An analytic proof of the theorems of Pappus and Desargues

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1 Note di Matematica 22, n. 1, 2003, An analtic proof of the theorems of Pappus and Desargues Erwin Kleinfeld and Tuong Ton-That Department of Mathematics, The Universit of Iowa, Iowa Cit, IA 52242, USA Received: 3/9/2003; accepted: 3/9/2003. Abstract. In this article we give an analtic proof of Pappus theorem and an analtic proof of Desargues theorem over a not necessaril commutative field. Kewords: analtic proofs, Pappus theorem, Desargues theorem, not necessaril commutative field. MSC 2000 classification: 51A30, 51A05. Introduction In this article we give an analtic proof of Pappus theorem and an analtic proof of Desargues theorem over a not necessaril commutative field. Both are listed as eercises in [4, p. 76 and p. 78]. The original version of Pappus theorem appeared in Pappus Συναγωγή or Collection (see, e.g., [2, pp ]). The original version of Desargues theorem appeared in A. Bosse s La Perspective de Mr. Desargues (Paris 1648, p. 340) as the First Geometrical Proposition (see also [1, Chapter VIII]). We refer to [3] for an ecellent comprehensive historical surve of these two theorems. 1 An Analtic Proof of Pappus Theorem For the terminolog and fundamental facts used in our proof we refer to [4]. Let K be a commutative field. Then the projective plane P 2 (K) is defined as the quotient of K 3 \ {0} b the equivalence relation : X, Y K 3 \ {0}, X Y if λ 0 in K such that Y = λx. Let π : K 3 \ {0} P 2 (K) denote the canonical projection, then a point P = π P 2 (K) is represented b the vector in three-space, not all

2 100 E. Kleinfeld, T. Ton-That,, = 0, and for λ 0 the vector abuse of language we identif P with λ λ λ represents the same point P. B and write P = if there is no possible confusion. We also call,, the homogeneous coordinates of P. Since an equation of a plane P in K 3 passing through the origin is of the form a + b + c = 0, not all a, b, c = 0, we sa that π (P \ {0}) is a line in P 2 (K). Thus a line l in P 2 (K) is the equivalence class of the row [ a b c ] and for λ 0 the row [ λa λb λc ] represents the same line l. We again identif the line l with the row [ a b c ] and call a, b, c the homogeneous coordinates of l. We see that a point P P 2 (K) lies on a line l P 2 (K) if and onl if [ a b ] c = a + b + c = 0. It can easil be shown that ever line is incident with at least three points. Also since we can find two triples of numbers [a, b, c], [,, ] such that a+b +c 0, we are assured that there eist a point and a line not incident. Further, using the theor of sets of linear homogeneous equations, we can prove that an two points are incident with one and onl one line. Then, appealing to the dualit of the definitions of point and line, it can be shown that an two distinct lines intersect in a unique point. The following Lemma 3 is an immediate consequence of the following two theorems on pages 73 and 74 of [4]. 1 Theorem. If P 1 2 3, P numbers i + λ i, i = 1, 2, 3, are not all = 0, and P with the first two points. are distinct points, then for an λ the 1 + λ λ λ 3 is collinear 2 Theorem. If X, Y, Z are coordinates of three distinct collinear points, then there eist λ, µ such that Z = λx + µy. 3 Lemma. Three distinct points P 1, P 2, P 3 P 2 (K) are collinear if and onl if for an one of these points, sa P 3, there eist nonero scalars λ, µ K

3 An analtic proof of the theorems of Pappus and Desargues 101 such that P 3 = λp 1 + µp 2. It follows that P 1, P 2, P 3 are collinear if and onl if [P 1, P 2, P 3 ] = 0, where [P 1, P 2, P 3 ] denotes the determinant formed b their coordinates. 4 Lemma. Let P 1, P 2, P 3 be noncollinear points of P 2 (K). If Q = a 1 P 1 + a 2 P 2 + a 3 P 3, R = b 1 P 1 + b 2 P 2 + b 3 P 3, and S = c 1 P 1 + c 2 P 2 + c 3 P 3 are points of P 2 (K), then Q, R, S are collinear if and onl if a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 = 0. Proof. Let i, i, i denote the homogeneous coordinates of P i, i = 1, 2, 3. We have [Q, R, S] = [a 1 P 1 + a 2 P 2 + a 3 P 3, b 1 P 1 + b 2 P 2 + b 3 P 3, c 1 P 1 + c 2 P 2 + c 3 P 3 ] = [a 1 P 1, b 2 P 2, c 3 P 3 ] + [a 1 P 1, b 3 P 3, c 2 P 2 ] + [a 2 P 2, b 1 P 1, c 3 P 3 ] + [a 2 P 2, b 3 P 3, c 1 P 1 ] + [a 3 P 3, b 1 P 1, c 2 P 2 ] + [a 3 P 3, b 2 P 2, c 1 P 1 ] = (a 1 b 2 c 3 a 1 b 3 c 2 a 2 b 1 c 3 + a 2 b 3 c 1 + a 3 b 1 c 2 a 3 b 2 c 1 ) [P 1, P 2, P 3 ] a 1 a 2 a 3 = b 1 b 2 b 3 c 1 c 2 c 3 [P 1, P 2, P 3 ]. The lemma follows from Lemma 3 and the assumption that P 1, P 2, P 3 are noncollinear. 5 Theorem. (Pappus) Let l and l be two distinct lines of P 2 (K), intersecting at a point O. Let A, B, E be three distinct points on l and C, D, F three distinct points on l such that each of these si points is different from O. Let P be the intersection of the lines AD and CB, Q the intersection of AF and CE, and R the intersection of BF and DE. Then P, Q, R are collinear. E B A Q R O P C D F (Figure 1)

4 102 E. Kleinfeld, T. Ton-That Proof. In drawing Pappus configuration, we represent the points A, B, and E as vectors in three-space (using homogeneous coordinates). We use the same letters A, B, and E to represent the vectors. B Lemma 3, for A, B, and E to be collinear it is necessar and sufficient that there eist nonero scalars s and t such that E = sa + tb. Since we are free to represent the point A b an nonero scalar multiple of the vector A, we choose to represent it b sa, and similarl we represent the point B b tb. Now we rename the vector sa as A and the vector tb as B. Thus we have the vector equation E = A + B in this notation. B the same reasoning there eist vectors C and D, such that C represents the point C, D represents the point D, and C + D represents the point F. Since A, B, O are distinct collinear points, and C, D, O are also distinct collinear points, there eist b Lemma 3 nonero scalars a, b, c, d such that O = aa + bb = cc + dd. (1) Since E is different from O we must have a b, and since F is different from O we must have c d. From (1) we get the relation aa dd = cc bb. (2) B Lemma 3, (2) represents a point which is on both the lines AD and BC, and this intersection is named P. From (1) we also get (a b) A d (C + D) = (c d) C b (A + B), (3) which is equivalent to the equation (a b) A df = (c d) C be. (3) Since a b, c d, d, and b are nonero scalars, Lemma 3 implies that (3) represents a point which is on both the lines AF and EC. This intersection is named Q. From (1) it follows that (b a) B c (C + D) = (d c) D a (A + B) (4) which is equivalent to the equation (b a) B cf = (d c) D ae. (4) Lemma 3 implies that (4) represents a point that is on both the lines BF and DE. This intersection is named R. B multipling (1) b the nonero scalar b we obtain aba + b 2 B bcc bdd = 0. (5)

5 An analtic proof of the theorems of Pappus and Desargues 103 Using the right-hand side of (3), we ma represent Q b ba bb + (c d) C which is equivalent to aba abb + (ac ad) C since a is different from 0. Adding the ero vector given b (5) to this representation of Q, we obtain the following representation of Q. Q: ( b 2 ab ) B + (ac ad bc) C (bd) D. (6) To prove that P, Q, and R are collinear we use (2), (6), and (4) to obtain the following representations: P : bb ( + cc, Q: b 2 ab ) B + (ac ad bc) C (bd) D, (7) R: (b a) B cc cd. Since B, C, and D are noncollinear points of P 2 (K), Lemma 4 implies that a necessar and sufficient condition for the points P, Q, and R to be collinear is that the determinant b c 0 b 2 ab ac ad bc bd b a c c = 0. An eas calculation confirms that this determinant is ero. It is well known that Pappus theorem implies the commutativit of the multiplication in the field K of segment arithmetic (see the discussion in [3] and a proof of this fact in [4, pp ], for eample). It is also well known (see [5]) that Pappus theorem implies Desargues theorem, but the converse is not true when the field K is not commutative. In [4, p. 75], Seidenberg gave a simple analtic proof of Desargues theorem when K is commutative; in the net section we adapt his proof to the case where the field K is not necessaril commutative. 2 Desargues Theorem over a Not Necessaril Commutative Field For terminolog, definitions, and basic facts, we refer to [4, Chapter III]. Let K be a not necessaril commutative field, i.e., we do not assume the rule ab = ba for all a, b K (if ab ba for some a, b K then K is called a noncommutative, or skew, field). If K is a not necessaril commutative field then the following hold: There is one and onl one element e K such that ae = ea for all a K. Such an element e is called the multiplicative neutral element of K.

6 104 E. Kleinfeld, T. Ton-That For ever a 0 in K there eists one and onl one element a K such that aa = a a = e. This element is called the inverse of a. The equation a = b, a 0, has one and onl one solution, namel = a b. Similarl, the equation a = b, a 0, has one and onl one solution, namel = ba. We now define the projective plane P 2 (K) b representing a point P P 2 (K) b a column vector, not all,, = 0 in K, and such that for λ λ 0 in K, λ λ represents the same point P. λ We identif P with. Similarl we represent a line l of P 2 (K) b a row vector [ a b c ], not all a, b, c = 0 in K, and such that for µ 0 in K, µ [ a b c ] [ µa µb µc ] represents the same line l. Again we identif the line l with the vector [ a b c ]. Then we see that P lies on l if and onl if for λ, µ 0 µ [ a b c ] 6 Lemma. If P 1 = λ = µaλ + µbλ + µcλ = µ (a + b + c) λ = and P 2 = are distinct points of P 2 (K) then for λ 1, λ 2 not both ero scalars, the point P 1 λ 1 + P 2 λ 2 lies on the line through P 2 and P 2. Proof. Obvious. 7 Lemma. If P 1, P 2, and P 3 are three distinct collinear points of P 2 (K), then for an one of these points, sa P 3, there eist nonero scalars λ, µ K such that P 3 = P 1 λ + P 2 µ. Proof. Let i, i, i be homogeneous coordinates of P i, i = 1, 2, 3. Since not all 1, 1, 1 equal ero we ma assume without loss of generalit that 1 0. Let l ( a b c ) be the line passing through P 1, P 2, and P 3. Then we have the sstem a 1 + b 1 + c 1 = 0, a 2 + b 2 + c 2 = 0, (8) a 3 + b 3 + c 3 = 0.

7 An analtic proof of the theorems of Pappus and Desargues 105 ( The equations 2 = 1 1 ) ( 2, 2 = 1 1 ) ( 2, and 2 = 1 1 2) cannot be all possible, for if 2 = 0 then 2 = 2 = 0 which is ecluded, and if 2 0 then and this would impl that ( P 1 and P 2 coincide, contradicting our hpothesis. Since obviousl 2 = 1 ( 1 2), we ma assume without loss of generalit that ). Now let us consider the sstem of linear equations with unknowns λ and µ { 1 λ + 2 µ = 3, (9) 1 λ + 2 µ = 3. Multipling (to the left) the first equation of (9) b 1 1 and subtracting the result from the second, we get ( ) 2 µ = Since we get µ = ( ) ( ). From the first equation of (9) we get 1 λ = 3 2 µ. Since 1 0 we obtain λ = 1 ( 3 2 µ). Thus the sstem (9) can be solved for λ and µ. Multipling (to the right) the first equation of the sstem (8) b λ, the second b µ, and subtracting their sum from the last, we get a [ 3 ( 1 λ + 2 µ)] + b [ 3 ( 1 λ + 2 µ)] + c [ 3 ( 1 λ + 2 µ)] = 0. Since λ and µ are solutions of the sstem (9), the latter is reduced to c [ 3 ( 1 λ + 2 µ)] = 0. Assuming that c 0, we then obtain the equation 1 λ+ 2 µ = 3, which together with the sstem (9) implies that P 3 = P 1 λ + P 2 µ. And since b hpothesis P 3 is distinct from P 1 and P 2, λ and µ must be different from 0. Thus it remains to show that c 0 to achieve the proof of the lemma. But if c = 0 then the sstem (8) is reduced to a 1 + b 1 = 0, a 2 + b 2 = 0, a 3 + b 3 = 0. (10) Multipling (to the right) the first equation of the sstem (10) b 1 2 and subtracting the result from the second, we get b ( ) = 0. Since , we get b = 0. But then the sstem (10) is reduced to a 1 = 0, a 2 = 0, and a 3 = 0. Since 1 0, we get a = 0. Thus a = b = c = 0, which is not possible b the definition of the line l. 8 Theorem. (Desargues) If two triangles ABC and A B C are in perspective from a point O distinct from the si points A, B, C, A, B, C (i.e., the joins of A and A, B and B, C and C intersect at O) then the intersections P : AB A B, Q: AC A C, and R: BC B C lie on a line.

8 106 E. Kleinfeld, T. Ton-That Q C O C P B R B A A (Figure 2) Proof. It follows from the hpothesis and Lemma 7 that O = Aa + A a = Bb + B b = Cc + C c, (11) for some nonero scalars a, a, b, b, c, c K. From (11) we deduce that Aa Bb = A a + B b, Aa Cc = A a + C c, Bb Cc = B b + C c. Since = ( ) for all, K, (12) is equivalent to Aa + B ( b) = A ( a ) + B b, Aa + C ( c) = A ( a ) + C c, Bb + C ( c) = B ( b ) + C c. (12) (12) It follows from Lemmas 6 and 7 that the first equation of (12) represents the intersection P of AB and A B, the second represents Q, and the third represents R. But then we obviousl have Q = P + R, and b Lemma 6, P, Q, R are collinear. References [1] J. V. Field, J. J. Gra: The geometrical work of Girard Desargues, Springer-Verlag, New York [2] A. Jones: Pappus of Aleandria, book 7 of the collection, Springer-Verlag, New York [3] E. A. Marchisotto: The theorem of Pappus: a bridge between algebra and geometr, Amer. Math. Monthl, 109 (2002), n. 6, [4] A. Seidenberg: Lectures in projective geometr, D. Van Nostrand Compan, Princeton [5] A. Seidenberg: Pappus implies Desargues, Amer. Math. Monthl, 83 (1976), n. 3,