Let $A B C D$ be a regular tetrahedron with side length 1 . Let $X$ be the point in triangle $B C D$ such that $[X B C]=2[X B D]=4[X C D]$, where $[\varpi]$ denotes the area of figure $\varpi$. Let $Y$ lie on segment $A X$ such that $2 A Y=Y X$. Let $M$ be the midpoint of $B D$. Let $Z$ be a point on segment $A M$ such that the lines $Y Z$ and $B C$ intersect at some point. Find $\frac{A Z}{Z M}$.
We apply three-dimensional barycentric coordinates with reference tetrahedron $A B C D$. The given conditions imply that \[ \begin{aligned} X & =(0: 1: 2: 4) \\ Y & =(14: 1: 2: 4) \\ M & =(0: 1: 0: 1) \\ Z & =(t: 1: 0: 1) \end{aligned} \] for some real number $t$. Normalizing, we obtain $Y=\left(\frac{14}{21}, \frac{1}{21}, \frac{2}{21}, \frac{4}{21}\right)$ and $Z=\left(\frac{t}{t+2}, \frac{1}{t+2}, 0, \frac{1}{t+2}\right)$. If $Y Z$ intersects line $B C$ then there exist parameters $\alpha+\beta=1$ such that $\alpha Y+\beta Z$ has zero $A$ and $D$ coordinates, meaning \[ \begin{aligned} \frac{14}{21} \alpha+\frac{t}{t+2} \beta & =0 \\ \frac{4}{21} \alpha+\frac{1}{t+2} \beta & =0 \\ \alpha+\beta & =1 \end{aligned} \] Adding twice the second equation to the first gives $\frac{22}{21} \alpha+\beta=0$, so $\alpha=-22, \beta=21$, and thus $t=\frac{7}{2}$. It follows that $Z=(7: 2: 0: 2)$, and $\frac{A Z}{Z M}=\frac{2+2}{7}=\frac{4}{7}$. $\fbox{\frac{4}{7}}$.