Conference rusure::math

From:	ROLL::USENET       "USENET Newsgroup Distributor" 16-JUL-1984 22:13
To:	HARE::STAN
Subj:	USENET net.math newsgroup articles

Newsgroups: net.math,net.puzzle
Path: decwrl!decvax!ittvax!dcdwest!sdcsvax!akgua!mcnc!ecsvax!pizer
Subject: Solution to Geometry II
Posted: Sun Jul 15 07:21:03 1984

References: <ecsvax.2878>,<decwrl.2523>

The following is the answer to the second geometry problem:

#2
>>This time, you have 5 circles, each externally tangent to the one next to it
>>and decreasing in size.  Each circle is also externally tangent to lines L1
>>and L2, which are not parallel.  If this is difficult to picture, imagine L1
>>and L2 being a funnel, with 5 marbles caught inside, each one touching the
>>funnel, and the one(s) next to it.  Anyway you are given that the radius of
>>the large circle is 18, the radius of the small circle is 8, and you are to
>>find the radius of the circle in the center.	HINT:  The answer is *NOT* 13!
>>Good luck!

In this problem, it appeared fairly obvious that the answer would either be the
arithmetic or geometric mean.  Since I told you it wasn't 13, logic would
indicate the answer to be twelve.  My answer involved setting up proportions
and then solving for a variable.  A more complete answer was recently posted by
Van Luong Nguyen <decwrl.2523> explaining how since the sequence is simply a
geometric progression, any circle's radii can be calculated in the sequence.
His explanation will also save me the trouble and space of going through and
solving it now.  Several other methods were mailed to me, but they all involved
setting up different proportions.


Since this is the shortest solution, I'll stick this here:

Credit must also be given to the people who got these questions together.  All
of the problems came off of this year's and last year's American High School
Math Exam (AHSME), sponsored by the Mathematical Association of America,
Society of Actuaries, Mu Alpha Theta, National Council of Teachers of
Mathematics, and Casualty Actuarial Society.  This exam is a qualifying exam
for the American Invitational Mathematics Exam (AIME).	 Questions about
ordering problem books can be directed to:

  Prof. Walter E. Mientka, Executive Director
  MAA Committee on High School Contests
  Department of Mathematics and Statistics
  University of Nebraska
  Lincoln, NE 68588-0322


Billy Pizer
({decvax,akgua,ihnp4,burl}!mcnc!ecsvax!pizer)

-- "Why is air speed different from ground speed?"
   "Because the Earth is round and the air is flat." --

Newsgroups: net.math
Path: decwrl!dec-rhea!dec-tonto!luong
Subject: Radius of third circle in series of five circles in a funnel.
Posted: Fri Jul 13 12:50:57 1984

There must be many possible ways to solve this problem. The solution below
gives the radius of not only the third circle, but of any circle in the series.

Terminology:
-----------
By symmetry, the tangents L1, L2, and the line L3 joining the centers of the
five circles are concurrent, meeting at the same point A.

Call the centers of the circles C1,C2,C3,C4,C5. Let line L3 meet the smallest
circle 1 at B1, the common point to circles 1 and 2 at B2, the common point
to circles 2 and 3 at B3,  etc...

Call the distance ABn = X(n), and call by R(n) the radius of circle n.

Thus, for example: 	AB1 = X(1)
			AC1 = AB1 + B1C1 = X(1) + R(1)
			AB2 = X(2) = AB1 + 2*R(1) = X(1) + 2R(1)  etc...

Proof:
-----

From centers C1, C2, C3, ... draw perpendiculars C1H1, C2H2, C3H3, ... to
the tangent L1. Thus C1H1 = R(1), C2H2 = R(2) ...

Let  X(1) = AB1 = d .

Since C1H1 and C2H2 are parallel, C2H2 = AC2   ie.  R(2) =  d + 2R(1) + R(2)
				  ----   ---        ----    ----------------
				  C1H1   AC1        R(1)      d + R(1)
Using R(1) = 8  for solving:
				R(2) = 8 [(d + 16)/d]
Also, X(2) = AB2 = d + 2R(1)
	therefore:		X(2) = (d + 16)

By recursion, it is easy to prove that, for n >= 2 : 

	R(n) = 8  [(d+16)/d]**(n-1)         (to the (n-1)th power)
and:
	X(n) = [(d + 16)**(n-1)] / [d**(n-2)]

(A recursion proof shows that if a statement or formula is true for n, it will
also be true for (n+1). Since the statement is true for n=2, it follows that
it is also true for all n>=2 . I will leave the recursion proof to the reader.)

For n=5, R(5) = 8 [(d+16)/d]**4 = 18    therefore  [(d+16)/d] = (18/8)**(0.25)

For n=3, R(3) = 8 [(d+16)/d]**2 = 8 [(18/8)**0.25]**2 = 8 [(18/8)**0.5]

		Thus:   R(3) = 12
			=========

For any general value n>=2 :

	R(n) = 8 [(18/8)**(0.25*(n-1))]

For example, if we draw four more larger circles 6,7,8,9, then:

	R(9) = 8[(18/8)**2] = 40.5


Van Luong Nguyen,
Digital Equipment Corp.

    Maybe I'm not a very fastidious and organized mathematician, but
    reply .2 seemed like a bit of overkill.
    
    I reasoned as follows:
    
    	Look at a figure containing the smallest two circles (C1 and
    	C2) and the two lines.  Compare that to a figure containing
    	C2 and C3 and the two lines.  In each figure, draw 4 radii
    	from the circles to the lines on both sides, and draw the line
    	between the two centers.   Notice that the lines and angles
    	between them are sufficient to determine the figure, and all
    	the angles are the same.  This means that the two entire figures
    	are similar (with a common ratio R3/R2), and from there I concluded
    	that R1/R2 = R2/R3, and that the radii are a geometric series.
    	
    	Now that I'm writing it down, this method looks awfully convoluted,
    	but it took me about 20 seconds to figure it out, and my answer
    	was not a guess.
    
    	Does this explanation give anybody any insights, or do I just
    	have weird thought processes?
                       
    	John
            

    Re .3:

    John H.:
    
    I don't see why R3/R2 and R2/R1 are equal from your argument, and
    I don't think you've used the fact that the circles are tangent to 
    one another.
    
    John M.

    Re: .3:
    
    John H.:
    
    I think you have not proved that the line joining C1 and C2 is the
    same line as the one joining C2 and C3.  You need to demonstrate
    this before you can argue the figures are similar.
    
      John H.

    The figure looks a bit like this:

	       /
	      /
	     /
	    /        C
	   /
	  /    B
	 / A
    	+--------------------

    I've used A, B and C to indicate the centers of the circles.

    Now, the claim is that

	the figure formed by the two lines and the A and B circles

    is similar to (i.e., same shape and angles, but different size)

	the figure formed by the two lines and the B and C circles.


    Now the similarity implies that lengths in the first figure are a
    constant factor larger (or smaller) than lengths in the second figure.

	radius of cicle A  x  constant  =  radius of circle B
	radius of cicle B  x  constant  =  radius of circle C


    Now the rest is trivial.  The result is that the radii of the circles
    are in a geometric progression.

    Suppose in reality the figure looks like this

	       /
	      /      C
	     /
	    /      			(skew exaggerated for effect)
	   /
	  /       B
	 / A
    	+----------------
    							   ___
    The similarity argument breaks down.  You need to show ABC is a
    single straight line before you get to the trivial part.
    
    In point of fact I'm certain (In E-space, anyway) that ABC is one
    straight line, but that has yet to be demonstrated.  You can't just
    draw the figure and then draw conclusions about its properties from
    visual inspection.
    
      John

    Sigh.  Points A, B, and C are colinear.  To see this, realize that
    the locus of points equidistant from the two lines are the two lines
    that bisect the angle between the lines.  Points A, B, and C are on
    one of these bisectors (the same one, in fact, since A, B, and C are
    all in the same 'quadrant'), because each is equidistant from the
    lines, by the construction of the problem (or drop perpendiculars
    to each line, and use the fact that all radii of a circle are equal).

    I haven't a quick way to show that the locus of points equidistant
    from two intersecting lines are the two lines bisecting the angle.
    However, this isn't really needed for the similarity argument, since
    a figure formed by the two lines and any one of the circles is similar
    to any other such figure, and given such a figure, the next outer
    circle is uniquely determined, independent of the proportions.  Thus,
    the figures with *two* circles must be similar.

    "Dropping perpendiculars" is (I think) a key step since you have
    to rely on the fact that the tangent to a circle is perpendicular
    to the radius at the point of intersection.  This gets back to the
    objection in .7 about using the tangent property.  So by doing this
    for any circle you end up with two right triangles with a common
    side and another side equal.  Invoke the Pythagorean theorem to
    prove the third sides are equal, and the two triangles are congruent by
    side-side-side.  So the centers of the circles are on the bisector of
    the angle at the intersection of the tangents, and are thus collinear.
    
    Unfortunately this only works in E-space because the Pythagorean
    theorem is equivalent to the parallel postulate.  (For that matter, so
    is side-side-side).  Too bad there doesn't seem to be a way to prove
    this without them... 

      John