In a regular square pyramid T.ABCD the base edge is AB = a cm and the height is t cm. P lies on the extension of AB with BP = AB, Q lies on the extension of CB with BQ = CB, and K lies on TB with TK = 2KB. The plane PQK cuts the pyramid into two pieces. Find the volume of each piece.

Step-by-step solution

1. 1

   Put the solid in coordinates. Place the square base in the plane $z=0$: $A(0,0,0)$, $B(a,0,0)$, $C(a,a,0)$, $D(0,a,0)$. In a regular pyramid the apex sits above the centre of the base, so $T\left(\tfrac a2,\tfrac a2,t\right)$. The whole pyramid has volume $V=\tfrac13 a^2t$.

2. 2

   Locate the three given points. Extending $AB$ past $B$ by another length $a$ gives $P(2a,0,0)$. Extending $CB$ past $B$ by $a$ gives $Q(a,-a,0)$. From $TK=2KB$ the point $K$ divides $TB$ with $TK:KB=2:1$, so $K=T+\tfrac23(B-T)=\left(\tfrac{5a}{6},\tfrac a6,\tfrac t3\right)$.

3. 3

   Find the equation of the cutting plane. With $a=t=1$ for the algebra, $\vec{PQ}=(-1,-1,0)$ and $\vec{PK}=\left(-\tfrac76,\tfrac16,\tfrac13\right)$; their cross product is proportional to $(1,-1,4)$. The plane is therefore $x-y+4z=2$, and substituting $Q$ and $K$ confirms both lie on it. Note the plane misses the base square entirely (on $z=0$ it is the line $x-y=2$, and $x-y\le 1$ over the base), so the section only meets the four lateral edges.

4. 4

   Find where the plane meets each lateral edge. Substituting the parametrisation of each edge into $x-y+4z=2$ gives: $M$ on $TA$ with $TM:TA=\tfrac12$; $K$ on $TB$ with $TK:TB=\tfrac23$; $N$ on $TC$ with $TN:TC=\tfrac12$; $L$ on $TD$ with $TL:TD=\tfrac25$. The section is the quadrilateral $MKNL$, and the upper piece is the solid $T.MKNL$.

5. 5

   Split the upper piece into two tetrahedra and use the ratio rule. For a tetrahedron, scaling three edges from a common vertex by factors $\lambda_1,\lambda_2,\lambda_3$ scales the volume by $\lambda_1\lambda_2\lambda_3$. Since $T.ABC$ and $T.ACD$ each have volume $\tfrac12 V=\tfrac16 a^2t$: $[T.MKN]=\tfrac12\cdot\tfrac23\cdot\tfrac12\cdot\tfrac16a^2t=\tfrac{1}{36}a^2t$ and $[T.MNL]=\tfrac12\cdot\tfrac12\cdot\tfrac25\cdot\tfrac16a^2t=\tfrac{1}{60}a^2t$.

6. 6

   Add the two tetrahedra and subtract from the whole. $\tfrac{1}{36}+\tfrac{1}{60}=\tfrac{5}{180}+\tfrac{3}{180}=\tfrac{8}{180}=\tfrac{2}{45}$, so the piece containing the apex has volume $\tfrac{2}{45}a^2t$. The other piece is $\tfrac13a^2t-\tfrac{2}{45}a^2t=\tfrac{15-2}{45}a^2t=\tfrac{13}{45}a^2t$.

7. 7

   Check by Monte Carlo integration. Sampling four million uniform points in the unit-size pyramid ($a=t=1$) and testing $x-y+4z>2$ gave $0.04436$ for the apex piece and $0.28864$ for the other, against the exact values $\tfrac{2}{45}=0.04444$ and $\tfrac{13}{45}=0.28889$.