Stacked spheres inscribed in a regular tetrahedron!

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In the regular tetrahedron above, extend the diagram to an infinite number of inscribed spheres. Along each radii O A , O B , O C OA, OB,OC and O P OP of the circumscribed sphere the inscribed stacked spheres are tangent to each other.

Let S S be the total volume of all the inscribed spheres and V T V_{T} be the volume of the tetrahedron.

If S V T = a b ∗ c c π \dfrac{S}{V_{T}} = \dfrac{a}{b * c\sqrt{c}}\pi , where a , b a,b and c c are coprime positive integers, find a + b + c a + b + c .


The answer is 28.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Oct 7, 2020

Using the diagram above we obtain the proportion:

a 1 2 3 r 1 = 3 a 1 2 ( 2 3 a 1 − r 1 ) ⟹ \dfrac{a_{1}}{2\sqrt{3}}r_{1} = \dfrac{\sqrt{3}a_{1}}{2(\sqrt{\dfrac{2}{3}}a_{1} - r{1})} \implies 4 r 1 = 2 3 a 1 − r 1 ⟹ r 1 = a 1 2 6 4r_{1} = \sqrt{\dfrac{2}{3}}a_{1} - r{1} \implies \boxed{r_{1} = \dfrac{a_{1}}{2\sqrt{6}}}

H 2 = H 1 − 2 r 1 = 2 3 a 1 − a 1 6 = a 1 6 H_{2} = H_{1} - 2r_{1} =\dfrac{\sqrt{2}}{\sqrt{3}}a_{1} - \dfrac{a_{1}}{\sqrt{6}} = \dfrac{a_{1}}{\sqrt{6}} and r 1 r 2 = H 1 − r 1 H 2 − r 2 \dfrac{r_{1}}{r_{2}} = \dfrac{H_{1} - r_{1}}{H_{2} - r_{2}}

After simplifying we obtain the proportion: 1 r 2 = \dfrac{1}{r_{2}} = 3 a 1 6 − r 2 ⟹ 4 r 2 = a 1 6 ⟹ \dfrac{3}{\dfrac{a_{1}}{\sqrt{6}} - r_{2}} \implies 4r_{2} = \dfrac{a_{1}}{\sqrt{6}} \implies r 2 = a 1 2 2 6 \boxed{r_{2} = \dfrac{a_{1}}{2^2\sqrt{6}}}

H 3 = H 2 − 2 r 2 = a 1 6 − a 1 2 6 = a 1 2 6 H_{3} = H_{2} - 2r_{2} =\dfrac{a_{1}}{\sqrt{6}} - \dfrac{a_{1}}{2\sqrt{6}} = \dfrac{a_{1}}{2\sqrt{6}} and r 2 r 3 = H 2 − r 2 H 3 − r 3 \dfrac{r_{2}}{r_{3}} = \dfrac{H_{2} - r_{2}}{H_{3} - r_{3}}

After simplifying we obtain the proportion: 1 r 3 = \dfrac{1}{r_{3}} = 3 a 1 2 6 − r 3 ⟹ 4 r 3 = a 1 2 6 ⟹ \dfrac{3}{\dfrac{a_{1}}{2\sqrt{6}} - r_{3}} \implies 4r_{3} = \dfrac{a_{1}}{2\sqrt{6}} \implies r 3 = a 1 2 3 6 \boxed{r_{3} = \dfrac{a_{1}}{2^3\sqrt{6}}}

In General: r n = a 1 6 ( 1 2 ) n \boxed{r_{n} = \dfrac{a_{1}}{\sqrt{6}}(\dfrac{1}{2})^n}

Note: H n = a 1 6 ( 1 2 ) n − 2 H_{n} = \dfrac{a_{1}}{\sqrt{6}}(\dfrac{1}{2})^{n - 2}

⟹ V n = 4 3 π r n 3 = 2 9 6 π a 1 3 ( 1 8 ) n \implies V_{n} = \dfrac{4}{3}\pi r_{n}^3 = \dfrac{2}{9\sqrt{6}}\pi a_{1}^3(\dfrac{1}{8})^{n}

⟹ S ∗ = ∑ n = 1 ∞ V n = 2 9 6 π a 1 3 ∑ n = 1 ∞ ( 1 8 ) n = \implies S^{*} = \sum_{n = 1}^{\infty} V_{n} = \dfrac{2}{9\sqrt{6}}\pi a_{1}^3 \sum_{n = 1}^{\infty} (\dfrac{1}{8})^n = 2 9 6 π a 1 3 ( 1 8 ( 8 7 ) \dfrac{2}{9\sqrt{6}}\pi a_{1}^3(\dfrac{1}{8}(\dfrac{8}{7}) = 2 63 6 π a 1 3 = \dfrac{2}{63\sqrt{6}}\pi a_{1}^3

and V T = ( 1 3 ) ( 1 2 3 2 a 2 ) ( 2 3 ) a = V_{T} = (\dfrac{1}{3})(\dfrac{1}{2}\dfrac{\sqrt{3}}{2}a^2)(\sqrt{\dfrac{2}{3} })a = 1 6 2 a 3 \dfrac{1}{6\sqrt{2}}a^3

⟹ S ∗ = ( 1 6 2 a 3 ) ( 2 63 6 π ) ( 6 2 ) = \implies S^{*} = (\dfrac{1}{6\sqrt{2}}a^3)(\dfrac{2}{63\sqrt{6}}\pi)(6\sqrt{2}) = 4 3 63 π ∗ V T \dfrac{4\sqrt{3}}{63}\pi * V_{T}

and after simplifying V 1 = π 6 3 ∗ V T V_{1} = \dfrac{\pi}{6\sqrt{3}} * V_{T} using r 1 = a 1 2 6 r_{1} = \dfrac{a_{1}}{2\sqrt{6}}

V d = S ∗ − V 1 = ( 4 3 63 − 1 6 3 ) π ∗ V T = V_{d} = S^{*} - V_{1} = (\dfrac{4\sqrt{3}}{63} - \dfrac{1}{6\sqrt{3}})\pi * V_{T} = 3 126 π ∗ V T ⟹ \dfrac{\sqrt{3}}{126}\pi * V_{T} \implies 3 V d = 3 42 π ⟹ 3V_{d} = \dfrac{\sqrt{3}}{42}\pi \implies

S = S ∗ + 3 V d = ( 4 3 63 + 3 42 ) π ∗ V T = S = S^{*} + 3V_{d} = (\dfrac{4\sqrt{3}}{63} + \dfrac{\sqrt{3}}{42})\pi * V_{T} = 11 3 126 π ∗ V T \dfrac{11\sqrt{3}}{126}\pi * V_{T}

⟹ S V T = 11 3 126 π = 11 π 14 ∗ 3 3 = a b ∗ c c π ⟹ a + b + c = 28 \implies \dfrac{S}{V_{T}} = \dfrac{11\sqrt{3}}{126}\pi = \dfrac{11\pi}{14 * 3\sqrt{3}} = \dfrac{a}{b * c\sqrt{c}}\pi \implies a + b + c = \boxed{28}

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