tag:blogger.com,1999:blog-6933544261975483399.post1027152449503144342..comments2024-03-19T00:02:30.728-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1189: Circle, Tangent Line, Secant, Chord, Collinear PointsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger13125tag:blogger.com,1999:blog-6933544261975483399.post-75915738598356812122019-05-07T21:30:44.142-07:002019-05-07T21:30:44.142-07:00See below for the link of problem 1189
https://ph...See below for the link of problem 1189<br /><br />https://photos.app.goo.gl/r2trcF4VxiqcWSjs8<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-73075160727715802352019-05-06T06:32:59.177-07:002019-05-06T06:32:59.177-07:00Unfortunately Peter Tran's link does not work,...Unfortunately Peter Tran's link does not work, so I don't know the identity of point P which is not on your drawing.<br />Thanks for helping.Gewürtztraminer68https://www.blogger.com/profile/00329980114313015297noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-78778837315138134612017-11-23T11:24:15.940-08:002017-11-23T11:24:15.940-08:00I would like to see a proof with elementary euclid...I would like to see a proof with elementary euclidean geometry without using poles and polars.<br /><br />Thank you very much.<br /><br />Manliomanliohttps://www.blogger.com/profile/08407240425765787841noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-46425143283091355542016-08-18T04:41:35.034-07:002016-08-18T04:41:35.034-07:00Let's denote the intersection of EG and DF as ...Let's denote the intersection of EG and DF as X. Clearly BC is the polar of A. Also from Brokard's theorem we have that HX is the polar of A. So B, C, H and X are all collinear since they lay on the polar of A. Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-64614251766926163742016-06-02T18:33:54.520-07:002016-06-02T18:33:54.520-07:00Dear Peter, following Ivan's solution using &...Dear Peter, following Ivan's solution using "Google Translate":<br /><br />Let P, the intersection of the circumscribed circle and the circumscribed circle FOE DOG, <DEG = a, and <FGE = b.<br /><EPG = <EPO + <GPO.<br />According to the properties of a circle, <EPO = <EFO = 90-b, and <GPO = <GDO = 90-a, then <EPG = 180-a-b = <EAG, then point E, P, A, G lies on a circle.<br />Thus <APG = <AEG = a because EPAG inscribed quadrilateral.<br /><OPA = <GPO + <APG = (90-a) + (a) = 90 on the previous findings.<br />This means that P also lies on a circle intersecting points B, A, C, O.<br />Regarding the circumference O, EF inversion circumference FOE, DG inversion DOG circumference, BC BACO circumference inversion and inversion H P.<br />If P lies on the circle BACO, should H It lies on BC.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-82128206302075274422016-06-02T18:19:23.264-07:002016-06-02T18:19:23.264-07:00can somebody translate the solution to English ?can somebody translate the solution to English ?Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-55704714330814267972016-06-01T12:43:17.647-07:002016-06-01T12:43:17.647-07:00is Mr. Bazarov's solution correct?is Mr. Bazarov's solution correct?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-9004541633919549752016-05-28T15:41:48.214-07:002016-05-28T15:41:48.214-07:00Пусть P, пересечение описанная окружность FOE и оп...Пусть P, пересечение описанная окружность FOE и описанная окружность DOG, <DEG=a, и <FGE=b. <br /><EPG=<EPO+<GPO.<br />По свойства окружность, <EPO=<EFO=90-b, и <GPO=<GDO=90-a, значит <br /><EPG=180-a-b=<EAG, значит точки E,P,A,G лежит на одной окружность.<br />Таким образом <APG=<AEG=a потому что EPAG вписанный четырехугольник. <br /><OPA=<GPO+<APG=(90-a)+(a)=90 по предыдущий выводы. <br />Значит что P также лежит на окружность пересекающий точки B, A, C, O.<br />Относительно окружность O, EF инверсии окружность FOE, DG инверсии окружность DOG, BC инверсии окружность BACO, и H инверсии P. Если P лежит на окружность BACO, H должен лежит на BC.Ivan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-15033674819134799452016-03-10T05:25:17.725-08:002016-03-10T05:25:17.725-08:00Some hints?Some hints?Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-15698834475318907242016-03-08T09:38:10.780-08:002016-03-08T09:38:10.780-08:00YesYesAntonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-56941339339407209912016-03-08T08:09:05.494-08:002016-03-08T08:09:05.494-08:00Dear Antonio - do u have another proof for this?
...Dear Antonio - do u have another proof for this?<br /><br />Perhaps extension of Pascal's Theorem?Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-32982248454008915472016-02-24T03:09:42.407-08:002016-02-24T03:09:42.407-08:00Polar point A is BC,Η is conjugate to A.AND A loca...Polar point A is BC,Η is conjugate to A.AND A located in BC.APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-5660841195900296152016-02-23T17:37:44.567-08:002016-02-23T17:37:44.567-08:00http://s10.postimg.org/i7q916o6x/pro_1189.png
See...http://s10.postimg.org/i7q916o6x/pro_1189.png<br /><br />See below for basic properties of complete quadrilateral and pole and polar of a circle . <br />1. ADEGFP is a complete quadrilateral with diagonals DG meet EF at H and PH meet DE and FG at N and M ( see sketch) then (ANDE)=(AMFG)=-1 or AD/AE=ND/NE and AF/AG=MF/MG .<br />2. A secant from A meet circle O at D and E and cut BC (polar of A) at N then (ANDE)=-1 or AD/AE= ND/NE.<br />Conversely if exist a point N on DE such that (ANDE)=-1 then N will be on BC.<br /><br />Applying these properties to our problem, PH cut DE at N and FG at M then (ANDE)=(AMFG)= -1 (property #1) and M and N will be on BC (polar of pole A) (property #2) . so B, C and H are collinear.<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com