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

	Answered by ajfour last updated on 10/Nov/21
	$((r^4 −(r+c)^2 )/c)=(r^2 /( r^2 −1)) r^3 −r=c ⇒ r^6 =(r+c)^2 ⇒ (1−r^2 )(r+c)^2 −r^4 +(r+c)^2 =cr^2 & (r+c)^2 −2cr^4 +r^2 =c^2 ⇒ (r+c)^2 +(r^2 −c^2 )=2cr^4 ⇒ (((r+c)/(r−c)))^2 +(((r+c)/(r−c)))=2c((r^2 /(r−c)))^2 ⇒ (((r+c)/(r−c))+(1/2))^2 =(1/4)+2c((r^2 /(r−c)))^2 {((r+c)/(r−c))+(1/2)−(√(2c))((r^2 /(r−c)))} ×{((r+c)/(r−c))+(1/2)+(√(2c))((r^2 /(r−c)))}=(1/4) ............$
	$\frac{r^{4} - {(r + c)}^{2}}{c} = \frac{r^{2}}{r^{2} - 1}$ $r^{3} - r = c \Rightarrow r^{6} = {(r + c)}^{2}$ $\Rightarrow (1 - r^{2}) {(r + c)}^{2} - r^{4} + {(r + c)}^{2}$ $= {c r}^{2}$ $& {(r + c)}^{2} - 2 {c r}^{4} + r^{2} = c^{2}$ $\Rightarrow {(r + c)}^{2} + (r^{2} - c^{2}) = 2 {c r}^{4}$ $\Rightarrow {(\frac{r + c}{r - c})}^{2} + (\frac{r + c}{r - c}) = 2 c {(\frac{r^{2}}{r - c})}^{2}$ $\Rightarrow {(\frac{r + c}{r - c} + \frac{1}{2})}^{2} = \frac{1}{4} + 2 c {(\frac{r^{2}}{r - c})}^{2}$ ${\frac{r + c}{r - c} + \frac{1}{2} - \sqrt{2 c} (\frac{r^{2}}{r - c})}$ $\times {\frac{r + c}{r - c} + \frac{1}{2} + \sqrt{2 c} (\frac{r^{2}}{r - c})} = \frac{1}{4}$ $. . . . . . . . . . . .$
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