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Question Number 176367 by mr W last updated on 17/Sep/22

Commented by mr W last updated on 17/Sep/22

$f i n d t h e s u m o f a r e a s o f a l l i n s c r i b e d$ $c i r c l e s i n a t r i a n g l e w i t h s i d e l e n g t h e s$ $a, b, c a s s h o w n .$
	Answered by mr W last updated on 18/Sep/22

	Commented by mr W last updated on 19/Sep/22

	$l e t Δ = a r e a o f t h e t r i a n g l e$ $R = r a d i u s o f c i r c u m c i r c l e$ $r_{0} = r a d i u s o f i n c i r c l e$ $Δ = \frac{b c \sin A}{2} = \frac{a b c \sin A}{2 a} = \frac{a b c}{4 R}$ $r_{0} = \frac{2 Δ}{a + b + c}$ $\frac{r_{0}^{2}}{Δ} = \frac{4 Δ}{{(a + b + c)}^{2}} = \frac{a b c}{R {(a + b + c)}^{2}}$ $\frac{r_{0}^{2}}{Δ} = \frac{2 \sin A \sin B \sin C}{{(\sin A + \sin B + \sin C)}^{2}} = \tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}$ $\frac{r_{n - 1} - r_{n}}{r_{n - 1} + r_{n}} = \sin \frac{A}{2}$ $\frac{r_{n}}{r_{n - 1}} = \frac{1 - \sin \frac{A}{2}}{1 + \sin \frac{A}{2}}$ $\frac{r_{n}^{2}}{r_{n - 1}^{2}} = {(\frac{1 - \sin \frac{A}{2}}{1 + \sin \frac{A}{2}})}^{2} = q_{A} < 1 \Rightarrow G . P .$ $1 - q_{A} = (1 + \frac{1 - \sin \frac{A}{2}}{1 + \sin \frac{A}{2}}) (1 - \frac{1 - \sin \frac{A}{2}}{1 + \sin \frac{A}{2}})$ $1 - q_{A} = \frac{4 \sin \frac{A}{2}}{{(1 + \sin \frac{A}{2})}^{2}}$ $\frac{1}{1 - q_{A}} = \frac{{(1 + \sin \frac{A}{2})}^{2}}{4 \sin \frac{A}{2}} = \frac{1}{4} (\sin \frac{A}{2} + \frac{1}{\sin \frac{A}{2}}) + \frac{1}{2}$ $A_{n} = π r_{n}^{2}$ $\underset{n = 0}{\sum^{\infty}} A_{n} = π \underset{n = 0}{\sum^{\infty}} r_{n}^{2} = \frac{π r_{0}^{2}}{1 - q_{a}}$ $s u m o f a r e a s o f a l l i n s c r i b e d c i r c l e s :$ $A_{i . c .} = \frac{π r_{0}^{2}}{1 - q_{A}} + \frac{π r_{0}^{2}}{1 - q_{B}} + \frac{π r_{0}^{2}}{1 - q_{C}} - 2 π r_{0}^{2}$ $A_{i . c .} = (\frac{1}{1 - q_{A}} + \frac{1}{1 - q_{B}} + \frac{1}{1 - q_{C}} - 2) π r_{0}^{2}$ $A_{i . c .} = [\frac{1}{4} (\sin \frac{A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2} + \frac{1}{\sin \frac{A}{2}} + \frac{1}{\sin \frac{B}{2}} + \frac{1}{\sin \frac{C}{2}}) + \frac{3}{2} - 2] π r_{0}^{2}$ $A_{i . c .} = (\sin \frac{A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2} + \frac{1}{\sin \frac{A}{2}} + \frac{1}{\sin \frac{B}{2}} + \frac{1}{\sin \frac{C}{2}} - 2) \frac{π r_{0}^{2}}{4}$ $\frac{A_{i . c .}}{Δ} = (\sin \frac{A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2} + \frac{1}{\sin \frac{A}{2}} + \frac{1}{\sin \frac{B}{2}} + \frac{1}{\sin \frac{C}{2}} - 2) \frac{π r_{0}^{2}}{4 Δ}$ $\frac{A_{i . c .}}{Δ} = \frac{π}{4} \tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2} (\sin \frac{A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2} + \frac{1}{\sin \frac{A}{2}} + \frac{1}{\sin \frac{B}{2}} + \frac{1}{\sin \frac{C}{2}} - 2)$ $e x a m p l e s :$ $A = 40 °, B = 60 °, C = 80 ° : \frac{A_{i . c .}}{Δ} \approx 0.826$ $A = 30 °, B = 40 °, C = 110 ° : \frac{A_{i . c .}}{Δ} \approx 0.812$ $A = 60 °, B = 60 °, C = 60 ° : \frac{A_{i . c .}}{Δ} = \frac{11 \sqrt{3} π}{72} \approx 0.831$
	Commented by Tawa11 last updated on 18/Sep/22

	$Wow, great sir .$
	Commented by behi834171 last updated on 18/Sep/22

	$s u p e r b!$ $f a n t a s t i c!$ $k u d o s t o y o u p r o p : m r W!$
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