tag:blogger.com,1999:blog-6933544261975483399.post1854086855600192254..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1168: Construction of the Inscribed Circle of the Arbelos, Semicircles, Diameter, Circle, Triangle, Circumcircle, TangentAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-67409830327520425172015-12-02T08:12:47.996-08:002015-12-02T08:12:47.996-08:00just change ´´ given ´´ to ´´ to prove ´´ at P ...just change ´´ given ´´ to ´´ to prove ´´ at P 638 c.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-36826426833189073682015-12-01T18:03:55.802-08:002015-12-01T18:03:55.802-08:00See below for minor correction and clarification ...See below for minor correction and clarification from previous comment<br />http://s21.postimg.org/aw9ih96yf/pro_1168.png<br /><br />Let 2R is the diameter of circle O , 2r1=diameter of circle O1 and 2r2=diameter of circle O2<br />Complete full circle O and let E’ is the midpoint of arc AC ( see sketch)<br />Let Au and Cv are the tangents or circle O at A and C<br />We have tri. O1O2O4 similar to tri. F’CB …( case AA)<br />So O2O4/BC= O1O2/CF’= ½ => CF’=2.O1O2= AC= 2.R<br />Similarly we also have AD’=AC= 2.R<br />Perform geometry inversion with inversion center at B and inversion power= -BA. BC=- BE.BE’=-BF.BF’=-BD.BD’<br />In this transformation Circle O → circle O <br />circle O2 → line Au <br />Circle O1 →Line Cv, <br />Circle O4→ Line AE’ <br />Circle O3 →line CE’<br />Circumcircle of tri. DEF → Circumcircle of tri. D’E’F’<br />But circumcircle of tri. D’E’F’ tangent to circle O, Lines Au and Cv ( images of circles O, O2 and O1)<br />So circumcircle of tri. DEF will tangent to circles O, O1 and O2<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com