Problem A. 503. (March 2010)

A. 503. In space, there are given some vectors u₁,u₂,...,u_n and v such that |u₁| $\ge$ 1, ..., |u_n| $\ge$ 1 and |v| $\le$ 1, and u₁+...+u_n=0. Show that

|u₁-v|+...+|u_n-v| $\ge$ n.

(5 pont)

Deadline expired on April 12, 2010.

Solution.

Lemma. If a,b are vectors such that |a| $\ge$ 1 and |b| $\le$ 1 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) $\ge$ 0.

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

|u₁-v|+...+|u_n-v| $\ge$ |1-u₁v|+...+|1-u_nv| $\ge$ |(1-u₁v)+...+(1-u_nv)|=|n-(u₁+...+u_n)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

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.

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