tag:blogger.com,1999:blog-6933544261975483399.post7093090038208130872..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 972: Equilateral Triangle, Vertices, Three Parallel, Equal Circles, Perpendicular BisectorAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-43110134529149365932016-08-31T00:42:28.265-07:002016-08-31T00:42:28.265-07:00Problem 972
Is the triangle AOQ equilateral.Let...Problem 972<br />Is the triangle AOQ equilateral.Let AC intersects the line L_2 and the circle Q in points F, C respectively.Then AFQO is cyclic(<FAO=90=<FQO, FA=FQ tangent parts), with <CFE=60=<QOA. Also the points G,Q, and O are collinear (<ΟΑΓ=90 ,ΓΟ is diameter) and <br />OQ=OG.But OD=DC so QD//GC or <FCO=<QDO. Is <OQE=90=<ODE so OQDE is cyclic.<br />Then <QDO=<QEO.So FCO=<FEO.Therefore OFCE is cyclic .So <COE=<CFE=60=<OCE.<br />Therefore triangle OCE is equilateral.<br />APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-40007809055764996052016-03-23T03:10:21.022-07:002016-03-23T03:10:21.022-07:00DA = DO and Tr. AQO is equilateral so < OQM = 3...DA = DO and Tr. AQO is equilateral so < OQM = 30 since Tr.s AQD and OQD are congruent (DQ, AO meet at M)<br /><br />Since OQDE is cyclic therefore < DEO = < OQM = 30 = < CED <br /><br />Hence CEO is equilateral <br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-70326102719773337412014-02-01T13:42:09.070-08:002014-02-01T13:42:09.070-08:00D is circumcenter of CAO, and QA=QO, so DQ is perp...D is circumcenter of CAO, and QA=QO, so DQ is perpendicular to AO and parallel to CA. Then <ODQ=<OCA, but <OEQ=<ODQ from cyclic quadrilateral OQDE. If L2 meets AC at P, then PCEO is cyclic because <OCA=<OEQ. <AOQ=<CPE=<COE=60Ivan Bazarovnoreply@blogger.com