[b]p1.[/b] If $3a + 1 = b$ and $3b + 1 = 2020$, what is a?


[b]p2.[/b] Tracy draws two triangles: one with vertices at $(0, 0)$, $(2, 0)$, and $(1, 8)$ and another with vertices at $(1, 0)$, $(3, 0)$, and $(2, 8)$. What is the area of overlap of the two triangles?


[b]p3.[/b] If $p$, $q$, and $r$ are prime numbers such that $p + q + r = 50$, what is the maximum possible value of $pqr$?


[b]p4.[/b] Points $A,B,C,D$ lie on a circle of radius $4$ such that $BC = 8$, $BD = 4$, and $m \angle ABC = 27^o$. If segments $\overline{AB}$ and $\overline{CD}$ do not intersect, what is the value of $m \angle ACD$? (Give your answer in degrees.)


[b]p5.[/b] Express  $\sqrt{14-5 \sqrt{52}}-\sqrt{14+5 \sqrt{52}}$ as a rational number.


[b]p6.[/b] Let $a_0 = 1$, and let $a_n = 1 + \frac{1}{a_{n-1}}$ for every integer $n \ge 1$. Find the value of the product $a_1a_2...a_9$.


[b]p7.[/b] Miki wants to distribute $75$ identical candies to the students in her class such that each students gets at least $1$ candy. For what number of students does Miki have the greatest number of possible ways to distribute the candies?


[b]p8.[/b] Let $ABCD$ be a rectangle. Let points $E$, $F$, $G$, and $H$ lie on the segments $\overline{AB}$, $\overline{AD}$, $\overline{BC}$, and $\overline{CD}$ (respectively) such that both $\overline{EF}$ and $\overline{GH}$ are parallel to $\overline{BD}$. If $\vartriangle AFE$ is congruent to $\vartriangle BEG$ and $\frac{AE}{HC} = \frac12$ , what is $\frac{AB}{BC}$ ?


[b]p9.[/b] If $a_1, a_2, a_3,...$ is a geometric sequence satisfying $$\frac{19a_{2019} + 19a_{2021}}{a_{2020}}= 25 +\frac{6a_{2006} + 6a_{2010}}{a_{2008}}$$ and $0 < a_1 < a_2$, what is the value of $\frac{a_2}{a_1}$?


[b]p10.[/b] Elizabeth has an infinite grid of squares. (Each square is next to the four squares directly above it, below it, to its left, and to its right.) She colors in some of the squares such that the following two conditions are met: 
(1) no two colored squares are next to each other; 
(2) each uncolored square is next to exactly one colored square. 
In a $20 \times 20$ subgrid of this infinite grid, how many colored squares are there?


[b]p11.[/b] Find the smallest whole number $N \ge 2020$ such that $N$ has twice as many even divisors as odd divisors and $N^2$ has a remainder of $1$ when it is divided by $15$.


[b]p12.[/b] We say that the sets A, B, and C form a "sunflower" if $A \cap B = A \cap C = B \cap C$. ($A \cap B$ denotes the intersection of the sets $A$ and $B$.) If $A$, $B$, and $C$ are independently randomly chosen $4$-element subsets of the set $\{1, 2, 3, 4, 5, 6\}$, what is the probability that $A$, $B$, and $C$ form a sunflower?


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. **Claim:**
   \[ a_n = \frac{F_{n+2}}{F_{n+1}} \text{ for all } n \in \mathbb{Z}_{\geq 0} \]
   where \( F_n \) denotes the \( n \)-th Fibonacci number.

2. **Proof by Induction:**

   - **Base Case:**
     \[ a_0 = 1 = \frac{1}{1} = \frac{F_2}{F_1} \]
     This holds true since \( F_2 = 1 \) and \( F_1 = 1 \).

   - **Inductive Hypothesis:**
     Assume that the desired result holds for some \( n = k \), i.e.,
     \[ a_k = \frac{F_{k+2}}{F_{k+1}} \]

   - **Inductive Step:**
     By the given recurrence relation, we have:
     \[
     a_{k+1} = 1 + \frac{1}{a_k}
     \]
     Substituting the inductive hypothesis:
     \[
     a_{k+1} = 1 + \frac{1}{\frac{F_{k+2}}{F_{k+1}}} = 1 + \frac{F_{k+1}}{F_{k+2}} = \frac{F_{k+2} + F_{k+1}}{F_{k+2}} = \frac{F_{k+3}}{F_{k+2}}
     \]
     This completes the inductive step.

   By the Principle of Mathematical Induction, we have proven that:
   \[ a_n = \frac{F_{n+2}}{F_{n+1}} \text{ for all nonnegative integers } n. \]

3. **Finding the Product:**
   \[
   a_1a_2 \cdots a_9 = \frac{F_3}{F_2} \cdot \frac{F_4}{F_3} \cdot \ldots \cdot \frac{F_{11}}{F_{10}}
   \]
   Notice that this is a telescoping product:
   \[
   a_1a_2 \cdots a_9 = \frac{F_3F_4 \cdots F_{11}}{F_2F_3 \cdots F_{10}} = \frac{(F_3F_4 \cdots F_{10})F_{11}}{F_2(F_3F_4 \cdots F_{10})} = \frac{F_{11}}{F_2}
   \]
   Since \( F_{11} = 89 \) and \( F_2 = 1 \):
   \[
   a_1a_2 \cdots a_9 = \frac{89}{1} = 89
   \]

### Final Answer:
\[
\boxed{89}
\]

### Self Correction:
The problem is solved completely.