## Παρασκευή, 24 Ιουνίου 2011

Let ABC be a triangle and D a point on AB such that inradius of CDA = inradious of CDB := d.

Find the primitive heronian right triangles ABC, A = 90 d. (ie a^2 = b^2 + c^2, a,b,c integers, gcd(a,b,c) = 1) with d integer.

ABC primitive heronian right triangle with A = 90 d. ==>

a = x^2 + y^2

b or c = x^2 - y^2

c or b = 2xy

where x > y > 0 integers.

We have inradius r = y(x-y) and c = 2d^2 / (2d - r) [see HERE].

1. c = 2xy, b = x^2 - y^2

==> xy = d^2 /(2d - y(x-y)) ==> d = xy +-ysqrt(xy) and since x>y,

d = xy + ysqrt(xy).

d integer ==> xy = z^2 and since gdc(x,y) = 1 ==> x = m^2, y = n^2, where m,n are integers.

Therefore
a = m^4 + n^4, b = m^4 - n^4, c = 2m^2*n^2, d = mn^2*(m+n)

2. c = x^2 - y^2, b = 2xy

==>

x^2 - y^2 = 2d^2 / (2d - y(x-y)) ==>

2d = x^2 - y^2 +- ((x-y)sqrt(x^2-y^2)) and since d is positive,

2d = x^2 - y^2 + ((x-y)sqrt(x^2-y^2))

d integer ==> x^2 - y^2 = z^2 ==> x = m^2 + n^2, y = m^2 - n^2 or 2mn

2.1 : x = m^2 + n^2, y = m^2 - n^2

==>

a = (m^2+n^2)^2 + (m^2-n^2)^2

b = 2(m^2+n^2)(m^2-n^2)

c = (m^2+n^2)^2 - (m^2-n^2)^2

d = 2mn^2*(m+n)

2.2 : x = m^2 + n^2, y = 2mn

==>

a = (m^2+n^2)^2 + (m^2-n^2)^2

b = 4mn(m^2+n^2)

c = (m^2-n^2)^2

d = m(m+n)(m-n)^2