Problem 3

The cube

ABCDA'B'C'D'

has upper face

ABCD

and lower face

A'B'C'D'

, with

A

directly above

A'

and so on. Point

X

moves at constant speed around

ABCD

, and point

Y

moves at the same speed around

B'C'CB

. They leave

A

toward

B

and

B'

toward

C'

simultaneously. Find the locus of the midpoint of

XY

Step 4 of 6: Third quarter: trace the segment from C to U

X=(1-x,1,0),\ Y=(1,1-x,0)\Longrightarrow\ M=(1-\tfrac{x}{2},1-\tfrac{x}{2},0)

Detailed analysis

While $X$ is on $CD$ and $Y$ on $CB$ , take $X=(1-x,1,0)$ and $Y=(1,1-x,0)$ . The midpoint joins $C$ to $U=(1/2,1/2,0)$ .