O caractericare remarcabila a unui triunghi neobtuzunghic.
Posted: Sun Aug 23, 2009 8:27 pm
Sa se arate ca triunghiul \( ABC \) este neobtuzunghic daca si numai daca \( \underline{\overline{\left\|\ p\ \ge\ 2R+r\ \right\|}} \) .
\( p\ \ge\ 2R+r\ \Longleftrightarrow\ p^2\ \ge\ 4R^2+4Rr+r^2\ \Longleftrightarrow\ 2p^2-8Rr-2r^2\ \ge\ 8R^2 \)
\( \Longleftrightarrow\ a^2+b^2+c^2\ \ge\ 8R^2\ \Longleftrightarrow\ \sin^{2}A+\sin^{2}B+\sin^{2}C\ \ge\ 2\ \Longleftrightarrow\ 3-\sum\cos 2A\ \ge\ 4 \)
\( \Longleftrightarrow\ 1+4\prod\cos A\ \ge\ 1\ \Longleftrightarrow\ \cos A\cos B\cos C\ \ge\ 0\ \Longleftrightarrow\ \triangle ABC \) este neobtuzunghic .
Egalitatea are loc pentru triunghiul dreptunghic (aici) .
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\( \bullet\ \ \)Astfel desprindem urmatoarea concluzie: \( \overline{\underline{\left\|\ \triangle ABC \ \begin{array}{ccc}
\nearrow & \normal\mbox{ascutitunghic} & \Longleftrightarrow & p & > & 2R & + & r \\\\\\\\
\rightarrow & \normal\mbox{dreptunghic} & \Longleftrightarrow & p & = & 2R & + & r \\\\\\\\
\searrow & \normal\mbox{obtuzunghic} & \Longleftrightarrow & p & < & 2R & + & r \end{array}\ \right\| \) .