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Question Number 124353 by ajfour last updated on 02/Dec/20

Commented by ajfour last updated on 02/Dec/20

$F i n d m a x i m u m r a d i u s o f t h e$ $b l u e c i r c l e s .$
	Answered by ajfour last updated on 02/Dec/20

	$C e n t e r o f l o w e r b i g c i c l e t o b o t t o m$ $p o i n t o f u p p e r b i g c i r c l e b e h .$ $A n g l e o f s e c t o r 2 θ .$ $\cos θ = h + \frac{1 - h}{2} = \frac{1 + h}{2}$ $(h - r) \sin θ = r$ $r = \frac{h \sin θ}{1 + \sin θ}$ $r = \frac{(2 \cos θ - 1) \sin θ}{1 + \sin θ}$ $\frac{d r}{d θ} = 0 \Rightarrow$ $(1 + \sin θ) (2 \cos^{2} θ - 2 \sin^{2} θ - \cos θ)$ $= \cos θ \sin θ (2 \cos θ - 1)$ $\Rightarrow$ $2 \cos^{2} θ - 2 \sin^{2} θ - \cos θ - 2 \sin^{3} θ = 0$ ${(4 \cos^{2} θ - \cos θ - 2)}^{2} = 4 {(1 - \cos^{2} θ)}^{3}$ $θ = 45 ° o r \approx 28.40406$ $r = 0.17157; r = 0.2447$
	Commented by MJS_new last updated on 02/Dec/20

	$I tried with the coordinate method with the$ $centers of the big circles \pm p from the origin$ $with \frac{1}{2} < p < 1$ $the radius of the blue circle is$ $r = \frac{2 p - 1}{p^{2}} \sqrt{(p^{2} - 1) (p^{2} - 2 + 2 \sqrt{1 - p^{2}}}$ $its maximum is at p being the real solution$ $of p^{3} + \frac{3}{2} p - 2 = 0$ $\Rightarrow p = \sqrt[3]{1 + \frac{3 \sqrt{2}}{4}} + \sqrt[3]{- 1 + \frac{3 \sqrt{2}}{4}} \approx . 879614880$ $\Rightarrow r \approx . 244737182$
	Commented by mnjuly1970 last updated on 03/Dec/20

	$m r a j f o r$ $p l e a s e e x p l a i n r e l a t i o n s o f$ $f i r s t a n d s e c o n d l i n e s . .$ $t h a n k y o u . . .$ $c o s (θ) \overset{? ? ?}{=} \frac{1 + h}{2}$ $s i n (θ) \overset{? ? ?}{=} \frac{r}{h - r}$
	Commented by ajfour last updated on 05/Dec/20

	$g r e a t w a y, t h a n k s M j S S i r .$
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