Stiv's Diabolical Instruments now offers a bundle of exactly 47 equal-length rods that can be joined by hinges at their ends – and only the ends – to form planar linkages (i.e. all hinge axes are normal to the plane containing the rods and rod thickness is neglected).
With 35 of these rods you can make a rigid linkage containing the vertices of a regular heptagon:
Now you want to do the same for a regular nonagon. But the best known bracing of a regular nonagon uses 51 rods, more than you have from one bundle:
Nevertheless it is possible to form a rigid linkage containing the vertices of a regular nonagon using only 47 rods by imagining that you have 7 more rods, making a fully symmetric ($D_{18}$ symmetry, same as that of the nonagon) rigid linkage with those 54 rods and removing 7 redundant rods. What does the 47-rod linkage look like?
Searching my MathWorld contributions will help (MathWorld Contributors -> Tan).
