Nested Spheres and Bipyramids.

Level pending

Let n > = 4 n >= 4 be a fixed positive integer and consider a bipyramid whose connected base is a regular n n - gon.

For each positive integer m m ,

Let V p ( m ) V_{p}(m) be the volume of the largest bipyramid that can be inscribed in a sphere of volume V s ( m ) V_{s}(m)

Let V s ( m + 1 ) V_{s}(m + 1) be the volume of the largest sphere that be inscribed in the bipyramid of volume V p ( m ) V_{p}(m) .

Let V s ∗ ( n ) = ∑ m = 1 ∞ V s ( m ) V^{*}_{s}(n) = \sum_{m = 1}^{\infty} V_{s}(m) and V p ∗ ( n ) = ∑ m = 1 ∞ V p ( m ) V^{*}_{p}(n) = \sum_{m = 1}^{\infty} V_{p}(m) and a a and b b be coprime positive integers.

If lim ⁡ n → ∞ V s ∗ ( n ) ∗ V p ∗ ( n ) = a 3 b 3 − a 2 a V s ( 1 ) 2 \lim_{n \rightarrow \infty} V^{*}_{s}(n) * V^{*}_{p}(n) = \dfrac{a^3 }{b^3 - a^2\sqrt{a}} V_{s}(1)^2 ,
find a + b a + b

Note: My intention is not to reduce the problem to nested spheres and right circular bicones, but you will get the correct result if you do so., since lim ⁡ n → ∞ V p ( n ) = V b i c o n e \lim_{n \rightarrow \infty} V_{p}(n) = V_{bicone} , where V p ( n ) V_{p}(n) is the volume of a bipyramid whose connected base is a regular n − n- -gon.


The answer is 5.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Dec 11, 2017

For area of n n - gon ):

Let B C = x BC = x be a side of the n n - gon, A C = A B = r ∗ AC = AB= r^{*} , A D = h ∗ AD = h^* , and ∠ B A D = π n \angle{BAD} = \dfrac{\pi}{n} .

x = 2 r ∗ sin ⁡ ( π n ) x = 2 r^{*} \sin(\dfrac{\pi}{n}) and h ∗ = r ∗ cos ⁡ ( π n ) ⟹ h^{*} = r^{*} \cos(\dfrac{\pi}{n}) \implies

A n − g o n = n 2 sin ⁡ ( 2 π n ) r ∗ 2 ⟹ V p ( 1 ) ( n ) = n 3 sin ⁡ ( 2 π n ) r ∗ 2 H A_{n-gon} = \dfrac{n}{2} \sin(\dfrac{2\pi}{n}) {r^{*}}^2 \implies V_{p(1)}(n) = \dfrac{n}{3} \sin(\dfrac{2\pi}{n}) {r^{*}}^2 H and the volume of the initial sphere V s ( 1 ) = 4 π 3 R 3 V_{s}(1) = \dfrac{4\pi}{3} R^3

For the bipyramid inscribed in the sphere r ∗ = R , H = 2 R ⟹ r^* = R, \:\ H = 2R \implies V p ( 1 ) ( n ) = n 2 π sin ⁡ ( 2 π n ) V s ( 1 ) V_{p(1)}(n) = \dfrac{n}{2\pi} \sin(\dfrac{2\pi}{n}) V_{s(1)} .

To find the inscribed sphere: Cut the bipyramid and inscribed sphere with a vertical plane passing through the center of the sphere and perpendicular to the connected base, the cross-section becomes a circle inscribed in a rhombus.

Let r r be the radius of the inscribed circle.

In △ A B C , D E ⊥ A B , D F ⊥ A C \triangle{ABC}, \:\ DE \perp AB, \:\ DF \perp AC .

The diagonal of the rhombus B C = 2 h ∗ = 2 r ∗ cos ⁡ ( π n ) , D E = D F = r , A B = A C = H 2 + 4 r ∗ 2 cos ⁡ ( π n ) 2 = s , BC = 2h^{*} = 2 r^{*} \cos(\dfrac{\pi}{n}), \:\ DE = DF = r, \:\ AB = AC = \dfrac{\sqrt{H^2 + 4 {r^{*}}^2 \cos(\dfrac{\pi}{n})}}{2} = s, and A D = H 2 . AD = \dfrac{H}{2}.

A △ A B C = h ∗ H s ⟹ r = h ∗ H 2 s = r ∗ cos ⁡ ( π n ) 1 + cos ⁡ 2 ( π n ) A_{\triangle{ABC}}= \dfrac{h^{*} H}{s} \implies r = \dfrac{h^{*} H}{2 s} = \dfrac{r^{*} \cos(\dfrac{\pi}{n})}{\sqrt{1 + \cos^2(\dfrac{\pi}{n})}}

Using r ∗ = R , h ∗ = 2 R r^{*} = R, \:\ h^{*} = 2R from above ⟹ r = R cos ⁡ ( π n ) 1 + cos ⁡ ( π n ) = R sec ⁡ 2 ( π n ) + 1 \implies r = \dfrac{R \cos(\dfrac{\pi}{n})}{\sqrt{1 + \cos(\dfrac{\pi}{n})}} = \dfrac{R}{\sqrt{\sec^2(\dfrac{\pi}{n}) + 1}}

Let r ∗ ( 1 ) = r ∗ = R , H ( 1 ) = H = 2 R , r ( 1 ) = r = R sec ⁡ 2 ( π n ) + 1 ⟹ r^{*}(1) = r^{*} = R, \:\, H(1) = H = 2R, \:\ r(1) = r = \dfrac{R}{\sqrt{\sec^2(\dfrac{\pi}{n}) + 1}} \implies

r ( m ) = ( 1 sec ⁡ 2 ( π n ) + 1 ) m − 1 ∗ R = r ∗ ( m ) r(m) = (\dfrac{1}{\sqrt{\sec^2(\dfrac{\pi}{n}) + 1}})^{m - 1} * R = r^{*}(m) and H ( m ) = 2 ( 1 sec ⁡ 2 ( π n ) + 1 ) m − 1 ∗ R H(m) = 2 (\dfrac{1}{\sqrt{\sec^2(\dfrac{\pi}{n}) + 1}})^{m - 1} * R , for each positive integer m m .

Let w n = sec ⁡ 2 ( π n ) + 1 ⟹ w_{n} = \sqrt{\sec^2(\dfrac{\pi}{n}) + 1} \implies

V s ( m ) = ( 1 w n 3 ) m − 1 V s ( 1 ) V_{s}(m) = (\dfrac{1}{{w_{n}}^3})^{m - 1} V_{s}(1) and V p ( m ) = n 2 π sin ⁡ ( 2 π n ) ( 1 w n 3 ) m − 1 V s ( 1 ) ⟹ V_{p}(m) = \dfrac{n}{2\pi} \sin(\dfrac{2\pi}{n}) (\dfrac{1}{{w_{n}}^3})^{m - 1} V_{s}(1) \implies

V s ∗ ( n ) = ∑ m = 1 ∞ V s ( m ) V^{*}_{s}(n) = \sum_{m = 1}^{\infty} V_{s}(m) = w n 3 w n 3 − 1 V s ( 1 ) = \dfrac{{w_{n}}^3}{{w_{n}}^3 - 1} V_{s}(1) and V p ∗ ( n ) = ∑ m = 1 ∞ V p ( m ) V^{*}_{p}(n) = \sum_{m = 1}^{\infty} V_{p}(m) = n 2 π sin ⁡ ( 2 π n ) ( w n 3 w n 3 − 1 ) V s ( 1 ) = \dfrac{n}{2\pi} \sin(\dfrac{2\pi}{n}) (\dfrac{{w_{n}}^3}{{w_{n}}^3 - 1}) V_{s}(1)

Let j ( n ) = n 2 π sin ⁡ ( 2 π n ) . j(n) = \dfrac{n}{2\pi} \sin(\dfrac{2\pi}{n}).

The inequality cos ⁡ ( x ) < sin ⁡ ( x ) x < 1 ⟹ cos ⁡ ( 2 π n ) < j n < 1 ⟹ lim ⁡ n → ∞ j ( n ) = 1 ⟹ \cos(x) < \dfrac{\sin(x)}{x} < 1 \implies \cos(\dfrac{2\pi}{n}) < j_{n} < 1 \implies \lim_{n \rightarrow \infty} j(n) = 1 \implies

lim ⁡ n → ∞ V p ∗ ( n ) = lim ⁡ n → ∞ V s ∗ ( n ) = ( 2 ) 3 ( 2 ) 3 − 1 V s ( 1 ) ⟹ \lim_{n \rightarrow \infty} V^{*}_{p}(n) = \lim_{n \rightarrow \infty} V^{*}_{s}(n) = \dfrac{(\sqrt{2})^3}{(\sqrt{2})^3 - 1} V_{s}(1) \implies

lim ⁡ n → ∞ V s ∗ ( n ) ∗ V p ∗ ( n ) \lim_{n \rightarrow \infty} V^{*}_{s}(n) * V^{*}_{p}(n) = 8 ( 2 2 − 1 ) 2 ( V s ( 1 ) ) 2 = \dfrac{8}{(2\sqrt{2} - 1)^2} (V_{s}(1))^2 = 8 9 − 4 2 V s ( 1 ) 2 = = \dfrac{8}{9 - 4\sqrt{2}} V_{s}(1)^2 = a 3 V s ( 1 ) 2 b 3 − a 2 a ⟹ a + b = 5 . \dfrac{a^3 V_{s}(1)^2}{b^3 - a^2\sqrt{a}} \implies a + b = \boxed{5}.

Using right circular bicones:

From above the volume of the bipyramid is V p ( n ) = n 3 sin ⁡ ( 2 π n ) r ∗ 2 H V_{p}(n) = \dfrac{n}{3} \sin(\dfrac{2\pi}{n}) {r^{*}}^2 H

Using the inequality cos ⁡ ( x ) < sin ⁡ ( x ) x < 1 ⟹ cos ⁡ ( 2 π n ) < n 2 π ∗ sin ⁡ ( 2 π n ) < 1 ⟹ 2 π 3 ∗ cos ⁡ ( 2 π n ) < n 3 sin ⁡ ( 2 π n ) < 2 π 3 \cos(x) < \dfrac{\sin(x)}{x} < 1 \implies \cos(\dfrac{2\pi}{n}) < \dfrac{n}{2\pi} * \sin(\dfrac{2\pi}{n}) < 1 \implies \dfrac{2\pi}{3} * \cos(\dfrac{2\pi}{n}) < \dfrac{n}{3} \sin(\dfrac{2\pi}{n}) < \dfrac{2\pi}{3}

⟹ lim ⁡ n → ∞ V p ( n ) = 2 π 3 r ∗ 2 H = V b i c o n e \implies \lim_{n \rightarrow \infty} V_{p}(n) = \boxed{\dfrac{2\pi}{3} {r^{*}}^2 H = V_{bicone}}

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