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Primitive Pythagorean Triangles (Posted on 2008-01-26) Difficulty: 3 of 5
A primitive Pythagorean triangle (PPT) is a right triangle whose side lengths are integers that are relatively prime.

1) Prove that the inradius of a PPT has a different parity than the mean of the hypotenuse and the odd leg.

2) Prove that there exists an infinite number of pairs of non-congruent PPTs such that both members of the pair have the same inradius.

  Submitted by Bractals    
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Solution: (Hide)
A triangle ABC with a2 + b2 = c2 is a PPT if and only if there exists a pair of integers (m,n) that are relatively prime with different parity such that

        c = m2 + n2
        b = m2 - n2
        a = 2mn

Let r and s be the inradius and semiperimeter respectively of the PPT. Then
 
             (m2 + n2) + (m2 - n2) + (2mn)
        s = ------------------------------ = m(m+n)
                            2
The area of the PPT is
                          ab     (2mn)(m2 - n2)
        rs = rm(m + n) = ---- = ----------------
                          2             2
Therefore, r = n(m-n).

Part I:

        Inradius is even           <==>      n(m - n) is even
                                          <==>      n is even
                                          <==>      m is odd
                                          <==>      m2 is odd
                                          <==>      [ (m2 + n2) + (m2 - n2) ] / 2 is odd
                                          <==>      Mean of the hypoteneuse and the odd side is odd.

Part II:

let k > 1 be an integer. Then the pairs (2k,1) and (2k,2k-1) generate non-congruent PPTs with the same inradius 2k - 1.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re(2): solutionDaniel2008-01-27 01:54:02
re: solutionBractals2008-01-26 17:00:04
solutionDaniel2008-01-26 16:31:54
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