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

Size: px

Start display at page:

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

Arthur Whitehead
6 years ago
Views:

1 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 in common, or they have one line and no further points in common. Encircle the correct commentaries about this statement, and scratch through the false assertions. Almost all axioms of incidence are used in the proof of this statement. The only exception is axiom (I.3). The statement implies that any three intersection points of two distinct planes lie on a line. The statement excludes the existence of any parallel planes. The statement implies neither existence nor uniqueness of a parallel plane, to a given plane through a given point. The statement implies that a third plane which intersects one of two parallel planes, intersects the other one, too. Remark. In order to prove Any two different planes have either no point in common, or they have one line and no further points in common. one needs axioms (I.1)(I.2)(I.5)(I.6)(I.7). Almost all axioms of incidence are used in the proof of the above statement. The exceptions are axioms (I.3)(I.4)(I.8). 1

2 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. 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

(f) a model with eight lines (c) a straight fan (d) the smallest affine plane

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

Figure 3: Two intersecting segments. Problem 1.5. We assume A B C and A X Y, and that not all five points lie on a line. Prove that the segments BY and CX intersect each other. Provide a drawing.

4 Figure 3: Two intersecting segments. Problem 1.5. We assume A B C and A X Y, and that not all five points lie on a line. Prove that the segments BY and CX intersect each other. Provide a drawing. (i) Why does the segment BY intersect the line CX? (ii) Why does the segment CX intersect the line BY? (iii) Why are these two intersection points equal? Why do the segments BY and CX intersect each other? Answer. (i) We use Pasch s axiom for triangle ABY and line CX. This line intersects side AY, but not side AC. Hence line CX intersects the side BY. (ii) We use Pasch s axiom for triangle ACX and line BY. This line intersects side AC, but not side AX. Hence line BY intersects the side CX. (iii) The two lines CX and BY are distinct, otherwise all points A, B, C, X, Y would lie on a line. Hence lines CX and BY have unique intersection point. By items (i) and (ii), this point is the intersection of the segments BY and CX. 4

5 Problem 1.6. Give exact definitions of the terms segment, ray, triangle in terms of the order relation. Clarify the obvious questions. Give illustrations. Why does one assign a definite order to the vertices and sides of a triangle? Answer. Definition 3 (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 4 (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 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 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 correct statement of the Theorems about congruence, similarity of two triangles, as well as for the Theorem of Desargues, and for the purpose of fixing an orientation, we assign a definite order to the vertices and sides of the relevant triangles. 5

Problem 1.7. Give exact definitions of the terms angle, interior and exterior of an angle. Clarify the obvious questions. Give illustrations. Definition 6 (Angle).

Figure 4: Interior and exterior of an angle Definition 7 (Interior and exterior domain of an angle).

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.

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.

6 Problem 1.7. Give exact definitions of the terms angle, interior and exterior of an angle. Clarify the obvious questions. Give illustrations. Definition 6 (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 7 (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. Problem 1.8. Prove the Four-point Theorem directly from the Three-point Theorem and the Plane separation Theorem Answer. Given are four different points on a line l. By the Three-point Theorem, one of them, which we name C lies between two others. We draw a line m through C different from l and use plane separation. In one half-plane of l lie two of the remaining points, in the other half-plane only the last remaining point, which we call D. Of the two points in the same half plane, one which we name B lies between C and the other one, finally named A. This statement follows from the three-point theorem. We have thus obtained the order relations A B C, A C D, B C D To confirm the fourth order relation, we draw a third line n through B different from l, and use plane separation once more. Point C and D lie in the same half-plane of n whereas A lies in the opposite half-plane from C. Hence A and D lie in opposite half-planes, confirming the order relation A B D. 6

Problem 1.9. Show that the interior domain of a triangle is nonempty. Which are the two theorems used in this reasoning. (i) Why do the points D and E exist?

Twice, we use Hilberts proposition 3, saying that the interior of a segment is nonempty. There exists a point D in the segment AB, and a point E in the segment CD.

7 Problem 1.9. Show that the interior domain of a triangle is nonempty. Which are the two theorems used in this reasoning. (i) Why do the points D and E exist? (ii) Why does point E lie in the interior of the triangle? Figure 5: Why is the interior of a triangle nonempty. Answer. Twice, we use Hilberts proposition 3, saying that the interior of a segment is nonempty. There exists a point D in the segment AB, and a point E in the segment CD. Finally, we use the plane separation theorem to confirm that point E lies in this interior of the triangle ABC. Indeed, points C and E lie on the same side of line AB, Points E, D, A lie on the same side of line BC, and finally points E, D, B lie on the same side of line AC. 7

Problem 1.10. Given is a triangle and a line through an interior point of the triangle. We suppose that the line does not intersect a vertex the triangle.

8 Problem Given is a triangle and a line through an interior point of the triangle. We suppose that the line does not intersect a vertex the triangle. Show that the line intersects two sides of the triangle. Put the appropriate names into the figure on page 8. Figure 6: A line through an interior point of a triangle. Answer. Let the line l go the point P in the interior of triangle ABC. We suppose 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. 8

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

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