There is a decagon with eight 150 degree angles and two 120 degree angles. The lengths of its sides are a set of ten consecutive integers. Maximize the length of its longest side.
Whatever the configuration, it will be possible to place a side at the top, horizontally, such that one of the 120-degree angles is among the 5 angles to the left (counterclockwise) from the top side, one the other is among the 5 angles to the right (clockwise) from the top side. The result will be that there will be a side at the bottom, also horizontal and therefore parallel to the top. Also, there will be 4 sides on the left and 4 sides on the right.
Each side will be placed at an angle that results in a lowering, from one end to the other, of .5, sqrt(3)/2 or 1 times its own length. In fact the sequence of multipliers, on one side of the decagon, of the sides' lengths will be .5, sqrt(3)/2, 1, sqrt(3/2), .5, but with one missing, after the sole 120-degree angle on that side of the decagon.
The left and right sides of the decagon cannot both have one of the sqrt(3)/2 multipliers left out, leaving one of them on each side, as each of the remaining ones would then be multiplied by a different value in the sequence of ten consecutive integers, and there would be no irrational value to offset it. This leaves the possibility that each side still has two sqrt(3)/2 multipliers, with two side lengths on one side that add up to the two side lengths on the other, or one side of the decagon has two sqrt(3)/2 multipliers and the other has one, with the corresponding side length on the one equalling the sum of the two corresponding side lengths on the other.
Similar considerations apply for the rational values of .5, 1 and the other .5 one either side.
Posted by Charlie
on 2005-01-30 22:27:41