tag:blogger.com,1999:blog-6933544261975483399.post465551089396544045..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1061: Triangle, Inradius (r), Circumradius (R), Circumcircle, Angle Bisector, Distance from a point to a Line, PerpendicularAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-55152127784383795482014-12-27T13:52:10.043-08:002014-12-27T13:52:10.043-08:00From property of incenter I, A1, B1, and C1 are ci...From property of incenter I, A1, B1, and C1 are circumcenters of BIC, AIC, and AIB respectively.<br />Let C' and A' be perpendiculars from C1 and A1 onto AB and BC respectively.<br />Due to first statement,<br />c1-r=IC1*sin(<C/2)=C1C'<br />a1-r=IA1*sin(<A/2)=A1A'<br />Let P be perpendicular from O to B1B2.<br /><OB1P=<C=<BOC', and OB1=OB=R, so triangles B1OP and OBC' are congruent. Therefore,<br />b1-OA'=B1P=OC',<br />b1=OA'+OC'<br />Summing up previous equations gives<br />(c1-r)+(a1-r)+(b1)=(C1C')+(A1A')+(OA'+OC')=2R.Ivan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-21772712700786522072014-11-17T07:37:40.628-08:002014-11-17T07:37:40.628-08:00Join A₁B, A₁C. Then A₁B=A₁C=2Rsin(A/2).
By Ptole...Join A₁B, A₁C. Then A₁B=A₁C=2Rsin(A/2). <br /><br />By Ptolemy Theorem on ABA₁C, <br />a×AA₁ = 2Rsin(A/2)×(b+c)<br />a×AA₁sin(A/2) = 2Rsin²(A/2)×(b+c)<br />a×a₁ = 2Rsin²(A/2)×(b+c)<br />sinA×a₁ = 2Rsin²(A/2)(sinB+sinC)<br />2sin(A/2)cos(A/2)×a₁ = 4Rsin²(A/2) cos(A/2)cos((B-C)/2)<br />a₁ = 2Rsin(A/2)cos((B-C)/2) = R(cosB+cosC)<br /><br />Similarly, <br />b₁ = R(cosA+cosC)<br />c₁ = R(cosA+cosB)<br /><br />a₁+b₁+c₁<br />= 2R (cosA + cosB + cosC)<br />= 2R + 8Rsin(A/2)sin(B/2)sin(C/2)<br />= 2R + 2r<br />= 2(R+r)Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.com