SHORT ANSWER: It is possible to inscribe a sphere inside a square pyramid whose edge lengths are all equal. If the edge lengths are $1$ then the center of the inscribed sphere is above the center of the square at a distance of $\frac{1}{\sqrt 6+\sqrt 2}$ from that square, and the radius of the sphere is $\frac{1}{\sqrt 6+\sqrt 2}$. If the sphere is to be inside the pyramid this is the only configuration. If the edge lengths are not equal then a sphere inside the pyramid that is tangent to all five faces may not exist.
LONG ANSWER: The problem is equivalent to finding a point that is equidistant from all five faces of the square pyramid. Let's use 3D analytic geometry to see if such a point exists for a general pyramid with a square base. To be particular, let's place the square base $ABCD$ in the first quadrant of the $xy$-plane with corner $A$ at the origin, sides aligned with the axes. Let's say the apex of the pyramid is at point $E(f,g,h)$. Here is a diagram looking at the pyramid from above.
We want to find the distance of point $P(x,y,z)$ to each of the side faces of the pyramid, labelled $F_1$ through $F_4$. The distance from $P$ to the square base is obviously $z$, but how do we find the distances to the faces defined by their vertices?
Here is one way. Consider face $ABE$. If we take the cross-product of two vectors defined by two sides of the triangular base, $\overrightarrow{AE}\times\overrightarrow{AB}$ we get a vector perpendicular to the face. Taking the right order of those vectors guarantees that the vector points from the face to the interior of the pyramid. Normalize that vector to one $\mathbf{F_1}$ in the same direction but with unit length. Then the distance of point $P$ to face $ABE$ is
$$(\mathbf{P}-\mathbf{A})\cdot\mathbf{F_1}$$
which uses the dot product.
For face $F_1$ we use $\overrightarrow{AE}\times\overrightarrow{AB}$ normalized to get the row vector
$$\mathbf{F_1}=\left[0,\ \frac{h}{\sqrt{g^2+h^2}},\ -\frac{g}{\sqrt{g^2+h^2}} \right]$$
For face $F_2$ we use $\overrightarrow{AD}\times\overrightarrow{AE}$ normalized to get the row vector
$$\mathbf{F_2}=\left[\frac{h}{\sqrt{f^2+h^2}},\ 0,\ -\frac{f}{\sqrt{f^2+h^2}} \right]$$
For face $F_3$ we use $\overrightarrow{DC}\times\overrightarrow{DE}$ normalized to get the row vector
$$\mathbf{F_3}=\left[0,\ -\frac{h}{\sqrt{(1-g)^2+h^2}},\ -\frac{1-g}{\sqrt{(1-g)^2+h^2}} \right]$$
For face $F_4$ we use $\overrightarrow{BE}\times\overrightarrow{BC}$ normalized to get the row vector
$$\mathbf{F_4}=\left[-\frac{h}{\sqrt{(1-f)^2+h^2}},\ 0,\ -\frac{1-f}{\sqrt{(1-f)^2+h^2}} \right]$$
Since the distance from $P(x,y,z)$ to face $F_1$ must equal $z$, we get the equation
$$(\mathbf{P}-\mathbf{A})\cdot\mathbf{F_1}=z$$
This can be written out and put into standard linear form. Doing this for all four faces we get these simultaneous linear equations.
$$\begin{array}{rrrr}
0\,x \ + &\frac{h}{\sqrt{g^2+h^2}}\,y \ + &\left(-1-\frac{g}{\sqrt{g^2+h^2}}\right)z= &0 \\
\frac{h}{\sqrt{f^2+h^2}}\,x \ + &0\,y \ + &\left(-1-\frac{f}{\sqrt{f^2+h^2}}\right)z= &0 \\\
0\,x \ + &\frac{-h}{\sqrt{(1-g)^2+h^2}}\,y \ + &\left(-1-\frac{1-g}{\sqrt{(1-g)^2+h^2}}\right)z= &\frac{-h}{\sqrt{(1-g)^2+h^2}} \\
\frac{-h}{\sqrt{(1-f)^2+h^2}}\,x \ + &0\,y \ + &\left(-1-\frac{1-f}{\sqrt{(1-f)^2+h^2}}\right)z= &\frac{-h}{\sqrt{(1-f)^2+h^2}} \\
\end{array}$$
If we use $f=g=\frac 12$, $h=\frac 1{\sqrt 2}$ we get the square pyramid with all edge lengths equal to $1$. Using those values in those four linear equations in three variables does give us a unique solution, namely
$$\left(\frac 12,\ \frac 12,\ \frac 1{\sqrt 6+\sqrt 2}\right)$$
That gives the first part of my short answer. However, if we use the apex point $f=\frac 12$, $g=h=1$ we get four inconsistent equations with no solution. We can find a sphere to be tangent to any three of the side faces as well as the square base, but none that fits all four side faces and the base. That is the last part of my short answer.
The result is clearly true in the case of a regular pyramid (think to egyptian pyramids).
We are going to show that one can bring back the general case to this special one.
The proof is based on two transformations (a) and (b):
- (a) due to the fact that there is an inscribed sphere with center $I$ and radius $r$, the issue can be converted into a question about volumes:
$$\tag{1}area(ABS)+area(CDS)=area(BCS)+area(ADS) \ ? \ \ \iff$$
$$\tag{2} vol(IABS)+vol(ICDS)=vol(IBCS)+vol(IADS) \ \ ?$$
Why is (1) equivalent to (2) ? In (2), the 4 tetrahedra $IABS, ICDS, ...$ have a common height $r$ ; therefore, applying the formula giving the volume of a tetrahedron: "$\tfrac13 r \times $ base area", one can proceed forward or backwards from (1) to (2) by multiplying or dividing by $\tfrac13 r$.
- (b) There exists a linear transform mapping pyramid $ABCDS$ onto a regular pyramid $A'B'C'D'S'$, making (2) equivalent to
$$\tag{3} vol(I'A'B'S')+vol(I'C'D'S')=vol(I'B'C'S')+vol(I'A'D'S')$$
which is known to be true, completing the proof.
Why is (2) equivalent to (3) ? Because a linear mapping $L$ transforms a polyhedron with volume $V$ into a polyhedron with volume $\det(L) \times V$.
Something remains to be established:
PROOF of the existence of the linear transform $L$:
Let us take the center of the parallelogram as origin $(0,0,0)$ of coordinates.
Let us give the following coordinates names:
$$\tag{2}A(a,b,0), \ B(c,d,0), \ C(-a,-b,0), \ D(-c,-d,0), \ S(e,f,g)$$
Let us consider the regular pyramid with vertices
$$\tag{3}A'(1,0,0), \ B'(0,1,0), \ C'(-1,0,0), \ D'(0,-1,0), \ S'(0,0,1)$$
Then the linear transform associated with matrix
$$M=\begin{pmatrix}a&c&e\\b&d&f\\0&0&g\end{pmatrix}$$
maps $A',B',C',D',S'$ onto $A,B,C,D,S$ resp.
Take $ L= M^{-1}$ for the inverse mapping.
Remarks :
1) The image $I'$ of the center $I$ of the sphere is not necessarily the center of the inscribed sphere in the regular pyramid.
2) About the initial condition (existence of an inscribed sphere), the center $I$ has to belong to the bisecting planes of the 8 dihedral angles. In other words, these bissecting planes have all to be concurrent in a point.
Best Answer
IMHO, it is hard to extract useful information using the bisectors. I 'll present an alternative criterion for the pyramid to admit an inscribed sphere and build up my answer based on that.
Part I - alternative criterion for a pyramid to admit an inscribed sphere.
Given any pyramid $\mathcal{V}$ with a planar parallelogram $ABCD$ as base and vertex $V$ as apex. Introduce following aliases of vertices $A,B,C,D$: $$\ldots,U_0 = D, U_1 = A, U_2 = B, U_3 = C, U_4 =D, U_5 =A, \ldots$$
For $i = 1, 2, 3, 4$ and $j = 0,1,2,3,4$, let
$e_i$ be the edge joining $U_i U_{i+1}$ and $\ell_i = |e_i|$ be its length.
Since $ABCD$ is a parallelogram, we have $\ell_1 = \ell_3$ and $\ell_2 = \ell_4$.
$d_i$ be the distance between $e_i$ and $V$ and $h$ be the height of $\mathcal{V}$.
$F_0$ be the face $ABCD$ and $F_i$ be the face $U_iU_{i+1}V$.
The criterion is
In terms of the half-spaces, the pyramid $\mathcal{V}$ can be regarded as the intersection of two triangular prisms.
$$\mathcal{V} = \bigcap_{j=0}^5 H_j = \left(H_0 \cap H_1 \cap H_3\right)\cap \left( H_0 \cap H_2 \cap H_4\right)$$
Let's look at the first prism $H_0 \cap H_1 \cap H_3$.
It axis is parallel to the direction $U_1U_2 = U_4U_3$. If one look at $\mathcal{V}$ along this direction, the edge $e_1$, $e_3$ becomes two points, the faces $F_1, F_3, F_0$ becomes three line segments of length $d_1$, $d_3$ and $\frac{\Delta_0}{\ell_1}$ respectively and $\mathcal{V}$ becomes a triangle. If $\mathcal{V}$ admits an inscribed sphere of radius $r$, it will become the incircle of this triangle. Recall inradius of a triangle equals to $\frac{2\verb/Area/}{\verb/perimeter/}$.
$$r = \frac{\frac{\Delta_0}{\ell_1} h}{d_1 + d_3 + \frac{\Delta_0}{\ell_1}} = \frac{\Delta_0 h}{2(\Delta_1 + \Delta_3) + \Delta_0} $$
For the second prism $H_0 \cap H_2 \cap H_4$, its axis is parallel to the direction $U_2U_3 = U_1U_4$. Look at $\mathcal{V}$ along this direction and repeat above argument, we obtain another formula for $r$.
$$r = \frac{\Delta_0 h}{2(\Delta_2 + \Delta_4) + \Delta_0}$$
In order for them to be compatible, we need $\Delta_1 + \Delta_3 = \Delta_2 + \Delta_4$. This justify the "only when" part of the criterion.
For the other direction. When $\Delta_1 + \Delta_3 = \Delta_2 + \Delta_4$, the two incenters of the two "triangles" obtained by viewing $\mathcal{V}$ from directions $U_1U_2$ and $U_2U_3$ describe two lines on a plane at a distance $r$ from the plane holding $F_0$. Since one of the line is in the direction $U_1U_2$ while the other line is in another direction $U_2U_3$. These two lines intersect in space. If one place a sphere of radius $r$ at the intersect point. It will touch all $5$ faces $F_j$. The justify the "when" part of the criterion.
Part II - answers to original 3 questions.
Question 1
Since I didn't use bisectors for anything, I cannot answer this part directly. However, assume all $\Delta_j$ are known and satisfy the criterion $\Delta_1 + \Delta_3 = \Delta_2 + \Delta_4$, the incenter can be determined as follows:
Let
When we look at $\mathcal{V}$ along direction $U_1U_2$, the barycentric coordinate of image of $I$ with respect to the triangle formed by image of $V$, $e_1$, $e_3$ has the ratios
$$\frac{\Delta_0}{\ell_1} : d_3 : d_1 = \Delta_0 : 2\Delta_3 : 2\Delta_1$$
This implies there are two real numbers $\alpha, \beta$ such that
$$I = \frac{1}{\Delta}\left[\Delta_0 V + 2\Delta_3(\alpha U_1 + (1-\alpha) U_2) + 2\Delta_1(\beta U_3 + (1-\beta)U_4)\right]$$
If we look at $\mathcal{V}$ along another direction $U_2U_3$ and repeat above argument, we find there are two real numbers $\gamma, \delta$ such that $$I = \frac{1}{\Delta}\left[\Delta_0 V + 2\Delta_4(\gamma U_2 + (1-\gamma) U_3) + 2\Delta_2(\delta U_4 + (1-\delta)U_1)\right]$$
Compare the coefficients for $U_i$ for these expression, we can express $I$ as following linear combination of the vertices:
$$I = \frac{1}{\Delta\Delta'}\left[V\Delta_0\Delta' + 2\left(U_1\Delta_2\Delta_3 + U_2\Delta_3\Delta_4 + U_3\Delta_4\Delta_1 + U_4\Delta_1\Delta_2\right)\right]$$
Question 2
Yes, you can move the projection of $V$ somewhere else so that $OH$ no longer parallel to $BC$.
When $ABCD$ is the rectangle $[-a,a] \times [-b,b]$ with $a > b$ and $V = (x,y,h)$. Condition $\Delta_1+\Delta_3 = \Delta_2 + \Delta_4$ implies $(x,y)$ belongs to following quartic curve:
$$((y^2-b^2)-(x^2-a^2))(a^2y^2-b^2x^2) + (a^2-b^2)h^2(y^2 - x^2) = 0$$
This quartic curve has $4$ branches. Two of them intersect at origin and lies within the sectors $|ay| \ge |bx|$. The other two branches looks like a hyperbola and belongs to the sectors $|ay| < |bx|$. The condition $\Delta_1 + \Delta_3 = \Delta_2 + \Delta_4$ is satisfied whenever $(x,y)$ falls on last two branches.
Question 3
Yes.
If you place the projection of $V$ on one of the diagonals $AC$ or $BD$, then $\Delta_1 + \Delta_3 = \Delta_2 + \Delta_4$ by symmetry. As a result, pyramid $\mathcal{V}$ admit an inscribed sphere. In fact, these are the only spots that works. This is because when $a \to b$, above quartic curve simplifies to $(y^2-x^2)^2 = 0$.