Determine the side length of a regular heptagon having the largest area which can fit inside a square with side length 1.
(In reply to
re: Some thoughts -- a solution by Charlie)
"shows that the optimal tilt is 45/7 of a degree"
We can rotate the square by 360/7 degrees and have a net effect of zero since the heptagon is rotationally symmetric at 360/7 degrees. What this rotation actually accomplishes is situating the square on a different set of vertices of the heptagon.
So then lets apply this rotation 6 times. Then the square is oriented 45/7 + 6*260/7 = 315 degrees = -45 degrees. This confirms what I suspected from the beginning - the optimal placement has the heptagon lined up with a diagonal of the square (one of the altitudes of the heptagon coincides with a diagonal of the square).
re(2): Some thoughts -- a solution
