Problem of the Week 1151

Construct a Regular 7-Gon

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Solution Notes

Solutions were received by Abijah Reed (Vermont), Michael Elgersma (Minneapolis), Barry Cox (Adelaide, Australia), and Aaron Laursen (Macalester student). One incorrect solution was received. Reed made a physical model, Elgersma worked hard on finding a system of equations that would reveal the algebraic structure, and Cox found equations that yielded numerical solutions quickly, thus allowing a neat computer animation:

Here are two views of the 3-D image:

And here is Mathematica code to generate those views:

pts = {{-(1/2), 0, 0}, {1/2, 0, 0}, {1/2, 1, 0}, {1/Sqrt[2], 1, 
    Sqrt[1/4 + 1/Sqrt[2]]}, 
     {0, 1 + Sqrt[(1/7)*(9 - 4*Sqrt[2])], Sqrt[13/28 + (4*Sqrt[2])/7]}, 
     {-(1/Sqrt[2]), 1, Sqrt[1/4 + 1/Sqrt[2]]}, {-(1/2), 1, 0}};

prs = Partition[Append[pts, pts[[1]]], 2, 1];
r = 0.1;
sol = Graphics3D[{EdgeForm[], (Cylinder[#1, r] &) /@ prs, 
   FaceForm[Yellow], (Sphere[#1, r] &) /@ pts}, Boxed -> False]

The coordinates of the 7 points can be taken to be: {{-(1/2), 0, 0}, {1/2, 0, 0}, {1/2, 1, 0}, {1/Sqrt[2],1, -Sqrt[1/4 + 1/Sqrt[2]]}, {0, 1 + Sqrt[(1/7)*(9 - 4*Sqrt[2])], -Sqrt[13/28 + (4*Sqrt[2])/7]}, {-(1/Sqrt[2]), 1, -Sqrt[1/4 + 1/Sqrt[2]]}, {-(1/2), 1, 0}}

Alternatively, the other symmetric solution is given by {{-(1/2), 0, 0}, {-(1/2), 1, 0}, {1/2, 1, 0}, {1/2, 0, 0}, {1/Sqrt[2], 0, (1/2)*Sqrt[1 + 2*Sqrt[2]]}, {0, (1/7)*Sqrt[63 - 28*Sqrt[2]], (3 + Sqrt[2])/(2*Sqrt[1 + 2*Sqrt[2]])}, {-(1/Sqrt[2]), 0, (1/2)*Sqrt[1 + 2*Sqrt[2]]}}

As always, there are many things to investigate. Michael Elgersma has been thinking about the general solution. And Barry Cox has also been working on efficient equations for generating all the solutions. The Elgersma notes can be downloaded from http://stanwagon.com/public/MichaelElgersmaNotes1151.pdf

An elegant equation approach (Barry Cox) is the following. We can specify three of the points a, b, g as follows:

   a = {0,0,0};   b = {1,0,0};   g = {0,1,0};

Then the rod that leaves b to a new point can be parametrized in terms of angle A, so that point c can be taken to be

   c = {1, Cos[A], Sin[A]};

One can get the remaining three points similarly.

   f = {Cos[C], 1, Sin[C]};
   d = {1 + Cos[B], Cos[A] - Sin[A] Sin[B], Sin[A] + Cos[A] Sin[B]};
   e = {Cos[C] - Sin[C] Sin[D], 1 + Cos[D], Sin[C] + Cos[C] Sin[D]};

Now each vector between points is guaranteed to be a unit vector, except for de, so one need only specify that condition and the remaining two right angles, as follows.

   (d-e) . (d-e) = 1          (c-d) . (d-e) = 0               (d-e) . (e-f) = 0

So when we view this system as having the parameter A and use a specific value such as A = π/2, we find that the solution involves a root of a degree-16 polynomial. But for general A, Barry is unable to get an algebraic expression.

He used a numerical technique to obtain two distinct one-dimensional families of solutions. In the first one, the angle A runs from 0 to 1.9 and down to -1.9 and back up to 0, in rough sine-curve form. The second family has A running from -2.51 to 2.51, but it has extra dips, so is more complicated than the simplest wave. One cannot get, physically, from a solution in one family to a solution in the other. Whether these two families exhaust all solutions is an open question.

I have prepared a Mathematica file which shows the motion through the two solution families in a very smooth way, and so that one can rotate the configuration in three-dimensions. It is hypnotizing to watch this, so if you have Mathematica available I encourage you to take a look. If you wish to see this but do not have Mathematica, just ask me for the data points and you can then use your favorite software to make a 3D movie from it.

The subject of right-angled configurations such as these is well studied; John Sullivan points to the following paper, which contains a discussion of the two families of solutions shown in the aforementioned movie.

S. D. Guest and P. W. Fowler, "Mobility of 'N-loops' : bodies cyclically connected by intersecting revolute hinges," Proceedings of the Royal Society A, 8 January 2010, vol. 466 no.2113, 63-77

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© Copyright 2012 Stan Wagon. Reproduced with permission.