Pappus’ Theorem for a Conic and Mystic Hexagonstexpower.sourceforge.net/gallery/Pappus.pdf ·...
Transcript of Pappus’ Theorem for a Conic and Mystic Hexagonstexpower.sourceforge.net/gallery/Pappus.pdf ·...
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Pappus’ Theorem for a Conicand Mystic Hexagons
Ross MooreMacquarie University
Sydney, Australia
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Pappus’ Theorem for a Conicand Mystic Hexagons
Ross MooreMacquarie University
Sydney, Australia
Pappus’ Theorem is a well-known result for triples of points on two lines in the(projective) plane:
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Theorem. (Pappus)
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Theorem. (Pappus) Given lines l and m in the plane,
l
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m
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Theorem. (Pappus) Given lines l and m in the plane,three distinct points A, B, C on l (but not on m),
l⊗
A ⊗B
⊗C ggggggggggggggggggggggggggggggggggggggggggggggggggggggg
m
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Theorem. (Pappus) Given lines l and m in the plane,three distinct points A, B, C on l (but not on m),and three distinct points D, E, F on m (but not on l),
l⊗
A ⊗B
⊗C ggggggggggggggggggggggggggggggggggggggggggggggggggggggg
m ⊗D ⊗
E⊗
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Theorem. (Pappus) Given lines l and m in the plane,three distinct points A, B, C on l (but not on m),and three distinct points D, E, F on m (but not on l),
construct the intersections L = BF ∧ CE,
l⊗
A ⊗B
⊗C ggggggggggggggggggggggggggggggggggggggggggggggggggggggg
m ⊗D ⊗
E⊗
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~~~~
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⊕L
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Theorem. (Pappus) Given lines l and m in the plane,three distinct points A, B, C on l (but not on m),and three distinct points D, E, F on m (but not on l),
construct the intersections L = BF ∧ CE, M = CD ∧ AF
l⊗
A ⊗B
⊗C ggggggggggggggggggggggggggggggggggggggggggggggggggggggg
m ⊗D ⊗
E⊗
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⊕M
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~~~~
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⊕L
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Theorem. (Pappus) Given lines l and m in the plane,three distinct points A, B, C on l (but not on m),and three distinct points D, E, F on m (but not on l),
construct the intersections L = BF ∧ CE, M = CD ∧ AFand N = AE ∧ BD.
l⊗
A ⊗B
⊗C ggggggggggggggggggggggggggggggggggggggggggggggggggggggg
m ⊗D ⊗
E⊗
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::::
::::
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⊕N
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⊕L
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Theorem. (Pappus) Given lines l and m in the plane,three distinct points A, B, C on l (but not on m),and three distinct points D, E, F on m (but not on l),
construct the intersections L = BF ∧ CE, M = CD ∧ AFand N = AE ∧ BD. Then L, M , N are collinear.
l⊗
A ⊗B
⊗C ggggggggggggggggggggggggggggggggggggggggggggggggggggggg
m ⊗D ⊗
E⊗
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::::
::::
::::
::
⊕N
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Definition. By a Pappus configuration we mean a set of 6 points in theplane, arranged as a pair of triples of points (not necessarily collinear) (A, B, C)and (D, E, F ), such that the intersections (L, M, N), of pairs of lines join-ing points taken as in Pappus’ Theorem, are collinear.
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Definition. By a Pappus configuration we mean a set of 6 points in theplane, arranged as a pair of triples of points (not necessarily collinear) (A, B, C)and (D, E, F ), such that the intersections (L, M, N), of pairs of lines join-ing points taken as in Pappus’ Theorem, are collinear.
Not so well-known is that if the six points A, B, C, D, E, F lie on anon-singular conic, then (A, B, C), (D, E, F ) form a Pappus configuration.
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Definition. By a Pappus configuration we mean a set of 6 points in theplane, arranged as a pair of triples of points (not necessarily collinear) (A, B, C)and (D, E, F ), such that the intersections (L, M, N), of pairs of lines join-ing points taken as in Pappus’ Theorem, are collinear.
Not so well-known is that if the six points A, B, C, D, E, F lie on anon-singular conic, then (A, B, C), (D, E, F ) form a Pappus configuration.
⊗A ⊗
B⊗
C
⊗
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E
⊗F
⊕N ⊕
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Remark. The usual Pappus’ Theorem is just the situation whereby the conicdegenerates into a pair of lines.
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Remark. The usual Pappus’ Theorem is just the situation whereby the conicdegenerates into a pair of lines.
Remark. When the six points are ordered as A, F , B, D, C, F the result-ing polygon is just Pascal’s “mystic hexagon”. Alternatively, given a mystichexagon, the Pappus configuration is obtained by taking the even and oddvertices for the groups of three.
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Remark. The usual Pappus’ Theorem is just the situation whereby the conicdegenerates into a pair of lines.
Remark. When the six points are ordered as A, F , B, D, C, F the result-ing polygon is just Pascal’s “mystic hexagon”. Alternatively, given a mystichexagon, the Pappus configuration is obtained by taking the even and oddvertices for the groups of three.
Remark. It is well known that a unique conic can be drawn through any 5points in general position.
Pappus configurations give a way to construct that conic, parametrised by thepoints on a line...
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Given five points A, B, C, D, E in the plane in sufficiently general position,
⊗A ⊗
B⊗
C
⊗
D⊗
E
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Given five points A, B, C, D, E in the plane in sufficiently general position,let N = AE ∧ BD.
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Given five points A, B, C, D, E in the plane in sufficiently general position,let N = AE ∧ BD. Pick P along CD.
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Given five points A, B, C, D, E in the plane in sufficiently general position,let N = AE ∧ BD. Pick P along CD. Let Q = NP ∧ CE.
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Given five points A, B, C, D, E in the plane in sufficiently general position,let N = AE ∧ BD. Pick P along CD. Let Q = NP ∧ CE.Then X = AP ∧ BQ lies on the conic.
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Given five points A, B, C, D, E in the plane in sufficiently general position,let N = AE ∧ BD. Pick P along CD. Let Q = NP ∧ CE.Then X = AP ∧ BQ lies on the conic.Vary P on CD for different points X on the conic:
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Given five points A, B, C, D, E in the plane in sufficiently general position,let N = AE ∧ BD. Pick P along CD. Let Q = NP ∧ CE.Then X = AP ∧ BQ lies on the conic.Vary P on CD for different points X on the conic:
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Given five points A, B, C, D, E in the plane in sufficiently general position,let N = AE ∧ BD. Pick P along CD. Let Q = NP ∧ CE.Then X = AP ∧ BQ lies on the conic.Vary P on CD for different points X on the conic:
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Given five points A, B, C, D, E in the plane in sufficiently general position,let N = AE ∧ BD. Pick P along CD. Let Q = NP ∧ CE.Then X = AP ∧ BQ lies on the conic.Vary P on CD for different points X on the conic:
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Given five points A, B, C, D, E in the plane in sufficiently general position,let N = AE ∧ BD. Pick P along CD. Let Q = NP ∧ CE.Then X = AP ∧ BQ lies on the conic.Vary P on CD for different points X on the conic:
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Given five points A, B, C, D, E in the plane in sufficiently general position,let N = AE ∧ BD. Pick P along CD. Let Q = NP ∧ CE.Then X = AP ∧ BQ lies on the conic.Vary P on CD for different points X on the conic:
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