Nested Square Pyramids and Spheres.

Level 2

For each positive integer n n ,

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

and

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

Let V s = ∑ n = 1 ∞ V s ( n ) V_{s} = \sum_{n = 1}^{\infty} V_{s}(n) and V p = ∑ n = 1 ∞ V p ( n ) V_{p} = \sum_{n = 1}^{\infty} V_{p}(n) and ϕ = 1 + 5 2 \phi = \dfrac{1 + \sqrt{5}}{2} is the golden ratio.

If V s ∗ V p V s ( 1 ) 2 = ( ( a ϕ ) a ( a ϕ ) a − b a ) b ( b 4 a a π ) \dfrac{V_{s} * V_{p}}{V_{s}(1)^2} = (\dfrac{(a\phi)^a}{(a\phi)^a - b^a})^b (\dfrac{b^4}{a^a\pi}) , where a a and b b are coprime positive integers,find a + b a + b .


The answer is 5.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Jan 22, 2019

V s ( 1 ) = 4 3 π R 1 3 V_{s}(1) = \dfrac{4}{3}\pi R_{1}^3 and V p ( 1 ) = 1 3 x 1 2 H 1 V_{p}(1) = \dfrac{1}{3}x_{1}^2H_{1}

R 1 2 = x 1 2 2 + H 1 2 − 2 H 1 R 1 + R 1 2 ⟹ x 1 2 = 4 H 1 R 1 − 2 H 1 2 R_{1}^2 = \dfrac{x_{1}^2}{\sqrt{2}} + H_{1}^2 - 2H_{1}R_{1} + R_{1}^2 \implies x_{1}^2 = 4H_{1}R_{1} - 2H_{1}^2

⟹ V p ( 1 ) = 1 3 ( 4 H 1 2 R 1 − 2 H 1 3 ) ⟹ d V 1 d H 1 = 2 H 1 3 ( 4 R 1 − 3 H 1 ) = 0 H 1 ≠ 0 ⟹ H 1 = 4 R 1 3 ⟹ x 1 = 4 R 1 3 \implies V_{p}(1) = \dfrac{1}{3}(4H_{1}^2R_{1} - 2H_{1}^3) \implies \dfrac{dV_{1}}{dH_{1}} = \dfrac{2H_{1}}{3}(4R_{1} - 3H_{1}) = 0 \:\ H_{1} \neq 0 \implies H_{1} = \dfrac{4R_{1}}{3} \implies x_{1} = \dfrac{4R_{1}}{3}

Using x 1 = 4 R 1 3 = H 1 x_{1} = \dfrac{4R_{1}}{3} = H_{1} above ⟹ s = 2 3 R 1 5 \implies s = \dfrac{2}{3}R_{1}\sqrt{5} and area of the isosceles triangle above with base x 1 x_{1} and congruent sides s s is A = 1 2 x 1 H 1 = 1 2 x 1 R 2 + s R 2 = ( x 1 + 2 s 2 ) R 2 ⟹ R 2 = x 1 H 1 x 1 + 2 s = 4 3 ( 1 1 + 5 ) R 1 A = \dfrac{1}{2}x_{1}H_{1} = \dfrac{1}{2}x_{1}R_{2} + sR_{2} = (\dfrac{x_{1} + 2s}{2})R_{2} \implies R_{2} = \dfrac{x_{1}H_{1}}{x_{1} + 2s} = \dfrac{4}{3}(\dfrac{1}{1 + \sqrt{5}})R_{1}

and

x 2 = H 2 = 4 3 R 2 = 4 3 ( 4 3 ( 1 + 5 ) ) R 1 x_{2} = H_{2} = \dfrac{4}{3}R_{2} = \dfrac{4}{3}(\dfrac{4}{3(1 + \sqrt{5})})R_{1}
R 3 = 4 3 ( 1 1 + 5 ) R 2 = ( 4 3 ( 1 + 5 ) ) 2 R 1 R_{3} = \dfrac{4}{3}(\dfrac{1}{1 + \sqrt{5}})R_{2} = (\dfrac{4}{3(1 + \sqrt{5})})^2R_{1} and x 3 = H 3 = 4 3 R 3 = 4 3 ( 4 3 ( 1 + 5 ) ) 2 R 1 x_{3} = H_{3} = \dfrac{4}{3}R_{3} = \dfrac{4}{3} (\dfrac{4}{3(1 + \sqrt{5})})^2R_{1} and R 4 = ( 4 3 ( 1 + 5 ) 3 R 1 R_{4} = (\dfrac{4}{3(1 + \sqrt{5}})^3R_{1}

In General:

R n = ( 4 3 ( 1 + 5 ) ) n − 1 R 1 R_{n} = (\dfrac{4}{3(1 + \sqrt{5})})^{n - 1} R_{1} and x n = H n = 4 3 ( 4 3 ( 1 + 5 ) ) n − 1 R 1 x_{n} = H_{n} = \dfrac{4}{3}(\dfrac{4}{3(1 + \sqrt{5})})^{n - 1}R_{1}

Using ϕ = 1 + 5 2 ⟹ \phi = \dfrac{1 + \sqrt{5}}{2} \implies

R n = ( 2 3 ϕ ) n − 1 R 1 R_{n} = (\dfrac{2}{3\phi})^{n - 1}R_{1} and x n = H n = 4 3 ( 2 3 ϕ ) n − 1 R 1 x_{n} = H_{n} = \dfrac{4}{3}(\dfrac{2}{3\phi})^{n - 1}R_{1}

⟹ V s ( n ) = 4 3 π R n 3 = ( 8 27 ϕ ) n − 1 V s ( 1 ) \implies V_{s}(n) = \dfrac{4}{3}\pi R_{n}^3 = (\dfrac{8}{27\phi})^{n - 1}V_{s}(1) and V p ( n ) = 1 3 x n 2 H n = 16 27 π ( 8 27 ϕ ) n − 1 V s ( 1 ) V_{p}(n) = \dfrac{1}{3}x_{n}^2H_{n} = \dfrac{16}{27\pi}(\dfrac{8}{27\phi})^{n - 1}V_{s}(1)

and

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

⟹ \implies

V s = ∑ n = 1 ∞ V s ( n ) = ( 3 ϕ ) 3 ( 3 ϕ ) 3 − 8 V s ( 1 ) V_{s} = \sum_{n = 1}^{\infty} V_{s}(n) = \dfrac{(3\phi)^3}{(3\phi)^3 - 8}V_{s}(1) and V p = ∑ n = 1 ∞ V p ( n ) = 16 27 π ( ( 3 ϕ ) 3 ( 3 ϕ ) 3 − 8 ) V s ( 1 ) V_{p} = \sum_{n = 1}^{\infty} V_{p}(n) = \dfrac{16}{27\pi}(\dfrac{(3\phi)^3}{(3\phi)^3 - 8})V_{s}(1)

⟹ V s ∗ V p V s ( 1 ) 2 = 16 27 π ( ( 3 ϕ ) 3 ( 3 ϕ ) 3 − 8 ) 2 = ( ( 3 ϕ ) 3 ( 3 ϕ ) 3 − 2 3 ) 2 ( 2 4 3 3 π ) = ( ( a ϕ ) a ( a ϕ ) a − b a ) b ( b 4 a a π ) ⟹ a + b = 5 \implies \dfrac{V_{s} * V_{p}}{V_{s}(1)^2} = \dfrac{16}{27\pi}(\dfrac{(3\phi)^3}{(3\phi)^3 - 8})^2 = (\dfrac{(3\phi)^3}{(3\phi)^3 - 2^3})^2(\dfrac{2^4}{3^3\pi}) = (\dfrac{(a\phi)^a}{(a\phi)^a - b^a})^b (\dfrac{b^4}{a^a\pi}) \implies a + b = \boxed{5}

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