mateforum.ro

Aratati ca in orice triunghi avem \( OH \geq R|\sin(B-C)| \). Cand are loc egalitatea ?

In triunghiul AOH : \( \frac{AH}{|\sin (B-C)|}=\frac{R}{\sin AHO} \)

\( \Longrightarrow \) \( OH=\frac{R|\sin (B-C)|}{\sin AHO}\ge R|\sin (B-C)| \)

\( OH\ \ge\ pr_{BC}OH=pr_{BC}OA=R\cdot \left|\cos\left(90^{\circ}-B+C\right)\right|=R\cdot \left|\sin (B-C)\right| \) .

Incearca sa gasesti asemanator o minorare pentru \( IH \) ... Poate iese ceva frumos, nu stiu. Succes !

Pentru \( IH \):

\( IH\ \ge\ pr_{BC}IH=pr_{BC}AI=\frac{b+c}{2p}\sqrt{b^2_a-h^2_a} \)

\( \Longleftrightarrow IH\ge\frac{b+c}{2p}\sqrt{4p(p-a)\left\(\frac{bc}{(b+c)^2}-\frac{(p-b)(p-c)}{a^2}\right\)} \)

Iata alta : \( \underline{\overline {\left\|\ \begin{array}{c}
A=60^{\circ}\ \Longleftrightarrow\ OI=2R\cdot\sin\frac {|B-C|}{4}\\\\\\\\

OI=R\sqrt {\left(2\sin\frac A2-1\right)^2+8\sin\frac A2\sin^2\frac {B-C}{4}}\\\\\\\\
OI\ \ge\ 2R\cdot \sin\frac {|B-C|}{4}\sqrt {2\sin\frac A2}\end{array}\ \right\|}} \)

Virgil Nicula wrote:\( A=60^{\circ}\ \Longleftrightarrow\ OI=2R\cdot\sin\frac {|B-C|}{4} \)

In identitatea \( \overline{\underline{\left\|\ \cos A+\cos B+\cos C=1+\frac rR\ \right\|}} \) particularizam cu \( A=60^{\circ} \) si obtinem :

\( \cos B+\cos C=\frac 12+\frac rR\ \Longleftrightarrow\ 2\cos\frac{B+C}{2}\cdot\cos\frac{B-C}{2}=\frac 12+\frac rR\ \Longleftrightarrow\ \cos\frac{B-C}{2}=\frac 12+\frac rR \)

\( \Longleftrightarrow\ 1-\cos\frac{B-C}{2}=\frac 12-\frac rR\ \Longleftrightarrow\ 2sin^2\frac{B-C}{4}=\frac{R-2r}{2R}\ \Longleftrightarrow\ 4R\sin^2\frac{B-C}{4}=R-2r \)

\( \Longleftrightarrow\ 4R^2\sin^2\frac{B-C}{4}=R^2-2Rr=OI^2\ \Longleftrightarrow\ \underline{\overline{\left\|\ OI=2R\cdot\sin\frac{|B-C|}{4}\ \right\|}} \)

Virgil Nicula wrote:\( OI=R\sqrt {\left(2\sin\frac A2-1\right)^2+8\sin\frac A2\sin^2\frac {B-C}{4}} \)

Deoarece \( \cos A=1-2\sin^2\frac A2 \)

\( \Longrightarrow \cos A+\cos B+\cos C=1+\frac rR \) \( \Longleftrightarrow \cos B+\cos C=\frac rR+2\sin^2\frac A2 \)

\( \Longleftrightarrow 2\cos\frac{B+C}2\cos\frac{B-C}2=\frac rR+2\sin^2\frac A2 \)

\( \Longleftrightarrow \left\ 2\sin\frac A2\cos\frac{B-C}{2}=\frac{r}{R}+2\sin^2\frac A2\ \right|\ \odot\ 2R^2 \)

\( \Longleftrightarrow 4R^2\sin\frac A2\left\(1-2\sin^2\frac{B-C}4\right\)=2Rr+4R^2\sin^2\frac A2 \)

\( \Longrightarrow OI^2=R^2-2Rr=R^2+4R^2\sin^2\frac A2-4R^2\sin\frac A2+8R^2\sin\frac A2\sin^2\frac{B-C}{4} \)

\( \Longleftrightarrow OI^2=R^2\left\(2\sin^2\frac A2-1\right\)^2+8R^2\sin\frac A2\sin^2\frac{B-C}{4} \)

\( \Longleftrightarrow OI=R\sqrt {\left(2\sin\frac A2-1\right)^2+8\sin\frac A2\sin^2\frac {B-C}{4}} \)

Virgil Nicula wrote:\( OI\ \ge\ 2R\cdot \sin\frac {|B-C|}{4}\sqrt {2\sin\frac A2} \)

Inlocuind pe \( OI \) si ridicand la patrat, inegalitatea de demonstrat devine:

\( \left(2\sin\frac A2-1\right)^2\ge 0 \), ceea ce este adevarat. Egalitatea are loc daca si numai daca \( A=60^{\circ} \)