If a triangle has rational coordinates, does it have rational area?

Question

Basically, the topic says it all. If a triangle has rational coordinates (say, in $\mathbb Q^2$), must it have rational area? I realize the side-lengths are usually irrational; that's fine.

Heron's formula seems pertinent: if the side-lengths are $a, b, c$, and you let $s=\frac{1}{2}(a+b+c)$, then the area is $\sqrt{s(s-a)(s-b)(s-c)}$. The side-lengths will always be square-roots of rational numbers, and when you simplify Heron's formula out, you get $\frac{1}{4}\sqrt{a^2b^2+a^2c^2+b^2c^2-a^4-b^4-c^4}$, and since all those inside terms have even degree, it's the square root of a rational number.

But this suggests it shouldn't be rational in general, just a square root of a rational. Is there some reason why it would be rational? Or, conversely, is there a "rational triangle" with irrational area?

Answer 1

If the coordinates of the triangle are $(A_x,A_y), (B_x, B_y), (C_x,C_y)$ then the area is:

$$Area=\left| \frac{A_x(B_y-C_y)+B_x(C_y-A_y)+C_x(A_y-B_y)}{2} \right|$$

This shows that your triangle must have rational area.

Answer 2

It works. You might as well put one vertex at the origin. Then the area is half the absolute value of the cross product of the two vectors, regarded as vectors in $\mathbb R^3.$

Answer 3

It is always rational. Denote the vertex coordinates by $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3, y_3)$ then the area is $$S = \frac{1}{2} \Bigl|(x_2 - x_1)(y_3-y_1) - (y_2-y_1)(x_3-x_1)\Bigr|.$$