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

Please note that the terminology and concept of this proof is based on that in Lyusternik

Why is Cauchy's Lemma true? Here is Cauchy's original proof:

CAUCHY'S LEMMA: Let us transform a convex polygon (plane or spherical) A1A2 .... An-1An into another convex polygon A1'A2' .... An-1'An' so that the lengths of the sides A1A2 , A2A3 , ...., An-1An are unchanged. If under this transformation the angles at vertices A2 , A3 , ...., An-1 either all increase or some of them increase and the remainder are unchanged, then the length of side AnA1 increases. On the contrary, if the angles at vertices A2 , A3 , ...., An-1 all decrease or if some of them decrease and the remainder are unchanged, then the length of side AnA1 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 A1A2 .... An-1An (please refer to the figure)

A convex polygon

Assume that the length of all sides except AnA1 stays constant, and that only one of the angles at vertices A2 , A3 , ...., An-1 changes. In particular, let the angle Ai-1AiAi+1 at vertex Ai increase. Form a triangle A1AiAn by drawing line segments A1Ai and AiAn.

But polygons A1A2 .... Ai-1Ai and AiAi+1 .... An-1An have enough of their edges and angles kept constant when angle Ai is increased to prove that the images of these polygons under the increase are congruent to the originals. In particular, the lengths of segments A1Ai and AiAn and the angles a and b do not change. But angle c = A1AiAn - a - b, and so, since angles a and b are constant, angle c must increase.

Triangle A1AiAn has two of its sides, A1Ai and AiAn constant, but an increasing angle c. Therefore, by the case for triangles, side A1An 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 Ai while keeping all the others constant, which causes side AiAn to increase. Then increase the angle at vertex Ak, again keeping other angles constant. Side AiAn must again increase. Repeat similarly with the angles at other vertices. So, after some or all of the angles at vertices A2 , ...., An-1 are increased while the others remain constant, side AiAn must have increased. By a similar argument, if some or all of these angles are decreased, then side AiAn 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

Please note that the terminology and concept of this proof is based on that in Singer

A convex polygon with its image and x and y axes

Given the original polygon, find the vertex Ak farthest from the line containing the segment A1An. Construct x and y axes, with segment A1An lying completely on the x - axis and vertex Ak on the y - axis. The x coordinate of An should be larger than that of A1.

Let (xi , yi) be the coordinates of vertex Ai.

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

The length of A1An (as it is on the x - axis) is

xn - x1 = (x2 - x1) + (x3 - x2) + .... + (xn - xn-1)

It can be seen that when we construct the new polygon, the length of A1'An', which is at least xn - x1, must be larger than A1An. To see this, note that the construction is like rotating each edge left of Ak clockwise about some origin, while those to the right of Ak are rotated counterclockwise. This must be true because no angles decrease. This always increases (makes more positive) the difference xi - xi-1.
An edge under rotation

We can illustrate this more fully by drawing a right triangle AiAi-1P, (see figure) where P = (xi , yi-1). Then the rotation is like increasing the angle Ai while keeping the hypotenuse of the triangle constant. It is a right triangle, so the opposite side PAi-1 = xi - xi-1 is proportional to cos(Ai), which increases. So each of the x differences increases similarly, and so does the length of A1An.

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


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

[2] Singer, David A. Geometry: Plane and Fancy. Springer, 1997