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Question Number 190405 by cortano12 last updated on 02/Apr/23

	Answered by cherokeesay last updated on 02/Apr/23


	Answered by mr W last updated on 02/Apr/23
	$((p−q)/(p+q))=((q−r)/(q+r))=k q=((1+k)/(1−k))r p=((1+k)/(1−k))q=(((1+k)/(1−k)))^2 r (p+q)^2 −(p−q)^2 =a^2 4pq=a^2 4(((1+k)/(1−k)))^3 r^2 =a^2 ...(i) (q+r)^2 −(q−r)^2 =b^2 4qr=b^2 4(((1+k)/(1−k)))r^2 =b^2 ...(ii) ⇒((1+k)/(1−k))=(a/b) 4×(a/b)×r^2 =b^2 ⇒r=(b/2)(√(b/a)) ⇒q=(b/2)((a/b))(√(b/a)) ⇒p=(b/2)((a/b))^2 (√(b/a)) R=p+q+r=[((a/b))^2 +((a/b))+1](b/2)(√(b/a)) A_(red) =(π/2)(R^2 −p^2 −q^2 −r^2 ) =(π/2){[((a/b))^2 +((a/b))+1]^2 −((a/b))^4 −((a/b))^2 −1}(b^2 /4)×(b/a) =π{((a/b))^3 +((a/b))^2 +((a/b))}(b^2 /4)×(b/a) ={((a/b))^2 +(a/b)+1}((πb^2 )/4) ={((8/4))^2 +(8/4)+1}((π×4^2 )/4) =28π$
	$\frac{p - q}{p + q} = \frac{q - r}{q + r} = k$ $q = \frac{1 + k}{1 - k} r$ $p = \frac{1 + k}{1 - k} q = {(\frac{1 + k}{1 - k})}^{2} r$ ${(p + q)}^{2} - {(p - q)}^{2} = a^{2}$ $4 p q = a^{2}$ $4 {(\frac{1 + k}{1 - k})}^{3} r^{2} = a^{2} . . . (i)$ ${(q + r)}^{2} - {(q - r)}^{2} = b^{2}$ $4 q r = b^{2}$ $4 (\frac{1 + k}{1 - k}) r^{2} = b^{2} . . . (i i)$ $\Rightarrow \frac{1 + k}{1 - k} = \frac{a}{b}$ $4 \times \frac{a}{b} \times r^{2} = b^{2}$ $\Rightarrow r = \frac{b}{2} \sqrt{\frac{b}{a}}$ $\Rightarrow q = \frac{b}{2} (\frac{a}{b}) \sqrt{\frac{b}{a}}$ $\Rightarrow p = \frac{b}{2} {(\frac{a}{b})}^{2} \sqrt{\frac{b}{a}}$ $R = p + q + r = [{(\frac{a}{b})}^{2} + (\frac{a}{b}) + 1] \frac{b}{2} \sqrt{\frac{b}{a}}$ $A_{r e d} = \frac{π}{2} (R^{2} - p^{2} - q^{2} - r^{2})$ $= \frac{π}{2} {{[{(\frac{a}{b})}^{2} + (\frac{a}{b}) + 1]}^{2} - {(\frac{a}{b})}^{4} - {(\frac{a}{b})}^{2} - 1} \frac{b^{2}}{4} \times \frac{b}{a}$ $= π {{(\frac{a}{b})}^{3} + {(\frac{a}{b})}^{2} + (\frac{a}{b})} \frac{b^{2}}{4} \times \frac{b}{a}$ $= {{(\frac{a}{b})}^{2} + \frac{a}{b} + 1} \frac{π b^{2}}{4}$ $= {{(\frac{8}{4})}^{2} + \frac{8}{4} + 1} \frac{π \times 4^{2}}{4}$ $= 28 π$
	Commented by mr W last updated on 02/Apr/23

	Answered by a.lgnaoui last updated on 03/Apr/23
	$Soit :C_1 (O1,R_1 ) ; C_2 (O_2 ,R_2 ) ;C_3 (O_3 ,R_3 ) R_1 =O_1 C R_2 =O_2 B R_3 =O_3 A 1•tan θ=((r_2 −r_1 )/4)=((r_3 −r_2 )/8)=((r_3 −r_1 )/(12))(1) 2• { ((4^2 +_1 (r_2 −r_1 )^2 =(r_1 +r_2 )^2 (2))),((8^2 +(r_3 −r_2 )^2 =(r_3 +r_2 )^2 (3))) :} (1) ⇒ { ((2(r_2 −r_1 )=(r_3 −r_2 ))),((3(r_3 −r_2 )=2(r_3 −r_1 ))) :} ⇒ r_3 =3r_2 −2r_1 (4) (2) { ((r_1 r_2 =4 (5))),((r_2 r_3 =16 (6))) :} (5)et(6)⇒ r_3 =4r_1 (7) (4)⇒4r_1 =3r_2 −2r_1 ⇒ r_2 =2r_1 (8) 3• (1)⇒12^2 +(r_3 −r_1 )^2 =(r_1 +2r_2 +r_3 )^2 36=r_2 ^2 +r_1 r_3 +r_2 (r_1 +r_3 ) =4r_1 ^2 +4r_1 ^2 +2r_1 (5r_1 )=18r_1 ^2 36=18r_1 ^2 ⇒r_1 =(√2) Rayon R=r_1 +2r_1 +4r_1 =7r_1 ⇒ R=7(√2) Aire totale=πR^2 Aire (Rouge)=((πR^2 )/2)−π(((r_1 ^2 +r_2 ^2 +r_3 ^2 )/2)) r_1 ^2 +r_2 ^2 +r_3 ^2 =42 { ((Aire=28π)),((Aire ≅87,96)) :}$
	$S o i t : C_{1} (O 1, R_{1}); C_{2} (O_{2}, R_{2}); C_{3} (O_{3}, R_{3})$ $R_{1} = O_{1} C R_{2} = O_{2} B R_{3} = O_{3} A$ $1 ∙ \tan θ = \frac{r_{2} - r_{1}}{4} = \frac{r_{3} - r_{2}}{8} = \frac{r_{3} - r_{1}}{12} (1)$ $2 ∙ {\begin{cases} 4^{2} +_{1} {(r_{2} - r_{1})}^{2} = {(r_{1} + r_{2})}^{2} (2) \\ 8^{2} + {(r_{3} - r_{2})}^{2} = {(r_{3} + r_{2})}^{2} (3) \end{cases}$ $(1) \Rightarrow {\begin{cases} 2 (r_{2} - r_{1}) = (r_{3} - r_{2}) \\ 3 (r_{3} - r_{2}) = 2 (r_{3} - r_{1}) \end{cases}$ $\Rightarrow r_{3} = 3 r_{2} - 2 r_{1} (4)$ $(2) {\begin{cases} r_{1} r_{2} = 4 (5) \\ r_{2} r_{3} = 16 (6) \end{cases}$ $(5) e t (6) \Rightarrow r_{3} = 4 r_{1} (7)$ $(4) \Rightarrow 4 r_{1} = 3 r_{2} - 2 r_{1}$ $\Rightarrow r_{2} = 2 r_{1} (8)$ $3 ∙ (1) \Rightarrow 12^{2} + {(r_{3} - r_{1})}^{2} = {(r_{1} + 2 r_{2} + r_{3})}^{2}$ $36 = r_{2}^{2} + r_{1} r_{3} + r_{2} (r_{1} + r_{3})$ $= 4 r_{1}^{2} + 4 r_{1}^{2} + 2 r_{1} (5 r_{1}) = 18 r_{1}^{2}$ $36 = 18 r_{1}^{2} \Rightarrow r_{1} = \sqrt{2}$ $R a y o n R = r_{1} + 2 r_{1} + 4 r_{1} = 7 r_{1}$ $\Rightarrow R = 7 \sqrt{2}$ $A i r e t o t a l e = π R^{2}$ $A i r e (R o u g e) = \frac{π R^{2}}{2} - π (\frac{r_{1}^{2} + r_{2}^{2} + r_{3}^{2}}{2})$ $r_{1}^{2} + r_{2}^{2} + r_{3}^{2} = 42$ ${\begin{cases} A i r e = 28 π \\ A i r e ≅ 87, 96 \end{cases}$
	Answered by Skabetix last updated on 10/Apr/23

	Commented by Skabetix last updated on 10/Apr/23

	Answered by ajfour last updated on 10/Apr/23

	$2 \sqrt{p q} = a$ $2 \sqrt{q r} = b$ $\frac{p}{q} = \frac{q}{r}$ $\Rightarrow 4 r p = 4 q^{2} = a b$ $R = p + q + r$ $\frac{2 S}{π} = {(p + q + r)}^{2} - p^{2} - q^{2} - r^{2}$ $= 2 (p q + q r + r p)$ $= 2 (\frac{a^{2}}{4} + \frac{b^{2}}{4} + \frac{a b}{4})$ $\Rightarrow S = \frac{π}{4} (a^{2} + b^{2} + a b)$ $= \frac{π}{4} (64 + 16 + 32) = 28 π$
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