mateforum.ro

Sa se arate ca daca triunghiul \( ABC \) este ascutitunghic atunci \( p^2\ge 2R^2+8Rr+3r^2 \) (Walker).

Inegalitatea este echivalenta cu
\( a^2+b^2+c^2 \geq 4(R+r)^2 \)
a carei demonstratie se gaseste aici.

Virgil Nicula wrote: Sa se arate ca daca triunghiul \( ABC \) este ascutitunghic atunci \( p^2\ge 2R^2+8Rr+3r^2 \) (Walker).

Demonstratie. Voi folosi identitatile liniar-unghiulare remarcabile \( \left\|\begin{array}{cc}
a^2+b^2+c^2=2\left(p^2-r^2-4Rr\right) & (1)\\\\\\\\
4S=\left(b^2+c^2-a^2\right)\tan A & (2)\\\\\\\\
\sin 2A+\sin 2B+\sin 2C=\frac {2S}{R^2} & (3)\\\\\\\\
\cos A+\cos B+\cos C=1+\frac rR & (4)\end{array}\right\| \). Prin urmare,

\( a^2+b^2+c^2=\sum\left(b^2+c^2-a^2\right)\ \stackrel {(2)}{=}\ 4S\sum \frac {\cos A}{\sin A}=8S\sum \frac {\cos^2A}{\sin 2A}\ \stackrel{C.B.S.}{\ge}\ 8S\frac {\left(\sum \cos A\right)^2}{\sum\sin 2A}\ \stackrel{(3)\wedge (4)}{=}\ 8S\frac {\left(1+\frac rR\right)^2}{\frac {2S}{R^2}}= \)

\( =4(R+r)^2\Longrightarrow \overline{\underline{\left\|a^2+b^2+c^2\ge 4(R+r)^2\\right\|}}\ \stackrel{(1)}{\Longrightarrow}\ 2\left(p^2-r^2-4Rr\right)\ge 4(R+r)^2\ \Longrightarrow\ \underline{\overline{\left\|p^2\ge 2R^2+8Rr+3r^2\right\|}}. \)