Leaning Pyramids

Geometry Level pending

Let ( 0 < r < 1 ) (0 < r < 1) and A : ( 0 , 0 , 0 ) A:(0,0,0)

The base of the square pyramid above has a side length of a a . The point P P with coordinates ( r a , r a , 0 ) (ra,ra,0) lies inside the square and the height of the pyramid is P Q PQ .

If m ∠ Q A R = λ m\angle{QAR} = \lambda is independent of r r , find the value of λ \lambda (in degrees) that minimizes the triangular face Q A R QAR when the volume is held constant.

If λ \lambda is not independent of r r , type 0 0 .


The answer is 60.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Oct 21, 2018

A = A △ Q A R = 1 2 a a 2 r 2 + h 2 A = A_{\triangle{QAR}} = \dfrac{1}{2} a\sqrt{a^2r^2 + h^2}

The volume V = 1 3 a 2 h = k ⟹ h = 3 k a 2 ⟹ A ( a ) = 1 2 r 2 a 2 + 9 k 2 a ⟹ d A d a = 1 2 ( 2 r 2 a 6 − 9 k 2 a 2 r 2 a 6 + 9 k 2 ) V = \dfrac{1}{3} a^2 h = k \implies h = \dfrac{3k}{a^2} \implies A(a) = \dfrac{1}{2}\dfrac{\sqrt{r^2 a^2 + 9k^2}}{a} \implies \dfrac{dA}{da} = \dfrac{1}{2}(\dfrac{2r^2 a^6 - 9k^2}{a^2\sqrt{r^2 a^6 + 9k^2}})

a ≠ 0 ⟹ a = ( 3 k 2 r ) 1 3 ⟹ h = ( 6 k r 2 ) 1 3 a \neq 0 \implies a = (\dfrac{3k}{\sqrt{2}r})^{\frac{1}{3}} \implies h = (6kr^2)^{\frac{1}{3}} .

tan ⁡ ( λ ) = r 2 a 6 + 9 k 2 a 3 r = 3 ⟹ λ = 6 0 ∘ \tan(\lambda) = \dfrac{\sqrt{r^2 a^6 + 9k^2}}{a^3 r} = \sqrt{3} \implies \lambda = \boxed{60^\circ} .

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