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Hajja's Rational Cosine (Posted on 2005-03-02) Difficulty: 2 of 5
Prove that the angles of a triangle all have rational cosines if and only if the triangle is similar to one with rational sides.

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Other Half Comment 2 of 2 |

Let the triangle have vertices A, B, and C with respective opposite sides of lengths a, b, and c.

Now cos C = <CB,CA> / a*b, and cos A = <AC,AB> / b*c, where <v,w> denotes the scalar (or "dot") product of the vectors v and w,  and AB, for example, denotes the vector from A to B. We have cos A = <-CA,CB-CA > / b*c = <-CA,CB > / b*c + <-CA,-CA > / b*c  and hence c*cos A = b - a*cos C since <-CA,-CA > = <CA,CA > = b^2. Thus we have the identities

a*cos C + c*cos A = b

a*cos B + b*cos A = c

b*cos C + c*cos B = a.

Substituting the value for a from the last one into the first, we obtain the relation

b*(1-cos^2 C) = c*(cos A + (cos B)*(cos C)) 

so that b and c are in a rational ratio to one another when the cosines are all rational.

 

 


 


  Posted by Richard on 2005-03-02 22:59:26
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