tag:blogger.com,1999:blog-6933544261975483399.post3085795135995405663..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 765: Triangle, Exradii, Inradius, Three Exradius, Harmonic MeanAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-64273907446322006002019-11-19T01:16:49.000-08:002019-11-19T01:16:49.000-08:00Thank you very much for that) I searched this answ...Thank you very much for that) I searched this answer all day))Anonymoushttps://www.blogger.com/profile/16001808638958742710noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-82352833185146559732012-06-15T10:30:06.542-07:002012-06-15T10:30:06.542-07:00Let BC touch incircle at X and excircle opp to B a...Let BC touch incircle at X and excircle opp to B at Y<br />Triangles BIX and BI₂Y are similar <br />r/r₂ = AX/AY = (s - b)/s<br />Similarly r/r₃ = (s - c)/s and r/r₁= (s - a)/s<br />r/r₁ + r/r₂ + r/r₃ <br />=(s - a)/s + (s - b)/s + (s - c)/s<br />= [3s - (a + b + c)] = (3s - 2s) /s = 1<br />Hence 1/r₁ + 1/r₂ + 1/r₃ = 1/rPravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-25487453581533442782012-06-14T18:49:38.581-07:002012-06-14T18:49:38.581-07:00Let S be the area of the triangle and X = (1/2)(a+...Let S be the area of the triangle and X = (1/2)(a+b+c), where a,b,c are the side lengths of the triangle.<br />By the properties of the exradii, S=(Ra)(X-a) ; S=(Rb)(X-b) ; S=(Rc)(X-c).<br />Therefore,<br />S/(Ra)=(X-a)<br />S/(Rb)=(X-b)<br />S/(Rc)=(X-c)<br />By summing up, S[(1/Ra)+(1/Rb)+(1/Rc)]=3X-(a+b+c) = X.<br />By the properties of the inradii, S=RX => X =S/R<br />So, <br />S/R = S[(1/Ra)+(1/Rb)+(1/Rc)]<br />1/R = (1/Ra)+(1/Rb)+(1/Rc)<br /><br />q.e.d.<br /><br />Now,W Fungnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-51065163794059434482012-06-14T18:22:49.762-07:002012-06-14T18:22:49.762-07:00http://img43.imageshack.us/img43/9397/problem765.p...http://img43.imageshack.us/img43/9397/problem765.png<br /><br />Denote S(XYZ)= area of triangle XYZ<br />Draw additional lines per attached sketch.<br />We have ∆AA’F similar to ∆AIE<br />So IE/A’F=r/rA =AI/AA’= S(IC’B’)/S(A’B’C’)<br />Similarly r/rB=S(IA’C’)/S(A’B’C’) and r/rC=S(IA’B’)/S(A’B’C’)<br />But S(A’B’C’)=S(IA’C’)+S(IA’B’)+S(IC’B’)<br />So r/rA+r/rB+r/rC= 1 or 1/r=1/rA+1/rB+1/rCPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com