TITLE: Proving the area is irrational for triangle with integer vertices
QUESTION [11 upvotes]: Question:

$A,B,C$ are three non-collinear points lying on a plane whose normal vector is $(\hat{i}+\hat{j}+\hat{k})$. If all the three coordinates of every point is an integer, then prove that the area of the triangle is necessarily an irrational number.

My attempt:
I don't understand how to do this problem. I know how such types of problems are dealt with in the 2D plane, but not in 3D space. Can anyone offer some help on the starting steps? Thanks!

REPLY [3 votes]: We'll show  that if three non-collinear points on the given plane have rational  coordinates, then the area of the triangle with those points as vertices is irrational.

Suppose $A',B',C'$ are three non-collinear points with rational coordinates on the given plane.

Let $G$ be the centroid of triangle $A'B'C'$.

Then $G = {\large{\frac{A'+B'+C'}{3}}}$, hence $G$ also has rational coordinates.

Let $A = A'-G,\;\;B=B'-G,\;\;C=C'-G$.

Then $A,B,C$ are nonzero (as vectors), have rational coordinates, and reside in the plane with equation
$$x+y+z = 0$$
so the normal vector of the plane is still the vector $n=\langle{1,1,1}\rangle$.

Since triangle $ABC$ is just a translation of triangle $A'B'C'$, the area stays the same.

Thus, it suffices to show the area of triangle $ABC$ is irrational.

But now we have $A + B + C = 0$, so $A = -B-C$.

Let $k\;$denote the area of triangle $ABC$. 
\begin{align*}
\text{Then}\;\;k
&={\small{\frac{1}{2}}}|AB\times AC|\\[4pt]
&={\small{\frac{1}{2}}}|(B-A)\times (C-A)|\\[4pt]
&={\small{\frac{1}{2}}}|(2B+C)\times (2C+B)|\qquad\text{[since$\;A = -B - C$]}\\[4pt]
&={\small{\frac{1}{2}}}|4B{\times}C\,+\,2B{\times}B\,+\,2C{\times}C\,+\,C{\times}B|\\[4pt]
&={\small{\frac{1}{2}}}|4B{\times}C\,+\,C{\times}B|\\[4pt]
&={\small{\frac{1}{2}}}|3B{\times}C|\\[4pt]
&={\small{\frac{3}{2}}}|B{\times}C|\\[4pt]
\end{align*}
so it suffices to show $|B\times C|$ is irrational.

But $B\times C$ is a nonzero vector with rational coordinates, and is parallel to $n$.

It follows that $B\times C = \langle{t,t,t}\rangle$, for some nonzero rational number $t$, hence $|B\times C| = \sqrt{3t^2} =  |t|\sqrt{3},\;$which is irrational.