Cylinders and Stacked Spheres.

Level 2

Let V c V_{c} be the volume of the largest right circular cone that can be inscribed in a sphere of radius R R .

Each sphere inscribed in the right circular cone above are tangent to each other and stacked vertically and is extended to an infinite number of inscribed spheres. Let V s ( n ) V_{s}(n) be the volume of the n n th stacked sphere and V T = ∑ n = 1 ∞ V s ( n ) V_{T} = \displaystyle\sum_{n = 1}^{\infty} V_{s}(n) .

If V T R 3 = a b π c d \dfrac{V_{T}}{R^3} = \dfrac{a^{b}\pi}{c^{d}} . where a , b , c a,b,c and d d are coprime positive integers, ,find a + b + c a + b + c .


The answer is 18.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Dec 22, 2019

Let V c = 1 3 π r 2 H 1 V_{c} = \dfrac{1}{3}\pi r^2H_{1} and V s = 4 3 π R 3 V_{s} = \dfrac{4}{3}\pi R^3 .

r 2 + H 1 2 − 2 H 1 R + R 2 = R 2 ⟹ r 2 = 2 H 1 R − H 1 2 ⟹ r^2 + H_{1}^2 - 2H_{1}R + R^2 = R^2 \implies r^2 = 2H_{1}R - H_{1}^2 \implies

V c = 1 3 π ( 2 H 1 2 R − H 1 3 ) ⟹ d V c d H 1 = π 3 ( 4 H 1 R − 3 H 1 2 ) ⟹ V_{c} = \dfrac{1}{3}\pi(2H_{1}^2R - H_{1}^3) \implies \dfrac{dV_{c}}{dH_{1}} = \dfrac{\pi}{3}(4H_{1}R- 3H_{1}^2) \implies

H 1 ( 4 R − 3 H 1 ) = 0 H 1 ≠ 0 ⟹ H 1 = 4 R 3 ⟹ r = 2 2 3 R H_{1}(4R - 3H_{1}) = 0 \:\ H_{1} \neq 0 \implies H_{1} = \dfrac{4R}{3} \implies r = \dfrac{2\sqrt{2}}{3}R

Let R 1 R_{1} be the radius of circle inscribed the triangle above.

The slant height s = 2 6 3 R s = \dfrac{2\sqrt{6}}{3}R

and the area of the triangle is A = r H 1 = R 1 ( r + s ) ⟹ R 1 = r H 1 r + s = A = rH_{1} = R_{1}(r + s) \implies R_{1} = \dfrac{rH_{1}}{r + s} = 4 R 3 ( 3 + 1 ) \dfrac{4R}{3(\sqrt{3} + 1)}

⟹ R = 3 ( 3 + 1 ) 4 R 1 ⟹ H 1 = ( 3 + 1 ) R 1 \implies R = \dfrac{3(\sqrt{3} + 1)}{4}R_{1} \implies H_{1} = (\sqrt{3} + 1)R_{1}

and H 2 = H 1 − 2 R 1 = ( 3 − 1 ) R 1 ⟹ H 1 H 2 = 3 + 1 3 − 1 ⟹ H 2 = 3 − 1 3 + 1 H 1 H_{2} = H_{1} - 2R_{1} = (\sqrt{3} - 1)R_{1} \implies \dfrac{H_{1}}{H_{2}} = \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1} \implies H_{2} = \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}H_{1}

Let j = 3 − 1 3 + 1 < 1 j = \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1} < 1 , then R 1 = 4 R 3 ( 3 + 1 ) ⟹ R 2 = j R 1 ⟹ R 3 = j R 2 = j 2 R 1 R_{1} = \dfrac{4R}{3(\sqrt{3} + 1)} \implies R_{2} = jR_{1} \implies R_{3} = jR_{2} = j^2R_{1} and in general

R 1 = 4 R 3 ( 3 + 1 ) R_{1} = \dfrac{4R}{3(\sqrt{3} + 1)} and for each positive integer n ≥ 1 ( R n + 1 = j n R 1 ) n \geq 1 \:\ (R_{n + 1} = j^n R_{1})

⟹ V T = 4 3 π R 1 3 + 4 3 π R 1 3 ∑ n = 1 ∞ ( j 3 ) n = \implies V_{T} = \dfrac{4}{3}\pi R_{1}^3 + \dfrac{4}{3}\pi R_{1}^3\displaystyle\sum_{n = 1}^{\infty} (j^3)^n = 4 3 π R 1 3 ∑ n = 1 ∞ ( j 3 ) n − 1 = \dfrac{4}{3}\pi R_{1}^3\displaystyle\sum_{n = 1}^{\infty} (j^3)^{n - 1} =

4 3 π R 1 3 ( 1 1 − j 3 ) = \dfrac{4}{3}\pi R_{1}^3(\dfrac{1}{1 - j^3}) =

4 3 π 4 3 3 3 ( 3 + 1 ) 3 R 3 ( ( 3 + 1 ) 3 ( 3 + 1 ) 3 − ( 3 − 1 ) 3 ) ) \dfrac{4}{3}\pi\dfrac{4^3}{3^3(\sqrt{3} + 1)^3}R^3(\dfrac{(\sqrt{3} + 1)^3}{(\sqrt{3} + 1)^3 - (\sqrt{3} - 1)^3)}) = 4 4 3 4 π ( 1 2 ∗ 3 2 ) R 3 = = \dfrac{4^4}{3^4}\pi(\dfrac{1}{2 * 3^2})R{3} =

2 7 3 6 π R 3 ⟹ V T R 3 = 2 7 3 6 π = a b π c d \dfrac{2^7}{3^6}\pi R^{3} \implies \dfrac{V_{T}}{R^3} = \dfrac{2^7}{3^6}\pi = \dfrac{a^{b}\pi}{c^{d}}

⟹ a + b + c + d = 18 \implies a + b + c+ d = \boxed{18} .

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