Nested Regular n-gons and Spheres

Geometry Level 4

Let n ≥ 4 n \geq 4 be a fixed positive integer.

For each positive integer m m ,

Let V p ( m ) V_{p}(m) be the volume of the largest n n -gonal pyramid inscribed in a sphere of volume V s ( m ) V_{s}(m)

and

Let V s ( m + 1 ) V_{s}(m + 1) be the volume of the largest sphere inscribed in the n n -gonal pyramid of volume V p ( m ) V_{p}(m) which is tangent to the faces of the n n -gonal pyramid.

Let V s = ∑ m = 1 ∞ V s ( m ) V_{s} = \sum_{m = 1}^{\infty} V_{s}(m) and V p = ∑ m = 1 ∞ V p ( m ) V_{p} = \sum_{m = 1}^{\infty} V_{p}(m) and

let ϕ ( n ) = 1 + 1 + 2 sec ⁡ 2 ( π n ) 2 \phi(n) = \dfrac{1 + \sqrt{1 + 2\sec^2(\dfrac{\pi}{n})}}{2} .

If V s ∗ V p V s ( 1 ) 2 = a a b b π n sin ⁡ ( 2 π n ) ( ( b ϕ ( n ) ) b ( b ϕ ( n ) ) b − a b ) a \dfrac{V_{s} * V_{p}}{V_{s}(1)^2} = \dfrac{a^a}{b^b\pi}n\sin(\dfrac{2\pi}{n})(\dfrac{(b\phi(n))^b}{(b\phi(n))^b - a^b})^a , where a a and b b are coprime positive integers, find a + b a + b .

Note: ϕ ( 4 ) \phi(4) is the golden ratio. Refer to previous problem


The answer is 5.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Jan 24, 2019

V s ( 1 ) = 4 3 π R 1 3 V_{s}(1) =\dfrac{4}{3}\pi R_{1}^3 .

x = 2 r 1 sin ⁡ ( π n ) x = 2r_{1}\sin(\dfrac{\pi}{n}) and h = r 1 cos ⁡ ( π n ) ⟹ A △ A B C = n 2 sin ⁡ ( 2 π n ) r 1 2 ⟹ V 1 ( p ) = n 6 sin ⁡ ( 2 π n ) r 1 2 H 1 h = r_{1}\cos(\dfrac{\pi}{n}) \implies A_{\triangle_{ABC}} = \dfrac{n}{2}\sin(\dfrac{2\pi}{n})r_{1}^2 \implies V_{1}(p) = \dfrac{n}{6}\sin(\dfrac{2\pi}{n})r_{1}^2H_{1}

R 1 2 = ( H 1 − R 1 ) 2 + r 1 2 ⟹ r 1 2 = 2 H 1 R 1 − H 1 2 ⟹ V 1 = n 6 sin ⁡ ( 2 π n ) ( 2 H 1 2 R 1 − H 1 3 ) ⟹ R_{1}^2 = (H_{1} - R_{1})^2 + r_{1}^2 \implies r_{1}^2 = 2H_{1}R_{1} - H_{1}^2 \implies V_{1} = \dfrac{n}{6}\sin(\dfrac{2\pi}{n})(2H_{1}^2R_{1} - H_{1}^3) \implies d V 1 ( p ) d H 1 = n 6 sin ⁡ ( 2 π n ) H 1 ( 4 R 1 − 3 H 1 ) = 0 ⟹ H 1 = 4 R 1 3 ⟹ r 1 = 2 2 3 R 1 \dfrac{dV_{1}(p)}{dH_{1}} = \dfrac{n}{6}\sin(\dfrac{2\pi}{n})H_{1}(4R_{1} - 3H_{1}) = 0 \implies H_{1} = \dfrac{4R_{1}}{3} \implies r_{1} = \dfrac{2\sqrt{2}}{3}R_{1}

s = h 2 + H 1 2 = 2 2 3 cos ⁡ 2 ( π n ) + 2 R 1 s = \sqrt{h^2 + H_{1}^2} = \dfrac{2\sqrt{2}}{3}\sqrt{\cos^2(\dfrac{\pi}{n}) + 2}R_{1}

The area of the above triangle is A = h H 1 = ( h + s ) R 2 ⟹ R 2 = h H 1 h + s A = hH_{1} = (h + s)R_{2} \implies R_{2} = \dfrac{hH_{1}}{h + s}

h H 1 = 8 2 9 cos ⁡ ( π n ) R 1 2 hH_{1} = \dfrac{8\sqrt{2}}{9}\cos(\dfrac{\pi}{n})R_{1}^2 and h + s = 2 2 3 ( cos ⁡ ( π n ) + cos ⁡ 2 ( π n ) + 2 ) R 1 ⟹ h + s = \dfrac{2\sqrt{2}}{3}(\cos(\dfrac{\pi}{n}) + \sqrt{\cos^2(\dfrac{\pi}{n}) + 2})R_{1} \implies R 2 = 4 cos ⁡ ( π n ) R 1 3 ( cos ⁡ ( π n ) + cos ⁡ 2 ( π n ) + 2 ) R_{2} = \dfrac{4\cos(\dfrac{\pi}{n})R_{1}}{3(\cos(\dfrac{\pi}{n}) + \sqrt{\cos^2(\dfrac{\pi}{n}) + 2})}

Let a n = cos ⁡ ( π n ) cos ⁡ ( π n ) + cos ⁡ 2 ( π n ) + 2 = a_{n} = \dfrac{\cos(\dfrac{\pi}{n})}{\cos(\dfrac{\pi}{n}) + \sqrt{\cos^2(\dfrac{\pi}{n}) + 2}} = 1 1 + 1 + 2 sec ⁡ 2 ( π n ) \dfrac{1}{1 + \sqrt{1 + 2\sec^2(\dfrac{\pi}{n})}}

Now let ϕ ( n ) = 1 + 1 + 2 sec ⁡ 2 ( π n ) 2 ⟹ j ( n ) = 4 3 a n = 2 3 ϕ ( n ) ⟹ R 2 = j ( n ) R 1 = 2 3 ϕ ( n ) R 1 \phi(n) = \dfrac{1 + \sqrt{1 + 2\sec^2(\dfrac{\pi}{n})}}{2} \implies j(n) = \dfrac{4}{3}a_{n} = \dfrac{2}{3\phi(n)} \implies R_{2} = j(n)R_{1} = \dfrac{2}{3\phi(n)}R_{1}

and ⟹ r 2 = 2 2 3 R 2 = 2 2 3 ( 2 3 ϕ ( n ) ) R 1 \implies r_{2} = \dfrac{2\sqrt{2}}{3}R_{2} = \dfrac{2\sqrt{2}}{3}(\dfrac{2}{3\phi(n)})R_{1} , H 2 = 4 3 R 2 = 4 3 ( 2 3 ϕ ( n ) ) R 1 , \:\ H_{2} = \dfrac{4}{3}R_{2} = \dfrac{4}{3}(\dfrac{2}{3\phi(n)})R_{1}, R 3 = 2 3 ϕ ( n ) R 2 = ( 2 3 ϕ ( n ) ) 2 R 1 , \:\ R_{3} = \dfrac{2}{3\phi(n)}R_{2} = (\dfrac{2}{3\phi(n)})^2R_{1}, r 3 = 2 2 3 R 3 = 2 2 3 ( 2 3 ϕ ( n ) ) 2 R 1 , \:\ r_{3} = \dfrac{2\sqrt{2}}{3}R_{3} = \dfrac{2\sqrt{2}}{3}(\dfrac{2}{3\phi(n)})^2R_{1}, H 3 = 4 3 R 3 = 4 3 ( 2 3 ϕ ( n ) ) 2 R 1 \:\ H_{3} = \dfrac{4}{3}R_{3} = \dfrac{4}{3}(\dfrac{2}{3\phi(n)})^2R_{1}

In General:

R m = ( 2 3 ϕ ( n ) ) m − 1 R 1 R_{m} = (\dfrac{2}{3\phi(n)})^{m - 1}R_{1}

r m = 2 2 3 ( 2 3 ϕ ( n ) ) m − 1 R 1 r_{m} = \dfrac{2\sqrt{2}}{3}(\dfrac{2}{3\phi(n)})^{m - 1}R_{1}

h m = 4 3 ( 2 3 ϕ ( n ) ) m − 1 R 1 h_{m} = \dfrac{4}{3}(\dfrac{2}{3\phi(n)})^{m - 1}R_{1}

⟹ V s ( m ) = ( 8 27 ϕ ( n ) ) m − 1 V s ( 1 ) \implies V_{s}(m) = (\dfrac{8}{27\phi(n)})^{m - 1}V_{s}(1) and V p ( m ) = 4 27 π n sin ⁡ ( 2 π n ) ( 8 27 ϕ ( n ) ) m − 1 V s ( 1 ) V_{p}(m) = \dfrac{4}{27\pi}n\sin(\dfrac{2\pi}{n})(\dfrac{8}{27\phi(n)})^{m - 1}V_{s}(1)

⟹ \implies

V s = V s ( 1 ) ∑ m = 1 ∞ ( 8 27 ϕ ( n ) ) m − 1 = 27 ϕ 3 ( n ) 27 ϕ 3 ( n ) − 8 V s ( 1 ) V_{s} = V_{s}(1)\sum_{m = 1}^{\infty} (\dfrac{8}{27\phi(n)})^{m - 1} = \dfrac{27\phi^3(n)}{27\phi^3(n) - 8}V_{s}(1)

and

V p = 4 27 π n sin ⁡ ( 2 π n ) ( 27 ϕ 3 ( n ) 27 ϕ 3 ( n ) − 8 ) V s ( 1 ) V_{p} = \dfrac{4}{27\pi}n\sin(\dfrac{2\pi}{n})(\dfrac{27\phi^3(n)}{27\phi^3(n) - 8})V_{s}(1)

⟹ V s ∗ V p V s ( 1 ) 2 = 4 27 π n sin ⁡ ( 2 π n ) ( 27 ϕ 3 ( n ) 27 ϕ 3 ( n ) − 8 ) 2 = 2 2 3 3 π n sin ⁡ ( 2 π n ) ( ( 3 ϕ ( n ) ) 3 ( 3 ϕ ( n ) ) 3 − 2 3 ) 2 = a a b b π n sin ⁡ ( 2 π n ) ( ( b ϕ ( n ) ) b ( b ϕ ( n ) ) b − a b ) a \implies \dfrac{V_{s} * V_{p}}{V_{s}(1)^2} = \dfrac{4}{27\pi}n\sin(\dfrac{2\pi}{n})(\dfrac{27\phi^3(n)}{27\phi^3(n) - 8})^2 = \dfrac{2^2}{3^3\pi}n\sin(\dfrac{2\pi}{n})(\dfrac{(3\phi(n))^3}{(3\phi(n))^3 - 2^3})^2 = \dfrac{a^a}{b^b\pi}n\sin(\dfrac{2\pi}{n})(\dfrac{(b\phi(n))^b}{(b\phi(n))^b - a^b})^a ⟹ a + b = 5 \implies a + b = \boxed{5}

Note: For nested cones and spheres we have: α = lim ⁡ n → ∞ ϕ ( n ) = 1 + 3 2 \alpha = \lim_{n \rightarrow \infty} \phi(n) = \dfrac{1 + \sqrt{3}}{2} and lim ⁡ n → ∞ n sin ⁡ ( 2 π n ) = 2 π \lim _{n \rightarrow \infty} n\sin(\dfrac{2\pi}{n}) = 2\pi .

Letting P ( n ) = V s ( n ) ∗ V p ( n ) V s ( 1 ) 2 ⟹ lim ⁡ n → ∞ P ( n ) = 8 27 ( 27 α 3 27 α 3 − 8 ) 2 P(n) = \dfrac{V_{s}(n) * V_{p}(n)}{V_{s}(1)^2} \implies \lim_{n \rightarrow \infty} P(n) = \dfrac{8}{27}(\dfrac{27\alpha^3}{27\alpha^3 - 8})^2 .

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