Triangles \(\Delta ABC\) and \(\Delta XYZ\) are Heronian triangles, which are triangles that have integer sides with integer areas. Points \(A,B,C\) are centers of circles, the radii of which are

Radius of circle \(A=1092=2^2\cdot 3\cdot 7\cdot 13\)

Radius of circle \(B=608=2^5\cdot 19\)

Radius of circle \(C=1140=2^2\cdot 3\cdot 5\cdot 19\)

The sides of triangle \( \Delta ABC\) are

\(AB=1700=2^2\cdot 5^2\cdot 17\)

\(BC=5491=17^2\cdot 19\)

\(CA=4437=3^2\cdot 17\cdot 29\)

Triangle \(\Delta XYZ\) has the maximum area of any triangle (Heronian or not) that has one vertex on the circumference of each of the circles \(A,B,C\)

Let \(\dfrac { \Delta XYZ }{ \Delta ABC } =\dfrac { p }{ q } \)

be the ratio of the areas of triangles \(\Delta XYZ\) and \(\displaystyle \Delta ABC\)

where \(p,q\) are \(4\) digit co-prime integers. The difference \(p-q\) is a \(4\) digit prime number.

Find that prime number.

For your convenience, area of triangle \(\displaystyle \Delta ABC\) works out to

\(3261654=2\cdot 3^3\cdot 11\cdot 17^2\cdot 19\)

Note: Triangle \(\Delta XYZ\) as drawn in graphic does not have the maximum area

Also, as you solve this, you'll come across more Herons. A helpful hint, I hope.

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