MathDB

perp. common in cube, D'M/D'C=DN/DA'=1/3 (2016 Romanian NMO grade VII P2)

Source:

June 1, 2020

3D geometrygeometrycubeperpendicularratio

Problem Statement

In a cube

ABCDA'B'C'D'

two points are considered,

M \in (CD')

and

N \in (DA')

. Show that the

MN

is common perpendicular to the lines

CD'

and

DA'

if and only if

\frac{D'M}{D'C}=\frac{DN}{DA'} =\frac{1}{3}.