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...

  • $\begingroup$ Are you sure that it is the same $\sqrt{k}$ for all triangles, or for all three angles within one particular triangle, or perhaps not required to be the same for different angles? I find puzzle problems and geometry problems like this very interesting and I will work on it for a while. You can post the original problem for me in French and I can see what the nuances of phrasing are. $\endgroup$ – victoria Mar 20 '17 at 2:53
  • $\begingroup$ (Version francaise) Es-tu certain que c'est la meme racine carree $\sqrt{k}$ pour tous triangles, ou peut-etre pour tous les trois angles d'un meme triangle, ou peut-etre pas necessaire d'etre la meme pour des angles differents? Je trouve les problemes de geometrie et les casse-tetes amusants alors je vais y travailler un bout de temps. Tu peux afficher le probleme original pour moi pour que je puisse regarder les nuances. $\endgroup$ – victoria Mar 20 '17 at 2:56
  • $\begingroup$ Is it the three vertices of the triangle that have integer coordinates? Also, the word space refers to $\mathbb R^3?$ $\endgroup$ – Will Jagy Mar 20 '17 at 19:07
  • $\begingroup$ Yeah, my mistake, I have problems with english vs geometry, don't use it enough to remember the vocabulary... $\endgroup$ – Nicolas FRANCOIS Mar 20 '17 at 20:23

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.)

  • $\begingroup$ My impression is that the vertices have integer coordinates in $\mathbb R^3,$ ( the question said edges), The vertices are $A,B,C,$ then $H$ is the result of dropping a perpendicular from $A$ to $BC,$ with foot $H,$ making $AH$ an altitude of the triangle. He is satisfied about $H,$ but is worried about the three angles, call them $\alpha, \beta,\gamma$ at vertices $A,B,C.$ Then the three edge lengths opposite, maybe $a,b,c,$ all have integer squared length, and we can use the law of sines and law of cosines. Does that sound right? $\endgroup$ – Will Jagy Mar 20 '17 at 19:05
  • $\begingroup$ You're right about me posting something not clear. I was able to draw similar conclusions to those of Victoria, but still can't connect them into something like $\tan\theta_A=r_A\sqrt k$, $\tan\theta_B=r_B\sqrt k$ and $\tan\theta_C=r_C\sqrt k$ with $r_A$, $r_B$ and $r_C$ all rationals, and $k$ a square-free integer. I think I need more on this... But it may be me having a black spot. I posted the subject (in french, sorry) and part of my work here : nicolas.francois.free.fr/fichiers/divers/cg2017-pbIII.pdf. $\endgroup$ – Nicolas FRANCOIS Mar 21 '17 at 1:33
  • $\begingroup$ Hi again. I will see if I can work on this a bit more tonight. I suspect it requires some intelligent use and playing around with algebra. $\endgroup$ – victoria Mar 21 '17 at 23:39

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