To my surprise there actually is another cheerful fact about the square of the hypotenuse. A Pythagorean triple is any integer solution of x^2+y^2=z^2 and, as we know, one gets all solutions by choosing any integers n>m. and defining T(m,n) by, x=n^2-m^2n, y=2mn, z=m^2+n^2. (T stands for triple and/or triangle) CHEERFUL FACT To see a given solution lay an mxn sheet of paper on the desk A_____ m ____ B n n C_____m_____D Fold and crease so that corner D falls on corner A. The triangle bounded by AB and the image of BD is (up to similarity) T(m,n). The proof is an exercise in 9th grade algebra. Surely this must have been noticed before, but where? Happy Holidays DG
A Pythagorean triple is any integer solution of x^2+y^2=z^2 and, as we know, one gets all solutions by choosing any integers n>m. and defining T(m,n) by, x=n^2-m^2n, y=2mn, z=m^2+n^2.
<pedant>Not quite. You need to allow for a scaling factor on x,y,z too. Scaling m,n only lets you scale x,y,z by squares.</pedant> -- g
On 12/11/05, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
A Pythagorean triple is any integer solution of x^2+y^2=z^2 and, as we know, one gets all solutions by choosing any integers n>m. and defining T(m,n) by, x=n^2-m^2n, y=2mn, z=m^2+n^2.
<pedant>Not quite. You need to allow for a scaling factor on x,y,z too. Scaling m,n only lets you scale x,y,z by squares.</pedant>
Er, not to mention permuting x and y. Or negative x (not excluded!). e.g. (x,y,z) = (6,8,10); (4,3,5); (-3,-4,5). Tut tut! Moving on a little, if D is superposed onto A, then its image D' = A lies on AB, and the "triangle" referred to appears to have zero area. Might this submission have benefitted from more careful presentation?
participants (3)
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David Gale -
Fred lunnon -
Gareth McCaughan