# When is it possible to inscribe sphere in parallelogram pyramid? Additional questions.

I have 3 questions to the reply (this one with an image) to the question which I link here: When is it possible to inscribe sphere in parallelogram pyramid?

Question 1: It is mentioned that a sphere can be inscribed in a pyramid if the bisectors of the dihedral angles formed by its lateral faces all meet on the same line. Is there any way to use that information and find the center of inscribed sphere without x?

Question 2: Is it possible to inscribe a sphere in a pyramid with the same base as in the image if the the projection of the vertex on the base in moved a bit somewhere else (OH is not parallel to DA and CB)?

Question 3: Is it possible to inscribe a sphere in any pyramid with a square base if the the projection of the vertex on the base is not in the center of the square? Let’s suppose that it would be in the vortex of the square.

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 $$U_iU_{i+1}V$$.

• $$\Delta_j$$ be the area of face $$F_j$$. In particular $$2\Delta_i = \ell_i d_i$$.
• $$H_j$$ be the half space containing $$\mathcal{V}$$ and supported on the plane holding $$F_j$$.

The criterion is

Pyramid $$\mathcal{V}$$ admit an inscribed sphere when and only when $$\Delta_1 + \Delta_3 = \Delta_2 + \Delta_4$$

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

• $$I$$ be the incenter of $$\mathcal{V}$$.
• $$\Delta'$$ be the common value of $$\Delta_1 + \Delta_3 = \Delta_2 + \Delta_4$$.
• $$\Delta = \sum_{j=0}^4 \Delta_j = \Delta_0 + 2\Delta'$$ be the surface area of $$\mathcal{V}$$.

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 $$(y^2-x^2)^2 = 0$$.

1. It is not clear what you mean by "without x". If you are asking for a purely geometrical solution, it can be done: just intersect the line where all the bisectors of the dihedral angles formed by the lateral faces of the pyramid meet with any plane bisecting the dihedral angle formed by a lateral face with the base. The resulting point is the center of the inscribed sphere (see also point 3. below).

2. If the vertex "is moved a bit somewhere else" there is no guarantee, in general, that an inscribed sphere still exists. But there could be some other, less symmetrical, solutions for a pyramid with rectangular base, even if at the moment I'm not aware of any.

3. Yes, it is possible to inscribe a sphere in any pyramid having a square base $$ABCD$$ and its vertex $$V$$ projecting on a vertex $$A$$ of the square (see figure below). By symmetry, plane $$VAC$$ bisects both dihedral angles with edges $$VA$$ and $$VC$$, while the bisectors of dihedral angles with edges $$VB$$ and $$VD$$ meet plane $$VAC$$ on the same line. To find that line, we can construct the bisector of $$\angle AFE$$ (where $$AF$$ and $$EF$$ are both perpendicular to edge $$VB$$), which intersects line $$AE$$ at $$H$$: line $$VH$$ is then the desired line where all dihedral bisectors meet. The centre $$L$$ of the inscribed sphere is then the intersection between line $$VH$$ and the plane bisecting the dihedral angle formed by base $$ABCD$$ with any of the lateral faces, for instance $$VAD$$. To find $$L$$ we can then construct the bisector $$AK$$ of $$\angle VAD$$: plane $$BAK$$ then bisects the dihedral angle with edge $$AB$$. If $$O$$ is the intersection of $$AK$$ with the projection $$VN$$ of line $$VH$$ on $$VAD$$, the line through $$O$$ parallel to $$AB$$ lies on that bisecting plane and it thus meet $$VH$$ at the desired centre $$L$$.