Let a, b, and c be the lengths of a side, a shortest diagonal, and a longest diagonal, respectively, of a regular nonagon (9 sided polygon).
Find an equation that relates a, b, and c.
This is very much in the form of a contest problem. So no calculators of any kind!
Take two triangles from the nonagon: one triangle consisting of two sides length a and one side length b; and one triangle consisting of two sides length b and one side length c.
In the first triangle the angle between the two a sides is 140 degrees. Apply the law of cosines to the triangle at this angle:
b^2 = 2a^2 - 2a^2*cos(140)
Isolate the cos(140):
1 - b^2/(2a^2) = cos(140)
In the second triangle the angle between the c side and one of the b sides is 40 degrees. Apply the law of cosines to the triangle at this angle:
b^2 = c^2 + b^2 - 2bc*cos(40)
Isolate the cos(40):
c/(2b) = cos(40)
One more identity: cos40 = -cos(140). Use this to equate the two results from earlier. Then
b^2/(2a^2) - 1 = c/(2b)
Simplify to the the solution:
b^3 = 2a^2*b + a^2*c.
Solution
