Question Number 197680 by mr W last updated on 26/Sep/23
Commented by mr W last updated on 26/Sep/23
$${unsolved}\:{old}\:{question}\:{Q}#\mathrm{197245} \\ $$
Answered by mr W last updated on 26/Sep/23
Commented by mr W last updated on 26/Sep/23
$${R}={radius}\:{of}\:{yellow}\:{semicircle} \\ $$$${r}={radius}\:{of}\:{green}\:{semicircle} \\ $$$${p}={radius}\:{of}\:{pink}\:{semicircle} \\ $$$$\sqrt{\left({R}+{r}\right)^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} }=\sqrt{\left({R}+{p}\right)^{\mathrm{2}} −{R}^{\mathrm{2}} }+\sqrt{\left({r}+{p}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$$\mathrm{2}\sqrt{{Rr}}=\sqrt{{p}\left(\mathrm{2}{R}+{p}\right)}+\sqrt{{p}\left(\mathrm{2}{r}+{p}\right)} \\ $$$$\mathrm{2}\sqrt{{Rr}}−\sqrt{{p}\left(\mathrm{2}{R}+{p}\right)}=\sqrt{{p}\left(\mathrm{2}{r}+{p}\right)} \\ $$$$\mathrm{2}{Rr}+\left({R}−{r}\right){p}=\mathrm{2}\sqrt{{Rrp}\left(\mathrm{2}{R}+{p}\right)} \\ $$$$\mathrm{4}{R}^{\mathrm{2}} {r}^{\mathrm{2}} +\left({R}−{r}\right)^{\mathrm{2}} {p}^{\mathrm{2}} +\mathrm{4}{Rr}\left({R}−{r}\right){p}=\mathrm{4}{Rrp}\left(\mathrm{2}{R}+{p}\right) \\ $$$$\Rightarrow\left(\mathrm{6}{Rr}−{R}^{\mathrm{2}} −{r}^{\mathrm{2}} \right){p}^{\mathrm{2}} +\mathrm{4}{Rr}\left({R}+{r}\right){p}−\mathrm{4}{R}^{\mathrm{2}} {r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{p}=\frac{−\mathrm{2}{Rr}\left({R}+{r}\right)+\mathrm{4}{Rr}\sqrt{\mathrm{2}{Rr}}}{\mathrm{6}{Rr}−{R}^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$$\frac{{yellow}\:{area}}{{green}\:\:{area}}=\left(\frac{{R}}{{r}}\right)^{\mathrm{2}} =\frac{\mathrm{2}\pi}{\pi/\mathrm{2}}=\mathrm{4}\:\Rightarrow{R}=\mathrm{2}{r} \\ $$$$\Rightarrow{p}=\frac{\mathrm{4}{r}}{\mathrm{7}} \\ $$$$\frac{{pink}\:{area}}{{green}\:\:{area}}=\left(\frac{{p}}{{r}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{4}}{\mathrm{7}}\right)^{\mathrm{2}} =\frac{\mathrm{16}}{\mathrm{49}} \\ $$$$\Rightarrow{pink}\:{area}\:=\frac{\mathrm{16}}{\mathrm{49}}×\frac{\pi}{\mathrm{2}}=\frac{\mathrm{8}\pi}{\mathrm{49}}\:\checkmark \\ $$