User:Wknight94/2

$$r_b=r_a{\sqrt{2}-1\over\sqrt{2}+1}$$

$$ r_a=r_b{\sqrt{2}+1\over\sqrt{2}-1} =r_b{\left(\sqrt{2}+1\right)^2\over\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)} =r_b{2+2\sqrt{2}+1\over2-1} =r_b\left(3+2\sqrt{2}\right) $$

Note that:

$$\sqrt{3+2\sqrt{2}}=\sqrt{2+2\sqrt{2}+1}=\sqrt{2}+1$$

$$ \begin{alignat}{2} 2\sqrt{r_a r_c}+2\sqrt{r_b r_c} & =r_a-r_b \\ 2\sqrt{r_c}\left(\sqrt{r_a}+\sqrt{r_b}\right)&=r_a-r_b \\ \sqrt{r_c}&={r_a-r_b\over2\left(\sqrt{r_a}+\sqrt{r_b}\right)} \\ &={r_b\left(3+2\sqrt{2}\right)-r_b\over2\left(\sqrt{r_b\left(3+2\sqrt{2}\right)}+\sqrt{r_b}\right)}\text{ ...substituted for }r_a \\ &={r_b\left(3+2\sqrt{2}-1\right)\over2\sqrt{r_b}\left(\sqrt{3+2\sqrt{2}}+1\right)} \\ &={r_b\left(2+2\sqrt{2}\right)\over2\sqrt{r_b}\left(\sqrt{2}+1+1\right)}\text{ ...per }\sqrt{3+2\sqrt{2}}\text{ note above} \\ &={\sqrt{r_b}\left(1+\sqrt{2}\right)\over\sqrt{2}+2} \\ r_c&={r_b\left(1+2\sqrt{2}+2\right)\over2+4\sqrt{2}+4}\text{ ...squared both sides...} \\ r_c&=r_b/2 \end{alignat} $$