Hi,

here my solution:

- the center C1, C2 , C3 of the 3 outer circles form a symmetric
  triangle C1-C2-C3 of side 2, and height h=sqr(3)
  because of  h**2 + 1**2 = 2**2

- due to the tangent and symmetry conditions,
  the center I of the inner small circle is a point on line h

- consider the new triangle C1-C2-I, and draw h from C3 to the middle
  point  C12 of line C1-C2 (angles are 30, 30, 120 degrees)

- For the rectangular triangle C1-C12-I the following equation is
  valid (x = radius of small circle):

    line C1-I    =  1 + x
    line C1-C12  =  1
    line C12-I   =  h - 1 - x

    (1 + x)**2     =  1**2  +  (h - 1 - x)**2
    1 + 2x + x**2  =  1  + (h-1)**2 - 2 (h-1) x  + x**2
        2x         =       (h-1)**2 - 2 (h-1) x

   x ( 2 + 2(h-1)) =    (h-1)**2


      x    =   (h-1)**2   /   2h

   Then,

      surface of circle I =  x**2 * pi  =
           =   ((h-1)**4 / 4 h**2 ) * pi

        where h = sqr(3)


   Evaluation in Qedit :

   /=3.14159 * (1.73205 - 1)**4 / (4*1.73205**2)
   Result=.751850762778E-01   = 0.075185...


Kurt Sager

_______________________________________________________________
Kurt Sager
SWS SoftWare Systems AG, Freiburgstr. 634, CH-3172 Niederwangen
Phones: 031 9810666  / 079 607 7085     Fax: +41 (0) 31 9813263
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-----Original Message-----
From: VANCE,JEFF (HP-Cupertino,ex1) [mailto:[log in to unmask]]
Sent: Montag, 11. Oktober 1999 19:44
To: [log in to unmask]
Subject: OT: High School math question


Hi,

This is off topic.  My wife was observing a high school math
class and the homework assignment was as follows.  She asked me if
I could answer it.  My answer is .075, but I could be wrong.

Here is the question:

Assume 3 unit circles (radius = 1).  A point on the perimeter of each
circle is tangent to the other two perimeters.  There is another circle
inscribed in the middle space defined by the three larger circles.
The perimeter of this smaller circle has points tangent to the
perimeter of each of the three larger circles.  Find the area of the
small circle.

Thanks for any help!

Jeff