KARAMATA'S PROOFS OF PAPPUS-PASCAL AND DESARGUES THEOREMSKARAMATA’S
PROOFS OF PAPPUS-PASCAL
AND DESARGUES THEOREMS
Faculty of Technical Sciences 21000 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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
1 Introduction
4 Conclusion
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
Jovan Karamata (1902-1967)
des Abelschen Statigkeitssatzes, Mathematische Zeitschrift, B. 32,
H. 2, 1930, pp. 319-320.
Foundation of the theory of regularly and slowly varying function
Sur un mode de croissance reguliere des
fonctions, Mathematica (Cluj), Vol. 4, 1930, pp. 38-53.
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
In paper Uber die anwendung der komplexen zahlen in der
elementar-
geometrie, and in the faculty textbook Complex Numbers, (1950),
Jovan Karamata maintains that, in planimetry, a complex number can
assume the role of the vector, whereby in definition and solving of
certain problems, there appears a product
ab = A + Bi
where a is a conjugated number of a, while a = |a|ei and b =
|b|eψi
are complex numbers which correspond to free vectors ~a and ~b.
Using trigonometric form of a and b, Karamata expresses A and B
as
A = |a||b| cosα and B = |a||b| sinα, α = ψ −
denotes them as
A = (a ⊥ b) i.e. B = (a | b)
and designated orthogonal product and parallel product.
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
By means of these symbols it is easy to express orthogonality and
paralellness 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 be reduced to another one. Of all the properties of
Karamata’s products we hereby mention the following:
λ(a | b) = (λa | b) = (a | λb), λ ∈ R.
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
Commutative law - which holds only for orthogonal product
(a ⊥ b) = (b ⊥ a),
(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.
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
1 Introduction
4 Conclusion
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
Blaise Pascal (1623-1662) wrote a treatise on conic sections Essai
pour
les coniques in 1639.
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.
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
Pappus theorem
If the conic section degenerates into two lines, that is, if the
hyperbola degenerates into its asymptotes, then one gets the case
described by Pappus, 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.
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
In his proof, Karamata considers a special case of the Pappus
theorem – which he called Pappus-Pascal theorem – when the
vanishing line lies in infinity:
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 .
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
In order to prove this theorem, Karamata divides figure into two
quadrangles and reduces the Pappus-Pascal theorem to the
following
Theorem
and
then
~a + ~b + ~c ~a ′ + ~b ′ + ~c ′.
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
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
and
(a + b | b′ + c ′) = 0, (b + c | a′ + b′) = 0,
then
(a + b + c | a′ + b′ + c ′) = 0.
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
Using only the assumptions of the theorem and the property of
distribution 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.
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
1 Introduction
4 Conclusion
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
Gerard Desargues (1591-1661)
of his friend, A.Bosse
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.
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
Karamata considers a special case of the inverse Desargues theorem,
when the homologous sides of two triangles are parallel, i.e. when
the vanishing 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.
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
In order to prove thus formulated theorem, Karamata conveniently
devides this figure and observes two quadrangles whose sides and
diagonals are considered as free vectors, and arrives at the
following form of Desargues theorem: If
~a ~a ′, ~b ~b ′, ~c ~c ′,
and
then
~a + ~b + ~c ~a ′ + ~b ′ + ~c ′,
i.e. the parallelness of three homologous sides and homologous
diagonals implies 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
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
Karamata’s formulation
and
(a + b | a′ + b′) = 0, (b + c | b′ + c ′) = 0,
then
(a + b + c | a′ + b′ + c ′) = 0.
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
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
sum members equal 0, there follows that to prove Desargues theorem,
one needs to prove that
(c + a | c ′ + a′) = 0.
As there exist two real numbers m and n such that ~c = m~a + n~b
for every (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
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
(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 the first 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.
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
But Karamata shows that, by introducing an intermediate quadrangle,
i.e. the three auxiliary vectors ~a ′′, ~b ′′ and ~c ′′, Desargues
theorem can be derived 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.
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
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 - this
time to vectors ~a ′, ~b ′ and ~c ′, there follows that
(a′′ + b′′ + c ′′ | a′ + b′ + c ′) = 0,
wherefrom directly follows
that is, the Desargues theorem.
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
1 Introduction
4 Conclusion
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
As Karamata himself notes in the paper, the terminology used in
formulation 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 five axioms
of incidence. If those 5 axioms are complemented with the Pappus
theorem, which is attributed the role of axiom, it is possible to
prove Desargues theorem, which G.Hassenberg showed in 1905
(Beweis
des Desarguesschen Satzes aus dem Pascalischen, Mathematische
Annalen, 61). Since his proof that the vanishing line lies in
infinity, was predominantly based on the distribution law of
parallel product, Karamata claims the following:
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
Indeed, if we want to get rid of parallelness in the proof, instead
of the line that lies in infinity, we should suppose that the
vanishing line lies in finiteness, and interpret the whole proof
projectively... In essence, this proof features the concept of
parallel product which is manifested in the form of coordination
between the set of four points and the set of real numbers, for
which the analogon of distributive law holds and it disappears when
the lines defined by these points are parallel, i.e. when they
intersect at infinity.
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
Introduction Proof of Pappus-Pascal theorem Proof of Desargues
theorem Conclusion
Originality of his approach to the topic, coupled with simplicity
and elegance of the proofs shown, speak best not only in favour of
Karamata’s exquisite giftedness for mathematics, but also of his
wide mathematical knowledge and his versatile mathematical
interest. However, his papers and results in geometry were somewhat
unjustly overlooked. 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-Pascal theorem three times.
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
Introduction