The third problem of the french "concours general" made students work on the problem of triangle of space which edges have integer coordinates.
First, they made prove that the coordinates of the orthogonal projection $H$ of $A$ (I don't know how to say this in english, sorry, so $H\in(BC)$, and $(AH)\perp(BC)$ if this is clearer) are rational numbers. This is OK by projection formula : $$\overrightarrow{AH} = \overrightarrow{AB} + \frac{\overrightarrow{BA}\cdot\overrightarrow{BC}}{BC^2}\overrightarrow{BC}$$ Then, you have to prove that any non right angle $\theta$ of the triangle verifies $\tan^2(\theta)$ is a rational which is clear because for example $$\tan(\theta_B)=\frac{AH}{BH}$$ But then you have to prove that there exists an integer $k$, square-free, such that for any non right angle $\theta$ of the triangle, $\tan(\theta)$ can be written $r\sqrt k$, with $r$ some rational.
I can see how one angle can verify this, but I don't see how all the angles must share the same square-free integer $k$.
For those who don't know, a square-free integer is an integer that cannot be divided by the square of a prime number (so it's a product of distinct primes).
Hope someone can help me on this, I have to produce a correction for my students...
Starting from where you are, $\tan \theta_C = AH/CH$
Coordinates of H are rational, $H = B + (p/q) \overline{BC}$ where $p,q \in Z$.
$\overline{CH} = \overline{CB} - \overline{HB}$ so we can relate $\tan \theta_C $ to $\tan \theta_B$
From C, draw a perpendicular CG to AB. G will be on AB. By the same argument as for H, G will have rational coordinates.
$\tan \theta_B = AH/BH = CG/BG$ so we can relate BG and AG and CG to AH and BH.
$\tan \theta_A = CG/AG$ so we an relate $\tan \theta_A$ to AH and BH
With some substitution, all these ratios can be put over the same denominator and $k$ can be deduced from there.
This is enough guide for advanced students to work on it.
I will try to get into more detail if time permits on Monday (5 or 6 hours earlier than you, probably.)