geometryproof-writing
There is given a pyramid ABCDS with parallelogram ABCD. It is also possible to inscribe a sphere inside ABCDS. Show that sum of areas ABS and CDS is equal to BCS and ADS.
I tried to consider five pyramids with the apex in the center of the sphere. But with no progress. Help me.
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 problem is essentially the motivation behind Hilbert 3rd Problem, which was solved by Dehn. The bottomline is: there is no way to use clever "cutting and putting together" pieces to find the volume of an arbitrary pyramid. You can look up "THE BOOK" by Aignier and Ziegler, there's a chapter on the Dehn invariant.
Now there are pyramids whose volume can be determined by cutting and taking multiple copies of pieces. I listed a few of those in this post
Which Pyramids have a volume which is computable by dissection?
On the other hand I don't think this is a complete list. But it is sufficient to get an idea that the formula should be right.
Lastly, the best proof which uses "the method of exhaustion" (which is a precursor to calculus; you basically need limits but not integration) is that of Euclid. See
https://mathcs.clarku.edu/~djoyce/elements/bookXII/bookXII.html The volume of pyramids is discussed in Proposition 3 to 5.
Best 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):
$$\tag{1}area(ABS)+area(CDS)=area(BCS)+area(ADS) \ ? \ \ \iff$$ $$\tag{2} vol(IABS)+vol(ICDS)=vol(IBCS)+vol(IADS) \ \ ?$$
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 $\tfrac13 r$.
$$\tag{3} vol(I'A'B'S')+vol(I'C'D'S')=vol(I'B'C'S')+vol(I'A'D'S')$$
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.