The lines divide the plane into different regions. We'll color them black and white such that no two adjacent regions (i.e. have a boundary segment in common) have the same color. (This can be achieved as follows - choose an orientation for each line, and color each point in the plane by the parity of the number of lines that have the point on their right side.) Assume w.l.o.g. that at first Turbo had a black region on her right. Now each time she turns either left or right, there will still be a black region on her right. Thus she cannot traverse the same segment in two different directions.

Is it even possible for Turbo to visit the same point twice? (I can't find any such examples.)

v_Enhance wrote: Is it even possible for Turbo to visit the same point twice? (I can't find any such examples.) Take a regular $n$-sided polygon ($n > 4$). Consider for each edge $e$ its two adjacent edges, and continue them past $e$. We'll have some sort of a star shape. Now put Turbo on one of the edges of the star (outside of the polygon), and let her move. She'll continue going in circles around the star (if she chooses her first turn wisely).

It seems that both the problem statement and solution (in particular the idea of coloring the regions and always having the same color on your right) were heavily inspired from ISL 2014 C9, which seems pretty unfair for participants who haven't seen it before...

I also solved this problem, but was inspired by TOT 2014.

The same argument of keeping colors on your right was also used in an alternate solution to part a of last IMO's P6.

Incredibly, the hypothesis of left/right alternating is not even necessary. The following argument is due to Meghal and Rachel from USA EGMO 2016: Color the regions black and white in alternating colors (as above). WLOG suppose the first turn is clockwise around a black region. Then Turbo will always move clockwise around black regions and counterclockwise along white regions. The end.

We can prove that Turbo cannot visit the same point twice: Assume that there is a path $v_1\rightarrow v_2\rightarrow ...\rightarrow v_n\rightarrow v_1$, $v_i$ is intersection point of some two lines. We note that on each segment $v_iv_{i+1}$($n+1\equiv 1$), there are no other intersection points except $v_i,v_{i+1}$. Then each line which consist a segment $v_iv_{i+1}$ does not intersect any edge of the path $\Rightarrow v_1v_2...v_n$ is a convex polygon. We can easily see a contradiction

Your statement is incorrect, see post 4 for a counterexample

v_Enhance wrote: Is it even possible for Turbo to visit the same point twice? (I can't find any such examples.) A star may be an good example ,Turbo can go in it repeatly

Let us consider these lines as roads. Since the city has a very low budget, the sidewalks were very very tiny. Therefore, the city enacted a law saying that you must stay to the right, even as a pedestrian. If Turbo turns right(red), then she does not cross any roads. If she turn left(blue) then she crosses two roads. Either way, she always crosses an even number of roads. When Turbo reaches her original road segment, she is on the same side of every road. Take, for example, the red line in the following image. Each time she crosses the red road, she ends up on a different side of it. Since she is on the same side of every road, she must've crossed each road, except her current one, an even number of times. Therefore, Turbo has crossed her current road an even number of times. Thus, she is on the same side of this current road, which means that she is going in the same direction.

Consider the areas that the $2017$ lines split the plane into. We will color these areas in red and blue such that no two areas that share at least one common side have the same colors. This is possible by induction on the number of lines. When there is $1$ line, we can just color it however we want. Assume that we can color the grid in the desired way when there are $k$ lines, then we can add a line in any place we want as long as it doesn't go through an intersection point. Each area the new line goes through is now split into two areas that share a common side (the new line), so we want to flip the colors of one side. Now, all adjacent areas along the line are colored in different colors, and every area not along the line already share no common color by the inductive hypothesis, so we are done. Note that turbo turning in alternating directions mean that the color on her sides remain the same. For example, if she starts with red on her left and blue on her right, she will always have those two colors on her left and right throughout her journey. Intersecting a line in two different directions produces a contradiction, thus the answer is no.

This is really nice! Solution. For each of the $2017$ lines, arbitrarily assign a weight of $0$ to one half-plane bounded by the line and $1$ to the other half-plane. For each region $\mathcal{R}$ of the plane, assign the sum of the weights given to all $2017$ half-planes containing $\mathcal{R}$. Color $\mathcal{R}$ red if this sum is even and blue if it's odd. Notice that no two adjacent regions have the same color, since their sums are entirely the same except for the single line separating the two regions. Without loss of generality, suppose Turbo begins by moving forward along a line segment which lies between a red region on the left and a blue region on the right. Assign to each line segment a direction oriented in such a way that matches with Turbo's initial segment, with a red region on the left and a blue region on the right. When Turbo turns left or right, the direction of her path is preserved, so Turbo can never cross the same segment in both directions. $\square$ Note that this proves a stronger result, since it doesn't rely on the alternating scheme given in the problem. Remark. My argument was inspired by the idea of coloring regions in the plane by counting weights associated to each half-plane, which I've seen in some other completely unrelated context. Going back to this idea actually reminded me of this problem (EGMO 2017/3), which I've seen before but only just now came up with a solution for.

Nice but too easy. The answer is no. Drop the condition on Turbo turning alternately left and right. Claim: We can colour each region bounded by lines red and blue so that no two regions of the same colour are directly opposite each other with respect to a line. Proof. Assign each line a direction, and for each region $\mathcal R$ and each line $\ell$ let \[f(\mathcal R, \ell) = \begin{cases} 0 & \text{if } \mathcal R \text{ lies to the left of }\ell \\ 1 & \text{otherwise.} \end{cases}\]Now colour $\mathcal R$ red if the value \[\bigoplus_{\text{lines } \ell} f(\mathcal R, \ell)\]is $0$, and blue if it is $1$. This construction works since if regions $\mathcal R_1$ and $\mathcal R_2$ are directly opposite each other with respect to a line $\ell$ then clearly $f(\mathcal R_1, m) = f(\mathcal R_2, m)$ for all lines $m \neq \ell$ and $f(\mathcal R_1, \ell) \neq f(\mathcal R_2, \ell)$, implying that $\mathcal R_1$ and $\mathcal R_2$ have different colours. $\square$ Now it's easy to see that Turbo always has the same colour to her left as she embarks on her endeavours. This is a contradiction if she travels the same segment both ways.

Chat is this where Turbo the Snail Originated from

Mr.Sharkman wrote: Chat is this where Turbo the Snail Originated from no it originated from 2014 c9