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.

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.

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.

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. *

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)

Assume that the length of all sides except
*A _{n}A_{1}*
stays constant, and that only one of the angles at vertices

But polygons *A _{1}A_{2}
.... A_{i-1}A_{i}* and

Triangle *A _{1}A_{i}A_{n}* has two of
its sides,

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

This completes Cauchy's proof of the lemma.

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

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.

Given the original polygon, find the vertex *A _{k}*
farthest from the line containing the segment

Let (*x _{i}* ,

Now construct the new polygon by keeping *A _{k}* in
place and keeping the

The length of *A _{1}A_{n}* (as it is on the

It can be seen that when we construct the new polygon, the
length of *A _{1}'A_{n}'*, which is at least

We can illustrate this more fully by drawing a right triangle
*A _{i}A_{i-1}P*, (see figure) where

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