Inegalitate geometrica 6
Posted: Mon Aug 17, 2009 7:29 pm
Daca Meste un punct interior triunghiului echilateral de latura a , atunci MA+MB+MC<2a.
M.Mainea
M.Mainea
Prin punctul \( M \) construim paralele la laturile triunghiului care intersecteaza \( [BC] \) in \( A_1,\ A_2 \), \( \ [CA] \) in \( B_1,\ B_2 \) si respectiv \( [AB] \) in punctele \( C_1,\ C_2 \) \( ( A_1\in [BA_2],\ B_1\in [CB_2],\ C_1\in [AC_2]\ ) \) .
Se formeaza 3 triunghiuri echilaterale \( (MA_1A_2,\ MB_1B_2,\ MC_1C_2\ )\ \Longrightarrow \left\| \begin{array}{ccc} MA_1=MA_2=A_1A_2\\\\\
MB_1=MB_2=B_1B_2\\\\\
MC_1=MC_2=C_1C_2 \end{array}\right\| \ (*) \) .
De asemenea se formeaza si 3 paralelograme \( (\ AC_1MB_2,\ BC_2MA_1\ CA_2MB_1\ ) \) .
In fiecare dintre acestea aplicam de doua ori ineg. triunghiului :
\( \Longrightarrow\ \left\| \begin{array}{ccc} 2MA & < & AB_2+AC_1+MB_2+MC_1 \\\\\\\\
2MB & < & BC_2+BA_1+MC_2+MA_1 \\\\\\\\
2MC & < & CA_2+CB_1+MA_2+MB_1 \end{array} \right\| \bigoplus\Longrightarrow \)
\( \Longrightarrow\ 2(MA+MB+MC)\ <\ (AB_2+MB_2+CB_1)+(CA_2+MA_2+BA_1)+(BC_2+MC_2+AC_1)+(MA_1+MB_1+MC_1) \)
\( \Longleftrightarrow^{(*)}\ 2(MA+MB+MC)\ <\ (AB_2+B_1B_1+CB_1)+(CA_2+A_1A_2+BA_1)+(BC_2+C_1C_2+AC_1)+(MA_1+MB_2+C_1C_2) \)
\( \Longleftrightarrow\ 2(MA+MB+MC)\ <\ AC+BC+AB+AB=4a \)
\( \Longleftrightarrow\ MA+MB+MC\ <\ 2a \)