In a regular square pyramid T.ABCD, P is the midpoint of AD and Q is the midpoint of AB. A plane is drawn through PQ parallel to the edge TA. Find the ratio of the volumes of the two pieces into which this plane cuts the pyramid.

Step-by-step solution

1. 1

   Set up coordinates. Take $A(0,0,0)$, $B(1,0,0)$, $C(1,1,0)$, $D(0,1,0)$ and apex $T\left(\tfrac12,\tfrac12,1\right)$; the ratio of volumes does not depend on the base edge or the height, so a unit pyramid of volume $\tfrac13$ suffices. Then $P\left(0,\tfrac12,0\right)$ and $Q\left(\tfrac12,0,0\right)$.

2. 2

   Build the plane from the two conditions. It must contain $\vec{PQ}=\left(\tfrac12,-\tfrac12,0\right)$ and be parallel to $\vec{TA}=\left(-\tfrac12,-\tfrac12,-1\right)$. Their cross product is proportional to $(1,1,-1)$, so the plane is $x+y-z=\tfrac12$. Checking: $\vec{TA}\cdot(1,1,-1)=-\tfrac12-\tfrac12+1=0$, so the plane really is parallel to $TA$, and both $P$ and $Q$ satisfy the equation.

3. 3

   Find the cross section. On the base $z=0$ the plane cuts the segment $PQ$. Edge $TB$: the plane meets it at its midpoint $R$; edge $TD$: at its midpoint $S$; edge $TC$: at the point $U$ with $TU:TC=\tfrac14$, i.e. $U\left(\tfrac58,\tfrac58,\tfrac34\right)$. Edge $TA$ is parallel to the plane and is never met. The section is the pentagon $Q\,R\,U\,S\,P$.

4. 4

   Decide which vertices land in which piece. With $f(x,y,z)=x+y-z$, we get $f(A)=0$ and $f(T)=0$, both below $\tfrac12$, while $f(B)=f(D)=1$ and $f(C)=2$ are above. So one piece contains the edge $TA$ and the other contains $B$, $C$ and $D$.

5. 5

   Compute the smaller piece by slicing horizontally. At height $z$ the pyramid's cross section is the square $x,y\in\left[\tfrac z2,1-\tfrac z2\right]$ of side $L=1-z$. Shifting to $u=x-\tfrac z2$, $v=y-\tfrac z2$ turns the condition $x+y<\tfrac12+z$ into $u+v<\tfrac12$, so the area is $\tfrac18$ while $L\ge\tfrac12$, then $L^2-\tfrac12\left(2L-\tfrac12\right)^2$, then $L^2$ once $2L<\tfrac12$.

6. 6

   Integrate the three pieces. $\int_0^{1/2}\tfrac18\,dz=\tfrac1{16}$; $\int_{1/2}^{3/4}\left[(1-z)^2-\tfrac12(1.5-2z)^2\right]dz=\tfrac{5}{192}$; $\int_{3/4}^{1}(1-z)^2dz=\tfrac{1}{192}$. The total is $\tfrac{12}{192}+\tfrac{5}{192}+\tfrac{1}{192}=\tfrac{18}{192}=\tfrac{3}{32}$.

7. 7

   Form the ratio. The whole pyramid has volume $\tfrac13=\tfrac{32}{96}$ and the smaller piece is $\tfrac{3}{32}=\tfrac{9}{96}$, so the other piece is $\tfrac{23}{96}$ and the ratio is $9:23$. A Monte Carlo check with six million samples gave $0.09378$ against the exact $\tfrac{3}{32}=0.09375$.