Inscribed Bonanza 2

Level pending

For each positive integer n n :

Let V s ( n ) , V p ( n ) , V c ( n ) V_{s}(n), V_{p}(n), V_{c}(n) and V m ( n ) V_{m}(n) be the volumes of the sphere, right circular cone, right circular cylinder and square pyramid respectively so that:

V p ( n ) V_{p}(n) is the volume of the largest right circular cone inscribed in the sphere of volume V s ( n ) V_{s}(n)

V c ( n ) V_{c}(n) is the volume of the largest right circular cylinder inscribed in the right circular cone of volume V p ( n ) V_{p}(n)

V m ( n ) V_{m}(n) is the volume of the largest square pyramid inscribed in the right circular cylinder of volume V c ( n ) V_{c}(n) .

and

V s ( n + 1 ) V_{s}(n + 1) is the volume of the largest sphere inscribed in the square pyramid of volume V m ( n ) V_{m}(n) .

If V p ∗ V c ∗ V m V s 3 = 1 π ( a b ) a ∗ c ( b a ∗ b ∗ α b b a ∗ b ∗ α b − a b ) b \dfrac{V_{p} * V_{c} * V_{m}}{V_{s}^3} = \dfrac{1}{\pi}(\dfrac{a}{b})^{a * c}(\dfrac{b^{a * b} * \alpha^{b}}{b^{a * b} * \alpha^{b} - a^{b}})^b , where α = 1 + 2 2 \alpha = \dfrac{1 + \sqrt{2}}{2} and a , b a,b and c c are distinct primes, find a + b + c a + b + c .


The answer is 12.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Jan 28, 2019

Let V s ( 1 ) = 4 3 π R 3 , V p ( 1 ) = 1 3 π r 2 H , V c ( 1 ) = π x 2 h V_{s}(1) = \dfrac{4}{3}\pi R^3, \:\ V_{p}(1) = \dfrac{1}{3}\pi r^2 H, \:\ V_{c}(1) = \pi x^2 h and V m ( 1 ) = 1 3 y 2 h V_{m}(1) = \dfrac{1}{3}y^2 h .

For the diagrams use R = R 1 , r = r 1 , H = H 1 x = x 1 h = h 1 R = R_{1}, \:\ r = r_{1}, H = H_{1} \:\ x = x_{1} h = h_{1} and y = y 1 y = y_{1}

R 1 2 = ( H 1 − r 1 ) 2 + r 2 = H 1 2 − 2 H 1 R 1 + R 1 2 + r 1 2 ⟹ r 1 2 = 2 H 1 R 1 − H 1 2 ⟹ V p ( 1 ) = 1 3 π ( 2 H 1 2 R 1 − H 1 3 ) ⟹ R_{1}^2 = (H_{1} - r_{1})^2 + r^2 = H_{1}^2 - 2H_{1}R_{1} + R_{1}^2 + r_{1}^2 \implies r_{1}^2 = 2H_{1}R_{1} - H_{1}^2 \implies V_{p}(1) = \dfrac{1}{3}\pi(2H_{1}^2R_{1} - H_{1}^3) \implies

d V p ( 1 ) d H 1 = π 3 R 1 H 1 ( 4 R 1 − 3 H 1 ) = 0 H 1 ≠ 0 ⟹ H 1 = 4 3 R 1 ⟹ r 1 = 2 2 3 R 1 ⟹ V p ( 1 ) = ( 2 3 ) 3 V s ( 1 ) \dfrac{dV_{p}(1)}{dH_{1}} = \dfrac{\pi}{3}R_{1}H_{1}(4R_{1} - 3H_{1}) = 0 \:\ H_{1} \neq 0 \implies H_{1} = \dfrac{4}{3}R_{1} \implies r_{1} = \dfrac{2\sqrt{2}}{3}R_{1} \implies \boxed{V_{p}(1) = (\dfrac{2}{3})^3 V_{s}(1)}

The two triangles in the above diagram are similar ⟹ r 1 − x 1 r 1 = h 1 H 1 ⟹ h 1 = ( r 1 − x 1 ) H 1 r 1 ⟹ V c ( 1 ) = π H 1 r 1 ( r 1 x 1 2 − x 1 3 ) ⟹ d V c ( 1 ) d x 1 = π H 1 x 1 r 1 ( 2 r 1 − 3 x 1 ) = 0 \implies \dfrac{r_{1} - x_{1}}{r_{1}} = \dfrac{h_{1}}{H_{1}} \implies h_{1} = \dfrac{(r_{1} - x_{1})H_{1}}{r_{1}} \implies V_{c}(1) = \dfrac{\pi H_{1}}{r_{1}}(r_{1}x_{1}^2 - x_{1}^3) \implies \dfrac{dV_{c}(1)}{dx_{1}} = \dfrac{\pi H_{1}x_{1}}{r_{1}}(2r_{1} - 3x_{1}) = 0 x 1 ≠ 0 ⟹ x 1 = 2 3 r 1 ⟹ h 1 = H 1 3 x_{1} \neq 0 \implies x_{1} = \dfrac{2}{3}r_{1} \implies h_{1} = \dfrac{H_{1}}{3}

r 1 = 2 2 3 R 1 r_{1} = \dfrac{2\sqrt{2}}{3}R_{1} and H 1 = 4 3 R 1 ⟹ x 1 = 4 2 9 R 1 H_{1} = \dfrac{4}{3}R_{1} \implies x_{1} = \dfrac{4\sqrt{2}}{9}R_{1} and h 1 = 4 9 R 1 ⟹ V c ( 1 ) = ( 2 3 ) 5 V s ( 1 ) h_{1} = \dfrac{4}{9}R_{1} \implies \boxed{V_{c}(1) = (\dfrac{2}{3})^5V_{s}(1)}

y 1 = 2 x 1 y_{1} = \sqrt{2}x_{1} and x 1 = 4 2 9 R 1 ⟹ y 1 = 8 9 R 1 ⟹ V m ( 1 ) = 1 π ( 2 3 ) 6 V s ( 1 ) x_{1} = \dfrac{4\sqrt{2}}{9}R_{1} \implies y_{1} = \dfrac{8}{9}R_{1} \implies \boxed{V_{m}(1) = \dfrac{1}{\pi}(\dfrac{2}{3})^6 V_{s}(1)}

From above y 1 = 8 9 R 1 y_{1} = \dfrac{8}{9}R_{1} and h 1 = 4 9 R 1 h_{1} = \dfrac{4}{9}R_{1}

⟹ \implies the slant height s = y 1 2 + 4 h 1 2 2 = 4 2 9 R 1 s = \dfrac{\sqrt{y_{1}^2 + 4h_{1}^2}}{2} = \dfrac{4\sqrt{2}}{9}R_{1}

and the area of the above triangle is A = 1 2 y 1 h 1 = ( y 1 + 2 s 2 ) R 2 ⟹ R 2 = y 1 h 1 y 1 + 2 s = 4 9 ( 1 + 2 ) R 1 A = \dfrac{1}{2}y_{1}h_{1} = (\dfrac{y_{1} + 2s}{2})R_{2} \implies R_{2} = \dfrac{y_{1}h_{1}}{y_{1} + 2s} = \dfrac{4}{9(1 + \sqrt{2})}R_{1}

Let α = 1 + 2 2 ⟹ R 2 = 2 9 α R 1 \alpha = \dfrac{1 + \sqrt{2}}{2} \implies R_{2} = \dfrac{2}{9\alpha}R_{1} .

In General:

For n ≥ 2 n \geq 2 :

R n = ( 2 9 α ) R n − 1 = ( 2 9 α ) n − 1 R 1 R_{n} = (\dfrac{2}{9\alpha})R_{n - 1} = (\dfrac{2}{9\alpha})^{n - 1}R_{1}

⟹ \implies

( r n , H n ) = ( 2 2 3 ( 2 9 α ) n − 1 R 1 , 4 3 ( 2 9 α ) n − 1 R 1 ) (r_{n},H_{n}) = (\dfrac{2\sqrt{2}}{3}(\dfrac{2}{9\alpha})^{n - 1}R_{1} ,\:\ \dfrac{4}{3} (\dfrac{2}{9\alpha})^{n - 1}R_{1})

( x n , h n ) = ( 4 2 9 ( 2 9 α ) n − 1 R 1 , 4 9 ( 2 9 α ) n − 1 R 1 ) (x_{n},h_{n}) = (\dfrac{4\sqrt{2}}{9}(\dfrac{2}{9\alpha})^{n - 1}R_{1} , \:\ \dfrac{4}{9} (\dfrac{2}{9\alpha})^{n - 1}R_{1})

( y n , h n ) = ( 8 9 ( 2 9 α ) n − 1 R 1 , 4 9 ( 2 9 α ) n − 1 R 1 ) (y_{n},h_{n}) = (\dfrac{8}{9}(\dfrac{2}{9\alpha})^{n - 1}R_{1}, \:\ \dfrac{4}{9} (\dfrac{2}{9\alpha})^{n - 1}R_{1})

⟹ \implies

V s ( n ) = ( 8 729 α 3 ) n − 1 V s ( 1 ) V_{s}(n) = (\dfrac{8}{729\alpha^3})^{n - 1}V_{s}(1)

V p ( n ) = ( 2 3 ) 3 ( 8 729 α 3 ) n − 1 V s ( 1 ) V_{p}(n) = (\dfrac{2}{3})^3 (\dfrac{8}{729\alpha^3})^{n - 1}V_{s}(1)

V c ( n ) = ( 2 3 ) 5 ( 8 729 α 3 ) n − 1 V s ( 1 ) V_{c}(n) = (\dfrac{2}{3})^5 (\dfrac{8}{729\alpha^3})^{n - 1}V_{s}(1)

V m ( n ) = 1 π ( 2 3 ) 6 ( 8 729 α 3 ) n − 1 V s ( 1 ) V_{m}(n) = \dfrac{1}{\pi}(\dfrac{2}{3})^6 (\dfrac{8}{729\alpha^3})^{n - 1}V_{s}(1)

⟹ \implies

V s = ( 729 α 3 729 α 3 − 8 ) V s ( 1 ) V_{s} = (\dfrac{729\alpha^3}{729\alpha^3 - 8})V_{s}(1)

V p = ( 2 3 ) 3 ( 729 α 3 729 α 3 − 8 ) V s ( 1 ) V_{p} = (\dfrac{2}{3})^3 (\dfrac{729\alpha^3}{729\alpha^3 - 8})V_{s}(1)

V c = ( 2 3 ) 5 ( 729 α 3 729 α 3 − 8 ) V s ( 1 ) V_{c} = (\dfrac{2}{3})^5 (\dfrac{729\alpha^3}{729\alpha^3 - 8})V_{s}(1)

V m = 1 π ( 2 3 ) 6 ( 729 α 3 729 α 3 − 8 ) V s ( 1 ) V_{m} = \dfrac{1}{\pi}(\dfrac{2}{3})^6 (\dfrac{729\alpha^3}{729\alpha^3 - 8})V_{s}(1)

⟹ V p ∗ V c ∗ V m V s 3 = \implies \dfrac{V_{p} * V_{c} * V_{m}}{V_{s}^3} = 1 π ( 2 3 ) 14 ( 729 α 3 729 α 3 − 8 ) 3 = 1 π ( 2 3 ) 2 ∗ 7 ( 3 2 ∗ 3 α 3 3 2 ∗ 3 α 3 − 2 3 ) 3 = 1 π ( a b ) a ∗ c ( b a ∗ b ∗ α b b a ∗ b ∗ α b − a b ) b ⟹ a + b + c = 12 \dfrac{1}{\pi}(\dfrac{2}{3})^{14}(\dfrac{729\alpha^3}{729\alpha^3 - 8})^3 = \dfrac{1}{\pi}(\dfrac{2}{3})^{2 * 7}(\dfrac{3^{2 * 3}\alpha^3}{3^{2 * 3}\alpha^{3} - 2^{3}})^3 = \dfrac{1}{\pi}(\dfrac{a}{b})^{a * c}(\dfrac{b^{a * b} * \alpha^{b}}{b^{a * b} * \alpha^{b} - a^{b}})^b \implies a + b + c = \boxed{12}

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