Relatie remarcabila în triunghi
Moderators: Laurian Filip, Beniamin Bogosel, Filip Chindea
-
Mateescu Constantin
- Newton
- Posts: 307
- Joined: Tue Apr 21, 2009 8:17 am
- Location: Pitesti
Relatie remarcabila în triunghi
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\|}} \) .
Last edited by Mateescu Constantin on Tue May 11, 2010 9:56 pm, edited 2 times in total.
-
Virgil Nicula
- Euler
- Posts: 622
- Joined: Fri Sep 28, 2007 11:23 pm
Re: Relatie remarcabila in triunghi
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\(\frac{\pi}{3}-A\right\)= \)
\( -2R\cdot\sum\left(\frac {\sqrt 3}{2}\cdot\cos A-\frac 12\cdot\sin A\right)=-2R\cdot\left(\frac {\sqrt 3}{2}\cdot\sum\cos A-\frac 12\cdot\sum\sin A\right)=-2R\cdot\left[\frac {\sqrt 3}{2}\cdot\left(1+\frac rR\right)-\frac 12\cdot\frac pR\right]=p-\sqrt 3 \cdot(R+r) \) .
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\(\frac{\pi}{3}-A\right\)= \)
\( -2R\cdot\sum\left(\frac {\sqrt 3}{2}\cdot\cos A-\frac 12\cdot\sin A\right)=-2R\cdot\left(\frac {\sqrt 3}{2}\cdot\sum\cos A-\frac 12\cdot\sum\sin A\right)=-2R\cdot\left[\frac {\sqrt 3}{2}\cdot\left(1+\frac rR\right)-\frac 12\cdot\frac pR\right]=p-\sqrt 3 \cdot(R+r) \) .