Prove the light will never go through a vertex In the equilateral triangle ABC we send a light beam through The vertex A so that it hits D on side BC so that: BD/BC= 1/√2
How many times does the light beam have to reflect so that it goes back through a vertex?
Now I know the answer is it will never go back through a vertex but I want the proof. 
To be more specific why is it when BD/BC=irrational number then the light will never go through a vertex?
 A: Consider the infinite triangular grid. A reflected light ray will pass through a vertex if and only if its straight continuation does. 
Remark: It is not fully sufficient that $\sqrt 2$ is irrational. If $\sqrt3$ were involved, the situation would be different.
A: The problem is equivalent to proving that there is no nonzero integer ordered pair $(a,b)$ such that $$\langle a \vec u + b \vec v , \vec w \rangle = \pm |a \vec u + b \vec v| {} |\vec w| \tag{*}$$ where $$\vec u = (1,0), \quad \vec v = \left(\tfrac{1}{2}, \tfrac{\sqrt{3}}{2}\right), \quad \vec w = \left(1 - \tfrac{1}{\sqrt{8}}, \sqrt{\tfrac{3}{8}}\right).$$  That is to say, there is no integer lattice point in the basis formed by $\{\vec u$, $\vec v\}$ that shares the same slope as the vector $\vec w$.  The condition (*) is equivalent to 
$$a^2+2ab+3b^2 = 2\sqrt{2}(a+b)b. \tag{**}$$  Note that the LHS is always an integer and the RHS is never an integer unless $a = -b$, or $b = 0$, in which case the RHS equals $0$.  But the discriminant of the quadratic in $a$ is $(2b)^2 - 4(3b^2) = -8b^2 \le 0$, hence there is a solution to (**) only when $a = b = 0$.  But this point is excluded since it is the origin of the ray.
