Conference rusure::math

                                   <<< Note 1277.1 by ALLVAX::JROTH "It's a bush recording..." >>>
                                                      -< but won't x3 = x1? >-


    I think I see the confusion. When the second and third
    inscribed circles are drawn, always draw the new one.
    In other words, the are always TWO circles which 

	o   pass through a fixed point on the inner circle     
 
	o   pass through the center of inner circle

	o   are tangent to outer circle

    Always pick the new circle at the second and third
    stages of the construction always moving in fixed
    direction around the inner circle. If this is done, 
    x1 and x3 will ALWAYS be distinct.

    -Franklin

Re .0

Actually not true, provided I've got it right. X1 is not always equal to X4.


Think about a special case, which is however included in your description:
The first circle (outer) and the second circle (inner) have the same center. 
(Does not really have to be, but it is easier to think.)

Then draw a third circle which is tangent to the outer circle and passes the 
center. Then imagine what happens to the passing point of the inner and the 
third circle (either one) if you change the size of the inner circle.

You see, as the size of the inner circle gets closer and closer to the outer 
circle, the amount of additional circles that must be drawn until their passing 
points (with the inner circle) overlap each other increases. So the number 4 
can  not have any special meaning (neither 3).

In other words: when the inner circle is very near the outer circle in size, 
you must draw very many additional circles until you would be even able to 
cover the whole 'word_I_do_not_know_but_is_two_times_pi_times_radius' of the 
inner circle with those points that are constructed by the passing points of 
the additional circles and the inner circle and the adjacent passing point.


Huh! My Geometrical English is rather poor (obvious, isn't it :-), but I hope I 
made myself clear.


P.S. Just by estimating from the figures I drawed with DECwrite, the number 4 
under discussion before is special number if the radius of the inner circle is 
about 2/3th of the radius of the outer circle. Is it? ;-).
    
    
    Best regards,
    Toffe

    re: -.1

    > Actually not true, provided I've got it right. X1 is not 
    > always equal to X4.

    That's the point; its NOT always true and in general won't be. 
    However, if one CAN find a configuration of two circles and a 
    starting point x1 on the inner circle with x1 = x4 then what I 
    am claiming is that x1 = x4 does not depend on which starting
    point you choose.

    As you correctly point out, one can easily find such a 
    configuration by choosing concentric circles with the appropriate 
    ratio of radii. Unfortunately, this example trivially satisfies 
    the independence of starting point choice. What is not as simple
    is how to construct examples with NON-concentric circles with 
    x1 = x4 and then to demonstrate that this condition is independent 
    of the choice of x1.

    -Franklin    

    Re: .0

	My apologies, the construction as stated .0 is not 
    correct and needs to be modified as follows: 

	Fix two circles in the plane with one inside the other,
    not necessary concentric and not tangent.

	 ASSUME HOWEVER THAT THE CENTER OF THE OUTER CIRCLE 
	 IS CONTAINED IN THE INTERIOR OF THE INNER CIRCLE. 

    Fix a point x1 on the inner
    circle. Draw one of the two circles that passes through
    x1 and the center of the *OUTER* circle and is tangent to 
    the outer circle. This new circle will intersect the inner
    circle at a new point x2. Now draw the new circle passing
    through x2 and the center of the *OUTER* circle and is 
    tangent to the outer circle. This new circle will intersect 
    the inner circle at a new point x3. Again draw a new third
    circle passing through x3 and the center of the *OUTER*
    circle and is tangent to the outer circle. Let x4 be
    the new point of intersection of the newly drawn third
    circle and the inner circle.

	Show that if x1 = x4 then if we repeated the 
    construction starting with a y1 on the inner circle
    to produce y2, y3 and y4, then y1 = y4 also.

	Said differently, if the construction has period
    3 for a given starting point then it always has 
    period 3 for any starting point.
	
	Having botched this up terribly the 1st time, let me
    attempt to make amends by describing a method for generating 
    lots of non-concentric examples of this configuration:

	Start with any circle A for our candidate outer circle. 
	Draw a new circle C1 which passes through the center of 
	A and is tangent to it. Draw a new circle C2 which passes 
	through the center of A, is tangent to A and intersects
	C1 transversally at the point x1 say. Now draw a circle
	C3 which passes through the center of A, is tangent to A and 
	intersects both C2 and C1 transversally at the points
	x2 and x3 respectfully. Our candidate inner circle is
	the unique circle containing x1, x2 and x3 with starting
	point x1.