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Incircles in equilateral triangles (Posted on 2014-01-18) Difficulty: 3 of 5
In an equilateral triangle ABC the lines AC'A', BA'B', CB'C' are drawn making equal angles with AB, BC, CA, respectively, forming the triangle A'B'C', and so that the radius of the incircle of triangle A'B'C' is equal to the radius of the incircle of triangle AA'B. Find A'B' in terms of AB

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Solution Comment 1 of 1

Let x = /BAA' and apply the law of sines to triangle AA'B:

         |AB|             |BA'|            |AA'|
     ------------  =  ------------  =  ------------ 
      sin(/AA'B)       sin(/BAA')       sin(/ABA')

                           or

        |AB|         |BA'|          |AA'|
     ----------  =  --------  =  -----------
      sin(120)       sin(x)       sin(60-x)

                           or

     |BA'|  =  2|AB|[sin(x)*sqrt(3)]/3  and

     |AA'|  =  |AB|[3cos(x) - sin(x)*sqrt(3)]/3

If [PQR] denotes the area of triangle PQR, s its 
semiperimeter, and r its inradius; then

     r = [PQR]/s

For triangle A'B'C':

          [|A'B'||A'C'| sin(/B'A'C')]/2
     r = -------------------------------
            [|A'B'|+|B'C'|+|C'A'|]/2

       = |A'B'|*sqrt(3)/6

       = (|AA'| - |BA'|)*sqrt(3)/6

       = |AB|*[cos(x)*sqrt(3) - 3sin(x)]/6        (1)

Note:  

   |A'B'| = [cos(x) - sin(x)*sqrt(3)]*|AB|        (2)
            All we need is x.

For triangle AA'B:

          [|AB||AA'| sin(/BAA')]/2
     r = --------------------------
            [|AB|+|BA'|+|AA'|]/2

          |AB|*sin(x)[3cos(x)- sin(x)*sqrt(3)]
       = --------------------------------------   (3)
              3 + 3cos(x) + sin(x)*sqrt(3)

Combining (1) and (2) we get

     sin(x)*[3 + 8cos(x)] = [1 + cos(x)]*sqrt(3)

Using MathCad we get

        cos(x) = [sqrt(3) + sqrt(7)]*sqrt(3)/8  and
     sin(x) = [3*sqrt(3) - sqrt(7)]/8

Plugging these into (2) gives

     |A'B'| = [sqrt(21) - 3]/4 * |AB|
           ~= 0.39593 * |AB|

QED

Edited on January 23, 2014, 4:29 am


  Posted by Bractals on 2014-01-23 05:08:06
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