Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
Already signed up?
New to KöMaL?
I want the old design back!!! :-)

Problem A. 503. (March 2010)

A. 503. In space, there are given some vectors u1,u2,...,un and v such that |u1|\ge1, ..., |un|\ge1 and |v|\le1, and u1+...+un=0. Show that

|u1-v|+...+|un-v|\gen.

(5 pont)

Deadline expired on April 12, 2010.


Solution.

Lemma. If a,b are vectors such that |a|\ge1 and |b|\le1 then

|a-b|\ge|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)\ge0.

 

To solve the problem, apply the lemma to a=ui, v=b and the apply the triangle inequality:

|u1-v|+...+|un-v|\ge|1-u1v|+...+|1-unv|\ge|(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