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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

484.0. "Here is the problem..." by CHAPLN::DAY () Thu May 08 1986 14:02

    The center of circle B is at an arbitrary point on the circumference
    of circle A.  The radius of circle A is 1 unit.  The radius of circle
    B is such that the area common to both circles is half the area of
    circle A.
    
    What is the radius of circle B?
    
T.RTitleUserPersonal
Name
DateLines
484.1CLT::GILBERTJuggler of NoterdomTue May 13 1986 19:521
    This is a well-known problem with no simple (closed form) solution.
484.2Oh yes there is!ZEPPO::DAYWed May 14 1986 09:492
    Re .-1  It does have a closed form solution.  - Bob Day
    
484.3use integrationSIERRA::OSMANand silos to fill before I feep, and silos to fill before I feepThu May 15 1986 14:5211
    I don't recall my integral claculus, but it seems to me that finding
    the area of a region formed by the overlap of two circles was a
    closed form expression of the equations of the two circles.
    
    One circle's radius is given as 1, and the other we can call R.
    Write down the expression in terms of R and set it equal to
    one half of pi (the area of circle A), and solve for R.
    
    Does this help ?
    
    /Eric
484.4solution to .0 and also to 496.0THEBUS::KOSTASKostas G. Gavrielidis <o.o> Tue Jun 03 1986 10:5351
    The difficulty in solving this problem arises from the fact that
    it can be solved only by successive approximations. Values should
    not be tried at random, but one should be quided by systematic
    interpolation. This is a good example of the use of the Regula
    Falsa.
    
    
    Half the area of the unit circle is   pi/2 = 1.5708

    Let  G  be the corresponding amount of area formed by the overlap
    of the two circles.
        
    if  R = 1.15  then the area  G = 1.5513   which is  0.0195 too small
    if  R = 1.16  then the area  G = 1.5740   which is  0.0032 too large
    
    after more interpolation
    
    we get  R = 1.1588    and   G = 1.5709
    
    Let see
    
            R = 1.1588
            sin x/2 = .5794
                x/2 = 35.41 degrees
    
                  y = 54.59 degrees
                  x = 70.82 degrees
              log R = 0.0640
    
            log R^2 = 0.1280
          log rad y = 9.9790-10
                      ----------
                      0.1070
    
                    1.2793
            rad x = 1.2217     
                    +  143
                    ------
                    2.5153
            sin x =  .9444
                    ------
        
                G = 1.5709   
    
    
    Note: Some of the computations are not included in here also
          no pictures were provided.
    
    Enjoy,
    
    Kostas G.
484.5Solution...ZEPPO::DAYWed Jun 11 1986 10:5711

    The answer is given by the equation:

    sin(theta) - theta * cos(theta) - (pi/2) = 0

    Theta is the angle between the center of circle B and the points
    where circle B intersects circle A.  Solve the equation for the
    value of theta that is between 0 and PI (the solution that makes
    physical sense).  Then the radius = 2 * cos(theta/2).

484.6History?STAR::BRANDENBERGCivilization is the progress toward a society of privacy.Wed Jun 11 1986 16:216
    
    I'm curious where the question comes from.  I know it from a math
    competition in High School ( I don't remember if I solved it ).
    
    				Monty
    
484.7re .6ZEPPO::DAYWed Jun 11 1986 17:048
    
    I heard it about 10 years ago from a friend.  He said that the
    only solution that exists is a series solution.  The equation
    given in .5 doesn't seem to be well known...  I'd also like to
    know more about the history and solutions to this problem.
    

    
484.8CLT::GILBERTJuggler of NoterdomWed Jun 11 1986 19:285
re .2,.5

The problem is that there is no closed form solution for theta.

That is, no equation: theta = some `simple' expression.
484.9Here are the numbers...MODEL::YARBROUGHThu Jun 12 1986 16:112
    Re .5: the principle value of the solution is theta = 1.9056957293...,
    and r = 1.1587284730... - in case anyone is interested.
484.10meaning of 'closed form'ISWISS::DAYFri Jul 25 1986 12:1513
re .1, .8

I checked in a number of mathematical dictionaries and encyclopedias
to try to find a definition of 'closed form solution', and it did not
appear in any of the references I checked.  So I guess it's one of those
terms that, to some degree, can mean what one means it to mean.  
My understanding is that a closed form solution is one that does not 
require a series to express it.

Be that as it may, 'explicit' is the term that is usually applied to a 
solution in which the independent variable is isolated on the left
side of an equation, and the independent variables are on the right.

484.11that reminds me of an anecdoteMODEL::YARBROUGHFri Jul 25 1986 14:214
    I knew of a Math prof. in college with a name that sounded like
    "Trijinski" but had several more letters in it. Once, when a Graduate
    student asked him to spell his name, he replied, " You can't spell
    'Trijinsky' in closed form; you must expand it in a Fourier series."