tag:blogger.com,1999:blog-6933544261975483399.post8965217866125490343..comments2024-06-15T03:59:07.072-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 92Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-72881864043898010992018-03-29T11:35:56.745-07:002018-03-29T11:35:56.745-07:00I saw just now your notification
the expression ( ...I saw just now your notification<br />the expression ( are on bisector of ang B is not correct)<br />but Circles O and O1 have the same tangente at the point B so they are collinear<br />c.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-89533134845845913972016-01-18T07:20:30.438-08:002016-01-18T07:20:30.438-08:00Here R=R1+R2+R3
Here R=R1+R2+R3<br />rakeshhttps://www.blogger.com/profile/08080860747596720415noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-55298081179110133812012-05-01T08:25:42.600-07:002012-05-01T08:25:42.600-07:00In solution proposed by c. t. e. o. I think that t...In solution proposed by c. t. e. o. I think that the sentence “B, O1, O are collinear (are on bisector of ang B)” is not true (am I wrong?). I suggest the solution above.<br />By their constructions, ABC, DBE, FGE and MGC are similar, so ang(BAO) = ang(CDO1) =<br />= ang(GFO2) = ang(GMO3). Thus R, R1, R2 and R3 are parallel.<br />Let t1 be the straight line tangent to C0 at B. OB and t1 are perpendicular. From the similarity of ABC and DBE, ang(DBO1)=ang(ABO), so O, O1 and B are collinear and O1B and t1 are perpendicular too. Hence, t1 is tangent to O1 at B and C0 and C1 are tangent at B.<br />The other cases are solved similarly.NIlton Lapanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-62071529739536605362010-03-01T10:57:54.074-08:002010-03-01T10:57:54.074-08:00B, O1, O are collinear ( are on bisector of ang B ...B, O1, O are collinear ( are on bisector of ang B )<br />=><br />O1 BD = BDO1 ( DO1 = BO1 = R )<br />DAO = ABO ( BO = AO = R )<br />=><br />BDO1 = DAO => ( as corresponding ang, DE//AC )<br />=> <br />R//R1<br /><br />join O1 to O2 ( O1, E, O2 are collinear, on bisect E )<br />=><br />O1DE = DEO1 = EO2F = FO2E ( DO1=O1E =R1,EO2=O2F=R2)<br />=><br />BDO1 = O2FG (as alternate ang BDE=EFG, AB//FG )<br />=> <br />R1//R2 <br /><br />in the same way (O2FG = O3MG)<br /><br />R2//R3<br />--------------------------------------<br />Tg at B to BO1 is tg for BO (B,O1, O are collinear)<br />=><br />circle O and circle O1 tg at Bc .t . e. onoreply@blogger.com