4

Mine is a very generic question. I believe the answer is yes for a triangle. I don't have a formal proof, just an image in my head.

Is it also true for a higher $n$-gon? Does anyone know of a theorem?

EDIT (following comments): The answer is NO for $n>3$. This is because the lengths of the sides of an $n$-gon do no specify its angles, for $n>3$. E.g., consider the square and the rhombus having the same sides.

FOLLOW-UP question: Why is it this so? What is special about $n=3$, that changes for higher $n$.

General guess to an answer: I am a physicist, so let me use a physicist's argument. The shape of an $n$-gon is specified by its $n$ sides and $n$ angles. That's $2n$ parameters. Somehow, for $n=3$, there are sufficient constraints such that all 3 angles are determined by the 3 sides. For higher $n$, that is not the case, so that we are left with extra parameters, and can get a range of shapes. But why is this so? Someone please fill in the blanks.

ap21
  • 253
  • 1
    It is true for a triangle and false for $n\gt 3$. For example, for $n=4$, making all sides equal to a given length still does not specify the angles, it may be a square or a rhombus. –  Dec 08 '17 at 19:02
  • For triangles, yes. For other polygons, no. For example, we can have a square or a rhombus with sides of length 1. – Xander Henderson Dec 08 '17 at 19:02
  • Great! Then what is the general argument for saying that the sides of a $n$-gon determine its angles only for $n=3$, and not for higher values? – ap21 Dec 08 '17 at 19:05
  • @stressed-out I am indeed talking of shapes up to rigid rotations+translations. But then again, a square and a rhombus do have different shapes. – ap21 Dec 08 '17 at 19:06
  • @ap21 I suppose that we could go by induction: if we have two non-congruent $n$-gons with congruent sides, we can replace any single side on each $n$-gon with a triangle. This gives two non-congurent $(n+1)$-gons with congruent sides. We can do this with $n=4$, which gives us a basis for our induction. – Xander Henderson Dec 08 '17 at 19:09
  • 6
    @ap21 I am a physicist For an informal argument, consider that $n$ points in the 2D plane have $2n$ degrees of freedom. But the "shape" of a polygon is invariant with respect to translations, rotations and reflections, so you are left with $2n-3$ degrees of freedom that define the "shape". In the case of the triangle, that means that the $3$ sides are enough, but the same no longer holds true for $n \gt 3,$. – dxiv Dec 08 '17 at 19:49
  • As a followup to dxiv's comment, note that it takes just one more constraint to determine a shape with $4$ points-one diagonal, for example. With $5$ points that is not enough, you need two. What you are doing is triangulating the shape. – Ross Millikan Dec 08 '17 at 20:48
  • @RossMillikan Is that why, when discretising a surface, we prefer to triangulate it? As opposed to using rhombuses or pentagons. Then fundamental units themselves are structurally stable? – ap21 Dec 08 '17 at 21:45
  • I think it is one reason. Another is that we know we can triangulate any set of points, while you can't make rhombuses because the sides may not match in length. Pentagons (and quadrilaterals) can become concave, which can make problems. – Ross Millikan Dec 08 '17 at 21:56

1 Answers1

3

The question is one of the first asked about the rigidity of bar and joint frameworks. For polygons in the plane the degree of freedom count in the comments settles the question (almost):

$n$ points in the plane have $2n$ independent degrees of freedom. If you subtract the three degrees of freedom corresponding to the Euclidean motions (translation and reflection) that leaves $2n-3$. So most of the time the configuration will be rigid if you add that many bars between pairs of points in some reasonable way. Adding diagonals to triangulate a convex polygon is one of those reasonable ways. The $n$ edges and $n-3$ diagonals do the job.

But there are both mechanical and combinatorial problems you have to avoid. For example, you can add $3$ diagonals to a hexagon but if they involve just $4$ of the vertices you'll still have a motion.

The mathematics to begin to formalize these notions and prove theorems is very elementary physics and a little linear algebra. Since I'm not really "filling in the blanks" I'll mark this answer community wiki.

You can start reading by searching the web for "bar and joint frameworks". Here are some links:

https://en.wikipedia.org/wiki/Structural_rigidity

http://www.math.cornell.edu/~connelly/Connelly-Jordan-Whiteley.pdf

http://maths.nuigalway.ie/~jc/docs/NUIG2.pdf

Ethan Bolker
  • 95,224
  • 7
  • 108
  • 199
  • Fascinating stuff! I also recommend the following two links I just dug up -- http://online.kitp.ucsb.edu/online/sheets16/guest/pdf/Guest_Sheets16_KITP.pdf, and https://arxiv.org/pdf/0803.2325.pdf – ap21 Dec 08 '17 at 21:51
  • 1
    @ap21 It is indeed fascinating. Shameless advertising: http://gallery.bridgesmathart.org/exhibitions/2018-joint-mathematics-meetings/ebolker – Ethan Bolker Dec 08 '17 at 22:03