Dedicatie : domnului profesor Mihai Miculita, un geometru de elita din invatamantul preuniversitar.
Se considera doua triunghiuri isoscele \( ABC \), \( MNP \) astfel incat \( AB=AC \),
\( MN=MP \) si \( N\in (AC) \), \( B\in (MP) \). Notam \( R\in AB\cap MN \), \( S\in BC\cap NP \).
Sa se arate ca \( BP=CN\ \Longleftrightarrow \) semidreapta \( [RS \) este una dintre bisectoarele unghiului \( \widehat {BRN}. \)
O problema de geometrie (proprie) care imi place f.f. mult.
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Virgil Nicula
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O problema de geometrie (proprie) care imi place f.f. mult.
Last edited by Virgil Nicula on Sun May 31, 2009 5:35 pm, edited 2 times in total.
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mihai miculita
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\( \mbox{Folosind teorema sinusurilor, obtinem din: }\\
\triangle{BPS}\Rightarrow\frac{|BP|}{\sin\angle{BSP}}=\frac{|BS|}{\sin\angle{MPN}}\ \ (1) \\
\triangle{CNS}\Rightarrow\frac{|CN|}{\sin{\angle{NSC}}}=\frac{|SN|}{\sin\angle{ACB}}\ \ (2)\\
\mbox{Impartind relatia (1) la (2), membru cu membru, obtinem: }
\frac{|BP|}{|CN|}=\frac{|BS|.\sin\angle{ACB}}{|SN|.\sin\angle{MPN}}.\ \ (3)\\
\mbox{Pe de alta parte, din:}\\
\triangle{BRS}\Rightarrow\frac{|RS|}{\sin\angle{ABC}}=\frac{|BS|}{\sin\angle{BRS}}\ \ (4)\\
\triangle{RSN}\Rightarrow\frac{|RS|}{\sin\angle{MNP}}=\frac{|SN|}{\sin\angle{SRN}}\ \ (5)\\
\mbox{Impartind relatiile (5) si (4), menbru cu membru, obtinem: } \frac{\sin\angle{ABC}}{\sin\angle{MNP}}=\frac{|SN|.\sin\angle{BRS}}{|BS|.\sin\angle{SRN}}\ \ (6)\\
\mbox{Tinand acum seama de congruenta unghiurilor de la baza celor doua triunghiuri isoscele, }\\
\mbox{din relatiile (3) si (6) }\Rightarrow \frac{|BP|}{|CN|}=\frac{\sin\angle{BRS}}{\sin\angle{SRN}}. \ \ (7)\\
\mbox{Asa ca, pe baza relatiei (7) avem: } |PB|=|CN|\Leftrightarrow \angle{BRS}\eq \angle{SRN}.
\)
FRUMOASA PROBLEMA !, mai putin solutia mea calculatorie...
\triangle{BPS}\Rightarrow\frac{|BP|}{\sin\angle{BSP}}=\frac{|BS|}{\sin\angle{MPN}}\ \ (1) \\
\triangle{CNS}\Rightarrow\frac{|CN|}{\sin{\angle{NSC}}}=\frac{|SN|}{\sin\angle{ACB}}\ \ (2)\\
\mbox{Impartind relatia (1) la (2), membru cu membru, obtinem: }
\frac{|BP|}{|CN|}=\frac{|BS|.\sin\angle{ACB}}{|SN|.\sin\angle{MPN}}.\ \ (3)\\
\mbox{Pe de alta parte, din:}\\
\triangle{BRS}\Rightarrow\frac{|RS|}{\sin\angle{ABC}}=\frac{|BS|}{\sin\angle{BRS}}\ \ (4)\\
\triangle{RSN}\Rightarrow\frac{|RS|}{\sin\angle{MNP}}=\frac{|SN|}{\sin\angle{SRN}}\ \ (5)\\
\mbox{Impartind relatiile (5) si (4), menbru cu membru, obtinem: } \frac{\sin\angle{ABC}}{\sin\angle{MNP}}=\frac{|SN|.\sin\angle{BRS}}{|BS|.\sin\angle{SRN}}\ \ (6)\\
\mbox{Tinand acum seama de congruenta unghiurilor de la baza celor doua triunghiuri isoscele, }\\
\mbox{din relatiile (3) si (6) }\Rightarrow \frac{|BP|}{|CN|}=\frac{\sin\angle{BRS}}{\sin\angle{SRN}}. \ \ (7)\\
\mbox{Asa ca, pe baza relatiei (7) avem: } |PB|=|CN|\Leftrightarrow \angle{BRS}\eq \angle{SRN}.
\)
FRUMOASA PROBLEMA !, mai putin solutia mea calculatorie...
Last edited by mihai miculita on Tue May 26, 2009 5:35 pm, edited 1 time in total.
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Virgil Nicula
- Euler
- Posts: 622
- Joined: Fri Sep 28, 2007 11:23 pm
Re: O problema de geometrie (proprie) care imi place f.f. mu
Dem. Notam \( X\in BC\cap MN\ ,\ Y\in AB\cap NP\ ,\ Z\in NB\cap RS \) . Aplicam teorema lui Menelaus transversalelor \( \overline {XBC}\ , \)Virgil Nicula wrote:Fie triunghiurile isoscele \( ABC \), \( MNP \) astfel incat \( AB=AC \), \( MN=MP \)
si \( N\in (AC) \) , \( B\in (MP) \). Notam \( R\in AB\cap MN \) , \( S\in BC\cap NP \). Sa se arate ca
\( BP=CN\ \Longleftrightarrow \) semidreapta \( [RS \) este una din bisectoarele unghiului \( \widehat {BRN}. \)
\( \overline {YNP} \) in triunghiurile \( ANR\ ,\ BMR \) respectiv : \( \left|\begin{array}{ccc}
\frac {XR}{XN}\cdot\frac {CN}{CA}\cdot\frac {BA}{BR}=1 & \Longrightarrow & \frac {XN}{XR}=\frac {CN}{BR}\\\\\\\\
\frac {YR}{YB}\cdot\frac {PB}{PM}\cdot\frac {NM}{NR}=1 & \Longrightarrow & \frac {YR}{YB}=\frac {NR}{PB}\end{array}\right| \) . Aplicam teorema lui Ceva
punctului \( S \) si triunghiului \( BRN\ : \ \ \frac {TB}{TN}\cdot\frac {XN}{XR}\cdot\frac {YR}{YB}=1 \Longrightarrow \frac {TN}{TB}=\frac {XN}{XR}\cdot\frac {YR}{YB}\ \Longrightarrow\ \frac {TN}{TB}=\frac {NC}{PB}\cdot\frac {RN}{RB} \) . In concluzie,
\( NC=PB\ \Longleftrightarrow\ \frac {TN}{TB}=\frac {RN}{RB} \Longleftrightarrow \) semidreapta \( [RS \) este una dintre bisectoarele unghiului \( \widehat {BRN} \) .