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.

let the sides be a,b and hypotenuse be c.  Then for a PPT we have

a=2mn  b=m^2-n^2  c=m^2+n^2 with m,n positive integers m>n and m,n coprime

1) the inradius of a triangle is equal to the area divided by the semiperimeter thus

r=mn(m+n)(m-n)/[m(m+n)]=n(m-n)

the mean of hypotenuse and odd side is m^2  now either m is odd and n even or vice versa.  In either case b has opposite parity of r

 

2)

pick n1,m1 with m1>n1 and m1,n1 co-prime

now set n2=(m1-n1)n1 and m2=n2+1  now obviously m2>n2 and m2,n2 are co-prime as well.  thus both (m1,n1) and (m2,n2) generate unique non-cogruent PPT's.  also their inradii are equal.  Since n1,m1 can be arbitrarily chosen then there exist an infinite number of pairs

Posted by Daniel on 2008-01-26 16:31:54