Problem A. 503. (March 2010)
A. 503. In space, there are given some vectors u1,u2,...,un and v such that |u1|1, ..., |un|1 and |v|1, and u1+...+un=0. Show that
|u1-v|+...+|un-v|n.
(5 pont)
Deadline expired on April 12, 2010.
Solution.
Lemma. If a,b are vectors such that |a|1 and |b|1 then
|a-b||1-ab|. | (1) |
Proof:
|a-b|2-|1-ab|2=(|a|2-2ab+|b|2)-(1-2ab+|ab|2)=(|a|2-1)(1-|b|2)+(|a2|.|b2|-|ab|2)0.
To solve the problem, apply the lemma to a=ui, v=b and the apply the triangle inequality:
|u1-v|+...+|un-v||1-u1v|+...+|1-unv||(1-u1v)+...+(1-unv)|=|n-(u1+...+un)v|=n.
Statistics:
8 students sent a solution. 5 points: Backhausz Tibor, Bodor Bertalan, Éles András, Nagy 648 Donát, Szabó 928 Attila. 4 points: Nagy 235 János. 0 point: 2 students.
Problems in Mathematics of KöMaL, March 2010