1989 AIME Problem 12

Question

Let

ABCD

be a tetrahedron with

AB=41,

AC=7,

AD=18,

BC=36,

BD=27,

and

CD=13,

as shown in the figure. Let

d

be the distance between the midpoints of edges

AB

and

CD.

Find

d^2.

Answer

Represent the vertices by vectors and write the vector between the two midpoints

Expand

\lVert A+B-C-D\rVert^2

in terms of the six edge lengths

Let the vertex names also denote their position vectors. The vector between the midpoints is

\frac{A+B-C-D}{2}.

Expanding squared lengths gives

\begin{aligned}4d^2={}&AC^2+AD^2\\&+BC^2+BD^2\\&-AB^2-CD^2.\end{aligned}

Therefore

\begin{aligned}4d^2={}&7^2+18^2+36^2+27^2\\&-41^2-13^2=548,\end{aligned}

so

d^2=137.

Problem 12 in other years