Question Number 64289 by ajfour last updated on 16/Jul/19
Commented by ajfour last updated on 16/Jul/19
$${Find}\:{s}\:{in}\:{terms}\:{of}\:{radii}\:{a}\:{and}\:{b}. \\ $$
Commented by behi83417@gmail.com last updated on 16/Jul/19
$$. \\ $$
Answered by mr W last updated on 16/Jul/19
$$\mathrm{cos}\:\alpha=\frac{{s}}{\mathrm{2}{a}} \\ $$$$\mathrm{cos}\:\beta=\frac{{s}}{\mathrm{2}{b}} \\ $$$$\alpha+\beta+\frac{\pi}{\mathrm{3}}=\pi \\ $$$$\alpha=\frac{\mathrm{2}\pi}{\mathrm{3}}−\beta \\ $$$$\mathrm{cos}\:\alpha=\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\beta\right)=−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{cos}\:\beta+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{sin}\:\beta \\ $$$$\frac{{s}}{\mathrm{2}{a}}=−\frac{{s}}{\mathrm{4}{b}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\sqrt{\mathrm{4}{b}^{\mathrm{2}} −{s}^{\mathrm{2}} }}{\mathrm{2}{b}} \\ $$$$\frac{\left(\mathrm{2}{b}+{a}\right){s}}{{a}}=\sqrt{\mathrm{3}\left(\mathrm{4}{b}^{\mathrm{2}} −{s}^{\mathrm{2}} \right)} \\ $$$$\frac{{b}^{\mathrm{2}} +{ab}+{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} }{s}^{\mathrm{2}} =\mathrm{3}{b}^{\mathrm{2}} \\ $$$$\Rightarrow{s}=\frac{\sqrt{\mathrm{3}}{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} }} \\ $$
Commented by ajfour last updated on 17/Jul/19
$${thank}\:{you}\:{sir}. \\ $$
Commented by Tawa1 last updated on 17/Jul/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{study}\:\mathrm{the}\:\mathrm{solution} \\ $$
Answered by mr W last updated on 17/Jul/19
Commented by mr W last updated on 17/Jul/19
$$\frac{{EC}}{{AC}}=\frac{{AC}}{{DC}} \\ $$$$\Rightarrow{EC}=\frac{{s}^{\mathrm{2}} }{\mathrm{2}{a}} \\ $$$${AE}^{\mathrm{2}} ={AC}^{\mathrm{2}} −{EC}^{\mathrm{2}} ={s}^{\mathrm{2}} −\frac{{s}^{\mathrm{4}} }{\mathrm{4}{a}^{\mathrm{2}} } \\ $$$${similarly} \\ $$$${CF}=\frac{{s}^{\mathrm{2}} }{\mathrm{2}{b}} \\ $$$${BF}^{\mathrm{2}} ={s}^{\mathrm{2}} −\frac{{s}^{\mathrm{4}} }{\mathrm{4}{b}^{\mathrm{2}} } \\ $$$${AB}^{\mathrm{2}} =\left({EC}+{CF}\right)^{\mathrm{2}} +\left({BF}−{AE}\right)^{\mathrm{2}} \\ $$$${s}^{\mathrm{2}} =\left(\frac{{s}^{\mathrm{2}} }{\mathrm{2}{a}}+\frac{{s}^{\mathrm{2}} }{\mathrm{2}{b}}\right)^{\mathrm{2}} +{s}^{\mathrm{2}} −\frac{{s}^{\mathrm{4}} }{\mathrm{4}{a}^{\mathrm{2}} }+{s}^{\mathrm{2}} −\frac{{s}^{\mathrm{4}} }{\mathrm{4}{b}^{\mathrm{2}} }−\mathrm{2}\sqrt{\left({s}^{\mathrm{2}} −\frac{{s}^{\mathrm{4}} }{\mathrm{4}{a}^{\mathrm{2}} }\right)\left({s}^{\mathrm{2}} −\frac{{s}^{\mathrm{4}} }{\mathrm{4}{b}^{\mathrm{2}} }\right)} \\ $$$$\frac{{s}^{\mathrm{4}} }{\mathrm{2}{ab}}+{s}^{\mathrm{2}} =\mathrm{2}\sqrt{\left({s}^{\mathrm{2}} −\frac{{s}^{\mathrm{4}} }{\mathrm{4}{a}^{\mathrm{2}} }\right)\left({s}^{\mathrm{2}} −\frac{{s}^{\mathrm{4}} }{\mathrm{4}{b}^{\mathrm{2}} }\right)} \\ $$$$\frac{{s}^{\mathrm{2}} }{\mathrm{2}{ab}}+\mathrm{1}=\mathrm{2}\sqrt{\left(\mathrm{1}−\frac{{s}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{{s}^{\mathrm{2}} }{\mathrm{4}{b}^{\mathrm{2}} }\right)} \\ $$$$\left(\frac{{s}^{\mathrm{2}} }{\mathrm{2}{ab}}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}\left(\mathrm{1}−\frac{{s}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{{s}^{\mathrm{2}} }{\mathrm{4}{b}^{\mathrm{2}} }\right) \\ $$$$\frac{{s}^{\mathrm{4}} }{\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }+\mathrm{1}+\frac{{s}^{\mathrm{2}} }{{ab}}=\mathrm{4}\left(\mathrm{1}−\frac{{s}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }−\frac{{s}^{\mathrm{2}} }{\mathrm{4}{b}^{\mathrm{2}} }+\frac{{s}^{\mathrm{4}} }{\mathrm{16}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }\right) \\ $$$$\frac{{s}^{\mathrm{2}} }{{ab}}=\mathrm{3}−\frac{{s}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{{s}^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$$\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{ab}}+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\right){s}^{\mathrm{2}} =\mathrm{3} \\ $$$$\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right){s}^{\mathrm{2}} =\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\Rightarrow{s}=\frac{\sqrt{\mathrm{3}}{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} }} \\ $$
Commented by Tawa1 last updated on 18/Jul/19
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{more} \\ $$