Classical Theorem

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## Monday, May 9, 2011

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## Monday, May 9, 2011

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

Online Geometry theorems, problems, solutions, and related topics.

Labels:
Brianchon theorem,
circumscribed,
concurrent,
corollary,
diagonal,
hexagon

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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.