[url=https://artofproblemsolving.com/community/c4h2372772p19406382]p1. [/url] In the figure, point $P, Q,R,S$ lies on the side of the rectangle $ABCD$. 
[img]https://1.bp.blogspot.com/-Ff9rMibTuHA/X9PRPbGVy-I/AAAAAAAAMzA/2ytG0aqe-k0fPL3hbSp_zHrMYAfU-1Y_ACLcBGAsYHQ/s426/2020%2BIndonedia%2BMO%2BProvince%2BP2%2Bq1.png[/img]
If it is known that the area of the small square is $ 1$ unit, determine the area of the rectangle $ABCD$.

p2. Given a quadratic function $f(x) = x^2 + px + q$ where $p$ and $q$ are integers. Suppose $a, b$, and $c$ are distinct integers so that $2^{2020}$ divides $f(a)$, $f(b)$ and $f(c)$, but $2^{1000}$ doesn't divide  $b-a$ nor $c-a$. Prove that $2^{1021}$ divides $b-c$.

p3. Find all the irrational numbers $x$ such that $x^2 +20x+20$ and $x^3-2020x+1$ both are rational numbers.

[url=https://artofproblemsolving.com/community/c6h2372767p19406215]p4.[/url] It is known that triangle $ABC$ is not isosceles with altitudes of $AA_1, BB_1$, and $CC_1$. Suppose $B_A$ and $C_A$ respectively points on $BB_1$ and $CC_1$ so that $A_1B_A$ is perpendicular on $BB_1$ and $A_1C_A$ is perpendicular on $CC_1$. Lines $B_AC_A$ and $BC$ intersect at the point $T_A$ . Define in the same way the points $T_B$ and $T_C$ . Prove that points $T_A, T_B$, and $T_C$ are collinear.	


p5. In a city, $n$ children take part in a math competition with a total score of non-negative round. Let $k < n$ be a positive integer. Each child $s$:  
(i) gets $k$ candies for each score point he gets, and
(ii) for every other child $t$ whose score is higher than $s$, then $s$ gets $ 1$ candy for each point the difference between the values ​​of $t$ and $s$.
After all the candy is distributed, it turns out that no child gets less candy from Badu, and there are $i$ children who get higher scores than Badu. Determine all values ​​of $i$ which may take.
1. We start by noting that the side length of the small square is 1 unit. Therefore, the area of the small square is \(1 \text{ unit}^2\).

2. Let the vertices of the small square be \(P, Q, R, S\) such that \(P\) and \(Q\) lie on side \(AB\), \(R\) lies on side \(BC\), and \(S\) lies on side \(CD\).

3. Since the side length of the small square is 1 unit, the distance \(PQ = QR = RS = SP = 1\).

4. By the Pythagorean Theorem, the diagonal \(QS\) of the small square can be calculated as:
   \[
   QS = \sqrt{PQ^2 + PS^2} = \sqrt{1^2 + 1^2} = \sqrt{2}
   \]

5. Now, consider the right triangle \(\triangle APQ\). Since \(P\) and \(Q\) are on the sides of the rectangle, \(\triangle APQ\) is similar to the right triangle with hypotenuse \(QS\). Therefore, the ratio of the sides of \(\triangle APQ\) to the sides of the small square is the same.

6. Let \(AQ = x\) and \(AP = y\). Since \(\triangle APQ\) is similar to the right triangle with hypotenuse \(QS\), we have:
   \[
   \frac{AQ}{QS} = \frac{x}{\sqrt{2}} = \frac{1}{\sqrt{2}} \implies x = 1
   \]
   \[
   \frac{AP}{QS} = \frac{y}{\sqrt{2}} = \frac{1}{\sqrt{2}} \implies y = 1
   \]

7. Similarly, consider the right triangle \(\triangle BSC\). Since \(S\) and \(C\) are on the sides of the rectangle, \(\triangle BSC\) is similar to the right triangle with hypotenuse \(QS\). Therefore, the ratio of the sides of \(\triangle BSC\) to the sides of the small square is the same.

8. Let \(SB = x\) and \(BC = y\). Since \(\triangle BSC\) is similar to the right triangle with hypotenuse \(QS\), we have:
   \[
   \frac{SB}{QS} = \frac{x}{\sqrt{2}} = \frac{1}{\sqrt{2}} \implies x = 1
   \]
   \[
   \frac{BC}{QS} = \frac{y}{\sqrt{2}} = \frac{2}{\sqrt{2}} \implies y = 2
   \]

9. Therefore, the dimensions of the rectangle \(ABCD\) are \(AQ + SB = 1 + 1 = 2\) and \(AP + BC = 1 + 2 = 3\).

10. The area of the rectangle \(ABCD\) is:
    \[
    \text{Area} = (AQ + SB) \times (AP + BC) = 2 \times 3 = 6
    \]

The final answer is \(\boxed{6}\)