Question Number 210694 by ajfour last updated on 16/Aug/24 | ||
Answered by mr W last updated on 17/Aug/24 | ||
Commented by mr W last updated on 17/Aug/24 | ||
$${R}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{sin}\:\alpha}{{b}}=\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\theta+\alpha\right)}{{R}}=\frac{\mathrm{cos}\:\theta\:\mathrm{cos}\:\alpha+\mathrm{sin}\:\theta\:\mathrm{sin}\:\alpha}{{R}} \\ $$$$\frac{{R}}{{b}}=\frac{\mathrm{cos}\:\theta\:}{\mathrm{tan}\:\alpha}+\mathrm{sin}\:\theta \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}=\left(\frac{{R}}{{b}}−\mathrm{sin}\:\theta\right)\frac{\mathrm{1}}{\mathrm{cos}\:\theta} \\ $$$$\frac{\mathrm{sin}\:\alpha}{{a}}=\frac{\mathrm{sin}\:\left(\theta+\alpha\right)}{{R}}=\frac{\mathrm{sin}\:\theta\:\mathrm{cos}\:\alpha+\mathrm{cos}\:\theta\:\mathrm{sin}\:\alpha}{{R}} \\ $$$$\frac{{R}}{{a}}=\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\alpha}+\mathrm{cos}\:\theta \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}=\left(\frac{{R}}{{a}}−\mathrm{cos}\:\theta\right)\frac{\mathrm{1}}{\mathrm{sin}\:\theta} \\ $$$$\left(\frac{{R}}{{b}}−\mathrm{sin}\:\theta\right)\frac{\mathrm{1}}{\mathrm{cos}\:\theta}=\left(\frac{{R}}{{a}}−\mathrm{cos}\:\theta\right)\frac{\mathrm{1}}{\mathrm{sin}\:\theta} \\ $$$$\frac{{R}}{{a}}\:\mathrm{cos}\:\theta−\frac{{R}}{{b}}\:\mathrm{sin}\:\theta=\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\Rightarrow\frac{\mathrm{cos}\:\theta}{\mathrm{cos}\:\phi}−\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:\phi}=\mathrm{cos}\:\mathrm{2}\theta \\ $$$$ \\ $$$${L}_{{min}} =\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} −\mathrm{2}{aR}\:\mathrm{cos}\:\theta}+\sqrt{{b}^{\mathrm{2}} +{R}^{\mathrm{2}} −\mathrm{2}{bR}\:\mathrm{sin}\:\theta} \\ $$$$\frac{{L}_{{min}} }{{R}}=\sqrt{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\phi−\mathrm{2}\:\mathrm{cos}\:\phi\:\mathrm{cos}\:\theta}+\sqrt{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\phi−\mathrm{2}\:\mathrm{sin}\:\phi\:\mathrm{sin}\:\theta} \\ $$ | ||
Commented by mr W last updated on 17/Aug/24 | ||
$${an}\:{exact}\:{solution}\:{seems}\:{not}\:{to}\:{be} \\ $$$${possible}. \\ $$ | ||
Commented by ajfour last updated on 17/Aug/24 | ||
$$\frac{\mathrm{sin}\:\alpha}{{b}}=\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}+\theta−\alpha\right)}{{R}}=\frac{\mathrm{cos}\:\theta\:\mathrm{cos}\:\alpha−\mathrm{sin}\:\theta\:\mathrm{sin}\:\alpha}{{R}} \\ $$$$\frac{{R}}{{a}}=\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\alpha}−\mathrm{cos}\:\theta \\ $$$$\left(\frac{\mathrm{sin}\:\theta}{{m}}−\frac{{R}}{{a}}\right)^{\mathrm{2}} =\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$${L}_{{min}} =\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} −\mathrm{2}{aR}\:\mathrm{cos}\:\theta}+\sqrt{{b}^{\mathrm{2}} +{R}^{\mathrm{2}} −\mathrm{2}{bR}\:\mathrm{sin}\:\theta} \\ $$$$\:\:=\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} −\mathrm{2}{R}\left(\frac{{a}\mathrm{sin}\:\theta}{{m}}−{R}\right)} \\ $$$$\:\:\:\:\:\:\:\:+\sqrt{{b}^{\mathrm{2}} +{R}^{\mathrm{2}} −\mathrm{2}{bR}\mathrm{sin}\:\theta} \\ $$$$=\sqrt{{a}^{\mathrm{2}} +\mathrm{3}{R}^{\mathrm{2}} −\frac{\mathrm{2}{aR}}{{m}}\mathrm{sin}\:\theta}+\sqrt{{b}^{\mathrm{2}} +{R}^{\mathrm{2}} −\mathrm{2}{bR}\mathrm{sin}\:\theta} \\ $$$$....... \\ $$ | ||