Polygons Inscribed In Polygons

Geometry Level pending

An analogy of the problem below is the octagon inscribed in the 16-gon above.

Let n n be a positive integer and n ≥ 3 n \geq 3 .

Let A 1 , A 2 , . . . , A 2 n A_{1}, A_{2}, ...,A_{2^n} be the vertices of a regular 2 n 2^n -gon.

Connect the midpoints m ( j ) = A 2 j − 1 + A 2 j 2 m(j) = \dfrac{A_{2j - 1} + A_{2j}}{2} , where ( 1 ≤ j ≤ 2 n − 1 ) (1 \leq j \leq 2^{n - 1}) , to form a 2 n − 1 2^{n - 1} -gon.

Let A 2 n A_{2^n} be the area of the 2 n 2^n -gon and A 2 n − 1 A_{2^{n - 1}} be the area of the 2 n − 1 2^{n - 1} -gon.

Find A 2 n − 1 A 2 n \dfrac{A_{2^{n - 1}}}{A_{2^n}} , then using n = 10 n = 10 find A 512 A 1024 \dfrac{A_{512}}{A_{1024}} to 9 9 decimal places.


The answer is 0.999971763.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Dec 30, 2017

For 2 n 2^n -gon:

Let B C = x BC = x be a side of the 2 n − g o n 2^n - gon , A C = A B = r AC = AB= r , A D = h AD = h , and ∠ B A D = π 2 n \angle{BAD} = \dfrac{\pi}{2^n} .

x 2 = r sin ⁡ ( π 2 n ) ⟹ r = x 2 sin ⁡ ( π 2 n ) ⟹ h = x 2 cot ⁡ ( π 2 n ) ⟹ A 2 n = 2 n − 2 c o t ( π 2 n ) x 2 \dfrac{x}{2} = r\sin(\dfrac{\pi}{2^n}) \implies r = \dfrac{x}{2\sin(\dfrac{\pi}{2^n})} \implies h = \dfrac{x}{2}\cot(\dfrac{\pi}{2^n}) \implies A_{2^n} = 2^{n - 2} cot(\dfrac{\pi}{2^n}) x^2 .

For 2 n − 1 2^{n - 1} -gon:

Let B C = L BC = L and A D = h ∗ AD = h^{*} .

A C = A B = h = x 2 cot ⁡ ( π 2 n ) AC = AB = h = \dfrac{x}{2}\cot(\dfrac{\pi}{2^n}) and ∠ B A C = π 2 n − 1 + π 2 n + π 2 n = π 2 n − 2 ⟹ ∠ B A D = π 2 n − 1 ⟹ \angle{BAC} = \dfrac{\pi}{2^{n - 1}} + \dfrac{\pi}{2^n} + \dfrac{\pi}{2^n} = \dfrac{\pi}{2^{n - 2}} \implies\angle{BAD} = \dfrac{\pi}{2^{n - 1}} \implies . L = x cot ⁡ ( π 2 n ) sin ⁡ ( π 2 n − 1 ) ⟹ h ∗ = x 2 cot ⁡ ( π 2 n ) cos ⁡ ( π 2 n − 1 ) ⟹ A 2 n − 1 = 1 2 ∗ ( 2 n − 3 cot ⁡ 2 ( π 2 n ) sin ⁡ ( π 2 n − 2 ) x 2 ) ⟹ L = x\cot(\dfrac{\pi}{2^n})\sin(\dfrac{\pi}{2^{n - 1}}) \implies h^{*} = \dfrac{x}{2} \cot(\dfrac{\pi}{2^n})\cos(\dfrac{\pi}{2^{n - 1}}) \implies A_{2^{n - 1}} = \dfrac{1}{2} * (2^{n - 3}\cot^2(\dfrac{\pi}{2^n})\sin(\dfrac{\pi}{2^{n - 2}}) x^2) \implies

A 2 n − 1 A 2 n = 1 4 ∗ ( cot ⁡ ( π 2 n ) sin ⁡ ( π 2 n − 2 ) ) \dfrac{A_{2^{n - 1}}}{A_{2^n}} = \dfrac{1}{4} * (\cot(\dfrac{\pi}{2^n}) \sin(\dfrac{\pi}{2^{n - 2}})) .

Rewriting this as A 2 n − 1 A 2 n = cos ⁡ 2 ( π 2 n ) cos ⁡ ( π 2 n − 1 ) \dfrac{A_{2^{n - 1}}}{A_{2^n}} = \cos^2(\dfrac{\pi}{2^n})\cos(\dfrac{\pi}{2^{n - 1}}) ⟹ lim ⁡ n → ∞ A 2 n − 1 A 2 n = 1. \implies \lim_{n \rightarrow \infty} \dfrac{A_{2^{n - 1}}}{A_{2^n}} = 1.

For n = 10 ⟹ A 512 A 1024 = 0.999971763 n = 10 \implies \dfrac{A_{512}}{A_{1024}} = \boxed{0.999971763}

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