1979 AMC 12 Problem 23

Question

The edges of a regular tetrahedron with vertices

A,

B,

C,

and

D

each have length one. Find the least possible distance between a pair of points

P

and

Q,

where

P

is on edge

AB

and

Q

is on edge

CD.

\frac12

\frac34

\frac{\sqrt2}{2}

\frac{\sqrt3}{2}

\frac{\sqrt3}{3}

Use symmetric coordinates for a regular tetrahedron with vertices

(\pm1,\pm1,\pm1)

Parameterize points on the two opposite edges and minimize the squared distance

Before scaling, take

\begin{aligned} A&=(1,1,1),\\ B&=(1,-1,-1),\\ C&=(-1,1,-1),\\ D&=(-1,-1,1). \end{aligned}

These edges have length

2\sqrt2,

so scale by

\frac{1}{2\sqrt2}.

Points on

AB

and

CD

have forms

P=(1,t,t)

and

Q=(-1,u,-u).

Before scaling,

\begin{aligned} PQ^2&=4+(t-u)^2+(t+u)^2\\ &=4+2t^2+2u^2, \end{aligned}

minimized at

t=u=0.

After scaling, the minimum squared distance is

\frac{4}{8}=\frac{1}{2},

so the distance is

\frac{\sqrt2}{2}.

Therefore, the correct answer is C.

Problem 23 in other years