[b]p1.[/b] Two players play the following game. On the lowest left square of an $8 \times 8$ chessboard there is a rook (castle). The first player is allowed to move the rook up or to the right by an arbitrary number of squares. The second  layer is also allowed to move the rook up or to the right by an arbitrary number of squares. Then the first player is allowed to do this again, and so on. The one who moves the rook to the upper right square wins. Who has a winning strategy?


[b]p2.[/b] Find the smallest positive whole number that ends with $17$, is divisible by $17$, and the sum of its digits is $17$.


[b]p3.[/b] Three consecutive $2$-digit numbers are written next to each other. It turns out that the resulting $6$-digit number is divisible by $17$. Find all such numbers.


[b]p4.[/b] Let $ABCD$ be a convex quadrilateral (a quadrilateral $ABCD$ is called convex if the diagonals $AC$ and $BD$ intersect). Suppose that $\angle CBD = \angle CAB$ and $\angle ACD = \angle BDA$ . Prove that $\angle ABC = \angle ADC$.


[b]p5.[/b] A circle of radius $1$ is cut into four equal arcs, which are then arranged to make the shape shown on the picture. What is its area?
[img]https://cdn.artofproblemsolving.com/attachments/f/3/49c3fe8b218ab0a5378ecc635b797a912723f9.png[/img]


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
To find the area of the shape formed by rearranging the four equal arcs of a circle of radius 1, we can use the following steps:

1. **Identify the Shape and Symmetry**:
   - The shape is formed by rearranging four equal arcs of a circle of radius 1.
   - By symmetry, the area of the figure not contained in the square equals the area of the square not contained by the figure.

2. **Determine the Side Length of the Square**:
   - The side length of the square is equal to the distance \( AB \) in the isosceles triangle \( \triangle AOB \), where \( O \) is the center of the circle, and \( A \) and \( B \) are points on the circle such that \( \angle AOB = \frac{\pi}{2} \).

3. **Use the Pythagorean Theorem**:
   - In the right triangle \( \triangle AOB \), the legs \( OA \) and \( OB \) are both equal to the radius of the circle, which is 1.
   - Therefore, the hypotenuse \( AB \) can be calculated using the Pythagorean theorem:
     \[
     AB^2 = OA^2 + OB^2 = 1^2 + 1^2 = 2
     \]
   - Thus, the length of \( AB \) is:
     \[
     AB = \sqrt{2}
     \]

4. **Calculate the Area of the Square**:
   - The area of the square with side length \( \sqrt{2} \) is:
     \[
     \text{Area of the square} = (\sqrt{2})^2 = 2
     \]

5. **Conclude the Area of the Figure**:
   - Since the area of the figure equals the area of the square, the area of the figure is:
     \[
     \boxed{2}
     \]