smc

This solution refers to the Diagram section. Let $O=(0,0),A=(3,3),B=(1,3),$ and $C=\left(\frac{3}{k},3\right).$ As shown below, note that $\overline{OA},\overline{OB},$ and $\overline{OC}$ are on the lines $y=x,y=3x,$ and $y=kx,$ respectively. By the Distance Formula, we have $OA=3\sqrt{2},OB=\sqrt{10},AC=3-\frac{3}{k},$ and $BC=\frac{3}{k}-1.$ By the Angle Bisector Theorem, we get $\frac{OA}{OB}=\frac{AC}{BC},$ or $$\begin{array}{rl}\frac{3\sqrt{2}}{\sqrt{10}}& =\frac{3-\frac{3}{k}}{\frac{3}{k}-1}\\ \frac{3\sqrt{2}}{\sqrt{10}}& =\frac{3k-3}{3-k}\\ \frac{\sqrt{2}}{\sqrt{10}}& =\frac{k-1}{3-k}\\ \frac{1}{5}& =\frac{(k-1{)}^2}{(3-k{)}^2}\\ 5(k-1{)}^2& =(3-k{)}^2\\ 4k^2-4k-4& =0\\ k^2-k-1& =0\\ k& =\frac{1\pm \sqrt{5}}{2}.\end{array}$$ Since $k>0,$ the answer is $k=\boxed{\frac{1+\sqrt{5}}{2}}.$ Remark The value of $k$ is known as the Golden Ratio: $\varphi =\frac{1+\sqrt{5}}{2}\approx \mathrm{1.61803398875.}$