The image below depicts a pyramid $ABCDE$ with a square base. It is to be cut by the plane $FGHI$ which is parallel to the face $ADE$, such that this plane cuts the pyramid in two halves. Find the ratio $\dfrac{\overline{AF}}{\overline{AB}}$.
I am having a hard time with this one, and I am still working on it. The straight forward idea is to assume that $\overline{AB} = 1 $, and to let $\overline{AF} = t $, then I have to express the volume of one of the two parts of the pyramid and equate that to the half the volume of the pyramid. The altitude can be assumed to be $h$, and left at that. The value of $t$ should be independent of $h$.
Best Answer
As David points out, $GH=AF=t$, because of Thales' theorem (applied twice: once in $BEA$, then in $BEC$).
Let's compute the volume of the "sand pile" $GFBCIH$.
Move the triangle $CHI$ to make $H$ and $G$ coincide. Then you get a smaller pyramid, with volume $V_1=(1-t)^3h/3$. The other part of the sand pile is a triangular prism. We need the area of the base triangle $CHI$, and the distance between the faces. To achieve this, we need the dihedral angle $\alpha$ between $CHI$ and the plane $BCIF$, which is the same as between $CED$ and $BCDA$. The altitude $h_f$ of $CED$ is $h_f=h/\sin\alpha$, so the area of the triangle $CHI$ is $(1-t)^2h/(2\sin\alpha)$. Then the distance between the parallel faces of the prism is $t\sin\alpha$, and the volume of the prism is $V_2=ht(1-t)^2/2$.
The volume of the sand pile is thus $V_1+V_2=h[(1-t)^3+t(1-t)^2/2]=h(t+2)(1-t)^2/6$. The ratio between this volume and the volume of the initial pyramid ($h/3$) is thus $(t+2)(1-t)^2/2$. Equate this to $1/2$ to get the equation for $t$, and this simplifies to $t^3-3t+1=0$.
Another solution to compute the volume, with calculus: cut the sand pile in horizontal slices, compute the area of the slice and integrate.
The height of the sand pile is $h_0=(1-t)h$. At height $h_0-s\in[0,h_0]$, the slice has dimension $L\times l$ with
$$\frac{L-t}{1-t}=\frac{s}{h_0}$$ $$\frac{l}{1-t}=\frac{s}{h_0}$$
Hence the area of the slice is $(t+\frac{(1-t)s}{h_0})\frac{(1-t)s}{h_0}$ and the volume is
$$V=\int_0^{h_0} (t+\frac{(1-t)s}{h_0})\frac{(1-t)s}{h_0}\;\mathrm ds\\=t(1-t)\frac{h_0^2}{2h_0}+(1-t)^2\frac{h_0^3}{3h_0^2}\\=t(1-t)^2h/2+(1-t)^3h/3\\=(1-t)^2\frac h6[3t+2(1-t)]\\=(1-t)^2(2+t)h/6$$