tag:blogger.com,1999:blog-6933544261975483399.post1451999865160227289..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 444: Tangent circles, Secant line, Chords, Angles, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-6933544261975483399.post-41413859012106819432016-06-18T23:01:55.313-07:002016-06-18T23:01:55.313-07:00Problem 444
I design the common external tangent ...Problem 444<br />I design the common external tangent at point A xAx’.Then <xAB=<AEB, <xAC=<ADC=<AED+<br /><DAE so <BAC=<DAE.<br />APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-72958256064611149462012-10-02T10:05:53.066-07:002012-10-02T10:05:53.066-07:00A new solution with inversion!
Let us take point ...A new solution with inversion!<br /><br />Let us take point A as pole and with any radius invert de figure.<br />You will see without problems that the inverted problem is "Let us consider PQR a triangle and K its circumcircle, take a point U in arc QR not containing A such that the tangent line to K through U is paralel to QR. Prove that AU is bisector of <A" witch si obviously true.Editorhttps://www.blogger.com/profile/18079120609888942700noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-71450054514630501942012-08-06T13:54:11.562-07:002012-08-06T13:54:11.562-07:00Yes, sorry abaut that mistake (:
PD: I wonder why...Yes, sorry abaut that mistake (:<br /><br />PD: I wonder why Blogger changed mi name randomly to "Editor", so far was "mathreyes"<br /><br />regards.Editorhttps://www.blogger.com/profile/18079120609888942700noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-66782721907842012462012-08-05T11:38:36.316-07:002012-08-05T11:38:36.316-07:00To Editor (problem 444), in your solution T should...To Editor (problem 444), in your solution T should be parallel to BEAntonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-72222436834015592052012-08-05T07:40:01.380-07:002012-08-05T07:40:01.380-07:00A lot simplier: Let T be a tangent line to smaller...A lot simplier: Let T be a tangent line to smaller circle wich is parallel to AB and let P be his touchng point with this circle.<br />AP bisects both <CAD and <BAE, so AC and AD are isogonals.Editorhttps://www.blogger.com/profile/18079120609888942700noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-40424604964508274762010-05-02T00:19:50.622-07:002010-05-02T00:19:50.622-07:00extend AC to K, and AD to L ( K, L on circle O )
j...extend AC to K, and AD to L ( K, L on circle O )<br />join K to L <br />join A and E to O', O<br />join C to P ( P point AE meet circle O' )<br />=><br />AO'P = AOE (1) (AO'P and AOE isoc and OAE common)<br />ACP = 1/2 AO'P<br />AKE = 1/2 AOE<br />from (1)<br />ACP = AKE (2)<br /><br />PCD = DAP = EKL = α' (3) ( have same arc EL, PD)<br />from (2) and (3)<br />ACE = AKL<br />=><br />BE // KL<br />=><br />BEK = α = EKL = α' as alternate angles<br />-----------------------------------------------c .t . e. onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-59928815265980532982010-05-01T09:24:45.758-07:002010-05-01T09:24:45.758-07:00Let AB,AE intercept circle O' at B',E'...Let AB,AE intercept circle O' at B',E'<br /><br />Method 1:<br />mAEB=mAAB=mACB' (can also be explained by homothety)<br />mEDA=mCB'A<br />thus α=α'<br /><br />Method 2:<br />consider homothety, B'E'//BE => B'C=E'D => α=α'Jankonyexnoreply@blogger.com