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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.

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Solution Product in terms of a,b,c only | Comment 9 of 10 |
First, a result needed below will be presented.  Consider the following figure.

                                                     E
                                                      *

                             A
                              *


      *                       *                       *
      B                      D                      C


ABC is a given triangle and AD is the bisector of its angle at A.  EC is constructed parallel to AD and intersects the line through AB at E.

It is to be shown that BD is to DC as BA is to AC (the Angle Bisector Theorem).

BD is to DC as BA is to AE. The angles BAD, DAC, AEC, and ACE are all equal which makes triangle AEC isoceles. Thus AC=AE, and the result is proved.


Now consider the right triangle with hypotenuse c and sides a and b.  Using the Angle Bisector Theorem just proved above, the squares of the lengths of the bisectors of the angles that are not the right angle, when multiplied together, yield

{b^2 + [a*b / (b+c)]^2} * {a^2 + [b*a / (a+c)]^2}.

The first factor in { } may be written as

{b^2 * [(b+c)^2 +  a^2]} / (b+c)^2

which by expanding the square and utilizing the theorem of Pythagorus becomes

2 * b^2 * (c^2 + b*c) / (b+c)^2

which simplifies to

2 * c * b^2 / (b+c).

Similarly, the other factor equals

2 * c * a^2 / (a+c).

Hence the product of the two factors equals

4 * c^2 * {a^2 * b^2 / [(b+c)*(a+c)]},

which, since a^2 * b^2 = (c^2 - b^2) * (c^2 - a^2), equals

4 * c^2 * (c-a) * (c-b).

But it is easily shown that 2*(c-a)*(c-b)=(a+b-c)^2.

Hence the product of the lengths of the bisectors of the angles that are not the right angle is given by

c * (a + b - c) * sqrt(2).

This quantity is irrational whenever the right triangle has integer (or rational) sides.Since no integer is irrational, but the product of integers is always an integer, we must conclude that at least one of the bisectors of the angle that is not the right angle is not an integer. As David Shin has pointed out, the bisector of the right angle is never an integer.  Thus the assertion of the problem has been established.

The 28,96,100 right triangle has an angle bisector of length 35, by the way.




  Posted by Richard on 2005-03-12 19:49:06
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