Classical Theorem
Click the figure below to see the complete classical theorem.
Monday, May 9, 2011
Geometry Problem 827: Brianchon Corollary, Circumscribed Hexagon, Concurrency lines
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Classical Theorem
Click the figure below to see the complete classical theorem.
Geometry Problem 827: Brianchon Corollary, Circumscribed Hexagon, Concurrency lines.
ReplyDeletelet GM,JK intersect at X.
ReplyDeleteSay GK and JM intersects at N. N lies on the polar of X.
Also, A is the pole of polar GM and D is the pole of polar JK.
By LaHire Theorem, AD is the polar of the intersection of GM,JK (i.e. Point X)
Therefore, A,N,D are collinear.
Q.E.D.
To W Fung
ReplyDeleteRefer to line 2 "Say GK and JM intersects at N. N lies on the polar of X."
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.
Please explain.
This would be a common properties for Pole and Polar application. If you insist, you may refer to Brokard's Theorem.
Deletewhat is meant by polar
DeleteThe 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.
Deletehttp://img844.imageshack.us/img844/6590/problem827.png
ReplyDeleteLet DE cut AF at P and AB cut DC at Q
Let N is the intersecting point of AD and PQ
1. Staring from hexagon ABCDEF ,Let B approach Q then H and C coincide to J
Then Hexagon ABCDEF become pentagon AQDEF
- Diagonal AD will stay the same
- Diagonal EB become EQ
- Diagonal FC become FJ
Per Brianchon’s theorem AD, EQ and JF are concurrent at point R ( see sketch)
2. Starting from pentagon AQDEF , let E approach P then L and F coincide to M
Then pentagon AQDEF become quadrilateral AQDP
- Diagonal QE become QP
- JF become JM
- EQ become PQ
Point of concurrent R will become N => PQ, AD and JM will concurrent at N
3. From hexagon ABCDEF , Let C approach Q then B and H coincide to G
Then hexagon become pentagon AQDEF
Similar to step 1, QF, GE and AD will concurrent at point S ( not shown)
4. From pentagon AQDEF, let F approach P then pentagon become quadrilateral QAPD and
L and E coincide to K
Similar to step 2 , point of concurrent S will become N
So PQ, AD, GK and JM will concurrent at point N
Let AB and CD meet at P, AF and DE meet at Q.
ReplyDeleteThen AHDQ is a tangential quadrilateral.
By a theorem, AD,GK,MJ,PQ are concurrent.
i.e. AD,GK,MJ are concurrent at N.
Remark.
The theorem I used can be view as a corollary of Brianchon theorem,
but it can be proved by other arguments.