Laser beams are known for reflecting off solid objects. Whenever a laser beam hits a straight, solid wall, it reflects off in the opposite direction, at an angle to the wall that is equal to the angle at which it hits, as shown.
[img]https://cdn.artofproblemsolving.com/attachments/8/2/e56b311d1cea61b999d36cbe9189b84586a70b.png[/img]

The path of the laser $L$ meets the wall, $W$, at an angle of $55.15$ degrees coming from the right. Then, $L$ reflects off $W$ to a redirected path, $L'$ , at an angle of $55.15$ degrees going off to the left. In particular, if $L$ were to meet $W$ at an angle of $90$ degrees, then $L' $ would follow $L$, going backwards (straight up).


[b]p6.[/b] Given a square $ABCD$, with $AB = 1$, mark the midpoints $M$ and $N$ of $AB$ and $BC$, respectively. A laser beam shot from $M$ to $N$, and the beam reflects of $BC$, $CD$, $DA$, and comes back to $M$. This path encloses a smaller area inside square $ABCD$. Find this area.

[b]p7.[/b] Given a rectangle $EFGH$, with $EF = 3$ and $FG = 40$, mark a point $P$ on $FG$ such that $FP = 4$. A laser beam is shot from $E$ to $P$, which then reflects off $FG$, then $EH$, then $FG$, etc. Once it reaches some point on $GH$, the beam is absorbed; it stops reflecting. How far does the beam travel?

[b]p8[/b]. Given a rectangle $EF GH$ with $EF = 3$ and $FG = 2010$, mark a point $P$ on $F G$ such that $FP = 4$. A laser beam is shot from $E$ to $P$, which then reflects off $FG$, then $EH$, then $FG$, etc. Once it reaches some point on $GH$, the beam is absorbed; it stops reflecting. How far does the beam travel?

[b]p9.[/b] Given a triangle $XYZ$ with $\angle Y = 90^o$, $XY = 1$, and $XZ = 2$, mark a point $Q$ on $YZ$ such that  $\frac{ZQ}{ZY} = \frac13$. A laser beam is shot from $Q$ perpendicular to $YZ$, and it reflects off the sides of $XYZ$ indefinitely. How many bounces does it take for the laser beam to get back to $Q$ for the first time (not including the release from $Q$ and the return to $Q$)?

[b]p10.[/b] Given a triangle $XYZ$ with $\angle Y = 90^o$, $XY = 1$, and $XZ = 2$, mark a point $Q$ on $YZ$ such that $\frac{ZQ}{ZY} = \frac13$. A laser beam is shot from $Q$ perpendicular to $YZ$, and it reflects off the sides of $XYZ$ indefinitely. How far has the laser traveled when it reaches its $2010$th bounce?


PS. You should use hide for answers.
1. **Problem 6:**
   - Given a square \(ABCD\) with \(AB = 1\), and midpoints \(M\) and \(N\) of \(AB\) and \(BC\) respectively.
   - A laser beam is shot from \(M\) to \(N\), reflects off \(BC\), \(CD\), \(DA\), and returns to \(M\).
   - By symmetry, the beam hits the midpoints of the sides before coming back to \(M\).
   - The inner figure formed by the path of the laser is a smaller square.

   Calculation:
   - The midpoints \(M\) and \(N\) divide the sides into segments of length \(\frac{1}{2}\).
   - The smaller square formed by the path of the laser has its vertices at the midpoints of the sides of the original square.
   - The side length of the smaller square is \(\frac{1}{\sqrt{2}}\) (since the diagonal of the smaller square is the side of the original square, and the diagonal of a square is \(\sqrt{2}\) times the side length).

   Area of the smaller square:
   \[
   \text{Area} = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}
   \]

   Conclusion:
   \[
   \boxed{\frac{1}{2}}
   \]

2. **Problem 7:**
   - Given a rectangle \(EFGH\) with \(EF = 3\) and \(FG = 40\), and a point \(P\) on \(FG\) such that \(FP = 4\).
   - A laser beam is shot from \(E\) to \(P\), reflects off \(FG\), then \(EH\), then \(FG\), etc., and is absorbed when it reaches some point on \(GH\).

   Calculation:
   - Path A: \(E \to P \to H\) (reflects off \(FG\) and \(EH\)).
     - Distance: \(EP + PH = \sqrt{3^2 + 4^2} + 3 = 5 + 3 = 8\)
   - Path B: \(H \to P' \to E'\) (reflects off \(FG\) and \(EH\)).
     - Distance: \(HP' + P'E' = 2 \times 8 = 16\)
   - The laser beam travels along Path A twice and Path B four times before being absorbed.

   Total distance:
   \[
   \text{Total distance} = 2 \times 8 + 4 \times 16 = 16 + 64 = 80
   \]

   Conclusion:
   \[
   \boxed{80}
   \]