In triunghiul ABC se considera cevienele \( [AA_1] ,[BB_1] ,[CC_1] \) concurente in \( O \).
a) Daca \( \overline{OA}+\overline{OB}+\overline{OC}=\overline{O} \) , atunci \( [AA_1] ,[BB_1] ,[CC_1] \) sunt mediane.
b) Daca \( \overline{OA_1}+\overline{OB_1}+\overline{OC_1}=\overline{O} \) , atunci \( [AA_1] ,[BB_1] ,[CC_1] \) sunt mediane.
c) Daca \( \overline{AA_1}+\overline{BB_1}+\overline{CC_1}=\overline{O} \) , atunci \( [AA_1] ,[BB_1] ,[CC_1] \) sunt mediane.
Conditii suficiente (2)
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Marius Mainea
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Mateescu Constantin
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a) Fie \( G \) centrul de greutate al triunghiului. Atunci \( \vec{OG}=\frac{\vec{OA}+\vec{OB}+\vec{OC}}{3} \) si din \( \vec{OA}+\vec{OB}+\vec{OC}=\vec{0}\ \Longrightarrow\ O=G \)
Deci \( O \) este intersectia medianelor si astfel rezulta concluzia.
b) Notam \( m=\frac{BA_1}{A_1C} \) , \( n=\frac{CB_1}{B_1A} \) , \( p=\frac{AC_1}{C_1B} \) , \( m \) , \( n \) , \( p\ >\ 0 \) . Evident \( mnp=1 \) .
Din teorema lui Van Aubel \( \Longrightarrow\ \frac{AO}{OA_1}=p+\frac{1}{n}\ \Longleftrightarrow\ \frac{OA_1}{AA_1}=\frac{n}{n+pn+1} \)
\( \Longrightarrow\ \vec{OA_1}=\frac{n}{n+pn+1}\vec{AA_1} \). Cum \( \vec{AA_1}=\frac{\vec{AB}+m\vec{AC}}{m+1}\ \Longrightarrow\ \vec{OA_1}=\frac{n}{(n+pn+1)(m+1)}\vec{AB}+\frac{nm}{(n+np+1)(m+1)}\vec{AC} \)
Prin urmare, avem : \( \left\|\ \begin{array}
\vec{OA_1} & = & \frac{n}{(n+np+1)(m+1)}\vec{AB} & + & \frac{mn}{(n+np+1)(m+1)}\vec{AC} \\\\\\\\
\vec{OB_1} & = & \frac{p}{(p+pm+1)(n+1)}\vec{BC} & + & \frac{np}{(p+pm+1)(n+1)}\vec{BA} \\\\\\\\
\vec{OC_1} & = & \frac{m}{(m+mn+1)(p+1)}\vec{CA} & + & \frac{pm}{(m+mn+1)(p+1)}\vec{CB}\end{array}\ \right|\ \bigoplus\ \Longrightarrow \)
\( \Longrightarrow\ \sum\vec{OA_1}=\vec{AB}\left(\frac{n}{(n+np+1)(m+1)}+\frac{pm}{(m+mn+1)(p+1)}-\frac{p}{p+pm+1}\right)+\vec{AC}\left(\frac{mn}{(n+np+1)(m+1)}+\frac{p}{(p+pm+1)(n+1)}-\frac{m}{m+mn+1}\right\)=\vec{0} \)
Insa vectorii \( \vec{AB} \) si \( \vec{AC} \) sunt necoliniari \( \Longrightarrow\ \)
\( \ \ \ \Longrightarrow\ \ \left\|\ \begin{array}
\frac{n}{(n+np+1)(m+1)} & + & \frac{pm}{(m+mn+1)(p+1)} & - & \frac{p}{p+pm+1} & = & 0 \\\\\\\\
\frac{p}{(p+pm+1)(n+1)} & + & \frac{mn}{(n+np+1)(m+1)} & - & \frac{m}{m+mn+1} & = & 0 \end{array}\ \right\|\ \)
\( \Longleftrightarrow^{mnp=1}\ \left\|\ \begin{array} \frac{mn}{(mn+m+1)(m+1)} & + & \frac{m}{(m+mn+1)(mn+1)} & - & \frac{1}{1+mn+n} & = & 0 \\\\\\\\
\frac{1}{(mn+n+1)(n+1)} & + & \frac{m^2n}{(mn+m+1)(m+1)} & - & \frac{m}{m+mn+1} & = & 0 \end{array}\ \right\| \)
\( \Longleftrightarrow\ \left\|\ \begin{array}
m^2n^2 & + & m^2 & = & m^2n & + & 1 & \ (1) \\\\\\\
m^2n^2 & + & 1 & = & m^2 & + & mn & \ (2) \end{array}\ \right\|\ \Longrightarrow^{(1)\ \ominus\ (2)}\ m^2-1=m^2n+1-m^2-mn \)
\( \Longleftrightarrow\ (m-1)(2m+2-mn)=0 \)
Daca \( 2m+2=mn \) atunci \( m^2n^2=4m^2+8m+4 \) si relatia \( (2) \) devine:
\( 4m^2+8m+5=m^2+2m+2\ \Longleftrightarrow\ (m+1)^2=0\ \Longleftrightarrow\ m=-1 \), fals deoarece \( m\ >\ 0 \).
Deci \( m-1=0 \), adica \( m=1 \). Din \( (2) \) rezulta imediat ca \( n=1 \) si apoi \( p=1 \), iar concluzia se impune.
c) Cu notatiile de la b) avem:
\( \left\|\ \begin{array}
\vec{AA_1} & = & \frac{\vec{AB}+m\vec{AC}}{1+m} \\\\\\\\
\vec{BB_1} & = & \frac{\vec{BC}+n\vec{BA}}{1+n} \\\\\\\\
\vec{CC_1} & = & \frac{\vec{CA}+p\vec{CB}}{1+p}\end{array}\ \right| \bigoplus\ \Longrightarrow\ \vec{AA_1}+\vec{BB_1}+\vec{CC_1}=\vec{AB}\left\(\frac{1}{1+m}+\frac{p}{1+p}-1\right\)+\vec{AC}\left(\frac{m}{1+m}+\frac{1}{1+n}-1\right\)=\vec{0} \)
\( \vec{AB} \) si \( \vec{AC} \) necoliniari \( \Longrightarrow\ \left\|\ \begin{array}
\frac{1}{1+m} & + & \frac{p}{1+p} & - & 1 & = & 0 & \Longleftrightarrow & m=p \\\\\\\\
\frac{m}{1+m} & + & \frac{1}{1+n} & - & 1 & = & 0 & \Longleftrightarrow & m=n \end{array}\ \right|\ \Longrightarrow\ AA_1\ ,\ BB_1\ ,\ CC_1 \) mediane \( (m=n=p=1) \) .
Deci \( O \) este intersectia medianelor si astfel rezulta concluzia.
b) Notam \( m=\frac{BA_1}{A_1C} \) , \( n=\frac{CB_1}{B_1A} \) , \( p=\frac{AC_1}{C_1B} \) , \( m \) , \( n \) , \( p\ >\ 0 \) . Evident \( mnp=1 \) .
Din teorema lui Van Aubel \( \Longrightarrow\ \frac{AO}{OA_1}=p+\frac{1}{n}\ \Longleftrightarrow\ \frac{OA_1}{AA_1}=\frac{n}{n+pn+1} \)
\( \Longrightarrow\ \vec{OA_1}=\frac{n}{n+pn+1}\vec{AA_1} \). Cum \( \vec{AA_1}=\frac{\vec{AB}+m\vec{AC}}{m+1}\ \Longrightarrow\ \vec{OA_1}=\frac{n}{(n+pn+1)(m+1)}\vec{AB}+\frac{nm}{(n+np+1)(m+1)}\vec{AC} \)
Prin urmare, avem : \( \left\|\ \begin{array}
\vec{OA_1} & = & \frac{n}{(n+np+1)(m+1)}\vec{AB} & + & \frac{mn}{(n+np+1)(m+1)}\vec{AC} \\\\\\\\
\vec{OB_1} & = & \frac{p}{(p+pm+1)(n+1)}\vec{BC} & + & \frac{np}{(p+pm+1)(n+1)}\vec{BA} \\\\\\\\
\vec{OC_1} & = & \frac{m}{(m+mn+1)(p+1)}\vec{CA} & + & \frac{pm}{(m+mn+1)(p+1)}\vec{CB}\end{array}\ \right|\ \bigoplus\ \Longrightarrow \)
\( \Longrightarrow\ \sum\vec{OA_1}=\vec{AB}\left(\frac{n}{(n+np+1)(m+1)}+\frac{pm}{(m+mn+1)(p+1)}-\frac{p}{p+pm+1}\right)+\vec{AC}\left(\frac{mn}{(n+np+1)(m+1)}+\frac{p}{(p+pm+1)(n+1)}-\frac{m}{m+mn+1}\right\)=\vec{0} \)
Insa vectorii \( \vec{AB} \) si \( \vec{AC} \) sunt necoliniari \( \Longrightarrow\ \)
\( \ \ \ \Longrightarrow\ \ \left\|\ \begin{array}
\frac{n}{(n+np+1)(m+1)} & + & \frac{pm}{(m+mn+1)(p+1)} & - & \frac{p}{p+pm+1} & = & 0 \\\\\\\\
\frac{p}{(p+pm+1)(n+1)} & + & \frac{mn}{(n+np+1)(m+1)} & - & \frac{m}{m+mn+1} & = & 0 \end{array}\ \right\|\ \)
\( \Longleftrightarrow^{mnp=1}\ \left\|\ \begin{array} \frac{mn}{(mn+m+1)(m+1)} & + & \frac{m}{(m+mn+1)(mn+1)} & - & \frac{1}{1+mn+n} & = & 0 \\\\\\\\
\frac{1}{(mn+n+1)(n+1)} & + & \frac{m^2n}{(mn+m+1)(m+1)} & - & \frac{m}{m+mn+1} & = & 0 \end{array}\ \right\| \)
\( \Longleftrightarrow\ \left\|\ \begin{array}
m^2n^2 & + & m^2 & = & m^2n & + & 1 & \ (1) \\\\\\\
m^2n^2 & + & 1 & = & m^2 & + & mn & \ (2) \end{array}\ \right\|\ \Longrightarrow^{(1)\ \ominus\ (2)}\ m^2-1=m^2n+1-m^2-mn \)
\( \Longleftrightarrow\ (m-1)(2m+2-mn)=0 \)
Daca \( 2m+2=mn \) atunci \( m^2n^2=4m^2+8m+4 \) si relatia \( (2) \) devine:
\( 4m^2+8m+5=m^2+2m+2\ \Longleftrightarrow\ (m+1)^2=0\ \Longleftrightarrow\ m=-1 \), fals deoarece \( m\ >\ 0 \).
Deci \( m-1=0 \), adica \( m=1 \). Din \( (2) \) rezulta imediat ca \( n=1 \) si apoi \( p=1 \), iar concluzia se impune.
c) Cu notatiile de la b) avem:
\( \left\|\ \begin{array}
\vec{AA_1} & = & \frac{\vec{AB}+m\vec{AC}}{1+m} \\\\\\\\
\vec{BB_1} & = & \frac{\vec{BC}+n\vec{BA}}{1+n} \\\\\\\\
\vec{CC_1} & = & \frac{\vec{CA}+p\vec{CB}}{1+p}\end{array}\ \right| \bigoplus\ \Longrightarrow\ \vec{AA_1}+\vec{BB_1}+\vec{CC_1}=\vec{AB}\left\(\frac{1}{1+m}+\frac{p}{1+p}-1\right\)+\vec{AC}\left(\frac{m}{1+m}+\frac{1}{1+n}-1\right\)=\vec{0} \)
\( \vec{AB} \) si \( \vec{AC} \) necoliniari \( \Longrightarrow\ \left\|\ \begin{array}
\frac{1}{1+m} & + & \frac{p}{1+p} & - & 1 & = & 0 & \Longleftrightarrow & m=p \\\\\\\\
\frac{m}{1+m} & + & \frac{1}{1+n} & - & 1 & = & 0 & \Longleftrightarrow & m=n \end{array}\ \right|\ \Longrightarrow\ AA_1\ ,\ BB_1\ ,\ CC_1 \) mediane \( (m=n=p=1) \) .