Teorema trisectiei
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Claudiu Mindrila
- Fermat
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Teorema trisectiei
Problema.(Teorema trisectiei) In triunghiul \( ABC \) fie medianele \( BB\prime \) si \( CC\prime \). Printr-un punct \( T \in (BC) \) se duc paralelele \( TD \) si \( TE \) la medianele \( BB\prime \) respectiv \( CC\prime, (D\in(AB), E\in(AC)) \). Atunci \( BB\prime \) si \( CC\prime \) impart \( DE \) in trei segmente congruente.
elev, clasa a X-a, C. N. "C-tin Carabella", Targoviste
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Virgil Nicula
- Euler
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ATENTIE, am schimbat notatiile din enuntul initial !
Demonstratie. Aplicam teorema Menelaus transversalelor/triunghiurilor mentionate corespunzator :
\( \left|\ \begin{array}{cccc}
\overline {BNX}/AEF\ : & \frac {BF}{BA}\cdot\frac {NA}{NE}\cdot\frac {XE}{XF}=1 & \Longrightarrow & \frac {XE}{XF}=\frac {BA}{BF}\cdot\frac {NE}{NA}\\\\\\\\\\
\overline {CPY}/AEF\ : & \frac {CE}{CA}\cdot\frac {PA}{PF}\cdot\frac {YF}{YE}=1 & \Longrightarrow & \frac {YE}{YF}=\frac {CE}{CA}\cdot\frac {PA}{PF}\end{array}\ \right|\ . \) Se observa usor ca :
\( \begin{array}{c}
\left|\ \begin{array}{c}
\frac {BA}{BF}=2\cdot\frac {BP}{BF}=2\cdot \frac {BC}{BM}\\\\\\\\
\frac {NE}{NA}=\frac {NE}{NC}=\frac {BM}{BC}\end{array}\ \right|\ \Longrightarrow\ \frac {BA}{BF}\cdot \frac {NE}{NA}=2\ \Longrightarrow\ \frac {XE}{XF}=2\\\\\\\\\\\\\\\\\\\\
\left|\ \begin{array}{c}
\frac {CE}{CA}=\frac 12\cdot\frac {CE}{CN}=\frac 12\cdot \frac {CM}{CB}\\\\\\\\
\frac {PA}{PF}=\frac {PB}{PF}=\frac {CB}{CM}\end{array}\ \right|\ \Longrightarrow\ \frac {CE}{CA}\cdot \frac {PA}{PF}=\frac 12\ \Longrightarrow\ \frac {YE}{YF}=\frac 12\end{array}\ \Longrightarrow\ FX\ =\ XY\ =\ XE\ . \)
Generalizare.
A cam trecut "neobservata" aceasta frumoasa problema. Chiar merita sa fie numita "teorema trisectiei".Fie \( \triangle\ ABC \) si mijloacele \( N \) , \( P \) ale lui \( [AC] \) , \( [AB] \) respectiv. Pentru \( M\in (BC) \) definim
\( \left|\ \begin{array}{ccc}
E\in AC & , & ME\ \parallel\ BN\\\\
F\in AB & , & MF\ \parallel\ CP\end{array}\ \ \wedge\ \ \begin{array}{c}
X\in BN\ \cap\ EF\\\\
Y\in EF\ \cap\ CP\end{array}\ \right|\ . \) Sa se arate ca \( FX\ =\ XY\ =\ YE \) .
Demonstratie. Aplicam teorema Menelaus transversalelor/triunghiurilor mentionate corespunzator :
\( \left|\ \begin{array}{cccc}
\overline {BNX}/AEF\ : & \frac {BF}{BA}\cdot\frac {NA}{NE}\cdot\frac {XE}{XF}=1 & \Longrightarrow & \frac {XE}{XF}=\frac {BA}{BF}\cdot\frac {NE}{NA}\\\\\\\\\\
\overline {CPY}/AEF\ : & \frac {CE}{CA}\cdot\frac {PA}{PF}\cdot\frac {YF}{YE}=1 & \Longrightarrow & \frac {YE}{YF}=\frac {CE}{CA}\cdot\frac {PA}{PF}\end{array}\ \right|\ . \) Se observa usor ca :
\( \begin{array}{c}
\left|\ \begin{array}{c}
\frac {BA}{BF}=2\cdot\frac {BP}{BF}=2\cdot \frac {BC}{BM}\\\\\\\\
\frac {NE}{NA}=\frac {NE}{NC}=\frac {BM}{BC}\end{array}\ \right|\ \Longrightarrow\ \frac {BA}{BF}\cdot \frac {NE}{NA}=2\ \Longrightarrow\ \frac {XE}{XF}=2\\\\\\\\\\\\\\\\\\\\
\left|\ \begin{array}{c}
\frac {CE}{CA}=\frac 12\cdot\frac {CE}{CN}=\frac 12\cdot \frac {CM}{CB}\\\\\\\\
\frac {PA}{PF}=\frac {PB}{PF}=\frac {CB}{CM}\end{array}\ \right|\ \Longrightarrow\ \frac {CE}{CA}\cdot \frac {PA}{PF}=\frac 12\ \Longrightarrow\ \frac {YE}{YF}=\frac 12\end{array}\ \Longrightarrow\ FX\ =\ XY\ =\ XE\ . \)
Generalizare.
Virgil Nicula wrote:Fie triunghiul \( ABC \) si punctele \( \left|\ \begin{array}{ccc}
N\in (AC) & , & \frac {NA}{NC}=n\\\\\\\
P\in (AB) & , & \frac {PA}{PB}=p\end{array}\ \right|\ . \) Pentru \( M\in (BC) \) definim punctele
\( \left|\ \begin{array}{ccc}
E\in AC & , & ME\ \parallel\ BN\\\\
F\in AB & , & MF\ \parallel\ CP\end{array}\ \ \wedge\ \ \begin{array}{c}
X\in BN\ \cap\ EF\\\\
Y\in EF\ \cap\ CP\end{array}\ \right|\ \Longrightarrow\ XY\ =\ \frac {FX}{n}\ =\ \frac {YE}{p}\ =\ \frac {EF}{n+p+1} \) .
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mihai miculita
- Pitagora
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- Location: Oradea, Romania
Alta solutie:
\( \mbox{Folosind notatiile initiale si punand: } \{M\}=BB^{\prime}\cap DE; \{N\}=CC^{\prime}\cap DE\mbox{, avem de aratat ca: } |DM|=|MN|=|NE|.\\
1).\mbox{Daca: } \{G\}=BB^{\prime}\cap CC^{\prime} \mbox{, atunci }\\
\frac{|GB^{\prime}|}{|GB|}=\frac{|GC^{\prime}|}{|GC|}=\frac{1}{2}. \ (1)\\
2).\mbox{Notand cu } \{P\}=DT\cap BB^{\prime}\mbox{, avem:}\\
\left \begin{\array} {\frac{|GC^{\prime}|}{|GC|}=\frac{1}{2}}\ {(1)} \\
{DT||CC^{\prime}} \right\}\Rightarrow \frac{|DP|}{|PT|}=\frac{1}{2}.\ (2) \\
3). \\
\left \begin{\array} {\frac{|DP|}{|PT|}=\frac{1}{2}}\ {(2)} \\
{TE||PM} \right\}\Rightarrow \frac{|DM|}{|ME|}=\frac{1}{2}\Rightarrow \frac{|DM|}{|DE|}=\frac{1}{3}.\ (3) \\
\mbox{In mod analog aratam ca: } \frac{|NE|}{|DE|}=\frac{1}{3}. \ (4)\\
\mbox{In fine, din (3) si (4) }\Rightarrow |DM|=|MN|=|NE|. \)
\( \mbox{Folosind notatiile initiale si punand: } \{M\}=BB^{\prime}\cap DE; \{N\}=CC^{\prime}\cap DE\mbox{, avem de aratat ca: } |DM|=|MN|=|NE|.\\
1).\mbox{Daca: } \{G\}=BB^{\prime}\cap CC^{\prime} \mbox{, atunci }\\
\frac{|GB^{\prime}|}{|GB|}=\frac{|GC^{\prime}|}{|GC|}=\frac{1}{2}. \ (1)\\
2).\mbox{Notand cu } \{P\}=DT\cap BB^{\prime}\mbox{, avem:}\\
\left \begin{\array} {\frac{|GC^{\prime}|}{|GC|}=\frac{1}{2}}\ {(1)} \\
{DT||CC^{\prime}} \right\}\Rightarrow \frac{|DP|}{|PT|}=\frac{1}{2}.\ (2) \\
3). \\
\left \begin{\array} {\frac{|DP|}{|PT|}=\frac{1}{2}}\ {(2)} \\
{TE||PM} \right\}\Rightarrow \frac{|DM|}{|ME|}=\frac{1}{2}\Rightarrow \frac{|DM|}{|DE|}=\frac{1}{3}.\ (3) \\
\mbox{In mod analog aratam ca: } \frac{|NE|}{|DE|}=\frac{1}{3}. \ (4)\\
\mbox{In fine, din (3) si (4) }\Rightarrow |DM|=|MN|=|NE|. \)