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I was working on a quite challenging problem (thought by myself):

Find all triangles (with rational sides), such that the lengths between the Fermat Point and its vertices are rational.

Note: The Fermat Point $F$ in a triangle is defined as a point such that $FA+FB+FC$ is minimized.

Fermat Point

My Attempt: (I will not prove my results here.)

  1. Let $f_a=FA$, $f_b=FB$ and $f_c=FC$. Utilize the fact that $\angle FAB=\angle FBC=\angle FAC=120^{\circ}$, we can get: \begin{equation} \begin{cases} a^2=f_b^2+f_c^2+f_bf_c \\ b^2=f_a^2+f_c^2+f_af_c \\ c^2=f_a^2+f_b^2+f_af_b \\ \end{cases} \end{equation} And by solving equation we get: \begin{equation} \begin{cases} f_a & = \dfrac{c^2+b^2-2a^2}{3\Sigma}+\dfrac{\Sigma}{3} \\ f_b & = \dfrac{c^2+a^2-2b^2}{3\Sigma}+\dfrac{\Sigma}{3} \\ f_c & = \dfrac{a^2+b^2-2c^2}{3\Sigma}+\dfrac{\Sigma}{3} \\ \end{cases} \end{equation} where $\Sigma=f_a+f_b+f_c$. By substitution, we can see that $\Sigma$ is the solution of: \begin{equation} x^4-(a^2+b^2+c^2)x^2+(a^4+b^4+c^4-a^2b^2-a^2c^2-b^2c^2)=0 \end{equation}
  2. Considering the discriminant, which is actually equal to $48[ABC]^2$, we know that $\sqrt{\Delta}=4\sqrt{3}[ABC]$ must be rational, to make $\Sigma^2$ to be rational. By doing Rational Parametrization for Conics, we get the general solution for that is: \begin{equation} \begin{cases} a=|r^2-3s^2|+|p^2-3q^2| \\ b=r^2+3s^2 \\ c=p^2+3q^2 \end{cases} \end{equation} (Although it might not be bijective) And all the solutions generated must satisfy the triangle inequality.

EDIT: I want to add 2 special cases here:

  1. Triangle with $60^{\circ}$: It remains to solve the set of equations: \begin{equation} \begin{cases} a^2=b^2+c^2-bc \\ \Sigma^2=b^2+c^2+bc \\ \end{cases} \end{equation} which does not have solutions that fit into a triangle.

  2. Triangle with $120^{\circ}$: The general solution is given by: \begin{equation*} \begin{cases} a = 2mn+m^2 \\ b = n^2-m^2 \\ c = m^2+mn+n^2 \end{cases} \end{equation*} with $n,m\in\mathbb{Q}$, $n>m$. Here's a table of examples: Examples of triangles with $120^{\circ}$

After this, I have no idea. I am unable to solve rational $\Sigma$, but only $\Sigma^2$. Can someone please help? I think this problem has a lot of generalization, such as the Weighted Fermat Point Problem.

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    (+1) Interesting question. A minor nitpick: the Fermat point is the solution of the Torricelli-Steiner problem only if the original triangle has no angle $>120^\circ$. Otherwise, the solution is given by a vertex. – 2017-01-24
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    Anyway, the problem boils down to understanding if in $\mathbb{Q}(\omega)$ there are three numbers $$f_a+\omega f_b,\quad f_b+\omega f_c,\quad f_c+\omega f_a$$ that are squares, i.e. if there is an equilateral triangle of squares. – 2017-01-24
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    We have $$(4\Delta)^2 = (a^2+b^2+c^2)^2-2(a^4+b^4+c^4) = 3\left(\sum_{cyc} f_a f_b\right)^2$$ and the primes represented by the quadratic form $x^2+y^2+xy$ over the integers are the primes $\equiv 1\pmod{3}$. I guess the problem has no solution by some tricky manipulation $\pmod{3}$ or $\pmod{9}$. – 2017-01-24
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    @JackD'Aurizio Thanks for your comments. However, I want to point out a solution: (there are some triangles without $120^{\circ}$ that has rational $f_a$, $f_b$, $f_c$) \begin{equation} a=43, b=147, c=152, [ABC]=1824\sqrt{3}, \Sigma=185 \end{equation} And in fact, the largest angle in the above is $88.395...^{\circ}$ which is smaller than $120^{\circ}$. – 2017-01-25
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    https://math.stackexchange.com/questions/968081/ seems to give the answers. – 2018-07-09

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