[b]p1.[/b] A teacher is setting up a demonstration before a class of $33$ students. She tells the students to close their eyes and not to look. Every boy in the class closes his left eye; one-third of the boys also close the right eye, but two-thirds of the boys keep the right eye open. Every girl in the class closes her right eye; one-third of the girls also close the left eye, but two-thirds of the girls keep the left eye open. How many students are watching the teacher despite her asking them not to?


[b]p2.[/b] The numbers from $1$ to $100$ are written on a $10\times 10$ board in the consecutive order. That is the first row is $1$, $2$,$ . . .$ , $10$, the second row is $11$, $12$,$ . . .$ , $20$ etc. You place $10$ rooks on the board so that no two of them are in the same vertical or horizontal row, and add up the numbers under the rooks. What is the largest sum you can get?


[b]p3.[/b] Cut the shape below into four equal pieces.
[img]https://cdn.artofproblemsolving.com/attachments/4/e/67c3f8da8b22e4f56a0b986f38df789a13e8e5.png[/img]


[b]p4.[/b] A landscaping company planted $2019$ trees in a circle around the K-State campus. $1010$ of them are oaks and $1009$ are maples. Prove that somewhere in this circle there are two oaks with exactly two other trees between them.


PS. You should use hide for answers.
1. Let the number of boys be \( b \) and the number of girls be \( g \). We know that the total number of students is \( b + g = 33 \).
2. Every boy closes his left eye. One-third of the boys also close their right eye, but two-thirds of the boys keep their right eye open. Therefore, the number of boys who have at least one eye open is:
   \[
   \frac{2}{3}b
   \]
3. Every girl closes her right eye. One-third of the girls also close their left eye, but two-thirds of the girls keep their left eye open. Therefore, the number of girls who have at least one eye open is:
   \[
   \frac{2}{3}g
   \]
4. The total number of students who can see the teacher is the sum of the boys and girls who have at least one eye open:
   \[
   \frac{2}{3}b + \frac{2}{3}g
   \]
5. Since \( b + g = 33 \), we can factor out \(\frac{2}{3}\):
   \[
   \frac{2}{3}(b + g) = \frac{2}{3} \times 33 = 22
   \]

The final answer is \(\boxed{22}\).