Perimeter and area are positive integers

Question

In Geometry 3D, How can I create a triangle whose perimeter and area are positive integers with Mathematica?

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I found three triangles. For example $A(-6,1,2)$, $B(-9,1,2)$, $C(-9,1,6)$ or $A(-2,-1,4)$, $B(-2,2,4)$, $C(-2,2,8)$ or $A(-1,-1,-2)$, $B(-1,-1,1)$, $C(-5,-1,1)$.

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And some another triangles: $A(-8,-9,-3)$, $B(4,-9,6)$, $C(-8,-9,1)$ or $A(0,3,3)$, $B(9,-9,3)$, $C(4,3,3)$ or $A(-7,5,7)$, $B(2,5,-5)$, $C(-3,5,7)$. Dear belisarius, I am sorry about that.

Answer 1

If we interpret the question to be a search for triangles with integer-valued perimeter and integer-valued area, then we can approach the problem by asking that a,b,c be the lengths of the sides with a+b+c=integer. The area needs to be calculated in terms of a,b,c. This can be done by solving the three equations:

c^2 = h^2 + b1^2;
a^2 = h^2 + b2^2;
b= b1 + b2;

Here h is the height of the triangle and the top two equations come from the pythagorean theorem. In Mathematica, we solve

Solve[{c^2 == h^2 + b1^2, a^2 == h^2 + b2^2, b == b1 + b2}, 
      {h, b1, b2}]

which gives two answers. One of the answers has h negative and one has h positive, so we can throw away the negative answer and find:

 h -> Sqrt[-a^4 + 2 a^2 b^2 - b^4 + 2 a^2 c^2 + 2 b^2 c^2 - c^4]/(2 b),
 b1 -> (-a^2 + b^2 + c^2)/(2 b), 
 b2 -> (a^2 + b^2 - c^2)/(2 b)}

The area is now 1/2 (b1+b2) h. Substituting these values in and simplifying gives

 area = 1/4 Sqrt[-(a - b - c) (a + b - c) (a - b + c) (a + b + c)]

We can now check to verify that the area is integer. Take for instance a right (3,4,5) triangle:

 area //. {a -> 3, b -> 4, c -> 5}

which has area 6.