p1. Inside a rectangle a point $P$ is marked. Segments are drawn that join the vertices with $P$. In an alternate way, the sectors that are formed are colored. Show that the sum of the areas of the painted sectors is equal to the sum of the unpainted sectors.


p2. Find the maximum number of different paths that can be built on a pool table to join two balls on the $n$-cushion table. No ball is touching a band. A band is defined when a ball bounces off one side of the table, with the angle of incidence equal to the angle of departure.
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p3. Using only two different digits $2$ and $d$, the following $90$-digit number is formed: $m= 2d22d222d...$ If $m$ is a multiple of $9$, determine all possible values of the digit $d$.


p4. Calculate all the solutions $x,y,z$ in the positive real numbers of the following system:
$$x(6- y) = 9\,\, ,\,\,y(6-z) = 9\,\, ,\,\ z(6-x) = 9$$

PS. Seniors p1 was also [url=https://artofproblemsolving.com/community/c4h2690911p23356729]Juniors[/url] p1.
1. Notice that the number \( m \) is formed by repeating a pattern of digits consisting of \( 2 \)'s and a single \( d \). The pattern is \( 2, 22, 222, \ldots, 2^{12}, d \), where the number of \( 2 \)'s increases by one each time, and there are 12 such groups.
2. The total number of digits in \( m \) is 90. We need to determine the number of groups \( x \) such that the sum of the digits in these groups equals 90.
3. The sum of the first \( x \) terms of an arithmetic sequence where the first term \( a = 1 \) and the common difference \( d = 1 \) is given by:
   \[
   S_x = \frac{x(x+1)}{2}
   \]
4. We need to find \( x \) such that:
   \[
   \frac{x(x+1)}{2} + x = 90
   \]
   Simplifying, we get:
   \[
   \frac{x(x+1)}{2} + x = 90 \implies \frac{x^2 + x + 2x}{2} = 90 \implies \frac{x^2 + 3x}{2} = 90 \implies x^2 + 3x - 180 = 0
   \]
5. Solving the quadratic equation \( x^2 + 3x - 180 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   x = \frac{-3 \pm \sqrt{9 + 720}}{2} = \frac{-3 \pm \sqrt{729}}{2} = \frac{-3 \pm 27}{2}
   \]
   This gives us two solutions:
   \[
   x = \frac{24}{2} = 12 \quad \text{and} \quad x = \frac{-30}{2} = -15
   \]
   Since \( x \) must be positive, we have \( x = 12 \).

6. Therefore, there are 12 groups of \( 2 \)'s and 12 \( d \)'s, making a total of 90 digits:
   \[
   78 \text{ (2's)} + 12 \text{ (d's)} = 90
   \]

7. For \( m \) to be a multiple of 9, the sum of its digits must be a multiple of 9. The sum of the digits is:
   \[
   2 \cdot 78 + 12d = 156 + 12d
   \]
   We need:
   \[
   156 + 12d \equiv 0 \pmod{9}
   \]
8. Simplifying modulo 9:
   \[
   156 \equiv 3 \pmod{9} \quad \text{(since } 156 = 17 \cdot 9 + 3\text{)}
   \]
   Thus:
   \[
   3 + 12d \equiv 0 \pmod{9} \implies 12d \equiv -3 \pmod{9} \implies 3d \equiv 6 \pmod{9}
   \]
9. Solving \( 3d \equiv 6 \pmod{9} \):
   \[
   d \equiv 2 \pmod{3}
   \]
   The possible values of \( d \) are \( 2, 5, 8 \).

10. Since \( d \) must be different from 2, the possible values of \( d \) are:
   \[
   d = 5 \quad \text{or} \quad d = 8
   \]

The final answer is \( \boxed{ d = 5 } \) or \( d = 8 \).