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Question Number 71184 by ajfour last updated on 12/Oct/19
Commented by ajfour last updated on 12/Oct/19
Each semicircle have radius unity. Find maximum value of x.
$${Each}\:{semicircle}\:{have}\:{radius} \\ $$$${unity}.\:{Find}\:{maximum}\:{value}\:{of}\:{x}. \\ $$
Commented by ajfour last updated on 12/Oct/19
Commented by mr W last updated on 13/Oct/19
(1−r)(√(1−(((b−1+1−r)/(1−r)))^2 ))= 1−(1−r)(√(1−(((1−1+r)/(1−r)))^2 )) let μ=1−b, λ=1−r λ(√(1−(((−μ+λ)/λ))^2 ))= 1−λ(√(1−(((1−λ)/λ))^2 )) (√(2μλ−μ^2 ))= 1−(√(2λ−1)) 2λ(1−μ)+μ^2 = 2(√(2λ−1)) 4(1−μ)^2 λ^2 −4(2−μ^2 +μ^3 )λ+μ^4 +4=0 λ=(((2−μ^2 +μ^3 )−(√((2−μ^2 +μ^3 )^2 −(1−μ)^2 (4+μ^4 ))))/(2(1−μ)^2 )) ⇒λ_(min) =0.5858 at μ=0.5858 ⇒r_(max) =1−0.5858=0.4142=(√2)−1
$$\left(\mathrm{1}−{r}\right)\sqrt{\mathrm{1}−\left(\frac{{b}−\mathrm{1}+\mathrm{1}−{r}}{\mathrm{1}−{r}}\right)^{\mathrm{2}} }=\:\mathrm{1}−\left(\mathrm{1}−{r}\right)\sqrt{\mathrm{1}−\left(\frac{\mathrm{1}−\mathrm{1}+{r}}{\mathrm{1}−{r}}\right)^{\mathrm{2}} } \\ $$$${let}\:\mu=\mathrm{1}−{b},\:\lambda=\mathrm{1}−{r} \\ $$$$\lambda\sqrt{\mathrm{1}−\left(\frac{−\mu+\lambda}{\lambda}\right)^{\mathrm{2}} }=\:\mathrm{1}−\lambda\sqrt{\mathrm{1}−\left(\frac{\mathrm{1}−\lambda}{\lambda}\right)^{\mathrm{2}} } \\ $$$$\sqrt{\mathrm{2}\mu\lambda−\mu^{\mathrm{2}} }=\:\mathrm{1}−\sqrt{\mathrm{2}\lambda−\mathrm{1}} \\ $$$$\mathrm{2}\lambda\left(\mathrm{1}−\mu\right)+\mu^{\mathrm{2}} =\:\mathrm{2}\sqrt{\mathrm{2}\lambda−\mathrm{1}} \\ $$$$\mathrm{4}\left(\mathrm{1}−\mu\right)^{\mathrm{2}} \lambda^{\mathrm{2}} −\mathrm{4}\left(\mathrm{2}−\mu^{\mathrm{2}} +\mu^{\mathrm{3}} \right)\lambda+\mu^{\mathrm{4}} +\mathrm{4}=\mathrm{0} \\ $$$$\lambda=\frac{\left(\mathrm{2}−\mu^{\mathrm{2}} +\mu^{\mathrm{3}} \right)−\sqrt{\left(\mathrm{2}−\mu^{\mathrm{2}} +\mu^{\mathrm{3}} \right)^{\mathrm{2}} −\left(\mathrm{1}−\mu\right)^{\mathrm{2}} \left(\mathrm{4}+\mu^{\mathrm{4}} \right)}}{\mathrm{2}\left(\mathrm{1}−\mu\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\lambda_{{min}} =\mathrm{0}.\mathrm{5858}\:{at}\:\mu=\mathrm{0}.\mathrm{5858} \\ $$$$\Rightarrow{r}_{{max}} =\mathrm{1}−\mathrm{0}.\mathrm{5858}=\mathrm{0}.\mathrm{4142}=\sqrt{\mathrm{2}}−\mathrm{1} \\ $$
Commented by ajfour last updated on 13/Oct/19
x_C =(1−r)sin β=1−(1−r)cos α y_C =b−(1−r)cos β=(1−r)sin α=r (1−r)[(√(1−(((b−r)/(1−r)))^2 ))+(√(1−((r/(1−r)))^2 ))=1 ........(i) (dr/db)=0 ⇒ (1/(2(√(1−(((b−r)/(1−r)))^2 )))){((2(b−r)(1−r)^2 −2(1−r)(b−r)^2 )/((1−r)^4 ))} =(1/(1−r)) ⇒ (((1−b)(b−r))/((1−r)^2 ))=(√(1−(((b−r)/(1−r)))^2 )) ..(ii) let b−r=λ(1−r) ⇒ 1−b=1−r−λ(1−r) from (ii) ⇒ λ(1−λ)=(√(1−λ^2 )) ⇒ λ^2 (1−λ)= 1+λ or λ^3 −λ^2 +λ+1=0 .....
$${x}_{{C}} =\left(\mathrm{1}−{r}\right)\mathrm{sin}\:\beta=\mathrm{1}−\left(\mathrm{1}−{r}\right)\mathrm{cos}\:\alpha \\ $$$${y}_{{C}} ={b}−\left(\mathrm{1}−{r}\right)\mathrm{cos}\:\beta=\left(\mathrm{1}−{r}\right)\mathrm{sin}\:\alpha={r} \\ $$$$\:\left(\mathrm{1}−{r}\right)\left[\sqrt{\mathrm{1}−\left(\frac{{b}−{r}}{\mathrm{1}−{r}}\right)^{\mathrm{2}} }+\sqrt{\mathrm{1}−\left(\frac{{r}}{\mathrm{1}−{r}}\right)^{\mathrm{2}} }=\mathrm{1}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……..\left({i}\right) \\ $$$$\frac{{dr}}{{db}}=\mathrm{0}\:\:\Rightarrow \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−\left(\frac{{b}−{r}}{\mathrm{1}−{r}}\right)^{\mathrm{2}} }}\left\{\frac{\mathrm{2}\left({b}−{r}\right)\left(\mathrm{1}−{r}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}−{r}\right)\left({b}−{r}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{r}\right)^{\mathrm{4}} }\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{1}−{r}} \\ $$$$\Rightarrow\:\frac{\left(\mathrm{1}−{b}\right)\left({b}−{r}\right)}{\left(\mathrm{1}−{r}\right)^{\mathrm{2}} }=\sqrt{\mathrm{1}−\left(\frac{{b}−{r}}{\mathrm{1}−{r}}\right)^{\mathrm{2}} }\:\:..\left({ii}\right) \\ $$$${let}\:{b}−{r}=\lambda\left(\mathrm{1}−{r}\right) \\ $$$$\Rightarrow\:\:\mathrm{1}−{b}=\mathrm{1}−{r}−\lambda\left(\mathrm{1}−{r}\right) \\ $$$${from}\:\left({ii}\right) \\ $$$$\Rightarrow\:\:\lambda\left(\mathrm{1}−\lambda\right)=\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }\: \\ $$$$\:\Rightarrow\:\:\:\lambda^{\mathrm{2}} \left(\mathrm{1}−\lambda\right)=\:\mathrm{1}+\lambda\: \\ $$$${or}\:\:\:\:\:\lambda^{\mathrm{3}} −\lambda^{\mathrm{2}} +\lambda+\mathrm{1}=\mathrm{0} \\ $$$$….. \\ $$
Answered by mr W last updated on 13/Oct/19
center of semicircle 1: (0,0) center of semicircle 2: (1,k) center of small circle: (h,r) h^2 +r^2 =(1−r)^2 ...(i) (1−h)^2 +(k−r)^2 =(1−r)^2 ...(ii) (i)−(ii): 2h−1+k(2r−k)=0 ⇒h=((1+k^2 )/2)−kr put into (i): (((1+k^2 )/2)−kr)^2 =1−2r k^2 r^2 −(k+k^3 −2)r−1+(((1+k^2 )^2 )/4)=0 ⇒r=((k(1+k^2 )−2+2(√((1−k)(1+k^2 ))))/(2k^2 )) (dr/dk)=0 ⇒ complicated eqn. for k ! numerically: r_(max) =0.4142 (=(√2)−1) at k=0.4142
$${center}\:{of}\:{semicircle}\:\mathrm{1}:\:\left(\mathrm{0},\mathrm{0}\right) \\ $$$${center}\:{of}\:{semicircle}\:\mathrm{2}:\:\left(\mathrm{1},{k}\right) \\ $$$${center}\:{of}\:{small}\:{circle}:\:\left({h},{r}\right) \\ $$$${h}^{\mathrm{2}} +{r}^{\mathrm{2}} =\left(\mathrm{1}−{r}\right)^{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$$\left(\mathrm{1}−{h}\right)^{\mathrm{2}} +\left({k}−{r}\right)^{\mathrm{2}} =\left(\mathrm{1}−{r}\right)^{\mathrm{2}} \:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\mathrm{2}{h}−\mathrm{1}+{k}\left(\mathrm{2}{r}−{k}\right)=\mathrm{0} \\ $$$$\Rightarrow{h}=\frac{\mathrm{1}+{k}^{\mathrm{2}} }{\mathrm{2}}−{kr} \\ $$$${put}\:{into}\:\left({i}\right): \\ $$$$\left(\frac{\mathrm{1}+{k}^{\mathrm{2}} }{\mathrm{2}}−{kr}\right)^{\mathrm{2}} =\mathrm{1}−\mathrm{2}{r} \\ $$$${k}^{\mathrm{2}} {r}^{\mathrm{2}} −\left({k}+{k}^{\mathrm{3}} −\mathrm{2}\right){r}−\mathrm{1}+\frac{\left(\mathrm{1}+{k}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}}=\mathrm{0} \\ $$$$\Rightarrow{r}=\frac{{k}\left(\mathrm{1}+{k}^{\mathrm{2}} \right)−\mathrm{2}+\mathrm{2}\sqrt{\left(\mathrm{1}−{k}\right)\left(\mathrm{1}+{k}^{\mathrm{2}} \right)}}{\mathrm{2}{k}^{\mathrm{2}} } \\ $$$$\frac{{dr}}{{dk}}=\mathrm{0}\:\Rightarrow\:{complicated}\:{eqn}.\:{for}\:{k}\:! \\ $$$${numerically}: \\ $$$${r}_{{max}} =\mathrm{0}.\mathrm{4142}\:\left(=\sqrt{\mathrm{2}}−\mathrm{1}\right)\:{at}\:{k}=\mathrm{0}.\mathrm{4142} \\ $$
Commented by mr W last updated on 13/Oct/19
Commented by ajfour last updated on 13/Oct/19
Thanks so much Sir, i yet think its possible to get an entire solution, am trying..
$${Thanks}\:{so}\:{much}\:{Sir},\:{i}\:{yet}\:{think} \\ $$$${its}\:{possible}\:{to}\:{get}\:{an}\:{entire} \\ $$$${solution},\:{am}\:{trying}.. \\ $$

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