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Circles within a Circle


Date: 10/12/2000 at 21:50:07
From: Brian Tinker
Subject: Geometric problem.

You have three circles: One has a diameter of 3 inches, another has a 
diameter of 2, and the third has a diameter of 1. The two smaller ones 
are put inside the larger one in a way so that the combined diameters 
of them make up the length of the largest circle's diameter. 

If you put a fourth circle in the leftover area, what is the largest 
area, or diameter, that this fourth circle could be? It has to touch 
only one point on the other three circles, but cannot cross the 
circles. 

I've figured out where the circle would be drawn, but can't figure out 
the whole thing using only math. Here is a graphic of this problem:

  


Date: 10/14/2000 at 07:03:12
From: Doctor Floor
Subject: Re: Geometric problem.

Hi, Brian,

Thanks for writing.

When you leave off the lower halves of the circles, the remaining 
figure of semicircles is called the arbelos, or shoemaker's knife. In 
general the smaller circles can have different radii, say a and b, and 
the greatest circle then of course has radius a+b. 

This figure of the arbelos was studied by Archimedes.

The circle you are looking for is called the incircle of the arbelos.

Let O1 be the center of the circle with radius a, O2 the center of the 
circle with radius b, and O the center of the circle around them. Let 
the incircle have center C and radius r.

                               ab(a+b)
Then we will find that r = --------------
                           a^2 + ab + b^2

Proof:

Let <COO2 = t. Note that O1C = a+r, O2C = b+r and OC= a+b-r. 
Note also that OO1 = b and OO2 = a.

By the Law of Cosines in triangles OO1C and OO2C we have:

   (a+r)^2 = (a+b-r)^2 + b^2 - 2b(a+b-r) cos(180-t)
           = (a+b-r)^2 + b^2 + 2b(a+b-r) cos t........[1]

   (b+r)^2 = (a+b-r)^2 + a^2 - 2a(a+b-r) cos t........[2]

From a*[1]  + b*[2] we get:

    a(a+r)^2 + b(b+r)^2 = (a+b)(a+b-r)^2 + ab^2 + ba^2

which reduces to

    a^3 + b^3 + 2(a^2 + b^2)r + (a+b)r^2 
          = (a+b)^3 - 2(a+b)^2r +(a+b)r^2 + ab^2 + a^2b
    a^3 + b^3 + 2(a^2 + b^2)r = (a+b)^3 - 2(a+b)^2r + ab^2 + a^2b
    4(a^2+ab+b^2)r = (a+b)^3 + a^2b + ab^2 - a^3 - b^3
    4(a^2+ab+b^2)r = 4(a^2b+ab^2)

          ab(a+b)
    r = ----------
        a^2+ab+b^2

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Conic Sections/Circles
High School Geometry

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