C3:Kẻ AH vuông góc với OB
Đặt OA =x => AH = x/2 ; OH =\(\frac{\sqrt{3}x}{2}\)
Do AB=1 áp dụng pitago cho HAB => HB =\(\sqrt{\frac{4-x^2}{4}}\)
=> OB = OH+HB =\(\frac{\sqrt{3}x}{2}\)+\(\frac{\sqrt{4-x^2}}{2}\)
=> 4OB2 =\(\left(\sqrt{3}x+\sqrt{4-x^2}\right)^2\le\left(3+1\right)\left(x^2+4-x^2\right)=16\)
OB2 </ 4 => OB</2 => OB max =2