Given a cube $(A'B'C'D'),(ABCD)$. The points have coordinates: \[A'(1,0,0),B'(1,1,0),C'(0,1,0),D'(0,0,0)\] and \[A(1,0,1),B(1,1,1),C(0,1,1),D(0,0,1)\] Question A. The direction of the line $AC$ is given by $(1,-1,0)$ The equation of line $AC$ is: \[\frac{x-1}{1}=\frac{y}{-1} \wedge z=1\] \[1-x=y \wedge z=1\] X is an arbitrary point of $AC$: X(1-t,t,1) with $0\leq t\leq 1$ The direction of the line $B'D'$ is given by $(1,1,0)$ The equation of line $B'D'$ is: \[\frac{x}{1}=\frac{y}{1} \wedge z=0\] \[x=y \wedge z=0\] Y is an arbitrary point of $B'D'$: Y(s,s,0) with $0\leq s\leq 1$ See drawing "cube" $M(x,y,z)$ is the midpoint of $[XY]$:$M(\frac{1-t+s}{2},\frac{t+s}{2},\frac{1}{2}).$ Then is: $x=\frac{1-t+s}{2}$ and $y=\frac{t+s}{2}$ \[ \left\{\begin{array}{l}{x+y=\frac{1}{2}+s}\\ {y-x=t-\frac{1}{2}}\end{array}\right. \Rightarrow \left\{\begin{array}{l}{x+y-\frac{1}{2}=s}\\ {-x+y+\frac{1}{2}=t }\end{array}\right. \] $0\leq t\leq 1$ and $0\leq s\leq 1$ For $s=t=o$, we find $x=\frac{1}{2}$,$y=0$, point $P(\frac{1}{2},0,\frac{1}{2})$ For $s=0$ and $t=1$, we find $x=0$,$y=\frac{1}{2}$, point $S(0,\frac{1}{2},\frac{1}{2})$ For $s=1$ and $t=0$, we find $x=1$,$y=\frac{1}{2}$, point $Q(1,\frac{1}{2},\frac{1}{2})$ For $s=t=1$, we find $x=\frac{1}{2}$,$y=1$, point $R(\frac{1}{2},1,\frac{1}{2})$ The locus of the midpoints $M$ is formed by all the points of the square $PQRS$ See drawing "cube1" Question B. The locus of the midpoints $Z$ with $\stackrel{\rightarrow}{ZY}=2\stackrel{\rightarrow}{XZ}$ \[\stackrel{\rightarrow}{Y}-\stackrel{\rightarrow}{Z}=2(\stackrel{\rightarrow}{Z}-\stackrel{\rightarrow}{X})\] \[\stackrel{\rightarrow}{Z}=\frac{2\stackrel{\rightarrow}{X}+\stackrel{\rightarrow}{Y}}{3}\] We know that $X(1-t,t,1)$ and $Y(s,s,0)$ Therefore: $Z(\frac{2-2t+s}{3},\frac{2t+s}{3},\frac{2}{3})$ and $0\leq t\leq 1$ and $0\leq s\leq 1$ The locus of the midpoints $Z$ is formed by all the points of the rectangle $P'Q'R'S'$: $P'(\frac{2}{3},0,\frac{2}{3})$,$Q'(1;\frac{1}{3},\frac{2}{3})$, $R'(0,\frac{2}{3},\frac{2}{3})$,$S'(\frac{1}{3},1,\frac{2}{3})$

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