Fie triunghiul \( ABC \) si punctele \( X\in (BC) \) , \( Y\in (CA) \) , \( Z\in (AB) \) . Notam \( P\in AX\cap YZ \) .
Sa se arate ca \( \underline {\overline {\left\|\ \frac {PZ}{PY}=\frac {XB}{XC}\cdot\frac {AZ}{AY}\cdot \frac {AC}{AB}\ \right\|}} \) si \( \underline {\overline {\left\|\ \frac {ZB}{ZA}\cdot XC+\frac {YC}{YA}\cdot XB=\frac {PX}{PA}\cdot BC\ \right\|}} \) .
Doua relatii uzuale precum Menelaus sau Ceva.
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Virgil Nicula
- Euler
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Doua relatii uzuale precum Menelaus sau Ceva.
Last edited by Virgil Nicula on Wed Jan 28, 2009 11:38 pm, edited 2 times in total.
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Claudiu Mindrila
- Fermat
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Omer Cerrahoglu
- Euclid
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Hai ca demonstrez eu prima relatie 
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Se stie ca \( \frac{PZ}{PY}=\frac{A[AZP]}{A[APY]}=\frac{\frac{AZ.AP.\sin(BAX)}{2}}{\frac{AY.AP.\sin(CAX)}{2}}=\frac{AZ.\sin(BAX)}{AY.\sin(CAX)} \)
La fel putem obtine si ca \( \frac{BX}{CX}=\frac{AB.\sin(BAX)}{AC.\sin(CAX)} \).
Facand inlocuirile, relatia data este echivalenta cu:
\( \frac{AZ.\sin(BAX)}{AY.\sin(CAX)}=\frac{AB}{AC}.\frac{AC}{AB}.\frac{AZ.\sin(BAX)}{AY.\sin(CAX)} \Longleftrightarrow \frac{AZ.\sin(BAX)}{AY.\sin(CAX)}=\frac{AZ.\sin(BAX)}{AY.\sin(CAX)} \), care este adevarata, si astfel identitatea este demonstrata
.Se stie ca \( \frac{PZ}{PY}=\frac{A[AZP]}{A[APY]}=\frac{\frac{AZ.AP.\sin(BAX)}{2}}{\frac{AY.AP.\sin(CAX)}{2}}=\frac{AZ.\sin(BAX)}{AY.\sin(CAX)} \)
La fel putem obtine si ca \( \frac{BX}{CX}=\frac{AB.\sin(BAX)}{AC.\sin(CAX)} \).
Facand inlocuirile, relatia data este echivalenta cu:
\( \frac{AZ.\sin(BAX)}{AY.\sin(CAX)}=\frac{AB}{AC}.\frac{AC}{AB}.\frac{AZ.\sin(BAX)}{AY.\sin(CAX)} \Longleftrightarrow \frac{AZ.\sin(BAX)}{AY.\sin(CAX)}=\frac{AZ.\sin(BAX)}{AY.\sin(CAX)} \), care este adevarata, si astfel identitatea este demonstrata
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Liviu Ornea
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Virgil Nicula
- Euler
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Iata o demonstratie pentru prima relatie (pentru Claudiu Mandrila !) :
Proof. Intr-adevar, \( \left\|\begin{array}{c}
\frac {PF}{PE}=\frac {AF}{AE}\cdot\frac {\sin \widehat {DAB}}{\sin \widehat {DAC}}\\\\
\frac {DC}{DB}=\frac {AC}{AB}\cdot\frac {\sin \widehat {DAC}}{\sin \widehat {DAB}}\end{array}\right\|\ \bigodot\ \Longrightarrow\ \frac {PF}{PE} = \frac {DB}{DC}\cdot\frac {AF}{AE}\cdot\frac {AC}{AB} \) .
Lema 1. \( \left\|\begin{array}{c}
\triangle ABC\\\\
=======\\\\
D\in (BC)\\\\
E\in (CA) \\\\
F\in (AB)\\\\
P\in EF\cap AD\end{array}\right\|\ \Longrightarrow\ \overline {\underline {\left\|\ \frac {PF}{PE} = \frac {DB}{DC}\cdot\frac {AF}{AE}\cdot\frac {AC}{AB}\ \right\|}} \) .
Proof. Intr-adevar, \( \left\|\begin{array}{c}
\frac {PF}{PE}=\frac {AF}{AE}\cdot\frac {\sin \widehat {DAB}}{\sin \widehat {DAC}}\\\\
\frac {DC}{DB}=\frac {AC}{AB}\cdot\frac {\sin \widehat {DAC}}{\sin \widehat {DAB}}\end{array}\right\|\ \bigodot\ \Longrightarrow\ \frac {PF}{PE} = \frac {DB}{DC}\cdot\frac {AF}{AE}\cdot\frac {AC}{AB} \) .
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Claudiu Mindrila
- Fermat
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Un rezultat asemanator si pentru tetraedru.
Teorema.
Fie \( M \) un punct pe baza \( BDC \) a tetraedrului \( ABCD \), iar \( M_{1} \) un punct pe segmentul \( AM \). Un plan arbitrar dus prin \( M_{1} \) taie muchiile \( AB, AC, AD \) in punctele \( B_{1},C_{1}, D_{1} \). Avem relatia:
\( \frac{B_{1}B}{B_{1}A} \cdot S_{MCD}+\frac{C_{1}C}{C_{1}A} \cdot S_{MDB}+ \frac{D_{1}D}{D_{1}A} \cdot S_{MBC}=\frac{M_{1}M}{M_{1}A} \cdot S_{BDC} \)
Bibliografie.
\( [1] \)-V. Cirtoaje, In legatura cu problema \( 9362 \), G.M.-B. \( 11/1970 \)
Teorema.
Fie \( M \) un punct pe baza \( BDC \) a tetraedrului \( ABCD \), iar \( M_{1} \) un punct pe segmentul \( AM \). Un plan arbitrar dus prin \( M_{1} \) taie muchiile \( AB, AC, AD \) in punctele \( B_{1},C_{1}, D_{1} \). Avem relatia:
\( \frac{B_{1}B}{B_{1}A} \cdot S_{MCD}+\frac{C_{1}C}{C_{1}A} \cdot S_{MDB}+ \frac{D_{1}D}{D_{1}A} \cdot S_{MBC}=\frac{M_{1}M}{M_{1}A} \cdot S_{BDC} \)
Bibliografie.
\( [1] \)-V. Cirtoaje, In legatura cu problema \( 9362 \), G.M.-B. \( 11/1970 \)
elev, clasa a X-a, C. N. "C-tin Carabella", Targoviste