TEORIE : Identitati intr-un triunghi.

[Virgil Nicula]

\( \odot\ \ \)Identitati liniare.

\( \bullet\ \begin{array}{c}
a^3+b^3+c^2=3abc+(a+b+c)\cdot \left[\ \left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)\ \right]\\\\
\left(a+b+c\right)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)\end{array}\ \Longrightarrow\ \begin{array}{c}
a^3+b^3+c^3\ \ge\ 3abc\\\\
(a+b+c)^3\ \ge\ a^3+b^3+c^3+24abc\\\\
\sum a\cdot\sum bc\ \ge\ 9abc\end{array} \)

\( \bullet\ (b+c)^2(p-b)(p-c)+(b-c)^2p(p-a)=a^2bc\ \Longrightarrow\ \sum (b+c)^2(p-b)(p-c)\ \le\ abc(a+b+c)\ . \)

\( \bullet\ \sum a(a-b)(a-c)+8(p-a)(p-b)(p-c)=abc\ \Longrightarrow\ \left\{\ \sum a(a-b)(a-c)\ \ge 0\ \Longleftrightarrow\ abc\ \ge\ \prod (b+c-a)\ \right\}\ . \)

\( \bullet\ \sum a^2(p-a)=abc+4\prod (p-a)\ ;\ \sum a^2(p-b)(p-c)=p\left[abc-4\prod (p-a)\right]\ . \)

\( \bullet\ \sum a(p-a)+2p^2=2\sum bc\ ;\ \sum (p-b)(p-c)+p^2=\sum bc\ ;\ \sum (b+c)(p-a)=\sum a^2\ . \)

\( \bullet\ abc+\prod (b+c)=(a+b+c)(ab+bc+ca)\ ;\ \sum a(p-a)^2=\sum a(p-b)(p-c)=abc-2\cdot\prod (p-a)\ . \)

\( \bullet\ \sum a(b+c)(p-a)=3abc\ ;\ \sum a(b+c)(p-b)(p-c)=abcp\ ;\ \sum a(p-a)=2\cdot\sum (p-b)(p-c)\ . \)

\( \bullet\ (p-b)(p-c)(b-c)^2\ +\ p(p-a)(b+c-2a)^2\ =\ 4\cdot\left[\ bcm_a^2\ -\ 2ap(p-a)^2\ \right]\ \Longrightarrow \)

\( 2ap(p-a)^2\ \le\ bc\cdot m^2_a\Longrightarrow 2a^2p(p-a)^2\ \le\ 4Rpr\cdot m_a^2\Longrightarrow a(p-a)\ \le\ m_a\cdot\sqrt {2Rr}\Longrightarrow \)

\( 2r(4R+r)\ \le\ \sum m_a\cdot \sqrt {2Rr}\ \Longrightarrow\ (4R+r)\cdot\sqrt {\frac {2r}{R}}\ \le\ \sum m_a\ \Longrightarrow\ \underline {\overline {\left\|\ \sqrt {\frac {2r}{R}}\ \le \frac {m_a+m_b+m_c}{4R+r}\ \le 1\ \right\|}}\ . \)

\( \bullet\ \begin{array}{ccccc}
p(p-a) & + & (p-b)(p-c) & = & bc\\\\
p(p-a) & - & (p-b)(p-c) & =& bc\cdot\cos A\end{array}\ \begin{array}{c}
+\\\\
-\end{array}\ \Longrightarrow\ \begin{array}{c}
p(p-a)=\frac 12\cdot bc(1+\cos A)=bc\cos^2\frac A2\\\\
(p-b)(p-c)=\frac 12\cdot bc(1-\cos A)=bc\sin^2\frac A2\end{array}\ \Longrightarrow \)

\( \left\|\ \begin{array}{c}
\sin\frac A2=\sqrt {\frac {(p-b)(p-c)}{bc}}\\\\
\cos\frac A2=\sqrt {\frac {p(p-a)}{bc}}\\\\
\tan\frac A2=\sqrt {\frac {(p-b)(p-c)}{p(p-a)}}\end{array}\ \right\|\ ;\ \begin{array}{ccc}
a=b\cdot\cos C+c\cdot\cos B & \ \odot & (-a)\\\\
b=c\cdot\cos A+a\cdot\cos C & \ \odot & b\\\\
c=a\cdot\cos B+b\cdot\cos A & \ \odot & c\end{array}\ \bigoplus\ \Longrightarrow\ \underline {\overline {\left\|\ b^2+c^2-a^2=2bc\cdot\cos A\ \right\|}}\ . \)

\( \odot\ \ \) Identitati liniar-unghiulare.

\( \bullet\ \sum\tan A=\prod\tan A\ \Longleftrightarrow\ \sum\cot B\cot C=1\ ;\ \sum\tan\frac B2\tan\frac C2=1\ . \)

\( \bullet\ \frac 1r\cdot\prod\sin\frac A2=\frac 1p\cdot\prod\cos\frac A2=\frac {1}{4R}\ ;\ \sum\sin A=4\cdot\prod\cos\frac A2\ ;\ 1+\prod\cos A=\frac {a^2+b^2+c^2}{8R^2}\ ;\ \sum a\cdot\tan\frac A2=2(2R-r)\ . \)

\( \bullet\ \sum\cos A=1+4\cdot\prod\sin\frac A2=1+\frac rR\ ;\ \sum \sin 2A=4\cdot\prod\sin A\ ;\ \sum\cos 2A=-1-4\cdot\prod\cos A=3-\frac {a^2+b^2+c^2}{2R^2}\ . \)

\( \odot\ \ \) Aria \( S \) a triunghiului \( ABC\ . \)

\( \bullet\ S=\frac {ah_a}{2}=\frac {bc\sin A}{2}=pr=(p-a)r_a=p(p-a)\tan\frac A2\ ;\ abc=4RS=4Rpr\ ;\ 16S^2+\left(b^2-c^2\right)^2=4a^2m_a^2\ . \)

\( \bullet\ bc=2Rh_a\ ;\ S=\sqrt {p(p-a)(p-b)(p-c)} \) (Heron) ; \( 4S\cdot\cot A=b^2+c^2-a^2\ ;\ S=2R^2\cdot\prod\sin A\ . \)

\( \bullet\ 16\cdot S^2=2\left(b^2c^2+c^2a^2+a^2b^2\right)-\left(a^4+b^4+c^4\right)=\sum a^2\left(b^2+c^2-a^2\right)=2abc\cdot\sum a\cdot\cos A\ . \)

[moldo]

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