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Se considera triunghiul ABC ascutitunghic, cu AC<AB<BC. Notam cu M,N,P mijlocul laturilor [BC],[AC], respcetiv [AB], iar cu D,E,F proiectiile varfurilor A,B, respectiv C pe laturile [BC],[CA] respectiv [AB]. Sa se demonstreze:
a) \( BC^2-AC^2=MD\cdot BC+NE\cdot AC+PF \cdot AB \)
b) \( BC-AC<MD+NE+PF \)

Cristi Popa wrote: Fie \( \triangle\ ABC \) ascutitunghic cu \( b\ <\ c\ <\ a\ . \) Notam mijloacele \( M \) , \( N \) , \( P \) ale laturilor \( [BC] \) ,

\( [AC] \) , \( [AB] \) respectiv si proiectiile \( D \) , \( E \) , \( F \) ale varfurilor \( A \) , \( B \) , \( C \) pe laturile opuse.

Sa se arate ca \( \underline {\overline {\left\|\ a^2-b^2=a\cdot MD+b\cdot NE+c\cdot PF\ \right\|}}\ (1)\ \ \) si \( \ \ \underline {\overline {\left\|\ a-b\ <\ MD+NE+PF\ \right\|}}\ (2)\ . \)

Demonstratie. Se arata usor ca \( \left|\begin{array}{c}
2a\cdot MD=c^2-b^2\\\\\\
2b\cdot NE=a^2-c^2\\\\\\
2c\cdot PF=a^2-b^2\end{array}\right|\Longrightarrow\ 2a\cdot MD+2b\cdot NE+2c\cdot PF=2\left(a^2-b^2\right)\Longrightarrow\ (1)\ . \)

Relatia \( (2)\Longleftrightarrow a-b\ <\ \frac {c^2-b^2}{2a}+\frac {a^2-c^2}{2b}+\frac {a^2-b^2}{2c}\Longleftrightarrow2abc(a-b)\ \ <\ \ cb\left(c^2-b^2\right)+ac\left(a^2-c^2\right)+ab\left(a^2-b^2\right)\ . \)

Se arata usor ca \( E(a,b,c)\equiv cb\left(c^2-b^2\right)+ac\left(a^2-c^2\right)+ab\left(a^2-b^2\right)=(a-b)(b+c)(a+c)(a+b-c)\ . \) Asadar,

relatia \( (2)\ \Longleftrightarrow\ 2abc\ <\ (b+c)(a+c)(a+b-c)\ \Longleftrightarrow\ 2abc\ <\ a^2(b+c)+ab(b+c)+c\left(b^2-c^2\right)\Longleftrightarrow \)

\( a^2(b+c)+ab(b-c)+c\left(b^2-c^2\right)\ >\ 0\ \Longleftrightarrow\ (a-c)\left[c^2+a(b+c)\right]+b^2(a+c)\ >\ 0 \) ceea ce este adevarat.

In concluzie, relatia \( (2) \) este adevarata, intotdeauna strict. Vom incerca sa o imbunatatim !

Virgil Nicula wrote: Fie \( \triangle\ ABC \) ascutitunghic cu \( b\le c\le a\ . \) Atunci exista inegalitatea

\( a-b\ \le\ (a-b)\ \left[1+\frac {b(a+c)}{2ac}\right]\ \le\ MD+NE+PF \) cu egalitate \( \Longleftrightarrow\ a=b\ \ \vee\ \ a=c\ . \)