Right Square Pyramid with a special base.

Geometry Level pending

In square base A B C D ABCD , B D BD and A Q AQ intersect at P P with Q D = 1 \overline{QD} = 1 and the area

A A B P = A B 2 A_{\triangle{ABP}} = \dfrac{\overline{AB}}{2} .

Let O O be the center of square A B C D ABCD and construct a circle centered at O O whose circumference is equal to the perimeter of the square. Folding the arc of the semi-circle at a right angle, as shown above, the radius of the semicircle becomes the height of the right square pyramid.

Let A s A_{s} be the lateral surface area and A A_{\square} be the area of the square base.

Find A s A \dfrac{A_{s}}{A_{\square}} .


The answer is 1.618993.

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1 solution

Rocco Dalto
Mar 4, 2020

I provided two solutions for the square base.

Solution using coordinate geometry:

For B D : y = a x BD: y = a - x and for A Q : y = 1 a x x = a 2 a + 1 AQ: y = \dfrac{1}{a}x \implies x = \dfrac{a^2}{a + 1} \implies

A A B P = a 3 2 ( a + 1 ) = a 2 A_{\triangle{ABP}} = \dfrac{a^3}{2(a + 1)} = \dfrac{a}{2} a ( a 2 a 1 ) = 0 a > 0 \implies a(a^2 - a - 1) = 0 \:\ a > 0 \implies

a = 1 + 5 2 = ϕ \boxed{a = \dfrac{1 + \sqrt{5}}{2} = \phi} .

Solution using similar triangles:

Since vertical angles are congruent and A B C D AB \parallel CD \implies alternate interior angles are congruent A B P D P Q \implies \triangle{ABP} \sim \triangle{DPQ} \implies

a 1 = x a x x = a 2 a + 1 A A B P = a 3 2 ( a + 1 ) = a 2 \dfrac{a}{1} = \dfrac{x}{a - x} \implies x = \dfrac{a^2}{a + 1} \implies A_{\triangle{ABP}} = \dfrac{a^3}{2(a + 1)} = \dfrac{a}{2}

a ( a 2 a 1 ) = 0 a > 0 a = 1 + 5 2 = ϕ \implies a(a^2 - a - 1) = 0 \:\ a > 0 \implies \boxed{a = \dfrac{1 + \sqrt{5}}{2} = \phi}

Let O R = r \overline{OR} = r \implies 2 π r = 4 ϕ r = 2 ϕ π 2\pi r = 4\phi \implies r = \dfrac{2\phi}{\pi}

Let T T be the midpoint of A D O T = ϕ 2 \overline{AD} \implies \overline{OT} = \dfrac{\phi}{2}

In R O T \triangle{ROT} the slant height is s = 4 ϕ 2 π 2 + ϕ 2 4 = s = \sqrt{\dfrac{4\phi^2}{\pi^2} + \dfrac{\phi^2}{4}} = ϕ 2 π 16 + π 2 \dfrac{\phi}{2\pi}\sqrt{16 + \pi^2} \implies

A s = ϕ 2 π 16 + π 2 A_{s} = \dfrac{\phi^2}{\pi}\sqrt{16 + \pi^2} A s A = 16 + π 2 π 1.618993 \implies \dfrac{A_{s}}{A_{\square}} = \dfrac{\sqrt{16 + \pi^2}}{\pi} \approx \boxed{1.618993} .

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