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Sastry's Evil Bisectors (Posted on 2005-03-04) Difficulty: 5 of 5
Prove that if a right triangle has all sides of integral length, then it has at most one angle bisector of integral length.

No Solution Yet Submitted by owl    
Rating: 4.2000 (5 votes)

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Bisector of the Right Angle Comment 10 of 10 |
The bisector of the right angle of a right triangle is easily handled with  linear equations. Orient the triangle on the usual (x,y) coordinate system as shown below.

     | \
     |   \
  a |     \
     |    / \
     |  /     \
     |/_____ \
    C      b

Then the bisector of the right angle at C has equation y=x, whereas the hypotenuse has equation x/b + y/a =1. These intersect at x=y=1/(1/a + 1/b)=a*b/(a+b).  The length of the bisector is therefore [a*b/(a+b)]*sqrt(2).

  Posted by Richard on 2005-03-14 01:53:42
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