There's a convex $3n$-polygon on the plane with a robot on each of it's vertices. Each robot fires a laser beam toward another robot. On each of your move,you select a robot to rotate counter clockwise until it's laser point a new robot. Three robots $A$, $B$ and $C$ form a triangle if $A$'s laser points at $B$, $B$'s laser points at $C$, and $C$'s laser points at $A$. Find the minimum number of moves that can guarantee $n$ triangles on the plane.
We claim that the minimum number of moves is \(\frac{9n^2 - 7n}{2}\). We will first show that it needs at most \(\frac{9n^2 - 7n}{2}\) moves.

1. **Expected Moves Calculation:**
   Consider a triangle chosen from the \(3n\) vertices uniformly at random. Then, it takes an expected value of \(\frac{3n-2}{2} \cdot 3\) moves to rotate every vertex of the triangle such that each robot points to the next robot in the triangle. 

   If \(a\) is the minimum number of moves such that the robots in the triangle are pointing to each other in counterclockwise order and \(b\) is the minimum number of moves such that the robots in the triangle are pointing to each other in clockwise order, then we must have:
   \[
   b \equiv a + 3n \pmod{3n-1}
   \]
   This implies:
   \[
   |b - a| \geq 1
   \]
   Therefore,
   \[
   \min(a, b) \leq \frac{a + b}{2} - \frac{1}{2}
   \]
   which means that the expected number of moves is at most:
   \[
   \frac{9n - 7}{2}
   \]
   So, it is possible to perform at most \(\frac{9n^2 - 7n}{2}\) moves to form \(n\) disjoint triangles.

2. **Lower Bound Construction:**
   Now, we will show that there exists a construction that requires at least \(\frac{9n^2 - 7n}{2}\) moves. Suppose that the vertices are labeled \(1, 2, \ldots, 3n\) in counterclockwise order. Suppose that \(a_i\) is the robot that the robot on vertex \(i\) faces. Then, let \(a_1 = 2\) and \(a_2 = a_3 = \ldots = a_{3n} = 1\).

   Suppose that \(b_i\) is the number of robots between \(i\) and \(a_i\) in counterclockwise order, including \(a_i\) but not including \(i\). If there are \(n\) disjoint triangles, then we must have:
   \[
   b_1 + b_2 + \ldots + b_{3n} = 3n^2
   \]
   Since at the beginning, we have \(b_1 = 1\) and \(b_i = 3n + 1 - i\) for \(i \neq 1\). Then, in each move, either \(b_i\) increases by \(1\) or decreases by \(3n - 2\). If \(b_i\) decreases by \(3n - 2\) twice in a sequence of moves, then \(b_i\) must have been moved at least \(3n - 1\) times, so by removing \(3n - 1\) of those moves, it is possible to decrease the number of moves.

   In each move, if \(b_i\) increases by \(1\), then \(a_i\) increases by \(1 \mod 3n\), and if \(b_i\) decreases by \(3n - 2\), then \(a_i\) increases by \(2 \mod 3n\). If there are \(n\) triangles, then all of the \(a_i\) must be distinct, so the sum of all of the \(a_i\) must be:
   \[
   \frac{3n(3n + 1)}{2}
   \]
   Therefore, if there are \(x\) moves that increase \(b_i\) by \(1\) and \(y\) moves that decrease \(b_i\) by \(3n - 2\), we have:
   \[
   x - y(3n - 2) = 3n^2 - 1 - \frac{(3n - 1)3n}{2} = -\frac{3}{2}n^2 + \frac{3}{2}n - 1
   \]
   and
   \[
   x + 2y \geq \frac{3n(3n + 1)}{2} - 3n - 1 = \frac{9}{2}n^2 - \frac{3}{2}n - 1
   \]

   Therefore, subtracting the first equation from both sides of the second inequality, we have:
   \[
   3ny \geq 6n^2 - 3n
   \]
   which means that:
   \[
   y \geq 2n - 1
   \]
   which means that:
   \[
   x = -\frac{3}{2}n^2 + \frac{3}{2}n - 1 + y(3n - 2) \geq -\frac{3}{2}n^2 + \frac{3}{2}n - 1 + (2n - 1)(3n - 2) = \frac{9}{2}n^2 - \frac{11}{2}n + 1
   \]
   Therefore, this means that:
   \[
   x + y \geq \frac{9}{2}n^2 - \frac{11}{2}n + 1 + 2n - 1 = \frac{9n^2 - 7n}{2}
   \]
   So the minimum number of moves is at least \(\frac{9n^2 - 7n}{2}\).

Therefore, the minimum number of moves that can guarantee \(n\) triangles is \(\boxed{\frac{9n^2 - 7n}{2}}\).