Geometry: can a triangle be equilateral... Problem from book: Can a triangle be equilateral, if distances from its vertices to two given perpendicular lines are expressed by whole number?
I do not understand this problem. Can someone perhaps restate it? Sorry for the bother but it seems interesting.
Edit:There is no figure in the problem. Image uploaded is just some random equilateral with random perpendicular line. 

 A: Think of the two given perpendicular lines as the $x$ and $y$ axes (you can just rotate the given lines). Then, the problem is asking if you can have an equilateral triangle $ABC$ where $A = (a_1, a_2), B = (b_1, b_2), C = (c_1, c_2)$ are points in $\mathbb Z^2$ (that is, both coordinates are integers).
A: First of all, by translating by $(A_1,A_2)$ you get a simpler but equivalent problem. So now we can assume that $A$ is the intersection of the two line.
Let $l$ be the length of a side of the triangle, $B_1$, $B_2$, $C_1$,$C_2$ be the signed distance.
By bijection with the complex plane you get :
$$\exists \alpha \in \mathbb{R}/\begin{cases} A = 0 \\ B=le^{i\alpha} \\ C = le^{i(\alpha+\pi/3)} \end{cases}$$
And you know that :
$$\begin{cases} le^{i\alpha} =  B_1+i B_2\\ le^{i(\alpha+\pi/3)}  = C_1+ i C_2  \end{cases}$$
$$e^{i\pi/3}= \frac{B_1+i B_2}{ C_1+ i C_2} = \frac{(B_1+i B_2)(C_1- i C_2)}{ C_1^2+  C_2^2}$$
Now, taking the imaginary part :
$$\frac{\sqrt{3}}{2} =\frac{-B_1C_2+ B_2C_1}{ C_1^2+  C_2^2} $$
but $\frac{\sqrt{3}}{2} \in \mathbb{R}\setminus\mathbb{Q}$, $\frac{B_1C_2+ B_2C_1}{ C_1^2+  C_2^2} \in \mathbb{Q}$
A: The problem is asking if there are equilateral triangles with their vertices in $\mathbb{Z}\times\mathbb{Z}$.
The answer is negative: if $\ell$ is the side length of such a triangle, $\ell^2$ is a positive integer by the Pythagorean theorem. Since the area $\Delta$ is given by $\frac{\sqrt{3}}{4}\ell^2$, it follows that $4\Delta\in\sqrt{3}\mathbb{Z}^+$.
On the other hand, by the shoelace formula we have $\Delta\in\mathbb{Z}^+$. 
$\sqrt{3}\mathbb{Z}^+$ and $\mathbb{Z}^+$ do not intersect since $\sqrt{3}\not\in\mathbb{Q}$, hence there aren't equilateral triangles made by lattice points.
