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.


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


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.

  • $\begingroup$ No comment ?... $\endgroup$ – Jean Marie Nov 5 '17 at 0:59

Let $E, F, G, H, T$ be the contact points of the inscribed sphere to the triangular faces $ABS$, $BCS$, $CDS$, $DAS$ and parallelogram $ABCD$ respectively.

Color triangles $ABS$, $CDS$, $BCT$, $DAT$ in black and triangles $BCS$, $DAS$, $ABT$, $CDT$, in white.

For each face, triangulate it into $3$ or $4$ triangles using the contact point in that face. For example, face $ABS$ becomes 3 triangles $ABE$, $BSE$ and $SAE$ and face $ABCD$ becomes 4 triangles $ABT$, $BCT$, $CDT$ and $DAT$. This give us totally $16 = 4\times 3 + 1 \times 4$ triangles.

We can group these $16$ triangles into $8$ pairs according to which edge of the original pyramid they attached to. Convince yourself the two triangle in such a pair has same area. As an example, triangle $ABE$ and $ABT$ are attached to the edge $AB$ and they have same area.

Notice for such a pair, one of the triangle is in black and the other in white. This means the total area in black equals to the total area in white. In other words,

$$\begin{align} & \verb/Area/(ABS) + \verb/Area/(CDS) + \verb/Area/(BCT) + \verb/Area/(DAT)\\ = & \verb/Area/(BCS) + \verb/Area/(DAS) + \verb/Area/(ABT) + \verb/Area/(CDT) \end{align}\tag{*1} $$ Since $ABCD$ is a parallelogram, it is not hard too see for any point $P$ inside $ABCD$, we have

$$\verb/Area/(ABP)+\verb/Area/(CDP) = \verb/Area/(BCP) + \verb/Area/(DAP) = \frac12\verb/Area/(ABCD)$$ In particular, set $P$ to $T$, we obtain

$$\verb/Area/(ABT)+\verb/Area/(CDT) = \verb/Area/(BCT) + \verb/Area/(DAT)$$ Substitute this back into $(*1)$, we obtain the desired equality:

$$\verb/Area/(ABS) + \verb/Area/(CDS) = \verb/Area/(BCS) + \verb/Area/(DAS)$$

  • $\begingroup$ Very clever solution : may I ask you how you have had the idea of such a reaoning ? $\endgroup$ – Jean Marie Oct 26 '17 at 18:17
  • $\begingroup$ @JeanMarie I learn this trick recently from this answer, $\endgroup$ – achille hui Oct 26 '17 at 18:43
  • $\begingroup$ Thanks for your answer. $\endgroup$ – Jean Marie Oct 26 '17 at 18:52
  • $\begingroup$ I like your solution. Thank you. $\endgroup$ – Stitch Oct 27 '17 at 18:37

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