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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?

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    $\begingroup$ perhaps if your theorem in question is "If a triangle has rational coordinates then it has a rational area", try proving the contrapositive...that is "if a triangle has irrational area, its coordinates are irrational". By the way, side lengths can be irrational if coordinates are rational. The distance from (1,0) to (0,1) is $\sqrt(2)$, so do you mean coordinates or sides? $\endgroup$ – Irish M Powers Sep 11 '13 at 3:17
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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.

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  • $\begingroup$ This is the sort of "elementary" argument I was looking for. Thank you. $\endgroup$ – Richard Rast Sep 11 '13 at 15:15
  • $\begingroup$ @RichardRast Well, this formula is easy to prove with elementary methods, but it is really the cross product formula written without cross product.... $\endgroup$ – N. S. Sep 11 '13 at 17:12
  • $\begingroup$ Sure. The context I was working in was wondering whether, in arbitrary ordered fields (which may or may not be a subfield of $\mathbb R$), a triangle in affine 2-space had a well-defined area. If so, it should have a fairly nice formula which can be proved by straightforward methods, and I couldn't remember what the cross product really used to prove it works. $\endgroup$ – Richard Rast Sep 11 '13 at 19:42
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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 $\mathbb R^3.$

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  • $\begingroup$ Oh, calculus. Why can't I remember to use you? $\endgroup$ – Richard Rast Sep 11 '13 at 3:16
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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 $$S = \frac{1}{2} \Bigl|(x_2 - x_1)(y_3-y_1) - (y_2-y_1)(x_3-x_1)\Bigr|.$$

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