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Congruent Incircles (Posted on 2017-01-01) Difficulty: 4 of 5
Let ABC be a 3-4-5 triangle with right angle C. Let D be a point on the hypotenuse. CD then partitions ABC into ACD and BCD.

Where is D located if ACD and BCD have congruent incircles?

No Solution Yet Submitted by Brian Smith    
Rating: 5.0000 (1 votes)

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Solution No Subject | Comment 1 of 6

  |CD|^2 = |AC|^2 + |AD|^2 - 2|AC||AD|cos(A)
    y^2  =   4^2  +   x^2  - 2*4*x*(4/5)  
         =   x^2-32*x/5+16                 (1)

The inradius of a triangle is equal to the
area divided by the semiperimeter.
     2*Area(ACD)        2*Area(BCD)
  ---------------- = ----------------
   Perimeter(ACD)     Perimeter(BCD)
   |AC||AD|sin(A)      |BC||BD|sin(B)
  ----------------- = -----------------
    |AC|+|CD|+|AD|      |BC|+|CD|+|BD|
   4*x*(3/5)     3*(5-x)*(4/5)
  ----------- = --------------- 
     4+y+x         3+y+(5-x)
       20-7*x
  y = --------                             (2)
        2*x-5
Combining (1) and (2) give
   (20-7*x)^2
  ------------ = x^2-32*x/5+16
    (2*x-5)^2
               or
  x*(5*x^3-57*x^2+210*x-250) = 0
               or
  x*(x-5)*[5*x-16+-sqrt(6)] = 0
Therefore,
  x = 0, [16-sqrt(6)]/5, [16+sqrt(6)]/5, or 5.

We can disregard 0 and 5. Plugging 
[16+sqrt(6)]/5 into (2) gives y < 0;
thus we can eliminate it also. Plugging
[16-sqrt(6)]/5 into (2) gives y = sqrt(6);
which is confirmed by Geometer's SketchPad.
Therefore,
  |AD| = x = [16-sqrt(6)]/5 and |CD| = y = sqrt(6),
  where A is the vertex of triangle ABC opposite
  the leg of length 3.
QED
  

  Posted by Bractals on 2017-01-01 11:00:32
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