$A; B; C; D; E$ and $F$ are points in space such that $AB$ is parallel to $DE$, $BC$ is parallel to $EF$, $CD$ is parallel to $FA$, but $AB \neq DE$. Prove that all six points lie in the same plane. (4 points)
Problem
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Tags: geometry, parallelogram
Ovchinnikov Denis
03.09.2010 12:53
Shis problem from Tournament Of Towns 2009 year, autumn It is easy problem. suppose that $A,B,C,D,,E,F$ don't lie in one plane. Then $A,B,E,D$ lie on one plane (because $AB || DE$) similarly $A,F,D,C$ and $B,C,F,E$ lie one one plane. $CD || AF; BC || EF => (BCD) || (AFE) => (BCD)\cap (ABDE) || (AEF)\cap (ABDE) => AE || DB => ABDE $ is parallelogram $=>AB=DE$ -contradiction , so $QED$
Amir Hossein
03.09.2010 13:04
Ovchinnikov Denis wrote: This problem from Tournament Of Towns 2009 year, autumn Yeah, I wrote in the title. TT=Tournament of Towns. You can see all of the problems here: http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=153&year=2009
Ovchinnikov Denis
03.09.2010 13:21
Sorry, i don,t understand