Distance from center of sphere to apex of pyramid?

Question

A sphere of radius 1 sits inside a container shaped as an inverted pyramid. The top of the pyramid is a square that is horizontal, and the other faces are equilateral triangles.

Answer 1

Let the square base be $BCDE$. Let $F,G$ respectively be the mid-point of $BC$ and $DE$.

Consider the cross-section containing triangle $AFG$. The triangle $AFG$ is equilateral, and a circle of radius $1$ which is a great circle of the sphere has tangents $AF,AG$. Let $AF,AG$ touch the circle at $X$ and $Y$ respectively.

Let $s$ be the edge length of the pyramid (they are all the same). Because $AF=AG$ is the altitude of an equilateral triangle with side length $s$, so $AF=AG=\frac12s\sqrt3$. Also let $P$ be the mid-point of $FG$, so $FP=\frac12s$.

Let the centre of the sphere (and thus the circle) be $O$. Consider that $\sin\angle XAO=\sin\angle PAF=\frac{FP}{FA}=\frac13\sqrt3$. So by the law of sines,

$$AO=\frac{\sin\angle AXO\cdot XO}{\sin\angle XAO}=\frac{\sin90^0\cdot XO}{\frac13\sqrt3}=\frac1{\frac13\sqrt3}=\sqrt3\text.$$

Therefore, the distance from the centre of the sphere to the apex of the pyramid is $\sqrt3$.

Answer 2

I am going to assume that the sphere is of radius 1.

Place the sphere at $(0,0,1)$

Place the vertices at $(k,0,0), (0,k,0), (0,0,k), (-k,0,0), (0,-k,0)$

I have turned it upside down, it is easier for me to visualize this way.

Find $k$ such that the sphere is tangent to the pyramid.

One of the faces is described by the plane $x + y + z - k = 0$

The point $(0,0,1)$ is one unit from the plane.

$\frac {k-1}{\sqrt 3} = 1$

$k = \sqrt 3 + 1$

The distance from $(0,0,1)$ to $(0,0,k) = \sqrt 3$ units.

For a sphere of radius $r$ the distance is $r\sqrt 3$