Solution: Given any line, it contains two points (Axiom 3), call them A and B. The other two points (since there are 4 all together - Axiom 1) will be called C and D. There is a line containing C and D (Axiom 2). This line and the first line contain all the points of the geometry.

24:10. Sketch a model for a geometry that satisfies Axioms 1 and 3 of Fano's geometry but not Axiom 2.

Solution:

24:15. For Fano's geometry, prove that the lines through any one point of the geometry contain all the points of the geometry.

Solution:Given a point P, by Axiom 4, any other point of the geometry is joined to P by a line. Thus, all the points are on lines through P.

30:7. Prove that the geometry of Pappus contains exactly nine points.

Solution:There is a line, l, (Axiom 1), with three points on it (Axiom 2). Every point not on l is joined to two of these three points by lines (Axioms 5 and 6). Since there are three lines through a point (Thm 1.10), there are exactly 6 lines, other than l, that go through points of l. Each of these lines has two points not on l. This gives 6(2) = 12 possible points, but we have counted each point twice, so there are only 12/2 = 6 points not on l. Plus the 3 on l, gives 9 points in total.

30:18. Prove that two lines parallel to the same line are not parallel to each other.

Solution: Let l be a fixed line with pole P (exists by Thm 1.11). Let m be any line parallel to l. If P is not on m, then by Axiom 6, m and l intersect. This contradicts the fact that l and m are parallel. So, any line parallel to l must pass through P. Therefore, two lines parallel to l must intersect at P, and so, can not be parallel to each other.