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Introduction Proof of Pappus-Pascal theorem Proof of Desargues theorem Conclusion

KARAMATA’S PROOFS OF PAPPUS-PASCAL

AND DESARGUES THEOREMS

Aleksandar M. NikolicUniversity of Novi Sad

Faculty of Technical Sciences21000 Novi Sad, Voivodina, Serbia

e-mail: [email protected]

Aleksandar M. Nikolic University of Novi Sad Faculty of Technical Sciences 21000 Novi Sad, Voivodina, Serbia e-mail: [email protected]

KARAMATA’S PROOFS OF PAPPUS-PASCAL AND DESARGUES THEOREMS


1 Introduction

2 Proof of Pappus-Pascal theorem

3 Proof of Desargues theorem

4 Conclusion




Jovan Karamata(1902-1967)

The Proof of Hardy-Littlewood TauberianTheoremUber die Hardy-Lttlewoodschen Umkehrungen

des Abelschen Statigkeitssatzes, MathematischeZeitschrift, B. 32, H. 2, 1930, pp. 319-320.

Foundation of the theory of regularly and slowlyvarying functionSur un mode de croissance reguliere des

fonctions, Mathematica (Cluj), Vol. 4, 1930, pp.38-53.




In paper Uber die anwendung der komplexen zahlen in der elementar-

geometrie, and in the faculty textbook Complex Numbers, (1950), JovanKaramata maintains that, in planimetry, a complex number can assumethe role of the vector, whereby in definition and solving of certainproblems, there appears a product

ab = A + Bi

where a is a conjugated number of a, while a = |a|eϕi and b = |b|eψi

are complex numbers which correspond to free vectors ~a and ~b. Usingtrigonometric form of a and b, Karamata expresses A and B as

A = |a||b| cosα and B = |a||b| sinα, α = ψ − ϕ

denotes them as

A = (a ⊥ b) i.e. B = (a | b)

and designated orthogonal product and parallel product.




By means of these symbols it is easy to express orthogonality andparalellness conditions for the two vectors ~a and ~b:

~a ⊥ ~b if (a ⊥ b) = 0 and ~a || ~b if (a | b) = 0,

and since(a ⊥ b) = (a | ib) and (ia ⊥ b) = (a | b)

where i is the imaginary unit, it follows that one product can always bereduced to another one.Of all the properties of Karamata’s products we hereby mention thefollowing:

λ(a | b) = (λa | b) = (a | λb), λ ∈ R.




Commutative law - which holds only for orthogonal product

(a ⊥ b) = (b ⊥ a),

while in the case of parallel product

(a | b) = −(b | a),

- and distributive law for addition, which holds for both products

((a + b) ⊥ c) = (a ⊥ c) + (b ⊥ c), ((a + b) | c) = (a | c) + (b | c).

It also holds relation between the three coplanar vectors

(b | c)a + (c | a)b + (a | b)c = 0,

where at least one of (b | c), (c | a), (a | b) is not equal 0.




1 Introduction



4 Conclusion




Blaise Pascal(1623-1662) wrotea treatise on conicsections Essai pour

les coniques in1639.

Theorem

If a hexagon is inscribed in a conic section, then, when

extended, the pairs of its opposite sides will intersect

in three collinear points.




Pappus theorem

If the conic section degenerates into two lines, that is, if the hyperboladegenerates into its asymptotes, then one gets the case described byPappus, a mathematician of the Alexandrian school.

Theorem

If the six vertices of a hexagon lie alternately on the two straight lines,

the three pairs AB’ and BA’, AC’ and CA’, BC’ and CB’ of opposite sides

intersect at three collinear points.




In his proof, Karamata considers a special case of the Pappus theorem –which he called Pappus-Pascal theorem – when the vanishing line lies ininfinity:

Theorem

If the points A, B ′ and C lie on the same line, and points A′, B and C ′

lie on another, and if

AB ‖ A′B ′ and BC ‖ B ′C ′

then also holds AC ′ ‖ A′C .




In order to prove this theorem, Karamata divides figure into twoquadrangles and reduces the Pappus-Pascal theorem to the following

Theorem

If ~a ‖ ~a ′, ~b ‖ ~b ′, ~c ‖ ~c ′,

and

~a + ~b ‖ ~b ′ + ~c ′, ~b + ~c ‖ ~a ′ + ~b ′,

then

~a + ~b + ~c ‖ ~a ′ + ~b ′ + ~c ′.




Karamata’s formulation

Knowing that~a ‖ ~b is equivalent to (a | b) = 0

Karamata claims that to prove proposition, one need to show

Theorem

If

(a | a′) = 0, (b | b′) = 0, (c | c ′) = 0

and

(a + b | b′ + c ′) = 0, (b + c | a′ + b′) = 0,

then

(a + b + c | a′ + b′ + c ′) = 0.




Using only the assumptions of the theorem and the property ofdistribution of the parallel product, he gets

(a + b + c | a′ + b′ + c ′) = (a | a′) + (a | b′) + (a | c ′) +

(b | a′) + (b | b′) + (b | c ′) +

(c | a′) + (c | b′) + (c | c ′) =

= ((a | b′) + (a | c ′) + (b | b′) + (b | c ′)) +

((b | a′) + (b | b′) + (c | a′) + (c | b′)) =

= (a + b | b′ + c ′) + (b + c | a′ + b′) = 0

where from the Pappus-Pascal theorem directly follows.




1 Introduction



4 Conclusion




Gerard Desargues(1591-1661)

In 1648, as anappendix to a book

of his friend,A.Bosse

(1602-1676)Maniere universelle

de M. Desargues,

pour pratiquer la

perspective.

Theorem

For two triangles which are perspective from a point,

there holds that the pairs of homologous sides AB and

A′B ′, BC and B ′C ′, AC and A′C ′ or their extensions,

intersect at three collinear points.




Karamata considers a special case of the inverse Desargues theorem,when the homologous sides of two triangles are parallel, i.e. when thevanishing line lies in infinity, which he formulats in a following way

Theorem

If the homologous sides of triangles ABC and A′B ′C ′ are parallel, i.e.

AB ‖ A′B ′, BC ‖ B ′C ′ and CA ‖ C ′A′

then the lines passing through homologous vertices AA′, BB ′ and CC ′

intersect at one point S.




In order to prove thus formulated theorem, Karamata convenientlydevides this figure and observes two quadrangles whose sides anddiagonals are considered as free vectors, and arrives at the following formof Desargues theorem:If

~a ‖ ~a ′, ~b ‖ ~b ′, ~c ‖ ~c ′,

and

~a + ~b ‖ ~a ′ + ~b ′, ~b + ~c ‖ ~b ′ + ~c ′,

then

~a + ~b + ~c ‖ ~a ′ + ~b ′ + ~c ′,

i.e. the parallelness of three homologous sides and homologous diagonalsimplies the parallelness of the fourth sides of the quadrangles considered.Expressed in symbols of parallel product and knowing that

~a ‖ ~b is equivalent to (a | b) = 0

the theorem reduces to the




Karamata’s formulation

Theorem

If

(a | a′) = 0, (b | b′) = 0, (c | c ′) = 0

and

(a + b | a′ + b′) = 0, (b + c | b′ + c ′) = 0,

then

(a + b + c | a′ + b′ + c ′) = 0.




In the direct proof he gets

(a+b+c | a′+b′+c ′) = (a+b | a′+b′)+(b+c | b′+c ′)+(c+a | c ′+a′),

wherefrom, based on assumptions that the first two right-hand side summembers equal 0, there follows that to prove Desargues theorem, oneneeds to prove that

(c + a | c ′ + a′) = 0.

As there exist two real numbers m and n such that ~c = m~a + n~b forevery (a | b) 6= 0, and as for the three coplanar vectors ~a, ~b and ~c holds

(a | b)c + (b | c)a + (c | a)b = 0,

Karamata took arbitrary real numbers p, q and r

p = (b | c), q = (c | a), r = (a | b), p 6= 0, r 6= 0

and finally gets




(pa + qb + rc | pa′ + qb′ + rc ′) =

pq(a + b | a′ + b′) + qr(b + c | b′ + c ′) + rp(c + a | c ′ + a′).

However, since for thus selected p, q, r , the left-hand side as well as thefirst two sum members of the right-hand side equal 0, there follows that

rp(c + a | c ′ + a′) = 0,

and since r 6= 0 and p 6= 0 there follows

(c + a | c ′ + a′) = 0,

which directly proves the Desargues theorem.




But Karamata shows that, by introducing an intermediate quadrangle,i.e. the three auxiliary vectors ~a ′′, ~b ′′ and ~c ′′, Desargues theorem can bederived by two consecutive applications of Pappus-Pascal theorem.Namely, if the auxiliary vectors are chosen so that, with regard to vectors~a, ~b and ~c , they satisfy the Pappus-Pascal theorem, i.e. that

(a | a′′) = 0, (b | b′′) = 0, (c | c ′′) = 0,

and(a + b | b′′ + c ′′) = 0, (b + c | a′′ + b′′) = 0,

then(a + b + c | a′′ + b′′ + c ′′) = 0.







However

(a′′ | a′) = 0 since (a′′ | a) = 0 and (a | a′) = 0,

so by analogous reasoning one can claim that

(b′′ | b′) = 0, (c ′′ | c ′) = 0,

(b′′ + c ′′ | a′ + b′) = 0, (a′′ + b′′ | b′ + c ′) = 0,

and from the second application of the Pappus-Pascal theorem - thistime to vectors ~a ′, ~b ′ and ~c ′, there follows that

(a′′ + b′′ + c ′′ | a′ + b′ + c ′) = 0,

wherefrom directly follows

(a + b + c | a′ + b′ + c ′) = 0,

that is, the Desargues theorem.




1 Introduction



4 Conclusion




As Karamata himself notes in the paper, the terminology used informulation and proof of the Pappus-Pascal and Desargues theorem,implies that it is possible to prove Desargues theorem for a planar case,using all 8 projective axioms of incidence.On the other hand, the proof cannot be completed with just the first fiveaxioms of incidence. If those 5 axioms are complemented with thePappus theorem, which is attributed the role of axiom, it is possible toprove Desargues theorem, which G.Hassenberg showed in 1905 (Beweis

des Desarguesschen Satzes aus dem Pascalischen, MathematischeAnnalen, 61).Since his proof that the vanishing line lies in infinity, was predominantlybased on the distribution law of parallel product, Karamata claims thefollowing:




Indeed, if we want to get rid of parallelness in the proof,instead of the line that lies in infinity, we should suppose thatthe vanishing line lies in finiteness, and interpret the wholeproof projectively... In essence, this proof features the conceptof parallel product which is manifested in the form ofcoordination between the set of four points and the set of realnumbers, for which the analogon of distributive law holds andit disappears when the lines defined by these points areparallel, i.e. when they intersect at infinity.




Originality of his approach to the topic, coupled with simplicity andelegance of the proofs shown, speak best not only in favour ofKaramata’s exquisite giftedness for mathematics, but also of his widemathematical knowledge and his versatile mathematical interest.However, his papers and results in geometry were somewhat unjustlyoverlooked. Let us mention that Karamata in his paper entitled Eine

elementare Herleitung des Desarguesschen Satzes aus dem Satze von

Pappos-Pascal, Elemente des Mathematik, 5, 1950, 9-10 (in German)gave planimetric proof of Desargues theorem applying the Pappus-Pascaltheorem three times.

