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Question Number 193464 by Mingma last updated on 14/Jun/23

Commented by Frix last updated on 14/Jun/23

$For any triangle$ $r_{n} = \frac{c δ}{2 ((a + b + c) c + (n - 1) δ)}$ $[δ = \sqrt{(a + b + c) (- a + b + c) (a - b + c) (a + b - c)}$ $For a rectangular triangle with c = \sqrt{a^{2} + b^{2}}$ $r_{n} = \frac{a b \sqrt{a^{2} + b^{2}}}{a^{2} + 2 (n - 1) a b + b^{2} + (a + b) \sqrt{a^{2} + b^{2}}}$ $With a = 8 \land b = 15 \Rightarrow c = 17$ $r_{n} = \frac{51}{6 n + 11}$
Commented by Mingma last updated on 15/Jun/23
Perfect ��
Commented by a.lgnaoui last updated on 15/Jun/23

$in terme a b c ?$
	Answered by a.lgnaoui last updated on 15/Jun/23
	$∡DCM=((∡C)/2) ∡EBN=((∡B)/2) BC=BMcos ((∡C)/2)+CNcos ((∡B)/2)+12r posons ∡C=θ ∡B=ϕ=90−θ ⇒cos (𝛉/2)=(√((a+c)/(2c))) cos (𝛟/2)=(√((b+c)/(2c))) ⇒BC=BM(√(((a+c)/(2c)) )) +CN(√((b+c)/(2c))) +12r(1) sin (𝛉/2)=(r/(BM)) BM=(r/(sin (𝛉/2)))=r(√((2c)/(c−a))) CN=r(√((2c)/(c−b))) (1)⇔ BC=r(√((2c)/(c−a))) (√((a+c)/(2c))) +r(√((2c)/(c−b))) (√((b+c)/(2c))) +12r BC =r(12+(√((a+c)/(c−a))) +(√(((b+c)/(c−b)) )) ) ABC triangld rectangle BC^2 =a^2 +b^2 =c^2 c=(√(a^2 +b^2 )) (√(a^2 +b^2 )) =r(12+(√((a+c)/(c−a))) +(√((b+c)/(c−b))) ) soit r=(c/(12+(√((a+c)/(c−a))) +(√((b+c)/(c−b))))) { ((a=8 b= 15 c=17)),(( r=((17)/(12+(√((25)/9)) +(√((32)/2))))=((17)/(16+(5/3))))) :} soit r =((51)/(53)) cas general pour n cercles de rayon r r =(c/(2(n−1)+(√((a+c)/(c−a))) +(√((b+c)/(c−b)))))$
	$∡ DCM = \frac{∡ C}{2} ∡ EBN = \frac{∡ B}{2}$ $BC = B Mcos \frac{∡ C}{2} + C Ncos \frac{∡ B}{2} + 12 r$ $posons ∡ C = θ ∡ B = φ = 90 - θ$ $\Rightarrow \cos \frac{θ}{2} = \sqrt{\frac{a + c}{2 c}} \cos \frac{φ}{2} = \sqrt{\frac{b + c}{2 c}}$ $\Rightarrow BC = BM \sqrt{\frac{a + c}{2 c}} + CN \sqrt{\frac{b + c}{2 c}} + 12 r (1)$ $\sin \frac{θ}{2} = \frac{r}{BM} BM = \frac{r}{\sin \frac{θ}{2}} = r \sqrt{\frac{2 c}{c - a}}$ $CN = r \sqrt{\frac{2 c}{c - b}}$ $(1) \Leftrightarrow BC = r \sqrt{\frac{2 c}{c - a}} \sqrt{\frac{a + c}{2 c}} + r \sqrt{\frac{2 c}{c - b}} \sqrt{\frac{b + c}{2 c}} + 12 r$ $BC = r (12 + \sqrt{\frac{a + c}{c - a}} + \sqrt{\frac{b + c}{c - b}})$ $ABC triangld rectangle$ ${BC}^{2} = a^{2} + b^{2} = c^{2} c = \sqrt{a^{2} + b^{2}}$ $\sqrt{a^{2} + b^{2}} = r (12 + \sqrt{\frac{a + c}{c - a}} + \sqrt{\frac{b + c}{c - b}})$ $soit$ $r = \frac{c}{12 + \sqrt{\frac{a + c}{c - a}} + \sqrt{\frac{b + c}{c - b}}}$ ${\begin{cases} a = 8 b = 15 c = 17 \\ r = \frac{17}{12 + \sqrt{\frac{25}{9}} + \sqrt{\frac{32}{2}}} = \frac{17}{16 + \frac{5}{3}} \end{cases}$ $soit r = \frac{51}{53}$ $cas general pour n cercles de rayon r$ $r = \frac{c}{2 (n - 1) + \sqrt{\frac{a + c}{c - a}} + \sqrt{\frac{b + c}{c - b}}}$
	Commented by a.lgnaoui last updated on 15/Jun/23

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