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Question Number 214100 by ajfour last updated on 28/Nov/24

	Answered by ajfour last updated on 29/Nov/24

	Commented by ajfour last updated on 29/Nov/24
	$sin α=(a/(2b+a))=(1/(2s+1)) ∀ s=(b/a), t=(r/a) A=2(√(ar)) B=(√((b+r)^2 −(p−r)^2 )) A+B=(a+b)cos α B^2 ={A−(a+b)cos α}^2 b^2 −p^2 +2r(b+p) =4ar+(a+b)^2 cos^2 α−4(√(ar))(a+b)cos α as p=bsin α 2rb(1+sin α)+4(a+b)(√(ar))cos α =4ar+a^2 cos^2 α+2abcos^2 α recalling s=(b/a), t=(r/a) 2t(s+ssin α−2)+4(1+s)(√t)cos α −(1+2s)cos^2 α =0 (√t)=−(((1+s)cos α)/(s+ssin α−2))±(√(((((1+s)cos α)/(s+ssin α−2)))^2 +(((1+2s)cos^2 α)/(2(s+ssin α−2))))) say a=1, b=2 ⇒ s=2, sin α=(1/5), cos α=((2(√6))/5) (√t)=−3(√6)+2(√(15)) r=t=114−36(√(10)) ⇒ r≈ 0.1580$
	$\sin α = \frac{a}{2 b + a} = \frac{1}{2 s + 1} \forall s = \frac{b}{a}, t = \frac{r}{a}$ $A = 2 \sqrt{a r}$ $B = \sqrt{{(b + r)}^{2} - {(p - r)}^{2}}$ $A + B = (a + b) \cos α$ $B^{2} = {A - (a + b) \cos α}^{2}$ $b^{2} - p^{2} + 2 r (b + p)$ $= 4 a r + {(a + b)}^{2} \cos^{2} α - 4 \sqrt{a r} (a + b) \cos α$ $a s p = b \sin α$ $2 r b (1 + \sin α) + 4 (a + b) \sqrt{a r} \cos α$ $= 4 a r + a^{2} \cos^{2} α + 2 a b \cos^{2} α$ $r e c a l l i n g s = \frac{b}{a}, t = \frac{r}{a}$ $2 t (s + s \sin α - 2) + 4 (1 + s) \sqrt{t} \cos α$ $- (1 + 2 s) \cos^{2} α = 0$ $\sqrt{t} = - \frac{(1 + s) \cos α}{s + s \sin α - 2} \pm \sqrt{{(\frac{(1 + s) \cos α}{s + s \sin α - 2})}^{2} + \frac{(1 + 2 s) \cos^{2} α}{2 (s + s \sin α - 2)}}$ $s a y a = 1, b = 2$ $\Rightarrow s = 2, \sin α = \frac{1}{5}, \cos α = \frac{2 \sqrt{6}}{5}$ $\sqrt{t} = - 3 \sqrt{6} + 2 \sqrt{15}$ $r = t = 114 - 36 \sqrt{10}$ $\Rightarrow r \approx 0.1580$
	Commented by mr W last updated on 29/Nov/24

	$p = \frac{a b}{a + 2 b}$ $\sqrt{{(a + b)}^{2} - {(a - p)}^{2}} = \sqrt{{(a + r)}^{2} - {(a - r)}^{2}} + \sqrt{{(b + r)}^{2} - {(p - r)}^{2}}$ $\sqrt{b^{2} + 2 a b + 2 a p - p^{2}} = 2 \sqrt{a r} + \sqrt{b^{2} - p^{2} + 2 (b + p) r}$ $\sqrt{b^{2} + 2 a b + 2 a p - p^{2}} - 2 \sqrt{a r} = \sqrt{b^{2} - p^{2} + 2 (b + p) r}$ $(2 a - b - p) r - 2 \sqrt{a (b + p) (2 a + b - p) r} + a (b + p) = 0$ $\sqrt{r} = \frac{\sqrt{a (b + p) (2 a + b - p)} - \sqrt{a (b + p) (2 a + b - p) - a (b + p) (2 a - b - p)}}{2 a - b - p}$ $\sqrt{r} = \frac{\sqrt{a (b + p)} (\sqrt{2 a + b - p} - \sqrt{2 b})}{2 a - b - p}$ $r = \frac{a (b + p) {(\sqrt{2 a + b - p} - \sqrt{2 b})}^{2}}{{(2 a - b - p)}^{2}}$ $r = \frac{a (b + \frac{a b}{a + 2 b}) {(\sqrt{2 a + b - \frac{a b}{a + 2 b}} - \sqrt{2 b})}^{2}}{{(2 a - b - \frac{a b}{a + 2 b})}^{2}}$ $r = \frac{a b (a + b) {(\sqrt{a^{2} + a b + b^{2}} - \sqrt{(a + 2 b) b})}^{2}}{{(a^{2} + a b - b^{2})}^{2}}$ $r = \frac{a b (a + b) [a^{2} + 2 a b + 3 b^{2} - 2 \sqrt{b (a + 2 b) (a^{2} + a b + b^{2})}]}{{(a^{2} + a b - b^{2})}^{2}}$
	Commented by ajfour last updated on 29/Nov/24

	Commented by ajfour last updated on 29/Nov/24

	$N o w b o t h o f u s h a v e d o n e b e t t e r!$
	Answered by mr W last updated on 28/Nov/24

	Commented by mr W last updated on 28/Nov/24
	$for maximum small circle: AF⊥FD FE=(√((a+r)^2 −(a−r)^2 ))=2(√(ar)) (a+b)CD^2 +b(a+r)^2 =(a+2b)[(b+r)^2 +(a+b)b] CD^2 =(((a+2b)[(b+r)^2 +(a+b)b]−b(a+r)^2 )/(a+b)) ED=(√((((a+2b)[(b+r)^2 +(a+b)b]−b(a+r)^2 )/(a+b))−r^2 )) FD=2(√(ar))+(√((((a+2b)[(b+r)^2 +(a+b)b]−b(a+r)^2 )/(a+b))−r^2 )) FD^2 =a^2 +(a+2b)^2 ⇒{2(√(ar))+(√((((a+2b)[(b+r)^2 +(a+b)b]−b(a+r)^2 )/(a+b))−r^2 ))}^2 =a^2 +(a+2b)^2$
	$f o r m a x i m u m s m a l l c i r c l e :$ $A F ⊥ F D$ $F E = \sqrt{{(a + r)}^{2} - {(a - r)}^{2}} = 2 \sqrt{a r}$ $(a + b) {C D}^{2} + b {(a + r)}^{2} = (a + 2 b) [{(b + r)}^{2} + (a + b) b]$ ${C D}^{2} = \frac{(a + 2 b) [{(b + r)}^{2} + (a + b) b] - b {(a + r)}^{2}}{a + b}$ $E D = \sqrt{\frac{(a + 2 b) [{(b + r)}^{2} + (a + b) b] - b {(a + r)}^{2}}{a + b} - r^{2}}$ $F D = 2 \sqrt{a r} + \sqrt{\frac{(a + 2 b) [{(b + r)}^{2} + (a + b) b] - b {(a + r)}^{2}}{a + b} - r^{2}}$ ${F D}^{2} = a^{2} + {(a + 2 b)}^{2}$ $\Rightarrow {2 \sqrt{a r} + \sqrt{\frac{(a + 2 b) [{(b + r)}^{2} + (a + b) b] - b {(a + r)}^{2}}{a + b} - r^{2}}}^{2} = a^{2} + {(a + 2 b)}^{2}$
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