To start with, we will simply practice computing cross ratio for 4 points on a line.
· Draw a line AB and put two points C and D on it.
· Measure ratio CA/CB as a 3-point ratio thus: Select in order, points C, B, A and the HOLDING DOWN THE SHIFT KEY choose Measure > Ratio.
· Measure DA/DB in the same way. Then use the calculator to compute the ratio of these ratios (CA/CB)/ DA/DB) = R(C, D, A, B) = R(A, B, C, D).
· Drag C and D around and see when the cross-ratio is positive and when it is negative. When is the ratio approximately 1? When is it approximately –1?
· Calculate the cross-ratio R(B, A, C, D) and check that it is the reciprocal of the previous cross-ratio.
Carry out the construction from class this morning, starting with collinear A, B, C and constructing a point D on the same line. (This is on the web if you have forgotten.) Then measure the cross-ratio R(A, B, C, D). It should be –1.
· Now in a new sketch, redo this figure starting from a different set of points. Begin with 4 non-collinear points W, X, Y, Z
· Construct (WITH LINES NOT SEGMENTS) all line connecting two of these points (6 in all). This makes a quadrilateral and two diagonals. Intersect these lines as in the figure. Do you recognize this morning's figure inside?
· On every line in the figure you should be able to find 4 points from intersecting with other lines. In each case, the 4 points should be harmonic. Check this by measuring cross-ratio for several examples.
Cross-ratio involves complicated relationships, but there are several basic points of view that it helps to visualize. Here is one. We will fix 3 points I, O, U and a variable point X and trace the value of the cross ratio R(I, O, U, X).
· In a new sketch, choose Show Axes. Let O and E denote the origin and unit point on the x-axis.
· Place a point P on the y-axis and construct the line through P parallel to the x-axis. Place a point I on this new line.
· Draw line OI. Construct a point X on this line. Draw line PX and let T be the intersection of PX and the x-axis.
· Draw line PE and let U be the intersection of PE and OI.
· Finally, measure the cross-ratio (in order) R(I, O, U, X). Then measure the coordinates of T. You should see a relationship. We will explore this relationship on Friday and beyond. Keep this figure for future reference.
Here is a remarkable ruler construction for tangents that we will want to explain.
· Start with a circle with center O and any point P outside the circle.
· Draw two secant lines PU and PV though P intersecting the circle in AB and CD.
· Then let X and Y be the intersections of AB and AC and BD and AD and BC.
· Draw line XY and intersect with the circle at S and T. Then PS and PT should be tangents! This is a tangent construction that only uses the straightedge.
· Also, the line XY should intersect OP at P', the inversion of P. (Do you see the harmonic construction in this?) Line XY is also perpendicular to OP.
· Move P around, even inside the circle. The line XY is still defined and still contains P'.
· This line XY is called the Polar line of P. An efficient way to construct it is simply as the perpendicular to OP through P'
These are experiments. We will be proving this.
· Start with a circle and a point P and its polar line p. Then construct any point Q on p and its polar line q. Check that P is on q for any choice of Q on p.
· Start with a circle and two points A and B. Construct the polar lines a and b of A and B. Let C be the intersection of a and b. Then the polar line of C is line AB.