Leaning Pyramid

Geometry Level pending

Let ( 0 < r < 1 ) (0 < r < 1) .

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 .

Find the value of r r for which m ∠ Q R S = 3 0 ∘ m\angle{QRS} = 30^\circ minimizes the triangular face Q R S QRS when the volume is held constant.


The answer is 0.75.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Oct 21, 2018

A = A △ Q R S = 1 2 a ( 1 − r ) 2 a 2 + h 2 A = A_{\triangle{QRS}} = \dfrac{1}{2} a\sqrt{(1 - r)^2 a^2 + h^2}

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

d A d a = 1 2 ( 2 ( 1 − r ) 2 a 6 − 9 k 2 a 2 ( 1 − r ) 2 a 6 + 9 k 2 ) \dfrac{dA}{da} = \dfrac{1}{2}(\dfrac{2(1 - r)^2 a^6 - 9k^2}{a^2\sqrt{(1 - r)^2 a^6 + 9k^2}}) a ≠ 0 ⟹ a = ( 3 k 2 ( 1 − r ) ) 1 3 ⟹ h = ( 6 k ( 1 − r ) 2 ) 1 3 \:\ a \neq 0 \implies a = (\dfrac{3k}{\sqrt{2}(1 - r)})^{\frac{1}{3}} \implies h = (6k(1 - r)^2)^{\frac{1}{3}} .

tan ⁡ ( 3 0 ∘ ) = 1 3 = ( 1 − r ) 2 a 6 + 9 k 2 a 3 r = 3 ( 1 − r r ) \tan(30^\circ) = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{(1 - r)^2 a^6 + 9k^2}}{a^3 r} = \sqrt{3}(\dfrac{1 - r}{r}) ⟹ r = 3 4 = 0.75 \implies r = \dfrac{3}{4} = \boxed{0.75} .

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