Is this related to Kepler?

Geometry Level 4

In a cuboid made of glass there are 12 spheres which perfectly fit in the cuboid in 3 × 2 × 2 3 \times 2 \times 2 orientation. All the spheres are identical. The dimensions of the cuboid ( l : b : h ) (l:b:h) are in 3 : 2 : 2 3:2:2 ratio. The cuboid has an inner volume V c V_c , the sphere has an outer volume V s V_s , inner volume v s v_s , outer radius R R , inner radius r r and width w w . The volume of the empty spaces in the cuboid (i.e. difference between the volume of the cuboid and the volumes of all the spheres) is equal to the inner volume of 30 such spheres. If the outer radius of a sphere is x x , find r r in terms of x x .

If for positive integers a , b , c , d , e a,b,c,d,e , we have r = x ( a b π − c d e ) r = x \left (\sqrt[e]{ \frac{a}{b \pi} - \frac{c}{d} } \right) , where gcd ( a , b ) = gcd ( c , d ) = 1 \text{gcd}(a, b)=\text{gcd}(c,d)=1 , with e e minimized. Find the value of a + b + c + d + e a+b+c+d+e


The answer is 27.

Muzaffar Ahmed by Muzaffar Ahmed , 21, India

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1 solution

Muzaffar Ahmed
Apr 1, 2014

Outer Radius of spheres ( R ) (R) = x x

Outer Diameter of spheres ( D = 2 × R ) (D = 2 \times R) = 2 x 2x

Dimensions of cuboid ( l , b , h ) (l, b, h) = 6 x , 4 x , 4 x 6x, 4x, 4x

Volume of cuboid ( V c ) (V_c) = 6 x × 4 x × 4 x = 96 x 3 6x \times 4x \times 4x = 96x^3

Volume of 12 spheres ( 12 × V s ) (12 \times V_s) = 12 × 4 3 π x 3 = 16 π x 3 12 \times \frac{4}{3} \pi x^3 = 16 \pi x^3

Empty spaces in cuboid ( V c − 12 V s ) (V_c - 12 V_s) = 96 x 3 − 16 π x 3 96x^3 - 16 \pi x^3

= 16 x 3 ( 6 − π ) = 16x^3 (6- \pi)

Inner raduis of sphere = r r

Inner volume of 1 sphere ( v s ) (v_s) = 4 3 π r 3 \frac{4}{3} \pi r^3

Inner volume of 30 spheres ( 30 × v s ) (30 \times v_s) = 30 × 4 3 π r 3 30 \times \frac{4}{3} \pi r^3

= 40 π r 3 = 40 \pi r^3

Empty spaces in cuboid ( V c − 12 V s ) (V_c - 12 V_s) = Inner volume of 30 spheres ( 30 × v s ) (30 \times v_s)

⟹ 16 x 3 ( 6 − π ) = 40 π r 3 \implies 16x^3 (6 - \pi) = 40 \pi r^3

⟹ 16 x 3 × ( 6 − π ) 40 × π = r 3 \implies \frac{16x^3 \times (6 - \pi)}{40 \times \pi} = r^3

= 16 x 3 40 × 6 − π π = r 3 = \frac{16x^3}{40} \times \frac{6 - \pi}{\pi} = r^3

= 2 x 3 5 × ( 6 π − 1 ) = r 3 = \frac{2x^3}{5} \times ( \frac{6}{\pi} - 1 ) = r^3

= x 3 × ( 2 5 ) ( 6 π − 1 ) = x^3 \times ( \frac{2}{5} )( \frac{6}{\pi} - 1)

= r 3 = x 3 × ( 12 5 π − 2 5 ) = r^3 = x^3 \times ( \frac{12}{5 \pi} - \frac{2}{5} )

⟹ r = x × 12 5 π − 2 5 3 \implies r = x \times \sqrt[3]{ \frac{12}{5 \pi} - \frac{2}{5} }

∴ a = 12 , b = 5 , c = 2 , d = 5 , e = 3 , a + b + c + d + e = 27 \therefore a=12, b=5, c=2, d=5, e=3, a+b+c+d+e= \boxed{27}

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