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Question Number 89618 by ajfour last updated on 18/Apr/20

Commented by ajfour last updated on 18/Apr/20

$F i n d r a d i i o f t h e c i r c l e s, i f$ $△ A B C i s e q u i l a t e r a l o f s i d e a .$ $(E i s t h e c e n t e r o f s m a l l e r c i r c l e)$
Commented by ajfour last updated on 19/Apr/20

$A n y s m o o t h w a y f o r t h i s Q .$ $S i r ?$
	Answered by ajfour last updated on 18/Apr/20

	$e q . A B : y_{B} = \frac{(2 r - R) x_{B}}{\sqrt{r^{2} - {(R - r)}^{2}}} + 2 r - R$ $x_{B}^{2} + {(2 r - R)}^{2} {(1 + \frac{x_{B}}{\sqrt{r (2 R - r)}})}^{2} = R^{2}$ ${(x_{B} + \sqrt{r (2 R - r)})}^{2} + y_{B}^{2} = a^{2} = 1 (s a y)$ $e q . o f B C :$ $y - y_{B} = \frac{m + \sqrt{3}}{1 - m \sqrt{3}} (x - x_{B})$ $m = \frac{2 r - R}{\sqrt{r (2 R - r)}} . . . . (i)$ $\frac{∣ y_{B} + R - r - (\frac{m + \sqrt{3}}{1 - m \sqrt{3}}) x_{B} ∣}{\sqrt[]{1 + {(\frac{m + \sqrt{3}}{1 - m \sqrt{3}})}^{2}}} = r . . . (i i)$ $\frac{∣ y_{B} - (\frac{m + \sqrt{3}}{1 - m \sqrt{3}}) (\sqrt{r (2 R - r)} + x_{B}) ∣}{\sqrt[]{1 + {(\frac{m + \sqrt{3}}{1 - m \sqrt{3}})}^{2}}} = \frac{\sqrt{3}}{2}$ $. . . . . . . (i i i)$ $. . . . .$
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