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Question Number 158893 by ajfour last updated on 10/Nov/21

Answered by ajfour last updated on 10/Nov/21

((r^4 −(r+c)^2 )/c)=(r^2 /( r^2 −1))  r^3 −r=c    ⇒  r^6 =(r+c)^2   ⇒ (1−r^2 )(r+c)^2 −r^4 +(r+c)^2               =cr^2   &   (r+c)^2 −2cr^4 +r^2 =c^2   ⇒  (r+c)^2 +(r^2 −c^2 )=2cr^4   ⇒  (((r+c)/(r−c)))^2 +(((r+c)/(r−c)))=2c((r^2 /(r−c)))^2   ⇒ (((r+c)/(r−c))+(1/2))^2 =(1/4)+2c((r^2 /(r−c)))^2   {((r+c)/(r−c))+(1/2)−(√(2c))((r^2 /(r−c)))}      ×{((r+c)/(r−c))+(1/2)+(√(2c))((r^2 /(r−c)))}=(1/4)  ............

$$\frac{{r}^{\mathrm{4}} −\left({r}+{c}\right)^{\mathrm{2}} }{{c}}=\frac{{r}^{\mathrm{2}} }{\:{r}^{\mathrm{2}} −\mathrm{1}} \\ $$$${r}^{\mathrm{3}} −{r}={c}\:\:\:\:\Rightarrow\:\:{r}^{\mathrm{6}} =\left({r}+{c}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\left(\mathrm{1}−{r}^{\mathrm{2}} \right)\left({r}+{c}\right)^{\mathrm{2}} −{r}^{\mathrm{4}} +\left({r}+{c}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:={cr}^{\mathrm{2}} \\ $$$$\&\:\:\:\left({r}+{c}\right)^{\mathrm{2}} −\mathrm{2}{cr}^{\mathrm{4}} +{r}^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\left({r}+{c}\right)^{\mathrm{2}} +\left({r}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)=\mathrm{2}{cr}^{\mathrm{4}} \\ $$$$\Rightarrow\:\:\left(\frac{{r}+{c}}{{r}−{c}}\right)^{\mathrm{2}} +\left(\frac{{r}+{c}}{{r}−{c}}\right)=\mathrm{2}{c}\left(\frac{{r}^{\mathrm{2}} }{{r}−{c}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\left(\frac{{r}+{c}}{{r}−{c}}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{2}{c}\left(\frac{{r}^{\mathrm{2}} }{{r}−{c}}\right)^{\mathrm{2}} \\ $$$$\left\{\frac{{r}+{c}}{{r}−{c}}+\frac{\mathrm{1}}{\mathrm{2}}−\sqrt{\mathrm{2}{c}}\left(\frac{{r}^{\mathrm{2}} }{{r}−{c}}\right)\right\} \\ $$$$\:\:\:\:×\left\{\frac{{r}+{c}}{{r}−{c}}+\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{2}{c}}\left(\frac{{r}^{\mathrm{2}} }{{r}−{c}}\right)\right\}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$............ \\ $$

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