More Pyramid Problems

For solution

First let's find a formula for the angle between two adjacent slant faces.

Let $0: (0,0,0), P:(x,0,0), R:(0,x,0), S:(\dfrac{x}{2},\dfrac{x}{2},H).$

$\vec{OS} = \dfrac{x}{2} \vec{i} + \dfrac{x}{2} \vec{j} + h \vec{k}$

$\vec{OP} = x \vec{i} + 0 \vec{j} + 0 \vec{k}$

$\vec{OR} = 0 \vec{i} + x \vec{j} + 0 \vec{k}$

$\vec{u} = \vec{OS} \times \vec{OR} = -hx \vec{i} + 0 \vec{j} + \dfrac{x^2}{2} \vec{k}$

$\vec{v} = \vec{OS} \times \vec{OP} = 0 \vec{i} + xh \vec{j} - \dfrac{x^2}{2} \vec{k}$

Let $\theta$ be the angle between the two adjacent faces $PSO$ and $RSO$

$\vec{u} \cdot \vec{v} = |\vec{u}| * |\vec{v}| \cos(\theta) \implies \dfrac{-x^4}{4} = \dfrac{4 h^2 + x^4}{4} \cos(\theta) \implies \cos(\theta) = \dfrac{-x^2}{4h^2 + x^2}$

Next we find the the height $H$ and base $x$ that will maximize the volume of the pyramid inscribed in a sphere of radius $R$ .

Let $x$ be the side of the square base and $AH$ be the height $h$ of the pyramid. Let $OA$ and $OC$ be radius $R$ of the sphere, so $OH$ is $h - R$ and half the diagonal $CH$ of the square is $\dfrac{x}{\sqrt{2}}$ . The volume of the sphere $V_{s} = \dfrac{4}{3} \pi R^3$ and the volume of the square pyramid $V_{p} = \dfrac{1}{3} x^2 h$ .

For right triangle $COH$ we have:

$$

$R^2 = \dfrac{x^2}{2} + (h - R)^2 = \dfrac{x^2}{2} + h^2 - 2hR + R^2 \implies x^2 + 2h^2 - 4hR = 0 \implies x^2 = 4hR - 2h^2 \implies$ $$

$V_{p}(h) = \dfrac{1}{3} (4h^2R - 2h^3) \implies \dfrac{dV_{p}}{dh} = \dfrac{2h}{3}(4R - 3h) = 0$ and $h \neq 0 \implies h = \dfrac{4R}{3}.$ $$

$h = \dfrac{4R}{3}$ maximizes $V_{p}(h)$ since $\dfrac{d^2V_{p}}{dh^2}|_{(h = \dfrac{4R}{3})} =\dfrac{-8R}{3} < 0$
$$

$h = \dfrac{4R}{3} \implies x^2 = \dfrac{16 R^2}{9} \implies x = \dfrac{4R}{3} = h.$

Using $\cos(\theta) = \dfrac{-x^2}{4h^2 + x^2}$ where $x = \dfrac{4R}{3} = h \implies \cos(\theta) = \dfrac{-1}{5} \implies \theta = \boxed{101.536959^\circ}$ to six decimal places.