Question Number 45630 by ajfour last updated on 14/Oct/18
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Oct/18
$$\frac{\mathrm{1}}{\mathrm{2}}×{l}×{x}=\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}\: \\ $$$${s}=\frac{{R}+{x}+{x}+{r}+{l}}{\mathrm{2}}={x}+\frac{{R}+{r}+{l}}{\mathrm{2}} \\ $$$${a}={R}+{x}\:\:\:\:{b}={x}+{r}\:\:\:\:{c}={l} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{lx}=\sqrt{\left({x}+\frac{{R}+{r}+{l}}{\mathrm{2}}\right)\left(\frac{{r}+{l}+{R}}{\mathrm{2}}−{R}\right)\left(\frac{{R}+{l}+{r}}{\mathrm{2}}−{r}\right)\left({x}+\frac{{R}+{r}+{l}}{\mathrm{2}}−{l}\right)}\: \\ $$$$\frac{{l}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{4}}=\left({x}+{M}\right)\left({N}\right)\left({P}\right)\left({x}+{Q}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\uparrow{here}\:{i}\:{forgot}\:{to}\:{putx} \\ $$$${l}^{\mathrm{2}} {x}^{\mathrm{2}} =\mathrm{4}\left({x}+{M}\right){NP}\left({x}+{Q}\right)\:\:{M}=\frac{{R}+{r}+{l}}{\mathrm{2}}\:\:{N}=\left(\frac{{R}+{r}+{l}}{\mathrm{2}}−{R}\right) \\ $$$${P}=\left(\frac{{R}+{r}+{l}}{\mathrm{2}}−{r}\right)\:\:\:{Q}=\frac{{R}+{r}+{l}}{\mathrm{2}}−{l} \\ $$$$\frac{{x}^{\mathrm{2}} }{\left({x}+{M}\right)\left({x}+{Q}\right)}=\frac{\mathrm{4}{NP}}{{l}^{\mathrm{2}} } \\ $$$${nw}\:{put}\:\:{M},\:\:{N},\:\:{P},\:\:\:{Q}\:{value} \\ $$$$\frac{{x}^{\mathrm{2}} }{\left({x}+{M}\right)\left({x}+{Q}\right)}={T} \\ $$$${x}^{\mathrm{2}} ={T}\left({x}^{\mathrm{2}} +{Qx}+{Mx}+{MQ}\right) \\ $$$${x}^{\mathrm{2}} \left({T}−\mathrm{1}\right)+{x}\left({Q}+{M}\right)+{MQT}=\mathrm{0} \\ $$$${x}=\frac{−\left({Q}+{M}\right)+\sqrt{\left({Q}+{M}\right)^{\mathrm{2}} −\mathrm{4}\left({T}−\mathrm{1}\right)\left({MQT}\right)}}{\mathrm{2}\left({T}−\mathrm{1}\right)}\: \\ $$$$ \\ $$
Commented by ajfour last updated on 15/Oct/18
$${thanks},\:{Tanmay}\:{Sir}. \\ $$
Answered by MrW3 last updated on 15/Oct/18
![(√((x+r)^2 −x^2 ))+(√((R+x)^2 −x^2 ))=l (√(r(2x+r)))+(√(R(2x+R)))=l r(2x+r)+R(2x+R)+2(√(rR(2x+r)(2x+R)))=l^2 2(√(rR(2x+r)(2x+R)))=(l^2 −R^2 −r^2 )−2(R+r)x let 2e^2 =l^2 −R^2 −r^2 or e=(√((l^2 −R^2 −r^2 )/2)) ⇒(√(rR(2x+r)(2x+R)))=e^2 −(R+r)x rR(2x+r)(2x+R)=e^4 +(R+r)^2 x^2 −2e^2 (R+r)x rR{4x^2 +2(R+r)x+Rr}=e^4 +(R+r)^2 x^2 −2e^2 (R+r)x 4Rrx^2 +2Rr(R+r)x+R^2 r^2 =e^4 +(R+r)^2 x^2 −2e^2 (R+r)x {(R+r)^2 −4Rr}x^2 −2(R+r)(e^2 +Rr)x−(R^2 r^2 −e^4 )=0 (R−r)^2 x^2 −2(R+r)(e^2 +Rr)x−(R^2 r^2 −e^4 )=0 for R=r: ⇒x=((e^4 −R^4 )/(4R(e^2 +R^2 )))=((e^2 −R^2 )/(4R))=((l^2 −4R^2 )/(8R)) for R≠r: ⇒x=((2(R+r)(e^2 +Rr)−2(√((R+r)^2 (e^2 +Rr)^2 +(R−r)^2 (R^2 r^2 −e^4 ))))/((R−r)^2 )) ⇒x=(((R+r)(2e^2 +2Rr)−2(√(Rr(2e^2 +2Rr)(R^2 +r^2 +2e^2 ))))/(2(R−r)^2 )) ⇒x=(((R+r)[l^2 −(R−r)^2 ]−2l(√(Rr[l^2 −(R−r)^2 ])))/(2(R−r)^2 ))](Q45643.png)
$$\sqrt{\left({x}+{r}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }+\sqrt{\left({R}+{x}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }={l} \\ $$$$\sqrt{{r}\left(\mathrm{2}{x}+{r}\right)}+\sqrt{{R}\left(\mathrm{2}{x}+{R}\right)}={l} \\ $$$${r}\left(\mathrm{2}{x}+{r}\right)+{R}\left(\mathrm{2}{x}+{R}\right)+\mathrm{2}\sqrt{{rR}\left(\mathrm{2}{x}+{r}\right)\left(\mathrm{2}{x}+{R}\right)}={l}^{\mathrm{2}} \\ $$$$\mathrm{2}\sqrt{{rR}\left(\mathrm{2}{x}+{r}\right)\left(\mathrm{2}{x}+{R}\right)}=\left({l}^{\mathrm{2}} −{R}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)−\mathrm{2}\left({R}+{r}\right){x} \\ $$$${let}\:\mathrm{2}{e}^{\mathrm{2}} ={l}^{\mathrm{2}} −{R}^{\mathrm{2}} −{r}^{\mathrm{2}} \:{or}\:{e}=\sqrt{\frac{{l}^{\mathrm{2}} −{R}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$\Rightarrow\sqrt{{rR}\left(\mathrm{2}{x}+{r}\right)\left(\mathrm{2}{x}+{R}\right)}={e}^{\mathrm{2}} −\left({R}+{r}\right){x} \\ $$$${rR}\left(\mathrm{2}{x}+{r}\right)\left(\mathrm{2}{x}+{R}\right)={e}^{\mathrm{4}} +\left({R}+{r}\right)^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{2}{e}^{\mathrm{2}} \left({R}+{r}\right){x} \\ $$$${rR}\left\{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}\left({R}+{r}\right){x}+{Rr}\right\}={e}^{\mathrm{4}} +\left({R}+{r}\right)^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{2}{e}^{\mathrm{2}} \left({R}+{r}\right){x} \\ $$$$\mathrm{4}{Rrx}^{\mathrm{2}} +\mathrm{2}{Rr}\left({R}+{r}\right){x}+{R}^{\mathrm{2}} {r}^{\mathrm{2}} ={e}^{\mathrm{4}} +\left({R}+{r}\right)^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{2}{e}^{\mathrm{2}} \left({R}+{r}\right){x} \\ $$$$\left\{\left({R}+{r}\right)^{\mathrm{2}} −\mathrm{4}{Rr}\right\}{x}^{\mathrm{2}} −\mathrm{2}\left({R}+{r}\right)\left({e}^{\mathrm{2}} +{Rr}\right){x}−\left({R}^{\mathrm{2}} {r}^{\mathrm{2}} −{e}^{\mathrm{4}} \right)=\mathrm{0} \\ $$$$\left({R}−{r}\right)^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{2}\left({R}+{r}\right)\left({e}^{\mathrm{2}} +{Rr}\right){x}−\left({R}^{\mathrm{2}} {r}^{\mathrm{2}} −{e}^{\mathrm{4}} \right)=\mathrm{0} \\ $$$${for}\:{R}={r}: \\ $$$$\Rightarrow{x}=\frac{{e}^{\mathrm{4}} −{R}^{\mathrm{4}} }{\mathrm{4}{R}\left({e}^{\mathrm{2}} +{R}^{\mathrm{2}} \right)}=\frac{{e}^{\mathrm{2}} −{R}^{\mathrm{2}} }{\mathrm{4}{R}}=\frac{{l}^{\mathrm{2}} −\mathrm{4}{R}^{\mathrm{2}} }{\mathrm{8}{R}} \\ $$$${for}\:{R}\neq{r}: \\ $$$$\Rightarrow{x}=\frac{\mathrm{2}\left({R}+{r}\right)\left({e}^{\mathrm{2}} +{Rr}\right)−\mathrm{2}\sqrt{\left({R}+{r}\right)^{\mathrm{2}} \left({e}^{\mathrm{2}} +{Rr}\right)^{\mathrm{2}} +\left({R}−{r}\right)^{\mathrm{2}} \left({R}^{\mathrm{2}} {r}^{\mathrm{2}} −{e}^{\mathrm{4}} \right)}}{\left({R}−{r}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{x}=\frac{\left({R}+{r}\right)\left(\mathrm{2}{e}^{\mathrm{2}} +\mathrm{2}{Rr}\right)−\mathrm{2}\sqrt{{Rr}\left(\mathrm{2}{e}^{\mathrm{2}} +\mathrm{2}{Rr}\right)\left({R}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{e}^{\mathrm{2}} \right)}}{\mathrm{2}\left({R}−{r}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{x}=\frac{\left({R}+{r}\right)\left[{l}^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} \right]−\mathrm{2}{l}\sqrt{{Rr}\left[{l}^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} \right]}}{\mathrm{2}\left({R}−{r}\right)^{\mathrm{2}} } \\ $$
Commented by ajfour last updated on 15/Oct/18
$${thank}\:{you}\:{Sir}! \\ $$