Răspuns:
3+2rad3 si 3-2rad3
Explicație pas cu pas:
Punctele M, N si P sunt varfurile unui triunghi echilateral daca MN=NP=MP
[tex]MN=\sqrt{(x_N-x_M)^2+(y_N-y_M)^2} =\sqrt{(x-1)^2+(-x+5)^2}[/tex]
[tex]NP=\sqrt{(x_P-x_N)^2+(y_P-y_N)^2} =\sqrt{(5-x)^2+(-1+x)^2}[/tex]
[tex]MP=\sqrt{(x_P-x_M)^2+(y_P-y_M)^2} =\sqrt{(5-1)^2+(-1+5)^2}=\sqrt{4^2+4^2}=\sqrt{32}[/tex]
MN=NP=MP =>
[tex]\Rightarrow \sqrt{(x-1)^2+(-x+5)^2}=\sqrt{(5-x)^2+(-1+x)^2}=\sqrt{32} \ |^2[/tex]
[tex]\Rightarrow (x-1)^2+(-x+5)^2=(5-x)^2+(-1+x)^2=32[/tex]
[tex]\Rightarrow (x-1)^2+(5-x)^2=(5-x)^2+(x-1)^2=32[/tex]
[tex](x-1)^2+(5-x)^2=32 \\ \\ \Rightarrow x^2-2x+1+25-10x+x^2=32\\ \\ \Rightarrow 2x^2-12x+26=32\\ \\ \Rightarrow 2x^2-12x+26-32=0 \\ \\ \Rightarrow 2x^2-12x-6=0 \ |:2\\ \\ \Rightarrow x^2-6x-3=0 \\ \\ \Delta=(-6)^2-4\cdot (-3) \cdot 1=36+12=48[/tex]
[tex]\Rightarrow x_1=\frac{6+\sqrt{48}}{2}=\frac{6+4\sqrt{3}}{2}=3+2\sqrt{3}\\ \\ \Rightarrow x_2=\frac{6-\sqrt{48}}{2}=\frac{6-4\sqrt{3}}{2}=3-2\sqrt{3}\\ \\[/tex]
[tex]\Rightarrow S=\{3+2\sqrt{3}, \ 3-2\sqrt{3}\}[/tex]
