Solid Geometry – Problem 23: Find the ratio

In triangular prism $ABC-A_1B_1C_1$, the top face is a translation of the base, i.e. $$\overrightarrow{AA_1}=\overrightarrow{BB_1}=\overrightarrow{CC_1}.$$ Plane $ABC_1$ cuts the prism into two pyramids: tetrahedron $ABCC_1$ and quadrilateral pyramid $C_1-ABB_1A_1$. Find the ratio $$V_{ABCC_1}:V_{C_1-ABB_1A_1}.$$

Use vectors (coordinates). Take $A$ as the origin. Let $$\vec u=\overrightarrow{AB},\qquad \vec w=\overrightarrow{AC},\qquad \vec v=\overrightarrow{AA_1}.$$ Then $B=A+\vec u$, $C=A+\vec w$, and $C_1=C+\vec v=A+\vec w+\vec v$.

The volume of tetrahedron $ABCC_1$ is $$V_{ABCC_1}=\frac16\left|\det(\vec u,\vec w,\overrightarrow{AC_1})\right|= \frac16\left|\det(\vec u,\vec w,\vec w+\vec v)\right|.$$ Since $\det(\vec u,\vec w,\vec w)=0$, this simplifies to $$V_{ABCC_1}=\frac16\left|\det(\vec u,\vec w,\vec v)\right|.$$ The prism volume is base-area $\times$ height (triple product form): $$V_{\text{prism}}=\frac12\left|\det(\vec u,\vec w,\vec v)\right|.$$ Hence the other pyramid has volume $$V_{C_1-ABB_1A_1}=V_{\text{prism}}-V_{ABCC_1}=\left(\frac12-\frac16\right)\left|\det(\vec u,\vec w,\vec v)\right|=\frac13\left|\det(\vec u,\vec w,\vec v)\right|.$$ Therefore $$V_{ABCC_1}:V_{C_1-ABB_1A_1}=\frac16:\frac13=1:2.$$

$V_{ABCC_1}:V_{C_1-ABB_1A_1}=1:2$.