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Question-201604




Question Number 201604 by ajfour last updated on 09/Dec/23
Commented by ajfour last updated on 09/Dec/23
Both circles in blue have equal  radii. Find.
$${Both}\:{circles}\:{in}\:{blue}\:{have}\:{equal} \\ $$$${radii}.\:{Find}. \\ $$
Answered by mr W last updated on 09/Dec/23
Commented by mr W last updated on 09/Dec/23
λ=(r/R)  sin β=((R−2r)/R)=1−2λ  tan (β/2)=((sin β)/(1+cos β))=((1−2λ)/(1+2(√(λ(1−λ)))))  α=(((π/2)−β)/2)=(π/4)−(β/2)  DC=DG=(r/(tan α))=((r(1+tan (β/2)))/(1−tan (β/2)))        =((r(1+((1−2λ)/(1+2(√(λ(1−λ)))))))/(1−((1−2λ)/(1+2(√(λ(1−λ)))))))=((r(1−λ+(√(λ(1−λ)))))/( (√(λ(1−λ)))+λ))  EF=(√((R+r)^2 −(R−r)^2 ))=2(√(Rr))  DH=((r(1−λ+(√(λ(1−λ)))))/( (√(λ(1−λ)))+λ))+2(√(Rr))  DH=2R tan β=((2R(1−2λ))/(2(√(λ(1−λ)))))=((R(1−2λ))/( (√(λ(1−λ)))))  ((r(1−λ+(√(λ(1−λ)))))/( (√(λ(1−λ)))+λ))+2(√(Rr))=((R(1−2λ))/( (√(λ(1−λ)))))  ((λ(1−λ+(√(λ(1−λ)))))/( (√(λ(1−λ)))+λ))+2(√λ)=((1−2λ)/( (√(λ(1−λ)))))  ⇒λ≈0.21995157
$$\lambda=\frac{{r}}{{R}} \\ $$$$\mathrm{sin}\:\beta=\frac{{R}−\mathrm{2}{r}}{{R}}=\mathrm{1}−\mathrm{2}\lambda \\ $$$$\mathrm{tan}\:\frac{\beta}{\mathrm{2}}=\frac{\mathrm{sin}\:\beta}{\mathrm{1}+\mathrm{cos}\:\beta}=\frac{\mathrm{1}−\mathrm{2}\lambda}{\mathrm{1}+\mathrm{2}\sqrt{\lambda\left(\mathrm{1}−\lambda\right)}} \\ $$$$\alpha=\frac{\frac{\pi}{\mathrm{2}}−\beta}{\mathrm{2}}=\frac{\pi}{\mathrm{4}}−\frac{\beta}{\mathrm{2}} \\ $$$${DC}={DG}=\frac{{r}}{\mathrm{tan}\:\alpha}=\frac{{r}\left(\mathrm{1}+\mathrm{tan}\:\frac{\beta}{\mathrm{2}}\right)}{\mathrm{1}−\mathrm{tan}\:\frac{\beta}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:=\frac{{r}\left(\mathrm{1}+\frac{\mathrm{1}−\mathrm{2}\lambda}{\mathrm{1}+\mathrm{2}\sqrt{\lambda\left(\mathrm{1}−\lambda\right)}}\right)}{\mathrm{1}−\frac{\mathrm{1}−\mathrm{2}\lambda}{\mathrm{1}+\mathrm{2}\sqrt{\lambda\left(\mathrm{1}−\lambda\right)}}}=\frac{{r}\left(\mathrm{1}−\lambda+\sqrt{\lambda\left(\mathrm{1}−\lambda\right)}\right)}{\:\sqrt{\lambda\left(\mathrm{1}−\lambda\right)}+\lambda} \\ $$$${EF}=\sqrt{\left({R}+{r}\right)^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{{Rr}} \\ $$$${DH}=\frac{{r}\left(\mathrm{1}−\lambda+\sqrt{\lambda\left(\mathrm{1}−\lambda\right)}\right)}{\:\sqrt{\lambda\left(\mathrm{1}−\lambda\right)}+\lambda}+\mathrm{2}\sqrt{{Rr}} \\ $$$${DH}=\mathrm{2}{R}\:\mathrm{tan}\:\beta=\frac{\mathrm{2}{R}\left(\mathrm{1}−\mathrm{2}\lambda\right)}{\mathrm{2}\sqrt{\lambda\left(\mathrm{1}−\lambda\right)}}=\frac{{R}\left(\mathrm{1}−\mathrm{2}\lambda\right)}{\:\sqrt{\lambda\left(\mathrm{1}−\lambda\right)}} \\ $$$$\frac{{r}\left(\mathrm{1}−\lambda+\sqrt{\lambda\left(\mathrm{1}−\lambda\right)}\right)}{\:\sqrt{\lambda\left(\mathrm{1}−\lambda\right)}+\lambda}+\mathrm{2}\sqrt{{Rr}}=\frac{{R}\left(\mathrm{1}−\mathrm{2}\lambda\right)}{\:\sqrt{\lambda\left(\mathrm{1}−\lambda\right)}} \\ $$$$\frac{\lambda\left(\mathrm{1}−\lambda+\sqrt{\lambda\left(\mathrm{1}−\lambda\right)}\right)}{\:\sqrt{\lambda\left(\mathrm{1}−\lambda\right)}+\lambda}+\mathrm{2}\sqrt{\lambda}=\frac{\mathrm{1}−\mathrm{2}\lambda}{\:\sqrt{\lambda\left(\mathrm{1}−\lambda\right)}} \\ $$$$\Rightarrow\lambda\approx\mathrm{0}.\mathrm{21995157} \\ $$
Commented by ajfour last updated on 09/Dec/23
Thank you sir. I got same. I will  analyse your way only now.
$${Thank}\:{you}\:{sir}.\:{I}\:{got}\:{same}.\:{I}\:{will} \\ $$$${analyse}\:{your}\:{way}\:{only}\:{now}. \\ $$
Answered by ajfour last updated on 09/Dec/23
Commented by ajfour last updated on 09/Dec/23
Eq. of line  y+xtan 2θ=1  cos 2θ=1−2r  ⇒ r=sin^2 θ  2(√r)+((rcos θ)/(sin θ))=(2/(tan 2θ))  2sin θ+sin θcos θ=((2cos^2 θ−1)/(sin θcos θ))  say  cos θ=t  (1−t^2 )t(2+t)=2t^2 −1  ⇒  t≈ 0.8832  r=sin^2 θ=1−t^2  ≈ 0.21995
$${Eq}.\:{of}\:{line} \\ $$$${y}+{x}\mathrm{tan}\:\mathrm{2}\theta=\mathrm{1} \\ $$$$\mathrm{cos}\:\mathrm{2}\theta=\mathrm{1}−\mathrm{2}{r} \\ $$$$\Rightarrow\:{r}=\mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$$\mathrm{2}\sqrt{{r}}+\frac{{r}\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}=\frac{\mathrm{2}}{\mathrm{tan}\:\mathrm{2}\theta} \\ $$$$\mathrm{2sin}\:\theta+\mathrm{sin}\:\theta\mathrm{cos}\:\theta=\frac{\mathrm{2cos}\:^{\mathrm{2}} \theta−\mathrm{1}}{\mathrm{sin}\:\theta\mathrm{cos}\:\theta} \\ $$$${say}\:\:\mathrm{cos}\:\theta={t} \\ $$$$\left(\mathrm{1}−{t}^{\mathrm{2}} \right){t}\left(\mathrm{2}+{t}\right)=\mathrm{2}{t}^{\mathrm{2}} −\mathrm{1} \\ $$$$\Rightarrow\:\:{t}\approx\:\mathrm{0}.\mathrm{8832} \\ $$$${r}=\mathrm{sin}\:^{\mathrm{2}} \theta=\mathrm{1}−{t}^{\mathrm{2}} \:\approx\:\mathrm{0}.\mathrm{21995} \\ $$
Commented by mr W last updated on 09/Dec/23
nice solution!
$${nice}\:{solution}! \\ $$
Commented by Frix last updated on 10/Dec/23
We can give the exact solution  (1−t^2 )t(2+t)=2t^2 −1  ⇔  t^4 +2t^3 +t^2 −2t−1=0  t=((−1+(√2)±(√(−1+2(√2))))/2)  t=((−1−(√2)±i(√(1+2(√2))))/2)
$$\mathrm{We}\:\mathrm{can}\:\mathrm{give}\:\mathrm{the}\:\mathrm{exact}\:\mathrm{solution} \\ $$$$\left(\mathrm{1}−{t}^{\mathrm{2}} \right){t}\left(\mathrm{2}+{t}\right)=\mathrm{2}{t}^{\mathrm{2}} −\mathrm{1} \\ $$$$\Leftrightarrow \\ $$$${t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{3}} +{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}=\mathrm{0} \\ $$$${t}=\frac{−\mathrm{1}+\sqrt{\mathrm{2}}\pm\sqrt{−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$${t}=\frac{−\mathrm{1}−\sqrt{\mathrm{2}}\pm\mathrm{i}\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$
Commented by ajfour last updated on 28/Oct/24
Thanks for the exact!
$${Thanks}\:{for}\:{the}\:{exact}! \\ $$

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