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Fie triunghiul A-dreptunghic \( ABC \) si punctele \( \{X,Y\}\subset [BC] \) astfel incat

\( \widehat {BAX}\equiv\widehat {XAY}\equiv\widehat {YAC} \) . Sa se arate ca \( \left\|\begin{array}{ccc}
a\cdot XY & = & 2\cdot XB\cdot YC\\\\\\\\
BX\cdot CY & \ge & bc\left(2-\sqrt 3\right)\\\\\\\\
XY & \le & a\left(2-\sqrt 3\right)\end{array}\right\| \) .

In \( \triangle\ XAY \) aplicam teorema cosinusului \( \Longrightarrow\ XY^2=AX^2+AY^2-\sqrt{3}\cdot AX\cdot AY\ (\ast) \) .

Cu teorema bisectoarei avem : \( \left\|\ \begin{array}{cc}
\frac{BX}{XY} & = & \frac{c}{AY} \\\\\\\\\\
\frac{CY}{XY} & = & \frac{b}{AX}\end{array}\ \right| \bigodot \ \Longrightarrow\ BX\cdot CY=\frac{bc\cdot XY^2}{AX\cdot AY} \)

\( \Longleftrightarrow^{(\ast)}\ BX\cdot CY=\frac{bc}{AX\cdot AY}\ \cdot\ \(AX^2+AY^2-\sqrt{3}\cdot AX\cdot AY\)\ \Longleftrightarrow\ BX\cdot CY=bc\left\(\frac{AX}{AY}+\frac{AY}{AX}-\sqrt{3}\right\) \)

Din inegalitatea \( x+\frac 1x\ge 2\ ,\ x\ >\ 0\ \Longrightarrow\ BX\cdot CY\ge bc(2-\sqrt{3}) \), cu egalitate cand \( b=c \).

Generalizare:


\( \left\|\ \begin{array}{cccc}
<<<\ \triangle\ ABC\ >>> \\\\\\\\
\{X\ ,\ Y\}\subset [BC] \\\\\\\\
\widehat {BAX}\equiv\widehat {XAY}\equiv\widehat {YAC}\ \end{array}\right\|\ \Longrightarrow\ \overline{\underline{\left\|\ BX\cdot CY\ \ge\ 2bc\left(1-\cos\frac A3\right)\ \right\|}} \) .


Cazuri particulare: \( \left\|\begin{array}{cccc}
\angle A & = & 45^{\circ} & \Longrightarrow & BX & \cdot & CY & \ge & \frac{bc(4-\sqrt 2-\sqrt 6)}{2} \\\\\\\\
\angle A & = & 90^{\circ} & \Longrightarrow & BX & \cdot & CY & \ge & bc(2-\sqrt{3}) \\\\\\\\
\angle A & = & 108^{\circ} & \Longrightarrow & BX & \cdot & CY & \ge & \frac{bc(3-\sqrt 5)}{2} \\\\\\\\
\angle A & = & 135^{\circ} & \Longrightarrow & BX & \cdot & CY & \ge & bc(2-\sqrt{2})\end{array}\ \right\| \)

F. frumoasa generalizarea ta. Imi place si scrierea ta in LaTeX.
Parca as fi scris-o eu. Te felicit, din partea mea ai nota +10.
Iti ofer o problema care adauga o inegalitate (relativ la \( XY \)) in generalizarea ta ...

C. Mateescu & V. Nicula wrote:Generalizare:

Fie \( \triangle ABC \) si punctele \( \{X,Y\}\subset BC \) astfel incat

\( \widehat {BAX}\equiv\widehat {XAY}\equiv\widehat {YAC} \). Sa se arate ca \( \left\|\begin{array}{ccc}
\ \ a\cdot XY & = & XB\cdot YC\cdot\left(4\cos^2\frac A3-1\right)\\\\\\\\
BX\cdot CY & \ge & 2bc\left(1-\cos\frac A3\right)\\\\\\\\
\ \ \ XY & \le & a\cdot\frac {2\cos\frac A3-1}{2\cos\frac A3+1}\ \ \ <\ \ \ \frac a3\end{array}\right\| \) .

Se poate arata ca \( \underline {\overline {\left\|\ \frac {2bc\left(1-\cos\frac A3\right)\left(4\cos^2\frac A3-1\right)}{a}\ \le\ XY\ \le\ a\cdot\frac {2\cos\frac A3-1}{2\cos\frac A3+1}\ \right\|}} \)

Observatie. \( \left(1-\co\frac A3\right)\left(2\cos\frac A3+1\right)^2=1-\cos A \) si \( \frac {2bc}{a}\le\frac {a}{1-\cos A} \) .

Ii multumesc foarte mult domnului profesor Nicula pentru problema oferita.

Sa demonstram mai intai urmatoarele:

Lema 1 (*)

Fie \( \triangle ABC \) si punctele \( \{X,Y\}\subset BC \) astfel incat \( \widehat {BAX}\equiv\widehat {XAY}\equiv\widehat {YAC} \).


Sa se arate ca : \( \fbox{\ \begin{array}{cc}
AX & = & \frac{bc\left\(4\cos^2\frac A3-1\right\)}{c+2b\cdot \cos\frac A3}\ ;\
AY & = & \frac{bc\left\(4\cos^2\frac A3-1\right\)}{b+2c\cdot \cos\frac A3}
\end{array}\ } \) . (Lungimile trisectoarelor intr-un triunghi)

Demonstratie : \( \ [ABX]+[AXC]=[ABC]\ \Longleftrightarrow\ \frac{c\cdot AX\cdot \sin \frac A3}{2}+\frac{b\cdot AX\cdot \sin\frac{2A}{3}}{2}=\frac{bc\cdot \sin A}{2} \)

\( \Longleftrightarrow\ AX=\frac{bc\cdot \sin A}{c\cdot \sin\frac A3+b\cdot \sin\frac{2A}{3}}\ \Longleftrightarrow\ AX=\frac{bc\left\(3\sin\frac A3-4\sin^3 \frac A3\right\)}{c\cdot \sin\frac A3+2b\cdot \sin\frac A3\cos\frac A3} \)

\( \Longleftrightarrow\ AX=\frac{bc\left\(3-4\sin^2 \frac A3\right\)}{c+2b\cdot \cos\frac A3}\ \Longleftrightarrow\ AX=\frac{bc\left\(4\cos^2\frac A3-1\right\)}{c+2b\cdot \cos \frac A3 \). Analog se calculeaza si \( AY \).

Lema 2 (**)

Daca in \( \triangle ABC \), \( M\ \in BC - \{B,C\} \) si \( \alpha=m(\angle MAB\)\ ,\ \beta=m(\angle MAC) \),

atunci exista relatia \( \fbox{\ \begin{array}\frac{MB}{MC} & = & \frac{c}{b} & \cdot & \frac{\sin\alpha}{\sin\beta}\end{array}\ }\ - \) caracterizarea pozitiei unui punct \( M\in BC \),

daca se cunosc masurile unghiurilor \( \widehat{MAB} \), \( \widehat{MAC} \) si raportul lungimilor laturilor \( (AB) \), \( (AC) \)

Demonstratie : \( \ \begin{array} \frac{MB}{MC} & = & \frac{[AMB]}{[AMC]} & = & \frac{c\cdot AM\cdot \sin\alpha}{b\cdot AM \cdot\sin\beta} & = & \frac cb & \cdot & \frac{\sin\alpha}{\sin\beta}\end{array} \)

==============================================================================

Virgil Nicula wrote:\( a\cdot XY = XB\cdot YC\cdot\left(4\cos^2\frac A3-1\right) \)

Demonstratie :

Aplicand Lema 2 (**) obtinem :

\( \left\|\ \begin{array}
\frac{BX}{XC} & = & \frac{c}{b} & \cdot & \frac{\sin\frac A3}{\sin\frac{2A}{3}} & = & \frac{c}{2b\cdot\cos\frac A3} & \Longleftrightarrow & \frac{BX}{a} & = & \frac{c}{c+2b\cdot\cos\frac A3}\\
\\\\\\
\frac{CY}{YB} & = & \frac{b}{c} & \cdot & \frac{\sin\frac A3}{\sin\frac{2A}{3}} & = & \frac{b}{2c\cdot\cos\frac A3} & \Longleftrightarrow & \frac{CY}{a} & = & \frac{b}{b+2c\cdot\cos\frac A3}\end{array}\right| \bigodot\ \Longrightarrow\ BX\cdot CY=\frac{a^2bc}{\left\(c+2b\cdot\cos\frac A3\right\)\left\(b+2c\cdot\cos\frac A3\right\)}\ (1)
\)


De asemenea avem relatiile :

\( \left\|\ \begin{array}{cc}
\frac{XY}{YC} & = & \frac{AX}{b}\ (\mbox{\normal T.Bis}) & \Longleftrightarrow & \frac{XY}{YC} & = & \frac{AX}{b} \\\\\\\\
\frac{YC}{YB} & = & \frac bc\ \cdot\ \frac{\sin\frac A3}{\sin\frac{2A}{3}}\ (\ast\ast)
& \Longleftrightarrow & \frac{YC}{a} & = & \frac{b}{b+2c\cdot \cos\frac A3}\end{array}\ \right|\ \bigodot\ \Longrightarrow\
\frac{XY}{a}=\frac{AX}{b+2c\cdot \cos\frac A3}\ \Longleftrightarrow\ a\cdot XY=\frac{a^2\cdot AX}{b+2c\cdot \cos\frac A3} \\\\
\\\ \\\
\\\ \\\
\\\ \\\
\Longleftrightarrow^{(\ast)}\ \fbox{\ a\cdot XY=\frac{a^2bc}{\left\(c+2b\cdot\cos\frac A3\right\)\left\(b+2c\cdot\cos\frac A3\right\)}\cdot \left\(4\cos^2\frac A3-1\right\)\ \ (2)\ }\ \Longleftrightarrow^{(1)}\ \fbox{\ a\cdot XY=BX\cdot CY\left(4\cos^2\frac A3-1\right\)\ } \)

Constantin Mateescu wrote:\( BX\cdot CY\ \ge\ 2bc\left(1-\cos\frac A3\right) \)

Demonstratie :

In \( \triangle XAY \) aplicam teorema cosinusului \( \Longrightarrow\ XY^2=AX^2+AY^2-2AX\cdot AY\cdot\cos\frac A3\ (3) \)

Cu teorema bisectoarei avem : \( \left\|\begin{array}{cc} \frac{BX}{XY} & = & \frac{c}{AY} \\\\\\\\\
\frac{CY}{XY} & = & \frac{b}{AX}\end{array}\right| \bigodot \ \Longrightarrow\ BX\cdot CY=\frac{bc\cdot XY^2}{AX\cdot AY} \)

\( \Longleftrightarrow^{(3)}\ BX\cdot CY=\frac{bc}{AX\cdot AY}\ \cdot\ \left\(AX^2+AY^2-2AX\cdot AY\cdot\cos\frac A3\right\)\ \Longleftrightarrow\ BX\cdot CY=bc\left\(\frac{AX}{AY}+\frac{AY}{AX}-2\cos\frac A3\right\) \)

Din inegalitatea \( x+\frac 1x\ge 2\ ,\ x\ >\ 0\ \Longrightarrow\ BX\cdot CY\ge 2bc\left(1-\cos\frac A3\right) \) , cu egalitate cand \( b=c \).

Virgil Nicula wrote:\( XY\ \le\ a\cdot\frac {2\cos\frac A3-1}{2\cos\frac A3+1}\ <\ \frac a3 \)

Demonstratie :

Din relatia \( (2)\ \Longrightarrow\ \fbox{\ XY=\frac{abc}{\left\(c+2b\cdot\cos\frac A3\right\)\left\(b+2c\cdot\cos\frac A3\right\)}\cdot \left\(4\cos^2\frac A3-1\right\)\ } \)

\( \Longleftrightarrow\ XY=\frac{abc\left\(2\cos\frac A3-1\right\)\left\(2\cos\frac A3+1\right\)}{bc+4bc\cdot\cos^2\frac A3+2\cos\frac A3(\underbrace{b^2+c^2}_{\small \ge 2bc})}\ \ \Longrightarrow\ \ XY\le\frac{abc\left\(2\cos\frac A3-1\right\)\left\(2\cos\frac A3+1\right\)}{bc\left\(4cos^2
\frac A3+4\cos\frac A3+1\right\)} \)

\( \Longleftrightarrow\ XY\le \frac{a\cdot \left\(2\cos\frac A3-1\right\)\left\(2\cos\frac A3+1\right\)}{\left\(2\cos\frac A3+1\right\)^2}\ \Longleftrightarrow\ \fbox{\ XY \le a\cdot\frac {2\cos\frac A3-1}{2\cos\frac A3+1}\ } \) , cu egalitate cand \( b=c \) .

Partea dreapta se reduce la : \( 6\cos\frac A3-3\ <\ 2\cos\frac A3+1\ \Longleftrightarrow\ \cos\frac A3\ <\ 1 \) adevarat .

Virgil Nicula wrote:\( \fbox{\ \frac {2bc\left(1-\cos\frac A3\right)\left(4\cos^2\frac A3-1\right)}{a}\ \le\ XY\ } \)

Demonstratie :

Intrucat \( \left\|\begin{array} XY=\frac{XB\cdot YC\cdot \left\(4\cos^2\frac A3-1\right\)}{a} \\\\\\\\\\\\\\
XB\cdot YC\ \ge\ 2bc\left\(1-\cos\frac A3\right\)\end{array}\ \right|\ \Longrightarrow\ \fbox{\ XY\ \ge\ \frac {2bc\left(1-\cos\frac A3\right)\left(4\cos^2\frac A3-1\right)}{a}\ } \)

Virgil Nicula wrote:\( \left(1-\cos\frac A3\right)\left(2\cos\frac A3+1\right)^2=1-\cos A \)

Demonstratie :

\( \left(1-\cos\frac A3\right)\left(2\cos\frac A3+1\right)^2=1-\cos A\ \Longleftrightarrow\ \left(1-\co\frac A3\right)\left\(4\cos^2\frac A3+4\cos\frac A3+1\right\)=1-\cos A
\\\ \\\
\\\ \\\
\Longleftrightarrow\ 4\cos^2\frac A3+4\cos\frac A3+1-4\cos^3\frac A3-4\cos^2\frac A3-\cos\frac A3=1-\cos A\ \Longleftrightarrow\ \cos A=4\cos^3\frac A3-3\cos\frac A3\ ,\ \mbox{\normal OK} \)

Virgil Nicula wrote:\( \frac {2bc}{a}\le\frac {a}{1-\cos A} \)

Demonstratie :

\( \frac {2bc}{a}\le\frac {a}{1-\cos A}\ \Longleftrightarrow\ 2bc-2bc\cdot \cos A\ \le\ a^2\ \Longleftrightarrow\ 2bc\ \le\ a^2+2bc\cdot\frac{b^2+c^2-a^2}{2bc}\ \Longleftrightarrow\ 2bc\ \le\ b^2+c^2\ ,\ \mbox{\normal OK} \)

Generalizare 2:


\( \left\|\ \begin{array}{ccc}
<<<\ \triangle\ ABC\ >>> \\\\\\\\
\{X\ ,\ Y\}\subset [BC]\ ;\ X\subset [BY] \\\\\\\\
\widehat{BAX}=x\ ;\ \widehat{XAY}=y\ ;\ \widehat{YAC}=z\end{array}\right\|\ \Longrightarrow \\\
\\\
\\\
\Longrightarrow\ \left\|\begin{array}{ccccccc}
\bullet & a\cdot XY&=&BX\cdot CY\ \cdot\ \frac{\sin y\cdot \sin A}{\sin x\cdot \sin z} \\\\\\\\\\
\bullet & \frac{BX\cdot CY}{BY\cdot CX}&=&\frac{\sin x\cdot\sin z}{\sin(A-x)\cdot\sin(A-z)} \\\\\\\\
\bullet & 2h_a\ \cdot\ \tan\frac y2 & \le & XY & \le & a & \cdot & \frac{\sqrt{\sin(A-x)\sin(A-z)}-\sqrt{\sin x\cdot\sin z}}{\sqrt{\sin(A-x)\sin(A-z)}+\sqrt{\sin x\cdot\sin z}} \\\\\\\\
\bullet & 2bc\ \cdot\ \frac{sin x\cdot\sin z}{1+\cos y} & \le & BX\cdot CY & \le & a^2 & \cdot & \frac{\sin x\cdot\sin z}{\left\[\sqrt{\sin x\cdot \sin z}+\sqrt{\sin(A-x)\cdot\sin(A-z)}\right\]^2} \\\\\\\\
\bullet & \frac{2bc\cdot\sin(A-x)\sin(A-z)}{1+\cos y} & \le & BY\cdot CX & \le & a^2 & \cdot & \frac{\sin(A-x)\cdot\sin(A-z)}{\left\[\sqrt{\sin x\cdot \sin z}+\sqrt{\sin(A-x)\cdot\sin(A-z)}\right\]^2} \\\\\\\\
\bullet & \frac{2}{1+\cos y}\ \cdot\ h_a^2 & \le & AX\cdot AY & \le & bc & \cdot & \frac{\sin^2 A}{\left\[\sqrt{\sin x\cdot \sin z}+\sqrt{\sin(A-x)\cdot\sin(A-z)}\right\]^2}\end{array}\ \right\| \) .

====================================================================================

Caz particular:

\( x=y=z=\frac{\angle A}{3}\ \Longrightarrow\
\left\|\begin{array}{cccc}
\bullet & \frac{2bc\left(1-\cos\frac A3\right)\left(4\cos^2\frac A3-1\right)}{a} & \le & XY & \le & a\cdot\frac {2\cos\frac A3-1}{2\cos\frac A3+1} \\\\\\\\
\bullet & 2bc\left\(1-\cos\frac A3\right\) & \le & BX\cdot CY & \le & \left\(\frac{a}{2\cos\frac A3+1}\right\)^2 \\\\\\\\
\bullet & 8bc\cdot \cos^2\frac A3\left\(1-\cos\frac A3\right\) & \le & BY\cdot CX & \le & \left\(\frac{2a\cdot\cos\frac A3}{2\cos\frac A3+1}\right\)^2 \\\\\\\\
\bullet & 2(1-\cos A)\left\[\frac{bc}{a}\left\(2\cos\frac A3-1\right\)\right\]^2 & \le & AX\cdot AY & \le & bc\left\(2\cos\frac A3-1\right\)^2\end{array}\ \right\| \) .


De aici deducem si urmatoarea inegalitate (mai tare !):

\( \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \fbox{\ 2bc(1-\cos A)\ \le\ 2\left\(1-\cos\frac A3\right\)\left\(b+2c\cdot\cos\frac A3\right\)\left\(c+2b\cdot\cos\frac A3\right\)\ \le\ a^2\ } \) .