Question

Let ABC be a triangle with $\angle BAC=2\pi /3$ and $AB=x$ such that $\left(AB\right)\left(AC\right)=1$ If x varies, then find the longest possible length of the angle bisector AD.

Answer 1

$AD=y=\frac{2bc}{b+c}\cos \frac{A}{2}=\frac{bx}{b+x}(asc=x)$

But $bx=1$ or $b=\frac{1}{x}$

$\therefore y=\frac{x}{1+x^2}=\frac{1}{x+\frac{1}{x}}$

Thus,$y_{max}=\frac{1}{2}$,since the minimum value of the denominator is 2 if $x>0$