alex2008 wrote: Fie un triunghi \( ABC \) si punctele \( D \in (BC) \) , \( E \in (CA) \) , \( F \in (AB) \) . Atunci
dreptele \( DA \) , \( EB \) si \( FC \) sunt concurente \( \Longleftrightarrow\ \frac{DB}{DC}\cdot\frac{EC}{EA}\cdot\frac{FA}{FB}=1 \) (teorema lui Ceva).
Problema propusa (clasa a VIII - a).
Fie un triunghi \( ABC \) si punctele \( D \in (BC) \) , \( E \in (CA) \) , \( F \in (AB) \)
definite prin \( \frac {DB}{DC}=\alpha \) , \( \frac {EC}{EA}=\beta \) , \( \frac {FA}{FB}=\gamma\ . \) Notam \( X\in BE\cap CF \) , \( Y\in CF\cap AD \) ,
\( Z\in AD\cap BE\ . \) Sa se arate ca \( [XYZ]=\frac {(1-\alpha\beta\gamma )^2}{(1+\alpha +\alpha \beta )(1+\beta +\beta \gamma)(1+\gamma +\gamma \alpha )}\cdot [ABC]\ . \)
Cazuri particulare.
\( \odot \) \( AD\cap BE\cap CE\ne\emptyset\ \Longleftrightarrow\ \alpha\beta\gamma=1\ \Longleftrightarrow\ \frac {DB}{DC}\cdot \frac {EC}{EA}\cdot\frac {FA}{FB}=1\ . \)
\( \odot\ \alpha =\beta =\gamma\ \Longrightarrow\ [XYZ]=\frac {(1-\alpha )^2}{1+\alpha +\alpha^2}\cdot [ABC]\ . \)
Problema propusa (clasa a X - a). Fie un triunghi \( ABC \) si dreptele \( d_k\ ,\ k\in\overline {1,3}\ . \) Notam :
\( \begin{array}{ccc}
M\in d_1\cap AB & , & \overline {MB}=m\cdot\overline {MA}\\\\
P\in d_2\cap BC & , & \overline {PC}=p\cdot\overline {PB}\\\\
R\in d_3\cap CA & , & \overline {RA}=r\cdot\overline {RC}\end{array}\ \ \ \wedge\ \ \ \begin{array}{ccc}
N\in d_1\cap AC & , & \overline {NC}=n\cdot\overline {NA}\\\\
Q\in d_2\cap BA & , & \overline {QA}=q\cdot\overline {QB}\\\\
S\in d_3\cap CB & , & \overline {SB}=s\cdot\overline {SC}\end{array} \)
Sa se arate ca \( d_1\cap d_2\cap d_3\ne\emptyset\ \Longleftrightarrow\ 1+mpr+nqs=mq+ps+rn\ . \)