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Question Number 213884 by mr W last updated on 20/Nov/24

Commented by mr W last updated on 20/Nov/24

find maximum radius of circle C

findmaximumradiusofcircleC

Commented by Frix last updated on 21/Nov/24

I think it′s at  r_A =(3/5)  r_B =(3/8)  r_C =(1/4)

IthinkitsatrA=35rB=38rC=14

Commented by mr W last updated on 21/Nov/24

any easy way to prove this?

anyeasywaytoprovethis?

Answered by Ghisom last updated on 21/Nov/24

r_A =p∧0<p<1  A= (((1−p)),(0) )  r_B =((4p(1−p))/((1+p)^2 ))     [ _(of circles)^(use tangents) ]  B= ((((1−3p)/(1+p))),(r_B ) )  now use Descartes to get  r_C =((4p(1−p))/(9p^2 −14p+9)) ⇒ max r_C  =(3/5)  C= (((((3−p)(3−5p))/(9p^2 −14p+9))),((3r_C )) )    we can find other interesting values like  properties of the triangle ABC etc

rA=p0<p<1A=(1p0)rB=4p(1p)(1+p)2[ofcirclesusetangents]B=(13p1+prB)nowuseDescartestogetrC=4p(1p)9p214p+9maxrC=35C=((3p)(35p)9p214p+93rC)wecanfindotherinterestingvalueslikepropertiesofthetriangleABCetc

Answered by mr W last updated on 21/Nov/24

((1/a)+(2/b)−(1/R))^2 =2((1/a^2 )+(2/b^2 )+(1/R^2 ))  (4/b)((1/a)−(1/R))=((1/a)+(1/R))^2   ⇒(1/b)=((((1/a)+(1/R))^2 )/(4((1/a)−(1/R))))  let α=(R/a), β=(R/b), γ=(R/c)  ⇒β=(((α+1)^2 )/(4(α−1)))  ((1/a)+(1/b)+(1/c)−(1/R))^2 =2((1/a^2 )+(1/b^2 )+(1/c^2 )+(1/R^2 ))  (R^2 /c^2 )−2((R/a)+(R/b)−1)(R/c)+1+((R/a)−(R/b))^2 +2((R/a)+(R/b))  γ^2 −2(α+β−1)γ+1+(α−β)^2 +2(α+β)=0  γ=α+β−1−2(√(αβ−α−β))  γ=((9α^2 −14α+9)/(4(α−1)))     =1+((9(α−1))/4)+(1/(α−1))≥1+2(√(9/4))=4  ⇒γ_(min) =4 ⇒c_(max) =(R/4) at a=((3R)/5)

(1a+2b1R)2=2(1a2+2b2+1R2)4b(1a1R)=(1a+1R)21b=(1a+1R)24(1a1R)letα=Ra,β=Rb,γ=Rcβ=(α+1)24(α1)(1a+1b+1c1R)2=2(1a2+1b2+1c2+1R2)R2c22(Ra+Rb1)Rc+1+(RaRb)2+2(Ra+Rb)γ22(α+β1)γ+1+(αβ)2+2(α+β)=0γ=α+β12αβαβγ=9α214α+94(α1)=1+9(α1)4+1α11+294=4γmin=4cmax=R4ata=3R5

Commented by mr W last updated on 21/Nov/24

Commented by mr W last updated on 21/Nov/24

Commented by mr W last updated on 21/Nov/24

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