Question Number 102769 by ajfour last updated on 11/Jul/20

Commented bybramlex last updated on 11/Jul/20

Commented bybramlex last updated on 11/Jul/20

(1)BG = (√((s^2 /4)−R^2 ))  (2)GD =  (√((r+R)^2 −R^2 ))= (√(r^2 +2rR))  (3)BD = (√((r+R)^2 +(s^2 /4)))  BD = BG + GD  (√((r+R)^2 +(s^2 /4))) =  (√((s^2 /4)−R^2 ))+(√(r^2 +2rR))

Commented byajfour last updated on 11/Jul/20

If △ABC is equilateral witb  side s, find radii of both circles  in terms of s.

Commented bybramlex last updated on 11/Jul/20

Commented bybramlex last updated on 11/Jul/20

AD = x+2r+R=((s(√3))/2) ...(1)  sin 30^o  = (r/(r+x)) = (1/2)⇒x=r...(2)  (1)&(2)⇒3r+R=((s(√3))/2) ...(3)  r+R=R(√2) ⇒r = R((√2)−1)...(4)  (3)&(4) ⇒3R((√2)−1)+R=((s(√3))/2)  R(3(√2)−2)=((s(√3))/2) ⇒R=((s(√3))/(2(3(√2)−2)))  then r = ((s((√6)−(√3)))/(2(3(√2)−2))) .⌣^• ∣⌣^•

Commented byajfour last updated on 11/Jul/20

how come R+r=R(√2)  ?

Commented bybramlex last updated on 11/Jul/20

Commented byajfour last updated on 11/Jul/20

the quadrilateral isn′t a   parallelogram,  wrong..

Commented bybramlex last updated on 11/Jul/20

no.

Answered by mr W last updated on 11/Jul/20

Commented bymr W last updated on 11/Jul/20

CF=(√((s^2 /4)−R^2 ))  ((OD)/(OC))=((OF)/(CF))  ⇒OD=((sR)/(√(s^2 −4R^2 )))  r=OD−R=((s/(√(s^2 −4R^2 )))−1)R  OA=R+3r=(((√3)s)/2)  R+3(((sR)/(√(s^2 −4R^2 )))−R)=(((√3)s)/2)  with λ=(R/s)  ⇒(3/(√(1−4λ^2 )))=2+((√3)/(2λ))  ⇒64λ^4 +32(√3)λ^3 +32λ^2 −8(√3)λ−3=0  ⇒λ=(R/s)≈0.3642  ⇒(r/s)=((1/(√(1−4λ^2 )))−1)λ≈0.1674

Commented bymr W last updated on 11/Jul/20

Commented byajfour last updated on 11/Jul/20

Very Nice solution Sir; Thanks!

Answered by 1549442205 last updated on 11/Jul/20

Commented by1549442205 last updated on 11/Jul/20

Since DKE^(�) =30°,KD=((DE)/(sinDKE^(�) ))=(r/(sin30°))  ⇒KD=2r.We have KC=KAcos30°=((s(√3))/2)  ,so KD=KC−CD.Hence,  2r=((s(√3))/2)−(R+r)⇔r=((s(√3)−2R)/6)(1)  On ther other hands,  AD^2 =(R+r)^2 +(s^2 /4)  KE=r(√3) and AE=s−KE=s−r(√3) ,  AE^2 +DE^2 =AD^2 =DC^2 +AC^2 ,so  (s−r(√3))^2 +r^2 =(R+r)^2 +(s^2 /4)  ⇔R^2 +2Rr−3r^2 +2sr(√3)−((3s^2 )/4)=0(due to (1))  ⇔R^2 +((s(√3)−2R)/3)(R+s(√3))−((3s^2 −4Rs(√3)+4R^2 )/3)−((3s^2 )/4)=0  4R^2 −12Rs(√3) +9s^2 =0  .Δ′=108s^2 −36s^2 =72s^2   R=((6s(√3)−6s(√2))/4)=((3s((√3)−(√2)))/2).Replace  into (1) we get   r=((s(√3)−3s((√3)−(√2)))/6)=(((3(√2)−2(√3))s)/6)  Thus, { (((R/s)=((3((√3)−(√2)))/2)≈0.47675586)),(((r/s)=((3(√2)−2(√3))/6)≈0.12975512)) :}