Question Number 222579 by ajfour last updated on 30/Jun/25
Answered by mr W last updated on 30/Jun/25
Commented by mr W last updated on 01/Jul/25
$${s}=\sqrt{\mathrm{3}}{R}+\frac{{R}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}\:\Rightarrow\frac{{s}}{{R}}=\sqrt{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}} \\ $$$${similarly} \\ $$$$\frac{{s}}{{r}}=\sqrt{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\beta}{\mathrm{2}}} \\ $$$$\frac{{R}}{{r}}=\frac{\sqrt{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\beta}{\mathrm{2}}}}{\:\sqrt{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}} \\ $$$${let}\:{t}=\mathrm{tan}\:\frac{\beta}{\mathrm{2}} \\ $$$$\alpha+\beta=\mathrm{60}° \\ $$$$\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\mathrm{tan}\:\left(\mathrm{30}°−\frac{\beta}{\mathrm{2}}\right)=\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}−{t}}{\mathrm{1}+\frac{{t}}{\:\sqrt{\mathrm{3}}}}=\frac{\mathrm{1}−\sqrt{\mathrm{3}}{t}}{\:{t}+\sqrt{\mathrm{3}}} \\ $$$$\frac{\sqrt{\mathrm{3}}+\frac{\mathrm{1}}{{t}}}{\:\sqrt{\mathrm{3}}+\frac{{t}+\sqrt{\mathrm{3}}}{\:\mathrm{1}−\sqrt{\mathrm{3}}{t}}}=\frac{{R}}{{r}}=\lambda \\ $$$$\left(\mathrm{2}\lambda−\mathrm{3}\right){t}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}\lambda{t}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{\sqrt{\mathrm{3}}\lambda−\sqrt{\mathrm{3}\lambda^{\mathrm{2}} −\mathrm{2}\lambda+\mathrm{3}}}{\mathrm{2}\lambda−\mathrm{3}} \\ $$$$\frac{{b}}{{a}}=\mathrm{tan}\:\beta=\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$${for}\:\lambda=\mathrm{2}: \\ $$$${t}=\mathrm{2}\sqrt{\mathrm{3}}−\sqrt{\mathrm{11}} \\ $$$$\frac{{b}}{{a}}=\frac{\mathrm{2}\sqrt{\mathrm{3}}−\sqrt{\mathrm{11}}}{\mathrm{2}\sqrt{\mathrm{33}}−\mathrm{11}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{11}}}\approx\mathrm{0}.\mathrm{302} \\ $$
Commented by ajfour last updated on 01/Jul/25
$${Yes}\:{Sir},\:{remarkable}\:{for}\:{an}\:{answer}. \\ $$
Commented by ajfour last updated on 01/Jul/25
https://youtu.be/zAjZiJvKOPk?si=hTaP4tnAAjFSPex6