Question Number 105992 by Algoritm last updated on 02/Aug/20
Answered by mr W last updated on 02/Aug/20
$${let}\:{A}={Radius}\:{of}\:{big}\:{semicircle} \\ $$$$\sqrt{\left({A}−\mathrm{3}\right)^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }+\sqrt{\left({A}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} }=\sqrt{\left(\mathrm{3}+\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{3}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\sqrt{{A}\left({A}−\mathrm{6}\right)}+\sqrt{{A}\left({A}−\mathrm{4}\right)}=\mathrm{2}\sqrt{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{23}{A}^{\mathrm{2}} −\mathrm{120}{A}−\mathrm{144}=\mathrm{0} \\ $$$$\Rightarrow{A}=\frac{\mathrm{12}\left(\mathrm{5}+\mathrm{4}\sqrt{\mathrm{3}}\right)}{\mathrm{23}}\approx\mathrm{6}.\mathrm{223} \\ $$$$ \\ $$$$\left(\frac{\mathrm{1}}{{R}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{{A}}\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{1}}{{R}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{{A}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{{R}^{\mathrm{2}} }−\frac{\mathrm{2}}{{R}}\left(\frac{\mathrm{5}}{\mathrm{6}}−\frac{\mathrm{1}}{{A}}\right)+\frac{\mathrm{1}}{{A}^{\mathrm{2}} }+\frac{\mathrm{5}}{\mathrm{3}{A}}+\frac{\mathrm{1}}{\mathrm{36}}=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{R}}=\frac{\mathrm{5}}{\mathrm{6}}−\frac{\mathrm{1}}{{A}}\pm\sqrt{\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−\frac{\mathrm{5}}{{A}}\right)} \\ $$$$\Rightarrow{R}=\frac{\mathrm{1}}{\frac{\mathrm{5}}{\mathrm{6}}−\frac{\mathrm{1}}{{A}}\underset{−} {+}\sqrt{\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−\frac{\mathrm{5}}{{A}}\right)}}\approx\begin{cases}{\mathrm{0}.\mathrm{966497}}\\{\mathrm{3}.\mathrm{219212}}\end{cases} \\ $$$$ \\ $$$${R}\approx\mathrm{0}.\mathrm{966497}\:{for}\:{the}\:{smaller}\:{yellow}\:{one} \\ $$
Commented by Algoritm last updated on 02/Aug/20
$$\mathrm{thanks} \\ $$