1. There exist exactly four lines.
2. Any two distinct lines have exactly one point on both of them.
3. Each point is on exactly two lines.
Prove that there are exactly 6 points in this geometry.
Solution:
Since there are 4 lines (Axiom 1) and every pair of them meet (Axiom 2), there are C(4,2) = 6 points of intersection. These 6 points are distinct, since if two of them coincided, there would be more than two lines through this point, contradicting Axiom 3. If there were a 7th point, it would have to be on two of these lines by Axiom 3, and therefore coincide with the point of intersection of these two lines which has already been counted. Therefore, there are exactly 6 points in this geometry.
2. Fano's Geometry can be described by the following axiom set (slightly different from the one in the text):
1. Every line of the geometry has exactly three points on it.
2. Every pair of distinct points is on exactly one line.
3. Every pair of distinct lines have exactly one point in common.
4. There exists a set of four points, no three of which are on the same line.
Show that for any set of three points which are not on the same line, there is a unique fourth point so that these four points satisfy axiom 4. (You may use without reproving, any theorem concerning Fano's geometry that was talked about in class).
Solution:
Let A,B, and C be a set of three points not on a common line. By axiom 2, each pair of these points is on exactly one line, so the three pairs of points determine exactly three lines. By axiom 1, each of these lines must contain a third point. None of these points can be identical, since otherwise there would exist two lines with two points in common contradicting axiom 3. We have now identified 6 points of the geometry. Since there are 7 points in Fano's geometry (shown in class), there is exactly one additional point, call it F. F does not lie on any line containing two of the three original points, so {A,B,C,F} satisfy the property that no three lie on a common line. Notice that no other point can be used in place of F.
3. (a) Write the equations for the glide reflection which consists of a reflection about the x-
axis followed by a translation by the vector (5, 0).
(b) Find the inverse of this glide reflection and write it's equations.
Solution:
(a) The equations for a reflection about the x-axis are x' = x and y' = -y. Translating (x', y') by the vector (5,0) is given by the equations x'' = x' + 5 and y'' = y' + 0 = y'. Therefore, the glide reflection, which is the composition of these maps has equations x'' = x + 5 and y'' = -y.
(b) Solving the equations for the glide reflection for x and y gives x = x'' - 5 and y = -y''. Therefore, the equations for the inverse of the glide reflection are x''' = x - 5 and y''' = - y.
4. Use coordinates to prove that a line and its image under a translation are parallel.
Solution:
Consider any non-vertical line with equation y = mx + k. If this line is translated by the vector (a,b), i.e., x' = x + a and y' = y + b, the image line is given by y' - b = m(x' - a) + k. Rewritten as y' = mx' + (k + b -am). As the original line and the image line both have slope m, they are parallel.
A vertical line has the equation x = k. The image under this translation is x' - a = k, or x' = k + a. As this is still a vertical line, it is parallel to the original line.
5. Find a matrix representing the product of a rotation of 270° about the origin followed by a reflection about the x-axis. What kind of motion does this product represent? [Cos 270° = 0 and Sin 270° = -1]
Solution:
The matrix representing a rotation of 270° about the origin is given by:
/ \ / \ | Cos 270 Sin 270 | | 0 -1 | | | = | | | -Sin 270 Cos 270 | | 1 0 | \ / \ /While the matrix representing a reflection about the x-axis is given by:
/ \
| 1 0 |
| |
| 0 -1 |
\ /
So, the composition has the matrix given by:
/ \ / \ / \
| 0 1 | | 1 0 | | 0 1 |
| | | | = | |
| -1 0 | | 0 -1 | | 1 0 |
\ / \ / \ /
And this matrix represents a reflection about the line y = x.