• P is a convex, near regular 2023-sided polygon.
• Precisely 2022 of its sides have length 1, but the remaining side has a length different from 1.
Determine the maximum area of the polygon.
Let X_1 be the polygon with the greatest area, with L being the length of the divergent side.
Construct a congruent polygon X_2. Lay X_1 and X_2 beside each other so that the divergent sides (both of length L) are aligned. This gives a 4044 sided regular polygon.
If follows that the area of X_1 is one half the area of a 4044 sided regular polygon.
That was the insight. Now for the calculation:
Let Y be a 4044 sided regular polygon with side length 1. Let Z be the circle circumscribing Y. Let W be the triangle with vertices at U (= the centre of Z) and any two adjacent points of the polygon. The angle W measured at U is v = 2 pi / 4044. Then sin( v/2 ) = (1/2) / r, where r is the radius of the circle. So r = 1/2 csc (v/2), and the area of W is given by
K_T = (1/2) * 1 * r cos (v/2) = (1/4) cot (v/2).
The area of the polygon is one half the area of 4044 such triangles
K_P = (1/2) 4044 (1/4) cot (pi / 4044) = (1011 / 2) cot (pi / 4044)
= 505.5 * (1287.245) = 650702 (near enough)
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Posted by FrankM
on 2023-05-31 13:42:04 |
Solution and proof
