mateforum.ro

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 \( d_1\cap d_2\cap d_3\ne\emptyset\ \Longleftrightarrow\ 1+mpr+nqs=mq+ps+rn\ . \)

Cred ca a doua problema e mai degraba afina decit metrica.
L.O.

Acum nu am timp pentru ca trebuie sa plec la scoala , dar va promit ca diseara o sa postez rezolvarea la doua problema domnule Nicula .

Am revenit ...

Comentariu : Sunt clasa a IX-a si cred ca stiu sa umblu pe Google . O sa tin cont de recomandarea dumneavoastra , domnule Nicula .

Acum a doua problema ...

Fie \( d_1\cap d_3=\{D\} , d_1\cap d_2=\{E\},\ d_3 \cap AB =\{X\} \Rightarrow \frac{XB}{XA}\cdot \frac{SC}{SB}\cdot \frac{RA}{RC}=1 \) (Menelaus) \( \Rightarrow \frac{XB}{XA}=\frac{s}{r} \) apoi vectorial se arata imediat ca \( \frac{XM}{XA}=\frac{s-mr}{r(1-m)} \) si \( \frac{RN}{RA}=\frac{r(1-n)}{1-nr} \) .
Apoi se aplica Teorema lui Menelaus in \( \Delta AMN \Rightarrow \) \( \frac{DM}{DN}\cdot \frac{RN}{RA}\cdot \frac{XA}{XM} =1 \Rightarrow \frac{DM}{DN}=\frac{(1-n)(s-mr)}{(1-m)(1-nr)} \)
Folosind un rationament asemanator de arata ca \( \frac{EM}{EN}=\frac{(1-n)(1-mq)}{(1-m)(p-nq) \) . Din aceste doua relatii se deduce ca \( \frac{DM}{DN}=\frac{EM}{EN}\ \Longleftrightarrow\ d_1\cap d_2\cap d_3\neq \emptyset \) adica \( (1-mq)(1-nr)=(p-nq)(s-mr) \).