Problem 5

The cube

ABCDA'B'C'D'

has

A

above

A'

B

above

B'

, and so on. Let

X

be any point of the face diagonal

AC

and

Y

any point of

B'D'

. (a) Find the locus of the midpoint of

XY

. (b) Find the locus of the point

Z

XY

such that

ZY=2XZ

Step 4 of 5: Compute the one-third point Z

Z=\dfrac{2X+Y}{3}=\left(\dfrac{2t+1-s}{3},\dfrac{2t+s}{3},\dfrac23\right)

Detailed analysis

The condition $ZY=2XZ$ means $Z$ is one-third of the way from $X$ toward $Y$ , so $Z=(2X+Y)/3$ . Hence its height is always $z=2/3$ .