Indonesia Regional [hide=MO]also know as provincial level, is a qualifying round for National Math Olympiad[/hide]  
Year 2019  [hide=Part A]Part B consists of 5 essay / proof problems, posted [url=https://artofproblemsolving.com/community/c4h2671394p23150636]here[/url][/hide]

Time: 90 minutes [hide=Rules]
$\bullet$ Write only the answers to the questions given. 
$\bullet$ Some questions can have more than one correct answer. You are asked to provide the most correct or exact answer to a question like this. Scores will only be given to the giver of the most correct or most exact answer.
$\bullet$ Each question is worth 1 (one) point.
$\bullet \bullet$ to be more exact:
$\rhd$ in years 2002-08 time was 90' for part A and 120' for part B
$\rhd$ since years 2009 time is 210' for part A and B totally 
$\rhd$ each problem in part A is 1 point, in part B is 7 points[/hide]


p1. In the bag there are $7$ red balls and $8$ white balls. Audi took two balls at once from inside the bag. The chance of taking two balls of the same color is ... 


p2. Given a regular hexagon with a side length of $1$ unit. The area of ​​the hexagon is ...


p3. It is known that $r, s$ and $1$ are the roots of the cubic equation $x^3 - 2x + c = 0$. The value of $(r-s)^2$ is ...


p4. The number of pairs of natural numbers $(m, n)$ so that $GCD(n,m) = 2$ and $LCM(m,n) = 1000$ is ...


p5. A data with four real numbers $2n-4$, $2n-6$, $n^2-8$, $3n^2-6$ has an average of $0$ and a median of $9/2$. The largest number of such data is ...


p6. Suppose $a, b, c, d$ are integers greater than $2019$ which are four consecutive quarters of an arithmetic row with $a <b <c <d$. If $a$ and $d$ are squares of two consecutive natural numbers, then the smallest value of $c-b$ is ...


p7. Given a triangle $ABC$, with $AB = 6$, $AC = 8$  and $BC = 10$. The points $D$ and $E$ lies on the line segment $BC$. with $BD = 2$ and $CE = 4$. The measure of the angle $\angle DAE$ is ...


p8. Sequqnce of real numbers $a_1,a_2,a_3,...$ meet $\frac{na_1+(n-1)a_2+...+2a_{n-1}+a_n}{n^2}=1$ for each natural number $n$. The value of $a_1a_2a_3...a_{2019}$  is .... 


p9. The number of ways to select four numbers from $\{1,2,3, ..., 15\}$ provided that the difference of any two numbers at least $3$ is ...


p10. Pairs of natural numbers $(m , n)$ which satisfies $$m^2n+mn^2 +m^2+2mn = 2018m + 2019n + 2019$$ are as many as ...


p11. Given a triangle $ABC$ with $\angle ABC =135^o$ and $BC> AB$. Point $D$ lies on the side $BC$ so that $AB=CD$. Suppose $F$ is a point on the side extension $AB$ so that $DF$ is perpendicular to $AB$. The point $E$ lies on the ray $DF$ such that $DE> DF$ and $\angle ACE = 45^o$. The large angle $\angle AEC$ is ...


p12. The set of $S$ consists of $n$ integers with the following properties: For every three different members of $S$ there are two of them whose sum is a member of $S$. The largest value of $n$ is .... 


p13. The minimum value of $\frac{a^2+2b^2+\sqrt2}{\sqrt{ab}}$ with $a, b$  positive reals is .... 


p14. The polynomial P satisfies the equation $P (x^2) = x^{2019} (x+ 1) P (x)$ with $P (1/2)= -1$ is ....


p15. Look at a chessboard measuring $19 \times 19$ square units. Two plots are said to be neighbors if they both have one side in common. Initially, there are a total of $k$ coins on the chessboard where each coin is only loaded exactly on one square and each square can contain coins or blanks. At each turn. You must select exactly one plot that holds the minimum number of coins in the number of neighbors of the plot and then you must give exactly one coin to each neighbor of the selected plot. The game ends if you are no longer able to select squares with the intended conditions. The smallest number of $k$ so that the game never ends for any initial square selection is ....
To solve the problem, we need to find the value of \((r-s)^2\) given that \(r\), \(s\), and \(1\) are the roots of the cubic equation \(x^3 - 2x + c = 0\).

1. **Identify the value of \(c\):**
   Since \(1\) is a root of the equation, we substitute \(x = 1\) into the equation:
   \[
   1^3 - 2 \cdot 1 + c = 0 \implies 1 - 2 + c = 0 \implies c = 1
   \]
   Therefore, the cubic equation becomes:
   \[
   x^3 - 2x + 1 = 0
   \]

2. **Use Vieta's formulas:**
   Vieta's formulas relate the coefficients of the polynomial to sums and products of its roots. For the polynomial \(x^3 - 2x + 1 = 0\), the roots \(r\), \(s\), and \(1\) satisfy:
   \[
   r + s + 1 = 0 \implies r + s = -1
   \]
   \[
   rs + r + s = -2 \implies rs + (-1) = -2 \implies rs = -1
   \]

3. **Calculate \((r-s)^2\):**
   We know that:
   \[
   (r-s)^2 = r^2 - 2rs + s^2
   \]
   We also know from Vieta's formulas that:
   \[
   (r+s)^2 = r^2 + 2rs + s^2
   \]
   Substituting \(r + s = -1\) and \(rs = -1\):
   \[
   (-1)^2 = r^2 + 2(-1) + s^2 \implies 1 = r^2 - 2 + s^2 \implies r^2 + s^2 = 3
   \]
   Now, substituting \(r^2 + s^2\) and \(rs\) into the expression for \((r-s)^2\):
   \[
   (r-s)^2 = r^2 - 2rs + s^2 = 3 - 2(-1) = 3 + 2 = 5
   \]

Therefore, the value of \((r-s)^2\) is \(\boxed{5}\).