The convex polygonal planar linkage problem

A special case of the problem of straightening a planar linkage is the case where the linkage is polygonal and convex. A polygonal linkage is a planar linkage where every section of the linkage is a straight line segment (as described by connected steel bars). A convex polygonal linkage is a linkage such that if you connect the two ends of the linkage a convex, simple (non-crossing) polygon is formed. A convex polygonal planar linkage will henceforth be called a convex linkage.





Convex (left) and non - convex (right) polygonal planar linkages

Cauchy's Lemma and its relation to the convex linkage problem

The answer to the question of whether a convex linkage can always be straightened is yes. The proof is a consequence of a result known as Cauchy's Lemma, formulated by the famous French mathematician Augustin Cauchy in 1813.

The lemma states that, given a convex polygon, transform it into another convex polygon while keeping all but one of the sides constant. If some or all of the internal angles at the vertices increase, then the remaining side gets longer. Conversely, if some or all of the internal angles decrease, the remaining side becomes shorter. (The lemma is formalized below)

What does this have to do with straightening a convex polygonal planar linkage? Well, every joint in a convex linkage must be bent at a convex angle when viewed from any portion of the linkage (this is equivalent to saying that the angles of a convex polygon are convex when seen from inside the polygon). So straightening a joint consists of increasing its angle. Also, the linkage has only one flexible side when seen as a polygon -- the side between the two ends (all other sides are steel bars).

Since, by Cauchy's Lemma, increasing internal angles makes a side of the polygon longer, the ends of the linkage will always get farther apart every time you straighten an angle. This means that the ends will never get caught on each other, and you can always completely straighten the linkage.

A convex polyhedron, the subject of Cauchy's Theorem

Cauchy's Lemma, despite its usefulness for linkages, is in fact part of a better known result known as Cauchy's Theorem. This theorem is a result for a similar problem about convex polyhedra (three-dimensional figures where each face is a polygon). The theorem states that:

Two convex polyhedra with corresponding faces equal (congruent) and equally arranged have equal dihedal angles between corresponding faces.

In other words, you cannot change any angles of a convex polyhedron without changing any of the dimensions of its faces.



Cauchy's proof of the lemma

Why is Cauchy's Lemma true? Here is Cauchy's original proof: CAUCHY'S LEMMA: Let us transform a convex polygon (plane or spherical) A 1 A 2 .... A n-1 A n into another convex polygon A 1 'A 2 ' .... A n-1 'A n ' so that the lengths of the sides A 1 A 2 , A 2 A 3 , ...., A n-1 A n are unchanged. If under this transformation the angles at vertices A 2 , A 3 , ...., A n-1 either all increase or some of them increase and the remainder are unchanged, then the length of side A n A 1 increases. On the contrary, if the angles at vertices A 2 , A 3 , ...., A n-1 all decrease or if some of them decrease and the remainder are unchanged, then the length of side A n A 1 decreases.

Cauchy's Proof:

The triangle case of the lemma

The lemma is obvious for triangles. If in triangle ABC side AB and BC are unchanged and the angle at B increases, then the length of side AC increases. (An application of the law of cosines). Similarly, decreasing the angle at B without changing the lengths of AB and BC will decrease AC's length.



Consider the polygon A 1 A 2 .... A n-1 A n (please refer to the figure)



A convex polygon

Assume that the length of all sides except A n A 1 stays constant, and that only one of the angles at vertices A 2 , A 3 , ...., A n-1 changes. In particular, let the angle A i-1 A i A i+1 at vertex A i increase. Form a triangle A 1 A i A n by drawing line segments A 1 A i and A i A n .

But polygons A 1 A 2 .... A i-1 A i and A i A i+1 .... A n-1 A n have enough of their edges and angles kept constant when angle A i is increased to prove that the images of these polygons under the increase are congruent to the originals. In particular, the lengths of segments A 1 A i and A i A n and the angles a and b do not change. But angle c = A 1 A i A n - a - b, and so, since angles a and b are constant, angle c must increase.

Triangle A 1 A i A n has two of its sides, A 1 A i and A i A n constant, but an increasing angle c. Therefore, by the case for triangles, side A 1 A n must increase.

Now, suppose that more than one of the angles of the polygon increase, while the others remain constant. Then we can increase the angles one at time, as follows:

Increase the angle at vertex A i while keeping all the others constant, which causes side A i A n to increase. Then increase the angle at vertex A k , again keeping other angles constant. Side A i A n must again increase. Repeat similarly with the angles at other vertices. So, after some or all of the angles at vertices A 2 , ...., A n-1 are increased while the others remain constant, side A i A n must have increased. By a similar argument, if some or all of these angles are decreased, then side A i A n decreases.

This completes Cauchy's proof of the lemma.

So that's all there is to it, right?

Well no, not exactly. It turns out that Cauchy's proof is wrong.

A counter-example to Cauchy's proof -- the problematic trapezoid

The incorrect assumption is that transforming a convex polygon into another like the lemma states cannot always be done angle by angle without producing a non-convex polygon in an intermediate stage. For example, (as in the figure) increasing the opposite angles of a (convex) trapezoid at the same time will produce another trapezoid with only one side increased. But if you try to increase either one of the opposite angles on its own by the necessary amount you might end up with a concave quadrilateral.

If you have a concave polygon at some intermidiate stage, of course, the lemma no longer applies, and Cauchy has not shown that the complete transformation from the original convex polygon to the final convex polygon obeys the lemma.

Cauchy's proof is a brilliant example of the difficulty of geometric reasoning, created by a world famous mathematician. The proof was corrected, by the German mathematician Steinitz, in 1934, but the corrected version is a horrendously technical induction proof. We will not attempt to give that version here; instead we now give a very simple version proof from Schoenberg and Zaremba that requires no more than high school mathematics.

Cauchy's Lemma: The correct proof

A convex polygon with its image and x and y axes

Given the original polygon, find the vertex A k farthest from the line containing the segment A 1 A n . Construct x and y axes, with segment A 1 A n lying completely on the x - axis and vertex A k on the y - axis. The x coordinate of A n should be larger than that of A 1 .

Let (x i , y i ) be the coordinates of vertex A i .

Now construct the new polygon by keeping A k in place and keeping the y coordinate of each vertex below or equal to y k . In other words, keep A k as an anchor point and increase angle A k symmetrically about the y - axis. If the polygon were to become a straight line segment, it would be horizontal with y = y k .

The length of A 1 A n (as it is on the x - axis) is

x n - x 1 = (x 2 - x 1 ) + (x 3 - x 2 ) + .... + (x n - x n-1 )

It can be seen that when we construct the new polygon, the length of A 1 'A n ', which is at least x n - x 1 , must be larger than A 1 A n . To see this, note that the construction is like rotating each edge left of A k clockwise about some origin, while those to the right of A k are rotated counterclockwise. This must be true because no angles decrease. This always increases (makes more positive) the difference x i - x i-1 .

An edge under rotation

We can illustrate this more fully by drawing a right triangle A i A i-1 P, (see figure) where P = (x i , y i-1 ). Then the rotation is like increasing the angle A i while keeping the hypotenuse of the triangle constant. It is a right triangle, so the opposite side PA i-1 = x i - x i-1 is proportional to cos(A i ), which increases. So each of the x differences increases similarly, and so does the length of A 1 A n .



An identical argument holds for the case where angles decrease, so the lemma is proved.

References

[1] Lyusternik, L.A. Convex Figures and Polyhedra . Trans. Barnett, D. D.C. Heath&co., Boston.