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Question-201707

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Question Number 201707 by ajfour last updated on 10/Dec/23
Answered by mr W last updated on 11/Dec/23
Commented by mr W last updated on 11/Dec/23
let (r/R)=λ cos 2α=((R−2r)/R)=1−2λ tan α=((√λ)/( (√(1−λ)))) ED=(r/(tan α))=((r(√(1−λ)))/( (√λ))) AD=2AB=2(√(R^2 −(R−2r)^2 ))=4(√((R−r)r)) cos 2α=((ED)/(AD))=(r/(4 tan α (√((R−r)r))))=((λ(√(1−λ)))/(4(√λ)(√((1−λ)λ))))=(1/4) 1−2λ=(1/4) ⇒λ=(3/8)
$${let}\:\frac{{r}}{{R}}=\lambda \\ $$$$\mathrm{cos}\:\mathrm{2}\alpha=\frac{{R}−\mathrm{2}{r}}{{R}}=\mathrm{1}−\mathrm{2}\lambda \\ $$$$\mathrm{tan}\:\alpha=\frac{\sqrt{\lambda}}{\:\sqrt{\mathrm{1}−\lambda}} \\ $$$${ED}=\frac{{r}}{\mathrm{tan}\:\alpha}=\frac{{r}\sqrt{\mathrm{1}−\lambda}}{\:\sqrt{\lambda}} \\ $$$${AD}=\mathrm{2}{AB}=\mathrm{2}\sqrt{{R}^{\mathrm{2}} −\left({R}−\mathrm{2}{r}\right)^{\mathrm{2}} }=\mathrm{4}\sqrt{\left({R}−{r}\right){r}} \\ $$$$\mathrm{cos}\:\mathrm{2}\alpha=\frac{{ED}}{{AD}}=\frac{{r}}{\mathrm{4}\:\mathrm{tan}\:\alpha\:\sqrt{\left({R}−{r}\right){r}}}=\frac{\lambda\sqrt{\mathrm{1}−\lambda}}{\mathrm{4}\sqrt{\lambda}\sqrt{\left(\mathrm{1}−\lambda\right)\lambda}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{1}−\mathrm{2}\lambda=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{3}}{\mathrm{8}} \\ $$
Commented by ajfour last updated on 10/Dec/23
let R=1 AB=(√(1−(1−2r)^2 ))=2(√(r(1−r))) (p/1)=(r/(1−2r)) ⇒ p+r=((2r(1−r))/(1−2r)) ((2AB−s)/(AB))=(r/(1−2r)) 2−(s/(2(√(r(1−r)))))=(r/(1−2r)) ......(i) s^2 =(p+r)(2−p−r) 2−(p+r)=2−((2r(1−r))/(1−2r))=((2−6r+2r^2 )/(1−2r)) s^2 =((2r(1−r))/((1−2r)))×((2(1−3r+r^2 ))/((1−2r))) but from (i) (s^2 /(4r(1−r)))=(2−(r/(1−2r)))^2 ⇒ (1−3r+r^2 )=(2−5r)^2 24r^2 −17r+3=0 r=((17±(√(289−4×3×24)))/(48)) r =((17±1)/(48)) ⇒ r=(1/3) or r=(3/8)
$${let}\:{R}=\mathrm{1} \\ $$$${AB}=\sqrt{\mathrm{1}−\left(\mathrm{1}−\mathrm{2}{r}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{{r}\left(\mathrm{1}−{r}\right)} \\ $$$$\frac{{p}}{\mathrm{1}}=\frac{{r}}{\mathrm{1}−\mathrm{2}{r}}\:\:\Rightarrow\:\:{p}+{r}=\frac{\mathrm{2}{r}\left(\mathrm{1}−{r}\right)}{\mathrm{1}−\mathrm{2}{r}} \\ $$$$\frac{\mathrm{2}{AB}−{s}}{{AB}}=\frac{{r}}{\mathrm{1}−\mathrm{2}{r}} \\ $$$$\mathrm{2}−\frac{{s}}{\mathrm{2}\sqrt{{r}\left(\mathrm{1}−{r}\right)}}=\frac{{r}}{\mathrm{1}−\mathrm{2}{r}}\:\:\:\:……\left({i}\right) \\ $$$${s}^{\mathrm{2}} =\left({p}+{r}\right)\left(\mathrm{2}−{p}−{r}\right) \\ $$$$\mathrm{2}−\left({p}+{r}\right)=\mathrm{2}−\frac{\mathrm{2}{r}\left(\mathrm{1}−{r}\right)}{\mathrm{1}−\mathrm{2}{r}}=\frac{\mathrm{2}−\mathrm{6}{r}+\mathrm{2}{r}^{\mathrm{2}} }{\mathrm{1}−\mathrm{2}{r}} \\ $$$${s}^{\mathrm{2}} =\frac{\mathrm{2}{r}\left(\mathrm{1}−{r}\right)}{\left(\mathrm{1}−\mathrm{2}{r}\right)}×\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{3}{r}+{r}^{\mathrm{2}} \right)}{\left(\mathrm{1}−\mathrm{2}{r}\right)} \\ $$$${but}\:{from}\:\left({i}\right) \\ $$$$\:\:\frac{{s}^{\mathrm{2}} }{\mathrm{4}{r}\left(\mathrm{1}−{r}\right)}=\left(\mathrm{2}−\frac{{r}}{\mathrm{1}−\mathrm{2}{r}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\left(\mathrm{1}−\mathrm{3}{r}+{r}^{\mathrm{2}} \right)=\left(\mathrm{2}−\mathrm{5}{r}\right)^{\mathrm{2}} \\ $$$$\mathrm{24}{r}^{\mathrm{2}} −\mathrm{17}{r}+\mathrm{3}=\mathrm{0} \\ $$$${r}=\frac{\mathrm{17}\pm\sqrt{\mathrm{289}−\mathrm{4}×\mathrm{3}×\mathrm{24}}}{\mathrm{48}} \\ $$$$\:{r}\:=\frac{\mathrm{17}\pm\mathrm{1}}{\mathrm{48}}\:\:\Rightarrow\:\:{r}=\frac{\mathrm{1}}{\mathrm{3}}\:\:{or}\:\:{r}=\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$ \\ $$
Commented by mr W last updated on 11/Dec/23
cos ∠EDA=((ED)/(AD))
$$\mathrm{cos}\:\angle{EDA}=\frac{{ED}}{{AD}} \\ $$

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