Hong Kong Preliminary Selection Contest

As $AD$ is the internal bisector of $\angle A$, point $C^{\prime }$ lies on $AB$. Since $△DB{C}^{\prime }$ is similar to $△ABC$, we have $\angle B{C}^{\prime }D=\angle ACB$. Together with $\angle A{C}^{\prime }D=\angle ACB$, we have $\angle B{C}^{\prime }D=\angle A{C}^{\prime }D$ and so each of $\angle B{C}^{\prime }D$, $\angle A{C}^{\prime }D$ and $\angle ACB$ is equal to $90^{\circ }$.

Let $BC=x$. Then $AC=\frac{2}{x}$ and $AB=\sqrt{x^2+\frac{4}{x^2}}$. Using the AM-GM inequality, the perimeter of $△ABC$ is



$$\begin{gathered} x+\frac{2}{x}+\sqrt{x^2+\frac{4}{x^2}}\ge 2\sqrt{2}+\sqrt{2\sqrt{4}} \\ =2+2\sqrt{2}. \end{gathered}$$

This minimum value is attained when $x=\frac{2}{x}$ and $x^2=\frac{4}{x^2}$, i.e. when $x=\sqrt{2}$.

In this case, $AB=\sqrt{{\sqrt{2}}^2+\frac{4}{{\sqrt{2}}^2}}=2$.