Monday, May 9, 2011

Geometry Problem 827: Brianchon Corollary, Circumscribed Hexagon, Concurrency lines

Classical Theorem
Click the figure below to see the complete classical theorem.

 Online Geometry: Brianchon Corollary, Circumscribed Hexagon, Concurrency lines.

8 comments:

  1. Geometry Problem 827: Brianchon Corollary, Circumscribed Hexagon, Concurrency lines.

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  2. let GM,JK intersect at X.
    Say 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.

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  3. To W Fung
    Refer 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.

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    Replies
    1. This would be a common properties for Pole and Polar application. If you insist, you may refer to Brokard's Theorem.

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    2. what is meant by polar

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    3. 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.

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  4. http://img844.imageshack.us/img844/6590/problem827.png

    Let 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

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  5. Let AB and CD meet at P, AF and DE meet at Q.
    Then 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.

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