Determine the side length of a regular heptagon having the largest area which can fit inside a square with side length 1.
Well sort of exact. I left them in terms of sines of angles with sevenths in them.
Call the heptagon ABCDEFG with center O and the square WXYZ. The optimum position is with one diagonal of the square WY containing A and bisecting side DE. B is on WX, D is on XY, E is on YZ and G is on ZW.
Let r be the radius of heptagon's circumcircle.
OB is not quite perpendicular to WX. Instead it is 2pi/7-pi/4 = pi/28 radians past perpendicular. OE is another 3*2pi/7 past this or 25pi/28.
The full width of the heptagon across the square is then r[cos(pi/28)-cos(25pi/28)] = 1.
So r=1/[cos(pi/28)+cos(3pi/28)] or about 0.5161
The side length is 2*r*sin(pi/14)
So the final solution is given by the expression
s =2*sin(pi/14)/[cos(pi/28)+cos(3pi/28)] = 0.4478579...
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Posted by Jer
on 2023-03-08 14:39:47 |
Exact solution
