1)In triunghiul ABC se considera bisectoarele \( AA_1,BB_1,CC_1 \) . Daca \( \overline{AA_1}+\overline{BB_1}+\overline{CC_1}=\overline{O} \) sa se arate ca triunghiul ABC este echilateral.
2) In triunghiul ascutitunghic ABC se considera inaltimile \( AA_1,BB_1,CC_1 \) . Daca \( \overline{AA_1}+\overline{BB_1}+\overline{CC_1}=\overline{O} \) sa se arate ca triunghiul ABC este echilateral.
Conditii suficiente (3)
Moderators: Laurian Filip, Beniamin Bogosel, Filip Chindea
-
Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
-
Mateescu Constantin
- Newton
- Posts: 307
- Joined: Tue Apr 21, 2009 8:17 am
- Location: Pitesti
1. Daca \( a,b,c \) sunt lungimile laturilor triunghiului avem:
\( \left\{ \begin \vec{AA_1}=\frac{b\vec{AB}+c\vec{AC}}{b+c} \\ \vec{BB_1}=\frac{a\vec{BA}+c\vec{BC}}{a+c} \\ \vec{CC_1}=\frac{a\vec{CA}+b\vec{CB}}{a+b}\right|\ \bigoplus\ \Longrightarrow \vec{AA_1}+\vec{BB_1}+\vec{CC_1}=\vec{AB}\left\(\frac{b}{b+c}+\frac{b}{a+b}-1\right\)+\vec{AC}\left(\frac{c}{b+c}+\frac{c}{a+c}-1\right\)=\vec{0} \)
\( \vec{AB} \) si \( \vec{AC} \) necoliniari \( \Longrightarrow \left\{ \begin \frac{b}{b+c}+\frac{b}{a+b}-1=0 \Longleftrightarrow b^2=ca\\ \frac{c}{b+c}+\frac{c}{a+c}-1=0\Longleftrightarrow c^2=ab \)
Impartind ultimele doua relatii \( \Longrightarrow b=c \) si apoi \( a=b \), deci triunghiul este echilateral.
\( \left\{ \begin \vec{AA_1}=\frac{b\vec{AB}+c\vec{AC}}{b+c} \\ \vec{BB_1}=\frac{a\vec{BA}+c\vec{BC}}{a+c} \\ \vec{CC_1}=\frac{a\vec{CA}+b\vec{CB}}{a+b}\right|\ \bigoplus\ \Longrightarrow \vec{AA_1}+\vec{BB_1}+\vec{CC_1}=\vec{AB}\left\(\frac{b}{b+c}+\frac{b}{a+b}-1\right\)+\vec{AC}\left(\frac{c}{b+c}+\frac{c}{a+c}-1\right\)=\vec{0} \)
\( \vec{AB} \) si \( \vec{AC} \) necoliniari \( \Longrightarrow \left\{ \begin \frac{b}{b+c}+\frac{b}{a+b}-1=0 \Longleftrightarrow b^2=ca\\ \frac{c}{b+c}+\frac{c}{a+c}-1=0\Longleftrightarrow c^2=ab \)
Impartind ultimele doua relatii \( \Longrightarrow b=c \) si apoi \( a=b \), deci triunghiul este echilateral.
Last edited by Mateescu Constantin on Mon Jun 29, 2009 5:08 pm, edited 1 time in total.
-
Mateescu Constantin
- Newton
- Posts: 307
- Joined: Tue Apr 21, 2009 8:17 am
- Location: Pitesti
2. Notam \( m=\frac{BA_1}{A_1C},\ n=\frac{CB_1}{B_1A},\ p=\frac{AC_1}{C_1B},\ m,\ n,\ p>0 \). Evident \( mnp=1 \). Avem:
\( \left\{ \begin \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}\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 \( \Longrightarrow \left\{ \begin \frac{1}{1+m}+\frac{p}{1+p}-1=0 \Longleftrightarrow m=p\\ \frac{m}{1+m}+\frac{1}{1+n}-1=0\Longleftrightarrow m=n \right|\ \Longrightarrow AA_1,\ BB_1,\ CC_1\ \mbox{mediane}\ (m=n=p=1) \)
Deci \( G=H \), ceea ce inseamna ca triunghiul este echilateral.
\( \left\{ \begin \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}\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 \( \Longrightarrow \left\{ \begin \frac{1}{1+m}+\frac{p}{1+p}-1=0 \Longleftrightarrow m=p\\ \frac{m}{1+m}+\frac{1}{1+n}-1=0\Longleftrightarrow m=n \right|\ \Longrightarrow AA_1,\ BB_1,\ CC_1\ \mbox{mediane}\ (m=n=p=1) \)
Deci \( G=H \), ceea ce inseamna ca triunghiul este echilateral.
Last edited by Mateescu Constantin on Mon Jun 29, 2009 5:09 pm, edited 1 time in total.
-
Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
Ambele afirmatii rezulta din urmatoarele 2 propozitii:
1 Teorema lui PAPPUS Fie ABC triunghi si \( M\in (AB) , N\in (BC) , P\in (CA) \). Atunci triunghiurile ABC si MNP au acelasi centru de greutate daca si numai daca \( \frac{AM}{MB}=\frac{BN}{NC}=\frac{CP}{PA} \)
2 Triunghiurile \( ABC \) si \( A^{\prime}B^{\prime}C^{\prime} \) au acelasi centru de greutate daca si numai daca \( \overline{AA^{\prime}}+\overline{BB^{\prime}}+\overline{CC^{\prime}}=\overline{O} \)
1 Teorema lui PAPPUS Fie ABC triunghi si \( M\in (AB) , N\in (BC) , P\in (CA) \). Atunci triunghiurile ABC si MNP au acelasi centru de greutate daca si numai daca \( \frac{AM}{MB}=\frac{BN}{NC}=\frac{CP}{PA} \)
2 Triunghiurile \( ABC \) si \( A^{\prime}B^{\prime}C^{\prime} \) au acelasi centru de greutate daca si numai daca \( \overline{AA^{\prime}}+\overline{BB^{\prime}}+\overline{CC^{\prime}}=\overline{O} \)