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Question Number 158079 by ajfour last updated on 30/Oct/21

Commented by ajfour last updated on 31/Oct/21

$S i r, t h e q u e s t i o n s h o u l d s t a t e :$ $f i n d s u m o f r a d i i (i n t e r m s o f$ $s i d e s o f △ A B C) t h a t t o u c h$ $t w o s i d e s o f t h e s a i d △ A B C$ $a n d o u t e r s i d e s o f r i g h t a n g l e d$ $△ c o n s t r u c t e d w i t h t h e t h i r d$ $s i d e o f △ A B C a s t h e h y p o t e n u s e .$ $q (a, b, c) = r_{A} + r_{B} + r_{C} .$
	Answered by mr W last updated on 31/Oct/21

	Commented by mr W last updated on 01/Nov/21

	$A G = A F = \frac{r}{\tan \frac{A}{2}}$ $C K = C G = b - A G = b - \frac{r}{\tan \frac{A}{2}}$ $B H = B F = c - A F = c - \frac{r}{\tan \frac{A}{2}}$ $C E = b - \frac{r}{\tan \frac{A}{2}} + r = b - r (\frac{1}{\tan \frac{A}{2}} - 1) = b - α r$ $B E = c - \frac{r}{\tan \frac{A}{2}} + r = c - r (\frac{1}{\tan \frac{A}{2}} - 1) = c + α r$ ${C E}^{2} + {B E}^{2} = {B C}^{2} = a^{2}$ ${(b - α r)}^{2} + {(c - α r)}^{2} = a^{2}$ $2 α^{2} r^{2} - 2 α (b + c) r + b^{2} + c^{2} - a^{2} = 0$ $r = \frac{α (b + c) + \sqrt{α^{2} {(b + c)}^{2} - 2 α^{2} (b^{2} + c^{2} - a^{2})}}{2 α^{2}}$ $r = r_{A} = \frac{(b + c) - \sqrt{2 a^{2} - {(b - c)}^{2}}}{2 α}$ $α = \frac{1}{\tan \frac{A}{2}} - 1 = \sqrt{\frac{1 + \cos A}{1 - \cos A}} - 1$ $\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$ $α = \sqrt{\frac{{(b + c)}^{2} - a^{2}}{a^{2} - {(b - c)}^{2}}} - 1$ $\Rightarrow r_{A} = \frac{b + c - \sqrt{2 a^{2} - {(b - c)}^{2}}}{2 (\sqrt{\frac{{(b + c)}^{2} - a^{2}}{a^{2} - {(b - c)}^{2}}} - 1)}$ $e x a m p l e : a = b = c$ $r_{A} = \frac{(2 - \sqrt{2}) a}{2 (\sqrt{3} - 1)} = \frac{(2 - \sqrt{2}) (\sqrt{3} + 1) a}{4} \approx 0.4 a$
	Commented by ajfour last updated on 01/Nov/21

	$A n E x c e l l e n t {s o l}^{n}! S i r . a n d$ $i^{'} v e r e p l a c e d m y {s o l}^{n} . p l e a s e$ $g o t h r o u g h . .$
	Commented by mr W last updated on 01/Nov/21

	$y e s, y o u a r e r i g h t!$
	Commented by ajfour last updated on 01/Nov/21

	$l e t s e q u a t e o u r r e s u l t s s i r!$
	Answered by ajfour last updated on 31/Oct/21

	Commented by ajfour last updated on 01/Nov/21
	$2×Area=2r^2 +(b+c)r +r{(acos θ−r)+(asin θ−r)} =2△_(ABC) +(a^2 sin θcos θ) c−(asin θ−r)=b−(acos θ−r) ⇒ a(sin θ−cos θ)=c−b ⇒ a^2 (1−2sin θcos θ)=(c−b)^2 a^2 (sin θ+cos θ)=(√((c−b)^2 +4a^2 sin θcos θ)) =(√(2a^2 −(c−b)^2 )) ⇒ (b+c)r+ar(sin θ+cos θ) = 2△+a^2 sin θcos θ hence ((2r)/a)=((4△+a^2 −(b−c)^2 )/(a(b+c)+(√(2a^2 −(c−b)^2 )))) e.g. say △ABC is eql. ((2r)/a)=(((√3)+1)/(2+(√2))) ⇒ r=((((√3)+1)/( (√2)+1)))(a/(2(√2))) or r=(a/4)((√3)+1)(2−(√2)) r≈0.400099577a$
	$2 \times A r e a = 2 r^{2} + (b + c) r$ $+ r {(a \cos θ - r) + (a \sin θ - r)}$ $= 2 △_{A B C} + (a^{2} \sin θ \cos θ)$ $c - (a \sin θ - r) = b - (a \cos θ - r)$ $\Rightarrow a (\sin θ - \cos θ) = c - b$ $\Rightarrow a^{2} (1 - 2 \sin θ \cos θ) = {(c - b)}^{2}$ $a^{2} (\sin θ + \cos θ) = \sqrt{{(c - b)}^{2} + 4 a^{2} \sin θ \cos θ}$ $= \sqrt{2 a^{2} - {(c - b)}^{2}}$ $\Rightarrow (b + c) r + a r (\sin θ + \cos θ)$ $= 2 △ + a^{2} \sin θ \cos θ$ $h e n c e$ $\frac{2 r}{a} = \frac{4 △ + a^{2} - {(b - c)}^{2}}{a (b + c) + \sqrt{2 a^{2} - {(c - b)}^{2}}}$ $e . g . s a y △ A B C i s e q l .$ $\frac{2 r}{a} = \frac{\sqrt{3} + 1}{2 + \sqrt{2}} \Rightarrow r = (\frac{\sqrt{3} + 1}{\sqrt{2} + 1}) \frac{a}{2 \sqrt{2}}$ $o r r = \frac{a}{4} (\sqrt{3} + 1) (2 - \sqrt{2})$ $r \approx 0.400099577 a$
	Commented by ajfour last updated on 01/Nov/21

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