$n$ vectors are on the plane. We can move each vector forward and backeard on the line that the vector is on it. If there are 2 vectors that their endpoints concide we can omit them and replace them with their sum (If their sum is nonzero). Suppose with these operations with 2 different method we reach to a vector. Prove that these vectors are on a common line
Problem
Source: Iran 2005
Tags: vector, geometry proposed, geometry
02.09.2005 00:54
Hello, You should clarify your wording. What do you mean by "if their sum is not zero" ? Do you mean that we then keep the two initial vectors or we just eliminate them without replacing them by their sum (which is just one point) ? Also can you please define what you call two "different" methods ? Thanks, -- Julien Santini
03.09.2005 17:48
We have some allowed operations in each operation we can add two vectors that their sum is not zero vector (with problem conditions); And we can move a vector on a line that it's on it. You see we can do these allowed operations in many methods to reach to a single vector. You must prove that all these vectors are on the same line I wonder why you don't understand what I said. I think everything is clear!!
06.09.2018 19:18
Omid Hatami wrote: I wonder why you don't understand what I said. I think everything is clear!! The grammar, for starters.
07.09.2018 07:02
I think this problem reduce to showing the operation become associative (eg. $n=3$), and this we can represent as a triangle geometry problem. After associativity then rest is simple
17.08.2020 17:57
The sum of the vectors is invariant. Right??? Or I'm wrong?