tag:blogger.com,1999:blog-6933544261975483399.post466300083448301895..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 827: Brianchon Corollary, Circumscribed Hexagon, Concurrency linesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-6933544261975483399.post-2878469372536184892013-01-18T02:22:42.786-08:002013-01-18T02:22:42.786-08:00Let AB and CD meet at P, AF and DE meet at Q.
The...Let AB and CD meet at P, AF and DE meet at Q. <br />Then AHDQ is a tangential quadrilateral. <br /><br />By a theorem, AD,GK,MJ,PQ are concurrent. <br />i.e. AD,GK,MJ are concurrent at N. <br /><br />Remark. <br />The theorem I used can be view as a corollary of Brianchon theorem, <br />but it can be proved by other arguments. Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-68550997673809352872012-12-16T10:12:51.620-08:002012-12-16T10:12:51.620-08:00http://img844.imageshack.us/img844/6590/problem827...http://img844.imageshack.us/img844/6590/problem827.png<br /><br />Let DE cut AF at P and AB cut DC at Q<br />Let N is the intersecting point of AD and PQ<br />1. Staring from hexagon ABCDEF ,Let B approach Q then H and C coincide to J<br />Then Hexagon ABCDEF become pentagon AQDEF<br />- Diagonal AD will stay the same <br />- Diagonal EB become EQ<br />- Diagonal FC become FJ <br />Per Brianchon’s theorem AD, EQ and JF are concurrent at point R ( see sketch)<br />2. Starting from pentagon AQDEF , let E approach P then L and F coincide to M <br />Then pentagon AQDEF become quadrilateral AQDP<br />- Diagonal QE become QP <br />- JF become JM<br />- EQ become PQ<br />Point of concurrent R will become N => PQ, AD and JM will concurrent at N<br />3. From hexagon ABCDEF , Let C approach Q then B and H coincide to G <br />Then hexagon become pentagon AQDEF<br />Similar to step 1, QF, GE and AD will concurrent at point S ( not shown)<br />4. From pentagon AQDEF, let F approach P then pentagon become quadrilateral QAPD and<br />L and E coincide to K<br />Similar to step 2 , point of concurrent S will become N<br /><br />So PQ, AD, GK and JM will concurrent at point N<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-16264186485361018212012-12-12T08:34:24.796-08:002012-12-12T08:34:24.796-08:00The polar line of a point is a line perpendicular ...The polar line of a point is a line perpendicular to line joining the point and the center of the circle, and it must contain the inverse of the point.Editorhttps://www.blogger.com/profile/18079120609888942700noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-76929034524013744182012-12-08T09:50:58.946-08:002012-12-08T09:50:58.946-08:00what is meant by polar
what is meant by polar<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-4595446195179681772012-11-30T04:43:03.965-08:002012-11-30T04:43:03.965-08:00This would be a common properties for Pole and Pol...This would be a common properties for Pole and Polar application. If you insist, you may refer to Brokard's Theorem.W Fungnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-57981360751026492502012-11-29T17:39:57.357-08:002012-11-29T17:39:57.357-08:00To W Fung
Refer to line 2 "Say GK and JM inte...To W Fung<br />Refer to line 2 "Say GK and JM intersects at N. N lies on the polar of X."<br />I am not sure how you get this. If this is the case, we can conclude that A,N, D are collinear since AD is the polar of intersecting point of GM and JK. <br />Please explain.Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-10061828285728057652012-11-28T21:50:56.095-08:002012-11-28T21:50:56.095-08:00let GM,JK intersect at X.
Say GK and JM intersect...let GM,JK intersect at X. <br />Say GK and JM intersects at N. N lies on the polar of X.<br />Also, A is the pole of polar GM and D is the pole of polar JK.<br />By LaHire Theorem, AD is the polar of the intersection of GM,JK (i.e. Point X)<br />Therefore, A,N,D are collinear.<br />Q.E.D.W Fungnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-45373790272292394772012-11-28T18:38:58.282-08:002012-11-28T18:38:58.282-08:00Geometry Problem 827: Brianchon Corollary, Circums...Geometry Problem 827: Brianchon Corollary, Circumscribed Hexagon, Concurrency lines.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.com