Sphere inscribed in a truncated pyramid.

Level pending

Let n n be a positive integer.

If the radius of a sphere inscribed in a 4 n 4n -gonal truncated pyramid is 4 4 , and the slant height of the 4 n 4n -gonal truncated pyramid is 10 10 , find the volume V n V_{n} of the 4 n 4n -gonal truncated pyramid.

Calculate V 100 V_{100} and express the result to five decimal places.

Refer to previous problem

Bonus: Given the radius of a sphere inscribed in a truncated cone is 4 4 , and the slant height of the truncated cone is 10 10 , find the volume V c V_{c} of the truncated cone and show that lim n V n = V c . \lim_{n \rightarrow \infty} V_{n} = V_{c}.


The answer is 703.73122.

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1 solution

Rocco Dalto
Mar 3, 2018

Cut the truncated 4 n 4n -gonal pyramid and inscribed sphere with a vertical plane passing through the center of the sphere and perpendicular to the two bases.

Let x x be the lower base of the 4 n 4n -gonal truncated pyramid and y y be the upper base of the 4 n 4n -gonal truncated pyramid, where x > y x > y .

From the diagram above the altitudes of the upper and lower bases are h = x 2 cot ( π 4 n ) h^{*} = \dfrac{x}{2} \cot(\dfrac{\pi}{4n}) and h = y 2 cot ( π 4 n ) h^{**} = \dfrac{y}{2} \cot(\dfrac{\pi}{4n}) respectively.

h = 10 h = x 2 cot ( π 4 n ) h^{*} = 10 - h^{**} = \dfrac{x}{2} \cot(\dfrac{\pi}{4n}) and 10 2 h = 10 y cot ( π 4 n ) 64 + ( 10 y cot ( π 4 n ) ) 2 = 100 10 - 2h^{**} = 10 - y\cot(\dfrac{\pi}{4n}) \implies 64 + (10 - y\cot(\dfrac{\pi}{4n}))^2 = 100 \implies 64 + 100 20 cot ( π 4 n ) y + cot 2 ( π 4 n ) y 2 = 100 cot 2 ( π 4 n ) y 2 20 cot ( π 4 n ) y + 64 = 0 y = tan ( π 4 n ) ( 10 ± 6 ) 64 + 100 - 20\cot(\dfrac{\pi}{4n})y +\cot^2(\dfrac{\pi}{4n})y^2 = 100 \implies \cot^2(\dfrac{\pi}{4n})y^2 - 20\cot(\dfrac{\pi}{4n})y + 64 = 0 \implies y = \tan(\dfrac{\pi}{4n})(10 \pm 6)

y 1 = 16 tan ( π 4 n ) \implies y_{1} = 16\tan(\dfrac{\pi}{4n}) and y 2 = 4 tan ( π 4 n ) y_{2} = 4\tan(\dfrac{\pi}{4n}) .

Letting the smaller of the two bases y 2 = 4 tan ( π 4 n ) h = 2 tan ( π 4 n ) cot ( π 4 n ) = 2 y_{2} = 4\tan(\dfrac{\pi}{4n}) \implies h^{**} = 2\tan(\dfrac{\pi}{4n}) \cot(\dfrac{\pi}{4n}) = 2 \implies

h = 10 h = 8 = x 2 cot ( π 4 n ) x = 16 tan ( π 4 n ) h^{*} = 10 - h^{**} = 8 = \dfrac{x}{2} \cot(\dfrac{\pi}{4n}) \implies x = 16\tan(\dfrac{\pi}{4n}) .

The volume of a 4 n 4n -gonal truncated pyramid is V n = 4 n h 12 ( x 2 + x y + y 2 ) cot ( π 4 n ) = 8 n 3 ( 256 + 64 + 16 ) tan 2 ( π 4 n ) cot ( π 4 n ) = 896 tan ( π 4 n ) n V_{n} = \dfrac{4nh}{12}(x^2 + xy + y^2) \cot(\dfrac{\pi}{4n}) = \dfrac{8n}{3}(256+64+16)\tan^2(\dfrac{\pi}{4n})\cot(\dfrac{\pi}{4n}) = 896\tan(\dfrac{\pi}{4n})n

V 100 = 89600 tan ( π 400 ) 703.73122 V_{100} = 89600\tan(\dfrac{\pi}{400}) \approx \boxed{703.73122}

To find the volume V c V_{c} of the truncated cone, given the radius of the sphere inscribed in the truncated cone is 4 4 , and the slant height of the truncated cone is 10 10

Let x x be the lower base of the truncated cone and y y be the upper base of the truncated cone, where x > y x > y .

Similar to the truncated 4 n 4n -gonal pyramid above we obtain:

64 + 100 40 y + 4 y 2 = 100 4 y 2 40 y + 64 = 0 4 ( y 2 10 y + 16 ) = 0 ( y 8 ) ( y 2 ) = 0 y 1 = 8 64 + 100 - 40y + 4y^2 = 100 \implies 4y^2 - 40y + 64 = 0 \implies 4(y^2 - 10y + 16) = 0 \implies (y - 8)(y - 2) = 0 \implies y_{1} = 8 and y 2 = 2 y_{2} = 2

Letting the smaller of the two bases y 2 = 2 x = 10 2 = 8 y_{2} = 2 \implies x = 10 - 2 = 8 \implies the volume of the truncated cone V c = π 3 h ( x 2 + x y + y 2 ) = 8 π 3 ( 84 ) = 224 π V_{c} = \dfrac{\pi}{3}h(x^2 + xy + y^2) = \dfrac{8\pi}{3}(84) = 224\pi

To show lim n V n = V c . \lim_{n \rightarrow \infty} V_{n} = V_{c}.

Let V n = 896 tan ( π 4 n ) n V_{n} = 896\tan(\dfrac{\pi}{4n})n .

Using the inequality cos ( x ) < sin ( x ) x < 1 π 4 < n tan ( π 4 ) < π 4 sec ( π 4 ) \cos(x) < \dfrac{\sin(x)}{x} < 1 \implies \dfrac{\pi}{4} < n\tan(\dfrac{\pi}{4}) < \dfrac{\pi}{4} \sec(\dfrac{\pi}{4}) lim n n tan ( π 4 ) = π 4 lim n V n = 896 ( π 4 ) = 224 π = V c \implies \lim_{n \rightarrow \infty} n\tan(\dfrac{\pi}{4}) = \dfrac{\pi}{4} \implies \lim_{n \rightarrow \infty} V_{n} = 896(\dfrac{\pi}{4}) = 224\pi = V_{c}

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