Projective treaty of conical/In a pappusien plan

the geometry is art to reason just on false figures. (author to be found)

In a plan arguésien one cannot define the concept of conical, only the concept of Bi-together of points of which all the alternate hexagrammes are pascaliens. Now, if one grants the right to use the axiom of Pappus, all is simplified, the conical ones exist, the tangent in a point also, the polar ones too.

Preliminary axioms

Anticipation: hexagon of Brianchon and hexagramme of Pascal

 <-----Return to the projective Treated of conical the

The definition of a pascalien unit

What one knows of the approach of Pascal himself.

(see projective Traité of conical the, §Ce that one knows approach of Pascal himself unrolling box)

Projective plan arguésien and absence of the conical ones.

to see  <------Return to the projective Treated of conical the

Summarized. In a projective plan arguésien the conical ones do not exist. One can only make there 6 permutations (out of 60) of a hexagramme pascalien.

Projective plan pappusien and conical.

The first step will be to show that all Permutation of a Hexagramme of Pascal preserves the property of pascalinity. Then all the consequences come rather easily (tangent in a point, intersection with a line, polar). The definition of conical in terms of Hexagramme of Pascal is close to those by the constancy of the Anharmonic ratio, by the quadratic equations, the homographic transformation of a beam of right-hand sides.

When we currently read handbooks of projective geometry of an absolute formalism, without figure, we think that the students cannot, for lack of adequate training, to include/understand the substance-Anne Boyé of it, For science n° November 21st, th and th 2004 - February 2005.

Substitution of two neighbors.

The principal theorem is: In a projective plan pappusien, if a hexagramme is pascalien, then the hexagramme obtained in substituent two close points is him also pascalien. On the figure opposite the conical one is simply evoked in form of " Patatoïde ", one works only on 6 of his points, one exchanges the points P3 and P4 to study the hexangle ordered P1 P2 P4 P3 P5 P6 P1. In other words if the plan is pappusien and if the product $\backslash \; begin\; \{pmatrix\}\; 1\; \backslash \; \backslash \; 3\; \backslash \; \backslash \; 5\; \backslash \; end\; \{pmatrix\}\; \backslash \; otimes\; \backslash \; begin\; \{pmatrix\}\; 4\; \backslash \; \backslash \; 6\; \backslash \; \backslash \; 2\; \backslash \; end\; \{pmatrix\}\; =\; \backslash \; begin\; \{pmatrix\}\; C\; \backslash \; \backslash \; B\; \backslash \; \backslash \; has\; \backslash \; end\; \{pmatrix\}$ gives a tripoint CBA then aligned the product $\backslash \; begin\; \{pmatrix\}\; 1\; \backslash \; \backslash \; 4\; \backslash \; \backslash \; 5\; \backslash \; end\; \{pmatrix\}\; \backslash \; otimes\; \backslash \; begin\; \{pmatrix\}\; 3\; \backslash \; \backslash \; 6\; \backslash \; \backslash \; 2\; \backslash \; end\; \{pmatrix\}\; =\; \backslash \; begin\; \{pmatrix\}\; Y\; \backslash \; \backslash \; Z\; \backslash \; \backslash \; has\; \backslash \; end\; \{pmatrix\}$ gives also an aligned tripoint YZA.

Of course, once shown the invariance of the property to a substitution of 2 neighbors, taking into account the invariance of the property to the circular shift of the 6 points, one can generate the 60 permutations of the hexagramme and deduce from it that the property is true for all. In other words, In a projective plan pappusien, if an ordered hexagramme is pascalien, then the 59 others hexagrammes ordered is it.

A complete squaring in a projective plan pappusien.

The figure of the hexagramme pascalien in which one distinguishes 2 points (P5 and P6) that one connects to the four other points forms a squaring 4x4 which does not fail to point out a tiled ground of certain paintings of the rebirth but which would be seen in skew. : Ni to point out to us the method of Leone Battista Alberti, with use of the diagonals with 45° which converge with the points of distance of the Perspective. The figure of Alberti is a particular case of projective squaring in which the " punctus lateralis" would be ad infinitum on the horizon, the diagonal average 12-34 would contain also the point P23, large diagonal P1-P4 would contain also the points P2 and P3 (idem for the other average diagonals and the other large diagonal), the conical one which contains the P1,2,3,4,5,6 points would be degenerated into a bidroite (P5-P6) - (P1-P4), which does not prevent the hexagramme 1-2-5-4-3-6-1 from being pascalien. Displeased in the figure of Alberti the 4 points are aligned on the horizon (centricus, lateralis ad infinitum, and pyramidis left and right).

Puisque we are in the universe of the projective plan not-Euclidean and not-closely connected (not of distance, not of angle, not of parallelism) we must give up the idea of paving by equal squares, give up the idea of squares at parallel sides and give up the idea of diagonals with 45°.Que remain-you it? A squaring not-parallel 4x4, and various diagonals (we will retain 12 of them) who have properties minimalistes.

In this figure, there is an ordered hexagramme: P5-1-2-6-4-3-5. Of two things one, either it is not pascalien, or it is it. If it is it, that can one say various diagonal lines of the squaring, for example 12-21, 13-31, 14-41, 23-32, 24-42, 34-43? or 13-24, 12-34, 2-3, 1-4, 21-43, 31-42? and well of autres

Does there exist a tangent in a projective plan pappusien?

Right/conical intersection in a projective plan pappusien.

Polar in a projective plan pappusien.

Hexagrammes mystics in Euclidean geometry.

A case relates to the Cercle of Euler, or Feuerbach, particular case of conical in connection with which one can show that the Droite of Euler is the line of Pascal of hexagramme mystical H1-I2-H3-I1-H2-I3-H1.

In the same spirit it is interesting to examine the relative properties of the 2 trigonal ones (ABC) and H1I2, H2I3, H3I1: are they in Arguesian prospect?

One can be posed the same kind of questions in connection with the Point of Lemoine and conical possible which would pass by the 6 feet of the céviennes concerned.

In a more general way one can consider 6 points: feet of the céviennes of a point P and feet of the céviennes of sound Combined isogonal Q; ref. biblio for example the geometry of the triangle, Yvonne and Rene Left, Hermann, 1997, chapters starting from " isogonalité".

Traditional Hexagramme and definitions of conical.

If, in a projective plan pappusien, we can define the concept of conical thanks to the hexagramme pascalien, it should be seen whether this concept corresponds with the traditional definitions of the conical ones.

Bond with the anharmonic ratios.

Starting from a hexagramme pascalien, can inférer that there is conservation of the anharmonic ratio on conical? The answer is yes if the plan is pappusien. The demonstration is done on the figure of the complete squaring seen previously. En employing the traditional notations of the anharmonic ratios in Euclidean geometry, one starts with birraport RO of the 4 lines of the P5 beam: (P5-P1), (P5-P2), (P5-P3) and (P5-P4), RO= ((P5-P1), (P5-P2), (P5-P3), (P5-P4))= ((P5-P1), (P5-12), (P5-13), (P5-14))= ((V-P1), (V-12), (V-13), (V-14))= ((V-P1), (V-21), (V-31), (V-41))= ((P6-P1), (P6-21), (P6-31), (P6-41))= ((P6-P1), (P6-P2), (P6-P3), (P6-P4)), V being illustrated by the question mark on the image.

We thus established the equality of two birapports of beams

((P5-P1), (P5-P2), (P5-P3), (P5-P4)) = ((P6-P1), (P6-P2), (P6-P3), (P6-P4))

One from of deduced that, knowing 5 points P1, 2 3 4 5, one describes the conical one in faisanr to move the sixth point with constant anharmonic ratio.

Bond with the second degree.

One can establish this bond in homogeneous coordinates initially, then to pass to the Cartesian coordinates. The small easy ways of calculation deserve explanations of linear algebra. In homogeneous coordinates we will write the equality of the anharmonic beams, the conical one being defined by 5 fixed points, the sixth being mobile. A Anharmonic ratio is expressed with 4 products of matrices (matriceligne * matricecolonne) or a scalar Produit if one prefers; a matrix-line represents a line of the beam, right-hand side which in fact is defined by a starting point and a point of arrival. Into homogeneous coordinates this results in a " vector Product ". If the two operations are connected, one thus (a vector product) multiplied scalairement by (a vector-column), which amounts making the Déterminant of the 3 vectors in question, put in column. Our equality of anharmonic ratios will be made up of an equation between 8 determinants 3x3.

In conical which contains the 6 P1 points, 2,3,4,5,6 (and thus the cords d51, d52, d53 etc) the anharmonic ratio of items 1,2,3,4 seen of item 5 can be expressed by

$\backslash \; rho\_5\; =\; \backslash \; frac\; \{(\backslash \; vec\; \{d\_\; \{51\}\}\; \backslash \; cdot\; \backslash \; vec\; \{P\_2\})\; *\; (\backslash \; vec\; \{d\_\; \{53\}\}\; \backslash \; cdot\; \backslash \; vec\; \{P\_4\})\}\; \{(\backslash \; vec\; \{d\_\; \{52\}\}\; \backslash \; cdot\; \backslash \; vec\; \{P\_3\})\; *\; (\backslash \; vec\; \{d\_\; \{54\}\}\; \backslash \; cdot\; \backslash \; vec\; \{P\_1\})\}$, that one can transform into

$\backslash \; rho\_5\; =\; \backslash \; frac\; \{\{\backslash \; rm\; Det\}\; (P\_5,\; P\_1,\; P\_2)\; *\; \{\backslash \; rm\; Det\}\; (P\_5,\; P\_3,\; P\_4)\}\; \{\{\backslash \; rm\; Det\}\; (P\_5,\; P\_2,\; P\_3)\; *\; \{\backslash \; rm\; Det\}\; (P\_5,\; P\_4,\; P\_1)\}$

In the same way the anharmonic ratio of same items 1,2,3,4 seen of the point can be expressed by

$\backslash \; rho\_6\; =\; \backslash \; frac\; \{\{\backslash \; rm\; Det\}\; (P\_6,\; P\_1,\; P\_2)\; *\; \{\backslash \; rm\; Det\}\; (P\_6,\; P\_3,\; P\_4)\; \}\; \{\{\backslash \; rm\; Det\}\; (P\_6,\; P\_2,\; P\_3)\; *\; \{\backslash \; rm\; Det\}\; (P\_6,\; P\_4,\; P\_1)\}$

Ecrire that these two birapports are equal amounts writing

$\{\backslash \; rm\; Det\}\; (P\_5,\; P\_1,\; P\_2)\; *\; \{\backslash \; rm\; Det\}\; (P\_5,\; P\_3,\; P\_4)\; *\; \{\backslash \; rm\; Det\}\; (P\_6,\; P\_2,\; P\_3)\; *\; \{\backslash \; rm\; Det\}\; (P\_6,\; P\_4,\; P\_1)\; =\; \{\backslash \; rm\; Det\}\; (P\_6,\; P\_1,\; P\_2)\; *\; \{\backslash \; rm\; Det\}\; (P\_6,\; P\_3,\; P\_4)\; *\; \{\backslash \; rm\; Det\}\; (P\_5,\; P\_2,\; P\_3)\; *\; \{\backslash \; rm\; Det\}\; (P\_5,\; P\_4,\; P\_1)$

By clarifying the homogeneous coordinates of the 6 points, the equation of equality of the two anharmonic ratios is

. If the first 5 points are fixed and the 6th mobile of variable coordinates X, Y, Z instead of X6, Y6, Z6, one obtains a homogeneous equation of the second degree, therefore conical is a curve of the second degree which one can develop and transform if it is wished.

Bond with the monofocale definition

The monofocale definition of conical, still called the definition by hearth and director (see Conical), can ^tre above deduced by calculations from the analytical definition from the second degree.

Bond with the homographic beams

Examples can be found in the pages:

The demonstration is done on the figure of the complete squaring seen previously. If the hexagramme 1254361 is pascalien, then diagonals 12-21, 13-31 and 14-41 convergent in a point V illustrated here by the question mark. However this property corresponds to a homographic correspondence between beams defined thus

the beam starting is of center P5,

it is cut by line P6-P14

starting from these intersections one defines a homographic beam of center V,

which in its turn is cut by line P5-P14

starting from these intersections one defines the beam of arrival of P6 center which is homographic for him.

the intersection of a line of the P5 beam starting and of its homographic of the P6 beam of arrival generates conical.

Remark. In any axiomatic rigor to show this property one needs also to be in a fundamental projective Plan, which will ensure us that the preceding definition of the unidimensional transformation between the faisceaus P5 and P6 is well an homography perfectly defined by the 5 points P1,2,3,6,5, item 4 being the variable intersection of a line of the P5 beam starting and of its homographic of the P6 beam.

Bond with the car-polarity

Projective plan fundamental and conical.

Projective plan homogeneous and conical.

In such a context, the properties of conical rise from those of the equations of the second degree in a commutative body. Conical specific is defined by a quadratic form of the homogeneous coordinates of a point. Conical tangential is defined by a quadratic form of the homogeneous coordinates of a line. The polarity is defined by the means of the harmonic report/ratio. Problems of intersection of conical and a line, or tangents, can be elegantly simplified by considerations of anharmonic ratio.

But the conical ones are not necessarily the continuous curves to which the Euclidean Géométrie accustomed us, the conical ones can be discrete whole of dotted lines, so that the concept of tangent (Tangente (geometry)) in M is not defined as the limiting position of the secant when the point NR tightens towards Mr. All depends on the Corps (mathematical) commutative used.

To deepen.

• Axioms of projective plans
• Axioms of projective/homogeneous plans
• Axioms of projective plans/Continuation of the axioms
• Treated projective of conical the (in a projective plan arguésien
• Treated of conical/In a plan pappusien
• Blaise Pascal
• a site where many developments on the Theorem of Pappus are given: Marvellous Pappus

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