The test this year was held on 13 September 2021. It consisted of 10 problems, 5 for Part 1 (each problem worth 4 points, or 1 point if the answer is correct but unjustified), and the other 5 for Part 2 (where each problem is worth 7 points). Part 1 takes 60 minutes to complete, whereas part 2 takes 150 minutes to complete. The contest requires complete workings on both parts. Of course you are not allowed to use a calculator, however protractors and set squares are also PROHIBITED.

[b]Part 1 (Speed Round: 60 minutes, 20 points)[/b]

[b]Problem 1.[/b] Determine the number of ways to distribute 8 distinct storybooks to 3 children, where each child receives at least 2 books.

[b]Problem 2.[/b] A point $P$ lies inside of a quadrilateral and the point is connected to the midpoints of all the sides of the quadrilateral, as shown in the figure below. From this construction, the quadrilateral is divided into 4 regions. The areas of three of these regions are written in each of the respective regions. Determine the area of the quadrilateral that is unknown (which is denoted by the question mark).
(The image lies on the attachments in this post!)

[b]Problem 3.[/b] Let $a,b,c$ be positive integers, and define $P(x) = ax^2 + bx + c$. Determine the number of triples $(a,b,c)$ such that $a, b, c \leq 10$ and $P(x)$ is divisible by 6 for all positive integers $x$.

[b]Problem 4.[/b] Determine all real solution pairs $(x, y)$ which satisfy the following system of equations:
\begin{align*}
(x^2 + y + 1)(y^2 + x + 1) &= 4 \\
(x^2 + y)^2 + (y^2 + x)^2 &= 2.
\end{align*}

[b]Problem 5.[/b] Given a triangle $ABC$ where $\angle{ABC} = 120^{\circ}$. Points $A_1, B_1,$ and $C_1$ lie on segments $BC, CA,$ and $AB$ respectively, so that lines $AA_1$, $BB_1$ and $CC_1$ are the bisectors of the triangle $ABC$. Determine the measure of $\angle{A_1B_1C_1}$.

1. Given the system of equations:
   \[
   (x^2 + y + 1)(y^2 + x + 1) = 4
   \]
   \[
   (x^2 + y)^2 + (y^2 + x)^2 = 2
   \]
   Let \( a = x^2 + y \) and \( b = y^2 + x \). Then the equations become:
   \[
   (a + 1)(b + 1) = 4
   \]
   \[
   a^2 + b^2 = 2
   \]

2. Expanding the first equation:
   \[
   (a + 1)(b + 1) = ab + a + b + 1 = 4
   \]
   \[
   ab + a + b + 1 = 4 \implies ab + a + b = 3
   \]

3. Substituting \( a \) and \( b \) into the second equation:
   \[
   a^2 + b^2 = 2
   \]

4. We now have the system:
   \[
   ab + a + b = 3
   \]
   \[
   a^2 + b^2 = 2
   \]

5. To solve this system, we can use the identity:
   \[
   (a + b)^2 = a^2 + b^2 + 2ab
   \]
   Substituting \( a^2 + b^2 = 2 \):
   \[
   (a + b)^2 = 2 + 2ab
   \]

6. Let \( s = a + b \) and \( p = ab \). Then:
   \[
   s^2 = 2 + 2p
   \]
   From \( ab + a + b = 3 \), we have:
   \[
   p + s = 3
   \]
   Substituting \( p = 3 - s \) into \( s^2 = 2 + 2p \):
   \[
   s^2 = 2 + 2(3 - s)
   \]
   \[
   s^2 = 2 + 6 - 2s
   \]
   \[
   s^2 + 2s - 8 = 0
   \]

7. Solving the quadratic equation:
   \[
   s^2 + 2s - 8 = 0
   \]
   Using the quadratic formula \( s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   s = \frac{-2 \pm \sqrt{4 + 32}}{2}
   \]
   \[
   s = \frac{-2 \pm 6}{2}
   \]
   \[
   s = 2 \quad \text{or} \quad s = -4
   \]

8. For \( s = 2 \):
   \[
   p = 3 - s = 1
   \]
   For \( s = -4 \):
   \[
   p = 3 - (-4) = 7
   \]

9. Considering \( a \) and \( b \) as roots of the quadratic equation \( t^2 - st + p = 0 \):
   For \( s = 2 \) and \( p = 1 \):
   \[
   t^2 - 2t + 1 = 0 \implies (t - 1)^2 = 0 \implies t = 1
   \]
   Thus, \( a = 1 \) and \( b = 1 \).

10. For \( s = -4 \) and \( p = 7 \):
    \[
    t^2 + 4t + 7 = 0
    \]
    The discriminant \( \Delta = 16 - 28 = -12 \), which is negative, so there are no real roots.

11. Therefore, the only real solutions are \( a = 1 \) and \( b = 1 \). Substituting back:
    \[
    x^2 + y = 1
    \]
    \[
    y^2 + x = 1
    \]

12. Solving these equations:
    \[
    y = 1 - x^2
    \]
    \[
    (1 - x^2)^2 + x = 1
    \]
    \[
    1 - 2x^2 + x^4 + x = 1
    \]
    \[
    x^4 - 2x^2 + x = 0
    \]
    \[
    x(x^3 - 2x + 1) = 0
    \]

13. Solving \( x(x^3 - 2x + 1) = 0 \):
    \[
    x = 0 \implies y = 1
    \]
    \[
    x = 1 \implies y = 0
    \]
    \[
    x^3 - 2x + 1 = 0
    \]
    Using the quadratic formula for \( x^3 - 2x + 1 = 0 \):
    \[
    x = \frac{-1 \pm \sqrt{5}}{2}
    \]
    \[
    y = \frac{-1 \pm \sqrt{5}}{2}
    \]

14. Therefore, the solutions are:
    \[
    (x, y) = (0, 1), (1, 0), \left( \frac{-1 + \sqrt{5}}{2}, \frac{-1 + \sqrt{5}}{2} \right), \left( \frac{-1 - \sqrt{5}}{2}, \frac{-1 - \sqrt{5}}{2} \right)
    \]

The final answer is \( \boxed{ (0, 1), (1, 0), \left( \frac{-1 + \sqrt{5}}{2}, \frac{-1 + \sqrt{5}}{2} \right), \left( \frac{-1 - \sqrt{5}}{2}, \frac{-1 - \sqrt{5}}{2} \right) } \)