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Question Number 79449 by mr W last updated on 25/Jan/20

Commented by mr W last updated on 25/Jan/20

$[R e p o s t o f a n o l d q u e s t i o n f r o m b . e . h . i . s i r]$ $F i n d t h e r a d i u s o f 20 t h c i r c l e .$
Commented by MJS last updated on 25/Jan/20
$corrected location of circle centers { ((x=t+(1/t^3 )−(1/t^3 )(√(t^4 +1)))),((y=(2/t)−(1/t)(√(t^4 +1)))) :}$
$corrected location of circle centers$ ${\begin{cases} x = t + \frac{1}{t^{3}} - \frac{1}{t^{3}} \sqrt{t^{4} + 1} \\ y = \frac{2}{t} - \frac{1}{t} \sqrt{t^{4} + 1} \end{cases}$
Commented by john santu last updated on 25/Jan/20

$w a w . . . .$
Commented by MJS last updated on 25/Jan/20

$sorry I made a mistake, I^{'} ll post a corrected$ $version soon$
Commented by MJS last updated on 25/Jan/20

$Can we exactly find the 2^{nd} circle at least ?$ $I {don}^{'} t think we can . . .$
Commented by MJS last updated on 25/Jan/20

$one possible path, but no exact solution$ $center of C_{n} = M_{n} = (\begin{matrix} t_{n} + \frac{1}{t_{n}^{3}} - \frac{1}{t_{n}^{3}} \sqrt{t_{n}^{4} + 1} \\ \frac{2}{t_{n}} - \frac{1}{t_{n}} \sqrt{t_{n}^{4} + 1} \end{matrix})$ $radius of C_{n} = r_{n} = t_{n} + \frac{1}{t_{n}^{3}} - \frac{1}{t_{n}^{3}} \sqrt{t_{n}^{4} + 1}$ $[t_{1} = 1 \Rightarrow M_{1} = (\begin{matrix} 2 - \sqrt{2} \\ 2 - \sqrt{2} \end{matrix}) \land r_{1} = 2 - \sqrt{2}]$ $∣ C_{n} C_{n + 1} ∣^{2} = {(r_{n} + r_{n + 1})}^{2}$ $starting with n = 1$ $let t_{2} = p$ $(\frac{4 - 2 \sqrt{2}}{p} - \frac{6}{p^{2}} + \frac{4 - 2 \sqrt{2}}{p^{3}} - \frac{2}{p^{6}}) \sqrt{p^{4} + 1} + 2 p^{2} - (4 - 2 \sqrt{2}) p - \frac{8 - 4 \sqrt{2}}{p} + \frac{8}{p^{2}} - \frac{4 - 2 \sqrt{2}}{p^{3}} + \frac{2}{p^{6}} + 12 - 8 \sqrt{2} = + (\frac{2}{p^{2}} + \frac{4 - 2 \sqrt{2}}{p^{3}} + \frac{2}{p^{6}}) \sqrt{p^{4} + 1} + p^{2} + (4 - 2 \sqrt{2}) p + \frac{3}{p^{2}} + \frac{4 - 2 \sqrt{2}}{p^{3}} + \frac{2}{p^{6}} + 6 - 4] 2$ $. . .$ $p {(p - 1)}^{2} (p^{7} - 2 (7 - 4 \sqrt{2}) p^{6} + 5 (11 - 8 \sqrt{2}) p^{5} - 4 (5 - 4 \sqrt{2}) p^{4} + (63 - 40 \sqrt{2}) p^{3} - 2 (23 - 16 \sqrt{2}) p^{3} + (73 - 48 \sqrt{2}) p - 8 (2 - \sqrt{2})) = 0$ $we know p \neq 0 \land p \neq 1$ ${we}^{'} re looking for 0 < p < 1 (there are several$ $possible circles, touching C_{1} from ‘ ‘ wrong$ ${sides}^{″})$ $\Rightarrow p \approx . 631658 = t_{2}$ $\Rightarrow r_{2} \approx . 327488$ $of course we can go on like this but I guess$ $if I calculate with an error of \pm 10^{- 14} at least$ $t_{14} will have an error of \pm 1 . . .$
	Answered by mr W last updated on 25/Jan/20

	$P a r t I$ $f i n d i n g t h e r a d i u s o f c i r c l e 1 :$ $r_{1} + \frac{r_{1}}{\sqrt{2}} = 1$ $\Rightarrow r_{1} = \frac{\sqrt{2}}{1 + \sqrt{2}} = 2 - \sqrt{2}$
	Commented by mr W last updated on 25/Jan/20

	Commented by mr W last updated on 25/Jan/20

	$P a r t I I (j u s t a n a t t e m p t . . .)$ $n - t h c i r c l e w i t h r a d i u s r_{n} a n d c e n t e r$ $a t A (r_{n}, y_{A}) .$ ${(x - r_{n})}^{2} + {(y - y_{A})}^{2} = r_{n}^{2}$ $i t t o u c h e s t h e c u r v e y = \frac{1}{x} a t C (x_{C}, \frac{1}{x_{C}})$ ${(x_{C} - r_{n})}^{2} + {(\frac{1}{x_{C}} - y_{A})}^{2} = r_{n}^{2}$ $\tan θ = \frac{\frac{1}{x_{C}} - y_{A}}{x_{C} - r_{n}} = - \frac{1}{y^{'}} = x_{C}^{2}$ $\Rightarrow y_{A} = \frac{1}{x_{C}} - x_{C}^{2} (x_{C} - r_{n})$ $\Rightarrow {(x_{C} - r_{n})}^{2} (1 + x_{C}^{4}) = r_{n}^{2}$ $\Rightarrow x_{C}^{5} - 2 r_{n} x_{C}^{4} + r_{n}^{2} x_{C}^{3} + x_{C} - 2 r_{n} = 0$ $. . . . . . {i t}^{'} s u n s o l v a b l e f o r m e u p o n h e r e$
	Commented by behi83417@gmail.com last updated on 25/Jan/20

	$wow! thanks for attemping dear masters :$ $sir : MJS and sir : mrW .$
	Commented by behi83417@gmail.com last updated on 25/Jan/20
	$x_c ^3 .r_n ^2 −2(1+x_c ^4 )r_n +x_c ^5 +x_c =0 r_n =((1+x_c ^4 ±(√((1+x_c ^4 )^2 −x_c ^3 (x_c ^5 +x_c ))))/x_c ^3 ) ⇒r_n = { (((1+x_c ^4 +(√(1+x_c ^4 )))/x_c ^3 )),(((1+x_c ^4 −(√(1+x_c ^4 )))/x_c ^3 )) :} ⇒^(x_c =x) (∂r_n /∂x_c )=(((4x^3 +((2x^3 )/(√(1+x^4 ))))x^3 −3x^2 (1+x^4 +(√(1+x^4 ))))/x^6 )=0 ⇒x^2 [4x^4 (√(1+x^4 ))+2x^3 )−3(1+x^4 +(1+x^4 )(√(1+x^4 )))]=0 ⇒ { (,(x=0 [not ok])),(,(4x^4 (√(1+x^4 ))+2x^3 −3−3x^4 −3(√(1+x^4 ))−3x^4 (√(1+x^4 ))=0)) :} ⇒(√(1+x^4 ))(x^4 −3)=3x^4 −2x^3 +3 ⇒(1+x^4 )(x^8 −6x^4 +9)=9x^8 +4x^6 +9−12x^7 +18x^4 −12x^3 ⇒x^(12) −5x^8 +3x^4 +9=9x^8 −12x^7 +4x^6 +18x^4 −12x^3 ⇒x^(12) −14x^8 +12x^7 −4x^6 −15x^4 +12x^3 =0 ⇒ { ((x^3 =0[not ok])),((x^9 −14x^5 +12x^4 −4x^3 −15x+12=0)) :} ⇒x=−2.51 ,0.733 ,1.747 ⇒r_n =((1+1.747^4 +(√(1+1.747^4 )))/(1.747^3 ))=2.537(×) ⇒r_n =((1+.733^4 +(√(1+.733^4 )))/(0.733^3 ))=6.155 (×) ⇒r_n =((1+(−2.51)^4 +(√(1+(−2.51)^4 )))/((−2.51)^3 ))=−2.97(×)$
	$x_{c}^{3} . r_{n}^{2} - 2 (1 + x_{c}^{4}) r_{n} + x_{c}^{5} + x_{c} = 0$ $r_{n} = \frac{1 + x_{c}^{4} \pm \sqrt{{(1 + x_{c}^{4})}^{2} - x_{c}^{3} (x_{c}^{5} + x_{c})}}{x_{c}^{3}}$ $\Rightarrow r_{n} = {\begin{cases} \frac{1 + x_{c}^{4} + \sqrt{1 + x_{c}^{4}}}{x_{c}^{3}} \\ \frac{1 + x_{c}^{4} - \sqrt{1 + x_{c}^{4}}}{x_{c}^{3}} \end{cases}$ $\overset{x_{c} = x}{\Rightarrow} \frac{\partial r_{n}}{\partial x_{c}} = \frac{({4 x}^{3} + \frac{{2 x}^{3}}{\sqrt{1 + x^{4}}}) x^{3} - {3 x}^{2} (1 + x^{4} + \sqrt{1 + x^{4}})}{x^{6}} = 0$ $\Rightarrow x^{2} [{4 x}^{4} \sqrt{1 + x^{4}} + {2 x}^{3}) - 3 (1 + x^{4} + (1 + x^{4}) \sqrt{1 + x^{4}})] = 0$ $\Rightarrow {\begin{cases} x = 0 [not ok] \\ {4 x}^{4} \sqrt{1 + x^{4}} + {2 x}^{3} - 3 - {3 x}^{4} - 3 \sqrt{1 + x^{4}} - {3 x}^{4} \sqrt{1 + x^{4}} = 0 \end{cases}$ $\Rightarrow \sqrt{1 + x^{4}} (x^{4} - 3) = {3 x}^{4} - {2 x}^{3} + 3$ $\Rightarrow (1 + x^{4}) (x^{8} - {6 x}^{4} + 9) = {9 x}^{8} + {4 x}^{6} + 9 - {12 x}^{7} + {18 x}^{4} - {12 x}^{3}$ $\Rightarrow x^{12} - {5 x}^{8} + {3 x}^{4} + 9 = {9 x}^{8} - {12 x}^{7} + {4 x}^{6} + {18 x}^{4} - {12 x}^{3}$ $\Rightarrow x^{12} - {14 x}^{8} + {12 x}^{7} - {4 x}^{6} - {15 x}^{4} + {12 x}^{3} = 0$ $\Rightarrow {\begin{cases} x^{3} = 0 [not ok] \\ x^{9} - {14 x}^{5} + {12 x}^{4} - {4 x}^{3} - 15 x + 12 = 0 \end{cases}$ $\Rightarrow x = - 2.51, 0.733, 1.747$ $\Rightarrow r_{n} = \frac{1 + 1 . 747^{4} + \sqrt{1 + 1 . 747^{4}}}{1 . 747^{3}} = 2.537 (\times)$ $\Rightarrow r_{n} = \frac{1 + . 733^{4} + \sqrt{1 + . 733^{4}}}{0 . 733^{3}} = 6.155 (\times)$ $\Rightarrow r_{n} = \frac{1 + {(- 2.51)}^{4} + \sqrt{1 + {(- 2.51)}^{4}}}{{(- 2.51)}^{3}} = - 2.97 (\times)$
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