[alogo] 1. Cross ratios of parameterized lines

Given is an angle XAY, a line L with a line-coordinate t on it, a fixed point O and the variable line OBt, where Bt is the point on L with coordinate t. Then the cross ratio of the four points (PX, QY, Bt, O) = f(t) is a Moebius function of t of the form.
                                                                                     f(t) = (at+b)/(ct+d).
Here PX, QY denote the intersections of line OBt with AX and AY correspondingly.
Inversely, given the two lines {AX, AY}, the fixed point O, and the Moebius function f(t)=(at+b)/(ct+d), introduce a coordinate t on AX and for each point Pt on it consider the intersection Qt of OPt with AY and the point Bt on line OPt such that (Pt, Qt, Bt,O) = f(t). Then Bt moves on a line L as Pt varies on AX.

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To prove this take the coordinates-origin at O, and consider the (variable) vector u = OBt. Consider also the unit vectors uX, uY along AX and AY correspondingly. Then the intersection points PX, QY are described on lines AX, AY by parameters r, r' and on line OBt by parameters s, s' such that.
         sOBt = A + ruX = PX           and                s'OBt = A + r'uY = QY.
From there, denoting by J(Z) the pi/2-rotated of Z and by (..., ...) the inner product we get.
         s(OBt,JuX) = (A,JuX)  =>            s = (A,JuX)/(OBt,JuX).
Analogously          s' = (A,JuY)/(OBt,JuY).
If L is described parametrically by  Bt = b0 + te, where e is a unit vector in the direction of L, then {s, s'} are easily seen to be of the form.
         s = 1/(mt+n)   and   s' = 1/(m't+n').
Since the cross-ratio is  (PX,QY,Bt,O) = [(1-s)/(1-s')]:[s/s'], introducing in this the previous expressions we find that it has indeed the desired form.
To prove the inverse consider a fixed line M through O and take the trace of the line-bundle at A.
A(X,Y,Bt,O) which has the same cross ratio given by the function f(t). Since the traces of OX, OY are fixed points on M, the trace of OBt on M is a point described by a function  g(t)m, where g(t) is a Moebius function (resulting easily from f(t)) and m is a unit vector along M. Then point Bt can be described as intersection of lines OPX and ABt.
       Bt = A + r'(g(t)m) = rPX, where  r, r' are line coordinates along lines ABt and M.
Taking inner products with J(m) we get.
                 (A,J(m)) = r(PX,J(m)).
Taking a parametrization along A:  PX = A + tuX implies that r is of the form                 r = 1/(at+b).
Hence     Bt = rPX = (1/(at+b))(A + tuX).
Putting  A' = A/b, u'X we see that  Bt = (1/(at + b))[ bA' + at u'X].
This shows that Bt is a point of the line  through  A' and  u'X.

[alogo] 2. Turning angles

Consider a fixed line L, a point A on it and a point O not lying on the line. Draw three lines through O intersecting line L at points B, C, D correspondingly and turn these three lines about O so that their mutual angles remain constant. Then the cross ratio (A,B,C,D) = f(t) is a Moebius function of the line coordinate t = AB on line L.

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This is essentially the first part of the previous exercise. But another argument results from the reduction of the cross ratio of bundles to ratios of sines (see Harmonic_Bundle.html ). (A,C,B,D) = (sin(v)/sin(v')):(sin(w)/sin(w')), where v = angle(OA,OB), v'=angle(OA,OD), w=angle(OC,OB), w'=angle(OC,OD).
In fact, by the assumed constancy of angles the second ratio sin(w)/sin(w') is constant.
Since the cross ratio is the same for all secants of the line bundle at O.
O(A,B,C,D), we can replace L with a line L' orthogonal to the fixed OA at A. Then the sin(v) is expressed through t = AB' (B' the trace of OB on L') through its tangent which is tan(v) = t/OA = t/a (putting a = |OA|). This implies sin2(v) = tan2(v)/(1+tan2(v)) = t2/(t2+a2),       cos2(v) = a2/(t2+a2).
Besides v'= v+w+w' = v+w'', where w''=w+w' implies  sin(v') = sin(v)cos(w'')+cos(v)sin(w'').
Thus the ratio (sin(v)/sin(v')) has the form   t/(tcos(w'')+asin(w'')) and this proves the claim.

[alogo] 3. Turning angles II

Consider a fixed angle XAY and a fixed angle X'OY' turning about the fixed point O. Let L denote the line joining the intersection points B=(AX,OX') and C=(AY,OY'). Let also D be the intersection point of L with a line OZ making constant angles with OX', OY'. Then D describes a line as X'OY' turns about O.

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Define the additional intersection points as shown in the figure. Introduce also a line coordinate t = AB along line AX.
By section-2 it suffices to show that the cross ratio f(t) = (O,D,E,F) is a Moebius function. Considering the bundle of lines at C: C(O,D,E,F) and taking its tracce on line AX this ratio transfers to (I,B,A,F), which by section-3 has the desired form of a Moebius function of t. Hence D moves on a line.

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