We have connected four metal pieces to each other such that they have formed a tetragon in space and also the angle between two connected metal pieces can vary. In the case that the tetragon can't be put in the plane, we've marked a point on each of the pieces such that they are all on a plane. Prove that as the tetragon varies, that four points remain on a plane. proposed by Erfan Salavati
Problem
Source: Iran 3rd round 2011-final exam-p3
Tags: geometry proposed, geometry
sumita
11.09.2011 14:49
Just 3D Menelaus' theorem.
goodar2006
11.09.2011 15:04
yep, and one can solve the problem using only this fact that the intersection of two planes is a line
vittasko
11.09.2011 20:36
It is also a direct application of the ERIQ theorem $($ = Equal Ratios In Quaqdrilateral $)$. Two elementary proofs of this powerful theorem, have been posted Here. Kostas Vittas.
Pathological
06.11.2019 06:11
Let our tetragon be $ABCD.$ Call points $E, F, G, H$ on $AB, BC, CD, DA$ respectively special if they are coplanar for some arbitrary orientation of the tetragon so that it's not on one plane. Let $d_1, d_2, d_3, d_4$ be the signed distances of $A, B, C, D$ to the plane. We have
$$\frac{AE}{EB} = \frac{-d_1}{d_2}$$
and similar relations, so multiplying yields:
$$\prod \frac{AE}{EB} = 1.$$
Conversely, for any other orientation of the tetragon, if we let $H'$ lie on $DA$ so that $E, F, G, H'$ are coplanar, and define $d_1', d_2', d_3', d_4'$ as the distances from $A, B, C, D$ to this plane, then the above relation implies that $ \frac{DH}{HA} = \frac{DH'}{H'A}$, so $H = H'.$ Hence we've shown that $E, F, G, H$ are also coplanar for any other orientation of the tetragon, and so we're done.
$\square$