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Question Number 88708 by ajfour last updated on 12/Apr/20
Commented by ajfour last updated on 12/Apr/20
If AE=BF , find r/R.
$${If}\:\:{AE}={BF}\:,\:{find}\:{r}/{R}. \\ $$
Commented by mr W last updated on 12/Apr/20
AE=FB=k (2R−k)^2 +r^2 =(k+r)^2 2R^2 =(2R+r)k ⇒k=((2R^2 )/(2R+r)) (r/(k+r))=((√((2R)^2 −(2r+k)^2 ))/(2R))=(√(1−((r/R)+(k/(2R)))^2 )) ((1/((k/r)+1)))^2 =1−((r/R)+(k/(2R)))^2 with λ=(r/R) ((1/((2/((2+λ)λ))+1)))^2 =1−(λ+(1/(2+λ)))^2 λ^2 (2+λ)^4 =(λ^2 +2λ+2)^2 [(λ+2)^2 −(λ+1)^4 ] (λ+1)^2 (λ^3 +3λ^2 +2λ−2)(λ^3 +3λ^2 +6λ+6)=0 ⇒λ=0.52138
$${AE}={FB}={k} \\ $$$$\left(\mathrm{2}{R}−{k}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} =\left({k}+{r}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{R}^{\mathrm{2}} =\left(\mathrm{2}{R}+{r}\right){k} \\ $$$$\Rightarrow{k}=\frac{\mathrm{2}{R}^{\mathrm{2}} }{\mathrm{2}{R}+{r}} \\ $$$$\frac{{r}}{{k}+{r}}=\frac{\sqrt{\left(\mathrm{2}{R}\right)^{\mathrm{2}} −\left(\mathrm{2}{r}+{k}\right)^{\mathrm{2}} }}{\mathrm{2}{R}}=\sqrt{\mathrm{1}−\left(\frac{{r}}{{R}}+\frac{{k}}{\mathrm{2}{R}}\right)^{\mathrm{2}} } \\ $$$$\left(\frac{\mathrm{1}}{\frac{{k}}{{r}}+\mathrm{1}}\right)^{\mathrm{2}} =\mathrm{1}−\left(\frac{{r}}{{R}}+\frac{{k}}{\mathrm{2}{R}}\right)^{\mathrm{2}} \\ $$$${with}\:\lambda=\frac{{r}}{{R}} \\ $$$$\left(\frac{\mathrm{1}}{\frac{\mathrm{2}}{\left(\mathrm{2}+\lambda\right)\lambda}+\mathrm{1}}\right)^{\mathrm{2}} =\mathrm{1}−\left(\lambda+\frac{\mathrm{1}}{\mathrm{2}+\lambda}\right)^{\mathrm{2}} \\ $$$$\lambda^{\mathrm{2}} \left(\mathrm{2}+\lambda\right)^{\mathrm{4}} =\left(\lambda^{\mathrm{2}} +\mathrm{2}\lambda+\mathrm{2}\right)^{\mathrm{2}} \left[\left(\lambda+\mathrm{2}\right)^{\mathrm{2}} −\left(\lambda+\mathrm{1}\right)^{\mathrm{4}} \right] \\ $$$$\left(\lambda+\mathrm{1}\right)^{\mathrm{2}} \left(\lambda^{\mathrm{3}} +\mathrm{3}\lambda^{\mathrm{2}} +\mathrm{2}\lambda−\mathrm{2}\right)\left(\lambda^{\mathrm{3}} +\mathrm{3}\lambda^{\mathrm{2}} +\mathrm{6}\lambda+\mathrm{6}\right)=\mathrm{0} \\ $$$$\Rightarrow\lambda=\mathrm{0}.\mathrm{52138} \\ $$
Commented by john santu last updated on 12/Apr/20
Commented by ajfour last updated on 12/Apr/20
Thanks for confirmation Sir. Nice solution.
$${Thanks}\:{for}\:{confirmation}\:{Sir}. \\ $$$${Nice}\:{solution}. \\ $$
Commented by john santu last updated on 12/Apr/20
let me try let AE = BF = x sin θ = sin θ (r/(r+x)) = ((BD)/(2R)) (i) BD = (√(4R^2 −(2r+x)^2 )) r+x = (√((2R−x)^2 −r^2 )) ⇒(r/( (√((2R−x)^2 −r^2 )))) = ((√(4R^2 −(2r+x)^2 ))/(2R)) (r^2 /((2R−x)^2 −r^2 )) = ((4R^2 −(2r+x)^2 )/(4R^2 ))
$${let}\:{me}\:{try}\: \\ $$$${let}\:{AE}\:=\:{BF}\:=\:{x}\: \\ $$$$\mathrm{sin}\:\theta\:=\:\mathrm{sin}\:\theta \\ $$$$\frac{{r}}{{r}+{x}}\:=\:\frac{{BD}}{\mathrm{2}{R}}\:\left({i}\right) \\ $$$${BD}\:=\:\sqrt{\mathrm{4}{R}^{\mathrm{2}} −\left(\mathrm{2}{r}+{x}\right)^{\mathrm{2}} } \\ $$$${r}+{x}\:=\:\sqrt{\left(\mathrm{2}{R}−{x}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{r}}{\:\sqrt{\left(\mathrm{2}{R}−{x}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }}\:=\:\frac{\sqrt{\mathrm{4}{R}^{\mathrm{2}} −\left(\mathrm{2}{r}+{x}\right)^{\mathrm{2}} }}{\mathrm{2}{R}} \\ $$$$\frac{{r}^{\mathrm{2}} }{\left(\mathrm{2}{R}−{x}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }\:=\:\frac{\mathrm{4}{R}^{\mathrm{2}} −\left(\mathrm{2}{r}+{x}\right)^{\mathrm{2}} }{\mathrm{4}{R}^{\mathrm{2}} } \\ $$$$ \\ $$
Answered by mahdi last updated on 12/Apr/20
2R=AF+FB=(√(AC^2 −CF^2 ))+FB =(√((AC+EC)^2 −CF^2 ))+FB⇒ (AE=FB=a,EC=CF=r) 2R=(√((r+a)^2 −r^2 ))+a⇒2R−a=(√(a^2 +2ar)) ⇒4R^2 +a^2 −4Ra=a^2 +2ar⇒ 2R^2 −2Ra=ar⇒a=((2R^2 )/(2R+r)) (i) AO^Δ D: OD^2 =AO^2 +AD^2 −2.AD.AO.cosA^� ⇒(cosA^� =((AF)/(AC))=((2R−a)/(a+r)))⇒ R^2 =R^2 +(a+2r)^2 −2.R.(a+2r).((2R−a)/(a+r)) ⇒(a+r)(a+2r)=2R(2R−a) (ii) ⇒paste a=((2R^2 )/(2R+r)) in (ii)⇒ ((2R^2 +r^2 +2Rr)/(2R+r))×((2R^2 +2r^2 +4Rr)/(2R+r))=2R((2R^2 +2Rr)/(2R+r)) (2R^2 +r^2 +2Rr)(R+r)=2R^2 (2R+r) ⇒get c=(r/R)⇒ (2+c^2 +2c)(1+c)=2(2+c)⇒ (1+c)^3 −(1+c)−2=0⇒1+c⋍1.5213 c=(r/R)⋍.5213
$$\mathrm{2R}=\mathrm{AF}+\mathrm{FB}=\sqrt{\mathrm{AC}^{\mathrm{2}} −\mathrm{CF}^{\mathrm{2}} }+\mathrm{FB} \\ $$$$=\sqrt{\left(\mathrm{AC}+\mathrm{EC}\right)^{\mathrm{2}} −\mathrm{CF}^{\mathrm{2}} }+\mathrm{FB}\Rightarrow \\ $$$$\left(\mathrm{AE}=\mathrm{FB}=\mathrm{a},\mathrm{EC}=\mathrm{CF}=\mathrm{r}\right) \\ $$$$\mathrm{2R}=\sqrt{\left(\mathrm{r}+\mathrm{a}\right)^{\mathrm{2}} −\mathrm{r}^{\mathrm{2}} }+\mathrm{a}\Rightarrow\mathrm{2R}−\mathrm{a}=\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{2ar}} \\ $$$$\Rightarrow\mathrm{4R}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} −\mathrm{4Ra}=\mathrm{a}^{\mathrm{2}} +\mathrm{2ar}\Rightarrow \\ $$$$\mathrm{2R}^{\mathrm{2}} −\mathrm{2Ra}=\mathrm{ar}\Rightarrow\mathrm{a}=\frac{\mathrm{2R}^{\mathrm{2}} }{\mathrm{2R}+\mathrm{r}}\:\:\:\left(\mathrm{i}\right) \\ $$$$\mathrm{A}\overset{\Delta} {\mathrm{O}D}:\:\mathrm{OD}^{\mathrm{2}} =\mathrm{AO}^{\mathrm{2}} +\mathrm{AD}^{\mathrm{2}} −\mathrm{2}.\mathrm{AD}.\mathrm{AO}.\mathrm{cos}\hat {\mathrm{A}} \\ $$$$\Rightarrow\left(\mathrm{cos}\hat {\mathrm{A}}=\frac{\mathrm{AF}}{\mathrm{AC}}=\frac{\mathrm{2R}−\mathrm{a}}{\mathrm{a}+\mathrm{r}}\right)\Rightarrow \\ $$$$\mathrm{R}^{\mathrm{2}} =\mathrm{R}^{\mathrm{2}} +\left(\mathrm{a}+\mathrm{2r}\right)^{\mathrm{2}} −\mathrm{2}.\mathrm{R}.\left(\mathrm{a}+\mathrm{2r}\right).\frac{\mathrm{2R}−\mathrm{a}}{\mathrm{a}+\mathrm{r}} \\ $$$$\Rightarrow\left(\mathrm{a}+\mathrm{r}\right)\left(\mathrm{a}+\mathrm{2r}\right)=\mathrm{2R}\left(\mathrm{2R}−\mathrm{a}\right)\:\:\left(\mathrm{ii}\right) \\ $$$$\Rightarrow\mathrm{paste}\:\mathrm{a}=\frac{\mathrm{2R}^{\mathrm{2}} }{\mathrm{2R}+\mathrm{r}}\:\:\mathrm{in}\:\left(\mathrm{ii}\right)\Rightarrow \\ $$$$ \\ $$$$\frac{\mathrm{2R}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} +\mathrm{2Rr}}{\mathrm{2R}+\mathrm{r}}×\frac{\mathrm{2R}^{\mathrm{2}} +\mathrm{2r}^{\mathrm{2}} +\mathrm{4Rr}}{\mathrm{2R}+\mathrm{r}}=\mathrm{2R}\frac{\mathrm{2R}^{\mathrm{2}} +\mathrm{2Rr}}{\mathrm{2R}+\mathrm{r}} \\ $$$$\left(\mathrm{2R}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} +\mathrm{2Rr}\right)\left(\mathrm{R}+\mathrm{r}\right)=\mathrm{2R}^{\mathrm{2}} \left(\mathrm{2R}+\mathrm{r}\right) \\ $$$$\Rightarrow\mathrm{get}\:\mathrm{c}=\frac{\mathrm{r}}{\mathrm{R}}\Rightarrow \\ $$$$\left(\mathrm{2}+\mathrm{c}^{\mathrm{2}} +\mathrm{2c}\right)\left(\mathrm{1}+\mathrm{c}\right)=\mathrm{2}\left(\mathrm{2}+\mathrm{c}\right)\Rightarrow \\ $$$$\left(\mathrm{1}+\mathrm{c}\right)^{\mathrm{3}} −\left(\mathrm{1}+\mathrm{c}\right)−\mathrm{2}=\mathrm{0}\Rightarrow\mathrm{1}+\mathrm{c}\backsimeq\mathrm{1}.\mathrm{5213} \\ $$$$\mathrm{c}=\frac{\mathrm{r}}{\mathrm{R}}\backsimeq.\mathrm{5213} \\ $$
Answered by ajfour last updated on 12/Apr/20
let ∠DAB=θ AE=BF=p = BD 2Rcos θ=p+2r ...(i) 2Rsin θ=p ...(ii) cos θ=((2R−p)/(p+r)) ...(iii) sin θ=(r/(p+r)) ...(iv) let (p/R)=μ , (r/R)=λ ⇒ 2cos θ=μ+2λ ...(I) 2sin θ=μ ...(II) 2cos θ=((4−2μ)/(μ+λ)) ...(III) 2sin θ=((2λ)/(μ+λ)) ...(IV) ⇒ μ+2λ=((4−2μ)/(μ+λ)) & μ(μ+λ)=2λ ⇒ λ=(μ^2 /(2−μ)) , hence μ^2 +((3μ^3 )/(2−μ))+((2μ^4 )/((2−μ)^2 ))+2μ−4=0 2μ^4 +3μ^3 (2−μ)+(2−μ)^2 (μ^2 +2μ−4)=0 ⇒ 2μ^4 +6μ^3 −3μ^4 +μ^4 +2μ^3 −4μ^2 −4μ^3 −8μ^2 +16μ+4μ^2 +8μ−16=0 ⇒ 4μ^3 −8μ^2 +24μ−16=0 or μ^3 −2μ^2 +6μ−4=0 ⇒ μ≈ 0.79321651 λ=(r/R) = (μ^2 /(2−μ)) ≈ 0.52138 .
$${let}\:\angle{DAB}=\theta \\ $$$${AE}={BF}={p}\:=\:{BD} \\ $$$$\mathrm{2}{R}\mathrm{cos}\:\theta={p}+\mathrm{2}{r}\:\:\:…\left({i}\right) \\ $$$$\mathrm{2}{R}\mathrm{sin}\:\theta={p}\:\:\:\:\:\:\:\:\:\:\:\:…\left({ii}\right) \\ $$$$\mathrm{cos}\:\theta=\frac{\mathrm{2}{R}−{p}}{{p}+{r}}\:\:\:\:\:\:\:…\left({iii}\right) \\ $$$$\mathrm{sin}\:\theta=\frac{{r}}{{p}+{r}}\:\:\:\:\:\:\:\:\:\:\:\:…\left({iv}\right) \\ $$$${let}\:\:\frac{{p}}{{R}}=\mu\:\:\:,\:\:\frac{{r}}{{R}}=\lambda \\ $$$$\Rightarrow\:\:\:\mathrm{2cos}\:\theta=\mu+\mathrm{2}\lambda\:\:\:\:\:…\left({I}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2sin}\:\theta=\mu\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\left({II}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2cos}\:\theta=\frac{\mathrm{4}−\mathrm{2}\mu}{\mu+\lambda}\:\:\:\:\:…\left({III}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{2sin}\:\theta=\frac{\mathrm{2}\lambda}{\mu+\lambda}\:\:\:\:\:\:\:\:\:\:…\left({IV}\right) \\ $$$$\Rightarrow\:\:\:\mu+\mathrm{2}\lambda=\frac{\mathrm{4}−\mathrm{2}\mu}{\mu+\lambda}\:\:\:\:\& \\ $$$$\:\:\:\:\:\:\:\:\mu\left(\mu+\lambda\right)=\mathrm{2}\lambda \\ $$$$\Rightarrow\:\:\lambda=\frac{\mu^{\mathrm{2}} }{\mathrm{2}−\mu}\:,\:\:\:{hence} \\ $$$$\:\:\:\:\:\:\mu^{\mathrm{2}} +\frac{\mathrm{3}\mu^{\mathrm{3}} }{\mathrm{2}−\mu}+\frac{\mathrm{2}\mu^{\mathrm{4}} }{\left(\mathrm{2}−\mu\right)^{\mathrm{2}} }+\mathrm{2}\mu−\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{2}\mu^{\mathrm{4}} +\mathrm{3}\mu^{\mathrm{3}} \left(\mathrm{2}−\mu\right)+\left(\mathrm{2}−\mu\right)^{\mathrm{2}} \left(\mu^{\mathrm{2}} +\mathrm{2}\mu−\mathrm{4}\right)=\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow\:\:\mathrm{2}\mu^{\mathrm{4}} +\mathrm{6}\mu^{\mathrm{3}} −\mathrm{3}\mu^{\mathrm{4}} +\mu^{\mathrm{4}} +\mathrm{2}\mu^{\mathrm{3}} −\mathrm{4}\mu^{\mathrm{2}} \\ $$$$\:−\mathrm{4}\mu^{\mathrm{3}} −\mathrm{8}\mu^{\mathrm{2}} +\mathrm{16}\mu+\mathrm{4}\mu^{\mathrm{2}} +\mathrm{8}\mu−\mathrm{16}=\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow\:\:\mathrm{4}\mu^{\mathrm{3}} −\mathrm{8}\mu^{\mathrm{2}} +\mathrm{24}\mu−\mathrm{16}=\mathrm{0} \\ $$$${or}\:\:\:\:\:\:\mu^{\mathrm{3}} −\mathrm{2}\mu^{\mathrm{2}} +\mathrm{6}\mu−\mathrm{4}=\mathrm{0} \\ $$$$\:\:\:\Rightarrow\:\:\mu\approx\:\mathrm{0}.\mathrm{79321651} \\ $$$$\:\:\:\:\lambda=\frac{{r}}{{R}}\:=\:\frac{\mu^{\mathrm{2}} }{\mathrm{2}−\mu}\:\approx\:\mathrm{0}.\mathrm{52138}\:. \\ $$$$ \\ $$
Commented by mr W last updated on 12/Apr/20
nicely solved sir! i got same result.
$${nicely}\:{solved}\:{sir}!\:{i}\:{got}\:{same}\:{result}. \\ $$

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