It's All Stars 2.

Calculus Level 5

Let n ≥ 6 n \geq 6 be a fixed even positive integer and extend the above hexagrams 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 A c ( m ) A_{c}(m) be the area of each circle C m C_{m} and A p ( m ) A_{p}(m) be the area of each n n gram P m P_{m} and let A c ( n ) = ∑ m = 1 ∞ A c ( m ) A_{c}(n) = \sum_{m = 1}^{\infty} A_{c}(m) and A c = lim ⁡ n → ∞ A c ( n ) A_{c} = \lim_{n \rightarrow \infty} A_{c}(n) and let A p ( n ) = ∑ m = 1 ∞ A p ( m ) A_{p}(n) = \sum_{m = 1}^{\infty} A_{p}(m) and A p = lim ⁡ n → ∞ A p ( n ) A_{p} = \lim_{n \rightarrow \infty} A_{p}(n) .

If A c ∗ A p A c ( 1 ) 2 = a b \dfrac{A_{c} * A_{p}}{A_{c}(1)^2} = \dfrac{a}{b} , where a a and b b are coprime positive integers, find a + b a + b .

Refer to previous problem for n n odd:


The answer is 17.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Feb 4, 2019

Let n ≥ 6 n \geq 6 be an even positive integer.

Note: This is simpler to type then using 2 n + 4 2n + 4 for each positive integer n n .

r 1 2 = r 2 cos ⁡ ( π n ) ) ⟹ r 2 = r 1 2 cos ⁡ ( π n ) \dfrac{r_{1}}{2} = r_{2}\cos(\dfrac{\pi}{n})) \implies r_{2} = \dfrac{r_{1}}{2\cos(\dfrac{\pi}{n})} ⟹ h 1 = r 1 2 tan ⁡ ( π n ) ⟹ h 2 = 1 2 tan ⁡ ( π n ) r 2 = \implies h_{1} = \dfrac{r_{1}}{2}\tan(\dfrac{\pi}{n}) \implies h_{2} =\dfrac{1}{2}\tan(\dfrac{\pi}{n})r_{2} = 1 2 tan ⁡ ( π n ) ( 1 2 cos ⁡ ( π n ) ) r 1 \dfrac{1}{2}\tan(\dfrac{\pi}{n})(\dfrac{1}{2\cos(\dfrac{\pi}{n})}) r_{1} and r 3 = 1 2 cos ⁡ ( π n ) r 2 = ( 1 2 cos ⁡ ( π n ) ) 2 r 1 ⟹ h 3 = h 2 = 1 2 tan ⁡ ( π n ) r 3 = r_{3} = \dfrac{1}{2\cos(\dfrac{\pi}{n})}r_{2} = (\dfrac{1}{2\cos(\dfrac{\pi}{n})})^{2}r_{1} \implies h_{3} = h_{2} =\dfrac{1}{2}\tan(\dfrac{\pi}{n})r_{3} = 1 2 tan ⁡ ( π n ) ( 1 2 cos ⁡ ( π n ) ) 2 r 1 \dfrac{1}{2}\tan(\dfrac{\pi}{n})(\dfrac{1}{2\cos(\dfrac{\pi}{n})})^{2} r_{1}

In General:

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

A c ( m ) = π ( 1 4 cos ⁡ 2 ( π n ) ) m − 1 r 1 2 ⟹ A c ( n ) = ∑ m = 1 ∞ A c ( m ) = ( 4 cos ⁡ 2 ( π n ) 4 cos ⁡ 2 ( π n ) − 1 ) A c ( 1 ) A_{c}(m) = \pi(\dfrac{1}{4\cos^2(\dfrac{\pi}{n})})^{m - 1}r_{1}^2 \implies A_{c}(n) = \sum_{m = 1}^{\infty} A_{c}(m) = (\dfrac{4\cos^2(\dfrac{\pi}{n})}{4\cos^2(\dfrac{\pi}{n}) - 1}) A_{c}(1)

and lim ⁡ n → ∞ A c ( n ) = 4 3 A c ( 1 ) \lim_{n \rightarrow \infty} A_{c}(n) = \dfrac{4}{3}A_{c}(1) .

and A p ( m ) = ( 2 n ) ( 1 2 ( 1 2 tan ⁡ ( π n ) ) ( 1 4 cos ⁡ 2 ( π n ) ) m − 1 r 1 2 A_{p}(m) = (2n)(\dfrac{1}{2}(\dfrac{1}{2}\tan(\dfrac{\pi}{n}))(\dfrac{1}{4\cos^2(\dfrac{\pi}{n})})^{m - 1}r_{1}^2 = n 2 tan ⁡ ( π n ) ( 1 4 cos ⁡ 2 ( π n ) ) m − 1 r 1 2 = \dfrac{n}{2}\tan(\dfrac{\pi}{n})(\dfrac{1}{4\cos^2(\dfrac{\pi}{n})})^{m - 1}r_{1}^2

⟹ A p ( n ) = ∑ m = 1 ∞ A p ( m ) = n 2 tan ⁡ ( π n ) ( 4 cos ⁡ 2 ( π n ) 4 cos ⁡ 2 ( π n ) − 1 ) r 1 2 \implies A_{p}(n) = \sum_{m = 1}^{\infty} A_{p}(m) = \dfrac{n}{2}\tan(\dfrac{\pi}{n})(\dfrac{4\cos^2(\dfrac{\pi}{n})}{4\cos^2(\dfrac{\pi}{n}) - 1})r_{1}^2

Let t n = n 2 tan ⁡ ( π n ) t_{n} = \dfrac{n}{2}\tan(\dfrac{\pi}{n}) and S n = 4 cos ⁡ 2 ( π n ) 4 cos ⁡ 2 ( π n ) − 1 S_{n} = \dfrac{4\cos^2(\dfrac{\pi}{n})}{4\cos^2(\dfrac{\pi}{n}) - 1}

lim ⁡ n → ∞ S n = 4 3 \lim_{n \rightarrow \infty} S_{n} = \dfrac{4}{3}

For lim ⁡ n → ∞ t n \lim_{n \rightarrow \infty} t_{n} :

Using the inequality cos ⁡ ( x ) < sin ⁡ ( x ) x < 1 ⟹ π < n sin ⁡ ( π n ) < π ⟹ π < n tan ⁡ ( π n ) < π sec ⁡ ( π n ) \cos(x) < \dfrac{\sin(x)}{x} < 1 \implies \pi < n\sin(\dfrac{\pi}{n}) < \pi \implies \pi < n\tan(\dfrac{\pi}{n}) < \pi\sec(\dfrac{\pi}{n}) and π lim ⁡ n → ∞ sec ⁡ ( π n ) = π ⟹ lim ⁡ n → ∞ t n = π 2 \pi\lim_{n \rightarrow \infty} \sec(\dfrac{\pi}{n}) = \pi \implies \lim_{n \rightarrow \infty} t_{n} = \dfrac{\pi}{2}

⟹ A p = r 1 2 lim ⁡ n → ∞ S n ∗ t n = ( 4 3 ) ( π 2 ) r 1 2 = 2 3 π r 1 2 = 2 3 A c ( 1 ) \implies A_{p} = r_{1}^2\lim_{n \rightarrow \infty} S_{n} * t_{n} = (\dfrac{4}{3})(\dfrac{\pi}{2})r_{1}^2 = \dfrac{2}{3}\pi r_{1}^2 = \dfrac{2}{3} A_{c}(1)

⟹ A c ∗ A p A c ( 1 ) 2 = ( 4 3 ) ( 2 3 ) = 8 9 = a b ⟹ a + b = 17 \implies \dfrac{A_{c} * A_{p}}{A_{c}(1)^2} = (\dfrac{4}{3})(\dfrac{2}{3}) = \dfrac{8}{9} = \dfrac{a}{b} \implies a + b = \boxed{17} .

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