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Relatie remarcabila în triunghi
Posted: Fri Sep 25, 2009 10:31 pm
by Mateescu Constantin
Sa se arate ca in orice \( \triangle ABC \) are loc relatia: \( \overline{\underline{\left\|\ p-\sqrt 3 \cdot(R+r)=8R\cdot\sin\left\(\frac{\pi}{6}-\frac A2\right\)\cdot\sin\left\(\frac{\pi}{6}-\frac B2\right\)\cdot\sin\left\(\frac{\pi}{6}-\frac C2\right\)\ \right\|}} \) .
Re: Relatie remarcabila in triunghi
Posted: Fri Sep 25, 2009 11:13 pm
by Virgil Nicula
Lema. \( x+y+z=0\Longrightarrow \prod\sin x=-\frac 14\cdot \sum \sin 2x \) (se arata usor de la dreapta spre stanga !).
Deoarece \( \sum\left\(\frac{\pi}{6}-\frac A2\right\)=0 \) putem aplica lema de mai sus. Asadar \( 8R\cdot\prod \sin\left\(\frac{\pi}{6}-\frac A2\right\)=-2R\cdot \sum\sin\left\(\frac{\pi}{3}-A\right\)= \)
\( -2R\cdot\sum\left(\frac {\sqrt 3}{2}\cdot\cos A-\frac 12\cdot\sin A\right)=-2R\cdot\left(\frac {\sqrt 3}{2}\cdot\sum\cos A-\frac 12\cdot\sum\sin A\right)=-2R\cdot\left[\frac {\sqrt 3}{2}\cdot\left(1+\frac rR\right)-\frac 12\cdot\frac pR\right]=p-\sqrt 3 \cdot(R+r) \) .