Vtyjkyuk

Let $ABCD$ a tetrahedron. $P$ and $Q$ be points on the sides $AB$ and $CD$ respectively. What is the locus of the midpoint of the line segment $PQ$?

The locus is the set of points $X$ satisfying the condition that $X$ is the midpoint of the line segment $PQ$. How do I go about finding and describing this set?

Say we fix $Q$ for a moment and move only $P$ on $AB$. Then $X$ describes a middle segment in $ABQ$. This segment is of lenght $AB/2$ no matter where is $Q$ on $CD$.

Let $K,L,M,N$ be a midpoints of $AC,BC,AD,BD$ respectively.

If $Q= C$ then this middle segment is $KL$ and if $Q=D$ then it is $NM$.

If $Q$ is in interior of $CD$ this middle segment has end points on segments $KN$ and $LN$.

So $X$ describes paralelogram (with it interior) $KLMN$.

Denote points of tetrahedron with $A(0, 0, 0)$, $B(a, 0, 0)$, $C(fraca2,fracasqrt 32, 0)$, $D(fraca2,fracasqrt36,fracasqrt63)$

Arbitrary points $Pin AB$ and $Qin CD$ have the following coordinates:

$$P(pa,0,0), Q(fraca2, fracasqrt36(3-2q),fracaqsqrt63)$$

...with:

$$p,qin[0,1]$$

Coordinates of the midpoint are:

$$M(fraca4(2p+1), fracasqrt312(3-2q),fracaqsqrt66)$$

or:

$$x=fraca4(2p+1)tag1$$

$$y=fracasqrt312(3-2q),space z=fracaqsqrt66tag2$$

Obviously, you can pick the value for $x$ independently of $y,z$.

The range for $x$ is:

$$x_p=0=fraca4,space x_p=1=frac3a4,$$

For any given $x$ the locus of points is given with (2) which is actually a straight line going from:

$$y_q=0=afracsqrt34, space z_q=0=0$$

...to:

$$y_q=1=afracsqrt312, space z_q=1=afracsqrt 66$$

So the locus of midpoints is actually a rectangle with the following coordinates:

$$(fraca4, afracsqrt34, 0),space(fraca4, afracsqrt312, afracsqrt 66)$$

$$(frac3a4, afracsqrt34, 0),space(frac3a4, afracsqrt312, afracsqrt 66)$$