Definition 1 (Hand-shake model). A hand shake model is an incidence geometry for which every line has exactly two points.

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1 Math 3181 Dr. Franz Rothe Name: All3181\3181_spr13t1.tex 1 Solution of Test I 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 1.1. 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 1)n 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 not on this line. There are n 1 lines with two points each of which connects P to a different point on the long line. 10 Problem 1.2. Which parallel property holds for the hand-shake model with 4 points. Which parallel property holds for the hand-shake model with n 5 points. Which parallel property holds for a straight fan. Answer. The hand-shake model with 4 points has the Euclidean parallel property. All the hand-shake models with n 5 points have the hyperbolic parallel property. For all straight fans, the elliptic parallel property holds. 1

10 Problem 1.3 (A six-point incidence geometry). Find an incidence geometry with six points and seven lines. Mark all three-point lines with different blue shades.

2 10 Problem 1.3 (A six-point incidence geometry). Find an incidence geometry with six points and seven lines. Mark all three-point lines with different blue shades. Describe the properties of the points and lines. Count the lines. Answer. A model easy to draw has three intersecting three-point lines forming a triangle and the in-circle as the fourth three-point line. There are three lines with two points each of which connects a vertex with the midpoint of the opposite side. Together we get = 7 lines. Figure 1: A six-point incidence geometry with seven lines 2

10 Problem 1.4 (Scheduling problem). Make a 4 day schedule for a group of nine participants. Each day the students are divided in a different way into 3 groups of 3 students.

3 10 Problem 1.4 (Scheduling problem). Make a 4 day schedule for a group of nine participants. Each day the students are divided in a different way into 3 groups of 3 students. Never are two students in the same group more than one time during the four days. Use the coordinate plane Z 3 Z 3, with addition and multiplication modulo 3. Make a picture of the schedule by drawing the 3 3 pattern of dots separately for every day. The main point is to show the partition into the groups clearly, in a separate drawing for every day. Answer. Figure 2: Scheduling four days with differently chosen groups

4 Definition 3 (Subplane). A subplane of an incidence plane is an incidence plane which has as "points" a subset P P from the points P of the given plane, and as "lines" the nonempty intersections of the lines of the original plane with the subset P of the remaining points. The incidence relation is the induced relation. The axioms of incidence (I.1), (I.2),(I.3) have still to hold for the subplane. 10 Problem 1.5. Given is a subplane of the affine coordinate plane Z 7 Z 7 which contains the 21 points on three parallel lines l 1, l 2, l 3. The lines of the subplane are these three parallel lines with 7 points on each, and a lot of lines with three points. How many lines k i with 3 points do exist? How many parallels through a given point P has a three point line l. Figure 3: In a subplane there exist multiple parallels. Answer. The original Z 7 Z 7 plane has = 56 lines. Four lines are deleted. There remain 52 lines, of which 49 have three points each one. In the subplane there exist many parallel to a given line through a given point. The parallels through point P to the three-point line l in the subplane are obtained by cutting down the original parallel m and the lines P A, P B, P C, P D. Here A, B, C, D be the four points deleted from the extended line l +. In the subplane, there are five parallels through a given point to a given three point line. 4

5 10 Problem 1.6. Give exact definitions of the terms segment, ray, triangle in terms of the order relation. Clarify the obvious questions. Give illustrations. Answer. Definition 4 (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 5 (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, 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. 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. 5

10 Problem 1.7. Give exact definitions of the terms angle, interior and exterior of an angle. Clarify the obvious questions. Give illustrations.

The point A is called the vertex of the angle. The rays AB and AC are called the sides of the 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

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

6 10 Problem 1.7. Give exact definitions of the terms angle, interior and exterior of an angle. Clarify the obvious questions. Give illustrations. Answer. Definition 7 (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 4: Interior and exterior of an angle Definition 8 (Interior and exterior 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. 6

Figure 5: A line through an interior point of a triangle intersects either two sides, or goes through a vertex and intersects the opposite side. Answer.

7 10 Problem 1.8. 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 either intersects exactly two sides of the triangle. Figure 5: 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 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 or 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. 7

Proposition 1 (Isosceles Triangle Proposition). [Euclid I.5, and Hilbert s theorem 11] An isosceles triangle has congruent base angles. 10 Problem 1.9.

8 Proposition 1 (Isosceles Triangle Proposition). [Euclid I.5, and Hilbert s theorem 11] An isosceles triangle has congruent base angles. 10 Problem 1.9. Formulate the theorem with specific quantities from a triangle ABC. Provide a drawing. Give a detailed proof. Figure 6: An isosceles triangle Answer. For the triangle ABC as in the drawing on page 8, we have to show: If a = b, then α = β. This is an easy application of SAS-congruence. Assume that the sides AC = BC are congruent in ABC. We need to show that the base angles α = BAC and β = ABC are congruent. Define a second triangle A B C by setting A := B, B := A, C := C (It does not matter that the second triangle is just on top of the first one.) To apply SAS congruence, we match corresponding pieces: (1) ACB = BCA = A C B because the order of the sides of an angle is arbitrary. By axiom III.4, last part, an angle is congruent to itself. Hence ACB = A C B. (2) AC = A C. Question. Explain why this holds. Answer. AC = BC because we have assumed the triangle to be isosceles, and BC = A C by construction. Hence AC = A C. (3) Similarly, we show that (3): BC = B C : Indeed, BC = AC because we have assumed the triangle to be isosceles, and congruence is symmetric, and AC = B C by construction. Hence BC = B C. Finally, we use axiom III.5. Items (1)(2)(3) imply BAC = B A C = ABC. But this is just the claimed congruence of base angles. 8

The statement implies that any three intersection points of two distinct planes lie on a line.

The statement implies that any three intersection points of two distinct planes lie on a line. Math 3181 Dr. Franz Rothe February 23, 2015 All3181\3181_spr15ts1.tex 1 Solution of Test Name: Problem 1.1. The second part of Hilbert s Proposition 1 states: Any two different planes have either no point