Question Number 186408 by normans last updated on 04/Feb/23
Answered by mr W last updated on 04/Feb/23
$${R}={radius}\:{of}\:{big}\:{circle} \\ $$$${r}={radius}\:{of}\:{small}\:{circles} \\ $$$$\left({R}−{r}\right)^{\mathrm{2}} =\left(\mathrm{3}{r}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} \\ $$$$\mathrm{9}{r}^{\mathrm{2}} +\mathrm{2}{Rr}−{R}^{\mathrm{2}} =\mathrm{0} \\ $$$${r}=\frac{\left(\sqrt{\mathrm{10}}−\mathrm{1}\right){R}}{\mathrm{9}} \\ $$$$\mathrm{sin}\:\frac{\theta}{\mathrm{2}}=\frac{{r}}{{R}+{r}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}+\mathrm{2}} \\ $$$$\theta=\mathrm{2}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}+\mathrm{2}} \\ $$$${n}=\lfloor\frac{\mathrm{2}\pi}{\theta}\rfloor=\lfloor\frac{\pi}{\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}+\mathrm{2}}}\rfloor=\mathrm{16} \\ $$$${i}.{e}.\:{max}.\:\mathrm{16}\:{small}\:{circles}\:{fit}\:{into}\:{the} \\ $$$${ring}. \\ $$
Commented by normans last updated on 04/Feb/23
$$\boldsymbol{{great}}\:\boldsymbol{{Sir}},\boldsymbol{{You}}\:\boldsymbol{{really}}\:\boldsymbol{{are}}\:\boldsymbol{{a}}\:\boldsymbol{{geometry}}\:\boldsymbol{{expert}} \\ $$