tag:blogger.com,1999:blog-6933544261975483399.post7700724899897557385..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 213: Triangle, Incircle, Inradius, Semicircles, Common TangentsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6933544261975483399.post-9670449741241451552009-04-02T00:16:00.000-07:002009-04-02T00:16:00.000-07:00Let AF = AD = 2r1, FC = FE = 2r2 & BE = BD = 2...Let AF = AD = 2r1, FC = FE = 2r2 & BE = BD = 2r3<BR/>Semisum of sides, s = 2r1+2r2+2r3<BR/>Thus, r^2= (Area/s)^2 =(s-a)(s-b)(s-c)/s<BR/>where a=2r1+2r2,b=2r2+2r3 & c=2r3+2r1 which gives us, r^2 = (2r1*2r2*2r3)/(2r1+2r2+2r3)<BR/> = (4r1*r2*r3)/(r1+r2+r3)<BR/>or 1/r^2 = (r1+r2+r3)/(4r1r2r3) --------(1)<BR/>It's easy to see that, d^2=(r3+r1)^2-(r3-r1)^2 =4r3r1. Likewise, e^2=4r2r3 & f^2=4r1r2. Now, 1/d^2+1/e^2+1/f^2 = 1/4r1r2+1/4r2r2+3/4r3r1 = (r1+r2+r3)/(4r1r2r3) <BR/>By equation (1), 1/r^2 = 1/d^2 + 1/e^2 + 1/f^2<BR/>QED<BR/>Ajit: ajitaathle@gmail.com<BR/>By (1),1/r^2 = (r1+r2+r3)/(4r1*r2*r3) (r1+r2+r3)Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com