Nested N-Grams and Circles

Geometry Level 3

Let n ≥ 5 n \geq 5 be a fixed odd positive integer and extend the above pentagrams to n n -grams.

For each positive integer m m , let P m P_{m} be the n n -gram inscribed in the circle C m C_{m} and C m + 1 C_{m + 1} be the circle inscribed in the n n -gram P m P_{m} .

Let r m r_{m} be the radius of each circle C m C_{m} .

If R ( n ) = ∑ m = 1 ∞ r m \displaystyle R(n) = \sum_{m = 1}^{\infty} r_{m} and R = lim ⁡ n → ∞ R ( n ) \displaystyle R = \lim_{n \to \infty} R(n) , find R r 1 \dfrac{R}{r_{1}} .


The answer is 1.5.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Mar 8, 2020

Let n ≥ 5 n \geq 5 be a fixed positive odd integer.

m ∠ A O P = π n m\angle{AOP} = \dfrac{\pi}{n} and by Inscribed Angle Theorem m ∠ E P F = 1 2 m ∠ E O F ⟹ m\angle{EPF} = \dfrac{1}{2}m\angle{EOF} \implies

m ∠ A P O = π 2 n m\angle{APO} = \dfrac{\pi}{2n}

A C O C = tan ⁡ ( π n ) ⟹ O C = A C tan ⁡ ( π n ) \dfrac{AC}{OC} = \tan(\dfrac{\pi}{n}) \implies OC = \dfrac{AC}{\tan(\dfrac{\pi}{n})} and

P C = A C tan ⁡ ( π 2 n ) = r 1 − O C = r 1 − A C tan ⁡ ( π n ) ⟹ PC = \dfrac{AC}{\tan(\dfrac{\pi}{2n})} = r_{1} - OC = r_{1} - \dfrac{AC}{\tan(\dfrac{\pi}{n})} \implies

r 1 = A C tan ⁡ ( π n ) + A C tan ⁡ ( π 2 n ) ⟹ r_{1} = \dfrac{AC}{\tan(\dfrac{\pi}{n})} + \dfrac{AC}{\tan(\dfrac{\pi}{2n})} \implies A C ( tan ⁡ ( π n ) + tan ⁡ ( π 2 n ) tan ⁡ ( π 2 n ) tan ⁡ ( π n ) ) = r 1 AC(\dfrac{\tan(\dfrac{\pi}{n}) + \tan(\dfrac{\pi}{2n})}{\tan(\dfrac{\pi}{2n})\tan(\dfrac{\pi}{n})}) = r_{1}

⟹ h = A C = tan ⁡ ( π 2 n ) tan ⁡ ( π n ) tan ⁡ ( π n ) + tan ⁡ ( π 2 n ) r 1 \implies h = AC = \dfrac{\tan(\dfrac{\pi}{2n})\tan(\dfrac{\pi}{n})}{\tan(\dfrac{\pi}{n}) + \tan(\dfrac{\pi}{2n})}r_{1}

tan ⁡ ( π n ) = tan ⁡ ( 2 π 2 n ) = 2 tan ⁡ ( π 2 n ) 1 − tan ⁡ 2 ( π 2 n ) ⟹ \tan(\dfrac{\pi}{n}) = \tan(\dfrac{2\pi}{2n}) = \dfrac{2\tan(\dfrac{\pi}{2n})}{1 - \tan^2(\dfrac{\pi}{2n})} \implies

h = 2 tan ⁡ ( π 2 n ) 3 − tan ⁡ 2 ( π 2 n ) r 1 h = \dfrac{2\tan(\dfrac{\pi}{2n})}{3 - \tan^2(\dfrac{\pi}{2n})}r_{1}

Let β ( n ) = tan ⁡ ( π 2 n ) 3 − tan ⁡ 2 ( π 2 n ) ⟹ h 1 = 2 β ( n ) r 1 = r 2 sin ⁡ ( π n ) \beta(n) = \dfrac{\tan(\dfrac{\pi}{2n})}{3 - \tan^2(\dfrac{\pi}{2n})} \implies h_{1} = 2\beta(n) r_{1} = r_{2}\sin(\dfrac{\pi}{n}) ⟹ r 2 = 2 β ( n ) sin ⁡ ( π n ) r 1 \implies \boxed{r_{2} = \dfrac{2\beta(n)}{\sin(\dfrac{\pi}{n})}r_{1}}

In General r m = ( 2 β ( n ) sin ⁡ ( π n ) ) m − 1 r 1 \boxed{r_{m} = (\dfrac{2\beta(n)}{\sin(\dfrac{\pi}{n})})^{m - 1}r_{1}} and R ( n ) = ∑ m = 1 ∞ r m = sin ⁡ ( π n ) sin ⁡ ( π n ) − 2 β ( n ) r 1 R(n) = \displaystyle\sum_{m = 1}^{\infty} r_{m} = \dfrac{\sin(\dfrac{\pi}{n})}{\sin(\dfrac{\pi}{n}) - 2\beta(n)}r_{1} .

For j ( n ) = 2 β ( n ) ⟹ j ′ ( n ) = − π n 2 sec ⁡ 2 ( π 2 n ) ( 3 + tan ⁡ 2 ( π 2 n ) ( 3 − tan ⁡ 2 ( π 2 n ) ) 2 ) j(n) = 2\beta(n) \implies j'(n) = -\dfrac{\pi}{n^2}\sec^2(\dfrac{\pi}{2n})(\dfrac{3 + \tan^2(\dfrac{\pi}{2n})}{(3 - \tan^2(\dfrac{\pi}{2n}))^2})

⟹ R = lim ⁡ n → ∞ R ( n ) = lim ⁡ n → ∞ − cos ⁡ ( π n ) − cos ⁡ ( π n ) + sec ⁡ 2 ( π 2 n ) ( 3 + tan ⁡ 2 ( π 2 n ) ( 3 − tan ⁡ 2 ( π 2 n ) ) 2 ) r 1 \implies R = \lim_{n \rightarrow \infty} R(n) = \lim_{n \rightarrow \infty} \dfrac{-\cos(\dfrac{\pi}{n})}{-\cos(\dfrac{\pi}{n}) + \sec^2(\dfrac{\pi}{2n})(\dfrac{3 + \tan^2(\dfrac{\pi}{2n})}{(3 - \tan^2(\dfrac{\pi}{2n}))^2})}r_{1}

= 3 2 r 1 ⟹ R r 1 = 3 2 = 1.5 = \dfrac{3}{2}r_{1} \implies \dfrac{R}{r_{1}} = \dfrac{3}{2} = \boxed{1.5}

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