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3 INCIRCLES (Posted on 2014-01-25) Difficulty: 4 of 5
In ΔABC, AB = BC. If one chooses D on side AB and J on CD such that AJ ⊥ CD and the incircles of ΔACJ, ΔADJ, and ΔBCD all have radius r, then r = AJ/4.

Prove this using (i)Geometry (ii)Trignometry

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 3.0000 (1 votes)

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Solution Comment 1 of 1

For the right triangles AJC and AJD to have an equal inradius,
the triangles must be congruent. WLOG let |AC| = AD| = 1 and
x = /CAJ and /DAJ. Then the following lengths and angles are 
easily found

   |AJ| = cos(x)
   |CJ| = |DJ| = sin(x)
   |BC| = 1/[2*cos(2x)]
   |BD| = [1 - 2*cos(2x)]/[2*cos(2x)]
   /ACJ = /ADJ = 90-x
   /CBD = 180-4x

From the configuration we have

   /ACD < /ACB < 90

        or

   30 < x < 45

        or

   sqrt(2)/2 < cos(x) < sqrt(3)/2                    (1) 

The inradius of a triangle is equal to its area divided by its
semiperimeter. For triangle AJC,

                |AJ||CJ|/2               cos(x)*sin(x)
     r = ------------------------ = -----------------------
          [|AJ| + |CJ| + |AC|]/2     |cos(x) + sin(x) + 1]

For triangle BCD,

           |DB||DC|*sin(/BCD)/2
     r = ------------------------ = 
          [|DB| + |DC| + |BC|]/2

          cos(x)*sin(x)*[1 - 2cos(2x)]
       = -------------------------------
          1 - cos(2x) + 2sin(x)*cos(2x)

Setting these two equal to each other gives,

              [1 - 2cos(2x)]
     -------------------------------
      1 - cos(2x) + 2sin(x)*cos(2x)

                  1
     = -----------------------                             (2)
        |cos(x) + sin(x) + 1]

Using MathCad cos(x) = 4/5 is the only solution of (2)
that satisfies (1). Therefore, sin(x) = 3/5 and

              (4/5)*(3/5)         1    
     r = --------------------- = --- = |AJ|/4
          [(4/5) + (3/5) + 1]     5    

QED
      

Edited on January 25, 2014, 9:23 pm
  Posted by Bractals on 2014-01-25 21:21:52

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