[tex]\displaystyle Fie ~\Delta ABC~cu~b=a+1~si~c=a+2. ~(a \in \mathbb{N^*})\\ \\ ~Stim~ca~in~orice~triunghi,~unghiul~opus~laturii~celei~mai~mare \\ \\ este~cel~mai~mare~unghi. \\ \\ Deci~unghiul~obtuz~este~cel~opus~laturii~de~lungime~a+2,~adica~C. \\ \\ Atunci~\cos C\ \textless \ 0.\\ \\ \cos C= \frac{a^2+b^2-c^2}{2ab}\ \textless \ 0. \\ \\ Dar~2ab\ \textgreater \ 0.~Asadar~a^2+b^2-c^2\ \textless \ 0. \\ \\ Adica~a^2+(a+1)^2-(a+2)^2\ \textless \ 0. \\ \\ a^2+a^2+2a+1-a^2-4a-4\ \textless \ 0 \\ \\ a^2-2a-3\ \textless \ 0 \\ \\ a^2-2a+1\ \textless \ 4 \\ \\ (a-1)^2\ \textless \ 4 \\ \\ -2\ \textless \ a-1\ \textless \ 2 \\ \\ -1\ \textless \ a\ \textless \ 3[/tex]
[tex]\displaystyle Stim~ca~ a \in \mathbb{N^*}.~Deci~a \in \{1,2 \}. \\ \\ \bullet Daca~a=1,~atunci~triunghiul~ar~avea~lungimile~laturilor~1,2,3. \\ \\ IMPOSIBIL,~deoarece~in~orice~triunghi~au~loc~relatiile~a+b\ \textgreater \ c,~ \\ \\ a+c\ \textgreater \ b,~b+c\ \textgreater \ a,~iar~prima~conditie~nu~este~respectata. \\ \\ \bullet Daca~a=2,~atunci~trunghiul~are~lungimile~laturilor~2,3,4.~ \checkmark \\ \\ (convine!) \\ \\ Deci~solutia~problemei~este~a=2.[/tex]
