Square Pyramid Problem 2

From the diagram above:

Let $N$ be the center of the circle and $r$ be the radius of the circle, then $r = DN = AN = NM$ and $AB = \dfrac{x}{2} = \dfrac{rj}{2} \implies BM = \dfrac{rj}{2}$ .

Let $CM = y$ .

$\triangle{NMC} \sim \triangle{ABC} \implies \dfrac{j}{2} = \dfrac{rj + 2y}{2(r + j)} \implies$ $y = \dfrac{j^2}{2}.$

For $\triangle{NMC}$ the Pythagorean theorem $\implies \dfrac{j^4}{4} + r^2 = r^2 + 2jr + j^2 \implies 4r^2 + j^4 = 4r^2 + 8rj + 4j^2 \implies r = j(\dfrac{j^2 - 4}{8})_(j > 2) \implies$ the height of the square pyramid $h = AC = 2r + j = \dfrac{j^3}{4}$ and the base $x = rj = (\dfrac{j^2 - 4}{8}) j^2 \implies \dfrac{x}{2} = (\dfrac{j^2 - 4}{16}) j^2 \implies$ the slant height $s = \sqrt{(\dfrac{x}{2})^2 + h^2} = \dfrac{j^2}{16} (j^2 + 4) = \dfrac{5j^2}{4} = \dfrac{20j^2}{16} \implies$ $\dfrac{j^2}{16} (j^2 - 16) = 0 \implies j = 4$ for $j > 2$ .

$j = 4 \implies r = 6, x = 24$ and $s = 20 \implies$ the lateral surface area of the square pyramid $A_{s} = 2xs = \boxed{960}$ .

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Note: $\triangle{ABC}$ is a $(12,16,20)$ triangle $\sim$ $(3,4,5)$ triangle.

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An alternative method is illustrated below:

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Using the above isosceles triangle we have:

$CG = BC = \dfrac{5j^2}{4}$ , $GB = x = rj$ , $AC = h = 2r + j$ and $r = OA = OE = OF = OD$ .

From the diagram the Area of the isosceles triangle is $A = \dfrac{(2r + j)rj}{2} = \dfrac{1}{2}r^2 j + \dfrac{5}{4}r j^2 \implies 4r^2j + 2rj^2 = 2r^2j + 5rj^2$

$\implies 3rj^2 - 2r^2j = 0 \implies rj(3j - 2r) = 0 \implies r = \dfrac{3j}{2}$ for $r > 0 \implies AB = \dfrac{x}{2} = \dfrac{3j^2}{4},$ $AC = h = 4j$ and $BC = s = \dfrac{5j^2}{4}$ .

$\therefore$ For right $\triangle{ABC}$ we have: $\dfrac{25j^4}{16} = \dfrac{9j^4}{16} + 16j^2 \implies j^2(j^2 - 16) = 0 \implies j = 4$ for $j > 2$ $\implies x = 24$ and $s = 20 \implies$ the lateral surface area of the square pyramid $A_{s} = 2xs = \boxed{960}$ .