There exists at most one parallel to a given line through a given point. Two lines can but need not have some points in common.

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1 Math 3181 Name: Dr. Franz Rothe February 6, 2014 All3181\3181_spr14t1.tex Test has to be turned in this handout. 1 Solution of Test 10 Problem 1.1. As far as two-dimensional geometry is concerned, Hilbert s Proposition 1 reduces to one simple statement: any two different lines either intersect in one point, or are parallel. This statement can be rephrased in many formulations. Encircle the statements below which are such equivalent reformulations; and scratch through the statements which are not. There exists at most one parallel to a given line through a given point. Any two different lines which are not parallel, have a unique point of intersection. On any given line lie two or more points. Two lines can but need not have some points in common. Two different lines have at most one point in common. Any two different lines which are parallel, have a unique point of intersection. If two lines have two or more points in common, they are equal. 1

2 10 Problem 1.2 (The four-point incidence geometries). Find all non isomorphic incidence geometries with four points. For each one of them provide a drawing; tell how many lines there exist; tell which parallel property (elliptic, Euclidean, hyperbolic, or neither) does hold. Figure 1: There are two four-point incidence geometries. Answer. There exist two non-isomorphic four-point geometries. (a) Six lines with each one two points. It has the Euclidean parallel property, and is the smallest affine plane. (b) There are four lines, one of which has three points. property. It has the elliptic parallel Definition 1 (Hand-shake model). A hand shake model is an incidence geometry for which every line has exactly two points. Definition 2 (Straight fan). A straight fan is an incidence geometry with all but one point lying on one line. 10 Problem 1.3. How many lines does the hand-shake incidence geometry with n points have. How many lines does the straight fan with n points have. Answer. The hand-shake incidence geometry with n points has (n 1)n (n 1) + (n 2) = 2 lines. We see this as follows. We can connect the first point to the other (n 1) points. Disregarding this point, we connect the second point to (n 2) different points, and so on. The last line to be drawn is between the (n 1)-th and the n-th point. The straight fan with n points has n lines. There is one long line with n 1 points, and only one point P does not lie on this line. There are n 1 lines with two points each of which connects P to a different point on the long line. 2

3 10 Problem 1.4. Match the figures on page 3 with the following descriptions: a handshake model!! (e) a straight fan!! (c) the smallest affine plane!! (d) the smallest projective plane!! (b) the number of lines is one more than the number of points!! (a) Figure 2: Five examples for incidence planes. 3

4 10 Problem 1.5. In a Cartesian plane over a field F, how are defined the "points" the "lines" the relation telling that "a point lies on a line" Answer. Definition 3 (The Cartesian plane over a field F). The "points" of the Cartesian plane are ordered pairs (x, y) of elements x, y F. The "lines" of the Cartesian plane of the field F are equations ax + by + c = 0 with coefficients a, b, c from the field, of which a and b are not both zero. A "point lies on a line" if and only if the coordinate pair (x, y) satisfies the equation of the line. 10 Problem 1.6. With addition and multiplication modulo 5, the set Z 5 is a field, because 5 is a prime number. How many points, and how many lines has the coordinate plane Z 5 Z 5. Answer. The coordinate plane Z 5 Z 5 is an affine plane with order 5. Each line has 5 points. An equivalence class of (equal or) parallel lines is called a pencil. The lines can be partitioned into n + 1 = 6 pencils, each one contains 5 parallel lines. Altogether, there exist 30 lines. 4

5 Definition 4 (Isomorphism of incidence planes). Two incidence planes are called isomorphic if and only if there exists a bijection between the points of the two planes, and a bijection between the lines of the two planes such that incidence is preserved. Figure 3: Two isomorphic six-point incidence geometries 10 Problem 1.7. Given are two incidence geometries, for which it is not obvious whether they are isomorphic. Show an isomorphism between the two six-point incidence geometries in the figure on page 5 in the following way: label corresponding points in both geometries by the same label; color the corresponding lines with three points by the same color. Four colors are needed. 5

6 Figure 4: Labels of the points and colors of the lines show the isomorphism. Answer. Such an isomorphism is easy to find as follows. One chooses in both geometries an arbitrary line with three points, and names the points of these two lines randomly by A, B, C. The correspondence of the remaining points is now uniquely determined. There exists exactly one line with two points, and point A one of them. The second point on this line is named D. Similarly, there is a line with two points, and point B among them. Now we name the second point E. Similarly we get a line with exactly two points C and F. The bijection among the lines is now given uniquely, and can be shown by the corresponding colors. 6

7 10 Problem 1.8. Give exact definitions of the terms segment, ray, triangle in terms of the order relation. Clarify the obvious questions. Give illustrations. Answer. Definition 5 (Segment). Let A and B be two distinct points. The segment AB is the set consisting of the points A and B and all points lying between A and B. The points A and B are called the endpoints of the segment, the points between A and B are called the interior points, and the remaining points on the line AB are called the exterior points of the segment. Definition 6 (Ray). Given two distinct points A and B, the ray AB is the set consisting of the points A and B, the points inside the segment AB, and all points P on the line AB such that the given point B lies between A and P. The point A is called the vertex of the ray. The axiom of order II.2 tells that the ray AB contains points not lying in the segment AB. Definition 7 (Triangle). We define a triangle to be union of the three segments AB, BC and CA. The three points A, B and C are assumed not to lie on a line. These three points are the vertices, and the segments BC, AC, and AB are the sides of the triangle. For a segment or ray, it is assumed that the two endpoints A and B are different. For a triangle ABC, it is assumed that the three vertices do not lie on a line. For the congruence, similarity of two triangles, and for the Theorem of Desargues, one has to consider the vertices and sides of these triangles in a definite order. 7

8 10 Problem 1.9. Give exact definitions of the terms angle, interior and exterior of an angle. Clarify the obvious questions. Give illustrations. Answer. Definition 8 (Angle). An angle BAC is the union of two rays AB and AC with common vertex A not lying on one line. The point A is called the vertex of the angle. The rays AB and AC are called the sides of the angle. Figure 5: Interior and exterior of an angle Definition 9 (Interior and exterior domain of an angle). The interior of an angle lying in a plane A is the intersection of two corresponding half planes bordered by the sides of the angle, and containing points on the other side of the angle, respectively. The exterior of an angle is the union of two opposite half planes -bordered by the sides of the angle, and not containing the points neither in the interior nor on the legs of the angle. Half planes, interior and exterior of an angle all do not include the lines or rays on their boundary. Thus the interior of BAD is the intersection of the half plane of AB in which D lies, and the half plane of AD in which B lies. The exterior of BAD is the union of the half plane of AB opposite to D, and the half plane of AD opposite to B. 8

9 10 Problem Given is a triangle and a line through an interior point of the triangle on which no vertex of the triangle lies. Show that the line intersects exactly two sides of the triangle. Figure 6: A line through an interior point of a triangle intersects either two sides, or goes through a vertex and intersects the opposite side. Answer. Let the line l go through the point P in the interior of triangle ABC. We have assumed that the line does not go through any vertex of the triangle. We draw the ray AP. Because point P lies in the interior of angle BAC, the crossbar theorem shows that the ray AP intersects the opposite side BC at some point, say Q. We now apply Pasch s axiom to the triangle ABQ and the line l. This line intersects side AQ at point P. Hence it intersects a second side, either BQ BC or AB, say at point D. In both cases, the line l intersects one side of the given triangle ABC. By Pasch s axiom it intersects a second side, say at point E. By Bernays lemma, the line l intersects exactly two sides of the given triangle. 9