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| Sastry's Evil Bisectors (Posted on 2005-03-04) |
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Prove that if a right triangle has all sides of integral length, then it has at most one angle bisector of integral length.
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Submitted by owl
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Rating: 4.2000 (5 votes)
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Bisector of the Right Angle
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Comment 10 of 10 | 
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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.
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a | \
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|/_____ \
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).
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Posted by Richard
on 2005-03-14 01:53:42 |
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