Problem 3.1 (Building up geometry). For an axiomatically built-up geometry, six groups of axioms needed:

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1 Math 3181 Dr. Franz Rothe September 29, 2016 All3181\3181_fall16h3.tex Names: Homework has to be turned in this handout. For extra space, use the back pages, or put blank pages between. The homework can be done in groups up to four students due Wednesday Oct 12 or Thursday Oct 13 3 Homework Problem 3.1 (Building up geometry). For an axiomatically built-up geometry, six groups of axioms needed: I. Axioms of incidence II. Axioms of order III. Axioms of congruence IV. Axiom of parallelism V. Axioms about circles VI. Axioms of continuity Which groups of axioms are valid for the following kind of planes 0 (a) an incidence plane; (c) neutral geometry; (e) a Euclidean plane; (b) an affine plane; (d) a Pythagorean plane; (f) a Cartesian plane. Incidence Order Congruence Parallelism Circles Continuity incidence yes no no no no no affine neutral Pythagorean Euclidean Cartesian 1

2 Incidence Definition 1 (Projective plane). A projective plane is a class of points, and a class of lines satisfying the axioms: P.1 Every two distinct points lie on exactly one line. P.2 Every two distinct lines intersect at exactly one point. P.3 There exist four points of which no three lie on a line. Problem 3.2. Convince yourself that in every point of a projective plane intersect at least three lines. 2

3 The Theorem of Desargues in parallel setting is a special case of the general theorem of Desargues in projective setting. This special case is produced by the extra assumption that the line witnessing the sides of the triangles are in perspective, is an improper line. A different special case may be produced by the assumption that the point witnessing the vertices of the triangles are in perspective, is an improper point. In that manner, we get the following less common case worth to be stated: Figure 1: Constructing the parallel to two given parallel lines. Lemma 1. Given is an affine plane having a projective completion for which Desargues Theorem holds. Assume the sides of the triangles ABC and A B C lie in perspective, and the lines AA and BB are parallel. Then the line CC is parallel to these two lines. Problem 3.3. Given is an affine plane of order n 4 having a projective completion for which Desargues Theorem holds. Given are two different parallel lines AA BB and the point C. Use Lemma 1 to construct the parallel to line AA through the point C. Explain the main outline for the procedure. (I do not insist that you consider all special cases.) Provide a drawing. Complete the construction for the example from figure 4. 3

4 . Figure 2: Another example for constructing the parallel to two given parallel lines. 4

5 Ordered incidence planes Problem 3.4. Given an angle (h, k) lying in a plane A. (a) Let points H h and K k be given, both different from the vertex of the angle. Explain why all interior points of segment HK lie in the interior of the angle (h, k). (b) Let A = h k be the vertex of the angle (h, k) and D be any point in the interior of the angle. Explain why all points of the ray AD except A lie in the interior of the angle (h, k). 5

6 Proposition 1 (About n points in a plane). Among any n 3 points lying in an ordered incidence plane but not on a line, there exist three points P, Q, R such that all n points lie in the interior or on the boundary of the angle P QR. Problem 3.5. Complete the reason for this little proposition given below. Induction start for n = 3: the angle P QR exists. The three given points P, Q, R do not lie on a line, hence Induction step n n + 1 : Assume that proposition holds for any n points. Given are now these points, and an extra point P n+1. We distinguish these cases: (i) The points P 1... P n lie on a line l. By the n-point proposition, they can be put into an ordered list P 1 P 2 P n. By assumption point P n+1 does not lie on the line l. We put P := P 1, the first item in the list; R := P n, the? list; and choose Q := P n+1 as? of the angle. Thus all n + 1 points lie? boundary of the angle P QR. (ii) The points P 1... P n do not lie on a line l. By the induction assumption, there are points A, B, C among them such that points P 1... P n lie in the interior or on the boundary of the angle ABC. In case that the extra point P n+1 lies in the interior or on the boundary of the angle ABC, we are ready. In case that the extra point P n+1 lies in the interior of a supplementary angle of the angle ABC, all points P 1... P n+1 lie in the interior or on the boundary of either angle ABP n+1 or angle P n+1 BC. 6

7 Figure 3: Any n points lie in the interior or on the boundary of an angle. We choose any points A B A and C B C. Assume that the extra point Q := P n+1 lies in the interior of the vertical angle A BC or on the ray BA. This case in shown in the figure on page 7. We draw the segments P i P n+1 for i = 1... n, obviously including the segments AP n+1, BP n+1 and CP n+1. Since the points P 1... P n either lie on the line AB, or the opposite side of AB as point P n+1, all these segments? the line? By the n-point theorem we may order these intersection points into a list S 1 S n. Let P be a point among P 1... P n on the first ray P n+1 S 1, and R be a point among P 1... P n on the? Now all points P 1... P n+1 lie in the interior or on the boundary of the angle P QR. 7

8 Neutral congruence Proposition 2 (Preliminary Converse Isosceles Triangle Proposition). If the two base angles of a triangle are congruent to each other, the triangle is isosceles. Problem 3.6. Formulate the proposition 2 above with specific quantities from a triangle ABC. Provide a drawing. Prove the theorem as stated. Remark. Later on we shall avoid the awkward assumption: If the two base angles of a triangle are congruent to each other. Hilbert proves the converse isosceles triangle theorem as his Theorem 24, after having the exterior angle theorem at his disposal. Problem 3.7. Give purely geometric definitions for the notions : right angle, acute angle, and obtuse angle. (Use only comparison of angles, not any measurements!) 8

9 Figure 4: Reflect a point Definition 2 (Reflection across a line). Given is a line l. The reflection across line l is the mapping r : P P which maps an arbitrary point P to a point P meeting the following requirements: If P lies on the symmetry axis l, we put P := P. If P does not lie on the symmetry axis, the point P is specified by requiring (i) P and P lie on different sides of l. (ii) the lines l and P P are perpendicular, intersecting at F. (iii) P F = P F. 9

10 Problem 3.8. Explain the construction of the reflected point. Use the notation from the figure on page 9. Choose any two points A and B on the line l. Problem 3.9. Give a justification for the construction of the reflected point, from the figure on page 9. Give drawings for the different cases that may occur.

11 Problem 3.. Use the drawing system available at to produce the figure from Euclid s proof showing that the uniqueness of parallels implies that the angle sum of any triangle equals 180. The figure has been drawn in the lecture, too. 11