Pyramids!

Geometry Level pending

Extend the above diagram to a pyramid whose base is a regular n n - gon.

Let O O be the center of the regular n n - gon and construct a circle centered at O O whose circumference is equal to the perimeter of the n n - gon. Folding the arc of the semi-circle at a right angle, as shown above, the radius of the semicircle becomes the height of the n n -gonal pyramid.

Let θ n \theta_{n} be the obtuse angle made between two adjacent faces .

(1) Find cos ⁡ ( θ n ) \cos(\theta_{n}) .

(2) Using n = 13 n = 13 find the obtuse angle(in degrees) made between two adjacent faces.


The answer is 160.321643935123.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Jan 1, 2021

Let each radii of the regular n n -gon be a a and A ( 0 , 0 , 0 ) , B ( a cos ⁡ ( 2 π n ) , − a sin ⁡ ( 2 π n ) , 0 ) , C ( a cos ⁡ ( 2 π n ) , a sin ⁡ ( 2 π n ) , 0 ) A(0,0,0), B(a\cos(\dfrac{2\pi}{n}), -a\sin(\dfrac{2\pi}{n}),0), C(a\cos(\dfrac{2\pi}{n}), a\sin(\dfrac{2\pi}{n}),0) and P ( 0 , 0 , r ) P(0,0,r) .

A B ⃗ = − 2 a sin ⁡ 2 ( π n ) i ⃗ − a sin ⁡ ( 2 π n ) j ⃗ + 0 k ⃗ \vec{AB} = -2a\sin^2(\dfrac{\pi}{n})\vec{i} - a\sin(\dfrac{2\pi}{n})\vec{j} + 0\vec{k}

A C ⃗ = − 2 a sin ⁡ 2 ( π n ) i ⃗ + a sin ⁡ ( 2 π n ) j ⃗ + 0 k ⃗ \vec{AC} = -2a\sin^2(\dfrac{\pi}{n})\vec{i} + a\sin(\dfrac{2\pi}{n})\vec{j} + 0\vec{k}

A P ⃗ = − a i ⃗ + 0 j ⃗ + r k ⃗ \vec{AP} = -a\vec{i} + 0\vec{j} + r\vec{k}

⟹ u ⃗ = A P ⃗ X A C ⃗ = − a r sin ⁡ ( 2 π n ) i ⃗ − 2 a r sin ⁡ 2 ( π n ) j ⃗ − a 2 sin ⁡ ( 2 π n ) k ⃗ \implies \vec{u} = \vec{AP} X \vec{AC} = -ar\sin(\dfrac{2\pi}{n})\vec{i} - 2ar\sin^2(\dfrac{\pi}{n})\vec{j} - a^2\sin(\dfrac{2\pi}{n})\vec{k}

and

v ⃗ = A P ⃗ X A C ⃗ = a r sin ⁡ ( 2 π n ) i ⃗ − 2 a r sin ⁡ 2 ( π n ) j ⃗ + a 2 sin ⁡ ( 2 π n ) k ⃗ \vec{v} = \vec{AP} X \vec{AC} = ar\sin(\dfrac{2\pi}{n})\vec{i} - 2ar\sin^2(\dfrac{\pi}{n})\vec{j} + a^2\sin(\dfrac{2\pi}{n})\vec{k}

⟹ u ⃗ ⋅ v ⃗ = 4 a 2 sin ⁡ 2 ( π n ) ( ( 2 r 2 + a 2 ) sin ⁡ 2 ( π n ) − ( r 2 + a 2 ) ) \implies \vec{u} \cdot \vec{v} = 4a^2\sin^2(\dfrac{\pi}{n})((2r^2 + a^2)\sin^2(\dfrac{\pi}{n}) - (r^2 + a^2))

and

∣ u ⃗ ∣ 2 = ∣ v ⃗ ∣ 2 = 4 a 2 sin ⁡ 2 ( π n ) ( ( r 2 + a 2 ) − a 2 sin ⁡ 2 ( π n ) ) |\vec{u}|^2 = |\vec{v}|^2 = 4a^2\sin^2(\dfrac{\pi}{n})((r^2 + a^2) - a^2\sin^2(\dfrac{\pi}{n}))

⟹ cos ⁡ ( θ n ) = u ⃗ ⋅ v ⃗ ∣ u ⃗ ∣ 2 = \implies \cos(\theta_{n}) = \dfrac{\vec{u} \cdot \vec{v}}{|\vec{u}|^2} = ( 2 r 2 + a 2 ) sin ⁡ 2 ( π n ) − ( r 2 + a 2 ) ( r 2 + a 2 ) − a 2 sin ⁡ 2 ( π n ) \dfrac{(2r^2 + a^2)\sin^2(\dfrac{\pi}{n}) - (r^2 + a^2)}{(r^2 + a^2) - a^2\sin^2(\dfrac{\pi}{n})}

Let a ∗ a^{*} be length of each side of the n n -gon ⟹ a ∗ = 2 a sin ⁡ ( π n ) \implies a^{*} = 2a\sin(\dfrac{\pi}{n})

and n a ∗ = 2 π r ⟹ r = n a ∗ 2 π = a n sin ⁡ ( π n ) π ⟹ na^{*} = 2\pi r \implies r = \dfrac{na^{*}}{2\pi} = \dfrac{an\sin(\dfrac{\pi}{n})}{\pi} \implies

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

⟹ cos ⁡ ( θ 13 ) ≈ − 0.9415978180419575 ⟹ θ 13 ≈ 160.32164393512 3 ∘ \implies \cos(\theta_{13}) \approx -0.9415978180419575 \implies \theta_{13} \approx \boxed{160.321643935123^{\circ}} .

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