A discrete problem about coordinates or angle? Suppose there are two straight lines. We call one of them line $a$ and another one line $b$ and they make a $60$ degree angle. Now we start from one point on $b$ and draw a line from that point to line $a$; we let this line be $d$ units long. Now from that new point on line $a$ we draw it back to line $b$ with the same distance $d$(can't take the same path that is used to come to the point). Now continue this process until we get back to the original point. A diagram of how it is done is shown below:( p0 is the original point)

Prove or disprove: no matter where we start on line $b$ we will get back to the original point. 
So I have tried to use trigonometry to bash out the coordinates but it quickly turns to hard and complicated to advance. I am just really confused where to start this off. Some hints will be appreciated. Thank you! 
 A: Set up a coordinate on line $b$ having $O$ as origin, and let $p_0$, $p_2$, $p_4$, ... be the coordinates of the points. Using the technique explained in this answer to another question, one finds that
$$
p_{n}=p_0\cos{n\pi\over3}\pm{\sqrt{{4\over3}d^2-p_0^2}}\sin{n\pi\over3},
$$
where the choice of sign depends on the direction taken at the first step of the construction.
It follows in particular that $p_6=p_0$.
A: Without loss of generality, let $OP_1=2$ and let $X$ be the foot of the perpendicular from $P_1$ to $OP_2$. Also let angle $XP_1P_0=\theta$.
It is sufficient to show that if $P_0P_1=P_1P_2=d$, and that $P_3$ is a point on $P_1O$ produced such that $OP_0=OP_3$, then $P_2P_3=d$. It will then follow by symmetry that all six red lines are equal, and that such a construction is always possible provided $\sqrt{3}\leq d\leq2$.
This claim follows because, following the construction described, we have $OP_3=1-d\sin\theta$ and $OP_2=1+d\sin\theta$ and the cosine rule in triangle $OP_2P_3$ gives $$P_2P_3^2=(1-d\sin\theta)^2+(1+d\sin\theta)^2-2(1-d^2\sin^2\theta)\cos120=3+d^2\sin^2\theta$$
However, $$\cos\theta=\frac{\sqrt{3}}{d}\implies\sin^2\theta=1-\frac{3}{d^2}$$
Hence $P_2P_3=d$ QED
A: It doesn't take much trigonometry; in a 30-60-90 right triangle, consider
the ratio of the short leg to the hypotenuse.
Suppose you have followed the construction of the points $p_0$ to $p_5$ in sequence but have not yet constructed the next point after $p_5.$ It is necessary to prove that that a segment of the same length as the others closes the loop, that is, $p_5p_0 = p_0p_1,$ and therefore that when you do plot the next point after $p_5$ it will coincide with $p_0$ and repeat the sequence.
From $p_3$ drop a perpendicular to $Op_4$; call the foot of this perpendicular $p_6.$
Place point $p_7$ on the line $Op_2$ so that $O$ is the midpoint of $p_4p_7.$
Let $d = p_6O$. That is, $d$ is the length of the shorter leg of the
30-60-90 right triangle $\triangle p_3 p_6 O.$
Let $r = Op_4 = Op_7.$
Find the distances $p_0p_6,$ $p_4p_6,$ $p_6p_2,$ and $p_2p_7$ in terms of $d$ and/or $r.$
Using congruent triangles, find $Op_0$ in terms of $d$ and/or $r.$
Show that $Op_3 = Op_0,$ and proceed to show (using congruent triangles)
that $Op_1 = Op_4$ and that $p_5p_0 = p_2p_3.$
A: I am not sure at all but here is my thoughts on maybe how to solve this; comments or corrections toward this solution would be greatly appreciated:
Let the point that $P_5$ goes be $P_6$. We need to prove $P_6$ coincide with $P_0$.
Note that with all line segments being equal, $P_4P_5$=$P_5P_6$. So $\triangle P_4P_5P_6$ is an isosceles triangle.  
Since we know $P_4P_5P_6$ is an isosceles triangle we also know that $\angle OP_4P_5$= $\angle OP_6P_5$.  But that will only happen when $P_6$ coincide with $P_0$ so we are done.
A: The following will prove that $\,\overrightarrow{OP_3}\,$ is $\,\overrightarrow{OP_0}\,$ rotated $\,120^\circ\,$ clockwise, meaning that $P_3$ is univocally determined by $\,P_0\,$ and does not depend on the choice of $\,P_1\,$. The conclusion of the problem then follows directly, since it means that the paths $\,P_0 \,\text{-}\, P_1 \,\text{-}\, P_2 \,\text{-}\, P_3\,$ and  $\,P_0 \,\text{-}\, P_5 \,\text{-}\, P_4 \,\text{-}\, P_3\,$ have $\,P_3\,$ as a common point, so the sequence $\,P_0 \,\text{-}\, P_1 \,\text{-}\, P_2 \,\text{-}\, P_3 \,\text{-}\, P_4 \,\text{-}\, P_5 \,\text{-}\, P_0\,$ is closed.
To prove the proposition, consider the complex plane with the origin at the intersection of the two lines, and the real axis along the blue line. Let $\,\color{blue}{p_0=a}\,$, $\,\color{blue}{p_1=b\omega}\,$ with $\,a,b \in \mathbb{R}\,$ and $\,\omega = e^{i \pi / 3}\,$ $=\cos \frac{\pi}{3}+ i \sin \frac{\pi}{3}\,$ where $|\omega|=1$ and $\,\omega + \bar \omega = 1\,$. The common distance $\,d\,$ is then given by:
$$
d^2=|p_0-p_1|^2 = (a- b \omega)(a - b \bar \omega)=a^2+b^2|\omega|^2-ab(\omega + \bar \omega) = a^2-ab+b^2
$$


*

*Point $\,p_2\,$ lies on the real axis, so it is $\,p_2 = c \in \mathbb{R}\,$. The condition that $\,|p_2-p_1|=s\,$ gives:
$$
\begin{align}
a^2-ab+b^2 = |c-b\omega|^2 = c^2 + b^2 -bc \;\;&\iff\;\; c^2 - bc + ab-a^2 = 0 \\
 &\iff\;\; (c-a)(c+a-b) = 0
\end{align}
$$
Since $p_2 \ne p_0$ it follows that $\,c-a\ne 0$, which leaves $\,c+a-b=0\,$, so $\,\color{blue}{p_2 = b-a}\,$.

*Point $\,p_3\,$ lies on the line $\,op_1\,$ so it is of the form $\,p_3 = \lambda \omega \mid \lambda \in \mathbb{R}\,$. The distance condition is:
$$
\begin{align}
&a^2-ab+b^2 = |\lambda\omega-(b-a)|^2 = \lambda^2 + (b-a)^2-\lambda(b-a) \\
&\quad\iff\quad \lambda^2 - \lambda(b-a) -ab = 0 \\
&\quad\iff\quad (\lambda - b)(\lambda+a) = 0 \\
\end{align}
$$
Since $p_3 \ne p_1$ it follows that $\,\lambda-b\ne 0$, which leaves $\,\lambda+a=0\,$, so $\,\color{blue}{p_3 = -a\omega}\;$ q.e.d.
