Selection Examination for Juniors

(i) We draw the line segment $K_{1}H\perp K_{2}Z$. In the triangle $K_{1}H{K}_2$ we have $K_1{K}_2=r_1+r_2$ and $K_{2}H=r_2-r_1$ and hence from Pythagorean theorem we find $K_{1}H=2\sqrt{r_1{r}_2}$. From the rectangular $K_{1}EZH$ we have $EZ=2\sqrt{r_1{r}_2}$. Moreover we have $$\begin{gathered} AB=AE+EZ+ZB\iff a=r_1+2\sqrt{r_1{r}_2}+r_2=(\sqrt{r_1}+\sqrt{r_2}{)}^2, \\ b=B\Gamma =Z\Theta =2r_2. \end{gathered}$$

(ii) We have $\Lambda E=\Lambda K=\Lambda Z=\sqrt{r_1{r}_2}$. Let $\Delta I=\Delta K=\Delta \Theta =x$. From the orthogonal triangle $AB\Delta$, since $x+r_2=a$ and $x+r_1=b$ we get $$\begin{gathered} r_1+2\sqrt{r_1{r}_2}=2r_2-r_1\iff r_1+\sqrt{r_1{r}_2}-r_2=0\iff {\omega }^2+\omega -1=0, \\ \text{where }\omega =\sqrt{\frac{r_1}{r_2}}>0.\text{ Hence }\omega =\sqrt{\frac{r_1}{r_2}}=\frac{\sqrt{5}-1}{2}\text{ and }\kappa =\frac{r_1}{r_2}=\frac{3-\sqrt{5}}{2}. \end{gathered}$$ Moreover $$\frac{a}{b}=\frac{(\sqrt{r_1}+\sqrt{r_2}{)}^2}{2r_2}=\frac{1}{2}{\left(\sqrt{\frac{r_1}{r_2}}+1\right)}^2=\frac{1}{2}\left(\frac{\sqrt{5}+1}{2}\right)=\frac{\sqrt{5}+1}{4}.$$

(iii) From the orthogonal triangle $AB\Delta$ we have $$(x+r_1{)}^2+(r_1+\sqrt{r_1{r}_2}{)}^2=(x+\sqrt{r_1{r}_2}{)}^2\iff x=\frac{(r_1+\sqrt{r_1{r}_2}{)}^2}{r_2-r_1}.$$