Assignment 10 (ANSWERS)

10-1 Cross-Ratio

(a)    Write the definition of the cross-ratio (AB, CD) of points A, B, C, D on a line.  (See Ogilvy 97-101)

(AC/AD)/(BC/BD). This can also be written as (AC/BC)/(AD/BD) or (AC BD)/(AD BC).

(b)    If on the real number line, A = 1, B = 2, C = 3, D = 4, what is the cross-ratio?

(AC/AD)/(BC/BD) = ((3-1)/(4-1))/((3-2)/(4-2)) = (2/3)/(1/2) = 4/3.

(c)    If the cross-ratio (AB, CD) = k, what is the cross-ratio (CD, AB)? What is the cross-ratio (BA, CD)?

(CD, AB) = (CA/CB)/(DA/DB) = (CA DB)/(CB DA) = (AC BD)/(BC AD) = k

(since AC = -CA etc, with an even number of mutliplications by -1)

(BA, CD) = (BC AD)/(BD AC) = 1/(AB, CD) = 1/k

(d)    Suppose CD divides AB harmonically.  Write down the definition that tells what this means.

Hamonic division means: AC/CB = AD/BD, which is the same as -(AC/BC) = AD/BD.

(e)    From the previous expression, tell what is the cross-ratio (AB, CD) if CD divides AB harmonically.

Dividing this equation to get a ratio of ratios, we get (AB, CD) = -1.

10-2 Composition of inversions in concentric circles

Suppose c1, c2, c3 are circles of radius r1, r2, r3 all with the same center O.  Let F, G, H denote inversion in c1, c2, c3 respectively. If P is a point, let P' = F(P) and P'' = G(F(P)) and P''' = H(G(F(P))). 

(a)    Based on the definition of inversion, each of the points P', P'', and P''' are on the ray OP.  Using the definition, write formulas for each of |OP'|, |OP''| and |OP'''| in terms of the circle radii and |OP|.

|OP'| = r12/|OP|

|OP''| = r22/|OP'| =( r22/r12)|OP|

|OP'''| = r32/|OP''| =( r32r12/ r22)(1/|OP|)

(b)    Based on your formula, tell what transformation of the plane is GF (tell the type and the defining details).

This is the formula for a dilation with center O and ratio of dilation r22/r12.

NOTE: The spelling of this word is DILATION not Dialation.

(c)    Based on your formula, tell what transformation of the plane is HGF (tell the type and the defining details).

This is the formulat for inversion in a circle with center O and ratio ( r3r1/ r2).

NOTE: The radius squared appears in the inversion formula, so the radius is the square root of the constant in the formula.

10-3 Composition of two other inversions

In the (x,y) plane, suppose J is inversion in the circle with radius 1 and center (0,0) and K is inversion in the circle with radius 3 and center (5,0).  We suppose a point P = (x,y).

(a)    Write a formula for J(x, y).

According to the formula for the inversion P' of P, in a circle of center O and radius R, |OP'| = R2/|OP|.
But P' is on ray OP, so vector OP' is a multiple of OP: OP' = kOP for some k > 0.
Putting these together, k|OP| = R2/|OP|, so k = R2/|OP|2

Applying this to J, we for O = (0,0), R = 1 and P = [x,y], J([x,y]) = (1/(x2+y2)) [x,y] = [x/(x2+y2), y/(x2+y2)]

(b)    Write a formula for K(x, y).

Applying the formula to K, with O = (5, 0) and R = 3, if [u,v] = K([x,y]),

[u,v] - [5,0] = (9/((x-5)2+y2)) [x-5,y]

[u,v] = K([x,y]) = [5,0] + (9/((x-5)2+y2)) [x-5,y]

(c)    Write a formula for KJ(x,y).

This is a substitution of one formula into another. So the unsimplified answer is easy to write. The simplified answer is quite complicated to compute.

Unsimplified: KJ(x,y) = [5,0] + (9/(({x/(x2+y2)}-5)2 +{y/(x2+y2)}2)) [{x/(x2+y2)}-5, y/(x2+y2)]

Simplified (but not so simple): (1/((5x-1)2+y2)) [80(x2+y2) - 41x + 5, 9y]

NOTE: This computation is very tough using (x,y) coordinates in this way. There is a nicer way -- use complex numbers. If we set z = x + iy and the conjugate z* = x - iy, then J(z) = 1/z* and K(z) = 5 + (9/(z*-5)). Then it is much easier to compute, since (1/z*)* = 1/z. KJ(z) = 5 + (9/(1/z) - 5) = 5 + (9z/1-5z) = (5 - 25z + 9z)/(1-5z) = (5-16z)/(1-5z) = (16z -5)/(5z -1).

In general the composition of two circle reflections has the formula (az + b)/(cz + d) for some complete constants a, b, c, d. If the centers of the circles are real, the a, b, c, d are real. Such transformations are called Mobius transformations.

To express this formula with a real denominator, multiply above a below by (5z* - 1) to get

KJ(z) = (1/|5z-1|2)(16z-5)(5z* - 1) = (1/|5z-1|2)(80zz*-16z - 25z* + 5) = (1/((5x-1)2+y2)) (80(x2+y2) - 41x + 5) +i(9y)).

This is quite a bit simpler than the straight (x,y) computation above.

10-4 The Poincaré Half-Plane Model

Since in inversive geometry, circles and lines are two special cases of I-circles or "circles", we can build another version of the Poincaré disk model by inverting the boundary circle of the disk into a line p, along with all the P-points and P-lines.  The P-points are the points on one side of line p (the points in a half-plane) and the P-lines in this model are either arcs of Euclidean circles orthogonal to p or rays on lines orthogonal to p.  Some things are easier to construct in this model and some things are harder, but all constructions can be figured out from basic circle constructions. (Note: You should be able to figure out the answers to these questions by reasoning, but you can also look up the half-plane model in many places.)

Explanations below. For Figures for 10-4 see this link. There are many illustrations of this at links on the Hyperbolic Links Page.

(a)    Draw a line m. Given two ("random") P- points A and B in the model, construct the P-line through A and B.  Give a second example of a P-line through points C and D for which the P-line is not an arc (i.e. a special case).

A P-line in this model is the arc of a circle orthogonal to p. Such a circle is simply a circle whose center is on p. So the problem is to construct a circle through A and B with center on p. The center is the intersection of the perpendicular bisector of AB with p. (Special case: If the perpendicular bisector is parallel to p, this means that line AB is perpendicular to p, so the P-line is a ray on this Euclidean line and not an arc.)

(b)    Given a P-point E and a P-point F, construct the P-circle with center E through F in the half-plane model.

The P-circle will be a Euclidean circle orthogonal to the P-lines through E. But the supporting circle of a P-line through E is not only a circle with center on p, it is also a circle through E and E', the reflection of E in p. This means that the supporting circles of the P-lines through E are the circles of the elliptic pencil of circles with base points E and E'. This means that the circles orthogonal to these circles are the circles of the hyperbolic pencil of circles with limit points E and E'. These circles are also the circles of Apollonius defined by E and E'.

To construct the P-circle through F, there are a couple of ways. One is to construct the circle of Apollonius through F with limit points E and E' (as done in 444 and explained in Ogilvy). To construct a diameter of the circle, intersect the interior and exterior angle bisectors of angle EFE' with line EE'.

A second way is to pick any two I-circles through E and E' and construct the circle through F that is orthogonal to these circles. (For simplicity, you can choose special circles that make this simpler, such as the line EE' (an I-circle) and the circumcircle of EFE'. But any two circles through E and E' will do.)

(c)    Given two "random" P- points G and H, construct the P-line that is the P-perpendicular bisector of GH (i.e., the reflection (circle inversion) of G in this P-line is H).

Let m be the supporting circle of the P-line that reflects G to H. Then the center of m has to be on the Euclidean line GH. Also, since m is orthogonal to p, the center has to be on p. Thus the center M of m is the intersection of p with line GH. To construct the radius, take any circle through G and H and construct a tangent segment MT from M to the circle. Then MT is a radius of m.

This construction works in every case except when line GH is parallel to p. This happens when G and H are the same Euclidean distance from p. For this case, let m = the Euclidean perpendicular bisector of GH. This is a line perpendicular to p (and this a support for a P-line that is a ray) and the P-reflection of G is the Euclidean line reflection = H.