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Question Number 63981 by ajfour last updated on 11/Jul/19

Answered by mr W last updated on 11/Jul/19

[(√(a^2 −b^2 −b^2 ))+(√((a−r)^2 −r^2 ))]^2 =(b+r)^2 −(b−r)^2 −b^2 +a(a−r)+(√((a^2 −2b^2 )a(a−2r)))=2br (√((a^2 −2b^2 )a(a−2r)))=(a+2b)r−(a^2 −b^2 ) (a^2 −2b^2 )a(a−2r)=(a+2b)^2 r^2 +(a^2 −b^2 )^2 −2(a+2b)(a^2 −b^2 )r (a+2b)^2 r^2 −2[2(a^3 −b^3 )+ab(2a−3b)]r+b^4 =0 ⇒r=((2(a^3 −b^3 )+ab(2a−3b)+(√([2(a^3 −b^3 )+ab(2a−3b)]^2 −(a+2b)^2 b^4 )))/((a+2b)^2 ))

[a2−b2−b2+(a−r)2−r2]2=(b+r)2−(b−r)2−b2+a(a−r)+(a2−2b2)a(a−2r)=2br(a2−2b2)a(a−2r)=(a+2b)r−(a2−b2)(a2−2b2)a(a−2r)=(a+2b)2r2+(a2−b2)2−2(a+2b)(a2−b2)r(a+2b)2r2−2[2(a3−b3)+ab(2a−3b)]r+b4=0⇒r=2(a3−b3)+ab(2a−3b)+[2(a3−b3)+ab(2a−3b)]2−(a+2b)2b4(a+2b)2

Commented by ajfour last updated on 12/Jul/19

NICE n WISE Sir, Thank you.

NICEnWISESir,Thankyou.

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