Fie \( \Gamma \) semicercul inchis avand centrul in origine si raza unitate, situat in semiplanul \( \{z \in \mathbb{C} \ : \ \mathrm{Im} z \ge 0\} \) si \( A_1, ..., A_{2n} \in \Gamma \).
Presupunem ca proiectia vectorului \( \mathbf{v} := \vec{OA_1} + \cdots + \vec{OA_{2n}} \) pe axa reala este un numar intreg impar.
Aratati ca proiectia lui \( \mathbf{v} \) pe axa imaginara este, in modul, mai mare sau egala cu \( 1 \).
[ DMO 2008, Problema 3 ]
Inegalitate trigonometrica non-standard
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Inegalitate trigonometrica non-standard
Life is complex: it has real and imaginary components.
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Avem \( \vec{OA_i}(x_i,y_i) \) cu \( x_i^2+y_i^2=1 \). Am ales semiplanul astfe incat \( y_i \) sunt pozitivi. Atunci \( \sum y_i=\sum \sqrt{1-x_i^2}\geq \sum (1-x_i^2)\geq \sum(1-|x_i|)=2n+\sum_{x_i<0}x_i -\sum_{x_i\geq 0}x_i=2n-A-B \).
Unde am notat cu \( A,B \) suma modulelor numerelor negative, respectiv pozitive. Atunci \( B-A=2k+1 \).
Caz 1 \( B> A\Rightarrow 2k+1>0 \Rightarrow 2n-A-B=2n+2k+1-2B\geq 1+2n-2A \geq 1 \)
Caz2 \( A<B \Rightarrow 2k+1<0 \Rightarrow 2n-A-B=2n-(2k+1)-2A\geq 1+2n-2A \geq 1 \).

Unde am notat cu \( A,B \) suma modulelor numerelor negative, respectiv pozitive. Atunci \( B-A=2k+1 \).
Caz 1 \( B> A\Rightarrow 2k+1>0 \Rightarrow 2n-A-B=2n+2k+1-2B\geq 1+2n-2A \geq 1 \)
Caz2 \( A<B \Rightarrow 2k+1<0 \Rightarrow 2n-A-B=2n-(2k+1)-2A\geq 1+2n-2A \geq 1 \).
Last edited by Beniamin Bogosel on Thu Nov 27, 2008 3:04 pm, edited 1 time in total.
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