tag:blogger.com,1999:blog-6933544261975483399.post4117524384932100735..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 885: Intersecting Circles, Three Tangent Lines, MidpointAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger10125tag:blogger.com,1999:blog-6933544261975483399.post-58786696956140898862016-08-26T22:25:22.593-07:002016-08-26T22:25:22.593-07:00< FHC = < FCE because EC tangent to circle A...< FHC = < FCE because EC tangent to circle A at C<br /><br />EC is the tangent to circle C at C so < CME = < FCE<br />Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-40904141586947024172016-08-11T07:57:18.435-07:002016-08-11T07:57:18.435-07:00everyone begins with equality of angles fhc, fce a...everyone begins with equality of angles fhc, fce and so on. I must be missing something trivial but what is the proof for that?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-26786656001375421992016-08-11T07:53:50.793-07:002016-08-11T07:53:50.793-07:00everyone starts with equality of angles fhc,fce an...everyone starts with equality of angles fhc,fce and so on. it must be trivial but I can't see the proof for that. illuminate me, please ?Anonymoushttps://www.blogger.com/profile/02270805610770006219noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-49037166870012124642016-07-22T00:46:48.820-07:002016-07-22T00:46:48.820-07:00< FHC = < FCE =< CME , < FCH = < F...< FHC = < FCE =< CME , < FCH = < FEC and < CFE = < ECM<br /><br />So easily Triangles HFC, EFC and ECM are similar<br /><br />Hence CE/EM = CF/CM….(1) from similar triangles EFC and ECM<br /><br />And CE/CF = EM/CH…(2) from similar triangles ECM and HFC<br /><br />Comparing (1) and (2) CM = CH<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-4565023695923287902016-07-20T20:11:45.944-07:002016-07-20T20:11:45.944-07:00AG and BC are perpendicular to FC, AC and GB perpe...AG and BC are perpendicular to FC, AC and GB perpendicular to CE, then ACBG parallelogram. So <GAC = <GBC = <GBE. <HAC = 2 <HFC = 2 <ECM = <EBM. <HAG = (<HAC = <EBM) + (<GAC = <GBE) = <GBM, HA = AC = GB, and AG = BC = BM, means that the triangles HAG and GBM congruent. GH = GM and GC perpendicular to HM, so that triangles GCH and GCM congruent.Ivan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-73773935912118524822016-07-20T19:53:04.340-07:002016-07-20T19:53:04.340-07:00AG и BC перпендикулярно FC, АС и GB перпендикулярн...AG и BC перпендикулярно FC, АС и GB перпендикулярно CE, то ACBG параллелограмм. Значит <GAC=<GBC=<GBE. <HAC=2<HFC=2<ECM=<EBM. <HAG=(<HAC=<EBM)+(<GAC=<GBE)=<GBM, HA=AC=GB, и AG=BC=BM, значит что треугольники HAG и GBM конгруэнтный. GH=GM и GC перпендикулярно HM, то есть треугольники GCH и GCM конгруэнтный.Ivan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-42370555021900796342015-01-11T12:11:50.143-08:002015-01-11T12:11:50.143-08:00To Jennifer
See above two solutions of problem 88...To Jennifer <br />See above two solutions of problem 885<br />http://gogeometry.blogspot.com/2013/06/problem-885-intersecting-circles.html <br />ThanksAntonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-23384897425874071692015-01-11T11:29:05.426-08:002015-01-11T11:29:05.426-08:00This might seem a bit strange but here goes. I am...This might seem a bit strange but here goes. I am a quilter and my guild has presented us with a challenge to create at 20" block using a specific set of requirements. We were given a color, a number and a noun. Mine is three green circles. When I googled that I came up with your proof as a possibility. What is the solution so that I can use that as part of my block? Thanks!Anonymoushttps://www.blogger.com/profile/18409973820918662158noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-57327816883100491562013-06-08T20:55:08.896-07:002013-06-08T20:55:08.896-07:00Let HF and EM intersect at P.
Apply inversion on C...Let HF and EM intersect at P.<br />Apply inversion on C with any inversion radius. <br />3 tangents are invariants.<br />Circle G becomes line through E'F' and parallel to H'M'<br />Similar on other two circles.<br />Triangle P'H'M' is similar to C'E'F'.<br />Remainings are straight forward. <br />QEDnakahttps://www.blogger.com/profile/11277356476170372732noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-51669827595474737602013-06-08T15:23:53.734-07:002013-06-08T15:23:53.734-07:00build :
EM, FE, FH
FHC=FCE=EMC
FEC=FHC => FHC~F...build :<br />EM, FE, FH<br />FHC=FCE=EMC<br />FEC=FHC => FHC~FCE => HC/CE=FC/FE => HC=CE*FC/FE<br />EFC=ECM => FEC~CEM => CM/FC=CE/FE => CM=CE*FC/FE<br />HC=CM => M midPoint of HMAnonymoushttps://www.blogger.com/profile/11694885671410014993noreply@blogger.com