I've got a problem to prove something, so I need your help ! :) Actually, we consider the complex plane and we have, for $z$, $z' \in \mathbb{C}$, and $r>0$ : $(zz') = \{ z+t(z'-z) \mid t \in \mathbb{R} \}$. Now, we consider for $k=1,2$ : $\alpha_k = a_k + ib_k $ and $\alpha'_k = a'_k + ib'_k $, and we have $(\alpha_1\alpha'_1)$, $(\alpha_2\alpha'_2)$ not parallel. So, let $z=x+iy \in (\alpha_1\alpha'_1) \cap (\alpha_2\alpha'_2)$. We want to show that $x,y \in \mathbb{Q}(a_1, a'_1, a_2, a'_2, b_1, b'_1, b_2, b'_2)$.

I've tried to say that we have :

$ \exists t, t' \in \mathbb{R} \; :$ $x=a_1 + t(a'_1 - a_1) = a_2+t'(a'_2-a_2)$ and $y=b_1 + t(b'_1 - b_1) = b_2+t'(b'_2-b_2)$, but of course it lead me nowhere, and after that I've tried several things but the fact is that I always have a problem cause $t$, or $t'$ in $\mathbb{R}$, and not $\mathbb{Q}$.

Someone could help me, please ?

  • $\begingroup$ Cramer's rule. The solution of a linear system of equations is a rational function of the coefficients. $\endgroup$
    – orole
    Feb 6, 2018 at 17:05
  • $\begingroup$ Why ? I don't understand... How do you see that ? $\endgroup$ Feb 6, 2018 at 17:18
  • $\begingroup$ @ChocoSavour Because the determinant of a matrix with rational coefficients is again rational. $\endgroup$ Feb 6, 2018 at 17:22
  • $\begingroup$ Wow... I don't understand at all... I know that if the matrix has rational coefficients, ofc the determinant will be also rational. But here, what's the matrix ? All I have is $x=a_1+t(a'_1-a_1)$ and $x=a_2+t'(a'_2-a_2)$, and same for $y$, but $t$ and $t'$ are not just real, not rational... I think i don't understand at all... $\endgroup$ Feb 6, 2018 at 17:25

1 Answer 1


First off, this is ruler only, no compass needed, which is crucial. Secondly, all this complex plane magic feels rather distracting, so I'd rather consider this a problem in good old $\mathbb R^2$.

When are $\alpha_i, \alpha_i', z$ collinear? When $\alpha_i'-\alpha_i$ and $z-\alpha_i$ are real multiples of one another. Which means their slope is the same. So


This has some nasty corner cases with vertical lines and zero distances, but all of those go away when you cross-multiply:


As a side note, you can multiply those terms and cancel the common one to obtain a more symmetric representation as a determinant of homogeneous coordinates:

$$b_i'x-b_ix-b_i'a_i-ya_i'+b_ia_i'+ya_i=-\begin{vmatrix} a_i & a_i' & x \\ b_i & b_i' & y \\ 1 & 1 & 1\end{vmatrix}=0$$

But for the purpose at hand, it's better to consider this an equation in $x$ and $y$:


Now you have two of these (for $i\in\{1,2\}$), which allows you to use Cramer's rule to compute the result as

$$x=\frac{\begin{vmatrix} b_1'a_1-b_1a_1'&a_1-a_1'\\b_2'a_2-b_2a_2'&a_2-a_2'\end{vmatrix}} {\begin{vmatrix}b_1'-b_1&a_1-a_1'\\b_2'-b_2&a_2-a_2'\end{vmatrix}}\qquad y=\frac{\begin{vmatrix} b_1'-b_1&b_1'a_1-b_1a_1'\\b_2'-b_2&b_2'a_2-b_2a_2'\end{vmatrix}} {\begin{vmatrix}b_1'-b_1&a_1-a_1'\\b_2'-b_2&a_2-a_2'\end{vmatrix}}$$

Now all the variables there are numbers inside your field extension. And a determinant can be computed using just additions, subtractions and multiplications. So unless you divide by zero (which is the case if your lines are parallel, as you can verify), all these operations will stay inside your field extension.


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