Răspuns:
Avem [tex]OA=v_1t, OB=v_2t, OM=vt[/tex].
In triunghiul OAM, avem [tex]m(OAM)=\alpha[/tex]. Din teorema cosinusurilor,
[tex]AM^2=OA^2+OM^2-2OA\cdot OM\cos\alpha[/tex].
Similar [tex]BM^2=OB^2+OM^2-2OB\cos OM\cos\beta[/tex]
[tex]AM^2+BM^2=OA^2+OB^2+2OM^2 - 2OM(OA\cos\alpha+OB\cos\beta)= v_1^2t^2+v_2^2t^2+2v^2t^2 - 2vt(v_1t\cos\alpha+v_2t\cos\beta)=t^2(v_1^2+v_2^2+2v^2-2vv_1\cos\alpha-2vv_2\cos\beta).[/tex]
Functia
[tex]f(x)=2x^2-2x(v_1\cos\alpha+ v_2\cos\beta)+v_1^2+v_2^2[/tex] are minimul in [tex]x_0=(-b/2a)=\frac{1}{2}(v_1\cos\alpha+v_2\cos\beta)[/tex] si valoarea minima [tex]f(x_0)=(-\Delta/4a)=-\frac{1}{2}((v_1\cos\alpha+v_2\cos\beta)^2-2v_1^2-2v_2^2) = \frac{1}{2}(v_1^2(1+\sin^2\alpha)+v_2^2(1+\sin^2\beta) - 2v_1v_2\cos\alpha\cos\beta)[/tex].
Dar [tex]x_0=v[/tex], iar distanta minima este [tex]t^2f(v)[/tex].
Pentru [tex]v_1=4\sqrt 3, v_2=4\sqrt 2,\alpha=30^o, \beta=45^o[/tex], avem
[tex]v=\frac{1}{2}(4\sqrt 3\cos(30)+4\sqrt 2\cos(45))=\frac{1}{2}(4\sqrt 3\frac{\sqrt 3}{2}+4\sqrt 2\frac{\sqrt 2}{2})=\frac{1}{2}(6+4)=5(m/s).[/tex]
Distanta minima este [tex]d=\frac{1}{2}(v_1^2(1+\sin^2\alpha)+v_2^2(1+\sin^2\beta) - 2v_1v_2\cos\alpha\cos\beta)\cdot t^2=\frac{1}{2}( (1+\frac{1}{4})(4\sqrt 3)^2 + (1+\frac{1}{2})(4\sqrt 2)^2 - 2(4\sqrt 3)(4\sqrt 2)\frac{\sqrt 3}{2}\frac{\sqrt{2}}{2})\cdot 5^2 =\frac{1}{2}( \frac{5}{4}\cdot 48 + \frac{3}{2}\cdot 32 - 8\cdot 6)\cdot 25 = \frac{1}{2}(60+48-45)\cdot 25 = 30\cdot 25 = 750(m^2)[/tex].
