p1. Points $E$ and $F$ divide sides $BC$ and $CD$ of the convex quadrilateral $ABCD$ into two equal parts. In addition, segments $AE$, $AF$, and $EF$ divide the quadrilateral into $4$ triangles, where their areas are $4$ consecutive integers.  Determine the largest possible area that the triangle $ABD$ can have.


p2. Find all the solutions of the system of inequalities:
$30x_1 -2003x_2 + 1973x_3  \ge 0$
$30x_2 -2003x_3 + 1973x_4 \ge 0$
$...$
$30x_{28} -2003x_{29} + 1973x_{30} \ge 0$
$30x_{29} -2003x_{30} + 1973x_1 \ge 0$
$30x_{30} -2003x_1 + 1973x_2 \ge 0$


p3. A sequence is defined $a_n$ by:
$a_1 = 8 \,\, ,  \,\,a_2 = 18$
$a_{n + 2} = a_n \cdot a_{n + 1}  \,\, (n\ge 1)$
Determine the values of $n$ for which $a_n$ is a perfect square.


p4. Maca's family has $4$ people: grandfather, mother, father, and herself. If we double the size of Maca's monthly scholarship, the family income will increase by $5\%$ . If instead of the Maca scholarship, her mother's salary would be doubled, the family income will grow by $15\%$ . The same procedure gives $25\%$ in the case of Dad now. In what percentage does the family income grow, if only the grandfather's salary is doubled?	



p5. Three positive numbers $a, b, c$ satisfy the equality $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}=\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b}.$$ Prove that at least two of the values are equal.


p6. Calculate the area of the region between the three semicircles, knowing that $CD$ is perpendicular to $AB$ and that $CD = 1$.
[img]https://cdn.artofproblemsolving.com/attachments/0/f/116606d89abfb10f8503669bfbe803e16a73a0.png[/img]

PS. Seniors P4 was also [url=https://artofproblemsolving.com/community/c4h2689598p23341563]Juniors [/url] P6.
1. Denote \( AC = 2x \) and \( BC = 2y \). 
2. The desired area is given by the formula for the area of the region between the three semicircles. This can be expressed as:
   \[
   \text{Area} = \frac{1}{2}((x+y)^2 - x^2 - y^2)
   \]
3. Simplify the expression inside the parentheses:
   \[
   (x+y)^2 - x^2 - y^2 = x^2 + 2xy + y^2 - x^2 - y^2 = 2xy
   \]
4. Therefore, the area is:
   \[
   \text{Area} = \frac{1}{2} \cdot 2xy = xy
   \]
5. Given that \(\triangle ABD\) is a right triangle, we use the property of right triangles to find the relationship between \(AC\) and \(BC\). Since \(CD\) is perpendicular to \(AB\) and \(CD = 1\), we have:
   \[
   AC \cdot BC = 4xy = 1
   \]
6. Solving for \(xy\):
   \[
   4xy = 1 \implies xy = \frac{1}{4}
   \]

The final answer is \(\boxed{\frac{1}{4}}\).