Answers to Homework Problems

(pg. 16, Smart, Modern Geometries, 5th Ed.)
1. For the three-point geometry, draw a pictorial representation different from those in Figure 1.4.

Solution: One possibility is:

3.

Solution: The two points on the bottom of the figure are on two lines, contradicting Axiom 2.

5. Rewrite the axioms, using the words student for point and committee for line.

Solution:

  1. There exist exactly three students in the geometry.
  2. Two different students are on exactly one committee.
  3. Not all the students of the geometry are on the same committee.
  4. Two different committees have at least one student in common.
7. Through a point not on a given line, is there at least one line not intersecting the given line?

Ans: No. Solution: Let P be a point not on the line m. Any line containing P is different from m. By Theorem 1.1, a line containing P must have a common point with m (since they are distinct lines) and so, it intersects m.

9. Through a point not on a given line, how many lines are parallel to the given line?

Ans: None. Solution: See problem 7.

11. Must lines be straight in the Euclidean sense?

Ans: No. Solution: Straightness is not a property given in the axioms. See the last two examples of Fig. 1.4.

13. Do any squares exist?

Ans: No. Solution: A square must have four distinct vertices (corners, points) and this geometry only has three points by Axiom 1.

15. Prove that there exists a set of two lines containing all the points of the geometry.

Solution: Let the three points of the geometry (Axiom 1) be called P, Q and R. By Axiom 2, there is a line containing P and Q. R is not on this line by Axiom 3. Again, by Axiom 2, there is a line containing P and R. This line can not be the same as the first line, since R is not on the first line. These two lines contain all the points, P, Q and R.

18. Discuss whether or not each of the following concepts from Euclidean geometry also appears as a concept in the three-point geometry: angle, congruence, circle, area, intersection, similarity, hexagon.

Solution: Any concept involving measurement can not appear as a concept in the three-point geometry since there are no axioms that are concerned with measurement (except counting the number of points and/or lines). All the concepts above, except intersection and hexagon, are concerned with measurement. A hexagon has six distinct vertices and there are not enough points in this geometry for it to contain a hexagon. The concept of intersection is a valid concept for the three-point geometry.

20. Check the three-point geometry to see that each of the eight desirable properties is satisfied.

Solution:

  1. The number of points is finite. Yes, three points.
  2. The number of lines is finite. Yes, three lines.
  3. Each line is on the same number s of points, where s >= 2. Yes, each line has 2 points.
  4. Each point is on the same number t of lines, where t >= 2. Yes, each point is on 2 lines.
  5. Each pair of distinct points is on at most one line. Yes, Axiom 2.
  6. Each pair of distinct lines is on at most one point. Yes, Theorem 1.1.
  7. Not all points are on the same line. Yes, Axiom 3.
  8. There exists at least one line. Yes, Theorem 1.2.