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Question-206484

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Question Number 206484 by sonukgindia last updated on 16/Apr/24
Answered by mr W last updated on 16/Apr/24
Commented by mr W last updated on 16/Apr/24
R=((AD)/2)=((m+n+x)/2) CF=(√(n^2 +R^2 −nR)) ((AE)/(FD))=((AC)/(CF)) ⇒a=(((m+x)R)/( (√(n^2 +R^2 −nR)))) similarly ⇒b=(((n+x)R)/( (√(m^2 +R^2 −mR)))) a^2 +b^2 =(2R)^2 (((m+x)^2 R^2 )/( n^2 +R^2 −nR))+(((n+x)^2 R^2 )/( m^2 +R^2 −mR))=4R^2 (((m+x)^2 )/( n^2 +R^2 −nR))+(((n+x)^2 )/( m^2 +R^2 −mR))=4 (((m+x)^2 )/( 4n^2 +(m+n+x)^2 −2n(m+n+x)))+(((n+x)^2 )/( 4m^2 +(m+n+x)^2 −2m(m+n+x)))=1 (((m+x)^2 )/((m+x)^2 +3n^2 ))+(((n+x)^2 )/((n+x)^2 +3m^2 ))=1 (1/(1+3((n/(m+x)))^2 ))+(1/(1+3((m/(n+x)))^2 ))=1 1=9((n/(m+x)))^2 ((m/(n+x)))^2 1=((3mn)/((m+x)(n+x))) (m+x)(n+x)=3mn ⇒x^2 +(m+n)x−2mn=0 ✓
$${R}=\frac{{AD}}{\mathrm{2}}=\frac{{m}+{n}+{x}}{\mathrm{2}} \\ $$$${CF}=\sqrt{{n}^{\mathrm{2}} +{R}^{\mathrm{2}} −{nR}} \\ $$$$\frac{{AE}}{{FD}}=\frac{{AC}}{{CF}} \\ $$$$\Rightarrow{a}=\frac{\left({m}+{x}\right){R}}{\:\sqrt{{n}^{\mathrm{2}} +{R}^{\mathrm{2}} −{nR}}} \\ $$$${similarly} \\ $$$$\Rightarrow{b}=\frac{\left({n}+{x}\right){R}}{\:\sqrt{{m}^{\mathrm{2}} +{R}^{\mathrm{2}} −{mR}}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left(\mathrm{2}{R}\right)^{\mathrm{2}} \\ $$$$\frac{\left({m}+{x}\right)^{\mathrm{2}} {R}^{\mathrm{2}} }{\:{n}^{\mathrm{2}} +{R}^{\mathrm{2}} −{nR}}+\frac{\left({n}+{x}\right)^{\mathrm{2}} {R}^{\mathrm{2}} }{\:{m}^{\mathrm{2}} +{R}^{\mathrm{2}} −{mR}}=\mathrm{4}{R}^{\mathrm{2}} \\ $$$$\frac{\left({m}+{x}\right)^{\mathrm{2}} }{\:{n}^{\mathrm{2}} +{R}^{\mathrm{2}} −{nR}}+\frac{\left({n}+{x}\right)^{\mathrm{2}} }{\:{m}^{\mathrm{2}} +{R}^{\mathrm{2}} −{mR}}=\mathrm{4} \\ $$$$\frac{\left({m}+{x}\right)^{\mathrm{2}} }{\:\mathrm{4}{n}^{\mathrm{2}} +\left({m}+{n}+{x}\right)^{\mathrm{2}} −\mathrm{2}{n}\left({m}+{n}+{x}\right)}+\frac{\left({n}+{x}\right)^{\mathrm{2}} }{\:\mathrm{4}{m}^{\mathrm{2}} +\left({m}+{n}+{x}\right)^{\mathrm{2}} −\mathrm{2}{m}\left({m}+{n}+{x}\right)}=\mathrm{1} \\ $$$$\frac{\left({m}+{x}\right)^{\mathrm{2}} }{\left({m}+{x}\right)^{\mathrm{2}} +\mathrm{3}{n}^{\mathrm{2}} }+\frac{\left({n}+{x}\right)^{\mathrm{2}} }{\left({n}+{x}\right)^{\mathrm{2}} +\mathrm{3}{m}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{3}\left(\frac{{n}}{{m}+{x}}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{3}\left(\frac{{m}}{{n}+{x}}\right)^{\mathrm{2}} }=\mathrm{1} \\ $$$$\mathrm{1}=\mathrm{9}\left(\frac{{n}}{{m}+{x}}\right)^{\mathrm{2}} \left(\frac{{m}}{{n}+{x}}\right)^{\mathrm{2}} \\ $$$$\mathrm{1}=\frac{\mathrm{3}{mn}}{\left({m}+{x}\right)\left({n}+{x}\right)} \\ $$$$\left({m}+{x}\right)\left({n}+{x}\right)=\mathrm{3}{mn} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\left({m}+{n}\right){x}−\mathrm{2}{mn}=\mathrm{0}\:\checkmark \\ $$

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