Pyramid Craze!

Geometry Level pending

Let n ≥ 3 n \geq 3 and V ( n ) V(n) be the volume of the largest n n -gonal pyramid that is inscribed in a sphere of radius R R .

If θ n \theta_{n} is the angle (in degrees) made between two adjacent faces of the n n -gonal pyramid and λ n \lambda_{n} is the slant height angle(in degrees) of the n n -gonal pyramid, find

tan ⁡ 2 ( λ n ) − sec ⁡ 2 ( λ n ) ∗ cos ⁡ ( θ n ) \tan^2(\lambda_{n}) - \sec^2(\lambda_{n}) * \cos(\theta_{n}) .


The answer is 5.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Feb 24, 2020

O P ⃗ − − 2 sin ⁡ 2 ( π n ) r i ⃗ + sin ⁡ ( 2 π n ) r j ⃗ + 0 k ⃗ \vec{OP} - -2\sin^2(\dfrac{\pi}{n})r\vec{i} + \sin(\dfrac{2\pi}{n})r\vec{j} + 0\vec{k}

O S ⃗ = − r i ⃗ + 0 j ⃗ + H k ⃗ \vec{OS} = -r\vec{i} + 0\vec{j} + H\vec{k}

O R ⃗ − − 2 sin ⁡ 2 ( π ( n − 1 ) n ) r i ⃗ + sin ⁡ ( 2 π ( n − 1 ) n ) r j ⃗ + 0 k ⃗ \vec{OR} - -2\sin^2(\dfrac{\pi(n - 1)}{n})r\vec{i} + \sin(\dfrac{2\pi(n - 1)}{n})r\vec{j} + 0\vec{k}

⟹ \implies

U ⃗ = O P ⃗ X O S ⃗ = sin ⁡ ( 2 π n ) r H i ⃗ + 2 sin ⁡ 2 ( π n ) r H j ⃗ + sin ⁡ ( 2 π n ) r 2 k ⃗ \vec{U} = \vec{OP} \:\ X \:\ \vec{OS} = \sin(\dfrac{2\pi}{n})rH\vec{i} + 2\sin^2(\dfrac{\pi}{n})rH\vec{j} + \sin(\dfrac{2\pi}{n})r^2\vec{k}

V ⃗ = O R ⃗ X O S ⃗ = sin ⁡ ( 2 π ( n − 1 ) n ) r H i ⃗ + 2 sin ⁡ 2 ( π ( n − 1 ) n ) r H j ⃗ + sin ⁡ ( 2 π ( n − 1 ) n ) r 2 k ⃗ \vec{V} = \vec{OR} \:\ X \:\ \vec{OS} = \sin(\dfrac{2\pi(n - 1)}{n})rH\vec{i} + 2\sin^2(\dfrac{\pi(n - 1)}{n})rH\vec{j} + \sin(\dfrac{2\pi(n - 1)}{n})r^2\vec{k}

After simplifying and using the fact that cos ⁡ ( π − w ) = − cos ⁡ ( w ) , sin ⁡ ( π − w ) = sin ⁡ ( w ) \cos(\pi - w) = -\cos(w), \sin(\pi - w) = \sin(w) and sin ⁡ ( 2 w ) = 2 sin ⁡ ( w ) cos ⁡ ( w ) \sin(2w) =2\sin(w)\cos(w) we obtain:

U ⃗ ∘ V ⃗ = − 4 r 2 sin ⁡ 2 ( π n ) ( cos ⁡ ( 2 π n ) H 2 + cos ⁡ 2 ( π n ) r 2 ) \vec{U} \circ \vec{V} = -4r^2\sin^2(\dfrac{\pi}{n})(\cos(\dfrac{2\pi}{n})H^2 + \cos^2(\dfrac{\pi}{n})r^2)

∣ U ⃗ ∣ = 2 r sin ⁡ ( π n ) H 2 + cos ⁡ 2 ( π n ) r 2 = ∣ V ⃗ ∣ |\vec{U}| = 2r\sin(\dfrac{\pi}{n})\sqrt{H^2 + \cos^2(\dfrac{\pi}{n})r^2} = |\vec{V}|

⟹ cos ⁡ ( θ n ) = U ⃗ ∘ V ⃗ ∣ U ⃗ ∣ ∣ V ⃗ ∣ = \implies \cos(\theta_{n}) = \dfrac{\vec{U} \circ \vec{V}}{|\vec{U}| |\vec{V}|} = − cos ⁡ ( 2 π n ) H 2 + cos ⁡ 2 ( π n ) r 2 H 2 + cos ⁡ 2 ( π n ) r 2 -\dfrac{\cos(\dfrac{2\pi}{n})H^2 + \cos^2(\dfrac{\pi}{n})r^2}{H^2 + \cos^2(\dfrac{\pi}{n})r^2}

For the n n gonal base x = 2 r sin ⁡ ( π n ) x = 2r\sin(\dfrac{\pi}{n}) and the height h = r cos ⁡ ( π n ) h = r\cos(\dfrac{\pi}{n})

⟹ \implies

The area of the n n -gon is A ( n ) = n 2 sin ⁡ ( 2 π n ) r 2 ⟹ A(n) = \dfrac{n}{2}\sin(\dfrac{2\pi}{n})r^2 \implies the volume of the n n -gonal pyramid is V ( n ) = n 6 sin ⁡ ( 2 π n ) r 2 H V(n) = \dfrac{n}{6}\sin(\dfrac{2\pi}{n})r^2 H

Let the above diagram represent an n n -gonal pyramid.

R 2 = H 2 − 2 H R + R 2 + r 2 ⟹ H 2 − 2 H R + r 2 = 0 ⟹ R^2 = H^2 - 2HR + R^2 + r^2 \implies H^2 - 2HR + r^2 = 0 \implies

r 2 = 2 H R − H 2 ⟹ V ( n ) = n 6 sin ⁡ ( 2 π n ) ( 2 H 2 R − H 3 ) ⟹ r^2 = 2HR - H^2 \implies V(n) = \dfrac{n}{6}\sin(\dfrac{2\pi}{n})(2H^2R - H^3) \implies

V ( n ) d H = n 6 sin ⁡ ( 2 π n ) H ( 4 R − 3 H ) = 0 H ≠ 0 ⟹ H = 4 R 3 \dfrac{V(n)}{dH} = \dfrac{n}{6}\sin(\dfrac{2\pi}{n})H(4R - 3H) = 0 \:\ H \neq 0 \implies H = \dfrac{4R}{3} ⟹ r 2 = 8 R 2 9 \implies r^2 = \dfrac{8R^2}{9}

⟹ \implies

cos ⁡ ( θ n ) = − 2 cos ⁡ ( 2 π n ) + cos ⁡ 2 ( π n ) 2 + cos ⁡ 2 ( π n ) \cos(\theta_{n}) = -\dfrac{2\cos(\dfrac{2\pi}{n}) + \cos^2(\dfrac{\pi}{n})}{2 + \cos^2(\dfrac{\pi}{n})} = − 5 cos ⁡ ( 2 π n ) + 1 5 + cos ⁡ ( 2 π n ) = -\dfrac{5\cos(\dfrac{2\pi}{n}) + 1}{5 + \cos(\dfrac{2\pi}{n})}

From above H = 4 R 3 , r = 2 2 3 ⟹ h = r cos ⁡ ( π n ) = 2 2 3 cos ⁡ ( π n ) R H = \dfrac{4R}{3}, r = \dfrac{2\sqrt{2}}{3} \implies h = r\cos(\dfrac{\pi}{n}) = \dfrac{2\sqrt{2}}{3}\cos(\dfrac{\pi}{n})R

⟹ tan ⁡ ( λ n ) = H h = 2 cos ⁡ ( π n ) \implies \tan(\lambda_{n}) = \dfrac{H}{h} = \dfrac{\sqrt{2}}{\cos(\dfrac{\pi}{n})} ⟹ cos ⁡ ( π n ) = 2 tan ⁡ ( λ n ) \implies \cos(\dfrac{\pi}{n}) = \dfrac{\sqrt{2}}{\tan(\lambda_{n})}

and

cos ⁡ ( θ n ) = − 5 cos ⁡ ( 2 π n ) + 1 5 + cos ⁡ ( 2 π n ) = − 10 cos ⁡ 2 ( π n ) − 4 4 + 2 cos ⁡ 2 ( π n ) \cos(\theta_{n}) = -\dfrac{5\cos(\dfrac{2\pi}{n}) + 1}{5 + \cos(\dfrac{2\pi}{n})} = -\dfrac{10\cos^2(\dfrac{\pi}{n}) - 4} {4 + 2\cos^2(\dfrac{\pi}{n})}

⟹ cos ⁡ ( θ n ) = \implies \cos(\theta_{n}) = tan ⁡ 2 ( λ n ) − 5 sec ⁡ 2 ( λ n ) ⟹ sec ⁡ ( λ n ) cos ⁡ ( θ n ) = tan ⁡ 2 ( λ n ) − 5 \dfrac{\tan^2(\lambda_{n}) - 5}{\sec^2(\lambda_{n})} \implies \sec^(\lambda_{n})\cos(\theta_{n}) = \tan^2(\lambda_{n}) - 5

⟹ tan ⁡ 2 ( λ n ) − sec ⁡ 2 ( λ n ) ∗ cos ⁡ ( θ n ) = 5 \implies \tan^2(\lambda_{n}) - \sec^2(\lambda_{n}) * \cos(\theta_{n}) = \boxed{5}

Note: cos ⁡ ( θ ) = lim ⁡ n → ∞ cos ⁡ ( θ n ) = − 1 ⟹ θ = 18 0 ∘ \cos(\theta) = \lim_{n \rightarrow \infty} \cos(\theta_{n}) = -1 \implies \theta = 180^{\circ} .

Also, d 2 V ( n ) d H 2 = − 2 3 n R < 0 ⟹ \dfrac{d^2V(n)}{dH^2} = -\dfrac{\sqrt{2}}{3}nR < 0 \implies max occurs at H = 4 R 3 H = \dfrac{4R}{3} .

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