Two Applications of Desargues Theorem

Transcription

1 Two pplications of Desargues Theorem Prof. Florentin Smarandache, University of New Mexico, U.S.. Prof. Ion Pătraşcu, The National College Fraţii Buzeşti, Craiova, Romania In this article we will use the Desargues theorem and its reciprocal to solve two problems. For beginning we will enunciate and prove Desargues theorem: Theorem (G.Desargues, 66, the famous perspective theorem : When two triangles are in perspective, the points where the corresponding sides meet are collinear.) Let two triangle BC and BC be in a plane such that BB CC = { }, B B = { N} BC BC = { M} C C = { P} then the points N, M, P are collinear. B B C C M N P Fig.

2 Proof Let { } = BB CC, see Fig... We ll apply the Menelaus theorem in the triangles C; BC; B for the transversals N,, C; M, B, C; P, B,, and we obtain N CC = () NC C MC BB C = () MB B CC PB B = () P BB By multiplying the relations (), (), and () side by side we obtain N MC PB =. NC MB P This relation, shows that N, M, Pare collinear (in accordance to the Menealaus theorem in the triangle BC ). Remark The triangles BC and BC with the property that, BB, CC are concurrent are called homological triangles. The point of concurrency point is called the homological point of the triangles. The line constructed through the intersection points of the homological sides in the homological triangles is called the triangles axes of homology. Theorem (The reciprocal of the Desargues theorem) ` If two triangles BC and BC are such that B B = { N} BC BC = { M} C C = P { } nd the points N, M, P are collinear, then the triangles BC and BC are homological. Proof We ll use the reduction ad absurdum method. Let BB= { } CC= { } BB CC = { } We suppose that. The Menelaus theorem applied in the triangles B, C, BC for the transversals N,, B; P,, C; M, B, C, gives us the relations NB B = (4) N B B

3 P CC = (5) PC C MC BB C = (6) MB B CC Multiplying the relations (4), (5), and (6) side by side, and taking into account that the points N, M, P are collinear, therefore P MC NB = (7) PC MB N We obtain that B C = (8) B C The relation (8) relative to the triangle BC shows, in conformity with Menelaus theorem, that the points,, are collinear. n the other hand the points, belong to the line, it results that belongs to the line. Because BB CC = { }, it results that { } = BB CC, and therefore = =, which contradicts the initial supposition. Remark The Desargues theorem is also known as the theorem of the homological triangles. Problem If BCD is a parallelogram, ( B), B ( BC), C ( CD), D ( D) such that the lines D, BD, BC are concurrent, then: a) The lines C, C and BD are concurrent b) The lines B, CD and C are concurrent. Solution B B C Q C D D Fig. P

4 Let { P} = D BC BD see Fig.. We observe that the sides D and BC ; CC and D ; and CB of triangles D and CBC intersect in the collinear points P, BD., pplying the reciprocal theorem of Desargues it results that these triangles are homological, that is, the lines: C, C and BD are collinear. Because { P} = D BC BD it results that the triangles DC D and BB are homological. From the theorem of the of homological triangles we obtain that the homological lines DC and BB ; DD and B ; DC and B intersect in three collinear points, these are CQ,,, where { Q} = DC B. Because Q is situated on C it results that B, CD and C are collinear. Problem Let BCD a convex quadrilateral such that B CB= { E} BC D= { F} BD EF = { P} C EF = { R} C BD= { } We note with G, H, I, J, K, L, M, N, Q, U, V, T respectively the middle points of the segments: ( B),( BF),( F),( D),( E),( DE),( CE),( BE),( BC),( CF),( DF),( DC ). Prove that i) The triangle PR is homological with each of the triangles: GHI, JKL, MNQ, UVT. ii) The triangles GHI and JKL are homological. iii) The triangles MNQ and UVT are homological. iv) The homology centers of the triangles GHI, JKL, PR are collinear. v) The homology centers of the triangles MNQ, UVT, PR are collinear. Solution i) when proving this problem we must observe that the BCDEF is a complete quadrilateral and if,, are the middle of the diagonals ( C),( BD ) respective EF, these point are collinear. The line on which the points,, are located is called the Newton- Gauss line [* for complete quadrilateral see []]. The considering the triangles PR and GHI we observe that GI R = { } because GI is the middle line in the triangle BF and then it contains the also the middle of the segment ( C ), which is. Then HI PR = { } because HI is middle line in the triangle FB and is evidently on the line PR also. GH P = { } because GH is middle line in the triangle BF and then it contains also the middle of the segment ( BD ). 4

5 The triangles GIH and RP have as intersections of the homological lines the collinear points,,, according to the reciprocal theorem of Desargues these are homological. G B J K I N Q H C T D M L U V E R F P Fig. Similarly, we can show that the triangle RP is homological with the triangles JKL, MNQ, and UVT (the homology axes will be,, ). ii) We observe that GI JK = { } GH JL = { } HI KL = { } are collinear and we obtain that the triangles GIH and JKL are homological then,, iii) nalog with ii) iv) pply the Desargues theorem. If three triangles are homological two by two, and have the same homological axes then their homological centers are collinear. v) Similarly with iv). Remark The precedent problem could be formulates as follows: The four medial triangles of the four triangles determined by the three sides of a given complete quadrilateral are, each of them, homological with the diagonal triangle of the complete 5

6 quadrilateral and have as a common homological axes the Newton-Gauss line of the complete quadrilateral. We mention that: - The medial triangle of a given triangle is the triangle determined by the middle points of the sides of the given triangle (it is also known as the complementary triangle). - The diagonal triangle of a complete quadrilateral is the triangle determined by the diagonals of the complete quadrilateral. We could add the following comment: Considering the four medial triangles of the four triangles determined by the three sides of a complete quadrilateral, and the diagonal triangle of the complete quadrilateral, we could select only two triplets of triangles homological two by two. Each triplet contains the diagonal triangle of the quadrilateral, and the triplets have the same homological axes, namely the Newton-Gauss line of the complete quadrilateral. pen problems. What is the relation between the lines that contain the homology centers of the homological triangles triplets defined above?. Desargues theorem was generalized in [] in the following way: Let s consider the points,...,n situated on the same plane, and B,...,Bn situated on another plane, such that the lines ibi are concurrent. Then if the lines ij and BiBj are concurrent, then their intersecting points are collinear. Is it possible to generalize Desargues Theorem for two polygons both in the same plane?. What about Desargues Theorem for polyhedrons? References [] Roger. Johnson dvanced Euclidean Geometry Dovos Publications, Inc. Mineola, New York, 007. [] F. Smarandache, Generalizations of Desargues Theorem, in Collected Papers, Vol. I, p. 05, Ed. Tempus, Bucharest,