstrate that the "midpoint" triangle is equilateral. Honsberger's proof shows that establishing similarity between triangles is a more fundamental and, hence, a more straightforward task.

In the above diagram, I placed the origin O at one of the vertices and identified the straight line segments as complex numbers directed (unconventionally) from a (small yellow) square to a (small yellow) circle. X, Y, Z are the vertices of the "midpoint" triangle. The four given triangles are similar to a generic triangle with sides 1, k, and (1-k), where k is a complex number. The factors a, b, c, and d define the rotation and size of the triangles. We have

1. X = ((ka) + (a - c + kd))/2,
2. Y = (a - (1-k)c - kb)/2,
3. Z = ((a - c + d) + (a - (1-k)c + (1-k)b))/2,

from where we can compute the sides

1. YZ = Z - Y = (a + b - c + d)/2,
2. YX = X - Y = k(a + b - c + d)/2,
3. XZ = Z - X = (1-k)(a + b - c + d)/2.

So that triangle XYZ is indeed similar to the generic triangle (1, k, (1-k)).

### References

Alex Bogomolny has started and still maintains a popular Web site Interactive Mathematics Miscellany and Puzzles to which he brought more than 10 years of college instruction and, at least as much, programming experience. He holds M.S. degree in Mathematics from the Moscow State University and Ph.D. in Applied Mathematics from the Hebrew University of Jerusalem. He can be reached at alexb@cut-the-knot.com