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

I'm learning about basics of solid geometry connected with pyramids. I've solved some problems with tetrahedrons, regular pyramids etc. and now I'm trying to solve problem about quite unusual setting - pyramid with parallelogram in base (it can be any pyramid, the only requirement is that it has parallelogram in base). I'm wondering when is it possible to inscribe a sphere in it that touches all five faces. I've tried to imagine it, but my imagination is not that good.

• It makes sense to me that the sphere would touch the pyramid at 5 points: the ground, and 4 points at the same height on the pyramid. I'm not sure that the last part is true, but if so, then you can look at only that cross-section and see if you can inscribe a circle in that similar parallelogram - if you can, you can put a sphere in the pyramid – Alex Li Sep 9 '17 at 15:03
• @AlexLi You are right in the case of a right pyramid: in that case the base must be a rhombus. But if you allow oblique pyramids you can have more general bases, for instance a rectangle, where it is not possible to inscribe a circle. – Aretino Sep 10 '17 at 12:59

In general, 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. This is always true, for any shape of the base.

If the base is a parallelogram and the projection of the vertex on the base is the center of the parallelogram, then a sphere can be inscribed only if the base is a rhombus. But in the case of an oblique pyramid, more general parallelograms are possible.

EDIT.

To show an example, let's see what kind of pyramids with rectangular base admit an inscribed sphere. Suppose we have a rectangle $ABCD$ as base, with $AB=2b$ and $AD=2a$, and let vertex $V$ be at a distance $VH=h$ over the base plane, with its projection $H$ lying on a median axis of the rectangle, at a distance $OH=x$ from the rectangle center (see diagram below).

The condition that the bisectors of the dihedral angles formed by the lateral faces meet on the same line is satisfied if we can find a quadrilateral section of the pyramid where a circle can be inscribed, and perpendicular to the line joining vertex $V$ with the center of the circle.

In our case, as $AB\perp VM$ (where $M$ is the midpoint of $AB$), we can conveniently search such a section in the form of an isosceles trapezoid with bases $AB$ and $EF$, with $EF$ at a distance $FL=h'$ over the base. A circle can be inscribed in $ABEF$ if $AB+EF=2AF$, while the condition that the trapezoid is perpendicular to the line joining $V$ with the center of the circle is satisfied if $VG=VM$ ($G$ midpoint of $EF$).

These two equations can be written in terms of $a$, $b$, $h$, $x$, $h'$ and can be solved for $h'$ and another variable, giving thus a constraint on the pyramid shape. Solving for $h'$ and $h$ gives, for example:

$$h={\sqrt{b^2+x^2-a^2}\over\sqrt{a^2-b^2}}b, \qquad h'={2(a^2-b^2)\over{a^2-b^2+ax}}h.$$ A sphere can be inscribed only in those pyramids having that particular value of $h$ (given $a$, $b$ and $x$). In addition, $a$, $b$ and $x$ must be such that the expressions under square roots be positive: $$b^2<a^2<b^2+x^2.$$ Once these quantities are known it is quite simple to find the center of the inscribed sphere, which lies on the line joining the vertex with the center of the inscribed circle. Of course I chose a particularly simple and symmetric setting, but this example should give an idea of how to handle more general cases.

In fact, your question is rather trivial: If the pyramid is non-degenerated, it is always possible to inscribe a sphere in it. More precisely, for every point that is inside the pyramid you can find infinitely many spheres that are centered in this point and are inside the pyramid.

So let $P$ be a point in the inner of the pyramid. Let $a_1, \ldots, a_5$ be the distances of point $P$ to the faces of the pyramid. Than any circle centered in $P$ with radius $r\leq \min\{a_1,\ldots,a_5\}$ is inside the pyramid.

A more interesting question would be if it is always possible to inscribe a sphere that touches all five faces of the pyramid. I think that this is not possible. In $\mathbb R^2$, there are parallelograms for which it is not even possible to inscribe a circle that touches every side of the parallelogram. My intuition tells me that the situation in $\mathbb R^3$ is even more complicated, but I may be wrong here.