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

Answer: 137

Concepts:3D geometry midpoint vector

Difficulty rating: 2560

Small Hint:

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

Big Hint:

Expand $\lVert A+B-C-D\rVert^2$ in terms of the six edge lengths

Solution:

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 $d^2=137.$

1989 AIME Problem 12

Answer: 137

Difficulty rating: 2560

Small Hint:

Big Hint:

Solution:

Problem 12 in Other Years