A cube ABCDEFGH is given. Determine the locus of all midpoints of segments MN, where M is any point on segment AC and N any point on segment FH.

Question

Question is:A cube $ABCDEFGH$ is given. Determine the locus of all midpoints of segments $MN$, where $M$ is any point on segment $AC$ and $N$ any point on segment $FH$.

My understanding : Here $M$ and $N$ are variable point, so its coordinates will have $1$ parameter of each. Now the midpoint of them will have those $2$ parameters.

My attempt : Assigning all the vertex of cube by 3d coordinates making $G$ as the origin (0,0,0) and each side have length a So $F$ is $(a,a,0)$, $A$ is $(a,0,a)$, $C$ is $(0,a,a)$ Now let $N$ divide the segment $FH$ by $n:1$ ratio, and $M$ divide the segment $AC$ by $m:1$ ratio.

So $N$ is $\left[\frac{a}{n+1}, \frac{a}{n+1}, 0\right]$,

$M$ is $\left[\frac{a}{m+1}, \frac{am}{m+1}, a\right]$.

So $L$ is $\left[\frac {a(m+n+2)} {2(m+1)(n+1)}, \frac {a(2m+mn+1)} {2(m+1)(n+1)}, \frac{a}{2}\right]$

Is this correct?

Should the locus be expressed in other form? Thanks in advance!

Answer 1

Point $M = A + t (C - A) = (a, 0, a) + t ( (0, a, a) - (a, 0, a) )$

Simplifying, we obtain,

$M = (a, 0, a) + t (-a, a, 0) = (a (1 - t) , a t , a ) $

Similarly,

Point $N = F + s (H - F) = (a, a, 0) + s ( (0,0,0) - (a, a, 0) )$

Simplifying,

$N = (1 - s) (a, a, 0)$

The midpoint is $L = \dfrac{1}{2} (M + N) $, so ,

$L = \dfrac{1}{2} ((a (1 - t) , a t , a ) + ( (1-s) a , (1-s) a , 0 ) )$

Simplifying,

$L = (a , \dfrac{a}{2}, \dfrac{a}{2}) + t ( -\dfrac{a}{2}, \dfrac{a}{2}, 0) + s ( -\dfrac{a}{2}, -\dfrac{a}{2}, 0 ) $

Remember that $t,s \in [0, 1]$

The locus is a horizontal patch that has the shape of a square with one vertex at $(a, \dfrac{a}{2}, \dfrac{a}{2})$ and with adjacent sides along vectors $( -\dfrac{a}{2}, \dfrac{a}{2}, 0)$ and $( -\dfrac{a}{2}, -\dfrac{a}{2}, 0 ) $

Answer 2

You can simplify your working and you need a few more steps to complete your working as what you have got is not locus of midpoints.

Any point $M$ on $AC$ is given by $(m, a - m, a), 0 \leq m \leq a$. Any point $N$ on $FH$ is given by $(n , n, 0), 0 \leq n \leq a$.

Midpoint of $MN$ is then given by $\left(x = \frac{m+n}{2}, y = \frac{a - m + n}{2}, z = \frac{a}{2} \right)$

$m+n = 2x$ and $m-n = a - 2y$. Solving,
$m = x-y + \frac{a}{2}$
$\implies x - y = m - \frac{a}{2}, - \frac{a}{2} \leq x - y \leq \frac{a}{2}$

Similarly, $x + y = n + \frac{a}{2}, \frac{a}{2} \leq x + y \leq \frac{3a}{2}$

And $ \ z = \frac{a}{2}$

Also note the set of lines $x - y = m - \frac{a}{2}, z = \frac{a}{2}$ are perpendicular to set of lines given by $x + y = n + \frac{a}{2}, z = \frac{a}{2}$.

So the locus is a square in plane $z = \frac{a}{2}$ bound by lines,
$x - y = - \frac{a}{2}, x - y = \frac{a}{2}, x + y = \frac{a}{2} $ and $x + y = \frac{3a}{2}$