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The longest side can be no more than 12 units. One way of making such a decagon is with sides of length 12, 11, 5, 3, 10, 9, 7, 8, 4, 6, going clockwise. The 120 degree angles would be between the sides with lengths 5 and 3, 4 and 6.
Proof (this will be elegant, but difficult):
First, rather than considering a decagon, consider an equiangular 12-sided polygon, except two of the sides have zero lengths. Now, rather than considering 12 sides, consider 12 vectors, and all their directions are multiples of 30 degrees. We can move the vectors so they all come from a common center. The sum of all the vectors must be zero.
We can divide the vectors into two separate sets of 6 vectors each, where each vector is a multiple of 60 degrees apart from the other vectors in the same set. Each set of vectors must have a sum of zero, since with integer length vectors, it is impossible for the sum of one set to cancel the other.
The diagram below represents either of the sets of vectors. If we increase the magnitude of vector A by 1, we must balance it out by either increasing the magnitude of D by one, or the magnitudes of E and C by one each.
A
F B
E C
D
The two zero vectors cannot be in the same set, since that would force two sides to have equal lengths. No matter which pair of vectors in the diagram above represent zero vectors, another pair of vectors are forced to have equal magnitudes.
Therefore, each set of 6 vectors must contain exactly one zero vector. We can assume without loss of generality that A is the zero vector 0.
F=D+C and B=E+D must be true for the vectors to sum zero. Vectors C, D, and E can have lenghths no greater than 9 because this is the greatest difference between the lengths of two sides. This is also true for the other set of vectors. So there must be at least 6 sides that are less than 10 units in length. The longest side can be no more than 13.
All that remains to show that the longest side cannot be 13. If the longest side were 13 then the two sums shown above would not be solvable for both sets of vectors. I leave it to the reader to confirm this.
The equations are solvable if the longest side is 12, so 12 is the maximum. |
