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 3 of 6: Second quarter: trace the segment from W to C

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

Detailed analysis

While $X$ is on $BC$ and $Y$ on $C'C$ , use $X=(1,x,0)$ and $Y=(1,1,1-x)$ . Then $M=(1,(x+1)/2,(1-x)/2)$ , which joins $W$ to $C=(1,1,0)$ .