Competition problem — Geometry — Olympiad Maths Prep

Library / /23 of 41

Geometry Difficulty 5.5 AIME, harder Prove it Romania

Consider a cube $ABCD A'B'C'D'$ and two points $M \in (CD')$ and $N \in (DA')$ . Prove that $MN$ is the common perpendicular of the lines $CD'$ and $DA'$ if and only if
$\frac{D'M}{D'C} = \frac{DN}{DA'} = \frac{1}{3}.$

Solution

If $\frac{D'M}{D'C} = \frac{DN}{DA'} = \frac{1}{3}$ , then $M$ and $N$ are the centroids of the triangles $DD'C$ and, respectively, $ADD'$ , hence the points $C'$ , $M$ , $P$ and $A$ , $N$ , $P$ are collinear,

where $P$ is the midpoint of the side $DD'$ .
From the triangle $APC'$ , $\frac{PM}{PC'} = \frac{PN}{PA}$ , hence $MN \parallel AC'$ .
From $CD' \perp (ADC')$ and $DA' \perp (AD'C')$ follows $AC' \perp D'C$ and $AC' \perp A'D$ . This shows that $MN$ is the common perpendicular of the lines $CD'$ and $DA'$ .

For the converse, if $MN$ is the common perpendicular of the two lines, due to the uniqueness of this line, it must coincide with the line from a), hence $\frac{D'M}{D'C} = \frac{DN}{DA'} = \frac{1}{3}$ .

Looking for a route rather than an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.