Question Number 193635 by mr W last updated on 17/Jun/23
Commented by mr W last updated on 17/Jun/23
$${find}\:{the}\:{largest}\:{circle}\:{and}\:{the}\:{largest} \\ $$$${square}\:{which}\:{you}\:{can}\:{completely} \\ $$$${cover}\:{with}\:{three}\:{circular}\:{plates}\:{with} \\ $$$${radius}\:\mathrm{1}\:{respectively}. \\ $$
Answered by mr W last updated on 18/Jun/23
Commented by mr W last updated on 18/Jun/23
$$\Delta{ABC}={equilateral}\:{with}\:{side}\:{length} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{s}=\mathrm{2}×{r}=\mathrm{2} \\ $$$${R}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{2}}=\frac{{s}}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\approx\mathrm{1}.\mathrm{155} \\ $$
Answered by mr W last updated on 18/Jun/23
Commented by mr W last updated on 18/Jun/23
$$\Delta{ABC}={equilateral}\:{with}\:{side}\:{length} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{s}=\mathrm{2}×{r}=\mathrm{2} \\ $$$$\sqrt{{s}^{\mathrm{2}} −{a}^{\mathrm{2}} }={a}−\frac{{s}}{\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} −\sqrt{\mathrm{2}}{sa}−\frac{{s}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{0} \\ $$$${a}^{\mathrm{2}} −\sqrt{\mathrm{2}}{a}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{a}=\frac{\sqrt{\mathrm{2}}+\sqrt{\mathrm{6}}}{\mathrm{2}}\approx\mathrm{1}.\mathrm{932} \\ $$