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Collinear Intersections (Posted on 2007-09-13) Difficulty: 3 of 5
Three circles A, B, and C have non-collinear centers, unequal radii, and pairwise the distance between their centers is greater than the sum of their radii.

Let P, Q, and R be the intersections of the external tangents to circles A&B, A&C, and B&C respectively.

Let L and M be the intersections of the internal tangents to circles A&B and A&C respectively.

Prove that P, Q, and R are collinear.

Prove that L, M, and R are collinear.

  Submitted by Bractals    
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Solution: (Hide)
Let XY denote the vector from point X to point Y.

If points X, Y, and Z are non-collinear and XW = aXY + bXZ, then
W lies on line YZ if and only if a + b = 1.

Let X and Y be the centers of circles with radii x and y ( |XY| > x + y ).
If E and I are the intersections of the external and internal tangents respectively, then
               x                         x
        XE = ----- XY     and     XI = ----- XY
              x-y                       x+y
For our problem we have
               a                         a
        AP = ----- AB     and     AL = ----- AB
              a-b                       a+b

               a                         a
        AQ = ----- AC     and     AM = ----- AC
              a-c                       a+c

               b                        
        BR = ----- BC  
              b-c                   

       -----------------------------------------------------

                              b
        AR = AB + BR = AB + ----- BC                             
                             b-c

                    b
           = AB + ----- (AC - AB)     
                   b-c

               c          b
           = ----- AB + ----- AC    
              c-b        b-c

              c(a-b)        b(a-c)
           = -------- AP + -------- AQ    
              a(c-b)        a(b-c)

              c(a+b)        b(a+c)
           = -------- AL + -------- AM    
              a(c-b)        a(b-c)

       -----------------------------------------------------
Since
            c(a-b)     b(a-c)
           -------- + -------- = 1  
            a(c-b)     a(b-c)
P, Q, and R are collinear. Since
            c(a+b)     b(a+c)
           -------- + -------- = 1  
            a(c-b)     a(b-c)
L, M, and R are collinear.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Monge's Circle TheoremBrian Smith2009-11-07 12:11:10
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